Mathematics (Standard) Class 10th

SAMPLE QUESTION PAPER 11

De taile d Solutions
1. (d) Circumference of circle, 2 pr = 30.80 2 sin q cos q 2 sin q cos q
= =
22 (1 - sin 2 q) cos 2 q
Þ 2´ ´ r = 30.80
7 [Q sin 2 A = 1 - cos 2 A]
7 2sin q
Þ r = 30.80 ´ = 4.9 cm = = 2 tan q
44 cos q
æ 22 ö 6. (a) Given, BD = 12 cm and AD = 6 cm
Area of the circle = pr 2 = ç ´ 4.9 ´ 4.9÷ cm 2
è7 ø
In DABD and DBDC,
= 75.46 cm 2 ÐBDA = ÐCDB [each 90°]
2. (c) Multiple of 3 in a single die are {3, 6}. ÐDBA = ÐDCB [each (90° - ÐA)]
2 1 DADB ~ DBDC [by AA similarity criterion]
\ Required probability = =
6 3 BD AD
Þ =
3. (a) Given, equation is 3x 2 - 4 3x + 4 = 0. CD BD
On comparing it with ax 2 + bx + c = 0, we get BD 2 12 2 144
Þ CD = = =
a = 3, b = -4 3 and c = 4 AD 6 6
\ D = b2 - 4ac = (-4 3 )2 - 4(3)(4) = 48 - 48 = 0 Þ CD = 24 cm
Which shows that the given quadratic equation has 7. (c) We know that, tangent at a point to a circle is
real and equal roots. perpendicular to the radius.
4. (c) Given, system of equation is \ ÐOAP = 90°
2x - 3y = 7 …(i) Q ÐAOP + ÐPOB = 180° [by linear pair]
5x + y = 10 …(ii) Þ ÐAOP + 105° = 180°
On multiplying Eq. (ii) by 3 and adding Eqs. (i) and (ii), Þ ÐAOP = 75°
we get
In DAOP,
2x - 3y = 7
ÐAPO + ÐAOP + ÐOAP = 180°
15x + 3 y = 30
37 Þ ÐAPO + 75° + 90° = 180°
17 x = 37 Þ x = Þ ÐAPO = 180° - 165° = 15°
17
On putting the value of x in Eq. (i), we get 8. (d) We have, ÐAPB = 60°
37 74 1 1
2´ - 3y = 7 Þ 3y = -7 \ ÐAPO = ´ ÐAPB Þ ÐAPO = ´ 60° = 30°
17 17 2 2
74 - 119 A
Þ 3y =
17
-45 15 60°
Þ y= Þ y=- P
M
O
17 ´ 3 17
5. (b) We have, B
cos q cos q cos q cos q In DAPM right angle at M,
+ = +
cosec q + 1 cosec q - 1 1 1
+1 -1 AM AM
sin q sin q sin ÐAPO = Þ sin 30° =
AP 10
sin q cos q sin q cos q 1 AM
= + \ = Þ AM = 5 cm
1 + sin q 1 - sin q 2 10
=
sin qcos q (1 - sin q + 1 + sin q) Now, AB = 2 AM = 2 ´ 5 = 10 cm
(1 + sin q) (1 - sin q)
 9. (c) Given, 7 sin 2 q + 3cos 2 q = 4 13. (c) The given graph intersects on the X-axis at 5 points.
Þ 7 sin 2 q + 3(1 - sin 2 q) = 4 So, The number of zeroes in the given graph is 5.
Þ 2 2
7 sin q + 3 - 3sin q = 4 14. (a) Plot the point (3, 11) in coordinate axes.
1 11 (3, 11)
Þ 4 sin 2 q = 1 Þ sin 2 q = 10
4
9
1 8
Þ sin q =
2 7
6
\ q = 30° 5
Now, sec q + cosec q = sec 30° + cosec 30° 4
2 3
= +2 2
3 1
X
10. (c) From the figure PQ = QB …(i) O 1 2 3
The distance between P and Q, Q and B are
From the figure above, the required distance is
PQ = (5 - 3)2 + (-3 - y )2 11 units.
and QB = (3 - 1)2 + ( y + 5)2 15. (a) The integers that end up with 1 are 51, 61, … , 341.
[using distance formula] Here, a = 51, d = 10 and l = 341
On putting the values in Eq. (i), we get So, 51 + (n - 1) 10 = 341
[Q nth term of an AP, an = a + (n - 1) d ]
(5 - 3)2 + (- 3 - y )2 = (3 - 1)2 + ( y + 5)2
Þ 10 n - 10 = 290 Þ n = 30
Þ (2 )2 + (3 + y )2 = (2 )2 + ( y + 5)2 30
\ Sn = [2 ´ 51 + (30 - 1) ´ 10 ]
[squaring on both sides] 2 n
Þ 4 + 9 + 6 y + y 2 = 4 + y 2 + 25 + 10 y [Q sum of n terms of an AP, S n = {2 a + (n - 1)d }]
2
Þ 6 y + 13 = 29 + 10 y
= 15[102 + 290 ] = 15 ´ 392 = 5880
Þ 4 y = - 16
16. (c) Given, P( y ) = 3 y 4 - 5 y 3 + y 2 + 8
Þ y=-4
For P(- 1,) on putting y = - 1, we get
11. (b) Let cost of one chair = ` x P(- 1) = 3(- 1)4 - 5(- 1)3 + (- 1)2 + 8
and cost of one table = ` y
P(- 1) = 3 + 5 + 1 + 8 = 17
Now, we have
17. (b) Let the length of the shadow be BC = x m
3x + 2 y = 1850 … (i)
A
and 5x + 3 y = 2850 … (ii)
On multiplying Eq. (i) by 5 and Eq. (ii) by 3, we get
15x + 10 y = 9250 … (iii) 8m

and 15x + 9 y = 8550 … (iv)

30º
On subtracting Eq. (iv) from Eq. (iii), we get B x C
y = 700
AB = Length of pole = 8 m
On substituting the value of y in Eq. (i), we get In right angled DABC,
3x + 1400 = 1850 AB é perpendicular ù
Þ 3x = 450 Þ x = ` 150
tan 30° =
BC êëQ tan q = base úû
The total cost of 1 chair and 1 table is 1 8
\ =
150 + 700 = ` 850. 3 x
1 é 1 ù
12. (c) Let one root be a, then another root be . Þ x=8 3m êëQ tan 30° = 3 úû
a
Given, equation is (k - 1)x 2 - 10 x + 3 = 0 Hence, the length of the shadow is 8 3 m.
1 10 1 3
Now, a + = and a × = 18. (a) Possible prime numbers in throwing of a die
a (k - 1) a (k - 1)
= {2, 3, 5}
3
\ 1= Þ k -1= 3 Þ k = 4 \ P (a prime number) =
3 1
=
k -1 6 2
19. (a) We have, P(E ) = 0.4, Þ 2 m = 18
where E = event of winning Þ m=9
15
P(not E) = 1 - P(E ) Hence, values of k and m are and 9, respectively.
2
= 1 - 0.4 = 0.6 (1)
Hence, both Assertion (A) and Reason (R) are true and Or
Reason (R) is the correct explanation of Assertion (A). Given, equation is
20. (a) If 3|93 means 3 divides 93 (k - 12 ) x 2 - 2 (k - 12 ) x + 2 = 0
Þ 93 = a ´ 3 [a is an integer] Here, a = k - 12, b = - 2 (k - 12 ) and c = 2

Similarly, if 5|0 means 5 divides 0 Since, the given equation has equal roots.

Þ 0 = 5´ b [b is an integer] \ D=0 (1/2)

2
Hence, both Assertion (A) and Reason (R) are true Þ [- 2 (k - 12 )] - 4 (k - 12 ) (2 ) = 0
and Reason (R) is the correct explanation of [Q D = b2 - 4ac ]
Assertion (A). Þ 4 (k - 12 )2 - 8 (k - 12 ) = 0 (1/2)
21. Since, XY is parallel to the side BC of DABC, by BPT, Þ 4 (k - 12 ) (k - 12 - 2 ) = 0
we have Þ 4 (k - 12 ) (k - 14) = 0
AY AX 3
\ = = Þ k - 12 = 0 or k - 14 = 0
YC XB 15 . (1)
AY 2 2 \ k = 12 or 14 (1/2)
Þ = 3´ =
YC 3 1 But when we put k = 12 in the given quadratic
AY 2 equation, the coefficients of x 2 and x become zero,
Þ = which is not possible.
YC 1 (1)
Hence, the required value of k is 14. (1/2)
22. Let first number = a = 180 4 4
cos a sin a
and second number = b 24. Given, + =1
cos 2 b sin 2 b
LCM = 360, HCF = 60
cos 4 a sin 4 a
We know that, Þ + = sin 2 a + cos 2 a
cos 2 b sin 2 b
LCM ´ HCF = First number ´ Second number [Q sin 2 q + cos 2 q = 1]
360 ´ 60 = 180 ´ b (1) cos 4 a sin 4 a
Þ 2
- cos 2 a = sin 2 a -
360 ´ 60 cos b sin 2 b
Þ b=
180 æ cos 2 a ö æ sin 2 a ö
Þ cos 2 a ç - 1÷ = sin 2 a ç1 - ÷
Þ b = 120 2
è cos b ø è sin 2 b ø
Hence, second number is 120. (1)
cos 2 a(cos 2 a - cos 2 b ) sin 2 a(sin 2 b - sin 2 a )
Þ =
23. Given that x = 2 and x = 3 are the roots of the cos 2 b sin 2 b
equation 3x 2 - 2 kx + 2 m = 0.
sin 2 b(1 - sin 2 a - 1 + sin 2 b )
\ 3(2 )2 - 2 k (2 ) + 2 m = 0 Þ
sin 2 a
Þ 12 - 4k + 2 m = 0 cos 2 b(1 - cos 2 b - 1 + cos 2 a )
=
Þ 4k - 2 m = 12 …(i) cos 2 a
2
and 3(3) - 2 k (3) + 2 m = 0 sin 2 b(sin 2 b - sin 2 a ) cos 2 b(cos 2 a - cos 2 b )
Þ =
Þ 27 - 6k + 2 m = 0 sin 2 a cos 2 a (1)
Þ 6k - 2 m = 27 …(ii) (1) sin 4 b - sin 2 b sin 2 a cos 2 b cos 2 a - cos 4 b
On subtracting Eq. (i) from Eq. (ii), we get Þ =
sin 2 a cos 2 a
15
2 k = 15 Þ k = sin 4 b sin 2 b sin 2 a cos 2 b cos 2 a cos 4 b
2 Þ - = -
sin 2 a sin 2 a cos 2 a cos 2 a
15
Put k = in Eq. (i), we get
2 sin 4 b cos 4 b
Þ + = cos 2 b + sin 2 b
15 sin 2 a cos 2 a
4 ´ - 2 m = 12
2 cos 4 b sin 4 b
\ + =1 Hence proved.(1)
Þ 30 - 2 m = 12 cos 2 a sin 2 a
 Or Þ 4x - 2 y = 8
2
sin q - 3sin q + 2 Þ 2x - y = 4 … (i) (1)
We have, =1
cos 2 q When, 4 students are removed from each row,
Þ sin 2 q - 3sin q + 2 = cos 2 q number of students in each row = ( y - 4)
Þ sin q + sin 2 q - 3sin q + 2 = cos 2 q + sin 2 q
2
and number of rows = (x + 4)
[adding sin 2 q both sides] Total number of students = (x + 4) ( y - 4)
Þ 2 sin 2 q - 3sin q + 2 = 1 [Q sin 2 x + cos 2 x = 1] Given, (x + 4) ( y - 4) = xy
2
Þ 2 sin q - 3sin q + 1 = 0 (1) Þ 4 y - 4x = 16
Þ 2 sin 2 q - 2 sin q - sin q + 1 = 0 Þ 4( y - x ) = 16
Þ 2 sin q(sin q - 1) - 1(sin q - 1) = 0
Þ (y - x) = 4 … (ii) (1)
Þ (2 sin q - 1) (sin q - 1) = 0
Adding Eqs. (i) and (ii),we get x = 8
Þ 2 sin q - 1 = 0 or sin q - 1 = 0
1 On putting x = 8 in Eq. (ii), we get
Þ sin q = or sin q = 1
2 y - 8 = 4 Þ y = 12
Þ q = 30° or q = 90° x = 8 and y = 12
\ q = 30° Total number of students in the class = 12 ´ 8 = 96
[Q q = 90° does not satisfy the given equation] (1) (1)
Or
25. According to the given information, figure is given Let the fixed charges = `x per month
below
and the cost of meals per day = `y
Then, we have x + 22 y = 1380 … (i)
x + 28 y = 1680 … (ii) (1)
On subtracting Eq. (i) from Eq. (ii), we get
O Q 6 y = 300 Þ y = 50
Substituting the value of y in Eq. (i), we get
O′ M
x + (22 ´ 50 ) = 1380
Þ x + 1100 = 1380 Þ x = 280 (1)
P
Hence, fixed charge = ` 280 per month,
Let us join OP, OQ and OM
and cost of food = ` 50 per day. (1)
Now, ÐOMP = 90°
[Q angle in a semi-circle is a right angle] (1)
27. Let AB be the building of height 60 m, CD be the tower
of height x m and distance between building and
and ÐOMQ = 180° - ÐOMP
tower be y m, i.e. BD = y m.
= 180° - 90°
A
X
= 90° [by linear pair] 30°
60°
In DOMP and DOMQ,
OP = OQ [each = 2r, radii of bigger circle] 30°
C E
ÐOMP = ÐOMQ [each 90°]
OM = OM [common sides] 60 m
Þ DOMP @ DOMQ [RHS congruence]
xm
\ PM = QM [by CPCT]
Hence, M bisect PQ. Hence proved.(1)
26. Let the number of rows = x 60°
D B
and the number of students in each row = y ym
(1)
Then, the total number of students = xy Draw CE || BD.
When, there are 4 more students in each row, number Then, CE = BD = y
of students in each row = y + 4 Also, ÐADB = ÐXAD = 60° [alternate angles]
and number of rows = x - 2 and ÐACE = ÐXAC = 30° [alternate angles]
Now, total number of students = (x - 2 ) ( y + 4) Now, in right angled DABD,
Given, (x - 2 ) ( y + 4) = xy
 BA é perpendicular ù Number of Mid-point
tan 60° = Q tan q = Class
DB ëê base ûú marks
schools ( xi ) di = x i - A fi di
60 (fi )
Þ 3= [Q tan 60° = 3] 20-39 526 29.5 -80 -42080
y
60 3 60 3 40-59 620 49.5 -60 -37200
Þ y= ´ =
3 3 3 60-79 674 69.5 -40 -26960
Þ y = 20 3 m …(i) (1) 80-99 717 89.5 -20 -14340
100-119 681 109.5 (A) 0 0
and in right angled DAEC,
120-139 612 129.5 20 12240
AE
tan 30° = 140-159 540 149.5 40 21600
CE
160-179 517 169.5 60 31020
1 60 - x
Þ = 180-199 552 189.5 80 44160
3 y
å f1 = 5439 å fi d i =
1 60 - x Total
-11560
Þ =
3 20 3 (2)
1 åf d
[Q tan 30° = and from Eq. (i)] \ Mean, x = A + i i
3 å fi
\ 60 – x = 20 Þ x = 40 = 109.5 -
11560
Hence, the height of the tower is 40 m. (1) 5439
= 109.5 - 2.13 = 107.37
28. Given, numbers are 404 and 96.
Hence, the average enrollment per higher secondary
Using prime factorisation method,
school is 107.37. (1)
404 = 2 ´ 2 ´ 101 and 96 = 2 ´ 2 ´ 2 ´ 2 ´ 2 ´ 3
Or
\ HCF of (404, 96) = 2 ´ 2 = 4
Let the frequency of class 20-30 be f1 and that of class
and LCM (404, 96) = 2 ´ 2 ´ 2 ´ 2 ´ 2 ´ 3 ´ 101
40-50 be f2 .
= 9696 (1½)
Since, the sum of all frequency is 170.
Verification HCF ´ LCM = 4 ´ 9696 = 38784 …(i)
The cumulative frequency table for given distribution is
and product of the given numbers
= 404 ´ 96 = 38784 …(ii) Class interval Frequency Cumulative frequency
From Eqs. (i) and (ii), we get 0-10 10 10
HCF ´ LCM = Product of the given numbers. 10-20 20 30
Hence proved. (1½) 20-30 f1 30 + f1 (c f )
29. Given, AD and CE are altitudes which intersect each 30-40 40 ( f ) 70 + f1
other at the point P. 40-50 f2 70 + f1 + f 2
(i) In DAEP and DCDP, 50-60 25 95 + f1 + f 2
60-70 15 110 + f1 + f 2
ÐAEP = ÐCDP [each 90°]
and ÐAPE = ÐCPD Here, median = 35, which belongs to the class 30-40.
[vertically opposite angles] (1)
So, the median class is 30-40.
\ DAEP ~ DCDP [by AA similarity criterion] (1)
N
Also, N = 170 Þ = 85
(ii)In DABD and DCBE, 2
ÐADB = ÐCEB [each 90°] \ l = 30, f = 40, cf = 30 + f1 and h = 10 (1/2)
and ÐABD = ÐCBE [common angle] ìN ü
- cf
DABD ~ DCBE ï ï
\ Now, median = l + í 2 ý´h
[by AA similarity criterion] (1) ï f ï
î þ
(iii) In DAEP and DADB,
ì 85 - (30 + f1 )ü
ÐAEP = ÐADB [each 90°] Þ 35 = 30 + í ý ´ 10
î 40 þ (1/2)
and ÐPAE = ÐBAD [common angle]
\ DAEP ~ DADB Þ 35 ´ 4 = 120 + (55 - f1 )
[by AA similarity criterion] (1) Þ 140 = 175 - f1 Þ f1 = 35 …(i)
Also, 110 + f1 + f2 = 170
30. Here, assumed mean, A = 109. 5
 [Q sum of all frequencies = 170, given] Consider the above figure, here AC is a chord of the
Þ f1 + f 2 = 60 circle.
Þ 35 + f 2 = 60 [from Eq. (i)] Since, OA = OC [radii of circle]
\ f 2 = 60 - 35 = 25 \ ÐOAC = ÐOCA = x [say] …(ii) (1)
Hence, the missing frequency of the class 20-30 is 35 Now, in DAOC,
and the class 40-50 is 25. (1) ÐAOC + ÐOAC + ÐOCA = 180°
[by angle sum property of triangle]
Common Be carefully to calculate the frequency
Þ 60°+ x + x = 180°
Mistake and cumulative frequency. Also, keep
mind the positive and negative sign in Þ x = 60° [from Eqs. (i) and (ii)]
calculaing the value of S f idi . Þ DAOC is an equilateral traingle.
3
\ Area of DAOC = . cm 2
´ (28)2 = 33972
31. Let breadth of rectangular park = x m 4
Then, length of rectangular park = (x + 3) m 3
Now, area of rectangular park = x(x + 3) (side)2
[Q area of an equilateral triangle =
4
= (x 2 + 3x ) m 2 and 3 = 1.732] (1)
[Q area = length ´ breadth] (1/2) qpr 2
Given, base of triangular park Also, area of sector OABCO =
360°
= Breadth of the rectangular park
60° ´ 22 ´ (28)2
\ Base of triangular park = x m =
7 ´ 360°
and also it is given that altitude of triangular park
22 ´ 4 ´ 28
= 12 m =
1 6
\ Area of triangular park = ´ x ´ 12 = 6x m 2
2 2464
=
1 6
[Q area of triangle = ´ base ´ altitude] (1/2)
2 = 410.67 cm 2 (1)
According to the question, \ Area of segment ABCA
Area of rectangular park = 4 + Area of triangular park = Area of sector OABCO – Area of DAOC
⇒ x 2 + 3x = 4 + 6x
= 410.67 - 339.72 = 71.19 cm 2
⇒ x 2 + 3x - 6x - 4 = 0 ⇒ x 2 - 3x - 4 = 0
Now, area of six segments = 6 ´ 71.19
Þ x 2 - 4x + x - 4 = 0 [by factorisation]
= 427.14 cm 2
Þ x(x - 4) + 1(x - 4) = 0 Þ (x - 4)(x + 1) = 0
Þ x - 4 = 0 or x + 1 = 0 Þ x = 4 or x = -1 (1) Since, the cost of making the design is
Since, breadth cannot be negative, so neglect x = - 1. ` 1.50 per cm 2 .
\ x = 4m \ Total cost of making the design is ` 1.50 per cm 2
Hence, breadth of the rectangular park = 4 m = 427.14 ´ 150
.
and length of the rectangular park = x + 3 = 4 + 3 = ` 640.71 (1)
=7m (1)
33. Let a and d be the first term and common difference of
32. We know that central angle of a circle is 360°.
an AP.
Then, a8 = 31 and a15 = 16 + a11
C Þ a + (8 - 1) d = 31
cm and a + (15 - 1) d = 16 + a + (11 - 1) d
28
60° D B
O [Q an = a + (n - 1) d ]
Þ a + 7d = 31 and a + 14 d = 16 + a + 10 d
A
Þ a + 7d = 31 and 4 d = 16 Þ d = 4
Þ a + 7(4) = 31
360° Þ a = 31 - 28 = 3 (2)
\ Angle of each sector = = 60° …(i)
6 (i) The AP series is a, a + d , a + 2d , ....
[Q six equal designs divide the circle in i.e., 3, 3 + 4, 3 + 2 ´ 4, ...... or 3, 7, 11, .... (1)
six equal sectors] (1) (ii) Sum of first 15 terms,
Now, let us determine the area of one segment with
angle 60°.
 15 = S1 + S 2 - S 3 = 249500 + 99500 - 49500
S15 = [2 ´ 3 + (15 - 1)4]
2 = 349000 - 49500 = 299500 (1)
é n ù Now, let S 4 be the sum of natural numbers less
êëQ S n = 2 [2 a + (n - 1) d úû than 1000.
15 15 ´ 62 n(n + 1) 999(1000 )
= [6 + 56] = = 465 (2) S4 = = = 499500
2 2 2 2
Or Now, sum of numbers less than 1000 which are
Let the AP of natural numbers less than 1000 neither divisible by 5 nor by 2.
which are divisible by 2 is 2, 4, 6, ... 998. = S 4 - (S1 + S 2 - S 3 )
Here, a = 2, l = 998 and d = 2 = 499500 - 299500 = 200000 (1)
\ a + (n - 1)d = l
34. A coin is tossed three times. Therefore, sample space
2 + (n - 1)2 = 998
is S = {HHT, HTH, THH, HTT, THT, TTH, HHH, TTT}
998
n= There are 8 possible outcomes, n(S ) = 8.
2
= 499 The game consists of tossing a coin 3 times.
n If one or two heads show, Sweta gets her entry fee
S1 = (a + l ) back.
2
499 If she throws 3 heads, she receives double the entry
= (2 + 998) fees. (2)
2
If she gets 3 tails, she loses the entry fees
499 ´ 500
= (i) loses the entry
2
= 249500 (1) Out of 8 possible outcomes, only one (TTT) is
Let the AP of natural numbers less than 1000 favourable.
which are divisible by 5 is 5, 10, 15, ...., 995. 1
\ P (she loses entry fee) = (1)
Here, a = 5, l = 995 and d = 5 8
Also, a + (n - 1)d = 1 (ii) gets double entry fee
Þ 5 + (n - 1) 5 = 995 Out of 8 possible outcomes, only one (HHH) is
Þ 5(n - 1) = 990 favourable.
n - 1 = 198 Þ n = 199
1
Þ \ P (she get double entry fee) = (1)
n 199 8
\ S 2 = (a + l ) = [5 + 995]
2 2 (iii) Let E be event that she just gets her entry fee.
199 ´ 1000 Then, E = {HHT, HTH, THH, HTT, THT, TTH}
= = 99500 (1)
2 \ n( E ) = 6
The AP of natural numbers less than 1000 which
P (she just gets her entry fee back)
are divisible by 10 is 10, 20, 30, ...., 990.
n( E ) 6 3
Here, a = 10, d = 10 and l = 990 = = =
n(S ) 8 4 (1)
\ l = a + (n - 1)d
Þ 990 = 10 + (n - 110
) 35. Given, Two circles meet at point A, tangents drawn at
980 A meet the circle at B and C.
Þ n - 1=
10 To prove P is the circumcentre of the DABC.
Þ n - 1 = 98 Construction Draw perpendicular lines from centre
Þ n = 99 O and Q to the point A.
99 Proof AQ ^ AB
\ S3 = (10 + 990 )
2 [Q a tangent to a circle is perpendicular to the
99 radius through the point of contact]
= ´ 1000
2 AQ || OP
Þ S 3 = 99 ´ 500 = 49500 (1) [Q opposite sides of a parallelogram are parallel]
Since, the sum of numbers divisible by 10 are Þ OP ^ AB
common in both the sum of numbers divisible by 2
[Q AQ || OP and AQ ^ AB, so OP ^ AB]
and 5.
Let OP intersects AB at M. (1)
Thus, the sum of numbers divisible by 2 and 5 is
 80 10
Þ x= = cm
A 24 3
10
Þ AE = cm
O 3
M Q 10
P Similarly, BE = cm
3
B C
10 10 20
Hence, AB = AE + BE = + = cm
3 3 3 (2)
\ OM ^ AB
36. (i) The air capacity of spherical balloon
Þ AM = BM (1)
= Volume of sphere
[Q perpendicular drawn from the centre of a circle to 4 4
a chord bisects the chord] = pr 3 = ´ 314
. ´ (6)3
3 3
\ OM and OP is the perpendicular bisector of AB. 4 ´ 314
. ´ 216
= = 904.32 m 3
Similarly, PQ is perpendicular bisector of AC. 3
Now, in DABC, OP is the perpendicular bisector of (ii) Total surface area of the conical balloon
side AB. = prl + pr 2
\ PA = PB = p(rl + r 2 ) = 314 . + (3.5)2 ]
. [3.5 ´ 610
[Q any point on the perpendicular bisector is
[\ l = r 2 + h 2 ]
equidistant from the fixed points]
Similarly, PA = PC = 314
. [2135
. + 12.25] = 314
. ´ 33.6
PA = PB = PC = 105.504 m 2
Þ P is equidistant from three vertices of DABC. (2) Or

Þ The circles with P as centre and its distance from According to the given condition.
any vertex as radius passes through the three vertices Volume of spherical balloon = Volume of cylinder
of DABC and the point P is the circumcentre of the container
4 3
DABC. Hence proved. (1) \ pr = pr 2 h
3
Or 4
Þ ´ (6)3 = (4)2 ´ h
Clearly, ÐOPT = 90° [Qradius is perpendicular to 3
the tangent at the point of contact] 4r ´ 216
Þ h= = 18 m
Now, from DOPT, we have 48
OT 2 = OP 2 + PT 2 [by Pythagoras theorem] Hence, height of the cylindrical container is 18 m.
Þ 2 2
PT = (13) - (5) 2 (iii) A cone is formed by revolving the plane figure of
triangle.
[Q OT=13 cm and OP = 5 cm]
Þ 2
PT = 169 - 25 = 144 37. (i) B
30º
Þ PT = 12 cm (1) 60º
Since, tangents drawn from an external point to a circle
are equal in lengths, therefore we have 1200 m
AP = AE = x cm (say) …(i)
Þ AT = PT - AP = 12 - x …(ii) (1)
Now, as OE is radius and AB is tangent to the circle at E. 30º C 60º
D A
\ OE ^ AB Þ ÐAET = ÐBET = 90°
In right DABC,
Now, in DAET, AT 2 = AE 2 + TE 2 AB
[by Pythagoras theorem] sin 60° =
BC
Þ (12 - x )2 = x 2 + (13 - 5)2 [Q TE = OT - OE] 3 1200
Þ =
[using Eqs. (i) and (ii)] (1) 2 BC
Þ 144 + x 2 - 24x = x 2 + 82 é perpendicular ù
Þ 144 - 24x = 64 êQsinq = hypotenuse ú
ë û
Þ 24x = 80 2400
Þ BC = m
3
 Or (iii) By using internal division formula.
AB æ 1 ´ 11 + 2 ´ 3 1 ´ 5 + 2 ´ 5 ö
In right DABC, tan 60° = Coordinate of P = ç , ÷
AC è 1+ 2 1+ 2 ø
éQ tanq = perpendicular ù æ 11 + 6 5 + 10 ö
êë base úû =ç , ÷
è 3 3 ø
1200 1200 3 æ 17 15 ö æ 17 ö
Þ 3= Þ AC = ´ = ç , ÷ = ç , 5÷
AC 3 3 è 3 3ø è 3 ø
1200
= ´ 3 = 400 3 m Or
3
Let the Jaspal is situated at the point E(x, y ).
AB
In DABD, tan 30° = Then, according to question, AE = BE = DE = CE
AD
1 1200 As shown in figure, the coordinate of A, B, C, D are
Þ = Þ AD = 1200 3 m (3, 5), (7, 9), (11, 5) and (7, 1) respectively.
3 AD
Q AE = BE
\ Distance between two ships
DC = AD - AC \ (x - 3)2 + ( y - 5)2 = (x - 7 )2 + ( y - 9)2
= 1200 3 - 400 3 = 800 3 m Þ x 2 + 9 - 6x + y 2 + 25 - 10 y
(ii) In right DABD, = x 2 + 49 - 14x + y 2 + 81 - 18 y
AB [on squaring both sides]
sin 30° =
BD Þ 34 - 6x - 10 y = 130 - 14x - 18 y
1 1200 Þ 8x + 8 y = 96
Þ =
2 BD Þ x + y = 12 …(i)
Þ BD = 2400 m Similarly, AE = CE
(iii) If the observer moves towards the perpendicular
Þ (x - 3)2 + ( y - 5)2 = (x - 11)2 + ( y - 5)2
line, then angle of elevation increases.
Þ x 2 + 9 - 6x + y 2 + 25 - 10 y = x 2 + 121
38. (i) The coordinates of points B and D are (7, 9) and
(7, 1.) Now, the coordinate of mid-point of BD - 22 x + y 2 + 25 - 10 y
æ 7 + 7 9 + 1ö 34 - 6x - 10 y = 146 - 22 x - 10 y
=ç , ÷ = (7, 5)
è 2 2 ø Þ 16x = 112
(ii) Here, the coordinates of points A and D are (3, 5) Þ x =7
and (7, 1.) Put x = 7 in Eq. (i), we get
Now, distance between points A and D y = 12 - 7 = 5
= (7 - 3)2 + (1 - 5)2 = (4)2 + (- 4)2 Hence, Jaspal should be place at point (7, 5).
= 16 + 16 = 4 2 units