ICSE Maths Solved Competency Focused Questions Solutions Class 10 Chapter 1 Multiple-Choice Questions

This page contains the complete ICSE Class 10 Maths MCQ Solutions from the Competency Focused Practice question bank. These multiple-choice questions are designed to test your quick thinking and conceptual clarity across the entire syllabus, from Commercial Mathematics like GST and Banking to key topics in Algebra, Geometry, and Trigonometry. In your Class 10 board exams, the objective section is crucial for building a strong score, and this chapter provides the perfect practice to master the format. Our solutions will guide you through the logic needed to select the correct option efficiently.

You have landed here because you need verified answers for the 49 MCQs in this important practice chapter. It’s not enough to just know the right choice; you must also understand the fastest way to get there. Our expert teachers have solved every question, showing the step-by-step working that follows the methods prescribed by the ICSE board. Here, you will find clear, accurate, and easy-to-follow solutions for all multiple-choice questions, helping you build both speed and confidence for your final exams.

Multiple Choice Questions (1 Mark Each)

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

For the retailer, the cost price includes GST.

C.P. = ₹ 2500

G.S.T. = 12%

C.G.S.T. = \dfrac{12}{2} % = 6%.

C.G.S.T. paid = \dfrac{6}{100} \times 2500 = ₹ 150.

The retailer marks up by 20%.

S.P. = ₹ 2500 + 20%

= ₹ 2500 + \dfrac{20}{100} \times 2500

= ₹ 2500 + ₹ 500

= ₹ 3000.

C.G.S.T. charged = \dfrac{6}{100} \times 3000 = ₹ 180.

Tax liability = ₹ 180 − ₹ 150 = ₹ 30.

Hence, Option 3 is the correct option.

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

GST is 7%, so SGST is half.

SGST = \dfrac{7}{2} = 3.5%.

M.P. = ₹ 800.

SGST = 3.5% of ₹ 800 = \dfrac{3.5}{100} \times 800 = ₹ 28.

Hence, Option 2 is the correct option.

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

The formula we use to calculate the interest on a recurring deposit account is:

Interest = P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

Based on this formula, we can see that the interest is calculated as simple interest for a period of one month.

Therefore, Option 4 is the correct option.

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

First, let’s calculate the interest earned in the initial scenario. Here’s the information we’re given:

P = ₹ 1000

r = 5%

n = 12 months

Using the formula for interest on a recurring deposit:

Interest = P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

So, the interest comes out to be ₹325.

Now, for the second case, we want to earn the same amount of interest (₹325), but with a different interest rate. Let’s call the new monthly deposit ₹x.

P = ₹ x

r = 4%

n = 12 months

Interest = ₹ 325

Let’s set up the equation with our new values:

Now, we just need to solve for x:

This tells us that the monthly installment in the second case is ₹1250.

Hence, Option 2 is the correct option.

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%

To compare the investments, let’s calculate the rate of return for both Mr. Das and Mr. Singh.

First, for Mr. Das:
* Face value of each share = ₹ 100
* Market value of each share = ₹ 60
* Dividend per share is 12% of the face value, so: 12% of ₹ 100 = ₹ 12.
* His rate of return is the dividend income based on his market value investment:
Rate of return = \dfrac{\text{Dividend per share}}{\text{M.V. of each share}} \times 100 = \dfrac{12}{60} \times 100 = 20%.

Next, for Mr. Singh:
* Face value of each share = ₹ 50
* Market value of each share = ₹ 40
* Dividend per share is 16% of the face value, which gives us: 16% of ₹ 50 = ₹ 8.
* His rate of return is calculated in the same way:
Rate of return = \dfrac{\text{Dividend per share}}{\text{M.V. of each share}} \times 100 = \dfrac{8}{40} \times 100 = 20%.

Look at that! Both Mr. Das and Mr. Singh have the same rate of return of 20%.

Hence, Option 4 is the correct option.

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

Let’s say the price Amit paid for each share is ₹ P.
Since he bought 10 shares, his total investment is ₹ 10P.

The dividend is 7.5%. Assuming a face value of ₹100 per share, the cash dividend on one share is:
Dividend per share = \dfrac{7.5}{100} \times 100 = ₹ 7.5

For his 10 shares, Amit’s total dividend income is:
Total dividend = ₹ 7.5 × 10 = ₹ 75.

We are told his rate of return is 10%. The formula that connects these ideas is:
Rate of Return × Investment = Dividend

Let’s substitute the values we know:
\Rightarrow 10% \times 10P = 75 \[1em] \Rightarrow \dfrac{10}{100} \times 10P = 75 \[1em] \Rightarrow \dfrac{100P}{100} = 75 \[1em] \Rightarrow P = ₹ 75.

This tells us that the market price per share was ₹75.
Amit’s total investment is for 10 shares, so:
Total investment = 10 × ₹75 = ₹750.

Hence, Option 2 is the correct option.

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}

Alright, let’s solve the given inequality step-by-step.

-3 ≤ -4x + 5

First, we’ll rearrange the terms to get the variable on one side and the numbers on the other. Let’s move -4x to the left and -3 to the right. Remember to flip their signs!

⇒ 4x ≤ 5 + 3

Simplifying the right side gives us:

⇒ 4x ≤ 8

Now, to find x, we just divide both sides by 4.

⇒ x ≤ \dfrac{8}{4}

⇒ x ≤ 2

The question specifies that x belongs to the set of whole numbers (x ∈ W). So, we need to list all the whole numbers that are less than or equal to 2. Those are 0, 1, and 2.

∴ The solution set is {0, 1, 2}.

Hence, Option 3 is the correct option.

Question 8

If -4x > 8y; then

• (a) x > 2y
• (b) x > -2y
• (c) x < -2y
• (d) x < 2y

Solving,

⇒ -4x > 8y

⇒ 4x < -8y

⇒ x < -\dfrac{8y}{4}

⇒ x < -2y.

Hence, Option 3 is the correct option.

Question 9

The value/s of ‘k’ for which the quadratic equation 2x^2 – kx + k = 0 has equal roots is (are) :

• (a) 0 only
• (b) 4, 0
• (c) 8 only
• (d) 0, 8

We’re given the quadratic equation 2x^2 - kx + k = 0.

A key concept for quadratic equations is that for the roots to be equal, the discriminant (D) must be zero. The formula for the discriminant is D = b^2 - 4ac.

First, let’s identify the coefficients from our equation:
a = 2, b = -k, and c = k.

Now, we’ll set the discriminant to zero and substitute these values:
b^2 - 4ac = 0
⇒ (-k)^2 - 4(2)(k) = 0

Simplifying this, we get:
⇒ k^2 - 8k = 0

We can solve this for k by factoring:
⇒ k(k - 8) = 0

This gives us two possible solutions:
k = 0 or k - 8 = 0

So, the final values are 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 x^2 + 3a – x = 0, then the value of ‘a’ is :

• (a) -8
• (b) -2
• (c) -\dfrac{1}{3}
• (d) \dfrac{1}{3}

We’re told that x = -2 is a solution for the quadratic equation x² + 3a – x = 0. This means if we substitute -2 in place of x, the equation will hold true. Let’s do just that!

∴ (-2)² + 3a – (-2) = 0

Now, we can simplify and solve for ‘a’:
⇒ 4 + 3a + 2 = 0
⇒ 6 + 3a = 0
⇒ 3a = -6
⇒ a = -6/3 = -2

So, the value of ‘a’ is -2.

Hence, Option 2 is the correct option.

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

x = 233.356

On rounding off to two significant figures

x = 230.

Hence, Option 4 is the correct option.

Question 12

In the adjoining diagram, AB = x cm, BC = y cm and x – y = 7 cm. Area of △ ABC = 30 cm^2. The length of AC is :

1. 10 cm
2. 12 cm
3. 13 cm
4. 15 cm

Let’s start with the formula for the area of a triangle.

Area of triangle = \dfrac{1}{2} \times \text{ base} \times \text{height}

We’re given that the area is 30, the base (BC) is y, and the height (AB) is x. Let’s plug those values in:

⇒ 30 = \dfrac{1}{2} \times y \times x

⇒ 30 = \dfrac{xy}{2}

Multiplying both sides by 2 gives us our first equation:

⇒ xy = 60 ……..(1)

We are also told that the difference between the sides is 7.

⇒ x – y = 7

Let’s rearrange this to express x in terms of y. This will be our second equation:

⇒ x = 7 + y ……..(2)

Now, let’s substitute the expression for x from equation (2) into equation (1):

⇒ (7 + y)y = 60

⇒ 7y + y^2 = 60

Rearranging this gives us a quadratic equation:

⇒ y^2 + 7y – 60 = 0

We can solve this by factoring. We’re looking for two numbers that multiply to -60 and add to +7. Those numbers are +12 and -5.

⇒ y^2 + 12y – 5y – 60 = 0

⇒ y(y + 12) – 5(y + 12) = 0

⇒ (y – 5)(y + 12) = 0

This gives two possible solutions for y:

⇒ y – 5 = 0 or y + 12 = 0

⇒ y = 5 or y = -12.

Since the length of a side cannot be negative, we must choose the positive value.

∴ y = 5 cm.

Now we can find x by substituting y = 5 back into equation (2):

⇒ x = 7 + y = 7 + 5 = 12 cm.

Finally, to find the hypotenuse (AC) of our right-angled triangle, we can use the Pythagoras theorem:

⇒ AC^2 = AB^2 + BC^2

⇒ AC^2 = x^2 + y^2

⇒ AC^2 = 12^2 + 5^2

⇒ AC^2 = 144 + 25

⇒ AC^2 = 169

Taking the square root of both sides:

⇒ AC = \sqrt{169} = 13 cm.

Hence, Option 3 is the correct option.

Question 13

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

• (a) p : q = p : r
• (b) q : r = p^2 : q^2
• (c) p : q^2 = r : p^2
• (d) p : r = p^2 : q^2

When three numbers like p, q, and r are in continued proportion, it means the ratio of the first to the second is equal to the ratio of the second to the third.

Now, let’s examine the proposed relationship: p : r = p^2 : q^2. We can write this as a fraction and rearrange it.

Since this result (equation 2) matches our definition of continued proportion (equation 1), the statement is true.

Hence, Option 4 is the correct option.

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

Hey there! Let’s break this down step-by-step. We’re given that the ratio of the diameter (d) to the height (h) of a cylindrical glass is 3:5, and we know the diameter is 6 cm.

First, let’s use this ratio to find the height:

Great! Now, we need the radius (r) for our next calculation. Remember, the radius is just half the diameter.

Radius = \dfrac{\text{Diameter}}{2} = \dfrac{6}{2} = 3 cm.

Perfect. Now we have everything we need to find the curved surface area using the formula 2πrh.

Curved surface area of glass = 2πrh

= 2π × 3 × 10

= 60π cm².

So, the curved surface area is 60π cm², which means Option 2 is the correct option.

Question 15

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

• (a) -3
• (b) -1
• (c) 1
• (d) 3

Let’s get started! We are given that the polynomial 2x^3 + 3x^2 - 2x - 3 is completely divisible by (2x + a), with the quotient being (x^2 - 1).

Since the division is exact (completely divisible), we know that:
Dividend = Divisor × Quotient
So, we can set up the equation:
2x^3 + 3x^2 - 2x - 3 = (2x + a)(x^2 - 1)

Next, let’s expand the expression on the right-hand side:
⇒ 2x^3 + 3x^2 - 2x - 3 = 2x^3 - 2x + ax^2 - a

Now we can simplify by cancelling the terms that appear on both sides of the equation (2x^3 and -2x).
⇒ 3x^2 - 3 = ax^2 - a

By comparing the coefficients of the terms on both sides, we can clearly see that a must be 3.

So, a = 3.

Hence, Option 4 is the correct option.

Question 16

A polynomial in x is x^3 + 5x^2 – 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)

To solve this, we can use the Factor Theorem. It tells us that if (x + 4) is a factor of the polynomial, then substituting x = -4 into the expression should result in zero. Let’s test the options to see which value of k makes this happen.

The polynomial is x^3 + 5x^2 - kx - 24.

Let’s try the value from Option 3, where k = 2. Our polynomial becomes:
x^3 + 5x^2 - 2x - 24

Now, let’s substitute x = -4 and check the result:
⇒ (-4)^3 + 5(-4)^2 - 2(-4) - 24
⇒ -64 + 5(16) + 8 - 24
⇒ -64 + 80 + 8 - 24
⇒ 88 - 88
⇒ 0

Since we got 0, it means that (x + 4) is indeed a factor when k = 2.

Hence, Option 3 is the correct option.

Question 17

If A = \begin{bmatrix} a & b \end{bmatrix} \text{ and } \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

First, let’s figure out the dimensions (or order) of each matrix.

Order of matrix A = 1 × 2
Order of matrix B = 2 × 1

Now, let’s see if we can multiply them. Remember, for the product AB to be possible, the number of columns in A must equal the number of rows in B. Here, that’s 2 and 2, so they match! For BA to be possible, the number of columns in B must equal the number of rows in A. That’s 1 and 1, so they match too. Great, both products are possible.

Let’s calculate AB:
AB = \begin{bmatrix} a \times c + b \times d \end{bmatrix} \\ = \begin{bmatrix} ac + bd \end{bmatrix}. \\

And now let’s calculate BA:
BA = \begin{bmatrix} c \times a & c \times b \\ d \times a & d \times b \end{bmatrix} \\ = \begin{bmatrix} ca & cb \\ da & db \end{bmatrix}.

As we can see, the results are not the same at all! AB is a 1×1 matrix, while BA is a 2×2 matrix.

∴ AB ≠ BA.

Hence, Option 3 is the correct option.

Question 18

Matrix A = \begin{bmatrix} 6 & 9 \\ -4 & k \end{bmatrix} \text{ such that } A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}. Then k is :

• (a) 6
• (b) -6
• (c) 36
• (d) ±6

We are given that A^2 is the zero matrix. So, let’s multiply matrix A by itself and set it equal to the zero matrix:

Performing the matrix multiplication on the left side gives us:

Let’s simplify the terms inside the resulting matrix:

Now, we can equate the corresponding elements. Let’s use the element in the first row, second column to find the value of k.

Hence, Option 2 is the correct option.

Question 19

If the sum of n terms of an arithmetic progression S~n = n^2 – n, then the third term of the series is :

• (a) 2
• (b) 4
• (c) 6
• (d) 9

We are given the formula for the sum of n terms of an arithmetic progression: S~n = n^2 – n.

First, let’s calculate the sum for the first few terms:
S~1 = 1^2 – 1 = 0
S~2 = 2^2 – 2 = 4 – 2 = 2
S~3 = 3^2 – 3 = 9 – 3 = 6

The sum of the first term (S~1) is simply the first term itself, which means the first term is 0.

Next, the sum of the first two terms (S~2) is 2. Since we know the first term is 0, the second term must be 2.

Finally, the sum of the first three terms (S~3) is 6. We know that this sum is made up of the first three terms added together:
First term + Second term + Third term = 6

Plugging in the values we found:
0 + 2 + Third term = 6

Solving for the third term, we get:
Third term = 6 – 2 = 4.

Hence, Option 2 is the correct option.

Question 20

Which of the following is not a geometric progression ?

• (a) \dfrac{1}{3}, 1, 3, 9
• (b) \dfrac{1}{5}, \dfrac{1}{5}, \dfrac{1}{5}, \dfrac{1}{5}
• (c) -2, 4, -8, 16
• (d) 2, 0, 4, 0, 8, 0

Let’s check the series: 2, 0, 4, 0, 8, 0. For a series to be a Geometric Progression (G.P.), the ratio between consecutive terms must be constant.

The ratio of the second term to the first is \dfrac{0}{2} = 0.
The ratio of the third term to the second is \dfrac{4}{0}, which is undefined.

Since the ratios are not the same, the series is not a G.P.

Hence, Option 4 is the correct option.

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 :

1. (2, -6)
2. (3, -6)
3. (6, -6)
4. (10, -6)

First, let’s remember the formula for the centroid of a triangle, which is the average of the vertex coordinates:

Centroid = \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)

We are given two vertices, A(3, -3) and B(2, -6), and the centroid at (5, -5). Let’s say the third vertex is C(x, y). Now we can plug all this information into our formula:

By comparing the x and y coordinates, we can set up two small equations to solve for x and y:

So, the coordinates of our third vertex C are (10, -6).

Next, we need to find point D. Since AD is a median of the triangle, D must be the midpoint of the opposite side, BC.

Using the midpoint formula for B(2, -6) and C(10, -6):

Hence, Option 3 is the correct option.

Question 22

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

1. y = x
2. 2y = x
3. y = 2x
4. y = -x

First, let’s identify the coordinates of points A and B from the graph provided. Point A is at (4, 0) and point B is at (0, 2).

The problem states that P is the mid-point of the line segment AB. We can find the coordinates of P using the midpoint formula:

P = \Big(\dfrac{4 + 0}{2}, \dfrac{0 + 2}{2}\Big) = \Big(\dfrac{4}{2}, \dfrac{2}{2}\Big) = (2, 1)

So, the coordinates of our midpoint P are (2, 1). Now, we need to find the equation of the line that joins this point P with the origin, O(0, 0).

We can use the two-point form to find the equation. Let’s use P(2, 1) and O(0, 0):

⇒ y – 1 = \dfrac{1 - 0}{2 - 0}(x - 2)

Now, let’s simplify this equation:

⇒ y – 1 = \dfrac{1}{2}(x - 2)

To clear the fraction, we can multiply both sides by 2:

⇒ 2(y – 1) = x – 2

⇒ 2y – 2 = x – 2

Adding 2 to both sides gives us our final answer:

⇒ 2y = x

Hence, Option 2 is the correct option.

Question 23

In the given figure line l~1 is a parallel to line l~2. If line l~3 is perpendicular to line l~1, then the slopes of lines l~2 and l~3 respectively are :

1. 1, 1
2. -1, -1
3. 1, -1
4. -1, 1

Alright, let’s figure this out step-by-step.

First, we need the slope of line l₁. We’re told it makes an angle of 45° with the x-axis. Remember that the slope is just the tangent of the angle, so:
Slope of line l₁ = tan 45° = 1.

Next up is line l₂, which is parallel to l₁. A key rule for us is that parallel lines always have the same slope. This means the slope of line l₂ is also 1.

Finally, we have line l₃, which is perpendicular to l₂. For perpendicular lines, their slopes have a special relationship: they multiply to give -1.
So, (Slope of line l₂) × (Slope of line l₃) = -1.

Let’s plug in the slope for l₂ that we just found:
1 × (Slope of line l₃) = -1
This makes it clear that the slope of line l₃ must be -1.

Hence, Option 3 is the correct option.

Question 24

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

• (a) 3x + 3y = 6
• (b) 5x + 10y = 10
• (c) -x + y = 1
• (d) 10x + 5y = 5

First, let’s find where our line crosses the y-axis. To do that, we can substitute x = 0 into the equation:

⇒ 3(0) + 3y = 6

⇒ 3y = 6

⇒ y = \dfrac{6}{3}

⇒ y = 2.

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

Next, let’s find where it crosses the x-axis by substituting y = 0:

⇒ 3x + 3(0) = 6

⇒ 3x = 6

⇒ x = \dfrac{6}{3}

⇒ x = 2.

This means the line touches the x-axis at the point (2, 0).

So, we can see that the line 3x + 3y = 6 cuts the positive x-axis and the positive y-axis at an equal distance from the origin (2 units).

Hence, Option 1 is the correct option.

Question 25

In the given diagram (not draw 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 :

1. 8 minutes
2. 12 minutes
3. 16.8 minutes
4. 20 minutes

To solve this, we’ll start by showing that the two triangles, △ APQ and △ ABC, are similar.

In △ APQ and △ ABC:
* ∠PAQ = ∠BAC (This is a common angle for both triangles)
* ∠APQ = ∠ABC (These are corresponding angles because PQ is parallel to BC)

Since two angles are equal, we can say that △ APQ ~ △ ABC by the A.A. similarity axiom.

Now that we know they’re similar, their corresponding sides must be in proportion. Let’s set the unknown distance AP to be ‘x’ km. We can write the ratio like this:

Let’s plug in the values from the diagram. The total length of AB is AP + PB, which is x + 18.

Now, we just need to solve for x. Let’s simplify the fraction first.

So, the distance AP is 12 km. This means the total distance AB = AP + BP = 12 + 18 = 30 km.

The final step is to calculate the time. The formula is Time = Distance / Speed.

Time = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{AB}{90} = \dfrac{30}{90} = \dfrac{1}{3} \text{ hr}

To get the answer in minutes, we multiply by 60:

Time = \dfrac{1}{3} \times 60 = 20 minutes.

Hence, Option 4 is the correct option.

Question 26

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

1. △ ABC ~ △ DEF
2. △ BCA ~ △ DEF
3. △ CAB ~ △ DEF
4. the similarity of given triangles cannot be determined.

We are given that ∠C = ∠F. For the triangles to be similar, the sides that form these equal angles must be proportional. However, we don’t have any information about the side lengths, so we don’t know if the ratio \dfrac{AC}{DF} is equal to the ratio \dfrac{BC}{EF}.

Since we can’t confirm this, the similarity of the triangles cannot be determined.

Hence, Option 4 is the correct option.

Question 27

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

1. ∠PQR = ∠STR
2. ∠QPR = ∠TSR
3. ∠PQR = ∠TSR
4. ∠PRQ = ∠RST

We’re given that the two triangles are similar: △ PQR ~ △ TSR.

From the diagram, we can see that both triangles share the angle at vertex R. This is their common angle, so we can say ∠PRQ = ∠SRT.

Now, the order of the letters in the similarity statement is a big clue. It tells us exactly which angles match up. Since Q is the second vertex listed for the first triangle and S is the second for the other, their angles must be equal.
That means ∠PQR = ∠TSR.

Because we have two pairs of matching angles, we can confirm their similarity by the A.A. axiom.

Hence, Option 3 is the correct option.

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

First, let’s get the units for our scale factor to match up. The given scale is 20 cm to 1.6 m. Since 1.6 m is the same as 160 cm, our scale is 20 cm : 160 cm.

This gives us the following relationship:

Now, we can plug in the person’s height in the picture, which is 18 cm, and solve for their actual height:

Converting this back to meters, we get 1.44 m.

Hence, Option 4 is the correct option.

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 :

1. 25°
2. 40°
3. 50°
4. 65°

First, we’ll use the key property that the angle in a semi-circle is a right angle.

∴ ∠OQA = 90°

Now let’s look at the larger triangle, △QAP. The sum of angles in a triangle is always 180°.

⇒ ∠PQA + ∠QAP + ∠APQ = 180°

Plugging in the values we know:

⇒ 90° + ∠QAP + 20° = 180°

Solving for ∠QAP, we get:

⇒ ∠QAP = 180° – 90° – 20° = 70°.

Next, let’s consider the smaller triangle, △OPA.

⇒ OA = OP because they are both radii of the same circle.

This makes △OPA an isosceles triangle, which means the angles opposite the equal sides are also equal.

⇒ ∠OAP = ∠OPA

Since ∠OPA is given as 20°, then ∠OAP = 20°.

Finally, we can see from the figure that ∠OAQ is just the difference between the two angles we’ve found.

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

Hence, Option 3 is the correct option.

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 :

1. 50°, 100°, 40°
2. 50°, 50°, 65°
3. 40°, 80°, 50°
4. 50°, 25°, 78°

Let’s start with a key circle property: the angle at the center subtended by an arc is double the angle at any point on the remaining part of the circle.
So, for angle y:
∴ ∠AOC (y) = 2∠ABC = 2 × 50° = 100°.

Now, let’s look inside the triangle △AOC.
OA and OC are both radii of the same circle, so they must be equal. This means △AOC is an isosceles triangle!
Because of this, the angles opposite these sides are also equal: ∠OAC = ∠OCA = z.

Using the angle sum property for a triangle (all angles add up to 180°):
⇒ ∠OAC + ∠OCA + ∠AOC = 180°
⇒ z + z + 100° = 180°
⇒ 2z = 180° – 100°
⇒ 2z = 80°
⇒ z = \dfrac{80°}{2} = 40°.

Almost done! To find x, we use the rule that a tangent is always perpendicular to the radius at the point of contact.
This tells us that the whole angle ∠OAQ = 90°.
From the diagram, we can see that x is the difference between ∠OAQ and ∠OAC.
⇒ ∠CAQ (x) = ∠OAQ – ∠OAC = 90° – 40° = 50°.

Hence, Option 1 is the correct option.

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 :

1. 30°
2. 45°
3. 75°
4. 105°

To solve this, we’ll use a key circle property: the Alternate Segment Theorem. This theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Applying this rule to our diagram, we find:
* ∠SQP = ∠SPT, which means ∠SQP = 45°.
* ∠SPQ = ∠SQT, which means ∠SPQ = 30°.

Now, let’s look at the triangle △SQP. We know the angles in any triangle add up to 180°.

∠SQP + ∠SPQ + ∠QSP = 180°

Let’s substitute the values we just found:
45° + 30° + ∠QSP = 180°

Combining the known angles, we get:
75° + ∠QSP = 180°

To find ∠QSP (which is our ‘x’), we simply subtract 75° from 180°:
x = 180° – 75° = 105°.

Hence, Option 4 is the correct option.

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

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

Hence, Option 3 is the correct option.

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 :

1. 21 cm
2. 42 cm
3. 21\sqrt{7} cm
4. 2\sqrt{7} cm

Let’s start by noting down what we’re given. The problem states that the height of the cone (h) is equal to its radius (r). We can call this common value ‘a’.

We are also given the volume of the cone, which is 9702 cm³.

Using the formula for the volume of a cone, we can set up our equation. Since both radius and height are ‘a’, we have:

Now, let’s rearrange to solve for ‘a’:

By taking the cube root, we find the value of ‘a’:

This tells us the radius is 21 cm. The question asks for the diameter, which is simply twice the radius.

Diameter = 2 × radius = 2 × 21 = 42 cm.

Hence, Option 2 is the correct option.

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.

1. 24π
2. 32π
3. 48π
4. 64π

Total surface area of semi-hemisphere = 2πr^2

= 2π × 4^2

= 2π × 16

= 32π.

Hence, Option 2 is the correct option.

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 :

1. 1 : 1
2. 3 : 2
3. 2 : 3
4. 2 : 1

Let’s say the radius of our hemisphere is ‘r’.

The formula for the total surface area of one hemisphere is 3πr².
So, for two hemispheres, the combined total surface area is 2 × 3πr², which gives us 6πr².

Now, the total surface area of a sphere with the same radius ‘r’ is 4πr².

We need to find the ratio of the total surface area of the two hemispheres to the total surface area of the sphere. Let’s set that up:

Ratio = 6πr² : 4πr²

The πr² terms cancel out, leaving us with 6 : 4.

Simplifying this by dividing both sides by 2, we get our final answer: 3 : 2.

Hence, Option 2 is the correct option.

Question 36

cosec^2 θ + sec^2 θ is equal to :

• (a) tan^2 θ + cot^2 θ
• (b) cot θ + tan θ
• (c) (cot θ + tan θ)^2
• (d) 1

Let’s simplify the expression cosec^2 θ + sec^2 θ.

First, we can use the Pythagorean identities to replace cosec^2 θ with 1 + cot^2 θ and sec^2 θ with 1 + tan^2 θ.

= (1 + cot^2 θ) + (1 + tan^2 θ)

Now, let’s combine the constant terms:

= cot^2 θ + tan^2 θ + 2

This expression is a perfect square trinomial, in the form of a² + b² + 2ab. So, we can factor it like this:

= (cot θ + tan θ)^2

Hence, Option 3 is the correct option.

Question 37

Given a = 3 sec^2 θ and b = 3 tan^2 θ – 2. The value of (a – b) is :

• (a) 1
• (b) 2
• (c) 3
• (d) 5

To find the value of a - b, we can substitute the given expressions:
a - b = 3 sec^2 θ - (3 tan^2 θ - 2)

First, let’s distribute the negative sign:
= 3 sec^2 θ - 3 tan^2 θ + 2

Now we can factor out the common 3 from the first two terms:
= 3(sec^2 θ - tan^2 θ) + 2

Using the trigonometric identity sec^2 θ - tan^2 θ = 1, this simplifies to:
= 3(1) + 2

= 3 + 2

= 5.

Hence, Option 4 is the correct option.

Question 38

At a certain time of day, the ratio of the height of the pole to the length of its shadow is 1 : \sqrt{3}, then the angle of elevation of the sun at that time of the day is :

• (a) 30°
• (b) 45°
• (c) 60°
• (d) 90°

Let’s use AB to represent the pole and BC for its shadow. The angle of elevation of the Sun will be our angle θ.

We’re told that the ratio of the pole’s height to its shadow’s length is 1 : √3. This means if we let the height of the pole AB be x, then the length of the shadow BC will be √3x.

From the right-angled triangle in the figure, we can use the tangent function:

⇒ tan θ = \dfrac{\text{Opposite}}{\text{Adjacent}} = \dfrac{AB}{BC}

Now, let’s substitute our values for AB and BC:

⇒ tan θ = \dfrac{x}{\sqrt{3}x}

We can cancel out the x from the numerator and denominator:

⇒ tan θ = \dfrac{1}{\sqrt{3}}

We know from our trigonometric tables that the angle whose tangent is \frac{1}{\sqrt{3}} is 30°.

⇒ tan θ = tan 30°

Therefore, the angle of elevation is 30°.

⇒ θ = 30°.

Hence, Option 1 is the correct option.

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

Let’s use the diagram to set up our problem. Point A represents the top of the lighthouse. The ship’s journey starts at point C and, after moving closer for 10 minutes, it reaches point D.

We can write the trigonometric ratios for the two angles of elevation:

From the initial position C, we have:
tan α = \dfrac{AB}{BC}

From the final position D, we have:
tan β = \dfrac{AB}{BD}

Now, let’s compare these. Since the ship is moving towards the lighthouse, the distance from its base gets shorter. This means the initial distance BC is greater than the final distance BD.

When you have a fraction, a larger denominator results in a smaller value. Therefore, \dfrac{AB}{BC} must be smaller than \dfrac{AB}{BD}.

This tells us that tan α < tan β.

For acute angles (like angles of elevation), the tangent function is an increasing function. This means that if tan α < tan β, then it must follow that α < β.

Hence, Option 2 is the correct option.

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.

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.

Let’s start by creating a cumulative frequency table to organize the data.

To find the median class, we first find the position of the median term. With a total frequency (N) of 16, this is the \dfrac{16}{2} = 8th term.
Looking at our cumulative frequency column, the 8th term falls in the class interval 50-60. This is our median class.

Next, the modal class is simply the class with the highest frequency. The highest frequency is 6, which also corresponds to the class 50-60.

Since the median class and the modal class are the same, Assertion (A) is true.

Now, let’s consider Reason (R). It claims that the modal and median classes are always the same for any distribution. This is not true; it just happens to be the case for this specific set of data.
Therefore, Reason (R) is false.

Hence, Option 3 is the correct option.

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.

Let’s check Assertion (A) first. We have 11 data points, and 11 is an odd number. To find the median for an odd number of terms, we use the formula:

Median = \dfrac{n + 1}{2} th term

For n = 11, the position is:
= \dfrac{11 + 1}{2} = \dfrac{12}{2} = 6th term.

The 6th term is indeed the middle number in a set of 11. So, Assertion (A) is true.

Now, let’s look at Reason (R). The numbers are 5, 9, 7, 13, 10, 11, 10. The first step to find the median is always to arrange the numbers in order. In ascending order, we get:
5, 7, 9, 10, 10, 11, 13.

Here, the number of terms (n) is 7, which is also odd. Using the same formula:
Median = \dfrac{n + 1}{2} = \dfrac{7 + 1}{2} = \dfrac{8}{2} = 4th term.

Counting along our sorted list, the 4th term is 10. The Reason states the median is 13, which is incorrect. So, Reason (R) is false.

Hence, Option 3 is the correct option.

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.

To figure out which investment is better, let’s calculate the rate of return, or the income we get for every rupee we invest in each company.

For Company A:
* Nominal Value (N.V.) = ₹ 100
* Market Value (M.V.) = ₹ 120
* Dividend = 7%

The income on one share is calculated on its N.V.:
Income = 7% of ₹ 100 = \dfrac{7}{100} \times 100 = ₹ 7.

To earn this ₹ 7, we must invest the M.V., which is ₹ 120. So, the income per rupee invested is:
Return = \dfrac{\text{Income}}{\text{Investment}} = \dfrac{7}{120} \approx ₹ 0.0583

For Company B:
* Nominal Value (N.V.) = ₹ 1000
* Market Value (M.V.) = ₹ 1620
* Dividend = 8%

The income on one share is:
Income = 8% of ₹ 1000 = \dfrac{8}{100} \times 1000 = ₹ 80.

To earn this ₹ 80, we must invest ₹ 1620. The income per rupee invested is:
Return = \dfrac{\text{Income}}{\text{Investment}} = \dfrac{80}{1620} \approx ₹ 0.0494

Conclusion:
Comparing the returns, we see that Company A (₹ 0.0583 per rupee) offers a higher rate of income than Company B (₹ 0.0494 per rupee).

Therefore, the Assertion and Reason are both true, and the Reason correctly explains the Assertion.

Hence, Option 1 is the correct explanation.

Question 43

Assertion (A) : x^3 + 2x^2 – 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.

Let’s check the polynomial x^3 + 2x^2 - x - 2.

According to the Factor Theorem, (x - a) is a factor of a polynomial f(x) if f(a) = 0. We want to see if (x + 2) is a factor.

To do this, we set x + 2 = 0, which gives us x = -2.

Now, let’s substitute x = -2 into the polynomial:

⇒ (-2)^3 + 2(-2)^2 - (-2) - 2
⇒ -8 + 2(4) + 2 - 2
⇒ -8 + 8 + 2 - 2
⇒ 0

Since we got 0, we’ve shown that (x + 2) is a factor.

Therefore, Option 4 is the correct option.

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.

Let’s break this down!

First, let’s look at Assertion (A). For the point (-2, 8) to be ‘invariant’ under reflection in the line x = -2, it means the point doesn’t move at all. This happens only if the point lies directly on the line of reflection.

The line x = -2 is a vertical line where every single point has an x-coordinate of -2. Our point is (-2, 8), and its x-coordinate is indeed -2. So, the point is on the line! This makes Assertion (A) true.

Now for Reason (R). It says a point with an x-coordinate of 0 is invariant when reflected in the x-axis. A point with an x-coordinate of 0, like (0, y), lies on the y-axis. The rule for reflection in the x-axis is (x, y) → (x, -y). So, a point like (0, 5) would reflect to (0, -5). It definitely moves! So, it’s not invariant under reflection in the x-axis. This makes Reason (R) false.

Since Assertion (A) is true and Reason (R) is false, Option 3 is the correct option.

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 …………… .

1. 2 : 3
2. 3 : 2
3. 1 : 3
4. 1 : 2

Let’s start by looking at the numbers from 1 to 6.

The prime numbers in this set are 2, 3, and 5.
The composite numbers are 4 and 6.

With 6 possible outcomes in total, we can find the probabilities.
The probability of getting a prime number is \dfrac{\text{Number of prime numbers}}{\text{Total outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.

And the probability of getting a composite number is \dfrac{\text{Number of composite numbers}}{\text{Total outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}.

The question asks for the ratio of the probability of getting a composite number to the probability of getting a prime number.
This ratio is \dfrac{1}{3} : \dfrac{1}{2}, which simplifies to 2 : 3.

Hence, Option 1 is the correct option.

Question 46

The product of A = \begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} 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

Let’s start by looking at the orders of the matrices we’re given:
Order of matrix A = 2 × 2
Order of matrix B = 2 × 1

We need to find the order of matrix M. Let’s say its order is a × b.

The problem tells us that AM = B. To figure out the order of M, we need to remember the rule for multiplying matrices. For the product AM to be possible, the number of columns in A must match the number of rows in M.

Since A is a 2 × 2 matrix, M must have 2 rows. So, we know that a = 2.

Now, what about the resulting matrix? The product matrix, AM, will have the same number of rows as A and the same number of columns as M. So, the order of AM is 2 × b.

Since AM = B, their orders must be identical. We know B is a 2 × 1 matrix. So, the order of AM must also be 2 × 1.

Comparing the order of AM (2 × b) to the order of B (2 × 1), we can see that b = 1.

So, the order of matrix M (a × b) is 2 × 1.

Hence, Option 2 is the correct option.

Question 47

Given, a~1, a~2, a~3, ….. and b~1, b~2, b~3, ….. are real numbers such that a~1 – b~1 = a~2 – b~2 = a~3 – b~3 = ……… are all equal.

a~1 – b~1, a~2 – b~2, a~3 – b~3, …….. forms a ……… progression.

• (a) Geometric (r = 1)
• (b) Arithmetic (d = 1)
• (c) Geometric (r < 1)
• (d) Arithmetic (d = 0)

We are given that:

a~1 – b~1 = a~2 – b~2 = a~3 – b~3.

This shows that the sequence a~1 – b~1, a~2 – b~2, a~3 – b~3, …….. is an arithmetic progression with a common difference of 0, as each term is the same.

Hence, Option 4 is the correct option.

Question 48

Locus of a moving point is …………… if it moves such that it keeps a fixed distance from a fixed point.

• (a) Circle
• (b) Line
• (c) Angle
• (d) Line segment

Locus of a moving point is circle if it moves such that it keeps a fixed distance from a fixed point.

Hence, Option 1 is the correct option.

Question 49

The point of concurrence of the angle bisectors of a triangle is called the …………… of the triangle.

• (a) centroid
• (b) incenter
• (c) circumcenter
• (d) orthocenter

The point of concurrence of the angle bisectors of a triangle is called the incenter of the triangle.

Hence, Option 2 is the correct option.