Multiple-Choice Questions

For retailer,

C.P. = ₹ 2500

G.S.T. = 12%

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

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

Given,

Retailer sells the article to the consumer in the same state after marking up the price by 20%.

S.P. = ₹ 2500 + 20%

= ₹ 2500 + $\frac{20}{100} \times 2500$

= ₹ 2500 + ₹ 500

= ₹ 3000.

C.G.S.T. = 6%

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

Tax liability = C.G.S.T. charged - C.G.S.T. given

= ₹ 180 - ₹ 150

= ₹ 30.

Hence, Option 3 is the correct option.

Since, GST rate is 7%, then SGST = $\frac{7}{2}$ = 3.5%.

M.P. of electrical fan = ₹ 800.

SGST = 3.5% of ₹ 800

= $\frac{3.5}{100} \times 800$

= ₹ 28.

Hence, Option 2 is the correct option.

In a recurring deposit account,

By formula,

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

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

Hence, Option 4 is the correct option.

In first case :

P = ₹ 1000

r = 5%

n = 12 months

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

$$\begin{aligned}= 1000 \times \frac{12 \times (12 + 1)}{2 \times 12} \times \frac{5}{100} \\[1em] = 1000 \times \frac{12 \times 13}{2 \times 12} \times \frac{1}{20} \\[1em] = 25 \times 13 \\[1em] = ₹ 325.\end{aligned}$$

In second case :

P = ₹ x (Let)

r = 4%

n = 12 months

Interest = ₹ 325

$$\begin{aligned}\therefore 325 = x \times \frac{12 \times 13}{2 \times 12} \times \frac{4}{100} \\[1em] \Rightarrow 325 = x \times \frac{13}{2} \times \frac{1}{25}\\[1em] \Rightarrow x = \frac{325 \times 25 \times 2}{13} \\[1em] \Rightarrow x = \frac{16250}{13} = \text{₹ 1250.}\end{aligned}$$

Hence, Option 2 is the correct option.

For Mr. Das,

Face value of each share = ₹ 100

Market value of each share = ₹ 60

Dividend per share = 12% of ₹ 100 = ₹ 12.

Rate of return = $\frac{\text{Dividend per share}}{\text{M.V. of each share}} \times 100 = \frac{12}{60} \times 100$ = 20%.

For Mr. Singh,

Face value of each share = ₹ 50

Market value of each share = ₹ 40

Dividend per share = 16% of ₹ 50 = ₹ 8.

Rate of return = $\frac{\text{Dividend per share}}{\text{M.V. of each share}} \times 100 = \frac{8}{40} \times 100$ = 20%.

∴ Both Mr. Das and Mr. Singh have the same rate of return of 20%.

Hence, Option 4 is the correct option.

Let ₹ P be the price per share.

Amit brought 10 shares, so total investment = ₹ 10P

Dividend = 7.5%

Dividend per share = $\frac{7.5}{100} \times 100$ = ₹ 7.5

Total dividend = ₹ 7.5 × 10 = ₹ 75.

Rate of return = 10%

By formula,

Rate of return × Investment = Dividend

$$\begin{aligned}\Rightarrow 10% \times 10P = 75 \\[1em] \Rightarrow \frac{10}{100} \times 10P = 75 \\[1em] \Rightarrow \frac{100P}{100} = 75 \\[1em] \Rightarrow P = ₹ 75.\end{aligned}$$

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

Hence, Option 2 is the correct option.

Solving,

⇒ -3 ≤ -4x + 5

⇒ 4x ≤ 5 + 3

⇒ 4x ≤ 8

⇒ x ≤ $\frac{8}{4}$

⇒ x ≤ 2.

Since, x ≤ 2 and x ∈ W.

∴ Solution set = {0, 1, 2}.

Hence, Option 3 is the correct option.

Solving,

⇒ -4x > 8y

⇒ 4x < -8y

⇒ x < $-\frac{8y}{4}$

⇒ x < -2y.

Hence, Option 3 is the correct option.

Given,

Equation : 2x² - 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.

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 = $-\frac{6}{3}$ = -2.

Hence, Option 2 is the correct option.

x = 233.356

On rounding off to two significant figures

x = 230.

Hence, Option 4 is the correct option.

By formula,

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

⇒ 30 = $\frac{1}{2} \times BC \times AB$

⇒ 30 = $\frac{1}{2} \times y \times x$

⇒ 30 = $\frac{xy}{2}$

⇒ 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 = $\sqrt{169}$ = 13 cm.

Hence, Option 3 is the correct option.

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

$$\begin{aligned}\therefore \frac{p}{q} = \frac{q}{r} \\[1em] \Rightarrow q^2 = pr \text{ ........(1)}\end{aligned}$$

Solving,

p : r = p² : q²

$$\begin{aligned}\Rightarrow \frac{p}{r} = \frac{p^2}{q^2} \\[1em] \Rightarrow q^2 = \frac{p^2 \times r}{p} \\[1em] \Rightarrow q^2 = pr \text{ .........(2)}\end{aligned}$$

Since, equation (1) and (2) are equal.

Hence, Option 4 is the correct option.

Given,

Ratio of diameter to height of a Borosil cylindrical glass is 3 : 5.

$$\begin{aligned}\Rightarrow \frac{d}{h} = \frac{3}{5} \\[1em] \Rightarrow \frac{6}{h} = \frac{3}{5} \\[1em] \Rightarrow h = \frac{6 \times 5}{3} \\[1em] \Rightarrow h = 10 \text{ cm}.\end{aligned}$$

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

By formula,

Curved surface area of glass = 2πrh

= 2π × 3 × 10

= 60π cm².

Hence, Option 2 is the correct option.

Given,

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

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

⇒ 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.

Hence, Option 4 is the correct option.

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, on substituting -4 in polynomial, we get remainder = 0.

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

Hence, Option 3 is the correct option.

Order of matrix A = 1 × 2

Order of matrix B = 2 × 1

Since, no. of columns in A is equal to the no. of rows in B and no. of columns in B is equal to the no. of rows in A.

∴ AB and BA are possible.

$$AB = \begin{bmatrix*}[r] a \times c + b \times d \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] ac + bd \end{bmatrix*}. \\[1em] BA = \begin{bmatrix*}[r] c \times a & c \times b \\ d \times a & d \times b \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] ca & cb \\ da & db \end{bmatrix*}.$$

∴ AB ≠ BA.

Hence, Option 3 is the correct option.

Given, $$A^2 = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}$$.

$$\therefore \begin{bmatrix*}[r] 6 & 9 \\ -4 & k \end{bmatrix*}\begin{bmatrix*}[r] 6 & 9 \\ -4 & k \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 \times 6 + 9 \times (-4) & 6 \times 9 + 9 \times k \\ -4 \times 6 + k \times (-4) & (-4) \times 9 + k \times k \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 36 + (-36) & 54 + 9k \\ -24 - 4k & -36 + k^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & 54 + 9k \\ -24 - 4k & -36 + k^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow 54 + 9k = 0 \\[1em] \Rightarrow 9k = -54 \\[1em] \Rightarrow k = -\frac{54}{9} = -6.$$

Hence, Option 2 is the correct option.

Given,

Sum of n terms of an arithmetic progression S_n = n² - n.

S₁ = 1² - 1 = 0,

S₂ = 2² - 2 = 4 - 2 = 2,

S₃ = 3² - 3 = 9 - 3 = 6.

Sum upto first term = First term = 0.

Given, sum upto 2 terms = 2 and first term = 0, second term = 2.

Sum upto third term = 6

∴ First term + Second term + Third term = 6

⇒ 0 + 2 + Third term = 6

⇒ Third term = 6 - 2 = 4.

Hence, Option 2 is the correct option.

In the series,

2, 0, 4, 0, 8, 0

$\frac{0}{2} \ne \frac{4}{0}$

∴ 2, 0, 4, 0, 8, 0 is not a G.P.

Hence, Option 4 is the correct option.

By formula,

Centroid of triangle = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

$$\begin{aligned}\therefore (5, -5) = \Big(\frac{3 + 2 + x}{3}, \frac{(-3) + (-6) + y}{3}\Big) \\[1em] \Rightarrow (5, -5) = \Big(\frac{x + 5}{3}, \frac{y - 9}{3}\Big) \\[1em] \Rightarrow \frac{x + 5}{3} = 5 \text{ and } \frac{y - 9}{3} = -5 \\[1em] \Rightarrow x + 5 = 15 \text{ and } y - 9 = -15 \\[1em] \Rightarrow x = 15 - 5 = 10 \text{ and } y = -15 + 9 = -6.\end{aligned}$$

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.

$$\begin{aligned}D = \Big(\frac{2 + 10}{2}, \frac{(-6) + (-6)}{2}\Big) \\[1em] = \Big(\frac{12}{2}, \frac{-12}{2}\Big) \\[1em] = (6, -6).\end{aligned}$$

Hence, Option 3 is the correct option.

From graph,

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

Given,

P is the mid-point of AB.

P = $\Big(\frac{4 + 0}{2}, \frac{0 + 2}{2}\Big) = \Big(\frac{4}{2}, \frac{2}{2}\Big)$ = (2, 1).

By two-point form,

⇒ y - y₁ = $\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

⇒ y - 1 = $\frac{1 - 0}{2 - 0}(x - 2)$

⇒ y - 1 = $\frac{1}{2}(x - 2)$

⇒ 2(y - 1) = x - 2

⇒ 2y - 2 = x - 2

⇒ 2y = x - 2 + 2

⇒ 2y = x.

Hence, Option 2 is the correct option.

Slope of line l₁ = tan 45° = 1.

We know that,

Slope of parallel lines are equal.

Slope of line l₂ = Slope of line l₁ = 1.

We know that,

Product of slope of perpendicular lines is equal to -1.

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

⇒ 1 × Slope of line l₃ = -1

⇒ Slope of line l₃ = -1.

Hence, Option 3 is the correct option.

Substituting x = 0 in first equation,

⇒ 3(0) + 3y = 6

⇒ 3y = 6

⇒ y = $\frac{6}{3}$

⇒ y = 2.

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

Substituting y = 0 in first equation,

⇒ 3x + 3(0) = 6

⇒ 3x = 6

⇒ x = $\frac{6}{3}$

⇒ x = 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 form the origin.

Hence, Option 1 is the correct option.

In △ APQ and △ ABC,

⇒ ∠PAQ = ∠BAC (Common angle)

⇒ ∠APQ = ∠ABC (Corresponding angles are equal)

∴ △ APQ ~ △ ABC (By A.A. axiom)

From figure,

Let AP = x km.

We know that,

Corresponding sides of similar triangle are proportional.

$$\begin{aligned}\Rightarrow \frac{AP}{AB} = \frac{PQ}{BC} \\[1em] \Rightarrow \frac{x}{AP + PB} = \frac{20}{50} \\[1em] \Rightarrow \frac{x}{x + 18} = \frac{2}{5} \\[1em] \Rightarrow 5x = 2(x + 18) \\[1em] \Rightarrow 5x = 2x + 36 \\[1em] \Rightarrow 5x - 2x = 36 \\[1em] \Rightarrow 3x = 36 \\[1em] \Rightarrow x = \frac{36}{3} \\[1em] \Rightarrow x = 12.\end{aligned}$$

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

Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{AB}{90} = \frac{30}{90} = \frac{1}{3} \text{ hr} = \frac{1}{3} \times 60$ = 20 minutes.

Hence, Option 4 is the correct option.

Given,

⇒ ∠C = ∠F

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

i.e., we don't know if $\frac{AC}{DF}$ and $\frac{BC}{EF}$ are equal or not equal.

∴ Similarity of given triangles cannot be determined.

Hence, Option 4 is the correct option.

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.

∴ ∠PQR = ∠TSR

∴ △ PQR ~ △ TSR (By A.A. axiom)

Hence, Option 3 is the correct option.

Scale factor of a picture and the actual height = 20 cm : 1.6 m = 20 cm : 160 cm

$$\begin{aligned}\therefore \frac{\text{Height in picture}}{\text{Actual height}} = \frac{\text{20 cm}}{\text{160 cm}} \\[1em] \Rightarrow \frac{18}{\text{Actual height}} = \frac{\text{20 cm}}{\text{160 cm}} \\[1em] \Rightarrow \text{Actual height} = \frac{18 \times 160}{20} = \text{144 cm} = 1.44 \text{ m}.\end{aligned}$$

Hence, Option 4 is the correct option.

We know that,

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°.

Hence, Option 3 is the correct option.

We know that,

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)

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ z + z + 100° = 180°

⇒ 2z = 180° - 100°

⇒ 2z = 80°

⇒ z = $\frac{80°}{2}$ = 40°.

We know that,

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°.

Hence, Option 1 is the correct option.

We know that,

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°

⇒ 45° + 30° + ∠QSP = 180°

⇒ 75° + ∠QSP = 180°

⇒ ∠QSP (x) = 180° - 75° = 105°.

Hence, Option 4 is the correct option.

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

Hence, Option 3 is the correct option.

Given,

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

Given,

Volume = 9702 cm³

$$\begin{aligned}\therefore \frac{1}{3}πr^2h = 9702 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times a^2 \times a = 9702 \\[1em] \Rightarrow a^3 = \frac{9702 \times 7 \times 3}{22} \\[1em] \Rightarrow a^3 = 441 \times 21 \\[1em] \Rightarrow a^3 = 9261 \\[1em] \Rightarrow a = \sqrt[3]{9261} = 21 \text{ cm}.\end{aligned}$$

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

Hence, Option 2 is the correct option.

Total surface area of semi-hemisphere = 2πr²

= 2π × 4²

= 2π × 16

= 32π.

Hence, Option 2 is the correct option.

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.

Hence, Option 2 is the correct option.

Solving,

    cosec² θ + sec² θ

= 1 + cot² θ + 1 + tan² θ

= cot² θ + tan² θ + 2

= (cot θ + tan θ)²

Hence, Option 3 is the correct option.

    a - b = 3 sec² θ - (3 tan² θ - 2)

= 3 sec² θ - 3 tan² θ + 2

= 3(sec² θ - tan² θ) + 2

= 3(1) + 2

= 3 + 2

= 5.

Hence, Option 4 is the correct option.

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

From figure,

⇒ tan θ = $\frac{AB}{BC}$

⇒ tan θ = $\frac{x}{\sqrt{3}x}$

⇒ tan θ = $\frac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, Option 1 is the correct option.

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 = $\frac{AB}{BC}$

tan β = $\frac{AB}{BD}$

Since, BC is greater than BD.

∴ tan α < tan β

⇒ α < β.

Hence, Option 2 is the correct option.

Cumulative frequency distribution table :

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

Median = $\frac{16}{2}$ = 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.

∴ Assertion (A) is true.

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

∴ Reason (R) is false.

Hence, Option 3 is the correct option.

For 11 arrayed data.

Median = $\frac{n + 1}{2}$ th term

= $\frac{11 + 1}{2}$

= 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 = $\frac{n + 1}{2} = \frac{7 + 1}{2} = \frac{8}{2}$ = 4th term = 10.

∴ Reason (R) is false.

Hence, Option 3 is the correct option.

In company A,

N.V. = ₹ 100

M.V. = ₹ 120

Dividend = 7% = $\frac{7}{100} \times 100$ = ₹ 7

∴ Investment = ₹ 120 and income = ₹ 7.

Income on ₹ 1 = $\frac{7}{120}$ = ₹ 0.0583

In company B,

N.V. = ₹ 1000

M.V. = ₹ 1620

Dividend = 8% = $\frac{8}{100} \times 1000$ = ₹ 80

∴ Investment = ₹ 1620 and income = ₹ 80.

Income on ₹ 1 = $\frac{80}{1620}$ = ₹ 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.

Hence, Option 1 is the correct 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.

Hence, Option 4 is the correct option.

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.

Hence, Option 3 is the correct option.

From numbers 1 to 6

Prime numbers = 2, 3, 5.

Composite numbers = 4, 6.

Probability of getting a prime number = $\frac{\text{No. of prime numbers}}{\text{No. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$.

Probability of getting a composite number

= $\frac{\text{No. of composite numbers}}{\text{No. of possible outcomes}} = \frac{2}{6} = \frac{1}{3}$.

Probability of getting a composite number to probability of getting a prime number = $\frac{1}{3} : \frac{1}{2}$ = 2 : 3.

Hence, Option 1 is the correct option.

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_{2 × 2} × M_{a × b} = B_{2 × 1}

⇒ a = 2 and b = 1.

Order of matrix M = 2 × 1.

Hence, Option 2 is the correct option.

Since,

a₁ - b₁ = a₂ - b₂ = a₃ - b₃.

∴ a₁ - b₁, a₂ - b₂, a₃ - b₃, ........ forms a arithmetic progression with common difference (d = 0).

Hence, Option 4 is the correct option.

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.

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.