ICSE Class 10 Mathematics Competency Focused Practice Questions Solution-Multiple Choice Questions

ICSE Class 10 Mathematics Competency Focused Practice Questions aims at developing critical and analytical thinking of students. The students should have a thorough understanding of the topics before attempting competency based questions.

Solution of ICSE Class 10 Mathematics Competency Focused Practice Questions are divided into different categories. This post contains Multiple choice Questions released by CISCE.

ICSE Class 10 Mathematics Competency Focused Practice Questions Solution-Multiple Choice Questions

Question 1
A retailer buys an article at its listed price from a wholesaler and sells it to a consumer in the same state after marking up the price by 20%. The list price of the article is ₹2500, and the rate of GST is 12%. What is the tax liability of the retailer to the central government?
a. ₹0
b. ₹15
c. ₹30
d. ₹60
Answer: c. ₹30
Explanation:

For retailer,
C.P. = ₹2500
G.S.T. = 12%
C.G.S.T. = 12/2 = 6%.
C.G.S.T. paid = 6/100×2500 = ₹150.
Given,
Retailer sells the article to the consumer in the same state after marking up the price by 20%.
S.P. = ₹2500 + 20% of 2500
= ₹2500 + 20/100×2500
= ₹ (2500 + 500)
= ₹3000.
C.G.S.T. = 6%
C.G.S.T. charged = 6/100×3000= ₹180.
Tax liability = C.G.S.T. charged – C.G.S.T. given
= ₹180 – ₹150
= ₹30.

Question 2
Dev brought an electrical fan which has a marked price of ₹800. If the GST on the goods is 7%, then the SGST is :
a. ₹ 24
b. ₹ 28
c. ₹ 56
d. ₹ 80
Answer: b. ₹ 28

Explanation:
Given, rate of GST rate = 7%, therefore, SGST = 7/2 = 3.5%.
M.P. of electrical fan = ₹800.
SGST = 3.5% of ₹800
= 3.5/100×800
= ₹28.

Question 3
₹P is deposited for n number of months in a recurring deposit account which pays interest at the rate of r% per annum. The nature and time of interest calculated is :
a. compound interest for n number of months
b. simple interest for n number of months
c. compound interest for one month
d. simple interest for one month
Answer: d. simple interest for one month
In a recurring deposit account,
By formula,

\[Interest=P\times\frac{n(n+1)}{2\times12}\times\frac r{100}\]

∴ The nature and time of interest calculated is simple interest for one month.

Question 4
Anwesha intended to open a Recurring Deposit account of ₹1000 per month for 1 year in a Bank, paying a 5% per annum rate of simple interest. The bank reduced the rate to 4% per annum. How much must Anwesha deposit monthly for 1 year so that her interest remains the same?
a. ₹12325
b. ₹1250
c. ₹1200
d. ₹1000
Answer: b. ₹1250
Explanation:

First case:
P = ₹1000
r = 5%
n = 12 months

\[Interest=P\times\frac{n(n+1)}{2\times12}\times\frac r{100}\\=1000\times\frac{12\times13}{2\times12}\times\frac5{100}\\=₹325\]

Second case :
P = ₹P
r = 4%
n = 12 months
Interest = ₹ 325

\[325=P\times\frac{12\times13}{2\times12}\times\frac4{100}\\P=\frac{325\times2\times12\times100}{12\times13\times4}\\\\P=₹1250\]

Question 5
Mr. Das invests in ₹100, 12% shares of Company A available at ₹60 each. Mr. Singh invests in ₹50, 16% shares of Company B available at ₹40 each. Use this information to state which of the following statements is true.
a. The rate of return for Mr. Das is 12%
b. The rate of return for Mr. Singh is 10%
c. Both Mr. Das and Mr. Singh have the same rate of return of 10%
d. Both Mr. Das and Mr. Singh have the same rate of return of 20%
Answer: d. Both Mr. Das and Mr. Singh have the same rate of return of 20%
Explanation:
For Mr. Das,
N.V. of each share = ₹ 100
M.V. of each share = ₹ 60
Dividend per share = 12% of ₹100 = ₹12.
Rate of return = Dividend per share/M.V. of each share×100=12/60×100= 20%.
For Mr. Singh,
N.V. of each share = ₹ 50
M.V. of each share = ₹ 40
Dividend per share = 16% of ₹ 50 = ₹ 8.
Rate of return = Dividend per share/M.V. of each share×100=8/40×100= 20%.
∴ Both Mr. Das and Mr. Singh have the same rate of return of 20%.

Question 6
Amit invested a certain sum of money in ₹100 shares, paying a 7.5% dividend. The rate of return on his investment is 10%. The money invested by Amit to purchase 10 shares is :
a. ₹250
b. ₹750
c. ₹900
d. ₹1100
Answer: b. ₹750
Explanation:

Let ₹ P be the price per share.
Number of shares bought by Amit =10, so total investment = ₹10P
Dividend = 7.5%
Dividend per share = 7.5/100 x 100 = ₹ 7.5
Total dividend = ₹7.5 × 10 = ₹75.
Rate of return = 10%
By formula,
Rate of return = Dividend/Investment
⇒10=75/10P x100

⇒100P =7500

⇒P=₹75.
Total investment = 10P = 10 × ₹75 = ₹750.

Question 7
If -3 ≤ -4x + 5 and x ∈ W, then the solution set is :
a. {…….-3, -2, -1, 0, 1, 2, 3, …….}
b. {1, 2}
c. {0, 1, 2}
d. {2, 3, 4, 5}
Answer: c. {0, 1, 2}
Explanation:

-3 ≤ -4x + 5
⇒ 4x – 5≤ 3
⇒ 4x ≤ 3 + 5
⇒ 4x ≤ 8
⇒ x ≤ 8/4
⇒ x ≤ 2

Since, x ≤ 2 and x ∈ W.
∴ Solution set = {0, 1, 2}

Question 8
If -4x > 8y; then
a. x > 2y
b. x > -2y
c. x < -2y

d. x < 2y

Answer: c. x < -2y

Explanation:

⇒ -4x > 8y
⇒ 4x < -8y
⇒ x < −8y/4
⇒ x < -2y.

Question 9
The value/s of ‘k’ for which the quadratic equation 2×2 – kx + k = 0 has equal roots is (are) :
a. 0 only
b. 4, 0
c. 8 only
d. 0, 8
Answer
Given,
Equation : 2×2 – kx + k = 0
a = 2, b = -k and c = k.
For equal roots,
⇒ Discriminant (D) = 0
⇒ b2 – 4ac = 0
⇒ (-k)2 – 4 × 2 × k = 0
⇒ k2 – 8k = 0
⇒ k(k – 8) = 0
⇒ k = 0 or k – 8 = 0
⇒ k = 0 or k = 8.
Hence, Option 4 is the correct option.

Question 10
If x = -2 is one of the solutions of the quadratic equation x² + 3a – x = 0, then the value of ‘a’ is :
a. -8
b. -2
c. −1/3
d. 1/3
Answer: b. -2
Explanation:

Given,
x = -2 is one of the solutions of the quadratic equation x² + 3a – x = 0.
∴ (-2)2 + 3a – (-2) = 0
⇒ 4 + 3a + 2 = 0
⇒ 3a + 6 = 0
⇒ 3a = -6
⇒ a = −6/3 = -2.

Question 11
In solving a quadratic equation, one of the values of the variables x is 233.356. The solution rounded to two significant figures is :
a. 233.36
b. 233.35
c. 233.3
d. 230
Answer: d. 230
x = 233.356
On rounding off to two significant figures
x = 230.

Question 12

In the adjoining diagram, AB = x cm, BC = y cm and x – y = 7 cm. Area of △ ABC = 30 cm². The length of AC is :

a. 10 cm

b. 12 cm

c. 13 cm

d. 15 cm

icse-class-10-mathematics-competency-focused-practice-questions-solution

Answer: c. 13 cm

Explanation:

By formula,

Area of triangle = 1/2× base×height

⇒ 30 = 1/2×BC×AB

⇒ 30 = 1/2×y×x

⇒ xy = 60 ……..(1)

Given,

⇒ x – y = 7

⇒ x = 7 + y ……..(2)

Substituting value of x from equation (2) in (1), we get :

⇒ (7 + y)y = 60

⇒ 7y + y² = 60

⇒ y² + 7y – 60 = 0

⇒ y² + 12y – 5y – 60 = 0

⇒ y(y + 12) – 5(y + 12) = 0

⇒ (y – 5)(y + 12) = 0

⇒ y – 5 = 0 or y + 12 = 0

⇒ y = 5 or y = -12.

Since, side cannot be negative,

∴ y = 5 cm.

Substituting value of y in equation (2), we get :

⇒ x = 7 + y = 7 + 5 = 12 cm.

In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = x² + y²

⇒ AC² = 12² + 5²

⇒ AC²= 144 + 25

⇒ AC² = 169

⇒ AC = √169 = 13 cm.

Question 13

If p, q, and r are in continued proportion, then :

a. p : q = p : r

b. q : r = p² : q²

c. p : q² = r : p²

d. p : r = p² : q²

Answer: d. p : r = p² : q²

Explanation:

Since, p, q, and r are in continued proportion.

∴ pq=qr

⇒ q²=pr ……..(1)

Putting the value of q² in d. , we get,

R.H.S.⇒ p² : q²

p² : pr

p : r = L.H.S.

Question 14

The ratio of diameter to height of a Borosil cylindrical glass is 3 : 5. If the actual diameter of the glass is 6 cm, then the curved surface area of the glass is :

a. 120π

b. 60π

c. 30π

d. 18π

Answer: b. 60π

Explanation:

Given, diameter : height = 3 : 5.

⇒ d/h=3/5

⇒ 6/h=3/5

⇒ h= 30/3

h=10 cm.

Radius = Diameter/2 = 6/2= 3 cm.

By formula,

Curved surface area of glass = 2πrh

= 2π × 3 × 10

= 60π cm².

Question 15

If the polynomial 2x³ + 3x² – 2x – 3 is completely divisible by (2x + a), and the quotient is equal to (x² – 1), then one of the values of a is :

a. -3

b. -1

c. 1

d. 3

Answer: d. 3

Explanation:

Given, The polynomial 2x³ + 3x² – 2x – 3 is completely divisible by (2x + a) and quotient is equal to (x² – 1).

As, f(x) = q(x) d(x) + r(x)

∴ 2x³ + 3x² – 2x – 3 = (2x + a)(x2 – 1) + 0

⇒ 2x³ + 3x² – 2x – 3 = 2x³ – 2x + ax² – a

⇒ 2x³ – 2x³ + 3x² – 2x + 2x – 3 = ax² – a

⇒ 3x² – 3 = ax² – a

From above equation,

a = 3.

Question 16

A polynomial in x is x³ + 5x² – kx – 24. Which of the following is a factor of the given polynomial so that the value of k is 2?

a. (x + 2)

b. (x – 3)

c. (x + 4)

d. (x – 4)

Answer: c. (x + 4)

Explanation:

By factor theorem,

(x – a) is a factor of f(x), if f(a) = 0.

Given,

Polynomial = x³ + 5x² – kx – 24

If k = 2, then Polynomial = x³ + 5x² – 2x – 24.

Dividing polynomial by (x + 4) or substituting -4 in polynomial, we get,

⇒ (-4)³ + 5(-4)² – 2(-4) – 24

⇒ -64 + 5(16) + 8 – 24

⇒ -64 + 80 + 8 – 24

⇒ 88 – 88

⇒ 0.

Since, remainder = 0.

∴ (x + 4) is the factor of x³ + 5x² – kx – 24, when k = 2.

Question 17

\[A=\begin{bmatrix}a&b\end{bmatrix}\;and\;B=\begin{bmatrix}c\\d\end{bmatrix}\;then,\]

a. only matrix AB is possible

b. only matrix BA is possible

c. both matrices AB and BA are possible

d. both matrices AB and BA are possible, AB = BA

Answer: c. both matrices AB and BA are possible

Explanation:

Order of matrix A = 1 × 2

Order of matrix B = 2 × 1

Since, no. of columns in A = no. of rows in B and no. of columns in B = no. of rows in A.

∴ AB and BA are possible.

\[AB=\begin{bmatrix}a&b\end{bmatrix}\;and\;B=\begin{bmatrix}c\\d\end{bmatrix}\;then,\\AB=\left[a\times c\;+b\times d\right]=\left[ac+bd\right]\\BA=\begin{bmatrix}c\times a&c\times b\\d\times a&d\times b\end{bmatrix}\\=\begin{bmatrix}ac&bc\\ad&bd\end{bmatrix}\\\therefore\;AB\;\neq\;BA.\]

Question 18

\[Matrix\;A=\begin{bmatrix}6&9\\-4&k\end{bmatrix}\;such\;that\;A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix},\;then\;k\;is:\]

a. 6

b. -6

c. 36

d. ±6

Answer: b. -6

Explanation:

\[Given,\;A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}\\\therefore\begin{bmatrix}6&9\\-4&k\end{bmatrix}\begin{bmatrix}6&9\\-4&k\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\\\Rightarrow\begin{bmatrix}6\times6\;+9\times(-4)&6\times9+9\times k\\(-4)\times6+k\times(-4)&(-4)\times9+k\times k\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\\\Rightarrow\begin{bmatrix}36-36&54+9k\\-24-4k&-36+k^2\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\\\Rightarrow54+9k=0\\\Rightarrow9k=-54\\\Rightarrow k=-6\]

Question 19

If the sum of n terms of an arithmetic progression S_n = n² – n, then the third term of the series is :

a. 2

b. 4

c. 6

d. 9

Answer: b. 4

Explanation:

Given, S_n = n² – n

S₁ = 1² – 1 = 0,

S₂ = 2² – 2 = 4 – 2 = 2,

S₃= 3² – 3 = 9 – 3 = 6.

S₁= First term = 0.

S₂ = 2 and first term = 0, second term = 2.

S₃ = 6

∴ First term + Second term + Third term = 6

⇒ 0 + 2 + Third term = 6

⇒ Third term = 6 – 2 = 4.

Question 20

Which of the following is not a geometric progression ?

\[a.\;\frac13,1,3,9\\b.\;\frac15,\frac15,\frac15,\frac15\\c.\;-\;2,4,-8,16\\d.\;2,0,4,0,8,0\]

Answer: d. 2, 0, 4, 0, 8, 0

Explanation:

\[In\;the\;series\;2,0,4,0,8,0\\\frac02\neq\frac40\\\therefore\;2,0,4,0,8,0\;is\;not\;a\;G.P.\]

Question 21

In the adjoining diagram, G is the centroid of △ ABC. A(3, -3), B(2, -6), C(x, y) and G(5, -5). The coordinates of point D are :

a. (2, -6)

b. (3, -6)

c. (6, -6)

d. (10, -6)

Answer: c. (6, -6)

Explanation:

Centroid of triangle = (x₁+x₂+x₃/3 , y₁+y₂+y₃/3)

∴(5,−5)= (x₁+x₂+x₃/3 , y₁+y₂+y₃/3))

⇒(5,−5)= (3 + 2 + x /3 , (-3)+(-6) + y /3)

⇒ x+5 /3= 5 and y−9 /3= −5

⇒ x=15 – 5 and y=−15 + 9

⇒x= 10 and y= −6.

C(x, y) = (10, -6).

Since, centroid is the point of intersection of all the three medians of a triangle.

∴ AD is the median.

∴ D is mid-point of BC.

D=(2+10 /2, (−6)+(−6) /2)

D= (12/2,−12/2)

D= (6,−6).

Question 22

In the given diagram, O is the origin and P is the mid-point of AB. The equation of OP is :

a. y = x

b. 2y = x

c. y = 2x

d. y = -x

Answer: b. 2y = x

Explanation:

From graph, A = (4, 0) and B = (0, 2).

Given, P is the mid-point of AB.

∴ P = (4+0 /2 , 0+2/2)=(4/2 , 2/2) = (2, 1).

\[origin\;O=\;(0,0)\;and\;P=\;(2,1)\\By\;two-point\;formula,\\y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\\y-0=\frac{1-0}{2-0}\left(x-0\right)\\y=\frac12x\\2y=x\]

Question 23

In the given figure line l₁ is a parallel to line l₂. If line l₃ is perpendicular to line l₁, then the slopes of lines l₂ and l₃ respectively are :

a. 1, 1

b. -1, -1

c. 1, -1

d. -1, 1

Answer: c. 1, -1

Explanation:

Slope of line l₁ = tan 45° = 1.

We know that,

Slope of parallel lines are equal.

Slope of line l₂(m₂)= Slope of line l₁(m₁)= 1.

We know that,

Product of slope of perpendicular lines = -1. [ m₂ x m₃ = -1]

⇒ Slope of line l₂ × Slope of line l₃ = -1

⇒ 1 × m₃ = -1

⇒ Slope of line l3 = -1.

Question 24

Which of the following lines cut the positive x-axis and positive y-axis at equal distances from the origin?

a. 3x + 3y = 6

b. 5x + 10y = 10

c. -x + y = 1

d. 10x + 5y = 5

Answer: a. 3x + 3y = 6

Explanation:

Putting x = 0 in first equation (a),

⇒ 3(0) + 3y = 6

⇒ 3y = 6

⇒ y = 6/3 = 2

The line touches y-axis at point (0, 2).

Putting y = 0 in first equation (a),

⇒ 3x + 3(0) = 6

⇒ 3x = 6

⇒ x = 6/3 = 2

The line touches y-axis at point (2, 0).

∴ Line 3x + 3y = 6 cuts positive x-axis and positive y-axis at equal distance i.e. 2 units from the origin.

Question 25

In the given diagram (not drawn to scale), railway stations A, B, C, P and Q are connected by straight tracks. Track PQ is parallel to BC. The time taken by a train travelling at 90 km/hr to reach B from A by the shortest route is :

a. 8 minutes

b. 12 minutes

c. 16.8 minutes

d. 20 minutes

Answer: d. 20 minutes

Explanation:

In △ APQ and △ ABC,

∠PAQ = ∠BAC (Common angle)

∠APQ = ∠ABC (Corresponding angles)

∴ △ APQ ~ △ ABC (By A.A. axiom)

From figure,

Let AP = x km.

We know that,

Corresponding sides of similar triangle are proportional.

⇒ AP/AB = PQ/BC

⇒ x/(AP+PB) = 20/50

⇒ x/x+18 = 2/5

⇒ 5x = 2(x+18)

⇒ 5x = 2x + 36

⇒ 5x − 2x = 36

⇒ 3x = 36

⇒ x = 36/3

⇒ x = 12

AB = AP + BP = 12 + 18 = 30 km.

Time = Distance/Speed

=AB/90

=30/90

=1/3 hr =1/3×60 = 20 minutes.

Question 26

In the given diagram, △ ABC and △ DEF (not drawn to scale) are such that ∠C = ∠F and AB/DE=BC/EF then

a. △ ABC ~ △ DEF

b. △ BCA ~ △ DEF

c. △ CAB ~ △ DEF

d. the similarity of given triangles cannot be determined.

Answer: d. the similarity of given triangles cannot be determined.

Explanation:

Given, ∠C = ∠F

Since, the sides containing the angle may or may not be proportional,

i.e., we don’t know if AC/DF = BC/EF or not equal.

∴ Similarity of given triangles cannot be determined.

Question 27

In the adjoining diagram, ST is not parallel to PQ. The necessary and sufficient conditions for △ PQR ~ △ TSR is :

a. ∠PQR = ∠STR

b. ∠QPR = ∠TSR

c. ∠PQR = ∠TSR

d. ∠PRQ = ∠RST

Answer: c. ∠PQR = ∠TSR

Explanation:

Given, △ PQR ~ △ TSR.

From figure,

⇒ ∠PRQ = ∠SRT (Common angle)

The order of vertices of two similar triangles are written in such a way that the corresponding vertices occupy the same position.

∴ ∠QPR = ∠STR and ∠PQR = ∠TSR

∴ △ PQR ~ △ TSR (By A.A. axiom)

Question 28

The scale factor of a picture and the actual height of Sonia is 20 cm : 1.6 m. If her height in the picture is 18 cm, then her actual height is :

a. 14.4 m

b. 2.25 m

c. 1.78 m

d. 1.44 m

Answer: d. 1.44 m

Explanation:

\[\frac{Height\;in\;picture}{Actual\;height}=\frac{20\;cm}{1.6\;m}\\\frac{18\;cm}{Actual\;height}=\frac{20\;cm}{160\;cm}\\Actual\;height\;=\frac{18\;\times160}{20\;}=144\;cm=1.44\;m\]

Question 29

In the adjoining figure, O is the center of the circle, and a semicircle is drawn on OA as the diameter. ∠APQ = 20°. The degree measure of ∠OAQ is:

a. 25°

b. 40°

c. 50°

d. 65°

Answer: c. 50°

Explanation:

We know by theorem,,

Angle in a semi-circle is a right angle.

∴ ∠OQA = 90°

In △QAP,

⇒ ∠PQA + ∠QAP + ∠APQ = 180°

⇒ 90° + ∠QAP + 20° = 180°

⇒ ∠QAP = 180° – 90° – 20° = 70°.

In △OPA,

⇒ OA = OP (Radii of same circle)

⇒ ∠OAP = ∠OPA (Angle opposite to equal sides are equal)

⇒ ∠OAP = 20°.

From figure,

⇒ ∠OAQ = ∠QAP – ∠OAP = 70° – 20° = 50°.

Question 30

In the given diagram, O is the center of the circle, and PQ is a tangent at A. If ∠ABC = 50°, then values of x, y and z respectively are :

a. 50°, 100°, 40°

b. 50°, 50°, 65°

c. 40°, 80°, 50°

d. 50°, 25°, 78°

Answer: a. 50°, 100°, 40°

Explanation:

We know by theorem,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOC = y = 2∠ABC = 2 × 50° = 100°.

In △ AOC,

⇒ OA = OC (Radii of same circle)

⇒ ∠OAC = ∠OCA = z (Angle opposite to equal sides are equal)

Now, ∠OAC + ∠OCA + ∠AOC = 180° (angle sum property of triangle)

⇒ z + z + 100° = 180°

⇒ 2z = 180° – 100°

⇒ 2z = 80°

⇒ z = 80°/2 = 40°.

We know by theorem,

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

⇒ ∠CAQ = x = ∠OAQ – ∠OAC = 90° – 40° = 50°.

Question 31

In the given figure, PT and QT are tangents to a circle such that ∠TPS = 45° and ∠TQS = 30°. Then, the value of x is :

a. 30°

b. 45°

c. 75°

d. 105°

Answer: d. 105°

We know by theorem,

The angle between a tangent and a chord through point of contact is equal to an angle in the alternate segment.

∴ ∠SQP = ∠SPT = 45° and ∠SPQ = ∠SQT = 30°.

In △ SQP,

⇒ ∠SQP + ∠SPQ + ∠QSP = 180° (angle sum property of a triangle)

⇒ 45° + 30° + ∠QSP = 180°

⇒ 75° + ∠QSP = 180°

⇒ ∠QSP = x = 180° – 75° = 105°.

Question 32

A cylindrical metallic wire is stretched to double its length. Which of the following will NOT change for the wire after stretching?

a. Its curved surface area

b. Its total surface area

c. Its volume

d. Its radius

Answer: c. Its volume

Explanation:

On changing the shape of a container, its volume remains same.

Question 33

A right circular cone has the radius of the base equal to the height of the cone. If the volume of the cone is 9702 cu. cm, then the diameter of the base of the cone is :

a. 21 cm

b. 42 cm

c. 21√7 cm

d. 2√7 cm

[Use π=22/7]

Answer: b. 42 cm

Explanation:

Given,

Height of cone (h) = Radius of cone (r) = x cm (let)

Volume = 9702 cm³

\[Volume\;=\;9702\;cm^3\\\frac13\mathrm{πr}^2\mathrm h=9702\\\frac13\mathrm{πx}^3=9702\\\mathrm x^3=\frac{9702\times3\times7}{22}\\\mathrm x^3=441\times3\times7\\\mathrm x=\sqrt[3]{441\times3\times7}\\\mathrm x=21\\\therefore\mathrm{radius}=21\;\mathrm{cm}\\\mathrm{Diameter}\;=\;2\;\times\;\mathrm{radius}\;=\;2\;\times\;21\;=\;42\;\mathrm{cm}.\]

Question 34

A solid sphere with a radius of 4 cm is cut into 4 identical pieces by two mutually perpendicular planes passing through its center. Find the total surface area of one-quarter piece.

a. 24π

b. 32π

c. 48π

d. 64π

Answer: b. 32π

One-quarter piece= half of a hemisphere

Therefore, Total surface area of semi-hemisphere = 2πr²

= 2π × 4²

= 2π × 16

= 32π.

Question 35

Two identical solid hemispheres are kept in contact to form a sphere. The ratio of the total surface areas of two hemispheres to the surface area of the sphere formed is :

a. 1 : 1

b. 3 : 2

c. 2 : 3

d. 2 : 1

Answer: b. 3 : 2

Explanation:

Let radius of hemisphere be r.

Total surface area of hemisphere = 3πr²

Total surface area of two hemisphere = 2 × 3πr² = 6πr²

Total surface area of sphere = 4πr²

Total surface area of two hemisphere : Total surface area of sphere

= 6πr² : 4πr²

= 6 : 4

= 3 : 2

Question 36

cosec²θ + sec²θ is equal to :

a. tan² θ + cot²θ

b. cot θ + tan θ

c. (cot θ + tan θ)²

d. 1

Answer: c. (cot θ + tan θ)²

Explanation:

cosec²θ + sec²θ

= 1 + cot²θ + 1 + tan²θ

= cot²θ + tan²θ + 2

= (cot θ + tan θ)²

Question 37

Given a = 3 sec²θ and b = 3 tan²θ – 2. The value of (a – b) is :

a. 1

b. 2

c. 3

d. 5

Answer: d. 5

Explanation:

a – b

= 3 sec²θ – (3 tan²θ – 2)

= 3 sec²θ – 3 tan²θ + 2

= 3(sec²θ – tan²θ) + 2

= 3(1) + 2

= 3 + 2

= 5.

Question 38

At a certain time of day, the ratio of the height of the pole to the length of its shadow is 1 : 33, then the angle of elevation of the sun at that time of the day is:

a. 30°

b. 45°

c. 60°

d. 90°

Answer: a. 30°

Explanation:

Let AB be the pole and BC be the shadow and angle of elevation of Sun be θ.

From figure,

⇒ tan θ = AB/BC

⇒ tan θ = x/√3x

⇒ tan θ = 1/√3

⇒ tan θ = tan 30°

⇒ θ = 30°

Question 39

A man standing on a ship approaching the port towards the lighthouse is observing the top of the lighthouse. In 10 minutes, the angle of elevation of the top of the lighthouse changes from α to β. Then:

a. α > β

b. α < β

c. α = β

d. α ≤ β

Answer: b. α < β

Explanation:

Let A be the top of the lighthouse, C the initial position of ship and D be the position after 10 minutes.

From figure,

tan α = AB/BC

tan β = AB/BD

Since, BC is greater than BD.

∴ tan α < tan β

⇒ α < β.

Question 40

Assertion (A) : The difference in class marks of the modal class and the median class of the following frequency distribution table is 0.

Class interval	Frequency
20-30	1
30-40	3
40-50	2
50-60	6
60-70	4

Reason (R) : Modal class and median class are always the same for a given frequency distribution.

a. Both A and R are correct, and R is the correct explanation for A.

b. Both A and R are correct, and R is not the correct explanation for A.

c. A is true, but R is false.

d. Both A and R are true.

Answer: c. A is true, but R is false.

Explantion:

Class interval	Class mark	Frequency	Cumulative frequency
20-30	25	1	1
30-40	35	3	4
40-50	45	2	6
50-60	55	6	12
60-70	65	4	16 = N

Median = N/2= 16/2 = 8th term.

The 8th term lies in the class 50-60.

∴ Median class = 50-60

Also, frequency of class 50-60 is highest.

∴ Modal class = 50-60.

∴ Their class marks will obviously be same

∴ Assertion (A) is true.

Modal class and median class are not always the same for a given frequency distribution.

∴ Reason (R) is false.

Question 41

Assertion (A) : For a collection of 11 arrayed data, the median is the middle number.

Reason (R) : For the data 5, 9, 7, 13, 10, 11, 10, the median is 13.

a. Both A and R are correct, and R is the correct explanation for A.

b. Both A and R are correct, and R is not the correct explanation for A.

c. A is true, but R is false.

d. Both A and R are true.

Answer: c. A is true, but R is false.

Explanation:

For a collection of 11 arrayed data.

Median = (n+1)/2 th term

= (11+1)/2

= 6th term, which will be the middle term.

∴ For a collection of 11 arrayed data, the median is the middle number.

∴ Assertion (A) is true.

Numbers = 5, 9, 7, 13, 10, 11, 10

Arranging in ascending order, we get :

5, 7, 9, 10, 10, 11, 13.

n = 7, which is odd.

Median = (n+1)/2 th term =(7+1)/2= 8/2 = 4th term = 10.

∴ Reason (R) is false.

Question 42

Ankit has the option of investing in company A, where 7%, ₹100 shares are available at ₹120 or in company B, where 8%, ₹1000 shares are available at ₹1620.

Assertion (A) : Investment in Company A is better than Company B.

Reason (R) : The rate of income in Company A is better than in Company B.

a. Both A and R are correct, and R is the correct explanation for A.

b. Both A and R are correct, and R is not the correct explanation for A.

c. A is false, but R is true.

d. Both A and R are false.

Answer: a. Both A and R are correct, and R is the correct explanation for A.

In company A,

N.V. = ₹100

M.V. = ₹120

Dividend = 7% = 7/100×100 = ₹ 7

∴ Investment = ₹ 120 and income = ₹ 7.

Income on ₹ 1 = 7/120 = ₹ 0.0583

In company B,

N.V. = ₹1000

M.V. = ₹1620

Dividend = 8% = 8/100×1000 = ₹ 80

∴ Investment = ₹1620 and income = ₹80.

Income on ₹ 1 = 80/1620 = ₹ 0.0494

Since, rate of income is greater in company A.

∴ Assertion and Reason both are true and Reason is the correct explanation of Assertion.

Question 43

Assertion (A) : x³ + 2x² – x – 2 is a polynomial of degree 3.

Reason (R) : x + 2 is a factor of the polynomial.

a. Both A and R are correct, and R is the correct explanation for A.

b. Both A and R are correct, and R is not the correct explanation for A.

c. A is true, but R is false.

d. Both A and R are true.

Answer: d. Both A and R are true.

Explanation:

x³ + 2x² – x – 2 is a polynomial of degree 3.

By factor theorem,

(x – a) is a factor of f(x) if f(a) = 0.

x + 2 = 0

x = -2

Substituting x = -2 in x³ + 2x² – x – 2, we get :

⇒ (-2)³ + 2(-2)² – (-2) – 2

⇒ -8 + 2(4) + 2 – 2

⇒ -8 + 8 + 2 – 2

⇒ 0.

Question 44

Assertion (A) : The point (-2, 8) is invariant under reflection in line x = -2

Reason (R) : If a point has its x-coordinate 0, it is invariant under reflection in both axes.

a. Both A and R are correct, and R is the correct explanation for A.

b. Both A and R are correct, and R is not the correct explanation for A.

c. A is true, but R is false.

d. Both A and R are true.

Answer: c. A is true, but R is false.

Explanation:

Point (-2, 8) lies on the line x = -2.

The point which lies on a line is invariant under reflection in the same line.

∴ (-2, 8) is invariant under reflection in line x = -2.

∴ Assertion (A) is true.

If a point has x-coordinate equal to 0 it means it lies on y-axis.

∴ It will be invariant under reflection in y-axis.

∴ Reason (R) is false.

Question 45

When a die is cast with numbering on its faces, as shown, the ratio of the probability of getting a composite number to the probability of getting a prime number is …………… .

a. 2 : 3

b. 3 : 2

c. 1 : 3

d. 1 : 2

Answer: a. 2 : 3

Explanation:

From numbers 1 to 6

Prime numbers = 2, 3, 5.

Composite numbers = 4, 6.

Probability of getting a prime number = No. of prime numbers/No. of possible outcomes

= 3/6 =1/2

Probability of getting a composite number = No. of composite numbers/No. of possible outcomes

=2/6 = 1/3

Probability of getting a composite number to probability of getting a prime number

= 1/3 : 1/2 = 2 : 3.

Question 46

\[The\;product\;of\;A\;=\begin{vmatrix}1&-2\\-3&4\end{vmatrix}\;and\;matrix\;M,\;AM\;=\;B\;\\where\;B\;=\begin{bmatrix}2\\24\end{bmatrix},\;then\;the\;order\;of\;matrix\;M\;is\;…..\;.\]

a. 2 × 2

b. 2 × 1

c. 1 × 2

d. 4 × 1

Answer: b. 2 × 1

Explanation:

Order of matrix A = 2 × 2

Order of matrix B = 2 × 1

Order of matrix M (let) = a × b

We know that,

Two matrix can be multiplied if the no. of columns of the first matrix is equal to the no. of rows of the second matrix and the resultant matrix has the no. of rows of first matrix and no. of columns of second matrix.

Given,

⇒ AM = B

⇒ A_{m × n} × M_{n × p} = B_{m × p}

⇒ A_{2 × 2} × M_{2 × 1} = B_{2 × 1}

a = 2 and b = 1.

Order of matrix M = 2 × 1.

Question 47

Given, a₁, a₂, a₃, ….. and b₁, b₂, b₃, ….. are real numbers such that a₁ – b₁ = a₂ – b₂ = a₃ – b₃ = ……… are all equal.