0% found this document useful (0 votes)
92 views52 pagesMath Set 2 Questions
Uploaded by
petpawsawesomeCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
0% found this document useful (0 votes)
92 views52 pagesMath Set 2 Questions
Uploaded by
petpawsawesomeCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 52
Date: 2023/01/23
Class: Standard X
Subject: Competitive Exams Time: 03:00 hrs
Instructions
Time Allowed: 3 Hrs. Maximum Marks : 80
General Instructions:
1. This Question Paper has 5 Sections A-E.
2. Section A has 20 MCQs carrying 1 mark each
3. Section B has 5 questions carrying 02 marks each.
4, Section G has 6 questions carrying 03 marks each
5, Section D has 4 questions carrying 05 marks each.
6. Section E has 3 case based integrated units of assessment (04 marks each) with
subparts of the values of 1,1 and 2 marks each respectively,
7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs
of 3 marks and 2 Questions of 2 marks has been provided, An internal choice has been
provided in the 2 marks questions of Section E.
8, Draw neat figures wherever required. Take 1 = 22/7 wherever required if not stated.
1, The graph of a polynomial p(x) passes through the points (-5, 0), (0.40), 14 gy
(8, 0) and (5, -30). Which among the following is a factor of p(x)?
A) (x- 5)
B) (x-8)
C) (x +30)
D) (x +40)
(1 Mark]
Answer: (B)
[1 mark]
Here, the x intercepts are (-5,0) and (8,0)
These are the zeroes of the polynomial.
So, two of the factors of the given polynomial are (x+5) and (x-8)
=. From the given options, (x-8) is the correct. BYJU'S
2. Shown below is a pair of linear equations
mx + 4y -6 =0
ny-12x+12=0
For which of the following values of m and n do the above equations have
infinitely many solutions?
[1.0]
D) m=6 andn=-2
(1 Mark]
‘Answer: (C)
Writing in standard form,
mx + 4y -6 =0...(i)
ny-12x+12=0
3-12x¢ny+12= 0...)
(1 mark]
On Compairing equations (i) and (i) with standard form of linear equation,
we get,
a, = ma, = —12
b=4b=n
y=
On solving the equations, we will get
m=6andn=-2 BYU's
3. Ateacher asks three students to complete the following statement about 14 o)
the nature of the roots of a quadratic equation.
If q?—4pr > 0, the roots of the quadratic equation pz? + gz +r = 0 will
be.
Zain answers, "always positive”
Vipul answers, "positive, if p, q, and r are positive"
Suman answers, "negative, if p, q, and rare positive".
Who answered correctly?
A) Zain
B) Vipul
C) Suman
D) None of them
[1 Mark]
Answer: (C)
[1 Mark]
pe? + qr +r=0
The discriminant of the quadratic equation is q?—4pr
Thus, D > 0 and roots are real and unequal.
Let the roots be @ and 8,
Sum of roots will be,
a+f=—4...(i)
Product of roots will be,
a = 4... (ii)
Now, if p, q, and r are positive, a + 8 will be negative, so one or both the
roots must be negative.
Similarly, «8 will be positive if p, q, and r are positive = both roots must be
negative.
Hence, we find that Suman's assertion is correct. BYJU'S
4. Two concentric circles are centered at O(-4, 3). The ratio of the area of
inner circle to that of the outer circle is 1:9. Points A and B lie on the
boundaries of the inner and outer circle, respectively, as shown below.
[1.0]
‘Answer: (B)
[1 Mark]
Given, The ratio of the area of inner circle to that of the outer circle is 1:9.
Hence, The ratio of the redius of inner circle to that of the outer circle is 1:3.
OA: OB
OA: AB = 1:
From section formula,
Ax, y)= (SE
Aw y)= (SC 8
Agcy) = ($52) 5. Sonaliis standing on one side of a 7 m wide road as shown below. She 14
wants to estimate the distance (D) between two light poles on the other
side without crossing the road.
Pole Pole 2
foot f4_
soni (@>
Which of the following expressions represent D in terms of p andr?
A) Em
Answer: (D)
(1 Mark
Distance between the two light poles = D
As the line joining the two poles is parallel to the road, the small triangle and
the big triangle will be similar to each other.
Thus, by using the property of correspondence of sides of similar triangles,
BYJU'S
0] BYU's
6. Shown below are two triangles such that length of two sides of each is.
1.0)
known. nol
6 unis
3 units
Kans L e Buns Q
Along with the given information, which of these is sufficient to conclude
whether AKLM is similar to APQR?
() ZKLM = 2PQR
(ii) Ratio of KM:PR
A) only (i)
B) only (i)
C) either (i) or (i)
D) the given information is enough
[1 Mark]
Answer: (C)
[1 Mark]
Using (i) along with given information, both triangles can be shown similar
by SAS criterion.
Using (i) along with given information, both triangles can be shown similar
by SSS criterion.
Hence, either (i) or (ji) is sufficient along with given information. BYJU'S
7. Shown below is a sector of a circle with centre P. All lengths are
[1.0]
measured in cm.
¥
Whats the length of PE?
A) 3m
Answer: (D)
[1 Mark]
As PR and PQ are the radius of the circle. So, from the figure,
PR= PQ =6cm,PA=3cm, and PB=4em.
In APAE and APQB
ZAPE = ZBPQ( Common angle)
ZPAE = ZPBQ = 90°
So, APAE ~ APQB by AA criterion, and we know that the corresponding
sides of similar triangles are in proportion. Therefore,
PE PA. PE 3 pg 8 8
PQ” PBS 6 i= 4
Hence, 4.5 cm is the required answer.
=4.50n weyu's
8. Acircle is drawn. Two points are marked outside the circle such that only 14 o)
3 tangents can be drawn to the circle using these two points.
Which of the following is true based on the above information?
A) All 3 tangents are equal in length.
B) Both the points lie on one of the tangents.
C) The tangents and the circle have two common points in total.
D) Such a situation is not possible as with 2 points, there will be 4
tangents to the circle
(1 Mark]
Answer: (B)
1 mark]
Copyright © Thnk and Lear Pt. Ld BYU's
9. Shown below is a circle with 3 tangents KQ, KP and LM. QL =2em and 14 o)
KL =6 om. PM = 2KL.
What is the measure of zLMK?
A) 50°
B) 65°
C) 80°
D) cannot be uniquely determined with the given information.
[1 Mark
Answer: (B)
[1 Mark]
We know that the tangents are drawn from exterior point of a circle are
always equal in length
Therefore, KQ = KP, LQ = LN, PM = PN.
KQ = KP shows that triangle KLM is an isoceles triangle such that Side KL &
KM are equal
So in triangle angle KLM = angle LMK
Therefore, Angle LMK + angle KLM tangle KLM = 180°
(sum of all angles of a triangle is 180°)
Angle LMK + angle LMK + 50° = 180°
2 Angle LMK = 180° - 50° = 130°
‘Angle LMK = 65° BYJU'S
10. In the figure shown below, lines AB and PQ are parallel to each other. All 14 oy
Measurements are in centimetres.
oa eB xa
Which of the following gives the value of cos 6?
Answer: (B)
[1 mark]
Itis give that lines PQ and AB are parallel to each other
= ZOPQ = ZOAB => ZOAB = 8
we know, Cos @ = base/hypotenuse
In triangle OAB
cos 6 = BA/OA
itis given that BA=c and OA=b
cos 6 = clb BYJU'S
11. The sine of an angle in a right triangle is 4. Which of these could be the
measures of the sides of the triangle?
A) 4m, 5 cm and 9 cm
B) 4 cm, 5 cm and V4 cm
C) 6 cm, 8 om and 10 cm
D) 8 cm, 10 cm and 4\/41 cm
[1.0]
[1 mark]
Answer: (C)
[1 Mark]
Given, Sin A= $= Sazendieuor
5~ “Typotemese
Let us assume that, Perpendicular is 4x and Hypotenuese is 5x.
Then using pythyagores theorem,
(42)? + Base? = (52)?
162 + Base” = 252”
Base” = 9x”
Base* = 32.
Therefore, Base : Perpendicular : Hypotenuese = 3:4:5
Hence, option (C) 6 cm, 8 cm and10 cm is correct BYU's
12. The marks obtained by a set of students in an exam are recorded in a
grouped frequency table. The maximum number of students are found to
be in the range of (70-80) marks.
Ifthe number of students in the ranges before and after the (70-80) range
are equal, which of the following is the mode of the data?
A) 70 marks
B) 75 marks
C) 80 marks
D) mode cannot be found as frequency is not given,
(1 Mark]
Answer: (B)
(1 Mark]
Modal class: (70 - 80)
L=70,h=10,fi=2, fo=fr=y
Mode xh
[1.0] BYU's
13. In the figure below, a unit square ROST is inscribed in a circular sector
with centre O,
[1.0]
|
ary °
Along with the above information, which of these is SUFFICIENT to find
the area of sector POQ?
A) area of the square ROST
B) radius of sector POQ
C) are length PQ
D) the given information is sufficient
[1 Mark]
Answer: (C)
[1 Mark
Here, ROST is unit square.
So, the side of a square is 1 unit.
The diagonal of a square = \/2xside = \/2 units
Thus, the radius of the sector is \/2 units
To find the area of sector, we should know central angle.
To find the central angle, we need the length of the arc PQ.
Hence, the length of arc PQ is required. 14. The radius of a wheel is 0.25 m. How many revolutions will it make in
covering 11 km ?
A) 2800
B) 4000
c) 5500
D) 7000
(1 Mark]
Answer : (D)
[1 mark]
Radius =0.25 m
Circumference = 2rir = 2x 220.25 m
total distance
No of revolutions =
= ftom
= liawxt_=7000 revolutions
[1.0] BYU's
15. Shown below is a solid made by joining a right circular cylinder anda 14 oy
hemisphere of equal radius (r cm). The total surface area of the solid is
equal to the surface area of a sphere with twice the radius of this solid.
a>
Which of the following gives the height of the cylinder in the above solid?
A) 6rem
B) 6.5r cm
C)7rem
D)7.5rcm
(1 Mark]
‘Answer: (B)
(1 Mark]
Total surface area of the solid = Curved Surface area of the cylinder + Total
surface area of the hemisphere
= Unrh + 3nr?
Given that,
Total surface area of the solid = Surface area of the sphere with twice the
radius of this solid
Thus, 2nrh 4 3mr? = An (2r)?
2arh + 3nr? = 16xr?
13nr?
h = (13/2)r = 6.57 BYU's
16. If ABC is right angled at C, then the value of cos(A+B) is
Ajo
B)1
cyt
p=
(1 Mark]
[1.0]
The answer is (A).
(1 mark}
We know that, in ABC, sum of three angles = 180°
ie, 2A+ZB+ 20 = 180°
But right angled at Ci..e, 20 = 90° [given]
2A+ 2B+ ZC = 180°
= A+B=90°
cos(A + B) = cos90° =0 BYJU'S
17. For the following distribution, the sum of lower limits of the median class
and modal class is
Class | 0-5 [5—10[10~15 | 15-20 20-25
[1.0]
Frequency | 10 15 12 20 9
A) 15
B) 25
c)30
D) 35
(1 Mark]
Answer: (B)
[1 mark]
Here,
Class | Frequency | Cumulative
frequency
0-5 10 10
5-10 15
10-15 12
15— 20 20
20 — 25 9
Ts
Now, = “= 33, which lies in the interval 10-15. Therefore, lower limit of
the median class is 10,
The highest frequency is 20, which lies in the interval 15-20. Therefore,
lower limit of modal class is 15.
Hence, sum of lower limits is 10+15=25. BYU's
18. The probability that a non-leap your selected at random will contains 53
1.0)
Sunday is mel
(1 Mark]
Answer: (A)
(1 Mark]
Anon-leap year has 365 days and therefore 52 weeks and 1 day. This 1 day
may be Sunday of Monday or Tuesday or Wednesday or Thursday or Friday
or Saturday. Thus, out of 7 possiblities. 1 favourable is the event that the
one day is Sunday
Required probability = 2
19. Assertion (A): 2 is a prime number.
Reason (R): The square of an irrational number is always a prime
number.
[1.9]
a) Both (A) and (R) are true and (R) is the correct explanation of (A).
b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
©) (A) is true but (R) is false.
d) (A) is false but (R) is true.
(1 Mark}
‘Answer: (C)
[1 mark
Itis given that 2 is a prime number.
We know that, prime numbers are numbers that have only two
factors, 1 and the number itself.
Since 2 has only two factors, so, the assertion is true,
Given reason statement is The square of an irrational number is always a
prime number.
Let us consider an irrational number /15, the square of 15 gives 15
Factors of 151, 2, 5, 15
So, we can say that 15 is a composite number.
So, the reason statement is false
Hence, the assertion is true and the reason is false. 20.
21.
Assertion (A): The origin is the ONLY point equidistant from (2, 3) and (-2,
-3)
Reason (R): The origin is the midpoint of the line joining (2, 3) and (-2,
-3)
*[1.0]
a) Both (A) and (R) are true and (R) is the correct explanation for (A)
b) Both (A) and (R) are true but (R) is not the correct explanation for (A).
c) (A) is true but (R) is false.
d) (A) is false but (R) is true.
(1 Mark]
Answer: (d)
(1 Mark]
Check whether the three lines represented by the equations given below
[2.0]
intersect at a common point.
2x+y-1=0
4x+3y+5=0
5x +4y+8=0
[2 Marks]
Given equations are
Let us find the point of intersection of equation (i) and (i) and check if it is
satisfying equation (iii) or not.
On solving equation (i) and (ii) values of
x=4, y=-7
(1 Mark]
Putting these values in equation (ii)
LHS = 5(4)+4(-7)+8
= 20-2848
=0=RHS
It satisfy eq (ii), hence intersect at a common point which is (4,-7)
(1 Mark] 22. In the given figure, AB and CD are common tangents to two circles of
unequal radii. Find the ratio of AB and CD.
A
B
[2 Marks]
D
c
On producing AB and CD such that they intersect at P.
A
t
PA=PC
[The length of tangents drawn from an internal point to a circle are equal]
Also, PB=PD.
[ The lengths of tangents drawn from an intemal point to a circle are equal]
[1 mark]
= PA-PB=PC-PD
= AB=CD
= ABICD = 1
[1 Mark]
23. Show that 12" cannot end with the digit 0 or 5 for any natural number n.
[2 Marks]
If any number ends with the digit 0 or 5, it is always divisible by 5.
f 12” ends with the digit zero or five, it must be divisible by 5.
This is possible only if prime factorization of 12 contains the prime number
5.
[1 Mark]
Now, 12=2x2x3=2x3
> 12 = (2x 2x 3)" =x Mx 3
12" is not divisible by 5 as 5 is not in the prime factorisation of 12”.
Hence, there is no value of n« NV for which 12" ends with digit zero or five.
[1 Mark]
BYJU'S
[2.0]
[2.0] BYJU'S
24. If cos (A+ 2B) = 0, 0° s (A+ 2B) $ 90° and cos (B - A) =
90°, then find cosec (2A +B).
[2 Marks]
0° (B-A)S 12.9]
oR
If 2 sin? @ — cos® @ = 2, then find the value of 6.
(Note: 0° < 8 90°)
[2 Marks]
Given that, cos (A + 2B) = 0
Thus, A+ 2B = 90° ...(i) (cos 90° = 0)
{0.5 mark]
Also, cos (B -A) =
Thus, B-A= 30° __....(i) (cos 30° = “4
{0.5 mark]
Subtracting (ii) from (i), we get
2A +B = 60°.
{0.5 mark]
Thus, cosec (2A + B) = cosec 60° = +,
we
{0.5 mark]
“lS
OR
Given, 2 sin? @— cos? @
= 2 sin? 9 — (1 — sin? 8) =
[0.5 Mark]
= 2 sin? 9+ sin? 9-1
{0.5 Mark]
=> 3 sin? = 3
> sin? @
=> sin =
[0.5 Mark]
>0= 90"
(0.5 Mark]
2
[- sin? 0-4 cos? @ = 1]
= sin 90° BYU's
25. A3.5 om chord subtends an angle of 60° atthe centre of a circle. What iS pp gy
the arc length of the minor sector?
[2 marks]
OR
Asemicircle MON is inscribed in another semicircle. Radius OL of the
larger semicircle is 6 om.
4 N
K 0 L
Find the area of the shaded segment in terms of x.
[2 Marks] BYJU'S
Ls
f
Since OM = ON (radii)
Therefore, ZOMN = ZONM = 60° (angles opposite equal sides) Hence,
AMON is an equilateral triangle.
{0.5 mark]
ie. OM = MN =3.5 em
{0.5 mark]
Now, Arc length of the minor sector
= 2x BT
wr TF
2
a
= Bom,
(1 Mark]
OR
On Joining O with M and N,
ZMON = 90° (Angle in a semicircle)
MO = NO = 6 cm (radii)
{0.5 mark]
‘Area of sector MON = (90°/360*) x xx (6/2) = 9 em?
{0.5 mark]
‘Area of AMON = (1/2) x 6 x 6 = 18 cm?
{0.5 mark]
Area of the shaded segment
= Area of sector MON - Area of AMON
= 9-18 cm?
= 9m - 2) em®
{0.5 mark] BYU's
26. The LCM of 6, 8* and kis 124 where kis a positive integer. Find the 4g gy
smallest value of k.
[3 Marks]
Given that the LCM of 6, 8?, and kis 124
124 = 28 x 34 (Prime factorization)
6* = 2* x 34 (Prime factorization)
8” — 2° (Prime Factorization)
(1 Mark}
On comparing Prime factorization of 6*, 8”, and 12 for finding smallest
value of k,
LCM should be 2° x 34 x &
[1 Mark]
which is 2° x 34 x k = 12* (Given)
at
k= wat 4
The smallest possible value of k is a positive integer, so that the smallest
value of kis 4
(1 Mark] BYJU'S
27. If mand nare zeroes of the polynomial 3x? — x — 2, find the values of the [3.0]
following without factorising the polynomial.
Wate
(ii) m? + n®
[3 Marks]
For polynomial 32? — x — 2 whose zeroes are m and n,
Thus, a= 3,b=-1 andc=-2
WF+t- EO
[0.5 mark]
Now,m+n
and mn = £
{0.5 mark]
Substituting the values of m +n and mn in (i), we get
z
ay
brie
[0.5 mark]
(ii) We know that, (m +n)? = m? +n? + 2mn_
oom? +n? = (m+n)? —2mn
[0.5 mark]
‘Substituting the values of m+ n and mn, we get
m? +n? = (3? (5)
[0.5 mark]
m4?
mn? =
m+n?
{0.5 mark] BYJU'S
28. The graph of a line represented by the equation ax + y + 8 = 0 is shown inrs gy
the figure below.
(i) Find the value of a.
(ii) Find the point of intersection of this line with the line represented by
the equation 4x - 3y - 14 = 0.
[3 Marks]
OR
Anuj and Safina started a new game zone consisting of two games -
shooting and bowling. They released the following rate card for the
customers:
Pack Shooting [Bowling | Price (inclusive taxes)
Solo [ - T round Rs 60,
Sola 2 Tround = Rs 75,
Combo 1 | 3 rounds | 2 rounds Rs 285
‘Combo 2 | 4 rounds | _5 rounds Rs 485
The price of shooting is the same in both the combos and the price of
bowling is the same in both the combos.
How much more is the price for one round of bowling in the solo pack
than in the combo packs?
[3 Marks] BYJU'S
(i) From the given graph,
Aty =0, line cuts x- axis at 4.
Hence, (4,0) lies on the line
Substituting x= 4 and y =0 in the given equation, we get
a(4)+0+8=0
4a+8=0
4
a=2
[1 mark]
(ii) Thus, we get the equation as -2x + y + 8= 0.
We have:
— 22 +y+8=0...(i)
4a: — By 14 = 0... (ii)
Therefore, solving the equations, we have:
Equation i x 2+ it,
—de +2y+16=0
= de 3y-14=
-y+2=0
Sy=2
(1 Mark]
Substituting the value of y in equation (i), we have:
=> -2n+24+8=0
>2=5
(1 Mark}
The point (x, y) = (5,2)
Hence, the point of intersection of both the lines is (5,2)
OR
Assuming the prices of one round of shooting and bowling in the combo
packs to be x and y respectively.
|_____—sThe pair of linear equations are framed as:
Copy Thnk and LaRue Dy = 285 ...(i) BYJU'S
4x + Sy = 485 ...(i)
[1 mark]
Multiplying the (i) by 5 and (ii) -2 to eliminate y:
15x + 1Oy = 1425
~8x - 10y = -970
Adding the two equations together:
7x = 455
x=65
Substitute x = 65 into (i:
3(65)+2y=285
195+2y=285,
2y=90
yas
Price of bowling in combo pack = Rs 45.
[1.5 marks}
Price of one round of bowling in the solo pack = Rs 60 (given)
Hence, Price of one round of bowling in the solo pack is 60 - 45 = Rs 15
more than that of the combo pack.
{0.5 mark] 29. Shown below is a circle and two congruent squares PQRS and QTUR.
ST, SU and UT are tangents to the circle. The side length of the square is
10cm.
s R wo
[\
8
Y
P 7 +
Find the radius of the circle.
[3 Marks]
OR
In the figure below, M and N are the centres of two semi-circles having
radii 9 cm and 16 cm respectively. ST is a common tangent.
7
Le
a
P
8
M N
Find the length of PQ.
[3 Marks]
BYJU'S
0] BYJU'S
Assuming the radius of the circle as x cm
Since NVUW is a square, WU = UV = x cm
SU = SR + UR=10+10= 20cm
(squares PORS and QTUR are congruent)
[0.5 mark]
Using the Pythagoras theorem in ASUT,
ST? = SU? +UT?
10? + 10”
ST = y100 F100 = 10/5 em
{0.5 mark]
Now, VT = UT-UV =10-xcm
As length of tangents drawn from an external point on the circle
Therefore, YT = VT
YT=10-x0m
[0.5 mark]
Now, S’ oT - YT = 105 — (10
and SW = SU - UW = 20 ~-... (i)
As length of tangents drawn from an external point on the circle
Therefore, SW = SY
[1 mark]
From (i) and (ii),
20—a=10y5 ~ (10~2)
z= 15—5y5em
[0.5 Mark]
OR
In the figure, Join MP, NQ and draw MR parallel to PQ.
M N
ZMPQ = ZPQR = 90° (At the point of tangency, the tangent is
perpendicular to the radius.)
Hence, PQRM is a rectengle.
PM = 9 cm (radius)
PM = QR = 9 cm (opposite sides of rectangle)
[1 Mark]
Now,
RN =QN-QR=6-9=7cm
and MN = sum of radii = 9 + 16 = 25 cm
[1 Mark]
In triangle MRN , By Pythagoras theorem,
MN? = MR? + RN?
25° = MR+7°
MR= 28-7
MR = 24 cm
Now since PQRM is a rectangle,
PQ = MR = 24cm
Copyright Tn ana Leas MEER] BYJU'S
30. Prove that: /sec? @ + cosec? 6 = tan @ + cot 0.
[3 Marks]
LHS = sec? 0+ cosec? 6
/:
Vineet
Bo]
me
nd)
[- see@ = Agand cosec 8 =
[0.5 mark]
in? oo
Af int Beost eV ink acont
[. sin? @ + cos @ = 1]
[1 mark]
sin? onto
= Fincor8 * inde
~ 1 = sin? 0+ cos? 6)
{1 mark]
= SFand cot 9 = 4)
tan @+ cot 0 = RHS
{0.5 mark] BYJU'S
31. Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3, respectively. They arep gy
thrown and the sum of the numbers on them is noted. Find the probability
of getting sum from 4 to 9, separately.
[3 Marks]
Number of total outcomes = n(S) = 36
Event of getting sum 4 = {(2,2), (2,2), (3,1), (3.1), (1.3), (1,3)}
{0.5 mark]
(ii) Let By
n(Ex)
©. P(Es) =
[0.5 mark]
(iii) Let 2; = Event of getting sum 6 = {(3,3), (3,3), (4,2), (4,2), (5,1), (5.1)}
n(Es)
PC
[0.5 mark]
(iv) Let Es = Event of getting sum 7 = {(4,3), (4,3), (5,2), (5,2), (6,1), (6,1)}
oo n(Es)
+, P(E) = “= $3
[0.5 mark]
(v) Let E; = Event of getting sum 8 = {(5,3), (5,3), (6,2), (6,2)}