Engineering I

Mathematics
Problem Sets
• EXERCISE NO. 1: Algebra - 1
• EXERCISE NO. 2: Algebra - 2
• EXERCISE NO. 3: Trigonometry
• EXERCISE NO . 4: Plane and Solid Geometry
• EXERCISE NO. 5: Analytic Geometry - 1
• EXERCISE NO. 6: Analytic Geometry - 2
• EXERCISE NO. 7: Differential Calculus
• EXERCISE NO. 8: Integral Calculus
• EXERCISE. NO. 9: Differential Equation
• EXERCISE .NO. ·10: ·Plane -and.: Spa.ce :Vectors
• EXERC.I SE .N.O ;•.-t .1 :.,:"S:tatistf~::5,;,i,~ / f).:,:ro~~biJ.ity.
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 r-:.:::..::__~-----------,, 15. Find the 121h term of the expen .
IIH IIJIMari,,.ay1sH . . I A 1365 4...11 s,on (x -y), "' .
.,. .-ower ouse 1
. - X y
3 12
B . - 1465 x y
C . - 1275 x4 11
D . - 1165 x J~1 2
• "T Team REVIEW and TRAINING CENTER 1
16. Find the middle term of the expansion f ( 2
-?:>~-' "Because the Power Is In the House" 5)8. 0 X -
A. 46940 xe C 44736 e
EXERCISE NO. 1 8. 45827 X 8 □: 43750 ~8
Algebra -1
17. Find the coefficient of the 6 1h term of the
1. Three greater than eight times a certain expansion (x - 3y)12
number is 35. Determine the number. A. -192456 C . - 174416
A. 4 8. 5 C. 6 D. 7 B. -183235 D. -165326
2. Find x if 6x - 2 = 2x + 6 . 18. What is the sum of the coefficients in the
A. 10 8. 5 C. 2 D. 4
expansion of (a - b + c) 8 ?
A. 1 B. 2 C. 3 D. 4
3. Solve for x: ✓(20 - x) = x
A 4,-5 C. - 4, 5
19. What is the sum of the coefficients for the
B. - 4, - 5 D. None of these
expansion of ( 4x - 3) 9 ?
A . 16842 C. 13824
4. What is/are the value of x in the expression x2
B. 14482 D. 19684
+ x-12 = O?
A. - 2,6 C. -4, 3
20. Find the sum of the exponents of the
B. 1,12 D. -3, 4
expansion (x + y) raised to 10.
5. If 42x" 1 = 1024, find the value of 4x_ A 45 B. 90 C. 110 D. 55
A. 16 B. 14 C . 18 D. 20
21 . Determine the sum of the exponents of the
expansion (2x + 1 )7.
6. Solve for x if log381x = 16.
A. 56 B. 28 C. 42 D . 21
A. 4 B. 5 C. 3 D. 6
22. If x4 - 2x3 - 3x2 - 4x - 8 is divided by (x - 2),
7. If log of 2 to the base of 2 plus log of x to the
the remainder is
base of 2 is equal to 2, find the value of x.
A. 20 B. 28 C . - 20 D. - 28
A 2 B. 4 C. 3 D. 5
23. ~i~en f(x) = (x+4)(x - 3) + 4, when f(x) is
8. In the equation x2 + 11x + 28 = 0, one root is 3
d1v1ded by (x - K) , the remainder is - K. Find
more than the other root. What are the roots
the value of K.
of the equation?
A. 2 B. 4 C. 6 D. 8
A -7,-4 C. 8,5
B. -6, -3 D. 9,6
24. When ax3 + 2x2 - 18x + 7 is divided by (x+1)
9. Determine the value of k so that the sum and the remainder is - 15. Find the value of a.
A. 42 B. 45 C. 48 D . 50
the product of the roots are equal from the
given equation 7x2 + (2k - 1 )x - 3k + 2 = 0.
25. Write a cubic equation whose roots are ( - 1
A - 2 B. - 1 C. 1 D. 2 2,4) I

10. Find the value of k from the given quadratic A. x3

- 5x + 2x + 8 = o
2

equation 2x2 - kx + 6 = 0 if the sum of the B. x3 - 4x2 + x - 6 = 0

roots is equal to 4. C. x3 + 5x2 - 3x + 10 = 0
A 3 B. 4 C. 6 D. 8 D. x3-4x 2 + 5x-12 = O

11 . Find the value of m that will make 4x2 - 4mx + 26. The radius of curvature of a giv~n curve
4m +5 is a perfect square trinomial. varies directly with x and inversely with the
A. 3 B. -2 C. 4 D. 5 square of y. When x = 2, y = 3, the radius of
curvature is 100. Find the radius of curvature
12. Determine the value of k so that 3x2 + kx + 12 when x = 4 and y = 6.
= 0 will have just one real solution. A. 50 B. 100 C. 80 D. 120
A. 9 B. 12 C. 6 D. 16
27 . The resistance of a wire varies directly with its
13. Simplify the expression: the square root of the length and inversely its area. If a certain wire
cube root of 64x30 . 1O m long and 0 .10 cm in diameter has a
A 4x5 B. 8x1 C. 2x5° D. 2x10 resistance of 100 ohms, what will be its
resistance if it is uniformly stretched so that its
14. Find the 4 th term of the binomial expansion (x length becomes 12 m. Assume diameter to be
+ y)10 constant after it is being stretched.
A. 210x6 y 4 C. 210x4y6 A. 80 B. 100 C. 120 D . 140
7 3
8. 120x y D. 120x3y7
 16 and x is
th
4 _ Determine the 5 term of the20~~qu e)nce whose
~
nal between sum of n terms is given by ( - 5 .

•
.
an proport 10 f A 258 B. 218 C. 128 D. 158
28. The rne Find the value o x. D 9
equal to 12. C. 6 ·
8 4
A 3 . . r s. Find the 100th term of the sequence. 1.01,
h.1 h a transmission ine
1.00, 0.99

•I
29 _ The electri~ !?ow~~ ~rt i~nal to the product _of A 0.04 B. 0.03 C. 0.02 D. 0.05
can transmit is P p d the current capacity
its design voltag~ afransmission distance. A
and invers_ely as J at 1500 amperes can 6. In an arithmetic sequence 3, 7, 11, the nth
240 kY hne
transm!t 25W 0
~V: a
over 1oo km. How much
kV line rated at 2000
term is 31 . Find the value of n.
A. 6 B. 8 C. 7 D. 4
ower in M can 500
~mperes transmit over 300 k~6o D. 270 7. What is the sum of all even integers from 1O
A 23 0 8 . 225 C.
to 500?

30
The average rate of production of PCB is 1
· unit for every 2 hours work by two w?rkers.
How many PCBs can be produced 1n one
month by 60 workers working 200 hours 8.
A. 87,950
B. 124,950
C. 62,730
D. 65,955

How many terms of the progression 3, 5, 7,

I
during the month?
A. 4,000
8. 5,000
C. 6,000
D. 3,000
should there be so that their sum will be 2600.
A. 60 B. 50 C. 52 D. 55 I
9. Four positive integers form an arithmetic
31. The value of pi (TT) is a _ _ _ number.
A. Rational
B. Prime
C. Irrational
D. Composite
progression. If the product of the 1at and the
last term is 70 and the 2 nd and the third term is
88, find the 1at term.
I
A. 5 B. 3 C. 14 D. 8
32. _ _ _ property states that if a < b and b <
c, then a< c.
A. Reflexive C. Symmetric
10. The 3rd term of a Geometric Progression is 20
and the 6 th term is 160, find the 1st term.
I
8 . Transitive D. Identity A. 5 B. 11 C. 22 D. 4

33. If the discriminant of the quadratic equation is

less than zero, the roots are said to be
11. A piece of paper is 0.05 inches thick. Each
time the paper is folded into half, the
I
A. Real, Equal thickness is doubled. If the paper was folded
8 . Real, Unequal; Rational
C. Real, Unequal, Irrational
12 times, how thick in feet the folded paper
be?
I
D. Imaginary A . 102.4 in C . 204.8 in
B. 51.2 in D. 409.6 in
34. If b2 - 4ac is greater than O and the quadratic
equation is NOT a perfect square, then the 12. Determine x, so that x, 2x + 7, 1Ox - 7 will
\
discriminant is form a geometric progression.
A. Real, Equal A. 7, -5/6 C. 7, -7/12
8. Real, Unequal, Rational B. 7, -14/5 D. 7, -7/6
C. Real, Unequal, Irrational
D. Imaginary · 13. Find the sum of the first 10 terms of the
arithmetic progression 1, 3, 9, .. .
35. What do you call the first and fourth terms in A. 29524 C. 11820
the proportion of four quantities? B. 31760 D. 11342
A. Numerators C. Means
8. Extremes D. Denominators 14. Find the ratio of an infinite geometric series if
the sum is 2 and the first term is ½ .
A. 1/3 B. ½ C. ¾ D. ¼
EXERCISE NO. 2
Algebra - 2 15. What is the 11 th term of the harmonic
progression if the first and the third terms are
1. Find_ the missing term-' 1 I 2 I 6 I 24 I ½ and 1/6 respectively?
720 A. 1/20 B. ¼ C. 1/12 D. 1/22
A. 250 B. 180 C. 120 D. 105
16. Find the 8th term: 0.5, 0.167, 0.1, .. .
2. Evaluate the sum of the first n odd numbers: A. 0.033 C. 0.025
1+3+5+ ... +an. B. 0.205 D. 0.412
A. n B. 2n C. n2 D. n(n+1)
17. The sum of the ages of Peter and Paul is 21 .
3. Find the sum of the first 5 terms in the Peter will be twice as old as Paul 3 years from
sequence whose general term is (2"-1) . now. What is the present age of Peter?
A. 16 B. 8 C. 31 D. 15 A. 8 B. 6 C. 18 D. 15

MPHT Review Center: REE Reviewer in MATHEMATICS Page 3

 18. Ana is 5 years older than Beth. In 5 years the distance. What is the speed of the boat in
product of their ages will be 1.5 times the calm water?
product of their present ages. How old is Beth A 193.45 mph C. 146.67 mph
now? 8. 73.33 mph D. 293.33 mph
A 27 B. 20 C. 25 D. 18
28. Bing and Ryan can jog around a circular park
19. Maria is 36 years old. Maria was twice as old in 8 and 12 minutes, respectively. If they start
as Arian was when Maria was as old as Arian at the same instant from the same place, in
is now. How old is Arian? how many minutes will they pass each other if
A 20 B. 18 C. 24 D. 16 they jog around the track in the same
direction?
20. For a particular experiment, you need 5 liters A 13 minutes C. 15 minutes
of 10% solution. You find 7% and 12% B. 24 minutes D. 8 minutes
solution on the shelf. How much of the 7%
solution you mix with the appropriate amount 29. How many minutes after 8:00 PM will the
of the 12% solution to get 5 liters of 10% minute hand of the clock overtakes the hour
solution? • hand?
A 1.5 B. 2.5 C. 2 D. 3 A 40/12 C. 43 7/11
B. 42 11/12 D. 35/11
21 . Ten liters of 25% salt solution and 15 liters of
35% salt solution are poured into a drum 30. What time after Midnight will the hands of the
originally containing 30 liters of 10% salt clock be perpendicular with each other for the
solution. What is the percent concentration in 181 time?
the mixture? A 12:16:36 am C. 12:16.54 am
A 10% C. 30% B. 12:16:21 am D. 12:16.21 am
B. 20% D. 40%

22. One pipe can fill a tank in 6 hours and another EXERCISE NO. 3
pipe can fill the same tank in 3 hours. A drain Trigonometry
pipe can empty the tank in 24 hours. With all
three pipes open, how long will it take to fill 1. Sin (B - A) is equal to when B = 270° and A is
the tank? an acute angle.
A 5.16 C. 2.18 A - cos A C. - sin A
B. 3.14 D. 1.48 B. cos A D. sin A

23. Lorna can finish a job in 5 hours. Fe can do it 2. If sin A = 4/5, A in quadrant II, sin B = 7/25, B
in 4 hours. If Lorna worked for 2 hours and in quadrant I, find the sin (A+B)
then Fe was asked to help her finish it, how A 3/5 B. 3/4 C. 4/5 D. 2/5
long will Fe have to work with Lorna to finish
the job? 3. If tan A= 1/3 and cot B = 2 , tan (A - B) is
A. 2/5 hours C. 28 hours A 11/7 B. -1fl C. -11fl D. 1/7
B. 4/3 hours D. 1.923 hours
4. Given that cos 0 = - 12/ 13 and the 0 is in the
24. A man can do a job three times as fast as a 3 rd quadrant, find the value of sin 0.
boy. Working together it would take them 6 A - 5/13 B. 12/13 C. 5/13 D. -12/13
hours to do the same job. How long will it take
the man to do the job alone? 5. Find the values of x, where 0° ~ x ~ 360°,
A g hours C. 7 hours such that 2 sin 2x + 5 cos x + 1 =O.
B. 8 hours D. 10 hours A. 120° or 240 ° C. 100° or 240°
8 . 115° or 320° D. 136° or 310°
25. A job could be done by twelve workers in 13
days. Five workers started the job and after 4 6. If tan 4A = cot 6A, then what is the value of
days, 3 more men were added. Find the total angle A?
number of days it took them to finish the job. A 9° B. · 10° C. 12° D. 14°
A 21 B. 12 C. 17 D. 15
7. A road makes an angle 4,8° with the
26. It takes an airplane one hour and forty-five horizontal. How far must an automobile go up
the track for it to gain 30 m in altitude?
I minutes to travel 500 miles against the wind
and covers the same distance in one hour and
fifteen minutes with the wind. What is the
A 458.64 m
B. 358.52 m
C. 285.68 m
D. 341 .56 m
speed of the airplane?
C. 450.50 mph
I A. 342.85 mph
B. 375.50 mph D. 285. 75 mph
8. A transmitter is located on top of a mountain,
which is 3 km high. What is the furthest
distance on the surface of the Earth that can
27. A speed boat can make a trip of 100 miles in be seen from the top of the mountain? Take
I one hour and 30 minutes if it travels
upstream. If it travels downstream, it will take
one hour and 15 minutes to travel the same
the radius of the Earth to be 6400 km.
A 205 km
B. 152km
C. 225 km
D. 196km

I MPHT Review Center: REE Reviewer in MATHEMATICS Page4

I
 20. If the angles of a triangle are 2~, x _+ 15, and
9. Two straight roads intersect to form an angle 2x + 15, find the smallest angle in mils._
of 75°. Find the shortest distance from one A 600 mils C. 900 m~ls
road to a gas station on the other road 1 km B. 800 mils D. 700 mils
from the junction.
A 3.732 km C. 4.365 km
21. A spherical triangle ABC has an angle C = 90
B. 5.325 km D. 2.856 km
degrees and the sides a= 50 d_e grees .and c =
80 degrees. Find the value of side b.
10. Find the height of a tree if the angle of A 51°4' C. 74°20'
elevation of its top changes from 20° to 40° as
the observer advances 23 meters toward the
B. 77°52' D. 68°36'
base.
A 13.78 meters C. 14.78 meters 22. If Greenwich Mean Time (GMT) is 7AM.
B. 16.78 meters D. 15.78 meters What is the time in a place located at 135° E
longitude?
11. Solve c of an oblique triangle ABC if a = 25, b A 3 P.M. C. 2 P.M.
= 16 and C = 94.1°. B. 4P.M. D. 6 P.M.
A 29 B. 31 C. 33 D. 35
23. If Greenwich Mean Time is 9 AM. What is the
12. Find the height of the lamp post if the angle of time in a place 45° W of longitude?
elevation of its top changes from 26° to 62° as A 7 AM. C. 6AM.
the observer 1.8 m tall advances 32 m toward B. 4AM. D. 2AM.
the base.
A 26.87 m C. 22.87 m 24. An isosceles spherical triangle has an angle A
B. 23.57 m D. 25.57 m = B = 54° and side b = 82°. Find the measure
of the third angle.
13. Two towers are 60 m apart. From the top of A 156°24'15" C. 155°35'43"
the shorter tower, the angle of elevation of the B. 158°18'43" D. 148°28'16"
top of the taller tower is 40°. How high is the
taller tower if the height of the smaller tower is 25. Considering the Earth as a sphere of radius
40 m? 6400 km. Find the area of a spherical triangle
A 75m C. 86m on the surface of the Earth whose angles are
B. 100 m D. 90m 50°,89° and 120°.
A 56,476,062 C. 45,065,746
14. A pole tilts toward the sun at an angle 10°
B. 64,754,034 D. 24,412,654
from the vertical casts a shadow 9 m long. If
the angle of elevation from the tip of the
26. Which of the following is true?
~hadow to the top of the pole is 43°, how tall
A sin (- 9) = sin 9
1s the pole?
B. tan (- 9) = tan e
A. 10.2 meters C. 11.3 meters
C. cos (- 9) = cos 9
B. 13.7 meters D. 12.6 meters
D. csc (- 9) = csc 9
15. Simplify the expression 4 cos y sin y ( 1 - 2
sin2y) 27. In what quadrants do the secant and cosecant
A. Sec 2y C. Tan 2y of an angle plotted on a Cartesian coordinat
have the same sign? es
B. Cos 2y D. Sin 4y
A. I, Ill B. I, 11 C. 11, IV D. II, Ill
16. If cos 9 = ✓3 / 2, find 1 - tan 29.
A. - 1 B. - 1/2 C. 2/3 28. Of wh~t quad~ant of A, if sec A is positive and
D. 2 csc A 1s negative?
17 · The angle of elevation of the top of a light A Ill B. IV C. I D. II
house from a boat 50 m from it is the
29. The ~orth - south location of a point on the
:~pl~ment of the angle of elevation of the Earth s surface measured as an an le .
it Fe light house when the boat is 11 o m from degrees from the equator g in
. tn d th e height of the lighthouse.
A· 56 .44m · A. Meridian c. Latitude
B. 67 .ae m C. 74.16 m B. Longitude o. Equator
D. 45.67 m
18. ~f the compleme 30. W~at is the angle which fine of sight to th
,ts supplement thnt of an a_ngle theta is 2/5 of obJect makes the horizontal is below th e
A. 45 0 en theta 1s
,
B 750 of the observer? e eye
19 . . C . 600 D. 300
. A certain angle h
A. Angle of depression c. Bearin
supplement, find ~hs an explement 5 times the B. Angle of elevation D. Acute g
A. 67 _5° e angle.
B. 108° C. 135° 31 . "Yhat ~o _you call an angle who .
D. 58.5° side co1nc1des with an axis? se terminal
A Reflex angle C
B. Quadrantal · Right angle
0 Co - terminal
MPll'r
Review Cent -
r: RP.P. Rev
iewer in MATHEMATICS
 32 Sin 28 is equal to
A 2 sine case 4. How many sides are there in a regular
C . sin9 case triacontagon?
B ½sine
D. 1 - sin 2 e A 20 8 . 30 C. 12 D. 1000
33. An angle more than TT radian but not less than
2rr radians is 5. The sum of the exterior angles of a polygon is
A Straight angle C. Reflex angle the sum of its interior angles. How many sides
8 . Obtuse angle D. Right angle does the polygon have?
A 3 B. 5 C. 4 D. 6
34 . Which of the following is NOT true about
spherical triangles? 6. How many sides has a polygon if twice the
sum of its interior angles equals thrice the
A The sum of the sides is less than 360° sum of its exterior angles?
8 . The sum of three angles is 180°
A. 3 B. 4 C. 5 D. 6
C. The sum of two sides is greater than the
third side
7. How many diagonals are there in a
D. If two sides are equal, the corresponding dodecagon?
angles opposite are equal.
A. 27 B. 20 C. 12 D. 54
35 . 50 gradients is
8. Compute for the number of diagonals of an
A 45 degrees C. ½ revolution icosagon.
8 . 90 degrees D. 16 radians A 20 B. 170 C . 85 D. 200

36. 174 degrees is equivalent to mils 9. A regular hexagon is inscribed in a circle

A. 3094 ~421 whose radius is 10. Find the area of the
B. 2084 D. 2800 hexagon.
A 307.12 C. 259.81
37 . Two angles are said to form a _ _ if they are B. 105.18 D. 315.72
adjacent angles and whose non-common
sides are opposite rays. 1O. Find the area of a regular octagon whose
A. Vertical angles sides measure 5 cm.
B. Linear pair A. 105.61 cm 2 C. 135.67 cm 2
C. Complementary pair B. 148.83 cm 2 D. 120.71 cm 2
D. Reference angles
11 . The apothem of a regular undecagon is 8.
38. Centroid is the intersection of the _ _ of the Determine its area.
triangle. A. 206.71 C. 261 .07
A. Sides 8. 170.62 D. 127.60
8 . Medians
C. Altitudes 12. Find the area of a pentagram inscribed in a
D. Angle bisectors circle of radius 6 cm.
A. 27.823 sq. cm
39. It is a measurement from the north or south, B. 35.173 sq. cm
clockwise or counterclockwise C. 40.428 sq. cm
A. Equator D. 54.176 sq. cm
B. Bearing
c. International Date Line 13. Find the area of a trapezoid whose median is
o. Prime Meridian 54 cm and whose altitude is 8.
A. 192 8 . 432 C. 342 D. 243
40 In the curve y = tan 3x, what is its period?
. A. 3rr 8 . rr/3 C. 2rr/3 D. TT 14. Find the area of a circle inscribed in a
rhombus whose perimeter is 56 1/3 in and
EXERCISE NO. 4 whose longer diagonal is 26 in.
A. 43.78 sq. in C. 78.54 sq. in
Plane and Solid Geometry
B. 25 sq. in D . 452 .39 sq . in
1. What type of polygon has ~II of its interior
15. The length of the side of a rhombus is 10 cm .
angles less than 180 degrees ·
If its shorter diagonal is of length 12 cm. What
A. concave C. irregular
is the area of the rhombus?
B. convex D. regular
A. 54 sq. cm C. 27 sq . cm
a. 48 sq. cm D. 96 sq . cm
2. Given a polygon , the sum of its exterior
angles is: 16. A rhombus is formed by two radii an~ two
A. oo B. 180° C. 360° D. goo
chords of a circle of radius 6 m. What Is the
to their area of the rhombus?
3. Polygons are named according
A. 15.5 sq. m C. 31 .2 sq . m
number of B. 52.3 sq. m D. 47 .8 sq. m
A. diagonals C. edges
B. exterior angles D. faces

·
MPHT Review Center: REE Reviewer in MATHEMATICS Page 6
 . . n ball 15 cm . in diameter
27 . A spherical woo: cm . in 8 certa in liquid
. of a quadrilateral th 12
17. f" ind the fourth I s,~e ving one of its sides sinks to a dep sed above the liquid .
Find the _area expo c 45 p I
inscribed in a circifs dfarneter, and the other
A 50 P! D. 15 pi
equal . to 13 dr:1a·cae~t to the diameter are 5 m. B. 25 pI
two sides a J
and 8 m., respectively. 5 781 m What is the area of a lune whose angle is as·
A 4.27 1 m g- 6.382 m
28 . on a sphere of radius 30 cm .C 1 782 .45 cm2
8 . 3.812 m · ·
A 1,832.25 cm22 o· 2 'a12 55 cm2
1a_The sides of a cyclic quadril~!e~~ a~n~ ~h!
B. 2,670.35 cm · ' ·
cm, b=3 cm, c=4 cm and d- . _. ·n it.
radius of the circle that can be inscribed I 29. Find the area of a ~pherical triangle . ABC ,
c 3.13 cm A= 115 o, B= 7 o•,C=92° In a sphere of radius 12
A. 1.71 cm o· 4 71 cm
8. 2.71 cm · ·

19. A circle having a radius . of 6.5 cm. is

r·
8
143.44 cm2
: 243 _78 cm2
c.
o.
343.65 cm~
117.86 cm
circumscribing a triangle having an ~rea of 30
sq. cm. If one side of the tri~ngle 1s 13 cm.,
30 _ Find the surface area of regular icosahedrons
find the shortest side of the triangle. when each edge is of length 5 .
A. 12cm 8. 10cm C. 5cm D. 8cm A 216.5 C. 126.6
20. An engineer places his transit along the line
a: 261 .5 o. 162.5
tangent to the circle at point A such that
PA=200 m. He locates another point 8 on the
circle and finds PB=80 m. If a third portion C, EXERCISE NO. 5
on the circle lies along PB, how far from point Anal tic Geometry - 1
B will it be?
A 500 m C. 480 m 1. On what quadrant does the coordinate (-2, 6)
B. 420 m D. 420 m lies?
A I B. 11 C. Ill D. IV
21. The perimeter of a sector is 9 units and its
radius is 3 units. What is the area of the 2. Given the points (3, 7) and (-4,-7). Solve the
sector? distance between them.
A. 5 B. 6.5 C. 4.5 D. 7 A 15.65 C . 16.65
8 . 17.65 D. 14.65
22. A cylinder is circumscribed about a right prism
having a square base one meter on an edge. 3. The distance between (2,-9) and (x, 3) is 15.
The volume of the cylinder is 6.283 cu. m. Find the X.
Compute its altitude. A. 10 B. 11 C. 12 D . 13
A. 3 8. 4 C. 5 D. 6
4. How far is the intersection of the lines x + 2y -
23. The volume of a truncated prism with an 12 = O and 3x- Sy+ 15 = 0 from the origin?
equilateral triangle as its horizontal base is A 10.2 8 . 6.3 C. 1.25 D. 5.38
equal to 1800 cu. cm. The vertical edges at
each corners are 3, 4, and 5 cm., respectively. 5. A point P(x,2) is equidistant from the points
Find one side of the base. (-2,9) and (4,-7) . The value of xis
A. 10.39 cm C. 32.23 cm A 11/3 B. 20/3 C . 19/3 D. 6

"'
B. 15.87cm D. 27 .17cm
6. The se_gmen~ from (-3, 4) to (1,-2) is extended
24 . A cone is inscribed in a hemisphere of radius three times its own length. Find the terminal
r. If the cone and the hemisphere share point.

"'
bases, find the volume of the region inside the A. (-13,-20)
hemisphere but outside the cone. C. (13, -20)
8 . (-13, 20) D (13, 20)

-Ill
A. 4w/3 C. m 2 h/3
8 . 2nr3/3 D . nr3/3
7. Determine the coordinates of the point which
25 . A wedge is cut from a cylinder of radius 3 m 1s three-fifths of the way from the point (2 -S)
to the point (-3, 5) '
by two planes, one perpendicular to the axis
A. (-2, 1)
of the cylinder and the other passing through C. (1, -1)
B. (2 , 1)
the diameter of the section made by the first D. (-1, 1)
plane and inclined to this plane at an angle of
45'. Find the volume of the wedge? 8. The line segment connecting ( -6) d

•
A. 16m
3
8 . 18m3 C. 20m 3 o. 22m3 is bisected by point (2 _1) F ~· an (-2, Y)
A 4 B 5 ' · in the value of x.
. C. 6 D. 7
26. If the edge of the cube is decreased by 12%
by what percent is the surface are~ 9. Locate the centroid of the tri
decreases? coordinates A(O -4) B( 2 S) angle ABC having
A. 23% 8 . 77% C . 73% D . 27% A 1,2 B. 2o , c' 1 and c (1.4 ).
,4

.."'
' · D. 1,-2

MPHT Review Center: REE Reviewer in MATHEMATICS

Page 7
10 Find the area of triangle whose vertices are A 22 . What is the equation of the line bisector off tthhe
(-3, -1 ), B (5, 3) and C ( 2, -8) . acute angle formed by the intersection o e
A 37 B. 28 C. 38 D. 17 lines 4x + 3y - 24 = 0 and 5x - 12y + 30 = O?
A. 9x + 33y = 154
11 . In a Cartesian coordinates the coordinates of B . 33x + 99y = 154
a quadrilateral are (1 , 1 ), (0, 8), (4, 5), and C . 9x - 33y = 154
(-3, 4) . What is the area? D . 33x - 9y = 154
A 16 8 . 20 C. 18 D. 25
23. Determine the acute angle between the lines
12. Find the area of the hexagon ABCDEF formed y - 3x = 2 and y - 4x = 9 .
by joining the points A(1,4), B(0,-3), C(2,3), A. 3.54 deg C . 7 .86 deg
0(-1,2), E{-2,-1) and F{3,0). B . 4.39 deg D. 5.87 deg
A. 24 B. 20 C. 22 D. 15
24. The distance between points (5, 30°) and (-8,
13. Are the lines represented by the equations y = -50°) is:
3x + 2 and 6x + 2y = 5 parallel? A. 9.84 C . 6 .13
A Nol B. 10.14 D. 12.14
B . Yes!
C Cannot be determined 25. Find the polar coordinates for the point whose
D . At only a given range rectangular coordinate of (-6, -8) .
A. (10, -233.23°)
14. What is the equation of the line that passes 8 . (10, 233.23°)
through (4, 0) and is parallel to the line x - y - C. (10,126.187°)
2 = O? D. (10,-53.13°)
A. x+y-2=0 C. x-y + 4 = 0
B. y-x + 4 = 0 D. x+y-4=0 26. Transform the equation below into Cartesian
coordinates:
15. What is the equation of the line that passes 3
through (-3 , 5) and is parallel to the line 4x -
r=-----
2y + 2 = 0?
3+2cos0
A 4x + 2y + 13 = O A. 5x2 -9y2 + 12x + 9 = 0
B. 2x+4y-17=0 B. 5x2 + 9y 2 - 12x - 9 = 0
C . x - 2y + 15 = 0 C . 5x 2 + 9y2 + 12x + 9 = 0
D. 2x - y + 11 = 0 D. 5x2 + 9y 2 + 12x- 9 = 0

16. Find the equation of the perpendicular

bisector of the line joining (4,0) and (6,3). EXERCISE NO. 6
A 4x + 6y - 29 = 0 Analytic Geometry - 2
B. 4x + 6y + 29 = 0
C. 4x -6y + 29 = 0 1. What conic section is represented by the
D. 4x-6y-29 = 0 equation 4x2 + 3y2 - ax + 16y +19 =O
A. Circle C. Hyperbola
17. A line 4x + 2y - 2 = 0 is coincident with line B. Ellipse D. Parabola
A. 4x + 4y - 2 =0
B. 4x + 3y + 3 =0 2. What is the nature of the curve represented
C. Bx+ 4y-2 =0 by the equation x2 + 4y2 + 4xy + 2x - 1 o = O?
D. 8x + 4y - 4 =0 A. Circle C . Hyperbola
8 . Parabola D. Ellipse
18. Wr.at is the distance of the line 4x - 3y + 5 = 0
from the point (4,2)? 3. What conic section is represented by the
A 5 B. 4 C. 2 D. 3 equation 4x2 + 8x - y2 + 4y - 15 = O?
A. Parabola C. Hyperbola
19. Find the value of k if the distance from the B. Ellipse D. Circle
point (2,1) to the line 5x + 12y + k = 0 is 2.
A 5 8. 2 C. 4 D. 3 4. Determine the nature of the curve represented
by the equation x 2 + y 2 - Bx+ 16y + 81 = 0 .
20. Find the distance between the lines, 3x + y - A. Circle C. Parabola
12 = 0 and 3x + y - 4 = 0 . 8. Hyperbola D. Empty set
A 2.53 B. 1.14 C. 3.17 D. 1.75
5. What is the center of the curve x2 + y 2 - 2x -
21 . What is the equation of the line through (-3, 5) 4y-31 = 0 .
which makes an angle of 45 degrees with the A. (-1, -2) C . (1, -2)
line 2x + y = 12? B. (-1, 2) D . (1, 2)
A . X + 2y - 12 = 0
B. X + 2y - 18 = 0 6. What is the center of the circle x2 + y2 + 12x -
C. x + 3y-12 = 0 By+16=0
D. x - 3y + 12 = 0 A (6, 4) C. (6, -4)
B. (-6, 4) D. (-6, -4)

MPHT Review Center. REE Reviewer in MATHEMATICS Page 8

 . f the circle whose center is 19. Find the ratio of the major axis to the minor
F
. d the equation o . 4 axis of the ell ipse: 9x2+4y2-72x-24y-144=0.
7. ,n r. and whose radius rs .
A 0.67 B. 0.33 C. 1.67 D . 1.5
4 y + 11 = 0
0
at (3 , - ) 2
8
A x2 + y2 + 6x: 1 Oy + 18 = 0
B x2 + y - x 0 20. Wh_ at is the equation of an ellipse, major axis
· x2 + y2 _ 1Ox - 2y + 1 =
g x2 + y2 + 14X + 16y + 117 = 0 ho~zontal, with center at ( 1, 2) and whose
maJor and minor axis are 6 and 4
. rea of the curve given the equation respectively? '
8 Find t1,ea 2 2
. x2 + y2 + 4x - 8y - 5 = 0 . A. x + 4y - 2x - 8y + 1 = o
A 24 _8 sq units C. 78.5 sq unrts 8 · 9 x2 + 4y 2 - 72x -24y- 144 = o
B. 81 . 7 sq units D. 95.2 sq unrts C. 4x2 + 9y2 -Bx - 36y + 4 = o
D. 9x2 + 25y2 - 36x -189 = o
_ Find the equation of the circle passing through
9
the points (-3, -1 ), (3, 1 ), and (5, 3). 21. Find !he eccentricity of an ellipse given the
A x2 + y 2 - 2x + 4y - 11 = 0 equation: 4x2 + 25y2 - ax - 1OOy - 296 = O
s . x 2 + y 2 + Bx - 1 Oy + 16 = 0 A 1.3 B. 0.92 C. 2.5 D. 0 .27
c. x2 + y2 + 6x - 18y - 10 = O
D. x 2 + y 2 - Bx - 12y + 3 = 0 22. The_ area of the ellipse 9x2 + 25y2 - 36x - 189
= 0 1s equal to:
1o. What is the equation of a circle whose ends of A 20rr B. 15rr C. 10rr D. 25rr
diameter are (10, 2) and (6,-4).
A x 2 + y2 - Bx + 2y - 48 = 0 23. Find the latus rectum of the ellipse 9x2 + 25y2
8. x 2 + y 2 - 16x + 2y + 52 = O - 36x-189 = 0.
C. x2 + y2 + 16x - 12y + 52 = O A. 1.8 B. 0.3 C. 2.8 D. 3.6
0. x2 + y2 + 16x- 2y + 48 = 0
24. Given an ellipse x 2/36 + y 2/32 = 1. Determine
11 . Compute the focal length and the length of the distance between foci.
latus rectum of parabola y 2 + 8x - 6y + 25 = o A. 2 B. 3 C. 4 D. 5
A. 2, 8 C. 4, 16
8. 16,64 D. 1,4 25. Find the center of the hyperbola: 4x2 - 9y2 +
Bx - 18y - 149 = 0.
12. Find the focal length of the curve 2x2 - Bx - A. (-1,1) C. (-1,-1)
4y + 16 = 0 . B. (2, -1) D. (-1, 2)
A 2 B. 1 C. 0 .5 D. 0.25
26. Find the transverse axis of the hyperbola: 4x2
13. The parabola x2 + x + y + 1 = O opens: - 9y2 + ax - 1By - 149 = o.
A To the left C. To the right A. 6 B. 4 C. 12 0. 8
8 . Upward D. Downward
27. Find the eccentricity of the curve 9x 2 - 4y2 _
14. Find the equation of the parabola that passes 36x + 8 y = 4.
through the points (3, 6), (-2, 1.5), (1, 1.2) . A. 1.8 B. 1.2 C. 1.5 D. 1.6
A 5x2 + 4x - 1 Oy + 3 = 0
8 . 2y2 -6x + Sy-25 = 0 28. What is the semi-conjugate axis of the
C. x2 -8x + 4y -15 = 0 hyperbola x2/9 - y2/4 = 1 .
D. 3x2 + 5x - 3y + 10 = 0 A. 1 B. 2 C. 3 D. 4

15. Find the value of the parabola whose axis is 29. What is the equation of the upward asymptote
vertical and passes through (-1, 0), (5, 0), (1 , of the hyperbola (x - 2}2/9 - (y + 4 )2/16 = 1.
8) and (4, y) . A. 4x + 3y - 20 = 0
A -5 B. 5 C. -6 D. 6 B. 4x - 3y - 20 = 0
C. 3x + 4y + 20 = O
16. An arch 18 m high has the form of parabola D. 3x-4y-20 = 0
wi th a vertical axis. The length of a horizontal
beam placed across the arch 8 m from the 30. Find the eccentricity of a hyperbola whose
top is 64 m. F ind the width of the arch at the transverse and conjugate axes are equal in
bottom. length?
A 106 m C. 54 m A 1.56 B. 1.41 C. 0.76 D. 2.31
B. 74 m 0. 96 m
31. Find the latus rectum of the curve 4x2 - 9y 2 +
17 . The center of the ellipse 4x2 + y 2 - 16x - 6y - ax - 1 ay - 149 = o.
A. 2.18 B. 5.33 C. 0.88 D. 3.16
43 = 0 is at:
A (1,1) C . (-3, 4)
8 . (2, 3) 0 . (2,1}

18. Find the major axis of the ellipse x2 + 4y

2
- 2x
- Sy+ 1 = 0.
A. 2 B. 4 C. 10 D. 6
 EXERCISE NO. 7 15. Find the equation of a fine normal to the cur.ta
Differential Calculus x 2 = 16y at ( 4 , 1 )
A. 4x - y = 8 C . 2x + y =9
Evaluate: lim (1-cos x)/x2 ; x-+0 B. 3x + y = 23 D. x + 2y = 7
A. 0 8. ½ C. 2 D. -½
16. The sum of two positive numbers Is 50. Wh at
2. lim (tan 2x - 2 sin x) / x 3 ; x-.0 are the numbers if their products is to be the
A. 0 B. 3 C. infinity D. ONE largest possible.
A 24 & 26 C . 25 & 25
3. lim (x 2 -1) / (x 2+3x-4) ; X-+1 B. 28 & 22 D. 20 & 30
A. 1/5 B. 2/5 C. 3/5 D. 4/5
17. A triangle has variable sides x,y,z subject to
4. lim (3x 4 - 2x3 + 7) I (5x3 + x - 3); X-+infinity the constraints such that the perimeter is fixed
A. 0 B. Infinity C. -1 D. ONE to 18 cm . What is the maximum possible area
for the triangle?
5. lim(x3 -x 2 -4)/(x2 -4);x-+2 A 15.59 cm 2 C. 17.15cm2
A. 1 8. 2 C. 3 D. 4 B. 18.71 cm 2 D. 14.03 cm 2

18. Two post one is 1 Om high and the other 1Sm

6. Differentiate the equation y =~
x+l high stand 30m apart. They are to be stayed
2
A. x +zx C 2x by transmission wires attached to a single
(x+t) 2 •
X zx 2 stake at ground level, the wires running to the
B. D. top of the posts. Where the stake should be
(x+l) x+l
placed to use the least amount of wire.
7. Differentiate y = ex cos x 2 A 12m B. 14m C. 18m D. 16m
A. -ex sin x2
8 . ex (cos x 2 - 2x sin x 2 ) 19. A fencing is limited to 20ft length. What is the
C. ex cos x2 -2x sin x2 maximum rectangular area that can be fenced
D. -2xexsin x in using two perpendicular comer sides of an
existing wall?
8. Differentiate y = log 10 (x2 + 1 )2 A. 120 B. 100 C. 140 D. 190
A. 4x (x2 + 1)
8 . (4x log 10 e)/ (x2 + 1) 20. A poster is to contain 300(cm square) of the
C. log e (x) (x 2 + 1) printed matter with margins of 10cm at the top
D. 2x(x2 +1) and bottom and 5cm at each sides. Find the
overall dimensions if the total area of the
9. Differentiate (x 2 + 2) 112 poster is minimum.
A. (x2 + 2) 112 I 2 C. 2x I (x2 + 2) 112 A. 27.76cm, 47.8cm
8 . X / (x2 + 2) 112 D. (x2 + 2) 312 8 . 20.45cm, 35.6cm
C. 22.24cm, 44.5cm
10. Find the equation of the normal line to x2 + y2 D. 25.55cm, 46.7cm
= 5 at the point (1,2)
A. 2x -y = 0 C. y + 2x =0 21 . Water is flowing in a conical reservoir 40 ft
8 . 3x + y = 0 D. 4x-y = 0 deep and 16 ft across the top at the rate of 2
ft~/min . Find how fast the surface is rising
11 . Determine the velocity when t = 4 given the when the deep is 4 ft deep.
equation D = 20t + (5 / t + 1) where D is in A 0.85 ft/min C. 0.83 ft/min
meters and t is in seconds B. 0.95 ft/min D . 0 .66 fUmin
A. 19.8 mis C. 13.4 m/s
B. 12.6 mis D. 14.5 m/s 22 . The cost per hour of running a motor boat is
proportio_nal to the cube of the speed . At what
12. If y = 4 cos x + sin 2x, what is the slope of the speed will the boat run against a current of s
curve when x = 2 radians? km/hr in order to go a given distance more
A. -2 .21 C. -3.25 economically?
B. -4 .94 D. 2.21 A 10 kph C . 11 kph
8 . 13 kph D. 12 kph
13. Locate the points of inflection of the curve y =
f(x) = x2 ex .
23. ~t an~ dista~ce ~ from the source of light, the
A. -2 ± ../3 C . -2 ± ../2 !ntens~ty of 1llum1nation varies directly as the
B. 2 t ../2 D. 2 ± ../3 1ntens1ty of the source and inversely as the
square of x. Suppose that there is a light at A
14. Given the curve y = 12 - 12x + x3 determine '.'3nd a_n othe_r at B, the one at B having a~
its maximum, minimum and inflecti~n point. 1ntens1ty 8 times that of A. The distance AB ·
A. (-2 ,28), (2,-4) , & (0, 12) 4 m: At w_hat p~int ~ram A on the line AB w'i~
B. (2,-28), (2,4), & (0,2) the 1ntens1ty of 1llumrnation be least?
C. (-2,-28} , (-2,-4), & (2, 12) A 2.15 m C. 1.50 m
D. (-2,28) , (-2,4), & (1 ,12) B. 1.33 m D. 1.92 m

MPHT Review Center. REE Reviewer in MATHEMATICS a

24 . There Is a constant inflow of liquid into a
conical vessel 15ft deep and 7 .Sft in diameter.
Water is rising at the rate of 2fUmin when the 4 . Evaluate f Jor2v (x2 + y2)dxdy
1
2

water is 4ft deep. What is the rate of inflow in A 16.5 B. 15.5 C. 17.5 D. 14.5
ft 3 per minute?
A 6 .28 ft 3/min C. 8.26 ft 3 /min r2 r3 rr dxdydz
3
B. 2.68 ft /min D. 8.88 ft 3 /min 5. Evaluate Jo Jo Jo
A13 B. 9 C.7 D. 5
25. Water drains from a hemispherical basin of
diameter 20 inch at the rate of 3 in 3 per 6. The integral of 3,ti( dx is equal to:
second. How fast is the water level falling A 4"11/(ln3) + C C. 3'4K/(ln81) + C
when the depth of water is 5 inch? B. 3-4x/(ln27) + C D. 3◄></(ln12) + C
A 0.012732 inch/sec
B. 0 .021732 inch/sec 7. What is the integral of (2sec2 x - sin x}dx?
C . 0.000232 inch/sec A. 2 cos x + tan x + C
D. 0 .000323 inch/sec B. 2 tan x + sin x + C
C. 2 sin x + cos x + C
26. A standard cell has an emf "E" of 1.2 volts. If D. 2 tan x + cos x + C
the resistance "R" of the circuit is increasing at
the rate of 0 .03 ohm/sec, at what rate is the
current "I" changing at the instant when the
resistance is 6 ohms? Assume Ohm's law E =
8. Evaluate fx ✓x 2 3
+ 3 dx

1 )-1/2
IR.
A. 9
2 (x'+3 )"' +C C. 3 (xJ +3 +C
A. -0.002 amp/sec
B. 0.004 amp/sec 1 ( x• +3 )1/l +C -X (x 3
+3
)312+ C
C. -0.001 amp/sec B. 3 D. 3
D. 0.003 amp/sec

27 . A point on the curve where the second

9. Evaluate: f yl J2y' + 1dy

3 3
derivative of a function is equal to zero is A. 1 ( 2y -1 )''' (3y +1 ) +C
30
called: 2

A Maxima B. ~(2y'-1)" (3y' +1)+c

B. Minima 2
...!_(2y +1)" (3y 2 +1)+c
C. Point of Inflection c. Jo
D . Point of intersection 2
~(2y 2 + 1)" (3y -1) +C
2

D. 3
28. The point on the curve where the first
derivative of a function is zero and the sin(1/ x)
second derivative is positive is called: 2
dx
A. Maxima 10. Integrate: x
B. Minima A. sec(1/ x)- cot 2 (1/ x)+ C
C . Point of Inflection B. tan(1/x)-x+C
D. Point of Intersection
C. cos(1/x)+C
29 . At the minimum point the slope from the y-
D. sin(1/ x)-csc 2 (1/x) + c
axis.
A. Negative C. Positive
B. Infinity D. Zero
Jev' 4y d Y
3
11 . Evaluate:
30. At the point of inflection where x=a A. 4yer' +C C . er' +y3 +C
A. f'(a) not equal 0 C. r(a} > 0
B . f'(a) = 0 D. f' (a)< 0 B. ev' +4y+C D. eY' +C

12. Find the area between the graphs of Y =

EXERCISE NO. 8 2x and y = x3.
Integral Calculus y

1. Evaluate the integral sin" 9d9 from O to pi/2.

A 2rr13 B. 3rr15 C. 3rr116 D. rr./4 X

2. What is the integral of cos xdx from pi/3 to

pi/2?
A.0.134 B.0.500 C. 0.707 D.0.293
A. 1 B. 2 0. 4
C. 3
3. What is the integral of sin 6 0 cos 4 0 d0
if the
13 - Compute the area of the plane region
upper limit is n/2 and the lower limit is O.
A. 0.0184 C. 0.1398 bounded by the curve x = y2 - 2 and the line y
B 1.0483 D. 0.9237 = -x.

MPHT Review Center: REE Reviewer In MATHEMATlrc

 y

A. 9/2 8. 17/5 C. 1217 D. 12.80

19. Determine the coordinates of the centroid of
14. Find the total area bounded by the curves y = the area bounded by the curves x"'2 = -y + 9 ,
x 3 - 4x and y = x2 + 2x. the co-ordinate axes of the 1 51 quadrant.
y y (0,9

A.28.12 8 . 18.02 C. 21 .08 D. 12.80

15. Find the area of one leaf of the four-leaved A. -1 .5, -1.8 C. 1.125, 3.6
clovers of r = 4 sin 20. B. -2.5, 6.5 D. 1.5, -3.4
y
20. The curve has an equation y = ex. Compute
the centroid from the y-axis of the area
X bounded by the curve from x = 0 to x = 1 .
A. 0.335 B. 0.146 C. 1.899 D. 0.582

21. The region in the first quadrant is bounded by

the curve y2 =4x and the lines x = 4 and the x-
A, TT 8. 2TT C. 4TT 0 . 8TT axis. How far is the centroid of the curve from
the x-axis?
16. Find the area inside the cardioid r=4( 1+cos9) A. 2.4 B. 1.5 C. 0.8 D. 1.8
but outside the circle r = 4.
22. Find the moment of inertia of the area
bounded by the curve y2 = 4x, the line x = 1 ,
the x-axis on the first quadrant with respect to
y-axis.
X
A. 0.571 B. 0.682 C. 0.436 D. 0.716

23. Find the moment of inertia of the area

bounded by the curve x2 = 4y, the line y = 1
and the y-axis on the first quadrant with
respect to x-axis.
A. 22.28 sq. units C. 11.14 sq. units A. 6/5 B. 7/2 C. 4fi D. 8/7
B. 44 .57 sq. units D. 66.71 sq. units
24. Find the moment of inertia of the area
bounded by the curve y2 = 4x, the line y = 2,
17. The curve has an equation of x2 + y2 = 25. the y-axis, on the first quadrant with respect to
Find the length of the arc on the first quadrant y-axis.
from x = 3 to x = 4. A. 0.095 B. 0.064 C. 0.088 D. 0.076
_ _y_
25. Given the area in the first quadrant bounded
by x2 = 8y, the line x = 4 and the x-axis. What
is the volume generated by revolving this area
about the y-axis?
A. 78.987 cu.units
B. 50.265 cu.units
X
D. 82.285 cu.units
A. 1.88 B. 1.42 C. 2.69 D . 3.44
26 . Find the volume by revolving the hyperbola xy
18. Find the length of the curve r = 4sin 9 from 9 = = 6 from x = 2 to x = 4 about the x-axis.
a
0 to = 90° and also the total length of curve. A. 28.27 cu.units
A. 2TT ; 4TT C. TT ;2 TT B. 25.53 cu.units
B. 3TT ; 6TT D. 4TT ; 8TT C. 23.23 cu.units
D. 30.43 cu.units

MPHT Review Center: REE Reviewer in MATHEMATICS Page 12

27 . T t1e area bounded by the graphs of y = 2x + 3 . the d1'fferential equation of the family of
and y = x2 revolves about the x-axis. 6. Obtain · t 1
straight lines with slope and y-rntercep equa .
Determine the volume generated.
A. ydy + (x+1 )dx =O
A. 228 B. 329 C. 255 D. 375
B. ydx + (x+1 )dy = 0
C. ydy - (x+1 )dx =O
28 . Given the area in the first quadrant bounded
by x2 = 8y, the line y - 2 and the y-axis. What
D. ydx - (x+1 )dy = 0
is the volume generated when this area is
revolved about the line y - 2 = O? 7. W rr·te a differential equation for
.
the fam ily of
circles with center on the x-axis.
A. 53.31 cu .units
B. 45.87 cu.units A. y" + y' + 1 = 0
C. 26.81 cu.units B. yy" + (y')"2 + 1 = 0
D. 33.98 cu.units c. y" + (y')"2 + 1 = 0
□ . yy" - (y')"2 = 0
29. Determine the volume generated by rotating
the curve 9x"2 + 4y"2 = 36 about the line 3x + 8. Obtain a general solution to the following
4y = 20. differential equation:
A.52n 2 B.96n2 C.26n2 D. 48n2 (1 + y 2 )dx + (1 + x 2 )dy = 0
A. tan x + tan y = C
30. The curve has an equation of x2 + y 2 = 25 . B. tan· 1x - tan- 1y + C
The area on the first quadrant of the curve is C. tan x - tan y = C
revolved about the line x = 10. Determine the D. tan- 1x + tan·1y = C
volume of the solid of revolution.
A. 674.16 C. 944.12 9. Obtain a particular solution for the following
B. 972.16 D. 887.14 differential equation, given x = 2 and Y = 3.
2xyy' = 1 + y
2

31 . Calculate the work done in pumping out the A. y = sqrt( 5x - 1)

water filling a hemispherical reservoir 3 m B. y = sqrt(2x + 1)
deep. C. y = sqrt(x - 1)
A. 550 kJ B. 325 kJ C. 450 kJ D. 624 kJ D. y = sqrt(3x + 2)

10. Solve for the general solution of the following

. EXERCISE NO. 9 differential equation:
Differential Equation 3(3x 2 + y 2 )dx - 2xydy = 0
A. x"3 = c(x"2 + 9y"2)
1. Determine the order of the following B. x"2 = c(9x"2 + y"2)
differential equation: C. x"3 = c(9x"2 + y"2)
d2
dx;
)3 Y(d~)
+ 3
d 1
+y
3
d 2
CB
= Sx
D. x"2 = c(x"2 + 9y"2)
(
11 . Which of the following differential equations is
A. 1 B. 2 C. 3 D. 4 EXACT?
A. (x + y)dx + (X - y)dy = 0
2. Determine the degree of the following B. (x + 2y)dx + (x - y)dy = o
differential equation: C. (x + y)dx + (2x - y)dy = 0
2
d2
3
d 7
_:!.) + 3y ( ~ + y 3 -
( dx 2 dx./ dx
(d~
= Sx D. All of the choices are correct

A. 1 B. 2 C. 3 D. 4 12. Solve for the general solution of the following

DE: 2xy dx + (1 + x 2 )dy = O
2
3. What is the order and degree of the following A. x y + Y C
2
= C. x3y2 + x C=
differential equation? B. xy + x = C D. x3y2 + y = C
y(y") 4 + 2x(y') 5 = 8xy 2
A. 1, 5 B. 3, 4 C. 4, 2 D. 2, 4 13. Whic_h of the following differential equations is
considered non-linear?
4. Eliminate the arbitrary constant in the A. (x"5 + 3y)dx - x dy = O
following equation: B. YY" - 2ycos x = sin x
x 3 - 3x 2 y =C C. y" - 2cosxy' = sin x
A. (X - 2y)dx - xdy = 0 0 . y' - (2/x)y = x
B. (x + 2y)dx - xdy = 0
C. (x- 2y)dx + xdy = 0 14. Determine the integrating factor for the
D. (x + 2y)dx + xdy = 0 following differential equation: y' - 3y = 6
A. 3x B. e•3x C. -3x o. e3"
5. Eliminate the arbitrary constants c1 and c2 in
the following equation:
y = C1e - 2X + C2e3X
15. S_
olve for the particular solution of the
d~fferent,al equation in the previous problem
A. y" - y' - 6y = 0 given that y = 6 when x = o.
B. y" - 2y' - 6y = 0 A. y = 8e3x - 2 c
C . 2y" - y' - 6y = O B. y = 4 8 3x + 2 · Y = 8e~ - 2
D . 2y " - y' - 6y = 0 D . y = 4e.;,x + 2

M PHT Review Center: REE Reviewer In MATHEMATICS

 16. Solve for the general solution of the following ,· EXERCISE NO. 10
differential equation : y' + xy = xy 2 Plane and Space Vectors
A y = Ce<x•2,12 + 1
B. y = Ce•<x•2>12 + 1 For problems # 1, 2, 3, 4 & 5 .
C. y· 1 = Ce<•• 2>12 + 1 Given: Two Vectors
D. y-1=ce-<x'2J12 + 1 A={1,2) 8=(3, 4)
1. What is the magnitude of A?
17. Solve for the general solution of the following A. 2.236 C . 6 .647
differential equation: y " - 3y' + 4y = O B. 4.189 D. 8.942
A. y= e1.sx(c1 cos ; x + c2 sin~ x) 2. What is the magnitude of B?
8. y = e - 1· 5 "(c 1 cos; x + c2 sin~ x) A 3 8. 5 C. 7 D. 9
C. y = e t. 5 x(c 1 cos../?x + c 2 sin ..fix)
3. Determine AB (dot product).
D. y = e -Lsx (c1 cos ../?x + c2 sin ../?x) D. 11
A. 18 B. 15 C . 20
18. Determine a third - order differential equation 4. Determine Ax B (cross product) .
with real numbers as coefficients if two A (4, 1, 7) C . (0, 0, -2)
solutions are known to be e•2x and sin 3x. B. (0, 5, -8) D. (6, 0, -5)
A y" ' + 2y" + 9y' + 18y = 0
8 . y" ' + 3y" + 4y' + 5y = 0
5. Determine the angle formed by A and 8 .
C y"' + 4y" + 3y' + Sy = 0
A 10.305° C. 16.189°
D . y" ' + 9y" + 2y' + 18y = 0
B. 13.267° D . 19.433°
19. Solve for y: (D 3 - 3D 2 + 3D - 1 )y = O
6. Suppose P is the point (-1, 8) and Q is the
A. y = e·"( c, + xc2 + x 2c3)
8 . y = ex(c, + xc2 + x2c3) point (3, 2). Find the vector A having PQ as a
C . y = ex(c1 + c2 + c3) representation.
D. y = e·•(c, + c2 + c3) A <3, -7> C. <-6, 16>
8. <4, -6> D. <4, -9>
20. A bacteria culture is known to grow at a rate
proportional to the amount present. After one 7. Given A = 3i + j and 8 = -2i + 4j, find the unit
hour, 1000 strands of the bacteria are vector having the same direction as A - 8.
observed in the culture; and after four hours, A _3_i __s_ C _s_ 1 _2.._J.
. ../34 ../34 1 . ,/34 ../34
3000 strands. Find the approximate number of 5. 3. D 3l s .
strands of the bacteria originally in the culture.
8 . ..m' + ,ml . ../34 + ,134J
A 496 B. 964 C. 694 D. 946
8. Given A = 3i + 2j and B = 2i + kj, where k is a
21 . A metal bar at a temperature of 100° F is scalar, find (a) k such that A and 8 are
placed in a room at a constant temperature of orthogonal; (b) k such that A and 8 are
0°F. If after 20 minutes the temperature of the parallel.
bar is 50° F, find the time it will take the bar to A. -3, 4/3 C. -8, 1/3
reach a temperature of 25° F 8. -7, 9/10 D. -2, 5/4
A. 39.6 min C. 49.6 min
B. 29.6 min D. 19.6 min 9. Given the vectors
A = -5i + j B = 4i + 2j
22. A body of mass 5 slugs is dropped from a Find: (a) the scalar projection of 8 onto A; (b)
height of 100 ft with zero velocity. Assuming the vector projection of 8 onto A
54
no air resistance, find the time required to A (a)-.E._ (b) i-...!.1·
,/'H,' 13 13
reach the ground. 72
A. 2.5 secs C. 3.0 ses B. (a)-~ (b) i -~J-
,/'H,' 14 13
B. 2.0 secs D. 1.5 secs C. (a)-~ (b) ~l _..,!_ .
,/'H,' 9 13 J
D. (a) - ~ (b)
45
23 . A tank initially holds 100 gal of a brine solution i - ~J-
../26' 13 13
containing 1 lb of salt. At t = 0 another brine
solution containing 1 lb of salt per gallon is 10. Find the area of a parallelogram with two
poured into the tank at the rate of 3 gal/min, sides identified by vectors from the origin to
while the well-stirred mixture leaves the tank the points (3,4) and (8,0).
at the same rate. Find the time at which the A 32 8 . 24 C. 12 D. 36
mixture in the tank contains 2 lb of salt.
A. 0.432 min C. 0.338 min 11 . Find the undirected distance between the
B. 0.823 min D. 0.243 min points P(-3, 4, -1) and Q(2, 5, -4) .

24 . Find the orthogonal trajectories of the family

A. v'45 B. ../35 C. 7 D. ../89
of curves y = cx 2
12. Given A= <5, -2, 6> and B = <8 -5 -4> fitnd
A. 0.5x"2 + y"2 = k A+ B. I ' '

B. x"2 + y"2 = k A <16, -4, -2>

C. x11 2 + 0.5y"2 = k C. <14,-6,1>
B. <15, -5, -1> D . <13, -7, 2>
D. x"2 - y"2 = k

MPHT Review Center: REE Reviewer In MATHEMATICS

Page 14
 13 Given A = <5, -2, 6> and B = <8 -5 -4> find ctor B onto tt,e
26. The scalar projection of the ve
A-8 . ' ' '
A <-5, 5, 8> C . <-3, 3, 1O> vector A is C. 5.46 D 6 78
A 3.73 8 . 4 .15 · ·
8. <-4, 4, 9> D. <-2, 2, 11 >
. . of t h e vector C onto the
27. The vector prolectIon
14. Given A= <-4, 7, -2>, find 3A and -SA vector Bis
A <-12, 21, -6>, <20, -35, 10> A - 1.143i + 4 .573j + 2.287 k
8 . <-13, 21, -6>, < 19, -35, 10> B. -2.182i -1 .309j + 1.528k
C. <-14, 21, -6>, <18, -35, 10> C. 6.080i-1.765j-0.196k
D. <-15, 21, -6>, <17, -35, 10> D. 1.961i+1 .177J-1.372k

15. Given R(2, -1, 3) and S(3, 4, 6), find the unit 28. Find the area of the triangle whose vertices
vector having the same direction as V(RS) . areA(3,1,2), 8(4,-2, 1)andC(1,1 ,3)
A 2-i+...!... · + _2_k A. 3.39 B. 4 .30 C. 5.38 D. 6 .74
. m ml vTI
B 1 ·+ s · + fil
. -JE I ill J
3 I
C For problems # 29 - 32 .
C 3 1· + 6 · + 4 k Given the position vectors A and B of a
· m ml m
D. 4 i + 7 j + 5 k rectangular coordinate system.
m vTs vTI A (2, 4, 3) 8 (1, - 5, 2) .
29. Compute the resultant in terms of unit vectors
16. Find the distance between the two points (3 4 i, j, and k.
-6) and (4, -7, 8) using vectors. ' A. 3i-j + 5k C. 4i-5j + 3k
A 17.83 B. 18.37 C. 13.78 D. 15.64 8. 2i-3j + 2k D. 5i-2j + 4k

17. Points C (5, 7, z) and D (4, 1, 6) are 7.28 cm 30. What is the magnitude of the resultant vector?
apart. What is the value of z? A. 5.92 8. 4.12 C. 7.07 D. 6 . 71
A 3 cm B. 2 cm C. 4 cm D. 1 cm
31 . The scalar projection of vector A onto vector B
18. Determine the scalar product at (1, 2, 3) is
A 144 B. 138 C. 132 D. 126 A. 2.19 B. -2.19 C. 2.22 D. -2.22

19. What is the cross product Ax B of the vectors 32. What is the vector projection of vector B onto
A = i + 4j + 6k and B = 2i + 3j + Sk? ' vector A?
A i- j- k C. 2i + 7j - Sk A < -.828i -1.655j -1.241k >
B. -i + j + k D. 2i + 7j + 5k B. < 4.27i 0.19j - 2.60k >
C. < 4.20i - 0.18j - 2.56k >
For problems# 20 - 27. D. < - 4.20i 0.18j 2.56k >
Assume the three force vectors intersect at a
single point 33. What is the angle between two vectors A and
A = i + 3j + 4k, B = - i + 4j + 2k, C = 2i + 7j - k B if A = 24i - 8j + 6k and B = 4i + 12j + 6k?
A 76°26' C. 87°19'
20. What is the magnitude of the resultant force B. 84°20' D. 92°6'
vector R?
A 15.0 B. 13.2 C. 14.7 D. 16.2 34. Comput~ the value of b if A and B are
= =
perpendicular. A 2i + bj + k, B 4i - 2j _ 2k
21 . Find the magnitude of displacement from A. 3 B. 2 C. 1 D. 4
vector A to vector B.
A 2 B. 3 C. 4 D. 5 Given a ve~tor V = (x2y)i - (xy)j + (xyz)k.
35. Determine the divergence of the t at
(3 ,2 , 1) vec or
22. Determine the scalar product of vectors Band
C. A. 9 8. 7 C. 12 D. 15
A 12 8 . 24 C. 36 D. 48
Given a ve~tor V = (x2y)i - (xy)j + (xyz)k.
23. Determine the magnitude of the vector cross 36. Determine the gradient of the vector at (
32
product Ax B.
A 11 .2 B. 12.4 C. 13.6 D. 14.8
A 12.6 B. 13.7 C. 14.6 D. •
15~~
Given a vector V = (x2y)i _ (xy)j + (xyz)k
24. At what angle does vector C makes with the x 37. Determine the curl of the vector ·
- axis? A (xy)i - (yz)j + (x2 - y)k ·
A 70°13' C. 74°12' B. (xz)i + (yz)j + (y - x 2 )k
8. 72°23' D. 76°9' C. (xy)i + (yz)j + (y _ x2)k
D . (XZ)i - (yz)j - (x 2 + y)k
25. Find the cross product of vector Bx C.
A. 15i - 3j + 18k
B. - 1Si - 3j - 18k
C. 18i - 3j - 15k
0 . - 18i + 3j - 15k
 22. A semiconductor will hire 7 men and
5 determfned the probability of shooting 5
women. In how many ways can the
company choose from 9 men and 6 out of 8 attempts.
women? A 13.21% c. 12.38%
B. 11 .44% D. 11.44%
A 678 B. 216 C. 324 D. 560
34. An items cost distribution has a given
23. In h?~ many ways can a committee
cons1st1ng of 3 mean and 2 women be function of the probability. What is th e
chosen from 7 men and 5 women? expected cost?
Cost in Pesos Probability
A 350 B. 400 C. 300 D. 200
1 0.2
24. In how many ways can you invite one of 2 0.28
your five friends in a party? 3 0.18
4 0.23
A 32 B. 30 C. 31 D. 29
5 0.11
25. The lotto uses numbers 1 to 42. A winning A 2.45 B. 2.77 C. 2.11 D. 2.89
number uses_ 6 d ifferent numbers in any
35. By investing in a particular stock, a
order. What Is your chance of winning if
you bet one ticket? person can make in one year P 40,000
A 1/4534568 C. 1/5245786 with a probability of O. 3 or take a loss of P
10,000 with a probabilityof 0.7. What is
B. 1/6580668 D. 1/2341668
the persons expected gain?
26. If you roll a pair of dice one time, what is A. 5000 C. 7000
the probability of getting a sum of 9? B. 4000 D. 8000
A 1/9 B. 1/4 C. 1/6 D. 1/3
36. A pack of cards contain 52 cards: 13
spades, 13 clubs, 13 hearts and 13
2 7. Three boys and 3 girls sit in a row. Find
diamonds. Of the 52 cards, 4 are aces
the probability that the 3 girls sit together.
one from each suit. The hearts and the
A 1/5 8 . 2/5 C. 3/5 D. 2/3
diamonds are colored red, the spades
and clubs are black. Four cards are drawn
28. What is the probability of a family with 5
from the pack, each card being returned
children of having 3 boys and 2 girls?
to the pack before the next card is drawn.
A 5/16 8 . 1/4 C. 3/16 D. 3/8
Find the probability that all are clubs.
A 1/4 C. 1/256
29. In a basketball game, the free throw
8. 1/100 D. 1/512
average is 0 .65. Find the probability that a
player misses one shot of the three free 37. Refer to the previous problem. If 5 cards
throws?
are drawn simultaneously, find the
A 0.441 C. 0.444 probability to get all the 4 aces.
8 . 0.422 D. 0.451 A 1/54145 C. 4/54145
B. 2/54145 D. 9/54145
30. A bag contains 3 white balls and 5 red
balls. If two balls are drawn at random in 38. A coin is tossed 5 times. What is the
succession without returning the first ball probability of getting 3 heads?
drawn, what is the probability that the A 0.3125 C. 0.6
balls drawn are both red? 8. 0.125 D. 0.2125
A 3/14 8 . 1/3 C. 5/14 D. 3/7
39. A fair coin is tossed 1O times. Compute
31 . One bag contains 4 white balls and 3 the probability of getting at least 7 heads.
black balls and a second bag contains 3 A 9/64 C. 7 /64
white and 5 black balls. One ball is drawn B. 11/64 D. 13/64
from the second bag and is placed
unseen in the first bag. What is the 40. A coin tossed 10 times. What is the
probability that the ball now drawn from probability of getting 4 tails and 6 heads?
the first bag is white? A 105/512 C. 13/64
A. 35/64 C. 31/64 8. 51/256 D. 25/128
B. 33/664 D. 19/64

32. Box A contains nine cards numbered 1 to EXERCISE NO. 12

9 and Box B contains 5 cards numbered
Advanced Engineering Mathematics_ 1
1 ' to 5. A box is chosen at random and a
card is drawn. If the number is even find 1. For the complex number z = 3 - 4i
the probability that it came from box A a. Find the absolute value.
A 9/19 C. 11/19 b. Find the argument.
B. 10/19 D. 21/19 A 5, -53.13°
B. 6, -61 .13° C. 7, -47.13°
33. If the probability that a basketball player D. 4, -70.13°
sinks the basket at a 3point range Is 2/5,

MPHT Review Center: REE Reviewer in MATHEMATICS

Page 17
2. Simplify (2 - 3i} (5 + 2i}
A 18 - 10i
B. 20-15i
C. 14 - 19i
D. 16-11i
A
[I
2
-1
1 ~] C.
[12 l 1
2
-1 ~1

[~1 JJ
-2 3
3.

4.
Simplify: jig+ j21 + j
A 3i B. 1 - i C.

Find the principal value of ln(4 + Si)

1+i D. 2i B.
[!
-1
0 !J D. -2
-2
17. Solve for x and y for the given equation
A. 1.86+0.90i C. 1.23-0.78i
B. 1.23 - 0.56i D . 1.67 + 0.45i [;] = H1 ! 21 [!]
A 5, -5 C. 15, -15
5. Find the principal value of lFJ 8 . 10, -10 D. 20, -20
A 0.1236 C . 0.1854

6.
B. 0.2078

(1 + i)7 is
D. 0.2979
18. Simplify 3 [ :
27 17
!] + 2 [! il 27 17
A. 7 + 81 C. 3 + Bi C. 20 5
B. 8-8i D . 4-81
A. 35 5
28 45 28 15
7. ln(3 + 4i) is? 27 7 27 17
A 1.61 - 0.88i C. 1.44 - 0.67i B. 35 5 D. 35 5
8 . 1.61 + 0 .93i D. 1.45 + 0.931 28 15 28 45
8. Find the value of x in 3x + 4y + 3yi + 15 -3i = 2 1 3
0 19. Find the inverse of a matrix of A =6 1 4
A --6.33 B. 2.33 C. 1.33 D. -4 .33 3 7 2
-3/5 17/65 1/65
9. The third principal root of -46 - 9i A 2 1/13 -2/13
A. 3 - 4i B. 2 - 3i C. 3 - Si D. 4 - 3i 4/5 -14/65 -6/65
-2/5 19/65 1/65
10. The first root of ( 1 + 1) 115 is B. 0 -1/13 2/13
A. 1.031 + 0.131 C. 0.168 + 1.061 3/5 -11/65 -4/65
B. -1 .031 + 0.231 D. 1.231 - 1.061 2/5 -19/65 9/65
c. 1 -1/13 2/13
11 . Find the value of sin (3 + 2j) 3/5 -25/65 -16/65
A 0 .531 + 3.59j C. 0.731 - 3.59j 3/5 14/65 5/65
8 . 0 .531 -3.59j D. 0.231 -3 .19j D. 0 -9/13 9/13
1/5 -10/65 -4/65
12. Find the value of cos (1 + 2j)
A -2.0327 - 3.0519i 20. Find the Laplace Transform of f(t) = t3
B. 3.0327 - 4.05191 A. 3/s 4 8 . 6/s 4 C. 12/s◄ D . 8/s 4
C . 2 .0327 - 3 .05191
D. 2 .0327 -4.0519i 21. Find the Laplace Transform of cos St
A. s/(s2 + 5) C. 1/(s2 + 5)
1 2 3 2
B. s/(s +25) D. 1/(s2 +25)
13. Evaluate the determinant:A = -2 -1 -2
3 1 4 22. Find the Laplace Transform of t sin t
A. 4 8. 5 C. 6 D. 7 A 2s/(s2 + 1)2 C. 1/(s2 + 1)2
B. s/(s2 + 1 )2 D. 2/(s2 + 1)2
14. Compute the value of x by determinant:
3 -2 4 2 23. Find the Laplace Transform of e31 cos 2t
2 1 -3 5 A. (s - 3)/((s - 3)2 + 4)
X -4= 0 -1 -2
B. ((s + 3)/((s + 3)2 + 4)
-5 3 2 4 C. 3/(s2 + 4)
A . -385 8 . -282 C. -427 D. -126
D. 3s/(s2 + 4)
15. Evaluate the determinant:
24. Find the Inverse Laplace Transform of 3/(s2 -
1 14 3 1
16)
X= 1 5 - 1 3
1 - 2 2 - 3 A. 3 sinh 4t C. 4 sinh 3t
B . ¾ sinh 4t D . 4/3 sinh 3t
3 -4 - 3 -4
A. 489 8. 373 C. 326 D. 452
1
25. Inverse Laplace Transform of - -

1s. If A •l~z
Matrix Ar.
~1 ~ ].
-1
find the transposed
A. 1/3 - 1/3e-3t
8 . 1/3 + 1/3e3t
s(s+3 )
C. 1/5 + 1 /5e 31
D. 3 - 3 e-31

MPHT Review Center: REE Reviewer in MAT1n ·u ".,..,,... ,..

 l~2 ~1\
1

ri ~;\
2 2
- 1
c. -1
A.
,, Simplify (2 - 3i) (5 + 2i) c. 14 -19i 1

F: 1J
"- · 3
A. 18-10i D. 16-11i

li ~J
-2 D.
-2
8 . 20 - 15i -1 -2
8.
Simplify ·. j211 + j21 +i D . 2i 0
3.
B. 1- i C . 1+i
A . 3i 17. Solve for x and y for the given equation
4. Find the principal value of ln(4 + 5i)
A . 1.86 + 0 .90i C . 1.23 -0.78t
.
A
l;1
5 _5
= H~ ! 2H!1
C. 15, -15
8 . 1.23 - 0.56i o. 1.67 + 0.451
s·. 10, -10 D . 20, -20

5. Find the principal value of

A . 0 .1236
8 . 0.2078
1A
C . 0 .1854
D . 0 .2979 18. Simplify 3 lI 11 l! 11
17
+ 2
27 17
27 c. 20 5
(1 + i)7 is A. 35 5 15
6. c. 3 + 8i
45
28
A . 7 + Bi o. 4-8i 28
8 . 8-8i 27 17
27 7 D. 35 5
ln(3 + 4i) is? 1.44-0.67i 8. 35 5 28 45
7. c. 15
A. 1.61 -0.88i o. 1.45 + 0.93i 28
8 . 1.61 + 0 .93i 2 1 3
6 1 4
8. Find the value of x in 3x + 4y + 3yi + 15 - 3i = 19. Find the inverse of a matrix of A = 3 7 2
0 17/65 1/65
A -6.33 B . 2 .33 C . 1.33 0 . -4 .33 -3/5
1/13 -2/13
A. 2
9. The third principal root of -46 - 9i 4/5 -14/65 -6/65
A. 3 - 4i B . 2 - 3i C . 3 - Si 0 . 4 - 3i 19/65 1/65
-2/5
0 -1/13 2/13
8.
10. The first root of (1 + i)
115
is -11/65 -4/65
3/5
A . 1.031 +0.13i C . 0 .168+1.06i -19/65 9/65
2/5
B . -1 .031 + 0 .23i 0 . 1.231 - 1.06i 1 -1/13 2/13
C.
3/5 -25/65 -16/65
11 . Find the value of sin (3 + 2j) 3/5 14/65 5/65
A. 0 .531 + 3.59j C . 0 .731 - 3 .59j -9/13 9/13
0. 0
B. 0 .531 - 3 .59j D . 0 .231 - 3 .19j -10/65 -4/65
1/5
3
12. Find the value of cos (1 + 2j) 20 . Find the Laplace Transform of f(t) = t 4
A . -2.0327 - 3.0519i A. 3/s4 8 . 6/s4 C. 12/s"" D . 8/s
8. 3.0327 -4.0519i
C . 2 .0327 - 3 .0519i 21 . Find the Laplace Transform of cos 5t
D. 2 .0327 - 4 .0519i 2
A. s/(s2 + 5) C. 1/(s + 5)
2
1 2 3 8. s/(s + 25)
2 D . 1/(s + 25)
13. Evaluate the determinant:A = -2 -1 -2
22. Find the Laplace Transform of t sin t
3 1 4
A. 4 8. 5 C. 6 0. 7 A. 2s/(t + 1)2 C. 1/(s2 + 1)2
s/(s + 1 )2 o.
\vr brr\inant:
B. 2/(s2 + 1 )2

14. Compute the ~x 23. Find the Lap\ace Transform of e3t cos 2t
2
A. (s - 3)/((s - 3) + 4)
X = -4 0 -1 -2 B . ((s + 3)/((s + 3)2 + 4)
-5 3 2 4 C . 3/(s2 + 4)
A . -385 B . -282 C . -427 0. -126 D. 3s/(s2 + 4)

15
::,\rf 1~·"Iint
3 -4 -3 -4\
24 · F ~ d the Inverse Laplace Transform of 3/(s2 _
1
A. 3 sinh 4t
B . ¼ sinh4t
C . 4 sinh 3t
D . 4/3 sinh 3t
A . 489 8 . 373 C. 326 D . 452
25. Inverse Laplace Transform of _1.-
A . 1 /3 - 1 /3e-3t
=l-~z ~ t 1·
s(s+3 )
1
16. If A 1 find the transposed B . 1/3 + 1/3e3l C. 15 + 1/5e3l
D . 3- 3 e-3t
-1
Matrix A1 .

MPHT Review Center~ RF."F o----: -

26 . Find the Inverse Laplace Transform of
A. 4cos 3t + sin 3t
4s+ 1
52 +9 Evaluate r(S)
11 . A 362880
c 40320
[) .181440
I
B. 4cos 3t + 1/3 sin 3t B. 5040 EXCEPT
C. 4sin 3t + ¼ cos 3t O) .s equal to
D. 4cos 3t + 3 sin 3t The expression n 1 I c. 1 0r(9)
12. A. 91 0 . 9(8) r(8)
9
27 . Find f(t) if L(f(t)) = (s+ z;~s - 6 ) B. 9 r< ) . · e form
rier series ,n cos1n .
A. e·21 + 3e61 C. 2e·21 + 3e61 one term of the Fou . the expression ,n
8 . e·21 + 2e61 D. 5e·21 + 3e61 13. is 6 cos20nt. Wnte
exponential form . 20irt)
28. Determine the Inverse Laplace Transform of A 5e0201Tll + seH
(s + 2)/(s2 + 4s + 20) a· 3eo20trt> + 3eH20m)
A e-31 cos 4t C. e·21 cos 2t c· 4e020lrt) + 4e<·J20n1)
B. e-21 cos 4t D. e·31 sin 4t D.. 6e0201TI) + 6e(•l201f1)

EXERCISE NO. 13 · eries e 01 om>+

14. Ev
aluate the terms of a Fourier s
Advanced Engineering Mathematics - 2
e<•11om> at t= 1. D. 4
A. 0 B. 1 C. 2
1. Find the eigenvalues of A = (! ~)
A. 5,2 C. -5,-2 15_ Sine function is said to be
B. 5,-2 D. -5,2 A. Odd function
s . Symmetry function
2. Determine the eigenvalues of the following c. Even function
matrix : [~ ~)
o. Unsymmetry function
A. 3,4 B. 2,5 C. 1, 6 D. 4, 1 16. The maximum period for the limits of the
Fourier series
3. Find an eigenvector of[! ;). A. 0° B. 90° C . 180° D . 360°

A l~1) B. [!1 C . [~) D. l-23) 17. The series expansion x + (1/2) x2 + (1/3)x3 +
(1/4 )x4 + ... is equal to the summation
4. How many sets of eigenvectors for the matrix • x"
A. L~=l x" C. Ln- 1 -n
1 2 3
-2 -1 - 2? x2n+1
BL- - • x"
3 1 4 · n • l 2n+l o.rn ..1-
nl
A. 1 B. 2 C. 3 D. 4
18. What is the equivalent function of f(x)
5. A matrix composed of eigenvectors. • xn =
A Eigen matrix C. Modal matrix Ln..,O;i?
B. Spectral matrix D. Inverse matrix A. eK B. sin x C. COS X D. In x

6. Find the modal matrix of A=(; ~). 19. Find the Maclaurin series of f(x) == x2ex_
A. r;,,.oxn+i C :,E• xn+z
A. (11 -3)
1
C. (34 -1)
1
· n=o-;;i-
B. (14 3)
1
D. (-31 1)
4
•
B. Ln=O (2n)t
x2n
D . ~• xn
£.n e o -
n
7. A diagonal matrix with eigenvalues. 20. What is the coefficient of x• ter
A. Eigen matrix C. Modal matrix series of cos x at x == Q? m of the Taylor
8 . Spectral matrix D. Inverse matrix A. 1/4 B. 1/6 . C. 1/12
D. 1/24
B. Inverse of modal matrix (M) times the given 21 . What is the coefficient of (x _ .
matrix (A) times the modal matrix (M) is equal 1 2
Taylor series expansion of ~( term in the
to expanded about x == 1 7 x) == In x
A. Eigen matrix C. Modal matrix
A. -1 B. 1/2 C. - 1/2
B. Spectral matrix D. Inverse matrix D. 1/6
22. Suppose that f"(x) == f(x) f
9. The equation (A - Al)x has zero as a and notice that by differ or ~II values of x
characteristic root if and only if A is _ _ 4 0
that f( > (x) = f'(x) , f(S>(x) = ?~at,ng found out
matrix
A. Inverse so f~rth. Suppose that f(3) == ( ), ... so on and
C. Singular
B. Modal 3. t1nd the coefficient of th 1 , f(3) = 2, f'(3) =
D. Diagonal 3) . e term having (x _
A. 3/2 B. 1/6
1 o. Find L[f(t)) of f(t) = t"( - 1/2 ) C. 1/12
A. ✓n I s C . ✓(n/s) D . 2/3
B. 1/2 ✓(n/s) D. 1/2 ✓n

MPHT Review Center. REE Reviewer In MATHEMATICS

 23. Let f(x) = ✓x and a = 16. Find the Taylor 35. He was very influential in the standardization
polynomial degree 2 for f(x) at a. of other mathematical terms and notations. He
A. 4 + 1/2 (x-16) + 1/16 (x-16) 2 named the square root of - 1 with the symbol
B. 4 + 1/4 (x -16)- 1/64 (x-16) 2 of i.
C . 4 + 1/8 (x - 16) + 1/256 (x- 16) 2 A. Euler C. Euclid
D. 4 + 1/8 (x - 16)- 1/512 (x- 16)2 B. Fermat D. Descartes

24 . The series is said to be if the limits of

~h~ sum as the number of terms approaches Most Valuable Assessment (MVA) Test In
inftnrty exists and is equal to the sum of the Mathematics - 1
given infinite series .
A . Divergent C. Dissonance MULTIPLE CHOICE
B. Convergent D. Assonance 1. Find the rectangle of largest area that can be
inscribed in an equilateral triangle of side 20.
25. An harmon ic series is said to be A. 45✓2 C. 50✓3
A. Divergent C. Dissonance 8 . 39✓5 D. 56✓7
B. Convergent D. Assonance
2. Bob is 2 years from being twice as old as
26 . Find the Fourier series expansion of Ellen. The sum of twice Bob's age and three
f(x) = 0 -5 <X<0 times Ellen's age is 66. How old is Ellen?
f(X) = 3 0 < X< 5 A. 15 8. 10 C. 18 D . 20
Period= 10
A.
. + :E~_,1(2._
;
nn (1 - cos nn]) sin(mtx)
S
3. What is the maximum area of a rectangle
whose base is on the x-axis and whose upper
~ + L~=1(;; n[1 - cos nn]) cos (";x)
3
B. two vertices lie on the parabola y = 12 - x 2 .
C. ~ + L~c 1 (~: (1 - sin nn]) sin(";") A. 16 B. 32 C. 64 D. 8
3
D. ¾+ :E~. 1(n n [1 - sin nn]) cos (";x) 4. Harold used a 3% iodine solution and a 20%
iodine solution to make a 85-ounce solution
27 . If n is any positive integer, then n(n - 1)(n - that was 19% iodine. How many ounces of the
2)(n - 3) .. . (3)(2)(1) = 3% iodine solution did he use?
A. e" (n - 1) C. n! A 5 B. 80 C. 60 D. 20
B. (n - 1 )! D. (n - 1)"(n)
5. The manager of a garden store ordered two
28 . Given the Fourier series in cosine form, f(t) = different kinds of marigold seeds for her
5 cos (20pi)t + 2 cos (40pi)t + cos (80pi)t. Find display. The first type cost her $1 per packet
the fundamental frequency. and the second type cost $1.26 per packet.
A. 20 B. 40 C. 10 D. 60 How many packets of the first type did she
purchase if she bought 50 more of the $1.26
29 . The symbol "/ " used in division is called packets than the $1 packets and spent a total
A Modulus C. Solidus of$402?
B. Minus D. Obelus A. 150 B. 200 C . 250 D. 100

30 . The constant Ue" is named in honor of: 6. Two angles are complementary. The larger
A. Euler C. Euclid angle is 15° more than twice the smaller. Find
B. Eigen D. Einstein the measure of the smaller angle.
A 25° B. 65° C. 90° D. 82.5°
31 . What do you call a radical expressing an
irrational number? 7. If 3x2 is multiplied by the quantity 2x3y raised
A. Surd C. Complex to the fourth power, what would this
B. Radix D. Index expression simplify to?
A. 48x 14y4 C. 6x9 y 4
32 . f is a function such that f(x) < 0. The graph of B. 1,296x16y" D. 6x 1"y4
the new function g defined by g(x) = lf(x)I is a
reflection of the graph of f 8. Sara's bedroom is in the shape of a rectangle.
A . on they axis C. on the line y = x The dimensions are 2x and 4x + 5. What is
B . on the x axis D. on the line y = - x the area of Sara's bedroom?
A. 18x C. Bx2 + 5x
33 . If f(x) is an odd function, then I f(x) I ls B. 18x2 D. 8x2 + 10x
A. an odd function
B . an even function 9. What Is the minimum possible perimeter for a
C. neither odd nor even rectangle whose area is 100 in 2?
D. even and odd A 10 B. 20 C. 30 D. 40

I 34. Find the 1Oth term of the series 1, 4, 9, 16, ...

A. 80 B. 88 C. 96 0. 100
10. :atrick has a rectangular patio whose length
~s 5 m less than the diagonal and a width that
1s 7 m less than the diagonal. If the area of his

I
 patio is 195 m2 what .
diagonal? ' is th e length of the . ed 15 feet from a
20. A 6-foot spruce tree 15 plant . feet above
18
A 10 m B. 8 m C. 16 m lighted streetlight whose lamp 15 _ •s the
D. 20m the ground. How many feet long '
11 . Samantha owns a rectan shadow of that tree? 6
an area of 3 280 gular field that has A. 5 B. 7.5 C. 7.8 D. 9 ·
~he field is 2 ~ore ~~~~r~reet. The_length of
is the width of the field? ce the width . What 21 . A 1-inch diameter coin is throw~ on a tab~e
A 40 ft . covered with a grid of lines two inches a~a ·
B. 82 ft C. 41 ft What is the probability the coin l~nds 1n a
D. 84 ft square without touching any of the lines of the
12. A river is 1 mile wide . Frank we t t grid?
· t A · n s o get from A 3/4 B. 1/2 C. 1/4 D. 0
~om to_point B on the opposite side of the
n~er, 3 miles downstream. If Frank can run 5 22. In a roomful of 30 people, what is the
miles per hour and can swi·m 3 .
.
wh at is m, 1es per hour probability that at least two people have the
the least amount of time in wh . h h ' same birthday? Assume birthdays are
can get from A to B? ,c e
A 52 min uniformly distributed and there is no leap year
C. 76 min complication.
B. 35 min
D. 27 min A 0.31 C. 0 .81
8 . 0.71 D. 0 .61
13. A garden in the shape of a rectangle is
surrounded .by a walkway of uniform width. For problems 23 to 25:
The d1mens1ons of the garden only are 35 by Let S be the triangle with vertices A = (2, 2, 2), B
2 4 . The area of the garden and the walkway = (4, 2, 1) and C = (2, 3, 1).
to_gether is 1,530 square feet. What is the 23. Find the cosine of the angle ABC at vertex A
width of the walkway in feet? A. 1/✓10 C. 1/✓7
A 4 ft B . 5 ft C . 34.5 ft D. 24 ft 8. 1/✓5 D. 1/✓11
14 . Bill and Ben can clean the garage together in 24. Find the area of the triangle ABC .
6 hours. If it takes Bill 1 O hours working alone, A 0.7 8. 2.4 C. 1.5 D. 3.2
how long will it take Ben working alone?
A 11 hours C. 16 hours 25. Find a vector that is perpendicular to the
B. 4 hours D. 15 hours plane that contains the points A, 8, C.
A <0, 1, -1 > C. <1, 2, 2>
15 . A rectangular garden has a width of 20 feet 8. <1,-1,0> D. <-1,0,2>
and a length of 24 feet. If each side of the
garden is increased by the same amount, how 26. Sand is being dumped from a dump truck at
many feet is the new length if the new area is the rate of 1O ft 3/min and forms a pile in the
141 square feet more than the original? shape of a cone whose height is always half
A. 23 8 . 24 C. 26 D. 27 its radius. How fast is its height rising when
the pile is 5 ft high?
16 . Find the rectangle of largest area that can be A. 0.025 ft/min C. 0.032 ft/min
inscribed in an equilateral triangle of side 20. 8. 0.011 fUmin D. 0.047 ft/min
A. 45✓2 C. 50✓3
8 . 39✓5 D. 56✓7 27. A stand~rd deck of 52 playing cards consists
of 4 su,ts (spaces, hearts, diamonds, and
17. In parallelogram ABCD, mLA = 3x + 10 and clubs) of 13 cards each. How many different
mLD = 2x + 30, find the mLA. 5-card hands can be formed?
A 700 B. 400 C. 860 D. 940 A 3,123,261 C. 3,838 380
a. 2,59a,sso o. 4 , 1 so'.ooa
18. If the radius of a right cylinder _is doubled and
the height is tripled, its volume 1s 28. An ~irplane is built to be able to fl on one
A multiplied by 12 engine. If the plane's two . Y
B. multiplied by 2 independently 8 d engines operate
f T . ' n each has 1% chance of
C. multiplied by 6 ~ ing in any given four-hour flight what is the
D. multiplied by 3 c ance the plane will fail tO '
hour flight to Cebu d ~omplete a four-
19. The dimensions of a . re~ngle are A. 0.019g ue to engine failure?
continuously changing; The width increases et B. 0.0001 C. 0.019
the rate of 3 in/sec while the length_decreases D . 0.81
t the rate of 2 in/sec. At one instant the 29. A point Is movin
1
a le - a 20-inch square. How fast 1s Its
rectang 15 r? in the first quad~aa t°':'g the circle x2 + y2 = 25
I
are a changing 3 seconds
. late coordinate change: ~ such a way that its x
A. 16 in2/sec, increas1':g How fast is its y a .th e rate of 2 cm/sec.
8 16 in2/sec, decrea~Ing point passes throu~~o~1•nate changing as the
c·. 27 in2/sec, increas1':g A -4/3 Cm/sec ' 4 )?
0 . 27 in2/sec, decreasing B. -7/4 cm/sec C. -3/2 cm/sec
D. -213 cm/sec
 43 For the 2 functions, f(x) and g(x), table s of
30. A pair of fair, standard dice are rolled. What is . values are shown below. What is the value of
the probability that the sum of the dice is 5?
A 1/9 B. 2/9 C. 1/36 D. 5/36 g(f(3))?
X fX X
-5 7 -2
For problems: 31 to 33
-2 -5 1
If P(A) = 1/2 and P(B) = 1/2 and P(B/A) = 1/3 .
1 3 2
31. Find P(A and 8)
3 2 3
A 2/3 B. 5/6 C. 1/6 D. 1/2

A -5 8 . -3 C. -1 D. 2
32. Find P(A or 8)
A. 2/3 8 . 5/6 C. 1/6 D. 1/2
44 . Jodi wishes to use 100 feet of fen~ing to
33. Find P(A/8)
enclose a rectangular garden. Determine the
A 2/3 B. 5/6 C. 1/6 D. 1/3 maximum possible area of her garden.
2
A 144 ft 2 C. 225 ft
2
34. Find the volume of the solid of revolution 8 . 625 ft 2 D. 96 ft
formed by rotating the region bounded by the 3
= =
parabola y x2 and the lines y 0 and x = 2 45. Find the area of the region bounded by y ~ X
about the x-axis. _ 3x2 + 2x + 1, the x-axis, and the vertical
A 64rr/15 C. Brr =
lines x O and x 2. =
8. 256rr/25 D. 32rr/5 A 1 B. 2 C. 3 D. 4

3t5. What are the odds in favor of rolling a sum of 46. Justin earned scores of 85, 92, and 95 on his
seven in one roll of a pair of fair standard science tests. What does he need to earn on
dice? his next science test to have an average
A 1:6 B. 1:5 C. 6:29 D. 6:36 (arithmetic mean) of 93%?
A 93 B. 100 C. 85 D. 96
36. Find ln(3 + 4i)
A 1.61 + 0.93i C. 1.01 + 3.56i 47. Brad's class collected 320 cans of food. They
8. 1.23 - 4.1 i D. 1.91 -1.23i boxed them in boxes of 40 cans each . How
many boxes did they need?
37. Find the area of the region above the x-axis A 280 8 . 10 C. 8 D. 5
bounded by the function y = 4x - x 2 - 3.
A 1/3 8 . 2/3 C. 1 D. 4/3 48. Find the volume of the solid revolution
obtained by revolving the region bounded by y
38. Listed below are 4 functions, each denoted = x- x2 and the x-axis about the x-axis.
g(x) and each involving a real number A rr/15 C. rr/60
constant c > 1. If f(x) = 2x, which of these 4 8. rr/30 D. rr/1 0
functions yields the greatest value for f(g(x)),
for all x > 1? 49. Joey participated in a dance-a-then. His team
A g(x) = ex C. g(x) = x/c started dancing at 10 A.M. on Friday and
8 . g(x) = c/x D. g(x) = loge>< stopped at 6 P.M. on Saturday. How many
hours did Joey's team dance?
39. If the function f satisfies the equation f(x + y) = A 52 8. 56 C. 30 D. 32
f(x) + f(x) for every pair of real numbers x and
y, what are the possible values of f(0)? 50. C~llie's wandmother pledged $0.50 for every
A Any real number mile Callie walked in her walk-a-thon. Callie
8. Any positive real number walked 9 miles. How much does her
C. 1 only grandmother owe?
D. 0 only A. $4.50 C. $5.00
8 . $18.00 D. $9.00
40. The imaginary number i is defined such that i2
= -1 . What does i + i2 + i3 + · ·· + i23 equal?
A i 8 . -i C. -1 D. 1

41. Find the area of the region bounded by y = x 2

- 5x + 6, the x-axis, and the vertical lines x =
0 and x = 4.
A 25/6 C. 16/3
B. 18/5 D. 12/7

42. lf the first term in an arithmetic series is 3, the

last term is 136, and the sum is 1,390, what
are the first 3 terms?
A. 3, 10, 17 C. 3, 36 1/3, 70
8. 3, 23, 43 D. 3, 69 ½ , 136
 Engineering
Mat hematics
Solutions and Answer Keys
• EXERCISE NO. 1: Algebra - 1
• EXERCISE NO. 2: Algebra - 2
• EXERCISE NO. 3: Trigonometry
• EXERCISE NO. 4: Plane and Solid Geometry
• EXERCISE NO. 5: Analytic Geometry - 1
• EXERCISE NO. 6: Analytic Geometry - 2
• EXERCISE NO. 7: Differential Calculus
• EXERCISE NO. 8: Integral Calculus
• EXERCISE NO. 9: -Diffe·r enti·a·I Equation
• EXERCl·S E N·O. 10: Plane . ·, and:Spa·c e:-,.~Vectors
• EXERC.I SE .N\O ., t1 .:...Statistic.S:':-& f:P,r.o:bability -
• EXE RGt:~:E ):N·O ~··.t2:: ·. Ady·a;n:~e'tf~
:f;:n·.gi:i.ri1·eeri ·n .g .'.· ·
. :.. · . ·. .· · .·_·: · ·. :·.:. ·.:·'.: M·-~tlt¢·~ -~·,:1~sft?\j_·i:_ .-_·: . ._.-.-. · -. - -.: :
• E),ERSJS'.8 ~~N:Q .i':.:'.1"3t;.!'.AQ~at1·.e:(fd1~E;r:,giH1,·e e·t i·ng;•''
1
. ·:
-. ·· .··.·.. ·:·· :.-.·::. -._·..·. ·_....·.,--:.·,-:_:M at· ~ :~ ·
R i, t1&~ ·
:; .:.
1

:·.:-:.2 ·:>·.·.'· . ·· _.·· · . . .

' . ;.. ,·.'·. i ,\. ,'. ·!:. ,~ ,_ :. ·.. ' .•. ~ .'t .. ,:·· _·· "·. ._· '
•·· .M ost,_V a I wa.b. ~-~-~.4~ls:s:e·~ s:~ )~.:n.~··i;(: MNtA)/:~.e st .:.in·-
·Ma'therriatic·s:··~.: 1' ·. · . . . . :: · · · · · ·
.,
~ . . . .
' '

.., ,._
. '
, .·' .l '. •·: -
.
.. ·. . . ' .
 2
11 . By Discriminant: 8 - 4AC "' 0
Marin.a~tsH · (-4m)2- 4(4)(4m+5) = 0
1,(4 Power ouse m, = -1 and mz =5

PH.T , Team REVIEWandTRAININGCENTER 12. By Discriminant:

~~-~ , "Because the Power Is In the HouseH B2 -4AC = 0
k 2 - 4(3)(12) = 0
SOLUTIONS TO THE PROBLEM SET k = 12

EXERCISE NO. 1
Algebra -1 13. !/.../64x 30 = 3.J../64 3..J-Jx3o
= 2 {(x30)113}112
1. By Shift - Solve: = 2x6
3 + 8x = 35
x=4 14. (x+y) 10 , 4 1h term (r = 4)
n = 10, r - 1 = 3, n - r + 1 = 7
2. By Shift - Solve Formula
1
6x-2= 2x + 6 rth term = (nCr-1) xn•r• 1 yr-
x=2 = (1 0C3) (><4+ )(y3 )
3

= 120x1 v'
3. By Shift - Solve
✓ zo - X =X 15. (x - y) 15 , 12th term (r = 12)
n = 15, r - 1 = 11, n - r + 1 = 4
x--4 and x = - 5

4. Quadratic Equation: Ax.2 + Bx + C = O Formula

1
rth term = (nCr-1) xn-r+ 1 yr-
Use Mode 5 3 and input all coefficients 11
= (15C11) (X°4)(-y)
and constants
b C
= - 1366x'y11
a
1 1 -12
16. (x2 - 12)6 , 51h term/middle term (r = 5)
x1 = 3 and x2 = - 4
= =
n 8, r - 1 4, n - r + 1 = 4
Formula
5. Solve the value of x rth term = (nCr-1) xn-r+ 1 yr-1
=
42x+ 1 1024 = (8C4) (x2 ) 4 (- 12)4
x=2 = 43760X-
Then, 4>< = 4 2 =16
17. (x-3y) 12 , 6 1h term (r = 6)
6. 1093 (81)() =16 n = 12, r - 1 = 5, n - r + 1 = 7
x=4 Formula
rth term = (nCr-1) xn-r+ 1 yr-1
log22 + log:v< = 2 5
7. Coeff. of rth term= (12C5) (1)7(- 3)
x=2 = -192.456

8. Enter Mode 5 3 18. Sum of coefficient (SOC)

a b C
(a+b-c) 6
1 11 28 soc = (1 - 1 + 1)6
x1 = - 7and xz = - 4 soc =1
9. A = 7; B = 2k - 1 ; C = -3k + 2 19. Sum of coefficients for (ax + by + ... + k)"
- B C
SOC= (a+b+ .... +k)" - (k)"
A A (4x - 3)0 , k= - 3
-8 = C
- (2k - 1) : -3k + 2
soc = (4 - 3)g - ( - 3)9
soc= 19684
k=1
20 . Sum of exponents for (ax+ by)"
10. A=2andB =- k SOE = n(n+t ) ( exp. ofx + exp. ofy)
-B 2
-=4
A For (x + y} 10 ,
- (-k) SOE=½(10){11)(1 +1)
-=4
2

I lc=B
SOE=lli

P :::i!Ye 24
I
 2 1. Sum of ex
ponents for (ax+ by)"
SOE :: n(n+t) 30. The man - hour is directly proportional to
2 (exp.ofx+exp.ofy) the production
2 hrs (Z men) == 60 men (200 hrs)
For (2x + 1)7,
1 unit U2
SOE = ½ (7)(6) ( 1 +O)
U2 = 3000
SOE= 28

22. Solve for x from the d IvIsor

' .
EXERCISE NO. 2
X-2::Q ➔ X=
2 Algebra - 2
Input the dividend
R = x" - 2 x3 - 3x2 - 4x - 8 1. 82 =1 X 2 =2
Press CALC ., then enter 2 83 =2 X 3 =6
R =-28 a,c = 6 X 4 = 24
as = 24 x 5 = 120
23. x- K =O ➔ x= K 815 = 120 X 6 = 720
= =
R f(x) (x + 4}(x _ 3)+ 4 Then the missing term is 120
-K = (K + 4)(K - 3)+4
K=2 2. a1 = 1
82 = 1 + 3 = 4
83 = 1 + 3 + 5 = 9
24. X+ 1 =0 ➔ X :: -1
84 = 1 + 3 + 5 + 7 = 16
Solve for A if x = - 1 and f(x} = _
3 2
15 as = 1 + 3 + 5 + 7 + 9 = 25
Ax + 2x - 1 Bx + 17 = -15
A =42
By inspection, Sn = n 2

3. For an= 2x- 1 ,

25. Enter Mode 5 4
Ss =Z:~= 1 (2x- 1 )
Assuming that the equation is x3 - Sx2 + Ss = 31
2x + 8
A B C D 4. Sn = 2°• 1 - 5
1 -5 2 8 The sum of the first 5 terms
X1 = -1 ; X2 = 4; X3 =2 Ss = 2s• 1 - 5 = 251
Correct Equation: x3 - 5x2 + 2x + 8 = o The sum of the first 4 terms
S◄ = 2◄+3 - 5 = 123
26. R = k yxz as= Ss- S4 = 251 - 123
8s=m,
R1Yf R2Yi
-- = --
X1 X2 5. Enter Mode 3 2
(1 00)(3) 2 (R 2 )(6) 2
X y
4
1 1.01
2 1.00

27 . R = p!:. Press AC, then 100 Shift Stat R •

A a100 = 1OOy = 0.02 eg Y
But the area remains constant, then
R t= kl ,
1000 = k(10m}
k = 10
R2 = k (12) = 120
6. rTO;l32
28. By ratio and proportion, Press AC, then 31 Shift Stat Reg R
16 : 12 :;: 12 : X n == 31~=!1,
16 12
- =-
12

29. P
x=9
X

=
kVI
--;i-
7. r-I~bi~2
Press AC,then 500 Shift S
(250 mW)100km - (300krn)P2
n :: soon - 2 tat Reg Sc
(240 mW)l,500A - 500kV(2,000A) I\ - 46 terms
p 1 = 230 lcV
The surn of 246 even ,·nt
~- (X · egers
L Y, 1, 246) = 82. 73.Q
 8. 1E•r7-r=-+--
2 3+5
15. For hannonic progression, enter Mode 3 2
to solve the arithmetic progression
a1 = 2, 83 = 6
~3 r 3 + s+ 7
Press AC, then 2600 Shift Stat Reg ~1 2
n = 26005' 1 = 50 terms 6

9. a◄ = a1 + 3d a3 = 01 + 2d Press AC, then 11 Shift Stat Reg y

a2 =a,+ d a11 =11y =22
In harmonic progression, a,, = 1122
(a1)(a1 + 3d) = 70
a,2 + 3a,d = 70
16. For harmonic progression,
a, = 1/2, a2 = 1/6
(a, + d)(a1 + 2d) = 88 Enter Mode 3 2
(a,2 + 3a1d) + 2d 2 = 88
70 + 2d 2 = 88
d=3
a, 2 + 3a,d = 70
a, 2 + 3a,(3) = 70
a,=§.
f±ffl
Press AC, then 8 Shift Stat Reg
as= By= 30
y

In harmonic progression,
10. Enter Mode 3 6 as = 1/30 = 0.033
X y
3 20 17. Let x = age of Peter and y = age of Paul
6 160
For their present ages
Press AC, then 1 Shift Stat Reg y X + y = 21
a,= 1y = §.
y = 21 -x ➔ Equation 1
3 years from now,
11 . a, = 0.05(2) = 0.1 x + 3 = 2(y + 3) ➔ Equation 2
a2 = 0.1 (2) = 0.2 Combining equations 1 and 2
Enter Mode 3 6 X + 3 = 2(21 - X + 3)
x= 15
1
2 18. Let B = Beth's age and A = Ana's age
Press AC, then 12 Shift Stat Reg y A+ 5 = Ana's age after 5 years
812 =12y =204.8 in. B + 5 = Beth's age after 5 years

12. For geometric progression, 81, a2, a3, ... Eq. 1: A= B + 5

a2 a3 Eq. 2: (A + 5)(8 + 5) = 1.5 (A)(B)
r=-= -
a1 a2 Combining Eq. 1 and 2
2x + 7 10x - 7 (8 + 5 + 5)(8 + 5) = 1.5 (B + 5)(8)
--=---
x 2x + 7 B = 20
Solve for x
x =I.or x = - 716 19. Let x = number of years ago and M = 36
36-x=A
13. Enter Mode 3 6 x=36-A

1 36 - X = 2 (A - x)
3 36 - (36 - A) = 2 (A - (36 - A)
A= 24
Press AC
The sum of the first 10 terms
L (Xy, 1, 10) = 29,524 20. Let x = amount of 7% solution
y = amount of 12% solution
14. For infinite geometric series
a1 For amount of solution,
S°' =-1-
-r
x+y=5 Eq. 1
l . Consider the concentration of solution
2 = _ 2_ 7x + 12y = 10(5) Eq. 2
1 - r
r = 3/4

MPHT Review Center: Solution and Answer key in MATH Page 26

 I against the wind
Airplane trave .
Enter Mode 5 1 soo mi == 400
R + C = (1 + G:)) hr
a b c
1 1 5
7 12 10(5) Enter Mode 5 1
b C
x = 2 L (amount of 7% solution) a
y = 3 L (amount of 12% solution) -1 200017
1
1 400
1
21 . Let x = % concentration of the mixture
c = 57 .14 mph
R = 342.85 mph
A B C A+B+C 27 Let R = speed of the boat
10L + 15L + 30L SSL . C = speed of the current
25% 35% 10% X
For upstream
100 200
25(10) + 35(15) + 10(30) = 55x
X = 19.55 %
x= 20 %
For downstream
22. Let T = total time it will take to fill the tank 100 mi 80
with all three pipes open R+C = (1 + G:)) hr =
Sum of individual rates = Combined rate
( +) rate for the pipe to fill the tank Enter Mode 5 1
( - ) rate for the pipe to empty the tank a b C
200/3
.6?: + .3?: + (-1-)
-24
=.?:T 1
1 - 1
1 80
T = 2.18 hrs R = 73.33 mph
C = 6.67 mph
23. For a job to finish,
Lorna's rate = 1/5 28. For motion around a circular path in the
Fe's rate= 1 /4 same direction,
Work done by Lorna= (1/5)(2) D = (R1 - R2)(T)
Work done by Loma & Fe= (1/5 + 1/4)(X) = =
For D 1, R1 1/8 and R2 1/12 =
(1/5)(2) + (1/5+1/4)(X) = 1
x = 413 hours (!8 - _2_)
12
(T) = 1
T = 24 minutes
24. Rate of boy= 1/B
Rate of the man = 1/M 29. For a = 0° and H = 8
= 60(H) ± 2(8)
~ = 3 (½) M --1-:--1--
1 1 1 M = 60(8) + 2(0)
-+ - = -
B M 6 11
1) 1 1 M = 480/11 minutes
3 ( ii + s=6 M = 43 7111 minutes
B = 24 hours, M = 8 hours
30. For a= 90° and H = o
25 Let X = number of days to finish the job M = 60(H) ± 2(0)
. Total Man - days= 12 men (13 days)
11
For 4 days, M = 60(0) + 2(90)
the man days is 5 (4 days)
11
For the remaining days, M = 16.36 minutes= 16'21 "
the man days is 8 (X - 4)
12(13) = 5(4) + 8(X - 4) Time: 12:16:21 A.M.
x = 21 days
26 Let R = speed of the ai~plane
· · C = speed of the win~
Airplane travel with the wind
500 2000
R- C = ( 1 + c::)) hr= - 7 -
 9. tan (75) = x/1
X = 3.732 km
1. When B = 270°, let A= 5°
Type the given function: sin (270 - A)
10. By sine law, _ a_== _c_ _ b
Press Cale. (A?) and enter 5 s ln A I - -
A - 140 B - s nc s tn B
sin (270 - A)= - 0 .9962 if A= 5 - · - 20 , C = 20
Among the choices, substitute A = 5 Y 23
sin140 - sln2Q
For choice A, solve - cos A
y = 43.23 m
- cos(5) = -0.9962
Then sin (270 - A) = - cos(AJ

2. sin A= 4/5 and sin B = 7/25

A= sin-1 (4/5) = 53.13
B = sin• 1(7/25) = 16.26
Since A is in the quadrant II
A= 180- 53.13 = 126.87
sin{A + B) = sin{A + B)
sin(A + B) = sin(126.87 + 16.26) = 315

3. tan A= 1/3 By right triangle approach

A= tan·1 (1/3) = 18.43 h
tan B = 1/2 _ = sin 20
43 23
B = tan· 1 (1/2) = 26.57 h =14.78m
tan {A - B) = tan{18.43 - 26.57)
=-0.142 11 . c2 = a 2 + b 2 - 2abcosC
tan (A - B) = - 1n c2 = 252 + 162 - 2(25)(16)cos(94.1)
C = 31
4. cos e = -12113
e = cos· 1 (-12/13) = 157.38 Another Solution: by Mode 2
sin 0 = sin (157 .38)= 5/13 c = IALC-81
But sine in QIII is ( - ) c = l23L 94.1 -161
Then sin 0 = - 5/13 C = 30.63 = 31

5. Enter the equation 12. ~_y Sine Law

2(sin(x)) 2 + 5 cos (x) +1 = O
x= 120 \. "\.~
Solve again the equation and enter 360 !.~o•~~/~
x=240

6. tan (4A) = cot (6A)

1
h \?{~r;:Jt( ,
tan(4X) = tan( 6 X) \_ !'-'~;;".,,> e 6L
X=9
Enter mode 5 1
30 m
7. sln(4.8) = -x- a b C

cos 118 cos 26 32

X = 358.52 m 0
sin 118 -sin 26
X = 23.87, y = 48.07
h = x sin(62)
= 23.87sin(62)
h = 21 .076
With the observer's height,
H = h + 1 .8 = 22.BB

8. Let the radius be one of the legs of the

right triangle formed and the distance
from the center of the Earth to the top of

-
the mountain be the hypotenuse.
(6400+3)2 = 64002 + d 2
d = 198 km ... -, n
 A • B ., C = 3200 ,nil s
ix • x • 1S • 2 x ~ 1 5 = 3200
'( 1-"" J.Q
2x 60
u
2Y 5 ::: 75
♦
The sma ·ie-s-t erx,io

X + 15
:: ,15
·
~(~-.!"~}
J60
= 800 mill

21 A = 50• 8 "' so•. C " 9Cf'

By sm - ·oo - op rule of NSP4er's circle
, ::-, 5 0 .3~ sin(90 _ c ) =- co, A cos B
h ~ > • ~c = ,o.p m &in( 90 - 80) = cos(SO ) cos B
a = 74 .32 = 1.r2o·u"
-1 Le-:_" bBJ ~ . ~~ '--e gh • a •~ po.e
-- - - -- ~
- --
#

, ',5" - .._ 22. FOf 2 ,4 hoc.T! , the Earth rotates 360•

ti l o ngitude 6 li m e
3 60 = ~
135 t:1 ti me
-==--
y f 360 24
1

k
'1 time= 9hrs
10 . l
From 7:00 AM (GMT), at the East
0
~ _43-._ - J4 7:00 + 9 = 16:00 houn

- ·- - - -- 9
---------- -------•h Time: 4 :00 P.M.

23 . From 9 AM. (GMT) ,at the West

£1 longitude £1 time
360 =~
45 £1 tlme
360 = 24
~$ 4 C06 yam y { 1 - 2 ( ti n y)2) 6 time= 3hrs
~ 2 s,r, 2y ((001 y }' + (sin y) 2 - 2 (sin y) 2) 9:00 - 3= 6 :00 hours
= 2 s,n 2y ((cos y) 2 - (sin y)2 ) Time: 8:00 A.M.
.. 2 sin 2y 001 2y
;; ,10 'fr 24. For oblique spherical tnangle,
cos (A)= - cos (8) cos (C) + sin ( 8) sin
(C) cos (a)
~( e ::: oos · 1 h 3t2) • 30•
1 - tan~ • 1 - (tan (30·)) ~ Z/1
~ (54) = - cos (54) cos ( C) + sin (54) sin
(C) cos (82)
., 7 ~L~ -~ be the !tetqt:1 .?" the I ghlhouse --1
C: 15So18'43''
; I
25 E ;:; A • B + C - 180
E e: 50• 8.9+ 120 - 180= 7g•
Atee "' " (• 1>{!]

~ ___s_o_~_
I AC

Area .. !.i~~::l'.,.,,!
l lcQ 1.80

Area ~ ~6.. 11.ti,.,Q(uUu uz

a,n A . 'JOI;; ... Y I 1 rn
.,, : 14. ff f/J 35 50 graaients (3001400)

, e oo - e • uo (, oo -~ o, 50 graa1ent1 0 ~§ dfo!lt,t

I• JD· 113 174 dev1oo a (t14 00rJc-oJ

17 4 ffOQr oc,1 • Jfllit?ll.ll.
19 300 - ~- .. S( 180-..:)
x • 135 40 I or an y tat'lQCnt C\Jf'<ltl Y = a t

20 3oo •.: 6 400 m.lla

f>611oc, , 1 ntm. th.e n f0< Y a: ta
Pun0<1 18 !l.Q. n X. the
a;
mx, the

For tnan~le , the sum of tho angloa ,s

=
s 180 3200 mds =
·"' w centen Solutlon and Aruwer k.e y In MATH
MPHT R ev,e

Page 29
 13. For the area of the trapezoid
Plane and Solid Geomet b1 + b2]
A= h [
2
=hm
5. For angles of ti1e polygon, A= 8(54)
Stnt = 180° (n - 2), Sext = 360° A= 432 cm 2
Stnt = Sext
180 (n - 2) = 360
n = 4 sides
14. For the rhombus,
6 . 2Stnt = 3Sext P = 4s
2 (180(n - 2)) = 3 (360) 56.33 = 4s
n = 5 sides s = 14.083
7. For dodecagon, n = 12 sides
n
d = 3)
2 (n -
d = 212 (12-3)
d = 54

8. For dodecagon, n = 20 sides

n
cl = 2 (n - 3)
20
d = (20- 3)
2
d = 170

9. For a polygon inscribed in a circle

R = 10, n = 6 sides
A= n~z sin
2
c:o)
A 6(10) sin (360)
= __;___:._ -
A= 259.81
2 6 13
By the similar right triangles,
13 -
10. For any polygon with "n" sides cos e = 14.083
n = 8 , x = 5 cm a= 22.s2°
nx 2
A = - - -0-
4tan
8(5) 2
e: )
r = 13 sine= 13 sin (22.62)
r = 5 in.
A = nr2 = TT(5) 2
A= 4tan ciao) A= 78.54 in 2
8
A= _12- ;-;~_c_m_ 2
15. The diagonals of the rhombus are
11 . Apothem - line segment from the center perpendicular, then the angle opposite the
sides is 90°
of regular polygon to the midpoint of one
of its sides. The radius of the inscribed
d1 = 12cm 10
circle
For a circle inscribed in a polygon, d2 = 2x
n = 11 sides, R = 8 s = 10 cm
180)
A= nR2 tan ( -;-
180
A = 11(8) 2 tan ( ) 11
A= 206.71

12. For a pentagram ( star) inscribed in a

circle , By Pythagorean theorem
52 + x2 = 102 I

r= 6 cm x=8cm
d2 = 16 cm
A=½ d1d2 = ½ (12 cm)(16 cm)
A= 96 cm2

A= l.123r 2
A = 1.123(6) 2
A = 40.428 cm 2
M O llT ,.. _ __, --·
 16.
Tt2
0
e are~ of rhombus is equal to the area
equilateral triangles By Pythagorean Theorem,
x2+ y2 = 132
,-,'.< ' :.,,-· ....._
6 ( ' ;,'!7• " ~ -" Y = /132 - x2
/,,,,.,,. .,,,.,- 6 . .u _"
"'~> The area of the triangle
A=½XY
( /'_,, ,.,..,. ''. 30 = ½ (X)(/ 13
X=5cm
2
- X
2
)

\ • , ✓•
,.------;:-
y = J13 2 - 5 2 = 12 cm
'"(·-1, ,Y x = distance from 8 to C
., 20. let
.......... . PA = 200 m. PB = BO m, PC = X + 80
PA(PA) = PB(PC)
A =2 (½ r2 sin 60) = (6)2 sin (60) 200(200) = 80 (X + 80)
A= 31.18 m 2
X = 420 m
PC = 80 + X = 600 m
17. let a = 5, b = 8, c ::: 13 and d = X
A
Solve right triangle whose legs are d and
1
8 and the hypotenuse is 13
2 2
13 = (d1) + (8)2
d1 = 10.247 m
Solve right triangle whose legs are d and
2
5 and the hypotenuse is 13
2 2
13 = (d2) + (5) 2
d2= 12 m C

21 . The arc length

s =re= 38

P = S + 2r
By Ptolemy's theorem, 9 = 30 + 2(3)
d1d2 = ac + bd 9 = 1 rad.
(10.247)(12) = (5)(8) + (13)(X)
A = ½ r 2 6 = ½ (3 )2( 1 rad)
X = 6.382 cm A=4.5
18. For a quadrilateral circumscribing a circle
where r = radius of the inscribed circle
22. The right prism is inscribed in a cylinder
Aquad = -v'abed
=
.Aquad ✓~(3-)-(3-)-,-(4-:-:-)-=-c4-:-:-
) =
12 cm 2
s = ½ (a + b + c + d) = ½ (3 + 3 + 4 + 4)
s = 7cm
.Aquac1 = rs
12 cm 2 = r ( 7 cm )
r = 1.714 cm
Find the diameter ( d) of the cylinder
d2 = 12 + 12 = 2
19. Property of the Circle: The largest triangle
inside the semicircle is a right trlangfe =
d 1.414, r = d/2 0.707
V = rrr2h
=
The diameter of the circle
d = 2(6.5) = 13 cm 6.283 = rr(O. 707)2h
h=4m
The hypotenuse of the inscribed triangle
is 13 cm
,,.----- 23. For truncated prism,
V = AahAVE
:~r ~~~eb~;~h~f;a~~lateral triangle , where
Aa = ½ x 2sin 60
hAvE == 1/3 (3 + 4 + 5) = 4 cm

1,800 = (½ x2sin 60)(4)

x = 32.24 cm

-MPHT Review Center: Solution and Answer key in MATH

Page 3 1
24 . For a cone inscribed in a hemisphere 30. Regular icosahedron is composed of 20
Vre,q = Vhemlephere - Vcone equilateral triangles
S = 20At.
S = 20(1/2) (x 2 )(sin 60)
S = 20 (1/2)(5}2(sin 60)
S = 216.51
R
·.. I •:
EXERCISE NO. 5
Analytic Geometry - 1
For the cone , h = r 1. ( - 2 , 6) lies on quadrant !J.
Vreq = 2/3 m 3 - 1/3 TTr 2h
Vreo, = 2/3 TTr3 - 1/3 TTr3
2. To find the distance between 2 points,
Vreq = 1/3 rrr3
Enter Mode 2
25 . For wedge cut from a cylinder d=IA-BI
V = 2/3 r3 tan 6 =
Let A 3 + 7i and B =-
4 - 7i
V = 2/3 (3) 3 tan 45 d = I 3 + 7i - ( - 4 - 7i) I
V = 18 m3 d = 15.66

26. For similar figures, surface area (S) is

directly proportional to the square of the 3. d = ✓ Cx2 - + (y2 - Y1) 2
X1)
2

edge (x) 15 = .J(x- 2) 2 + (3 - (-9)) 2

X2 = (1 - 0 .12)X1 = 0.88 X1
X = 11

S1 = (X1) 2 4. Solve the intersection of two lines

S2 Xz
2 Enter Mode 5 1
S1 ( x1 )
S2 = 0.88x 1 a b c
S2 = 0.7744S1 = (1 - 0.2256)81 1 2 12
The surface area is decreased by 22.56% 3 -5 - 15
X = 30/11 Y=51/11
27 . The area exposed above the liquid
Zone of the sphere (2) = 2mh Enter Mode 2
d=IA-BI
r! 30
Let A = 11 + 51
11
I and B = 0
3
d=l~+~il
l ., },, , . ',;~;-\~·/,/·•·~:).\\\·:.-,.;~· 11 11
- .. _ - > ' ,
·.. ~ , ....I••,., .. " • ' --••• -·1
.J'
;:,_-:_ , >>"•
·,\ ~ d = 5.38

'<~?tJfJJ~~~t
5. By distance formula

T d1 = d2
✓ ex+ 2) 2 + (2 - 9) 2 =✓ (x - 4) 2 + (2 + 7) 2
For r = 7 .5 cm, X =3.667 11/3 =
Z = 2TT(7 .5 cm)(3 cm)
Z = 45 rrsq. cm. 6. Since the line segment extend 3 times its
length
28 . For the area of a lune,
k = k = P1P / P1P2 = 4P1P2 / P1P2 = 4
Al,.me = 4TTr2 (6/360)
Ai.xi, = 4TT(30 cm}2(85/360) Enter Mode 2
At.- = 2.670.35 cm 2 P1 = - 3 + 4i, P2 = 1 - 2i
P = P1 + k (P2- P1)
29 . For the area of a spherical triangle P = ( - 3 + 4i ) + 4((1 - 2i) - ( - 3 + 4i))
nR 2 E P = 13- 20i
AsT = 180 P (13. - 20)
For the excess (E)
7. k = P1P / P1P2 = 3/5
E = A + B + C - 180
E = 115 + 70 + 92 - 180 = 97 Enter Mode 2
n(12)2(97) P1 = 2 - 51, P2 = - 3 + 5i
As,· = 180 P=P1+k(P2-P1)
A s T = 243.78 cm 2 P = ( 2 - Si ) + (3/5)((- 3 + Si) - ( 2 - Si))
P= - 1 +i
P(-1.1)
 The equation will be
8. By midpoint formula
1/4 (x) + ( - 1/4)(y) = 1
-
x=---
X1 + X2 '
2 x-v=4
2 =x+(-2) +2=0
2 16. From the line 4 x - 2 Y
x=~ _ 2y = - 4x- 2,
= 2x + 1· m1 = 2 h ( 3
9. For centroid, Y s Iope 'of the line passing throug
The - ,

X = X1 + X2 + X3 = 0 .+ 2 + 1 5) is equal to 2
3 3 Enter Mode 5 1
x=1 a b C
1
y= Y1 +y2 +y3 -4 +6+4 -3 5
3 5+2 1
3 -3+1
y=2 X = -2/11 y = 1/11
Centroid (x, y) = ( 1. 2) The equation will be
(- 2/11 )(x) + ( 1/11 )(Y) =1
10. Enter Mode 6 -2x+y=11
Mat A, 3 x 3 (m x n) 2x- v+ 11 =O

MatA = [ -3
5
-1
3
1]
1
17. Enter Mode 3 2
2 -8 1 X y
Use Shift 5 to enter 0.5 det (Mat A)
4 0
IAl=M!
6 3
Midpoint of the line segment
11. Use matrix to solve the area
Shift Stat Var x = 5
A=; [I~~¼~~~] Shift Stat Vary = 1.5
1 Shift Stat Reg B = 1.5 = 3/2 ➔ m1
A= 2 [(4- 24 + 4) - (-3 + 32 + 5)]
For perpendicular bisector that passes
A= 1-251 = 25 through ( 5, 1.5), its slope is m2 = - 2/3
Enter Mode 5 1
12. Use matrix to solve the area
a b c
A =; [I!~~ ~3Jti~:>tt!] 5+1
5 1.5
1.5 + ( -2/3)
1
1
1
A = -((3
2
- 9 - 4 - 4) - (8 + 9 + 6 + 1 + 2)) X = 4/29 Y = 6/29
The equation will be
A= 1- 20 I= 20 (4/29)x + (6/29)y = 1
4x + 6y = 29
13. By slope - intercept form, y = mx + b 4x+ 6y-29 = O
y = 2x- 3
The general form will be 2x- v = 3
18. For the equation 4x + 2y _
2
=
0
14. Two equations of line are parallel if their Multiply 2 by both sides of the equation .
2 (4x + 2y - 2) = o
slopes are equal Bx+4y-4 =o
=
y 3x + 2; m1 3 =
6x + 2y = 5
19. Distance of ( 4, 2) to 4 x _ Y + =
2y = - 6x + 5 3 5 0
d = A(x 1 ) + B(y1 ) + c
y =-
3x + 5; m2 3 =-
Then m 1 is not equal to m2, they are not ✓A2+a2
parallel.
d =
(4)(4) + (-3)(2) + 5
✓ 42+(-3)2
15. From the line x - y - 2 = 0
y = x-2, m1 = 1
The slope of the line passing through ( 4, 20. A = 5, B = 12 C = k _
d = A(x1) + '
B(y1) +' X1C - 2, Y1 =1
0) is equal to 1
Enter Mode 5 1 ✓AZ+ 82
a b c 2 = (5)(2) + (12)(1) + k
4 0 1 ,/52 + 122
4+ 1 0+1 1 k=~
X = 1/4 Y = -1/4

MPHT Review Center: Solution and Answer key in MATH

Page 33
 Square both sides
2
21 . Distance between two parallel lines 9r2 =
g - 12x + 4x 2
= =
A 3, B 1 , C1 =- 12, C2=- 4 g(x2 + y2) = g - 12x + 4x
C2 - C1 -4 - (-12) 5x2 + 9y2 + 12X 9 = 0
cl = -::::=:::::::;;: = -;::::;::==--
✓ A2 + 02 ✓ 32 + 12
ct=W EXERCISE NO. 6
22. For the line 2x + y = 12 Anal tic Geometry - 2
y =-2x + 12; m1 2 =- 1. For a conic 4x2 + 3y2 - Bx+ 16y +19 =O,
The angle between the lines
111 2 - 1111 A= 4, C = 3 . . .
tan0 =----
1 + m 2 m1
For Ai- C with same sign, the come 1s an
1112 - (-2) Ellipse
tan 45 :::: 1 + (-·2)mz
2. A = 1, B = 4, C = 4
rn 2 :::: - 0.333 ... = - 1/3 Find the discriminant
Enter Mode 5 1 D = B 2 -4AC
a b c = 4 2 - (4)(1 )(4) = 0
-3 5 1 The conic is a Parabola
-3 +1 5 -1/3 1
3. For a conic 4x2 + Bx - y2 + -4y - 15 =O,
X = 1/12 Y = 1/4
A= 4, C = -1
The equation will be
(1/12)x + (1/4)y = 1 For A and C with opposite signs, the
X + 3y = 12 conic is an Hyperbola
x+3y-12=0

23. Using the distance of a point to a line, 4. The conic is a circle since A = C = 1
Convert the equation into standard form
dz= d1
4x + 3y - 24 Sx - 12y + 30 (x - h) 2 + (y - k) 2 = r 2
--===-- = --;::::======
.Js 2 + c-12) 2
✓4 2 + 3 2
(x 2 - Bx +16) + (y 2 +16y+64) = -81+16+64
(x - 4) 2 + (y + 8) 2 = - 1
13(4x + 3y- 24) = 5 (5x-12y + 30) The radius is imaginary, the conic is an
Emptvset
52x + 39y- 312 = 25x- 60y + 150
5. x 2 +y 2 -2x-4y-31=O
27x + 99y = 462
(x 2 -2x + 1)+(y2 -4y+4) = 31 + 1 + 4
9x+ 33v= 154 (x - 1) 2 +(y - 2) 2 = 36
For standard form of equation of a circle
24. For y = 3x + 2, m1 = 3 (x - h) 2 + (y - k) 2 = r 2 ,
For y = 4x + 9, m2 = 4 C (1 , 2), R = ✓36 = 6
m 2 - m1
tan8=-:-.--
1 + m2 m1 6. x 2 + y 2 + 12x - By + 16=0
4-3 A= 1, C = 1, D = 12, E = - 8
tan0 h = - D/2A = -12/2(1)
1 + (4)(3)
h=6
0 = 4.399°
k = - E/2C = - ( - 8)/2(1)
k=4
25 . Enter Mode 2 Then the center is at C (- 6, 4)'
d=IA-BI
Where A= 5 L 30 and B = - (8 L - 50) 7. For standard form of equation of a circle
d = 10.14 (x - h) 2 + (y - k) 2 = r 2 ,
(x- 3) 2 + (y+ 5) 2
2 2
=
42
26. Enter Pol ( - 6, - 8) x + Y - 6x + 1 Oy + 9 + 25 - 16 = o
r = 10, 0 = -126.87° x 2 + v2 - 6x + 1 Oy + 18 = o
r = 10, 0 = -126.87° + 360°
r = 10. 8 = 233.23°
8. x2 + y2
+ 4x - By - 5=0
(x + 4x + 4)+(y 2 - 8y+16) = 5+ 4+ 16
2

(x + 2) 2 +(y - 8) 2 = 25
27 . Using x = r cos0 and x2 + y2 = r2 For standard form of equation of a circle
3
r=----
(x - h)2 + (y - k)2 = r2,
3 + 2 cos0 C (- 2, 8), R = ✓25 = 5
3
r=---
3 + 2 (;)
A= TTR2 = TT(5) 2
A= 78.5
3r+ 2x = 3
3r = 3 - 2x

MPHT Review Center: Solution and Ans1ATPr 1r.,,... :- •• •

 9. By Trial and error, Find the coefficie~ts 3 == 3/10
Substitute the given points in the Shift Stat Reg A: g-
4
:: 4/10
equations from the choices. Shift Stat Reg B : o· 5 :: 5/10
Let x = - 3, y = - 1 Shift Stat Reg C - ·
For letter A,
2 2
x + Y - 2x + 4y - 11 = O The equation formed will be
2
2
(-3) .+(-1)2-2(-3)+4(-1)-11 =O
1 ;c O y:
y - A + Bx + Cx
(3/10) + (4/10~X + (5/10)X
2

For letter B 1oy = 3 + 4x + Sx

x 2 + y2 + 8x - 10y + 16 = o 5x2 + 4X - 10V + 3 = O
2
(- 3 ) + (- 1)2 + 8( - 3) - 10(- 1) + 16 = 0
12 ;c 0 15. Enter Mode 3 3
For letter C X
y
2
x + y 2 + 6x - 18y - 1O = O 0
5
(- 3) 2 + (- 1)2 + 6(- 3) - 18(- 1) - 10 = 0
O=O -1 0
For letter D
x2 + y 2 - ax - 12y + 3 = o 1 8
(- 3)2 + (- 1 )2 - 8(- 3) - 12(- 1) + 3 = 0
49 ;c 0 Find the value of y if 4 x:
Enter 4 Shift Stat Reg Y
Then the equation of the circle passing
through ( - 3, - 1) is y = 4y = §.
x 2 + y2 + 6x- 18v- 10 = o law for the equation of
16. By Square
10. The center is between the endpoints of parabola,
diameter. X12 Xz2
h = ½ (10 + 6) = 8 -=-
Y1 Y2 _
k = ½ (2 - 4) = - 1 Let X2 = ½ (64) = 32 at y2 - 8
The diameter will be If y1 = 18,
D = I A - B I by Mode 2 x2 32 2
Where A = 10 + 2i and B = 6 - 4i 18= 8
D = 7.21 X = 48 m.
Then, r = 3.61, r2 = 13 2x = 96 m.
By the standard form o( equation of circle Then the width of the arch at the bottom is
(x - h) 2 + (y - k) 2 = r 2 , 96m
(x - 8) 2 + (y + 1) 2 = 13
x2 + y 2 - 16x + 2y + 64 + 1 - 13 = O 17. 4x 2 +y 2 -16x - 6y- 43 = 0
x 2 + v2 - 16x + 2v + 52 = o (4x 2 -16x) + (y 2 - 6y) = 43
4(x 2 -4x+4)+(y 2 -6y+9) = 43+16+9
11 . y 2 + 8x - 6y + 25 = 0 1 1
yz - 6y = -8x - 25 68 [4(x - 2)2 + (y - 3)2 = 68] 68
y 2 - 6y + 9 = -8x - 25 - 9 (x - 2) 2 (y - 3)2
(y - 3) 2 = -8x - 16 17 + 68 =l
(y - 3) 2 = -8(x + 2) For standard form of equation of ellipse
LR= 4a = !l. (x - h) 2 (y - k) 2
a =1 b2 + a2 = 1
C(h, k ) = ~
12. 2x 2 - Bx - 4y + 16 = 0
Multiply both sides by 1/2
18. x + 4y 2 - 2x - By + 1 = 0
2
~2 [2x2 - Bx+ 16 = 4y]~2 (x2 - 2x) + (4y2 - By) = -1
x 2 -4x+ 8 = 2y (xz - 2x+1 )+4(y2 - 2y+1) = 1 +4 - 1
x 2 - 4x + 4 = 2y - B + 4 (x - 1)2 + 4(y - 1)2 = 4
(x - 2) 2 = 2(y - 2) (x - 1) 2 (y- 1)2
LR= 4a = 2 4 + 1 =1
a= 1/2 = 0.5 a2 = 4, a= 2
Major axis: 2a = ~
13. X2 +X+y+ 1=0
x 2 +x=-y-1
Since the coefficient of y is ( - ) then the 19. A9x~ + 4y2 - 72x- 24y- 144 =0
- 9, C = 4
parabola opens downward. For the major axis
a2 = 9, a = 3, 2a = '5
14. Enter Mode 3 3 For the minor axis
X y
b2 = 4, b = 2, 2b = 4
3 6 2a/2b = 6/4
-2
1
1.5
1.2 =u

MPHT Review Center: Solution and Answer key in MATH

 h = - D/2A = - 8/2(4)
20. For standard form of equation of ellipse h = -1
k = - E/2C = - ( - 18)/2(9)
whose major axis is horizontal
2 k = -1
(x - h) 2 (y - k) center (h,k) = [: 1, -tl
a2 + b2 = 1
C (h,k) = 1,2
2
26. 4x2 - 9y2 + Bx - 1 BY - 149 = O
2a = 6, a = 3, a = 9
2
4x2 + Bx- (9y2 + 18y) = 149
2b = 4, b = 2, b = 4 4(x2 + 2x+1) - 9(y2 + 2y+1) = 149 + 4 - 9
2
(x - 1) 2 (y - 2) 1 4(X + 1 )2 - 9(Y + 1 )2 = 144
9 + 4 2 = (x + 1) 2 (y + 1)
2
4(x - 1 )2 + 9(Y - 2) =.36 36 16 = l
4x2 - Bx + 4 + 9y2 - 36y + 36 = 36
4x7 + 9y2 - Bx- 36y + 4 = 0 For the transverse axis,
21. 4x 2 + 25y 2 - 8x - 100y- 296 = 0 a 2 = 36, a= 6
4x2 - Bx+ 25y2 - 100y = 296 2a =11
4(x2 - 2x+1 )+25(y2 - 4y+4) = 296+4+100
2
4(x-1)2 + 25(y-2) = 400 27 . 9x2 - 4y 2 - 36x+ Sy = 4
2 9x2 - 36y - ( 4y2 - By) = 4
(x - 1) 2 (y - 2) 1 9(x2 - 4x+4) -4(y2 -2y+1) = 4 + 36 - 4
100 + 16 = 9(x - 2)2 - 4(Y - 1 )2 = 36
For the major and minor axis 2
= =
a 2 100, a 1O (x - 2) 2 (y - 1)
b 2 =16,b=4
--4-- - --9-- =l
For the transverse and conjugate axis ,
c=~
C =✓ 100 -16 = 9.165
a2 = 4, a= 2
C 9.165 =
b 2 9, b 3 =
e=-= - -
a 10
e = 0.92
c =./a2 + b2
22. 9x 2 + 25y 2 - 36x - 189 = o C =./22+ 32 = ./13
9x2 - 36x + 25y2 = 1 89 C ./13
9(x2 - 4x+4 )+25(y2 ) = 189+36 e=-=--
a 2
9(x - 2)2 + 25(y)2 = 225 e = 1.8
(x - 2)2 (y)2
25 +9 = 1 28. For the equation of hyperbola
x2 y2
For the major and minor axis 9 -4 = 1
a2 = 25, a= 5
b 2 = 9, b = 3 For the transverse and conjugate axis
b2 = 4, a2 = 9 '
A =
nab = n(5)(3) a=:_ 3 (sem~ -transverse axis)
A= :J.M ~ (semi - conjugate axis)
23. 9x2 + 25y 2 - 36x - 189 = o
29. The equation of the hyperbola
For the major and minor axis
(x - 2) 2 (y + 4 )2
a2 = 25, a= 5 .
b 2 = 9, b = 3
9 16 =1
a2 = 9,b2 = 16
2b 2 2(3) 2
LR=-=- a=3,b=4
a 5 C (h, k) = (2, -4)
LR= 3.6 For the equ a t·,on of the asymptotes
Y-:- k = ±(b/a)(x - h)
24. For the given equation of ellipse Smee upward asym t t
x2 y2 Y + 4 = (4/3)(x - 2) P o e , use ( + )
36 + 32 = l
3y + 12 = 4x-B
For th e major and minor axis
4x-3v 20=0
a2 = 36, a= 6
b 2 = 32, b = 5.66
30. The eccentricity of the hyperbola
c ✓ a2+b2
c=..../a2=b2 e= - = - - -
C =✓ 62-5.662 = 2 a a
Since a= b
The distance between foci is e= &+a2 ✓ 2a 2

2c=~ a a
=
e ✓2 1.41 =
25. 4x~ - 9y2 + Bx -18y- 149 = 0
A - 4, C = - 9, D = 8, E = - 18

MPHTRev1ew
· Center: Solution and swer key in M ATl-l
 I
log1oe (2)(xz + 1)(2,Q
31 . 4x22 -gy2 + 8 X - 1By - 149 = 0
4x + Bx - (9y2 + 1 By) = 149
4(x2 + 2x+1) - 9(y2 + 2y+1) = 149 + 4 - 9
4(x + 1 )2 - 9(y + 1 )2 = 144
y'

y'
= (xz + 1)2
4xlog1oe
(x2 + 1)
I
(x + 1) 2 (y + l)2
36 - 16 =1 9. By Power rule,
y= (x 2 + 2)z
1 I
F~~ th e transverse and conjugate axis 1
y' = !. (x 2 + Z)i- (2x)
a - 36, a= 6
I
• 2 1
b2 = 16, b = 4 y' = (x 2 + 2)-z(x)
X
LR= 2b2/a = 2(16)/6 y'= 1
(x 2 + 2)2
LR= 5.33
10. From the equation x
2 2
+y = 5, I
Let: y = ✓s - x
2
EXERCISE NO. 7
Differential Calculus y' = :x(✓s-x2 )lx=l = -¼
1. Set the calculator in Radian mode
For the normal line, m = 2 \
Using point slope form,
. 1 - cos(x) (y - yo) = m(x - Xo)
l 1m-----=-...:..
x-+O x2 y-2 = 2(x-1)
T 1-cos(X)
ype xz th~n CALC (X?) 0.0001 y-2 = 2x-2
= 1/2 2x-y=0

2. Set the calculator in Radian mode 11 . At t = 4 seconds,

tan(2X) - 2 sin(X) V = D'
X3
CALC (X?) 0.0001 = ~
v = :t(zot+ (t+ 1)) lt=4
v = 19.68 mis
3. As x approaches 1
(X.2 - 1) ' 12. In Radian mode,
d
(X. 2 + 3X- 4) y' = dx ( 4cos(X) + sin(2X)) lx.,z
CALC (X?) 1.0001 = 0.4 y' = -4.94

4. As x-infinity 13. y = x2ex

3X4 - 2x 2 + 7 y' = 2xex + x2 ex
SX 3 +X-3 For the point/s of inflection, solve for y"
9
CALC (X?) 9 x 10 y'' = 2ex + '4xex + x2 ex
= 5.4 X 10 9
y'' = ex (x2 + 4x + 2)
As x approaches 00 , the limit approaches Set y" = 0
infinity O = ex (x2 + 4x + 2)
e>< = 0
5. As x-2 x=O
X 3 -X 2 - 4 x2 + 4x + 2 = 0
ByMode53wherea= 1, b=4, c= 2
X 2 -4
CALC (X?) 2 .0001 = 1. x, = -2 +ft
x2=-2-..J2
6. Find the derivative by quotient rule
x2 14. y = 12 -12x+x 3
y=~ y' = 0 - 12 + 3x 2
(x + 1)(2x)-x2 (1) 0 = 0-12 + 3x 2
y' (x + 1) 2 x= 2 and-2
2
2x2 + 2x-x If X = - 2,
y' = (x + 1) 2 Y = '. 2 - 12( - 2) + ( _ 2)3 = 28
x 2 +2x Maximum point r_
2 , 281
y' = (x + 1) 2 If X = 2,
Y ~ ~ 2 - 12( 2) + (2)3 = _ 4
7. Find the derivative by Product rule Minimum point ( 2. _ 41
y= ex cos(x 2 ) 2
y'= ex(-sin(x 2 ))(2x) + cos(x ) ex For the point of inflection
2
2
y' = fr ( cos (x ) - 2x sin (x )) y" = 6x
Set Y" = 0, then x = o
8. For derivative of logarithmic function If X = 0
d log 3 e du Y = 12_:..12(0) + (0)3 = 12
y' = dxlogau = -u-dx Inflection point ( o. 12)

MPHT Review Center: Solution and· Answer ke YID

• M ATH
Page 37
 X = 10; y = 20 - X = 10
15. From the equation x2 = 16y,
x2 A= xy = (10)(10)
Let: y = -16 A= 100 ft2
d (x
y' = d.J: 16 lx-=..f = z
1
) I
20. Let x, Y = dimensions of the printed matter
For the normal line, m = - 2

-r-
xy = 300
Using poirrt slope form , y = 300/x
(y - Yo) = m(x - x,,)
y - 1 = - 2(x - 4) 10 cm
y - 1= - 2x +8
2:x+ y=9

16. The sum of two numbers is 50,

T y
! I·:'
" ''
,.
,,
,
' (.J1

n
3
y + 20
X + )' = 50
,I

l_
'
y = 50 - X ''

The prodLJc1 of lwo numbers

p = x(y)
J X
', ·, ••, •• I
'

P = x-{ 50 - x) X + 10
p = 50.x - x 2
For the area of the poster
d P = 50 - Zx
ch A= (x + l0)(y + 20)
dP
Sel - = 0, then x = 25 300 )
dx A= (x + 10) ( --;z- + 20
The number s are 25 and 25
Set dA
dx
= o, solve for x
17 . For maximum are-a, the tri angle is an
equilateral
x = 12.25 cm
300
a+b+c=18 y = - = 24.49 cm
X .
a+a+a=18 x + 10 = 22.25 cm
a=6cm=b=c y + 20 = 44.25 cm

A =.: ab si ne 21 . Enter Mod

2
A=.:2 (6)(6)sl n60 x h
A = 15,6 sq. cm
40
0
2
18. Let x = location of stake from 10 m post -40 TC 8
r = 8 ft
- - - - - - ~ 40 ft

15'
- - - ---JE--- - - -o

le----- 30' - - -------+t

r = 8 ft - 40 ft
/, Jx + 100 + J( 30 - x + 225
c:::: 2 2
)
dL 2x 2(30 - x) Press AC, then
-dx = --;=:;==== + ---:;:::::::=====
z ✓:x 2 + 100 z J(30 - x 2 ) + 225 dV = hydy
Set dUdx = 0, then solve for x =
At h 4, dV 2, =
X= 12 ft. dy = dV / hy
dy = 2149
19. Let x and y = dimensions of the fence dy = 0.995 ft/min.

X 22. For the total cost

C = (Cost per ho~r)(t)
C = (kV3)(t)
y Since the boat run against current,
C = (kV 3 ) ( ~ )

X + y = 20 C = (kS) (-t
_;_B)
V- 0
y = 20- X
A= xy _dC = (kS) _(V_-_D)_(_3V_2_)_-_;(_V3. .:. ). :. .(1..:. .)
A= x(20 -x) dV • (V-8)2
dA = 20-2x O = (V - B)(3V 2 ) - (V3)(1)
Set dA = 0, then (V - 8) 2
V = 12 kph
 2 J
For; (x 3 + 3)2, 13. Solve for the intersection of x = Y
2
-
2 0nd

= ¾((fJ + 3)~ - (0 3 + 3)i) = 0.623 x=-y

-y = y2 - 2
Then,
y 2 +y- 2 = 0
f x 2 ✓x 3 + 3 dx = 2/9 (x3 + 3) 312 + C y = 1, -2

9. fy 3 J2y 2 + ldy A = f( XR - XL)dy

Set limits o to 1 f
A= 2 ( -y - (y 2
f yl✓2y
- 2))dx
2 + 1dy = o.3797 A= 9/2

Among the choices, using the same limits

For
1
(2y 3 -
3
l)'i(3y 3 + 1), 14. Solve for the intersection of y = x 3 - 4x
30 and y =
x 2 + 2x
1
= 30 ( (2 - l)i(3
3
+ 1) - (-1)z(1)
3 ) x - 4x = x 2 + 2x
3

x 3 - 4x-x 2 - 2x = 0
= Error x 3 -x 2 - 6x 0 =
1 1
For (2y 2 - 1)z(3y 3 + 1), By Mode 5 4 where a= 1, b = - 1, c = - 6 ,
30
d=O
=;0 ( (2 - 1)i(3 + 1) - (-1)½(1)) X = 3,-2,0

= Error A= i( b
Yu - YL) dx
For:~ (2y + 1)i(3y + 1),
1
= 3 0 ( (2 + 1)i(3 + 1) - (1)i(1))
2

A1 = f
-2
0
(x 3 - 4x - (x 2 + 2x))dx
A1 = 16h
= 0 .1976
For
30
1
(2y 2 + 1)z(3y 2
3
- 1),
A2 = L( 3
x2 + 2x - (x 3 - 4x))dx

= 30
1
( (2 + 1)i(3 - 1) - (1)i(-1)) A2=
6
¾
A=A1+A2=16h+6¾
=0.3797 A= 21.08
Then
- -+-1 dy
I y3.Jzy2 15. Enter Mode 7
= 1130 f2v2 + 1) 312(3y2- 1) + C f(x) = 4 sin(2x)
Start?= 0
10. Let u = 1/x, du= (-1/x2 ) dx End?= 360
Isin u du = - cos u + C Step?= 15

f sin(~):: X(9)
0
f(X) (R)
0

= -x du
sin(u) ~
2
= - Jsin(u) du 30 3.46
J
= - (- cos u) + C
45
60
4
3.46
= cos {1/x} + C 90 0
135 4
11 . Let u = y 4 and du= 4y3 180 0
feu du= eu + C 225 4
f ei' 4y3 dy = f eudu = eu +C 270
315
0
-4
360 0
,,, \ey•+c\
3
Set into Radian mode,
12. Solve for the intersection of y = 2x, Y = x For one leaf, the limits is from Oto rr/2
2x
x 3
= x3
- 2x =0 A=1
2
Jr 2 dB f
= 21 ¥(4sln(28))~ds ?

x(x 2
-· 2) = 0 0
A= 2rr
X = 0,-Jz, - .Jz
16. Enter Mode 7
A = f ( Yu - YL)dx f(x) = 4 (1 + cos(x))
Start?= 0
= J/2c 2x -
/\ 1
3
x )dx End?= 360
A1 =1 Step'/= 15
Az = J~J:!( x 3 - 2x)dx
/\2 = 1
AT = A1 + A2 =1+1
AT =1
 ,_ X(8)
0
f(X) (R) f
Ay = YcdA
30
8
7.46 _f
Ay=
Yu+YLdA
~ z I
45
60
6
9 - x2 _ J~ (9 - x
2
) dx
90
135
2
4
A-y-
-
J-d 2
A-
J
2

180
0.5 lSy =! ( 2
(9 - x )2 dx
0 2 Jo
225 1.2 y=3.6
270 4
315 7.46 20. A= f ydx
360 8
A= fo\ex) dx = 1.718
The limits is from Oto rr/2 for one region
r2 =
4(1 + cos(O)); r 1 = 4 Ax= J xcdA
1

A= "if (r2 2 - r 1 2 ) d9 1.718x = fu xex dx

A1 =; fl(4(1 + cos(0)))
11'.
2
- 4 2 ) dO x = 0.582
A 1 = 22.28 21 . Using vertical strip (dy)
AT = 2Ai = 2(22.28) y2
AT= 44.57 Xt =4
At X = 4, y = 4
17. x 2 + y 2 = 25 x2 =4
2x
X
+ 2yy' = 0
+ yy' = 0 A J
= (X2 - X1)dy
-x
y'=-
y
xz
A = i4 ( :2)
4- dy = ~2
=
(y ' )2
But = 25 - x 2 ,
y2
y2
Ay = f YcdA
2
x2
(y
')2 _ _ _.....,.
- 25 - x 2
3Y = Jo(4 y ( 4 - 4Y
32 )
dy

S f.
= b✓ 1 + (y')2 dx y = 1.5
22. The moment of inertia with respect to y -
r4 x2 axis
S = )3 1 + 25 - x2 dx y=Kx
s = 1.419 Iy = f 1x 2 dA
Iy = J0 x 2 (y)dx
1
18. r = 4sin e ly = J0 x 2 (Kx) dx
_dr = 4cos0 Iy =0.571
d8 1t
81 = 0, 82= 2 23. The moment of inertia with respect to x-
axis
S = t rz+(*fd0 x=fo
S = ~ /(4sin(0))2+(4cos(0))2 d0
Ix= f y 2 dA
s = 11!
The total length of the cu
rve Ix= J01 y 2 (x)dy
$-r =2(2TT) = £I Ix= folYz (.,/4y)dy
Ix= 0.6714 or4n
2
19. y = 9-X
A = J ydx
24. The moment of inertia with respect toy-
axis
A = iJ ( 9 - x 2) dx = 18 Y1 = fu,Y2 = 2
ly= x 2 dAJ
Ax= f xc dA 1
ly = J0 x 2 (2 - ffx)dx

18X
-= f 0
3
x( 9- x 2Jdx ly = 0,095

x = 1.125
..'.!.>. Uy Shell method
30. The area of the quarter circle
V == 2n J x(yu - ~L)dx n(5) 2
nr 2
A= 4 = - 4 - = 4 11
25
xz
Yu = - ; YL =0 The centroid of the quarter circle
8 4r 4(5)
x = y = 3TC = 3n = 2.12
2n Lx(x:) dx
4

V = r = 10 - 2.12 = 7.88

V = 50.265
26. By Disk method,
V=2mA

V
V
= 2n(7.88)
= 972.18
e: TT)

V = TI J (y )dx 2
31. For the work done in pumping out water
For hyperbola y = 6/x, W=yVfi. where y = 9.81 x 103 N/m
r4 For hemisphere
V = TT )
2
(-x)6 2
dx 2n
V=-r 3
V = 28.27 3
Zn
V = -(3)
3
3
= 18n
27. By Ring Method,
V f
= TT (Yu - YL )dx 2 2
_ 3R 3(3) 9
h=-=-=-
8 8 8
Yu = 2x + 3; YL = x 2 W = (9.81 X 10 3 )(18TT) ( i)
For the intersection,
w = 624x 10' J
2x + 3 = x 2
x 2 -2x-3 = O
EXERCISE NO. 9
By Mode 5 3, x = - 1 , 3
3 Differential Equation
V = TI J_ ((2x + 3)
1
2
- (x 2 )2) dx
1. The order of a differential equation is the
V = 227.87 highest occurring derivative in the
differential equation
28. For the functions
y, = x2/8, y2 = 2
Yz -y1 =2-
xz
--
8
(::;y + 3y(1f +y
Order of the differential equation = 2.
3
(:~f = Sx

Solve for their intersection,

x2
2. The degree of a differential equation is
z =s the power of the highest occurring
X= 4,-4 derivative in the differential equation

(:7zY + 3y(:~f + y(::f =

By Disk method,
V =n f (y 2 )dx
3
Sx

V
= ,{ ( 2 -
= 26.81
•:r dx
Order of the differential equation = 2
Degree of the differential equation = J

3. y(y") 4 + 2x(y') 5 = 8xy2

29. The area of ellipse
A= rrab y( :7zy (:~f =+ 2x 8xy
2

9x 2 + 4y 2 = 36 Order of the differential equation = g_

xz y2 Degree of the differential equation = ~
-+-=1
4 9
a2 = 9 ; a= 3 4. Differentiate the equation: x 3 - 3x 2 y =o
b 2 = 4; b = 2
A = rr(3)(2) = 6TT
2
3x dx - 3[x 2 dy + y(2xdx)] = 0
For center (0,0) 3x 2 dx - 3x 2 dy - 6xydy = 0
A(x 1 ) + B(y1 ) +C Divide both sides by 3x,
r=
✓A2 + sz xdx - xdy - 2ydx =0
(3)(0) + (4)(0) - 20 fx- 2y) dx-xdy = 0
r=
-J32 + (4)2
r =4
5. Solve for the first and second derivatives
y = C1 e-Zx + Cze3x Eq. ~
V =
2nrA 2TT(4)(6rr) =
V = 473.74 or48rr y' = -2c1 e- 2 x + 3c 2 e 3 x Eq . 2
y" =
4c1 e- 2x + 9c2e 31 Eq 3
 Then.
From Eq. 1 In (1 +y2 ) = In X + In 5
• - 2X Eq.4
CzC '' " = y - C1C
1 + y 2 = Sx
Substitute Eq. 4 in Eq. 2
Y = ../Sx-1
y , = - 2 Ct e -2,c + 3 [y - C 1 C-2X]
y' = -2c1 e-2x + 3y - 3c1 e-2x
10. Let: y = wx
y' = 3y - 5c1 e-zx
Eq. 5 dy = wdx + xdw
5c 1 e - 2 " = 3y-y'
Substitute Eq. 4 in Eq. 3 9x2dx + 3y2dx - 2xydy = o
y"= 4c1e- 2x + 9[y- C1e-2x] 9x2dx + 3w2x2dx - 2x(wx)(wdx+xdw) = 0
y" = 4c1 e - 2 " + 9y- 9c 1 e- 2 x 9x2dx + 3w2x2dx - 2w2x2dx - 2wx3dw = o
y" =
9y - Sc1 e- 2 x Eq.6 9dx + 3w2dx - 2w2dx-2wxdw = o
Substitute Eq. 5 in Eq. 6 (9+w2)dx - 2wxdw = O
y" =
9y - [3y-y']
y" = 9y -3y+y' By variable separable,
y" = 6y +y'
v"- v'- 6y = o f J+
In
dx _
X
X -
2wdw =
9 w2
In (9+w2) = In
J
C
O

I
6. For equation of straight line,y = mx + b e "9+w2
X
= eln c
Since m = b, X

y = mx+ m - Eq. 1 9 +w2 =C

dy = mdx - m = dy/dx x = c(9 + w 2 ) but w = y/x
Substitute m in Eq. 1
Y = (dy/dx) x + dy/dx x=c[9+::J
ydx = xdy + dy Multiply both sides by x 2
ydx = (x+1 )dy x3=cf9x2+ y2)
ydx - (x+1 Jdy = O
11 . For differential equation, Mdx + Ndy = o
7. For the circles with center on the x - axis oM iJN
(x-h)2+y2=r2 ' oy = ax
x2 - 2hx - h 2 + y 2 = r2
A. (x + y)dx + (x - y)dy = O
Differentiate the equation to solve for y'
2x - 2h + 0 + 2yy' = 0 aM = a(x + y) = l
2x - 2h + 2yy' = O
ay ay
x-h + yy' = 0 oN = o(x - y) =1
ox ax
Solve for y"
Then rx
+ y)dx + fx- y)dy = 0 is an
EXACT differential equation
1 + [yy" + y'(y')] = 0
1 + yy" + (y')2 = 0
12. By variable separable,

f ~+J
yy" + (y') 2 + 1 = 0
1 +x 2
dy
Y
0 =f
2
8. (1 + y )dx + (1 + x 2 )dy =O In (1 + x2) + In Y = In C
dx dy
1 + x + 1 + y2 = O
2 (1 + x2 )y
=C
Integrate both sides
y+.x2y= C
tan - 1 x + tan - 1 y = C _ A differential equation is linear if the
13
derivative of y is independent.
9. 2xydy = (1 + y2 } dx
Then, yv" - 2v cosx = slnx is a nonlinear
Divide both sides by x,
differential equation
2ydy = (1 +y2 } (dx/x)

14
. For differential equation y' +Py= a
Separate variables and integrate

J2ydy = dx
1 + y2
J X
y' _ 3x = 6 where P = - 3 and Q = 6
The integrating factor
In ( 1 +y 2 ) = In x + In C v=e JPdx -- ef -3dx
If X = 2, y = 3 V = e-Jx
ln(1 + 3 2 ) = In 2 + In C
C=5 _ y' _ 3x = 6 where P = - 3 and a =6
15
The integrating factor l = e - 3x
 If the roots are real and equa l,

The solution
y =C1 em x + Cz x e m x + C3 X2 e mx
y( V ) = f Q ( V )dx + C Since m = 1
ye - 3x :: JSe - 3x + C y = c1e x + Cz Xe x + C3X 2ex
ye -3x = - 2e- 3
x +C y =eK (c1 + x c z + x 2cu
=
If X 0, y 6, =
6=-2+C ➔ C=B 20. Enter Mode 3 5 where x = time and Y =
amount of strands of bacteria
Then,
ye - 3 • = - 2e - 3 )( + 8 X y
y=-2+8elx
1 1000
4 3000
2
16. y' + xy = xy Press AC
=
p X, Q X, n = 2= 2
Initially, the number of strands of bacteria
Divide both sides by y Press O then Shift Stat Reg y=
y· 2y' + xy - 1 :: X
Let =
z y1 - n = y1 -2 =y -1 09=~
z' = -y-2y' 21 . Enter Mode 3 5
z Let x = time in minutes, y == T - T room
y= -y-2
-
T = temperature of the bar
In terms of z, the given equation will be Troom = 0
- z' + XZ:: X Find x if y == 25 - 0 = 25
z' - XZ:: - X X y
where P = -x and Q = -x 0 100-0
Solve the transformed linear equation 20 50-0
v :::: ef -xdx = ex 2
/2
Press AC,
zV = f QVdx+C
Then 25 Shift Stat Reg i =

,,z
ze2=
f -xe-zdx+C ,,z
25i = 39.tJ min

_,,z -x2
22. At Vo= 0
ze_z_ = e-2- + C h = Vot + ½ gt
2
2
-x2 100 = O(t) + ½ (32.2)t
Divide both sides by e-2 t = 2.5 seconds
z = Ce x•m + 1
Butz= y-1,
y-1 = cex"2/2 + 1
23. Let: a- amount of salt
dQ
dt = Qin - Qout
17 . y" -3y' +4y = 0 dQ lb ( gal ) Q ( 3 gal )
3
In terms of differential operators, dt = 1 gal min 100 gal min
(D 2 -3D+4)y=b dQ
m2 -3m + 4 = o dt = 3- 0.03Q
Solve by Mode 5 3 dQ
dt + 0.03Q = 3
m= !
2 -
+ ./72 i
Solving linear equation; p = 0 _03 , R = 3
a= 3/2 = 1.5; b = ..fi
2 V = ef Pdt = ef o.o3dt
Solution: y = eax (c1 cos bx+ c2 sin bx) V = eo.o3t

Y =e · 1sxflc 1 cos ./7 x + c2 sin 2~ x

2
1 QV = f RVdx+c

18. For solutions e -21< and sin 3x

Qeo.o3t = f 3eo.o3tdx + c

Y = c,e- 2K + (c2 cos 3x + C3 3x) sin Qeo.o3t = 3eo.o3t (-1-)

0 .03 + C
The roots (m) are - 2, 3i and - 3i
(m + 2)(m + 3i)(m - 3i) = O
Qeo.o3t = 100eo.o3t + c
(m + 2)(m2 + 9) = O Q = ce-o.o3t + 100
m3 + 2m2 + 9m + 18 = o Solve C if t = o, Q =1
Transform into differentials 1 = ce 0
+ 100
v'" + 2y" + 9y' + 1sv = o · C = -99
Q = -99e-o.o3t + 100
19. (D 3 - 30 2 + 30 - 1 )y = 0 If a = 2, solve t
m3 - 3m 2 + 3m - 1 = o 2 = _gge-o.o3t + 100
Solve using Mode 5 4 , the value of m is 1 t = 0.338 min

MPHT Review Center. Solution and Answer kev in M A -r1-1

 7. A= 3i + j, B = - 2i +4j
~
24 G . Find unit vector A - 8
. iven the curves: Y = cx2
C =: y/x2
~terentiate the equation
A_ B = (3 - (-2))i + (1 - 4)j
A-8 = 5i-3j
I
f-5
IA - BI = 2_+_(__-3 2
=--=) == ~
dx
dy
= 2cx
dx = 2(x2)x
y u
A-8
=IA-Bl
l
5 3
dy 2y u = ./341 - ffeJ
dx--;z-
By orthogonal trajectory,
dy -x
= =
8. A 3i + 2j, B 2i + kj where k = scalar
a.) If A and B are orthogonal, A • B = 0
dx 2y
(3i•2i) + (2j•kj) = 0
2ydy = - xdx
3(2) + 2k = 0
J 2ydy + J
xdx = Jo 6 + 2k::: 0
k =-3
x2
y z +- =k b.) If A and Bare parallel, A= cB
2
(3,2) = c (2,k) = (2c, ck)
0.5x2 + y2-k
3 = 2c 2 = ck
c~ 2 = (2/3)k
2
EXERCISE NO. 1 O k=413
Plane and Space Vectors
9. A = -Si + j, B = 4i + 2j
For# 1, 2, 3, 4 and 5 a.) By Dot Product
Set the calculator in Degree Mode (D) A • B -5(4) + 1(2) -18
Enter MODE 8 7AI = ✓52 + 12 = ../26
Choose Vet A (1 ), m = 2 (2)
b.)
DISPLAY -
Shift Vet (5) Data (2)
[1 ,2]
A • B) -5(4) + 1(2)
52 + 12 (51 + J)
( l1All2 A=
Choose Vet B (2), m = 2 (2)

(A• B) 45 9
DISPLAY - [3,4]
AC llAll2 A= 131-13j
1. Shift, hyp, Shift, Vet (5), Vet A (3), =
Abs (Vet A) = 2.24 10. Using Mode 8
Input the Vet A, Band C
2. Shift, hyp, Shift, Vet (5), Vet B (4), = Vet A { 0, O] Vet. B [ 3, 4]
Abs (Vet B) = §. Vet C [ 8, O]
Solve using Shift 5
3. Shift, Vet (5), Vet A (3), Shift, Vet (5), Dot
For a triangle,
(7) , Shift, Vet (5), Vet B (4) 1
Vet A• Vet B = 11 Ati =
2 l(VctA- Vet B) x (Vet A- Vet C)\
Ati = 16
4. Shift, Vet (5), Vet A (3), x (multiplication
sign), Shift, Vet. (5), Vet B (4)
Vet Ax Vet B = (0, 0, -2]
Aparallelogram = 2At,
Aparallelogram = 2(16) = 32

5. By Dot Product
11. p (-3,4,-1 ),Q (2,5,-4)
A • B = IAIIBI cos0
cos0 = A • B / IAIIBI IPOl=✓ Cx2 X1)2 + (yz -y + (z - )2
)2
IPQ - 2 1 2 Z1
Vet A•Vct B + (Abs(Vet A) Abs(Vct B)) I-✓ (2 + 3) + cs - 4)2 + (-4 + 1)2
IPOl=fil
cos e = 11 + ((2 .236)(5))

6.
e= 10.305"

P = (-1,8) . a = (3,2)
12. A = 5i - 2j + 6k , B -- 81. - SJ. - 4k
A + B = (5 + 8)"
- . i+(-2-S)j+(6-4)k
I
A+ B - 131 - 71 + 2k
PQ = Q-P
PQ = <3 - ( - 1 ),
PQ = <4,-6>
2 - 8> 13. A= Si-2•
A-B
J + 6k, B = 8i - SJ. - 4k
= (5- B)I + (- 2 + S)j + (6 + 4)k
I
A- B::

~
-JI +31 + 10k
MPHT Review Center: Solution and Answe k eylnMATH
 22 . ByDot Product (Shift. Vet (5) , Dot (7 ))
14 . A = - 4i + 7j - 2k Vet 8 • Vet C = 24
3A = 3(- 4i + 7j - 2k) = -12i + 211- Bk
- 5A = - 5(- 4i + 7j - 2k) = 201- 35/ + 10k 23. By Shift, hyp and Shift, Vet (5)
IA x Bl = Abs(V ct A x Vet 8)
15. R (2,-1,3), S (3,4,6) IA x Bl= 13.6
IR°sl = (3 - 2)i + (4 + 1)j + (6 - 3)k
24. Solve for the unit vector of C by Shift, hyp
IR°sl = i + 5j + 3k and Shift, Vet (5)
IR°sl = ✓ 1 2 +s 2 +32 = ...JE Uc = Vet C + Abs (Vet C)
A-B Uc= [0.272 0 .153 - 0 . 136]
u = -:----
IA- Bl
i + Sj + 3k The angle of Vet C with x - axis
ll = ----
v'Js cos8x = u)(
cos 8,c = 0.272
16. A (3, 4, - 6), 8(4, - 7, 8) ex = 74.22° = 74•13•
d = A- B
d = ✓ (3 - 4) 2 + (4 + 7)2 + (-6 - 8)2 25. By Shift, Vet (5)
d = 17.83 8 x C = Vet B x Vet C
8 x C = Shift 5 4 x Shift 5 5
17. C (5, 7, z) , D (4, 1, 6) B x C = • 1 Bl + 3/ - 15k
d=C-O
-
7.28 = ✓cs 4) 2 + (7 - 1) 2 + (z- 6)2 26. Scalar projection of B on A
B • A Vet B Dot Vet A
z =1
!Al= Abs (Vet A)
18. By Dot Product = 3.73
A • B = (xy)(yz) + (2yz)(2zx) + (3zx)(3xy)
A • B = (xy 2z) + ( 4xyz2 ) + (9x2yz) 27. Vector projection of Con B
If X = 1, y = 2, Z = 3, C• B Vet C • Vet B ( Vet B )
A • B = (1 )(2)2(3)+4(1 )(2)(3) 2 +9(1 }2(2)(3) 7si2 B = Abs(Vet B) Abs(Vct B)
A • B = 138 = Vet C Dot Vet Bx Vet B + (Abs (Vet 8))2
= - 1.1431 + 4.763/ + 2.287k
19. Mode 8
Vet A, m = 3 [1 4 6] 28. Using Mode 8
Shift Vet (5 Data Input the Vet A, B and C
Vet B, m= 3 [2 3 5] Vet A [ 3, 1, 2] Vet B [ 4, - 2, 1]
Vet C [ 1 , 1, 3]
AC
Shift 5 3 x Shift 5 4 Solve using Shift 5
Vet A x Vet 8 = 2i + 7i - 6k 1
At,.= l(Vet A-Vet B) x (Vet A- Vet
2 C)I
At.= 3.39
For #20 to 27
Enter Mode 8
For# 29 to 32
Vet A , m = 3 [1 3 4]
Mode 8,Vet A, m = 3 [2 4 3]
Shift, Vet (5), Data (2)
2]
Shift Vct(S) Data
Vet B, m = 3 [- 1 4
Shift, Vet (5), Data (2) Vet B, m = 3 [1 -5 2]
AC
Vet C , m = 3 {2 7 - 1]
AC
29. By Shift 5,
20. By Shift, hyp and Shift, Vet (5) R=A+B
!RI = Abs(Vct A+ Vet B + Vet C) R = Shift 5 3 + Shift 5 4
IRI = 15 R=[ 3 -1 5]
R = 3/-/+ 5k
21 . By Shift, hyp and Shift, Vet (5)
IABI = Abs(Vet A- Vet 8) 30. R = 3i - j + Sk
Ab s = 1 IRI = Abs (3i - j + Sk)
IRI = 5.92

MPHT Review Ce nter. Solution and Answer key in MATH

 :.i ) 0 (-,cy)
oxyi l + ~(x ~; + -g; - k
31 . Scalar projection of A on B VxV = Oy Oz 0( - KY) O(xyz) .
By Dot Product , O( xzy)k
______ _~
uZ I - Ox l
A• B Vet A Dot Vet B
Oy z1c - 0 - y zj
lsf = Abs (Vet B) 0 yk- X
= - 2.19 V x V = xzt : - ._ (V + x21lf
vx V -- xz1- vzt
32 . Vector projection of Bon A
8 •A Vet 8 • Vet A f _ Vet A )
ENO 11
EXERCIS . b'l't
IAl 2 A= Abs(VctA) \Abs(VctA) . & proba 11
statistics
= Vet B Dot Vet A x Vet A + (Abs (Vet A)) 2
= - 0.828i- 1.655i- 1.241k 1. Enter Mode 3 ~ es 12 34 45, 23 , 87, 91,
Input all the va u , ,
33. Mode 8 and 121 in the column X
Vet A, m = 3 [ 24 -8 6] AC Shift Stat var X
Shift Vct(5) Data x = §.2
Vet B, m = 3 [4 12 6]
AC 3 1
2. Enter Mode 4 45 23, 87 , 91 ,
Input all the va 1ues 12 ' 3 , ,
By Dot Product and 121 in the column X
A • B = IAIIBI eos8 AC Shift Stat var OX
VctADotVctB ax= 37.75
cos 8 = ,--.,--------:-------:----:--
( Abs (Vet A)Abs (Vet B))
cos e = 0 .0989 3. Enter Mode 3 1 4 45 23 87 91
8 = 84.32°, = 84°19' 27" Input all the values 12, 3 , , , • ,
and 121 in the column X
34. By Dot Product, the angel between A and AC Shift Stat Var sx
B is90 sx = 40.78
A • B = (2i•4i) + (bj • - 2j) + (k • - 2k)
cos (90) = 2(4) + b(-2) + 1(-2) 4. Enter Mode 3 1
b=1 To ON the frequency, Shift/ Mode/ Down
Key I Stat / ON
35. For divergence of V X FREQ
8Vx iJVy 8V7.
V•V=-+-+- 30 1
ax ay az 42 2
8(x 2 y) 8(-xy) 8(xyz)
V•V=--+ ay +-a- 50 3
ax z 60 6
V•V=2xy-x+xy AC Shift Stat Var x
x.= 52
=
At x 3, y 2, z = 1, =
V • V = 2(3)(2)-3 + 3(2) = 15
5. Enter Mode 3 1
36. For gradient of V To ON the frequency, Shift/ Mode/ Down
Key / Stat / ON
8Vx . 8Vy. 8V7.k
'iJV=-1+-1+- X FREQ
ax ay az
2
8(x y) . 8(-xy) . a(xyz) l 30 1
VV=--1+--1+--{
iJx ay az 42 2
vv = 2xyi - xj + xyk 50 3
At x =3 , y = 2, z = 1, 60 6
VV = 12 i - 3j + 6k AC Shift Stat Var ox
ax== 9.49
1vv1 = .J 12 2 + (-3) 2 + 6 2
1vv1 = 13.7
6. Enter Mode 3 1
37 . For curl of V , To ON the frequenc S .
Key I Stat/ ON Y. hift /Mode/ Down

~1
i
15 X FREQ
VxV=l/x2y ISy ISz 1.5 5
- xy xyz 1.75 3
1.25 2
2.5 4
1
MPHT Rev1ew Center: Solution and Answer ke 2
1
 AC Shift Stat Var x 9. Two different doors use for entry and exit
x = 1.703 No of ways to enter by one door = 8
No. of ways to leave by another door = 7
7. Let X = like to smoke both Marlboro and N :: (8)(7) = 68
Philips Moms
33 - X = like to smoke Marlboro ONLY 10 . If 1 digit is used only once using 6 digits
20 - X = like to smoke Philips Morris N =6x5x4x3
ONLY N = 300
By Venn diagram,
11 . From city A to cit y B,
M PM
n = 5 roads
r = 2 different roads used back and fort h
From city B to city C,
n ""4 roads
r = 2 different roads used back and fortt,
N = (5P2)(4P2)
N = 240
(33 - X) + X + (20 - X) = 40
X = 13 12. The positions of boys and girls sitting in
alternate seats
8 Let X = no of viewers who watch both B G B G B G
games G B G B G B 2 ways
Bowing
No. of ways to arrange girls = 3 1
F oo lbafl
No. of ways to arrange boys = 3 !
N = (3!)(3!)(2)
N = 72

13. The positions of boys and girls where the

girls are always together
GGGBBB
B G G G B B
\" I 20 +X I BBGGGB
Basketball B B B G G G 4 ways
No. of ways to arrange girls = 31
4 5- X = Viewers who watch football and No. of ways to arrange boys= 3!
basketball N = (31}(3!)(4) = 144
70 - X = Viewers who watch football and
boxing 14. n = 6 digits
50 - X = Viewers who watch boxing and r = 4 different digits
basketball N = 6P4 = 380
45 - X + 70 - X + 50- X = 165 -3X
15. n = 4 different flags
V iewers who watch football only r = 1, 2 , 3 or 4 flags used for signal
= 285- (45 - X + 70-X + X) = 170 + X
Viewers who watch basketball only N = 4P1 + 4P2 + 4P3 + 4P4
= 115 - (45 - X + 50 - X + X) = 20 + X N = 64
Viewers who watch boxing only
= 195 - (70 - X + 50 - X + X) = 75 + X 16. n = 11 letters
170 + X + 20 + X + 75 + X = 265 + 3X l's =4, S's =4, P's = 2
nl 11 !
N = - - -=----
Viewers who watch any of the 3 games p! q! r! ... (41)(4!)( 2 1)
::: 500 - 50 = 450 N == 3,f,060
265 + 3X + 165 - 3X + X = 460 17. n ::: 2(4) = 8
X = 20 p"' 2, q ::i 2, r = 2, s i;:: 2
n! O!
Vi0w ers who watch boxi ng only N :: - - - = - - - - --
p l q! r ! ... (2 !)(2!)(21) ( 2!)
75 + X = 75 + 20 :: 95 N ~ 2...lli

MPHT Review Center: Solution and Answer key In MATH P;-io&> .d.A
 B repeated trial probability,
= =
18. N (n-1)! (10-1)! Y )n r
p = (nCr)(p)r (q -
3
N = 362,880 p = (5C2)(1/2)2(1/2)
p = 5/16
19. n = 13 teams
r = 2 teams per game 29. Free throw average (p) = 0 .65
N = nCr = 13C2 = 78 q = 1 -p = 0.35
Minimum no. of days for the tournament
n = 3, r = 2, n - r = 1
= 78/6 = 13 By repeated trial probability,
P = (nCr)(p)f (q) n- r
20. n = 10, r = 6
p = (3C2)(0.65)2(0.35)
N = nCr = 10C6
P = 0.444
N = 210

30. Probability of getting red ball

21 . n = 10, r = 3 points for triangle
N = nCr = 10C3 P1 = 5/8
Without returning the first red ball, the
N = 120
probability of getting another red ball
22. n1 = 9 men, r1 = 7 men P2 = 4/7
n2 = 6 women, r1 = 5 women P = (5/8)(4/7)
N = (9C7)(6C5) P = 6/14
N = 216
31. If white ball is drawn from the second bag
23. n1 = 7 men, r1 = 3 men P1 = 3/8
n2 = 5 women, r1 = 2 women If this ball is placed in the first bag and
N = (7C3)(5C2) white ball is drawn
N = 350 P2 = 5/8
If black ball is drawn from the second bag
24. n = 5, r = 1, 2, 3, 4, or 5 p3 = 5/8
N = SC 1 + 5C2 + 5C3 + 5C4 + 5C5 If this ball is placed in the first bag and
N = 31 white ball is drawn
p4 = 4/8
25. n = 42, r = 6 P = P1P2 + p3p4
N = 42C6 = 5,245,786 P = (3/8)(5/8) + (5/8)(4/8)
Probability of winning if you bet a ticket P = 35/64
P = 115,245,786
32. Even numbers (box A) = 2, 4, 6, 8
26 . For sum of 9, the possible outcomes are Even numbers (box B) = 2, 4
(4 , 5) (5 , 4) (6, 3) and (3, 6) = 4 outcomes Probability of getting an even number
The total outcomes is 36 from box A,
= =
P 4/36 1/9 P(A) = 1 /2 (4/9) = 2/9
Probability of getting an even number
27 . The positions of boys and girls where the from box B,
girls are always together =
P(B) 1/2 (2/5) 1/5 =
G G G B B B
B G G G B B P(T) = P(A) + P(B) = 2/9 + 1/5
BBGGGB P(T) = 19/45
B B B G G G 4 ways Probability that the even numbered card
No. of ways to arrange girls= 3! came from box A
No. of ways to arrange boys = 31 p _ P(A) _ 2/9
N = (3!)(31)(4) = 144 - P(T) - 19/45
Possible outcomes = 144 P = 10/19
Total outcomes = 61 = 720
P = 144/720 = 1/5
33 . p = 2/5 I q =1- p 3/5=
n = 8, r = 5, n - r = 3
28 . Probability of having a boy (p) = 1/2
By repeated trial probability,
Probability of having a girl (q) = 1/2
P = (nCr)(PY (q)n-r
n = 5, r = 2 , n - r = 3
P = (8C5)(2/5) 5 (3/5)3
P = 0.1239

MPHTRev\ew
nter: Solution and Answer kPv In M " .... •
 34 Enter Mode 3 2
X y
2. Enter Mode 2
1 0 .2
(2 -3i) (5 + 2i) = 16-11/
2 0 .28
3 0 .18
4 0.23 3. i =..r::::f. ; i2 = -1 i3 =- i
i21l + i21 + i = (i26)i + (i20)i + i
5 0 .11
= (j4(7))i + (i"<5l)j + j
Press AC, then Shift Stat Sum rxy = i + i + i = 3i
LXY = 2.77
4. Shift Mode 4 (Radian Mode)
35. Expected gain = Investment - Loss z = 4 + Si
Expected gain= 40000(0.3)-10000(0.7) Pol (4, 5) =
Expected gain = 6000
r = 6.40, 8 = 0 .896
In (4 + Si) = In (6.40 L 0.896)
36. Probability of getting a card of club
= In (6.40) + j0.896
P1 = 13/52 = 1/4
= 1.856 + /0.896
For four cards with replacement,
=
P (1/4)(1/4)(1/4)(1/4) 111.§§. =
5. 'A = (0 - 1
J) /i
37. Ways of getting 4 aces out of 5 cards TT 1/J
N = (4C4)(48C1)
=(1 L -
2)
p = (4C4)(48Cl) ff 1;1
= ( e- 12)
(52CS)
P = 1/54145 = e_nh
',A= 0.2078
38. Probability of getting head (p) = ½ 6. Enter Mode 2
Probability of getting tail (q} = ½ (1 + i)7 = (1 + i) 3 (1 + i) 3 (1 + i)
n =5, r = 3 (1 + i)7 = 8- Bi
By repeated trial probability,
p = (nCr)(PY (q)n-r 7. Shift Mode 4 (Radian Mode)
P = (5C3}(1/2)3(1/2) 5 - 3 z = 3 + 4i
P = 5/16 = 0.3125 Pol (3, 4) =
r = 5, 8 = 0.927
39 . Probability of getting head (p) = ½ In (3 + 4i) = In (5 L 0.927)
Probability of getting tail (q) = ½ = In (5) + j0.927
n = 10, r = 7, 8, 9 or 10 = 1.60 + 10.927

By repeated trial probability, 8. 3x + 4y + 3yi + 15 - 3i = 0

P = L (nCr)(PY (q) n-r 3x + 4y + (3y - 3)i = -15 + jO

P = f[
Xaaa7
(lOCX) (½f (½f o-x]
Consider the imaginary parts
3y-3 = 0
3x+4y=-15-+
- y=1
3x+4(1}=-15
P = 11164 X = • 6.33

40. Probability of getting head (p) = ½ 9. Shift Mode 4 (Radian Mode)

Probability of getting tail (q) = ½ z = -46-9i
n = 10, r = 4, n - r = 6 Pol (- 46, - 9) =
By repeated trial probability, r = 46.872, 0 = -2.948
P = (nCr)(PY (q)n-r (-46 - 9i) 113 = (46.872 L - 2 .948) 113
P = (10C4)(1/2) 4 (1/2)6 = 46.872 113 L (1/3 X - 2 .948)
P = 1051512 = 3.61 L - 0.983
= 2-3i

EXERCISE NO. 12 10. Shift Mode 4 (Radian Mode)

Advanced Engineering Mathematics - 1 z = 1+ i
Pol (1, 1) =
1. z = 3 - 4i, x = , y = - 4
r = 1.414, 8 = 0.785
Pol (3, - 4) =
a. r = §
b . 9 = 53.13•

MPHT Revtew Center. Solution and Answer key in MATH

Pa!le sn
 t- or the first root, k = 1
( 1 + i)115 - (1 1
- .4 4L(0.785+2rr(1))115
::: 1.4141/S L (1/5 X 7.068)
::: 1.072 L 1.4136
15.
X-
-[i
1 -2 2
1:
-3
!1 ;J
::: 0.168 + 1.059i 3-4 -3 -
Solve this matrix using cofactor,
11. Enter Mode 2, Shift Mode 4 For cofactor of 1et row, 1-' column
sin(x + J'y)
.
= si·n x cos h Y + j cos x sinh y Enter Mode 6
sin(3 + 2J) = sin 3 cosh 2 + j cos 3 slnh 2 Choose Mat A (1 ), 3 x 3 (1)
sln(3 + 2j) = 0.531 - i3.S9 5
A -2
- 1 3
2 -3
J
12. Enter Mode 2, Shift Mode 4
f
-4 -3 -4
Shift, Matrix (4), Data (2)
cos( x + Jy) = cos x cosh y - J sin x sinh y
cos(l + 2J) == cos 1 cosh 2 - j sin 1 sinh 2 For cofactor of 2 nd row, 11t column
cos(l + 2j) = 2.0327 - i3.0519 Choose Mat 8 (2), 3 x 3 (1)

13. Enter Mode 6

14
B -2 2 -3
3 11
[
Choose Mat A (1), 3
1
3
2
A [ -2 -1 -2
1
3
4
l x 3 (1) -4 -3 -4
Shift, Matrix (4), Data (2)

For cofactor of 3 rd row, 1•1 column

AC Shift 4 7 Shift 4 3 = Choose Mat C (3), 3 x 3 (1)
det (Mat A) = 5
14 3 1 ]
C[ 5 -1 3
14.

l
-4 -3 -4

x::: [
-4
~ -: ~3
0 -1 2
~ Solve the partial determinant using Shift 4
= 1(-1 )1• 1 det (Mat A)+ 1(-1 ) 2• 1 det (Mat
-5 3 2 4 8) + 1(-1)3 • 1 det (Mat C)
Solve this matrix using cofactor, = 352
For cofactor of 1st row, 1st column Shift, Matrix (4), Data (2)
Enter Mode 6 For cofactor of 4 th row, 1at column
Choose Mat A (1 ), 3 x 3 (1) Choose Mat A (1), 3 x 3 (1)
1 -3 5 1 4
A [ O -1 -2 J
3 2 4
A[~-2 !12 -3~ ]
Shift, Matrix (4), Data (2)
For cofactor of 1•t row, 2 nd column X = 352 + 3(-1) 4 • 1 det (Mat A)
Choose Mat B (2), 3 x 3 (1) X=ill
B[!4 =i !2]
-5 2 4 16. A=[}z _\ i1
Shift, Matrix (4), Data (2) 0 2 -1
For cofactor of 1at row, 3 rd column BY transpose, interchange the elements
of rows and columns
Choose Mat C (3), 3 x 3 (1)

C[!4 ~ !2 l AT= r~ =i ~]
~ 0 -1
- 5 3 4
17. Multiply the·matrices at the right side
Solve the partial determinant using Shift 4
= 3(-1 )1• 1 det (Mat A) + (-2)(-1 ) 1• 2 det (Mat r;J = ½ [ ~ -1iJ [!]
B) + (4)(-1) 1 • 3 del (Mat C)
~] = ~ [2(2) 1(6) ]
= - 275 2 1(2) -2(6)

r;J = ½ [-1~0] = [lsJ

Shift, Matrix (4), Data (2)
For cofactor of 111t row, 4 th column
Choose Mat A (1 ), 3 x 3 (1) X•6, V-=-6

A[-\- 5
~3 =~
2
·1
18 3 r: u ♦ r; !Ht ;sl. f;:
2 2J
x = - 275 + (2)(-1 )1 • 4 det (Mat A) 27 16 1~
X = - 385
31} !]•2[! }[!: ~71 9 28 45
MPHT Review Center: Solution and Answer key In MATH

Page St
19. Mode 6
Ifs= 0,
Mat A (1), 3 x 3 (1) A =~
1 = 3A
2 1 3] J

[3 7 2
A= 6 1 4 Then,
1/3 -1/3
= -s +s-+-3
l
F(s)
AC, Shift, Matrix (4), Mat A (3), x- 1

-2/5 19/65 1/65 f(t) = L- 1 (1/3) + L-1 (-1/3)

A·1 = [ 0 -1/13 2/13 s s+3
3/5 -11/65 -4/65 From L (e•') = 1 / (s - a),
f(t) = 113- 113 e-"
20. For f(t) = t"
nl 26. F(s) = ♦s+t
L (t 11 ) = --
5n + l
2
s +9

31 f(t) = L-1 (4s + 1)

L(t 3 ) =- ·
53+1
s2 + 9
L (t 3 ) = 6/s4 f(t)=L-1(
sz
4s
+ 32
)+L- 1(
s2
1
+ 32
)

21 . For f(t) = cos at Since L (sin at) = s 2 +a

a 2 where a = 3
s s
L (cos at) =- --
s2 + a2
and L (cos at)=
s 2 +a 2
L (cos St) =-
s
-- f(t)=4L-1 s )+.!.L- 1
3 )
52 + 52
( (
s2 + 32 3 s + 32
2
L (cos St) = s I (s 2 + 25) f(t) = 4 cos 3t + 113 sin 3t
22. For f(t) = sin t and n = 1 27. Resolve F(s) into partial fractions
d" 4s A B
L [t 11 f(t)] = (-1)" ds" (L [f(t)]) F(s) =- --- =--+--
(s + 2)(s - 6) s + 2 s - 6
L [t5in t] = (-1) 1 ~(L (5int)) Multiply both sides by (s + 2)(s - 6)
ds 4s = A (s - 6) + B(s + 2)
a
But, L [sin at] = where a = 1 Ifs= - 2
s 2 +a 2
1 4(-2) = A(- 2 - 6) -+ A= 1
L [tsint] = -~(- 2 -) Ifs= 6,
ds s + 1
2 4(6) = 8(6 + 2) -+ B= 3
. _ -((s + 1)(0) - 1 (2s))
L [tsm t] - (s 2 + l) 2 Then,
1 3
L [tsin t] = 2s /(s 2 + 1) 2 F(s) =- -+--
s+2 s-6
1 3
23. For f(t) = cos 2t f(t) = L-1 ( -- )
s+2
+ L-1 (- -)
s-6
s
L(cosat) = z+ a2 From L (e81 ) = 1 / (s - a),
s f(t) = e- 21 + 3e1"
s s
L (cos 2t) = s2 + 22 = s2 + 4
By First Shifting theorem, replace s by s - 28. By reverse of First Shifting theorem ,
3, since L (e31 ) = 1 / (s - 3) F(s) (s + 2) (s + 2)
L (e 3 t cos 2t) = F (s - a) (s 2 + 4s + 20) (s + 2) 2 + 4 2
L (e 3 t cos 2t) = (s- 3)/ ({s- 3) 2 + 4) Then a = - 2 for eat and k = 4 for cos kt
The inverse Laplace transform
24. F(s) = 3 / (s 2 -16) f(t) = e .. 21 cos 4t

f(t) = L-1(s2 ~ 16) = ¾L-1 (s2 ~ 42) EXERCISE NO . 1 ;~

a Advanced Engmeermq Mathematics - 2
Since L (sinh at) = s 2 -a 2
where a =4
f(t) = ¾ sinh (4t) 1. Given A= G ~) ,subtract >.. from the
elements of the principal diagonal
25. Resolve F(s) into partial fractions 0 = A-Al
1 A 8
F(s) = s(s + 3) = + s + 3 s O = (2-A
4
3 )
1-J\
Multiply both sides by s(s + 3)
0=(2->..)(1->.)-3(4)
1 = A (s + 3) + Bs
A. =5 and A, = -2
If s= - 3
1= B (-3)
- B= _.!.
3

M PHT Review Center: Solution and Answer key in MATH Page 52

 , \ " · r·
17 . f(X) = X + ( 1. / 2) x~ + ( l I ~ P·
2. (5-A. 3 ) =0
0 2-A.
By inspection,
(5-A.)(2-A.) - 3(0)=0
A1=2 andA1 =5
f(x) = I,,
- x"

n-o
•n
3. (1-A. 2 ) =0 1a. t(x)= l:~c0;;
3 2-A.
f(x) = e"
(1-A.)(2-A.) - 3(2)=0
A.=4,-1
19. f(x) = x2 e•
If A. = - 1
f(x) = X2 ( ~- ~)
(x
1 x3)= (1-(-1) 2 ) (2 2) £mrO 11 1
xn•1
x2 x4 3 2 - (-1) = 3 3 f(x) = l:;-o-;;,
c~) c;~) c;)
= =
20. f(0)=cos(0)=1
4. 3
Since 3 x matrix, there will be 3 sets of f'(0)= - sin(0)=0
eigenvectors 3 x 1 matrix f'(0)=-cos(0)= - 1
f"(0)=-sin(0)=0
6. Refer to# 1 f(lv>(Q)=cos(0)=1
Eigen values A.1 =5 and A2= - 2 r(iv)(Q) 1
coefficient of X◄ = ~ --
- 4!
lfA.1 =5, (2-5
4
3_ ) = (34 -43)
15 coefficient of X
4
= 24
1

If A2=- 2, (
2-(-2) 3
-(- ) -
)- ( 4
\4
33)
4 1 2 21. f(x) = In x
x1 111
Eigenvectors: x2
1
x31-31
x
4 rcx> = x
1
4
Modal Matrix: 1
11 -31
4
-1
f'(x) = -
x2 2 - 1
For the coefficient of (x - 1) at a -
112
10. f(t) = t" -1
f '( 1) = 2 = - 1
£ [f(t)] = £ (f 112 )=
("1)1
1 = C1)1
21 1
-1 -1
s~1 S2 coefficient of (x - 1)
4
= 21 = 2
Where (~-)i= r(-~+1)
22. For the coefficient of the (x - a) 4 at a =3
(~) 1= rG) ~ -frr f(tv>(x) = f(x)
f(lv)(3)= f(3) = 2
Then, L (r112 ) = :
L(r112 )= !s coefficient of (x - 3)◄ =
f(tv)(3) 2
coefficient of (x - 3)4 = -- -= =
1
41
1
41 12

12
11 . r(8)=r(7+1)=7!
r(8)= 5040 23. For f(x)='fX , where a = 16
f(16) =4
12. r(10) = r(9+1)= 91 f'(x) = !x·½ 2
= 9 r(9)
f(16) =.!.x·i = .!.
= g r(8+1) = 9 (8)r(8) 2 8
Then r( 1 0) is not equal to 10 r(9) f {x)= 1(- 1) X-~
11
2
2 2
f'(16)= 2(-.!.) (16)"~ = __ 1
2 2 256
eJ8 •• -j8
13. cose = The power series will be
2

6cos20rrt =6 • )20nt + e -J20nt) 1 _1

( 2 f(x)= 4 + -(x-16) - ~ (X-16)2
6cos20rrt = 3el20"'+3e·JZom 8 21
1 1
f{x)=4 + - (x-16)- - (><-16 ll
8 512 F
-I10lTI
14. For the term"' +e-j10n1 ,
At t=1 (set into radian mode) 26. Assume x = - 1, n =1
e11om+e•J101rt = 2 ( •
)10nt + e
2
~10nt) ~~~~:~a~d error, (assume 1oo for infinity)
= 2cos10rrt 3 ~ 3
= 2cos(1 Orr(1 )) f(x)= 2 + L (;; [1 - cos nn)) sin(~)
= z.
-· s

-MPHT Review Center: Solution and Answer key in MATH

f>age 53
 MODE 5-1
f(-1) = -3
2
+ L (-nn [1- cos mr]) sin (-rm)
100
3 -
5
a
1
b
-2
c
-2
11 ~ 1
f(-1)=0 2 3 66
B = 18 years old ; E = 10 years old
Assume x = 1, n = 1
3. A= 2xy
f(1) = -+
3
2
L nn
100
3
(-(1 - cos nn])stn (nn)
-
5
A= 2x(12 - x 2 ) = 24x - 2:x3
dA= 24-6x
n= l
f(1)=3 0 = 24-6x
x=4
Then , the function is A= 2(4)(12 - 4 2 ) = - 32

3
f(x)=-+
2
L. . (-[1- cosnn])sln (nmc)
3
llll
-
5 4.
A = 32 sq. units

Let x = amount of 3% iodine solution

u :,:: l
y = amount of 20% iodine solution
27 . n(n - 1)(n - 2)(n - 3) ... (3)(2)(1) = n/ The amount of the solutions
x + y = 85 - Equa~ion 1
28. Fundamental frequency = 20rr The concentrations of the solutions
w =2rrf 3x + 20y = 85(19) - Equation 2
20lT =2ni
f= 10 MODE 5-1
a b C
34 . The terms are perfect square, then for the 1 1 85
10th term, 3 20 85(19)
a10 =(10) =100
2 x=5;y=80
Harold used 5 ounces with 3% iodine
solution
Most Valuable Assessment (MVA) Test In
Mathematics - 1 5. Let x = number of first type seeds ($1)
y = number of second type seeds ($1 .26)

1. y = X + 50
($1 )(x) + ($1.26)(y) = $402
X+ 1.26 (X + 50) = 402
x = 150 packets

6. Let x = smaller angle

90 - x = larger angle
90-x = 2x + 15
X = 25°

Let 2x = length of rectangle, y = width of 7. By laws of exponents

rectangle (3x2 )(2x3 y) 4 = (3x2 )(16x12 y4)
lfxo= 10, ho=20, = 4Bx 14 y'
Yo= ✓ 202 - 102 = 10 ../3
8. For the area of rectangle
of similar A= length (width)
By ratio and proportion
A= (2x)(4x + 5)
triangles,
A= 8x2 + 10x
y Yo 10 ../3
--=-=
10 - X Xo 10 9. Let x and y be the dimensions of the
rectangle
y = ✓3 (1 O - x)
A = 2xy = 2x ( ✓3) ( 10 - x) A=xy
A= 2,/3 (10x - X2 } 100 = xy
Differentiate A with respect to x y = 100/x
<lA = 2../3 (10 -2X)
dx
For the perimeter
0 = 2../3 (10-2x)
x=S
P = 2 (x + y)
P = 2 (x + 100/x)
A = 2(5)( ✓3) (10 - 5)
A= 5CN3
Differentiate P with respect to x
2. B = 2E-2
dP = 2(1 - 1OO/x2 )
B-2E = -2 ➔ Equation 1 o = 2(x2 - 100)
2B + 3E = 66 ➔ Equation 2 X = 10, y = 10
Then the minimum possible perimeter
P = 2(10 + 10) = 40

MPHT Review Center: Solution and Answer key in MATH

 1O. Let x = diagonal 15. Let x = Jncreased dimension of the garden
x- 5 = length
x- 7 = width
A= length (width) X

3x
195 = (x-S)(x- 7)
x= 20m
X 24+2X
11 . Let L = 2W + 2
A=LW 2
3280 = LW X
3280 = W ( 2W + 2 )
W= 40ft 20+2x

12. ~et C be lhe point on the other side of the ~ = Aor1g + 141
nv~ directly oppoSite A Let D be the
po,_nt between C and B that Frank should (20 + 2x) (24 + 2x) = (20) (24) + 141
sw,m to; he will run the rest of the way to X = 1.5 ft.
B. Let x be the distance from to D. c
C X
The new length of the garden
0
3- x 8 Lnew = 24 + 2x = 24 +2(1 .5) = 27 ft

16. Refer to #1 for solution

A= 5<N3

A
17. In ABCD, LA= LC and LB= LD, then LA
and LO are supplementary
T = tsw1M + lRuN LA+ LD = 180
T = dswim + drun 3x + 10 + 2x + 30 = 180
Vswlm Yrun x=28
-.ll+x 2 3-x LA= 3(28) + 10 = 94°
T=---+--
3 S
Differentiate T with respect to x 18. For initial volume; V 1 = rrr2 h
dT 1 For final volume, let r2 = 2r
-1
dx =6(1 + xz)z(2x)+
(
-51) V2 = TT (2r)2(3h) = 12 TTr2h
2x -1 1
O=
6c1 + x2 )z - 5 V2 = 12V1
x = 0. 75 miles The volume is multiplied by 12

19. A= LxW
✓1 + (0.75) 2 3 - 0.75
T = - - -3 - - + - 5- - dA dW dL
dt = L dt+ Wdt
T = 0.87 hr ( 60 mins I hr)
T = 52 mins After t = 3 sec.
13. Let x = width of the walkway = =
W 20 + 3t 20 + 3(3) = 29 in
L = 20-2t = 20 - 2(3) = 14 in
X
!~ = (-z sec
x □4x
(14 in) ( 3 ~ ) + (29 in) in)
dA sec
d"r = - 16 in2/sec
24+2x dA
dt = 16In 1/sec fdecreaslnql
35
X
20.
35+2x

rn
For the area of the garden surrounded by
walkway
10
( 35 + 2x ) ( 24 + 2x ) = 1530
X = 5 ft.
6

14. Let B1 = time for Ben to work alone and B1

= time for Bill to work alone
1 l 1 x- - - 1
Bi+ B2 = T Let x = length of the sh
By similar triangles adow of the tree
1 1 1 18 6 I
- +- =- ---==
81 10 6 + 15 -X
l(
8 1 = 15 hours X::: 7.5 ft.

- --- - , .. ,.. , , .... u Arru

 21 . The area of the grid square is 2 in x 2 in = 4n dh
4 sq . in. 10=-(5) 2 -
3 dt
dh/dt = 0.032 ft/min

27 . Let N = no. of combinations of 5 card

2-d hands from the deck
N = 52C5 = 2,698,960

28. If the plane's two engines operate

When a coin's center is within the smaller independently,
square, it won't touch any line P = P1P2 = {1/100)(1/100)
The area of the smaller square is (2 - 1 )(2 P = 0.0001
- 1 ) = 1 sq. in.
Approximately, 29. x 2 + y 2 = 25
P = 1 in 2 / 4 in 2 = 1/4 Differentiate the equation
dx
2x-+2y-
dt
dy
=Odt
22. Let Q = probability that every 2 people X dx + y dy = O
have different birthdays lit dt
For 30 people,
365 364 363 336 At x = 3, y = 4 and dx/dt = 2
Q = 365 X 365 X 365 X ... X 365 3 (2) + 4 :~ = 0
365P30 dy/dt = - 3/2 cm/sec
Q = 365 30
Q =
0.29 30. If the sum of the dots of pair of dice is 5,
Let P = probability that at least 2 people Possible Outcomes: (1,4) (2,3) (3,2) (4, 1)
have the same birthday Possible Outcomes = 4
P = 1 - a= 1 - 0.2s = 0.11 Total Outcomes: 6(6) = 36
P = 4/36 = 119
For problems 23, 24 and 25
23. Enter MODE 8 and use m = 3 For problems 31, 32 and 33
31 . By conditional probability,
AB= B-A P (A and 8) = P(A) P (8/A)
AB = < 4 - 2 2- 2 1 - 2> - VCT A P (A and B) = (1/2)(1/3)
P (A and B) = 1/6
AC= C-A
AC = < 2 - 2 3- 2 1 - 2> - VCT B 32. P (A or B) = P (A) + P (B) - P (A and B)
P (A or B) = 1/2 + 1/2 - 1/6
Using SHIFT 5 7 to solve cos 9 by Dot P (A or B) = 5/6
Product
VctA DotVctB 33. The probability of A under B, P(A/8)
cos 8 = Abs(Vct A)Abs(Vct B) P (A and B) = P(B) P (A/8)
case= 11 ✓ 10 1/6 = (1/2) P (A/B)
P (A/8) = 113
24. Area= 0.5 (AB)(AC)
Under Mode 8, solve the area by Shift 5 34. The boundaries involved ere yI = 0, Yu= x2
Using Ring method,
A= 0.5 Abs ( Vet Ax Vet B) V = nf(yu 2 - Y1 2 ) dx
A= 1.5 V = n fo2((x 2 ) 2 - 0) dx
V = 32rr/5
25. The vector that is perpendicular to the
plane is the cross product of the two
35. There are 6 ways to roll a sum of 7
vectors on the plane
(1,6) (2,5) (3,4) (4,3) (5,2) (6, 1)
By Shift 5 under Mode 8 ,
P (sum of 7) = 6/36 = 1/6
AB x AC == Vet A x Vet 8
There are 6 favorable outcomes with 36
AB x AC = < 1 2 2>
equally likely outcomes, so 30 are
unfavorable. The odds in favor of a sum
26. Given: dV/dt == 10 ft 3 /min
of 7 are 6: 30 or 1:5
Find dh/dt at h = 5 ft
Since r = 2h
36. Enter Mode 2 (Complex), Shift Mode 4
V = ~r 2 h = ~(2h) 2 h
(Rad)
3 3
4TT Convert 3 + 4i into polar farm
V=-h 3 Enter 3 + 4i then Shift 2 3
3
= 5 L 0.927r
The derivative of V with respect to t In (r L 8) = In (re Je) = In r + j8
dV 2
41t dh In (3 + 4i) = In (5) + j0.927
-=-h - In (3 + 4i) = 1.61 + i0.93
ctt 3 dt
37. Solve the values of x at y = 0
4x -x2 - 3 = O
Enter Mode 5 3
a b c
-1 4 -3
x1 = 3, X2 = 1 For the enclosed area

A= f (yu - Y1) dx A = xy )- 50 - x2
A = x ( 50 - x - . ect to x
A= J/(4x - x 2 - 3) dx Differentiate A with resp
A=4/3 dA = 50-2X
h x-25
Set dA = O,t ~n - ea for her garden
40. Let i + i2 + i3 + i4 = i - 1 - i + 1 = 0 Then the maximum ar
Then, A= 25 (50-25)
j + j2 + j3 + . . . + j23 = j21 + j22 + i23 A = 826 sq. ft.
j + j2 + j3 + . . . + j23 : j + j2 + i3
i + i2 + i3 + . . . + i23 = i - 1 - i = - 1 45. A= f ydx
41 . Solve for the x if y = 0,
A= fo\x3- 3x2 + 2x + 1) dx
0 = x 2 - Sx + 6 A=j
Enter Mode 5 3
a b c 46. For the average of 93% for 4 subjec~s,
1 -5 6 Total scores of 4 subjects= ( 93 )( 4 ) - 372
X1 = 3, X2 = 2
Missing grade = 372 - (85 + 92 + 95 )
The area bounded from x = 0 to x = 2 Missing grade = 100
A1 = f y dx = f0\x 2 - Sx + 6) dx
47. Given: 320 cans, 40 cans per box
A1 = 14/3
No. of boxes = 320 cans I 40
The area bounded from x = 2 to x = 3
No. of boxes = !!.
A2 = f y dx = fz3 (x 2 - Sx + 6) dx
A2 = - 1/6 48. Solve x at y = 0 (x-axis)
The area bounded from x = 3 to x = 4 x-x2 =0
4
J
A3 = f y dx = 3 (x 2 - Sx + 6) dx X (1 - X) = 0
A3 = 5/6 X1 =0, X2 = 1
The total area Using Disk method
A= IA1I + IA21 + IA31 V = nf(y 2 ) dx
A= 14/3 + 1/6 + 5/6 = 1713 f1
V = tt 0 (x- x 2 ) 2 dx
V = rr/30
42. Given: a 1 = 3 an = 136
n
Sn= 2 (a 1 + an) 49. From 10 AM (Fri) to 10 AM (Sat)
n t, = 24 hours
1390 = 2 (3 + 136) From 10 AM (Sat) to 6 PM (Sat)
n = 20 tz = 8 hours
T = t1 + tz = 24 + 8 = 32 hours
Enter Mode 3 - 2
X y 50. If she pledged 0.50/mile, then
1 3 Total amount= (9 miles)($ 0 50/ ·1 )
20 136 Total amount= $4.SO · m, 8
Solve for a2 and a3 where y = Shift 1 5 5
a2 = 29 = 10
83 = 3y = 17
The first 3 terms are 3 1 10 and 17

43 . Based on the table,

f(3) = 2
=
g(f(3)) g(2) =-
3

44 . Let x and y = dimensions of the fencing

The perimeter of the garden
2x + 2y = 100
x+y=50
y=S0-x

MPHT Review Center: Solution and Answer key In MATH

 ANSWER KEYS 24. B. 8 hours
25. A. 21
EXERCISE NO. 1: Algebra _ Part 1 26. A 342 .85 mph
27. B. 73.33 mph
1. A. 4 28. B. 24 minutes
2. C. 2 29. C. 43 7/11 minutes
3. A. 4 , - 5 30. B. 12:16:21 am
4. D. 4 , 3
5. A. 16
EXERCISE NO. 3: Trigonometry
6. A. 4
7 . A. 2 1. A -cosA
8 . A. -4,-7 2. A 3/5
9. C. 1 3. B. -1/7
10. D . 8 4. A 5/13
11 . D. 5 5. A 120° or 240°
12. B. 12 6. A 9
13.C. 2x5 7. B. 358.52 m
14. B. 120x7 y3 8. D. 196 km
15. A. -1365x 4y 11 9. A 3.732 km
16. D. 43750x8 10. C. 14.78 meters
17. A. -192456 11. B. 31
18. A. 1 12. C. 22.87 m
19. D. 19684 13. D.90m
20 . C. 110 14. A 10.2 meters
21. B. 28 15. D. Sin 4y
22. D. -28 16. C 2/3
23. A.2 17. C. 74.16 m
24. A 42 18. D. 30°
19. C. 135°
25. A . x3 - 5x2 + 2x + 8 = 0
20. B. 800mils
26. A. 50
21 . C. 7 4°20'
27. C . 120
22. B. 4 P.M.
28 . D. 9
23. C. 6AM.
29. A 230
24. B. 158°18'43"
30. D . 3000
25. A 56,476,062 km 2
31 . C. irrational
26. C. cos(- 9) = cos(S)
32 . B. transistive
27. A 1,111
33. D . imaginary 28. B. IV
34. 8 . real,unequal,rational
29. C. latitude
35. B. extremes 30. A angle of depression
31. B. quadrantal
EXERCISE NO. 2: Algebra - Part 2 32. A 2sin8cos8
33. C. reflex angle.
1. C . 120 34. B. The sum of three angles is 180°
2. C. n2 35. A 45 degrees
3 . C. 31
36. A. 3094
4 . C. 128
5. C. 0 .02 37. B. linear pair
6 . B. 8 38. B. Medians
7 . C. 62730 39. B. bearing
8. B. 50 40. A. rr/3
9. A. 5
10. A . 5
11 . C . 204 .8 in EXERCISE NO. 4: Plane and Solid
12. D. 7 , - 7/6 Geometry
13. A 29524
14. C . 3/4 1. A concave
15. D. 1/22 2. C. 360
16. A. 0 .033 3. C. edges
17. D. 15 4. B. 30
18. B. 20 5. C. 4
19. C. 24 6. C. 5
20. C. 2 7. D. 54
21 . B . 20% 8. B. 170
22.C . 2.18 9. C. 259.81
23. B. 4/3 hours 10. D. 120.71 cm 2

MPHT Review Center: Solution and Answer key in MATH Page 58

 11 . A. 206.71 12. C. 0 .5
12. B. 40.428 sq. cm 13. D. Downward +3- 0
13. 8 . 432 14 . A.Sx2 +4x-10y -
14. C. 78 .54 sq. in 15. B. 5
15. D . 96 sq. cm 16. D. 96 m
16. C . 31 .2sq. m 17. 8 . (2, 3)
17. D . 6 .382 m 18. B. 4
18. A.1.71cm 19. D. 1.5 + 4 ::: 0
19. C . 5 cm 20. C. 4x2 + 9y2 - 8x - 3 6 y
20. A. 500 m 21 . 8 . 0.92
21 . C. 4 .5 22 . B. 15TT
22 . B . 4 23. D. 3.6
23. C . 32 .23 cm 24. c. 4
24 . D . nr'/3 25. C. (-1, -1)
25. B. 18 m 3 26. c. 12
26 . A 23% 27. A 1.8
27 . C. 45 Pl 28. 8 . 2
28. B. 2.670.35 cm2 29. B. 4x - 3y - 20 =0
29. B. 243.78 cm2 30. B. 1.41
30 . A 216.5 31 . B. 5.33
EXERCISE NO. 7: Dlfferentlal Calculus
EXERCISE NO. 5: Analytic Geometry - 1
1. B. ½
26. B. II
2. 8 . 3
27 . A. 15.65
3. B. 2/5
28 . B . 11 4. 8. Infinity
29. D. 5.38
5. 8.2
30 . A . 11/3
6. A. x2+z:
31 . C. (13, - 20) (x+l)
32 . D. (- 1, 1) 7. 8. e)( (cos x 2 -2x sin x 2 )
33. C. 6 8. 8. (4x 10910 e)/ (x2 + 1)
34 . A 1,2 9. 8. X /(X2 + 2) 112
35. C . 38 10. A 2x-y = 0
36. D. 25 11. A. 19.8 mis
37 . B. 20 12. B. -4.94
38. D. 2x - y = 3 13. C. - 2 ± ../2
39 . A Nol 14. A. ( -2, 28) , (2, -4) , & (0, 12)
40. 8 . y - X + 4 = 0 15. C. 2x + y = 9
41 . D. 2x - y + 11 = 0 16. C.25&25
42 . A. 4x + 6y - 29 = 0 17. A. 15.59 cm 2
43 . 0 . Bx + 4y - 4 = 0 18. A. 12 m
44 . D. 3 19. 8 . 100
45 . C. 4 20. C. 22.24 cm, 44.5 cm
46. A 2 .53 21. 8 . 0.95 ft/min
47 . B. 4 .39 deg. 22. D. 12 kph
48 . C . x+3y-12=0 23. B. 1.33 m
49 . A. 9x + 33y = 154 24. A 6.28 ft3/min
50. B. 10.14 25. A. 0.012732 inch/sec
5 1 . B . ( 10, 233.23°) 26. C. -0.001 amp/sec
52. D. 5x2 + 9y2 + 12x - 9 = 0 27. C. Point of Inflection
28. 8. Minima
29. D. Zero
30. B. r(a) = o
EXERCISE NO. 6: Analytic Geometry - 2
1. B . Ellipse EXERCISE NO. 8: Integral Calculus
2. B. Parabola
3 . C . Hyperbola 1. C. 3n/16
4 . D . Empty set 2. A. 0 .134
5. D. (1,2) 3. A. 0 .0184
6 . B . ( - 6 , 4) 4. C. 17.S
7 . B . x2 + y 2 - 6x + 10y +18 = 0 5. 8. 9
8 . C . 78 .5 sq units 6. C. 3._ (In 81) + C
9. C. x 2 + y 2 + 6x - 18y - 10 = 0
7. D. 2 tan x + cos x + C
10. B. x2 + y 2 - 16x + 2y + 52 = 0
11. A . 2 ,8

MPHT Review Center. Solution and Answer key ln MATH

 8. A 9
-~ ( X
3
+3 t 2
+C
4.
5.
C . (0, 0 , -2)
A. 10.305°
6. B. (4, -6>
1 ' 12
7
- (2y + 1r (3y 2 - 1)+c 7. c-s
. v'34
i-- 3 J
v'34
9. D. 30
8. A. -3, 4/3
10 _ C. cos(1/x)+C 9. D.(a)-..!!. (b)~i-~j
../26' 13 13

11 . D. ev' +C 10. A. 32
11. B . ../35
12. B. 2
12. D. <13, -7, 2>
13. A 9/2
13. C. <-3, 3, 10>
14. C. 21 .08
14. A. <-12, 21, -6> , <20, -35, 10>
15. B. 2TT 1 S 3
16. B. 44.57 sq. units 15. B . ./351 + v"35j + ./35 k
17. B. 1.42 16. A. 17.83
18. A. 2TT, 4TT 17. B. 2 cm
19. C. 1.125, 3.6 18. B. 138
20. D. 0.582 19. C. 2i+7j-5k
21 . B. 1.5 20. A. 15.0
22. A . 0.571 21. B. 3
23. C. 4/7 22. B. 24
24 . A 0.095 23. C. 13.6
25. B. 50.265 cu. units 24. C. 74°12'
26. A. 28.27 cu. units 25. D. -18i+3j-15k
27 . A. 228 26. A. 3.73
28. C. 26.81 cu. units 27. A. -1.143i + 4.573j + 2.287k
29. D. 48TT2 28. A. 3.39
30. B. 972.16 29. A. 3i - j + Sk
31. D. 624 30. A. 5.92
31 . B. -2.19
EXERCISE NO. 9: Differential Equation 32. A. < -.828i -1.655j -1.241k >
33. B. 84°20'
1. B. 2 34. A. 3
2. C. 3 35. D. 15
3. D. 2, 4 36. B. 13.7
4 . A (x - 2y)dx - xdy = 0 37. D. (xz}i - (yz)j - (x2 + y)k
5. A. y" -y' - 6y = 0
6. D. ydx - (x + 1)dy = 0 EXERCISE NO. 11: Statistics & Probablllty
7 . B. yy'' + (y')"2 + 1 = 0
8. D. tan- 1 x + tan- 1 y = C 1. B. 59
9. A. y = sqrt(Sx-1) 2. C. 37.75
1o. C . x"3 = c(9x11 2 + y"2) 3. A. 40.78
11 . A. (x + y)dx + (x - y)dy = 0 4. B. 52.00
12. A. x 2y + y = C 5. B. 9.49
13. B. yy" - 2ycos x = sin x 6. C.1 .7
14. B. e3 x 7. B. 13
1 5 . A. y = 8e3 x - 2 8. C. 95
16. C. y- 1 = Ce<x•z)/2 + 1 9. A. 56
10. A. 360
17 . A. y = e 1 ·5 x(c 1 cos 2,/7 x + c2 sm
. ,/7 )
2 x 11 . B. 240
18. A . y" ' + 2y" + 9y' + 18y = 0 12. C. 72
19.
2
B . y = e><(c, + xc2 + x C3) 13. A. 144
20 . C . 694 14. A. 360
21 . B. 39.6 min 15. A. 64
22 . A. 2.5 secs 16. C. 34650
23. C . 0 .338 min 17. B. 2520
24 . A. 0 .5x"2 + y"2 = k 18. A. 362 ,880
19. C. 13
EXERCISE NO.10: Plane and Space 20. A. 210
Vectors 21 . A.120
22 . B. 216
1. A. 2.236 23. A. 350
2. B. 5 24. C. 31
3 . D . 11 25. C. 1/5245786

MPHT Review Center: Solution and Answer key in MATH

 2 7 . A 1/5
ti. t,, ..:>µt:,vtl QI I•''-" ••.,.
2 8 . A 5/16
9. C. Singular
29. C . 0 .444
10. C. ✓(rr/s)
30. C . 5/14
11 . 8 . 5040
31 . A 35/64 12. C. 10r{9)
32. 8 . 10/19 13. 8. 3e()20lrt> + 3e<·J20n1)
33. C . 12.38% 14. C . 2
34 . B . 2 .77 15. A Odd function
35. A 5000 16. D. 360°
36. C . 1/256 .. x"
17. C. Ln•l-
n
37. A 1/54145 18. A ex
38. A. 0.3125 .. xn+z
19. C. }: 0 • 0 -
39. B . 11/64 nl

40. A 105/512 20. D. 1/24

21. C. -1/2
22. C. 1/12
EXERCISE NO. 12: Advanced Engineering 23. D. 4 + 1/8 (x- 16)- 1/512 (x - 16)
2

Mathematics - 1 24. B. Convergent

25. A Divergent
1. A. 5, 53.13° 26. A~+
2
r:.. 1
(~
nn
[1 - cos mr]) sin(";")
2. D. 16-11i 27. C. n!
3. A. 3i 28. C. 10
4. A. 1.86 + 0 .90i 29. C. Solidus
5. 8 . 0.2078 30. A Euler
6. 8. 8-8i 31 . A Surd
7. B. 1.61 + 0 .93i 32. 8. on the x axis
8. A -6.33 33. B. an even function
34. D. 100
9. 8 . 2-3i
35. A Euler
10. C . 0.168 + 1 .06i
11 . 8. 0 .531 - 3.59j Most Valuable Assessment (MVA) Test In
12. C . 2 .0327 - 3 .0519i Mathematics - 1
13. 8. 5
14. A -385 1. C 21. C 41. C
15. 8 . 373 2. B 22. 8 42. A
3. B 23.A 43. 8
16. 0.[~
2 0
=i ~] -1
4. A
5. A
24. C
25. C
44. B
45. B
17. A. s. - 5 6.A 26. C 46. B
27 17 7.A 27. 8 47. C
18. A. 35 5 8.D 28.B 48. B
28 45 9.D 29. C 49. D
-2/5 19/65 1/65 10. D 30. A SO.A
19 _ B. o -1/13 2/13 11. A 31. C
3/5 -11/65 -4/65 12.A 32. B
20. 8 . 6/s"' 13.B 33.D
21 . B . s/(s2 + 25) 14. D 34.D
2
22 . 2s/(s2 + 1 ) 15.D 35. B
2
23. A (s - 3)/((s - 3 ) + 4 ) 16. C 36.A
24 . B . ¼ sinh 4t 17. D 37. D
25. A 1/3- 1/3e-3t 18. A 38. A
26. B. 4cos 3t + 1/3 sin 3t 19. B 39. D
27. A e· 2t + 3am 20. B 40.C
28. B. e-2t cos 4t
. E NO 13· Advanced Engineering
EXERCIS · ·
Mathematics - 2

1. B . 5, - 2
2. 8. 2,5
3. B. [;]
4. C. 3 t ·x
,:; C . Modal ..,man
,