Official Paper

CDS-I 2025 (Elementary Mathematics) Official Paper (Held On: 13 Apr, 2025) (Previous Year Paper)

100 questions · 120 minutes · with answers · free

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Elementary Mathematics (100 questions)

A real number M is squared to give the value N. What is the minimum value of (M + N) ?

((a))
-0·25
((b))
-0·50
((c))
0
((d))
0·25

Show Answer

Answer: ((a))

-0·25

Given:

A real number M is squared to give the value N. Find the minimum value of (M + N).

Formula used:

N = M²

Minimize (M + N) = M + M²

Calculation:

Let f(M) = M + M²

To find the minimum value, take the derivative and set it to zero:

f'(M) = 1 + 2M

⇒ 1 + 2M = 0

⇒ M = -0.5

Substitute M = -0.5 into f(M):

f(M) = M + M²

⇒ f(-0.5) = (-0.5) + (-0.5)²

⇒ f(-0.5) = -0.5 + 0.25

⇒ f(-0.5) = -0.25

∴ The minimum value of (M + N) is -0.25, and the correct answer is option 1.

What is the sum of all 3-digit numbers that give a remainder of 5 when they are divided by 50 ?

((a))
9005
((b))
9540
((c))
9600
((d))
9640

Show Answer

Answer: ((b))

9540

Given:

We need to find the sum of all 3-digit numbers that give a remainder of 5 when divided by 50.

Formula used:

Sum of arithmetic progression (AP): S_n = (n/2) × (a + l)

Where,

a = First term, l = Last term, n = Number of terms

Calculation:

The 3-digit numbers that give a remainder of 5 when divided by 50 are: 105, 155, 205, ..., 955.

a = 105, l = 955

Common difference (d) = 50

n = (l - a)/d + 1

⇒ n = (955 - 105)/50 + 1

⇒ n = 850/50 + 1

⇒ n = 17 + 1

⇒ n = 18

Sum of all terms:

S_n = (n/2) × (a + l)

⇒ S₁₈ = (18/2) × (105 + 955)

⇒ S₁₈ = 9 × 1060

⇒ S₁₈ = 9540

∴ The correct answer is option (2).

If the average of 64, 69, 72, 75, x lies between 62 and 76 (excluding 62 and 76), then what is the number of possible integer values of x ?

((a))
68
((b))
69
((c))
70
((d))
71

Show Answer

Answer: ((b))

Given:

The average of 64, 69, 72, 75, x lies between 62 and 76 (excluding 62 and 76).

Formula used:

Average = $\dfrac{\text{Sum of all numbers}}{\text{Total numbers}}$

Calculation:

Let the sum of the given numbers = 64 + 69 + 72 + 75 + x = 280 + x

The average = $\dfrac{280 + x}{5}$

Since the average lies between 62 and 76:

⇒ $62 < \dfrac{280 + x}{5} < 76$

⇒ 310 < 280 + x < 380

⇒ 30 < x < 100

The possible integer values of x are from 31 to 99 (both inclusive).

Total number of integers = 99 - 31 + 1 = 69

∴ The correct answer is option (2).

Let x, y, z be variables such that (x + y + z) = k, where k is a constant. If (x + z – y) × (x – z + y) is proportional to yz, then (y + z – x) is proportional to :

((a))
x
((b))
y
((c))
yz
((d))
xz

Show Answer

Answer: ((c))

Given:

(x + y + z) = k

(x + z – y) × (x – z + y) ∝ yz

Formula used:

Algebraic manipulation and properties of proportionality.

From (x + y + z) = k, we can derive:

x + z – y = (x + y + z) – 2y = k – 2y

x – z + y = (x + y + z) – 2z = k – 2z

Calculations:

Given that (x + z – y) × (x – z + y) ∝ yz

⇒ (k – 2y) × (k – 2z) = C × yz (where C is the constant of proportionality)

⇒ k² – 2kz – 2ky + 4yz = C × yz

⇒ k² – 2k(z + y) + 4yz = C × yz

⇒ k² – 2k(y + z) = C × yz – 4yz

⇒ k² – 2k(y + z) = (C – 4)yz

Now, let's rearrange to find what (y + z – x) is proportional to.

From (x + y + z) = k, we have y + z = k – x.

Substitute (y + z) in the equation k² – 2k(y + z) = (C – 4)yz:

⇒ k² – 2k(k – x) = (C – 4)yz

⇒ k² – 2k² + 2kx = (C – 4)yz

⇒ –k² + 2kx = (C – 4)yz

⇒ k(2x – k) = (C – 4)yz

We are interested in (y + z – x).

We know (y + z – x) = (k – x) – x = k – 2x

From k(2x – k) = (C – 4)yz, we can write:

⇒ –k(k – 2x) = (C – 4)yz

⇒ k – 2x = $\dfrac{(C – 4)yz}{-k}$

⇒ k – 2x = $\dfrac{(4 – C)yz}{k}$

Since k, C are constants, $\dfrac{(4 – C)}{k}$ is also a constant.

Therefore, (k – 2x) is proportional to yz.

Since (y + z – x) = k – 2x,

∴ (y + z – x) is proportional to yz.

Let p be the remainder when 7⁸⁴ is divided by 342 and q be the remainder when 7⁸⁴ is divided by 344. What is (p – q) equal to ?

((a))
0
((b))
1
((c))
2
((d))
6

Show Answer

Answer: ((a))

Given:

p = remainder when 7⁸⁴ is divided by 342

q = remainder when 7⁸⁴ is divided by 344

Formula used:

Modular arithmetic and properties of remainders.

If a ≡ b (mod m), then aⁿ ≡ bⁿ (mod m)

aⁿ - bⁿ is divisible by a - b

Calculations:

For p:

We need to find the remainder when 7⁸⁴ is divided by 342.

Notice that 342 = 7³ - 1 (since 7³ = 343).

We know that aⁿ - bⁿ is divisible by a - b.

Consider 7³ ≡ 1 (mod 342) because 7³ - 1 = 343 - 1 = 342, which is divisible by 342.

Now, we have 7⁸⁴ = (7³)²⁸.

⇒ 7⁸⁴ ≡ 1²⁸ (mod 342)

⇒ 7⁸⁴ ≡ 1 (mod 342)

So, p = 1.

For q:

We need to find the remainder when 7⁸⁴ is divided by 344.

Notice that 344 = 7³ + 1 (since 7³ = 343).

We know that aⁿ + bⁿ is divisible by a + b when n is odd.

We also know that aⁿ - bⁿ is divisible by a + b when n is even.

Consider 7³ ≡ -1 (mod 344) because 7³ + 1 = 343 + 1 = 344, which is divisible by 344.

Now, we have 7⁸⁴ = (7³)²⁸.

⇒ 7⁸⁴ ≡ (-1)²⁸ (mod 344)

⇒ 7⁸⁴ ≡ 1 (mod 344)

So, q = 1.

Value of (p – q):

⇒ p – q = 1 – 1

⇒ p – q = 0

∴ (p – q) is equal to 0.

N is the smallest 5-digit number which when divided by 2, 2², 2³, 2⁴, ..., 2ⁿ leaves a remainder 1. What is the value of n ?

((a))
12
((b))
13
((c))
14
((d))
15

Show Answer

Answer: ((c))

Given:

N is the smallest 5-digit number which leaves a remainder of 1 when divided by 2, 2², 2³, 2⁴, ..., 2ⁿ

Formula used:

N ≡ 1 (mod 2^k) for all k = 1 to n

⇒ N - 1 is divisible by LCM of all 2^k for k = 1 to n

LCM(2, 4, 8, ..., 2ⁿ) = 2ⁿ

Calculations:

N - 1 = multiple of 2ⁿ

N = 2ⁿ × k + 1

We want the smallest N ≥ 10000

Try values of n:

n = 13 ⇒ 2¹³ = 8192

⇒ N = 8192 × 2 + 1 = 16385

n = 14 ⇒ 2¹⁴ = 16384

⇒ N = 16384 × 1 + 1 = 16385

n = 14 follows the condition.

∴ The value of n is 14.

What is the minimum value of p for which $\frac{1}{532900} + \frac{p^{2}}{266450} + \frac{p^{4}}{532900}$ is an integer ?

((a))
729
((b))
243
((c))
27
((d))
1

Show Answer

Answer: ((c))

Given:

The expression is:

$\frac{1}{532900} + \frac{p^{2}}{266450} + \frac{p^{4}}{532900}$

Calculation:

$\frac{1}{532900} + \frac{p^{2}}{266450} + \frac{p^{4}}{532900}$

⇒ $\frac{1}{532900} + \frac{p^2}{266450} + \frac{p^4}{532900} = \frac{1}{532900} + \frac{2p^2}{532900} + \frac{p^4}{532900}$

⇒ $\frac{1 + 2p^2 + p^4}{532900}$

⇒ $\frac{(p^2 + 1)^2}{(730)^2}$

Since both numerator and denominator both are perfect square and the result is an integer, minimum possible value of expression is 1

So. p² + 1 = 730

p2 = 730 - 1 = 729

p = 27

The minimum value of p for which the expression is an integer is 27.

If α and β are the roots of the equation x+a+b= $\frac{abx}{ab+ax+bx} $ ,then what is (αβ + α + β) equal to ?

((a))
ab + a + b
((b))
ab – a – b
((c))
a + b – ab
((d))
– (ab + a + b)

Show Answer

Answer: ((b))

ab – a – b

Given:

Equation: x + a + b = $(\dfrac{abx}{ab + ax + bx})$

Roots of the equation are α and β.

Formula used:

For a quadratic equation of the form Ax2 + Bx + C = 0, with roots α and β:

Sum of roots: α + β = $-\dfrac{B}{A}$

Product of roots: αβ = $\dfrac{C}{A}$

Calculations:

First, rearrange the given equation into the standard quadratic form Ax2 + Bx + C = 0.

x + a + b = $(\dfrac{abx}{ab + ax + bx})$

Multiply both sides by (ab + ax + bx):

⇒ (x + a + b)(ab + ax + bx) = abx

⇒ x(ab) + x(ax) + x(bx) + a(ab) + a(ax) + a(bx) + b(ab) + b(ax) + b(bx) = abx

⇒ abx + ax² + bx² + a²b + a²x + abx + ab² + abx + b²x = abx

⇒ ax² + bx² + abx + a²x + abx + abx + b²x − abx + a²b + ab² = 0

⇒ (a + b)x² + (a²x + b²x + 2abx) + (a²b + ab²) = 0

⇒ (a + b)x² + (a² + b² + 2ab)x + ab(a + b) = 0

Recognize that (a² + b² + 2ab) = (a + b)²:

⇒ (a + b)x² + (a + b)²x + ab(a + b) = 0

⇒ x² + (a + b)x + ab = 0

Sum and product of the roots (α and β):

Sum of roots: α + β = $(-\dfrac{B}{A}) = -\dfrac{(a+b)}{1}$ = −(a + b)

Product of roots: αβ = $\dfrac{C}{A} = \dfrac{ab}{1}$ = ab

Finally, calculate (αβ + α + β):

⇒ (αβ + α + β) = ab + (−(a + b))

⇒ (αβ + α + β) = ab − a − b

∴ The correct answer is option 2.

Consider a 2-digit number N. Let P be the product of the digits of the number. If P is added to square of the digit in the tens place of N, we get 84. If P is added to the square of the digit in the unit place of N, we get 60. What is the value of P + N ?

((a))
100
((b))
110
((c))
115
((d))
120

Show Answer

Answer: ((b))

110

Given:

Let the 2-digit number be N = 10×a + b, where:

a = digit in tens place

b = digit in units place

P = a × b

Formula used:

Condition 1: P + a² = 84

Condition 2: P + b² = 60

Calculations:

From (1): P = 84 − a²

From (2): P = 60 − b²

⇒ 84 − a² = 60 − b²

⇒ 24 = a² − b²

⇒ (a − b)(a + b) = 24

Try factor pairs of 24: (1,24), (2,12), (3,8), (4,6)

Try a − b = 2 and a + b = 12

⇒ Add: 2a = 14 ⇒ a = 7, b = 5

Check P = a × b = 7 × 5 = 35

P + a² = 35 + 49 = 84

P + b² = 35 + 25 = 60

N = 10×7 + 5 = 75

P = 35

P + N = 35 + 75 = 110

∴ The correct answer is: 110

A mixture of 100 L contains kerosene and turpentine oil in the ratio 3 : 2. What is the minimum quantity of kerosene in litres (whole number) that should be mixed in the mixture so that the resulting mixture has 20% of kerosene ?

((a))
10 L
((b))
20 L
((c))
25 L
((d))
Not possible

Show Answer

Answer: ((d))

Not possible

Given:

The initial mixture contains kerosene and turpentine oil in the ratio 3:2.

Total mixture = 100 L

Kerosene = (3/5) × 100 = 60 L

Turpentine oil = (2/5) × 100 = 40 L

Final kerosene percentage = 20%

Formula used:

Final kerosene in the mixture = Total mixture × (Final kerosene percentage / 100)

Additional kerosene to be added = Final kerosene - Initial kerosene

Calculation:

Let the additional kerosene to be added = x L.

Total mixture after adding kerosene = 100 + x L

Kerosene in the new mixture = 60 + x L

For 20% kerosene in the final mixture:

⇒ (60 + x) / (100 + x) = 20 / 100

⇒ (60 + x) = (20 / 100) × (100 + x)

⇒ 60 + x = 20 × (100 + x) / 100

⇒ 60 + x = (2000 + 20x) / 100

⇒ 100(60 + x) = 2000 + 20x

⇒ 6000 + 100x = 2000 + 20x

⇒ 6000 - 2000 = 20x - 100x

⇒ 4000 = -80x

⇒ x = -4000 / 80

⇒ x = -50

∴ It is not possible to achieve 20% kerosene in the final mixture. The correct answer is option (4).

A lamp is kept on a vertical pole. The height of the top of the lamp above the ground is $\frac{5\sqrt{3}}{2} \text{ m.}$ The perpendicular distances of the bottom of the pole from two adjacent walls meeting perpendicularly are 0.7 m and 2.4 m. What is the distance of the top of the lamp from the corner point of the walls on the ground?

((a))
3 m
((b))
5 m
((c))
6 m
((d))
7 m

Show Answer

Answer: ((b))

5 m

Given:

Height of pole = PQ = $\dfrac{5\sqrt{3}}{2}$ m.

BQ = 0.7m

CQ = 2.4m

We need to find length AP.

Formula used:

Distance from the corner point = $\sqrt{(x^2 + y^2 + h^2)}$

Where, x = 0.7 m, y = 2.4 m, and h = height of the lamp = $\dfrac{5\sqrt{3}}{2}$ m.

Calculation:

We can see AQ² = BQ² + AB² and AB = CQ (Shown in fig.)

AP² = AQ² + PQ²

AP² = BQ2 + AB2 + PQ2

AP = $\sqrt{BQ^2 + AB^2 + PQ^2}$

The distance of the top of the lamp from the corner point (AP) is:

⇒ Distance = $\sqrt{(0.7^2 + 2.4^2 + \left(\dfrac{5\sqrt{3}}{2}\right)^2)}$

⇒ Distance = $\sqrt{(0.49 + 5.76 + \dfrac{75}{4})}$

⇒ Distance = $\sqrt{(0.49 + 5.76 + 18.75)}$

⇒ Distance = $\sqrt{25}$

⇒ Distance = 5 m.

∴ The distance of the top of the lamp from the corner point is 5 meters.

C is the centre of a circle of radius 20 cm. AB is a chord of length 32 cm. E is a point on AB such that CE = 13 cm. What is AE × EB equal to?

((a))
231 square cm
((b))
256 square cm
((c))
272 square cm
((d))
297 square cm

Show Answer

Answer: ((a))

231 square cm

Given:

C is the center of a circle with radius = 20 cm.

AB is a chord of length = 32 cm.

CE = 13 cm, E is a point on AB.

We need to find AE × EB.

Formula used:

In a circle, for any chord AB and point E on AB:

AE × EB = (Radius² - Distance from center to chord²)

Calculation:

Radius = 20 cm, so Radius² = 20² = 400

Distance from center to chord (CE) = 13 cm, so CE² = 13² = 169

⇒ AE × EB = Radius² - CE²

⇒ AE × EB = 400 - 169

⇒ AE × EB = 231

∴ The correct answer is option (1).

The inside of a bowl is part of a sphere. When water is put into the bowl to a depth d, the water surface becomes a circle of radius 2d. What is the radius of the sphere?

((a))
2.5d
((b))
2.75d
((c))
3d
((d))
3.25d

Show Answer

Answer: ((a))

2.5d

Given:

The inside of the bowl is part of a sphere.

Depth of water (d) = d

Radius of the water surface = 2d

Formula used:

Using the geometry of spheres, the radius of the sphere (R) is given by:

R = $\frac{(r^2 + d^2)}{2d}$

Where r = radius of the water surface = 2d

Calculation:

R = $\frac{(r^2 + d^2)}{2d}$

⇒ R = $\frac{((2d)^2 + d^2)}{2d}$

⇒ R = $\frac{(4d^2 + d^2)}{2d}$

⇒ R = $\frac{5d^2}{2d}$

⇒ R = 2.5d

∴ The correct answer is option (1).

Alternate Method

Given:

Depth of water = d

Radius of water surface = 2d

Let Radius of sphere = R

Formula used:

By geometry of sphere cross-section:

$(R - d)^2 + (2d)^2 = R^2$

Calculations:

⇒ (R - d)² + (2d)² = R²

⇒ (R² - 2Rd + d²) + 4d² = R²

⇒ R² - 2Rd + 5d² = R²

⇒ -2Rd + 5d² = 0

⇒ 2Rd = 5d²

⇒ R = (5d / 2) = 2.5d

∴ Radius of sphere = 2.5d

In a triangle ABC, AB = 2 cm, BC = 4 cm and AC = 3 cm. The bisector of angle A meets BC at D and the bisector of angle B meets AD at E. What is AE : ED equal to?

((a))
5 : 4
((b))
5 : 3
((c))
4 : 3
((d))
3 : 2

Show Answer

Answer: ((a))

5 : 4

Given:

AB = 2 cm, BC = 4 cm, AC = 3 cm

Formula used:

Angle bisector theorem: If a bisector of an angle divides the opposite side of a triangle, the ratio of the divided parts equals the ratio of the other two sides.

BD/DC = AB/AC

Calculations:

In Δ ABC,

AD is the bisector of ∠A

So, BD/DC = AB/AC = 2/3

Since, BC = 4 cm

So, BD = 2/5 × 4 = 8/5 cm

Now, In Δ ABD

BE is the bisector of ∠B

So, BD/AB = ED/AE

⇒ (8/5) / 2 = ED/AE

⇒ 4/5 = ED/AE

⇒ 5/4 = AE / ED

So, AE : ED = 5 : 4

Hence, the correct answer is option (1).

In a triangle ABC, the bisector of angle A cuts BC at D. If AB + AC = 10 cm and BD : DC = 3 : 1, then what is the length of AC?

((a))
2.5 cm
((b))
6 cm
((c))
7.5 cm
((d))
8 cm

Show Answer

Answer: ((a))

2.5 cm

Given:

AB + AC = 10 cm

BD : DC = 3 : 1

Formula used:

Angle bisector theorem: If a bisector of an angle divides the opposite side of a triangle, the ratio of the divided parts equals the ratio of the other two sides.

BD/DC = AB/AC

Calculation:

Since AD is the angle bisector of ∠A,

So, AB/AC = BD/DC

⇒ AB/AC = 3/1

Since AB + AC = 10 cm

So, AC = 1/(1 + 3) × 10 = 10/4 = 2.5 cm

Thus, the length of AC is 2.5 cm.

The correct answer is option (1).

In a triangle ABC, AB + BC = 7.1 cm, BC + CA = 12.1 cm and CA + AB = 7.2 cm. What is the area of the triangle?

((a))
3 square cm
((b))
32 square cm
((c))
33 square cm
((d))
3.3 square cm

Show Answer

Answer: ((d))

3.3 square cm

Given:

AB + BC = 7.1 cm

BC + CA = 12.1 cm

CA + AB = 7.2 cm

Calculation:

Let the sides of the triangle be:

AB = a

BC = b

CA = c

We are given the following system of equations:

a + b = 7.1

b + c = 12.1

c + a = 7.2

(a + b) + (b + c) + (c + a) = 7.1 + 12.1 + 7.2

2a + 2b + 2c = 26.4

a + b + c = 13.2

c = 13.2 - 7.1 = 6.1

a = 13.2 - 12.1 = 1.1

b = 13.2 - 7.2 = 6.0

Heron’s formula for the area of a triangle is given by:

$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$

Where s is the semi-perimeter of the triangle, given by:

s = $\frac{a + b + c}{2} = \frac{1.1 + 6.0 + 6.1}{2} = \frac{13.2}{2} = 6.6$ cm

$\text{Area} = \sqrt{6.6(6.6 - 1.1)(6.6 - 6.0)(6.6 - 6.1)}$

$\text{Area} = \sqrt{6.6 \times 5.5 \times 0.6 \times 0.5}$

$\text{Area} = \sqrt{6.6 \times 5.5 \times 0.3}$

$\text{Area} = \sqrt{10.89}$

$\text{Area} = 3.3 , \text{cm}^2$

Therefore, the area of the triangle is: 3.3 cm²

The adjacent sides of a parallelogram are 10 cm and 8 cm and the angle between them is 150^∘ . What is the area of the parallelogram?

((a))
$40\sqrt{3} \text{ square cm}$
((b))
40 square cm
((c))
$20\sqrt{3} \text{ square cm}$
((d))
20 square cm

Show Answer

Answer: ((b))

40 square cm

Given:

Adjacent sides of a parallelogram: 10 cm and 8 cm

Angle between them: 150^∘

Formula used:

Area of parallelogram = a × b × sin(θ)

Where:

a = one side, b = adjacent side, θ = angle between sides

Calculation:

Area = 10 × 8 × sin(150^∘)

⇒ sin(150^∘) = sin(30^∘) = 1/2

⇒ Area = 10 × 8 × (1/2)

⇒ Area = 40 cm²

∴ The correct answer is option (2).

The measure of and angle formed by the bisectors of the angles A and C of the triangle ABC is 130^o. What is the measure of the angle B?

((a))
65^o
((b))
75^o
((c))
80^o
((d))
85^o

Show Answer

Answer: ((c))

80^o

Given:

Triangle ABC

Bisectors of angles A and C intersect and form an angle = 130°

Find angle B

Formula used:

Angle between bisectors of A and C inside triangle = 90° + B/2

Calculations:

Given, angle between bisectors = 130°

Using formula: 90 + B/2 = 130

⇒ B/2 = 130 - 90

⇒ B/2 = 40

⇒ B = 80°

∴ Angle B = 80°.

What is log₁₀2000 + log₁₀400 + 4log₁₀25 + 5log₁₀20 equal to?

((a))
10
((b))
16
((c))
18
((d))
20

Show Answer

Answer: ((c))

Given:

$\log_{10}2000 + \log_{10}400 + 4\log_{10}25 + 5\log_{10}20$

Formula used:

$\log a + \log b = \log(ab)$

$n\log a = \log(a^n)$

Calculation:

$\log_{10}2000 + \log_{10}400 + 4\log_{10}25 + 5\log_{10}20$

= $\log_{10}(2000 \times 400 \times 25^4 \times 20^5)$

⇒ $2000 = 2^4 \times 5^3$ , $400 = 2^4 \times 5^2$ , $25^4 = (5^2)^4 = 5^8$ , $20^5 = (2^2 \times 5)^5 = 2^{10} \times 5^5$

⇒ Product = $(2^4 \times 5^3) \times (2^4 \times 5^2) \times (5^8) \times (2^{10} \times 5^5)$

⇒ $= 2^{4+4+10} \times 5^{3+2+8+5}$

⇒ $= 2^{18} \times 5^{18}$

⇒ $= (2 \times 5)^{18} = 10^{18}$

Therefore,

$\log_{10}(10^{18}) = 18$

∴ The value is 18.

If $\frac{\log_{10}(100001 - 4^x)}{5 - x} = 1$ , then what is x equal

((a))
0
((b))
1
((c))
10
((d))
100

Show Answer

Answer: ((a))

Given:

$\frac{\log_{10}(100001 - 4^x)}{5 - x} = 1$

Formula used:

log a = b ⇒ a = 10^b

Calculation:

⇒ log₁₀(100001 - 4^x) = 5 - x

⇒ 100001 - 4^x = 10^5-x

Check all the options:

x = 0

⇒ 100001 - 40 = 105-0

⇒ 100001 - 1 = 100000

x = 0 satisfies the condition.

x = 1

⇒ 100001 - 41 = 105 - 1

⇒ 100001 - 4 = 10000

⇒ 99997 ≠ 10000

x = 1 does not satisfy the condition.

x = 10

⇒ 100001 - 410 = 105 - 10

⇒ 100001 - 410 ≠ 10- 5

x = 10 does not satisfy the condition.

∴ The correct answer is option 1.

If 2 sin⁴ $\alpha$ + 2 cos⁴ $\alpha$ - 1 = 0, where 0 < $\alpha$ < $\pi$ /2, then what is sin 2 $\alpha$ + cos 2 $\alpha$ equal to ?

((a))
0
((b))
1
((c))
$\frac{\sqrt{3}+1}{2}$
((d))
$\frac{\sqrt{3}-1}{2}$

Show Answer

Answer: ((b))

Given:

2 sin⁴(α) + 2 cos⁴(α) - 1 = 0, where 0 < α < π/2

Formula used:

sin⁴(α) + cos⁴(α) = (sin²(α) + cos²(α))² - 2sin²(α) cos²(α)

We know that sin²(α) + cos²(α) = 1.

Calculation:

The given equation is:

2 sin⁴(α) + 2 cos⁴(α) - 1 = 0

⇒ 2 (sin⁴(α) + cos⁴(α)) = 1

⇒ sin⁴(α) + cos⁴(α) = 1/2

Now, using the identity:

sin⁴(α) + cos⁴(α) = (sin²(α) + cos²(α))² - 2sin²(α) cos²(α)

⇒ (1)² - 2sin²(α) cos²(α) = 1/2

⇒ 1 - 2sin²(α) cos²(α) = 1/2

⇒ 2sin²(α) cos²(α) = 1/2

⇒ sin²(α) cos²(α) = 1/4

Now, take the square root of both sides:

⇒ sin(α) cos(α) = 1/2

Using the identity sin(2α) = 2 sin(α) cos(α):

⇒ sin(2α) = 2 × 1/2 = 1

sin(2α) = 1 implies that: 2α = 90°

Therefore: α = 45°

Now, find sin(2α) + cos(2α):

Since α = 45°:

sin(2α) = sin(90°) = 1

cos(2α) = cos(90°) = 0

Thus: sin(2α) + cos(2α) = 1 + 0 = 1.

sin(2α) + cos(2α) = 1.

∴ The correct answer is option 2.

Consider the following:

(I): 1 - sin⁶α = cos²α ( cos⁴α − 3 cos²α + 3)

(II): cos⁸α−sin⁸α = 2sin²α(1 − cos⁴α + sin²α cos²α)

Which of the above is/are identities?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((a))

I only

Given:

(I): 1 − sin⁶α = cos²α (cos⁴α − 3cos²α + 3)

(II): cos⁸α − sin⁸α = 2sin²α (1 − cos⁴α + sin²α cos²α)

Check (I):

LHS = 1 − sin⁶α = (1 − sin²α)(1 + sin²α + sin⁴α)

⇒ = cos²α (1 + (1 − cos²α) + (1 − cos²α)²)

⇒ = cos²α (3 − 3cos²α + cos⁴α)

⇒ RHS ✔

So, (I) is true.

Check (II):

LHS = cos⁸α − sin⁸α = (cos⁴α − sin⁴α)(cos⁴α + sin⁴α)

⇒ (cos²α − sin²α)(cos²α + sin²α)(cos⁴α + sin⁴α)

⇒ (cos 2α)(1)(cos⁴α + sin⁴α)

RHS = 2sin²α(1 − cos⁴α + sin²α cos²α)

This form does not simplify to LHS.

So, (II) is false.

Final Answer:

Only (I) is an identity.

If p = $1\over{cosecθ \ + \ \cotθ}$ and q = cosec θ, then what is p² - 2pq equal to?

((a))
-1
((b))
0
((c))
1
((d))
2

Show Answer

Answer: ((a))

-1

Given:

p = 1/(cosecθ + cotθ)

q = cosecθ

Calculations:

We have,

p = 1/(cosecθ + cotθ) = 1/ (1/sinθ + cosθ/sinθ)

⇒ 1/{(1 + cosθ)/sinθ}

⇒ sinθ / (1 + cos θ)

q = cosec θ = 1/sinθ

So, p² - 2pq

⇒ {sinθ / (1 + cos θ)}² - 2 × {sinθ / (1 + cos θ)} × {1/sinθ}

⇒ {sin²θ / (1 + cos θ)2} - 2 / (1 + cos θ)

⇒ {sin²θ / (1 + cos θ)2 - 2 / (1 + cos θ)

⇒ (1 - cos²θ) /(1 + cos θ)2 - 2 / (1 + cos θ)

⇒ (1 - cosθ)(1 + cosθ) /(1 + cos θ)2 - 2 / (1 + cos θ)

⇒ (1 - cosθ) / (1 + cos θ) - 2 / (1 + cos θ)

⇒ (1 - cosθ - 2) / (1 + cos θ)

⇒ (-1 - cosθ) / (1 + cos θ)

⇒ - (1 + cosθ) / (1 + cos θ)

⇒ - 1

∴ The correct answer is option (1)

Consider the following statements:

I. (cosec α - sec α) is always positive in the first quadrant.

II. (tan α - cot α) is always negative in the first quadrant.

Which of the statements given above is/are correct?"

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((d))

Neither I nor II

Given:

Statement I: (cosec α - sec α) is always positive in the first quadrant.

Statement II: (tan α - cot α) is always negative in the first quadrant.

Formula used:

In the first quadrant, all trigonometric functions are positive.
cosec α = 1/sin α, sec α = 1/cos α, tan α = sin α/cos α, cot α = cos α/sin α.

Calculation:

For Statement I:

(cosec α - sec α) = 1/sin α - 1/cos α

⇒ Taking LCM:

(cosec α - sec α) = (cos α - sin α)/(sin α × cos α)

Since cos α and sin α are both positive in the first quadrant, the numerator (cos α - sin α) can be positive, zero, or negative depending on the specific value of α. Thus, Statement I is not always true.

For Statement II:

(tan α - cot α) = (sin α/cos α) - (cos α/sin α)

⇒ Taking LCM:

(tan α - cot α) = (sin² α - cos² α)/(sin α × cos α)

In the first quadrant, sin α and cos α are positive. However, sin² α - cos² α is negative because cos α > sin α for α in the first quadrant. Hence, (tan α - cot α) is always negative in the first quadrant.

∴ The correct answer is option (4).

A tower subtends an angle 60° at a point A on the same level as the foot of the tower. B is a point vertically above A and AB = h. The angle of depression of the foot of the tower, measured from B is 30°. What is the height of the tower ?

((a))
2h
((b))
2.5h
((c))
3h
((d))
3.5h

Show Answer

Answer: ((c))

Given:

AB = h

Calculation:

Let OP be the height of the tower,

In Δ ABP,

tan 30° = p/b = AB / AP

⇒ 1/√3 = h / AP

⇒ AP = h √3

Now, In Δ AOP,

tan 60° = p/b = OP/AP

⇒ √3 = OP / h √3

⇒ OP = h √3 × √3

⇒ OP = 3 h

Hence, the height of the tower is 3h.

What is $\frac{\sin\theta}{1 - \cot\theta} + \frac{\cos\theta}{1 - \tan\theta}$ (θ $\ne$ π/4) equal to?

((a))
sin θ + cos θ
((b))
sin θ − cos θ
((c))
cos θ − sin θ
((d))
− (sin θ + cos θ)

Show Answer

Answer: ((a))

sin θ + cos θ

Given:

Expression = (sin θ) / (1 − cot θ) + (cos θ) / (1 − tan θ), where θ ≠ π/4.

Formula used:

cot θ = cos θ / sin θ, tan θ = sin θ / cos θ.

Simplify each term separately and then add.

Calculation:

First term:

(sin θ) / (1 − cot θ) = (sin θ) / (1 − cos θ/sin θ)

= (sin θ) / ((sin θ − cos θ)/sin θ)

= (sin²θ) / (sin θ − cos θ)

Second term:

(cos θ) / (1 − tan θ) = (cos θ) / (1 − sin θ/cos θ)

= (cos θ) / ((cos θ − sin θ)/cos θ)

= (cos²θ) / (cos θ − sin θ)

= − (cos²θ) / (sin θ − cos θ)

Now, add both terms:

(sin²θ)/(sin θ − cos θ) − (cos²θ)/(sin θ − cos θ)

= (sin²θ − cos²θ)/(sin θ − cos θ)

= −(cos²θ − sin²θ)/(sin θ − cos θ)

= −(cos θ − sin θ)(cos θ + sin θ)/(sin θ − cos θ)

= −(−1)(cos θ + sin θ)

= cos θ + sin θ

∴ The correct answer is (sin θ + cos θ).

The length of an arc of a circle of radius 4 cm is π cm. What is the magnitude of the angle subtended by the arc at the centre?

((a))
π
((b))
π/2
((c))
π/3
((d))
π/4

Show Answer

Answer: ((d))

π/4

Given:

Radius (r) = 4 cm

Length of arc (l) = π cm

Formula used:

Length of arc (l) = r × θ

Where θ = Angle subtended by the arc at the centre (in radians)

Calculation:

l = r × θ

π = 4 × θ

⇒ θ = π / 4

Hence, the magnitude of the angle subtended by the arc at the center is π/4.

The correct answer is option (4).

If cot²θ - $3\sqrt{3}$ cotθ + 6 = 0,where $\frac{\pi}{6}$ < θ < $\frac{\pi}{2}$ , then what is the value of sin θ + cos 2θ ?

((a))
0
((b))
1
((c))
$\sqrt{3}$
((d))
$1 + \sqrt{2}$

Show Answer

Answer: ((b))

Given:

cot2θ - $3\sqrt{3}$ cotθ + 6 = 0

Formula used:

cot2θ - $3\sqrt{3}$ cotθ + 6 = 0

put cotθ = x

So, x² - $3\sqrt{3}$ x + 6 = 0

⇒ x² - 2 $\sqrt3$ x - $\sqrt3$ x + 6 = 0

⇒ x(x - 2 $\sqrt3$ ) - $\sqrt3$ (x - 2 $\sqrt3$ ) = 0

⇒ (x - 2 $\sqrt3$ ) (x - $\sqrt3$ ) = 0

so, x = $\sqrt3$ & 2 $\sqrt3$

i.e. cot θ = $\sqrt3$ = cot 30°

So, θ = 30°

Now,

sin θ + cos 2θ

⇒ sin 30 + cos 60

⇒ 1/2 + 1/2 = 1

∴ The correct answer is option (2).

Which of the following equations is/are possible?

I. sin²θ= $\frac{(x + y)^2}{4xy}$ , where x, y are positive unequal real quantities.

II. sinθ+cosθ=x+ $\frac{1}{x}$ , where x is a positive real quantity.

Select the correct answer using the code given below :

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((d))

Neither I nor II

Given:

I. sin²θ = ((x + y)²) / (4xy), where x, y are positive unequal real numbers.

II. sin θ + cos θ = x + 1/x, where x is a positive real number.

Formula used:

For any real angle θ, sin²θ ≤ 1 and |sin θ + cos θ| ≤ √2.

Check whether the given equations satisfy these constraints.

Calculation:

For I: sin²θ = ((x + y)²) / (4xy)

Since x, y > 0 and x ≠ y, (x + y)² > 4xy ⇒ ((x + y)²)/(4xy) > 1

But sin²θ ≤ 1 always.

Hence equation I is not possible.

For II: sin θ + cos θ = x + 1/x

By AM–GM inequality, x + 1/x ≥ 2 for x > 0

Also, |sin θ + cos θ| ≤ √2 ≈ 1.414

So x + 1/x ≥ 2 > √2 ⇒ equation II is also not possible.

∴ Neither I nor II is possible.

If m²(sin θ−1) + n²(sin θ + 1)=0 , where 0 < θ < $\frac{\pi}{2}$ , then what is (m²+ n²) cosθ−(m²− n²) cot θ equal to?

((a))
4 mn
((b))
2 mn
((c))
1
((d))
0

Show Answer

Answer: ((d))

Given:

Equation: m²(sin θ - 1) + n²(sin θ + 1) = 0

Range: 0 ≤ θ ≤ π/2

Expression to find: (m² + n²) cos θ - (m² - n²) cot θ

Formula used:

Basic trigonometric identities:

cot θ = cos θ / sin θ

Calculation:

From the given equation:

⇒ m²(sin θ - 1) + n²(sin θ + 1) = 0

⇒ m² sin θ - m² + n² sin θ + n² = 0

⇒ (m² + n²) sin θ + (-m² + n²) = 0

⇒ (m² + n²) sin θ = (m² - n²)

⇒ sin θ = (m² - n²) / (m² + n²)

Now, substitute sin θ into the given expression:

(m² + n²) cos θ - (m² - n²) cot θ

⇒ (m² + n²) cos θ - (m² - n²) × (cos θ / sin θ)

⇒ (m² + n²) cos θ - (m² - n²) × (cos θ / [(m² - n²) / (m² + n²)])

⇒ (m² + n²) cos θ - (m² - n²) × (cos θ × (m² + n²) / (m² - n²))

⇒ (m² + n²) cos θ - (m² + n²) cos θ

⇒ 0

∴ The correct answer is option (4).

If sin α + cos α = $\sqrt{2}$ , where 0 < α < $\frac{\pi}{2}$ , then what is sin³α − cos³α equal to?

((a))
1
((b))
1/2
((c))
1/4
((d))
0

Show Answer

Answer: ((d))

Shortcut Trick

sin α + cos α = $\sqrt{2}$

It is valid when α = π/4

As, sin π/4 + cos π/4 = 1/ $\sqrt{2}$ + 1/ $\sqrt{2}$ = 2/ $\sqrt{2}$ = $\sqrt{2}$

So, sin α - cos α = sin π/4 - cos π/4 = 1/ $\sqrt{2}$ - 1/ $\sqrt{2}$ = 0

So, sin3α − cos3α

⇒ (sin α - cos α) (sin² α + cos²α - sinα cosα)

⇒ 0 (sin2 α + cos2α - sinα cosα)

⇒ 0

Hence, the correct answer is 0.

Alternate Method

Given:

sin α + cos α = √2, where 0 < α < π/2

Formula used:

sin³α - cos³α = (sin α - cos α)(sin²α + sin α cos α + cos²α)

Calculation:

sin α + cos α = √2

⇒ (sin α + cos α)² = (√2)²

⇒ sin²α + cos²α + 2sin α cos α = 2

⇒ 1 + 2sin α cos α = 2

⇒ 2sin α cos α = 1

⇒ sin α cos α = 1/2

Now, sin α - cos α = √(sin²α + cos²α - 2sin α cos α)

⇒ sin α - cos α = √(1 - 1)

⇒ sin α - cos α = 0

Plugging into the formula:

sin³α - cos³α = (sin α - cos α)(sin²α + sin α cos α + cos²α)

⇒ sin³α - cos³α = 0 × (1 + 1/2)

⇒ sin³α - cos³α = 0

∴ The correct answer is option (4).

What is (1 + cot α−cosec α)(1 + tan α + sec α) equal to?

((a))
1/2
((b))
1
((c))
2
((d))
4

Show Answer

Answer: ((c))

Given:

Expression: (1 + cot α − cosec α)(1 + tan α + sec α)

Formula used:

cot α = cos α / sin α

cosec α = 1 / sin α

tan α = sin α / cos α

sec α = 1 / cos α

(x − 1)(x + 1) = x² − 1

sin²α + cos²α = 1

Calculations:

(1 + cot α − cosec α)(1 + tan α + sec α)

⇒ (1 + cos α / sin α − 1 / sin α)(1 + sin α / cos α + 1 / cos α)

⇒ ((sin α + cos α − 1) / sin α) × ((cos α + sin α + 1) / cos α)

Multiply numerators:

⇒ ((sin α + cos α − 1)(sin α + cos α + 1)) / (sin α × cos α)

Use difference of squares:

⇒ ((sin α + cos α)² − 1) / (sin α × cos α)

Expand and apply identity:

⇒ (sin²α + cos²α + 2 sin α cos α − 1) / (sin α × cos α)

⇒ (1 + 2 sin α cos α − 1) / (sin α × cos α)

⇒ 2 sin α cos α / (sin α × cos α)

⇒ 2

∴ (1 + cot α − cosec α)(1 + tan α + sec α) = 2

If tan θ = $\frac{sin\alpha - cos\alpha}{sin\alpha + cos\alpha}$ where θ and α are $\left( \alpha \neq \frac{\pi}{4} \right)$ acute angles, then what is $\sqrt{2}$ sin θ equal to ?

((a))
sin α − cos α
((b))
sin α + cos α
((c))
cos α − sin α
((d))
± (sin α − cos α)

Show Answer

Answer: ((a))

sin α − cos α

Given:

tan θ = (sin α - cos α) / (sin α + cos α)

θ and α are acute angles, α ≠ π/4.

Formula used:

√(2)sin θ = ?

Calculation:

tan θ = (sin α - cos α) / (sin α + cos α)

⇒ θ = tan^-1[(sin α - cos α) / (sin α + cos α)]

Using trigonometric identity: sin θ = tan θ / √(1 + tan²θ)

⇒ sin θ = [(sin α - cos α) / (sin α + cos α)] / √(1 + [(sin α - cos α) / (sin α + cos α)]²)

Let k = sin α - cos α and m = sin α + cos α.

⇒ sin θ = k / m / √[1 + (k / m)²]

⇒ sin θ = k / √(m² + k²)

⇒ √(2)sin θ = √(2) × (sin α - cos α) / √[(sin α + cos α)² + (sin α - cos α)²]

⇒ √(2)sin θ = √(2) × (sin α - cos α) / √[2(sin²α + cos²α)]

⇒ √(2)sin θ = √(2) × (sin α - cos α) / √(2)

⇒ √(2)sin θ = sin α - cos α

∴ The correct answer is option (1).

For how many values of α does the expression (sin α + 2)(sin α + 4)(sin α − 2)(sin α − 4) become zero?

((a))
No value
((b))
One
((c))
Two
((d))
Four

Show Answer

Answer: ((a))

No value

Given:

(sin α + 2)(sin α + 4)(sin α − 2)(sin α − 4) = 0

Formula used:

Expression = 0 ⇔ at least one factor = 0. But sin α ∈ [−1, 1]

Calculations:

⇒ sin α + 2 = 0 ⇒ sin α = −2 (not possible)

⇒ sin α + 4 = 0 ⇒ sin α = −4 (not possible)

⇒ sin α − 2 = 0 ⇒ sin α = 2 (not possible)

⇒ sin α − 4 = 0 ⇒ sin α = 4 (not possible)

Number of real values of α = 0.

∴ The correct answer is option 1.

What is the value of x, where 0 ≤ x ≤ 30^∘, satisfying tan3x tan 6x = 1 ?

((a))
0^∘
((b))
10^∘
((c))
12^∘
((d))
15^∘

Show Answer

Answer: ((b))

10^∘

Given:

tan3x × tan6x = 1, 0 ≤ x ≤ 30°

Formula used:

tanA × tanB = 1 ⇒ tanA = cotB = tan(90° − B + k × 180°)

Calculations:

⇒ tan3x = tan(90° − 6x + n × 180°)

⇒ 3x = 90° − 6x + n × 180°

⇒ 9x = 90° + n × 180°

⇒ x = 10° + n × 20°

Check n values for 0 ≤ x ≤ 30°:

n = 0 → x = 10° (valid)

n = 1 → x = 30° (but 3x = 90° → tan3x undefined → reject)

n = −1 → x = −10° (out of range)

∴ x = 10°.

What is $\frac{(a-b)^2}{(b-c)(c-a)} + \frac{(b-c)^2}{(c-a)(a-b)} + \frac{(c-a)^2}{(a-b)(b-c)}$ - 3 equal to, where a $\neq$ b $\neq$ c?

((a))
0
((b))
3
((c))
a + b + c
((d))
3 (a−b) (b−c) (c−a)

Show Answer

Answer: ((a))

Given:

Expression to evaluate:

$\frac{(a - b)^2}{(b - c)(c - a)} + \frac{(b - c)^2}{(c - a)(a - b)} + \frac{(c - a)^2}{(a - b)(b - c)} - 3$

Step 1: Let

x = $\frac{(a - b)^2}{(b - c)(c - a)}$

y = $\frac{(b - c)^2}{(c - a)(a - b)}$

z = $\frac{(c - a)^2}{(a - b)(b - c)}$

So expression becomes: x + y + z − 3

Step 2: Take LCM of all 3 terms:

LCM = (a − b)(b − c)(c − a)

Write all terms with common denominator:

x = $\frac{(a - b)^2}{(b - c)(c - a)} = \frac{(a - b)^3}{(a - b)(b - c)(c - a)}$

y = $\frac{(b - c)^2}{(c - a)(a - b)} = \frac{(b - c)^3}{(b - c)(c - a)(a - b)}$

z = $\frac{(c - a)^2}{(a - b)(b - c)} = \frac{(c - a)^3}{(c - a)(a - b)(b - c)}$

Numerators:

⇒ x + y + z = $\frac{(a - b)^3 + (b - c)^3 + (c - a)^3}{(a - b)(b - c)(c - a)}$

Use identity:

If x + y + z = 0 ⇒ x³ + y³ + z³ = 3xyz

Here, let: u = (a − b), v = (b − c), w = (c − a)

Then: u + v + w = 0

So: u³ + v³ + w³ = 3uvw

Apply:

Numerator = (a − b)³ + (b − c)³ + (c − a)³ = 3(a − b)(b − c)(c − a)

Denominator = (a − b)(b − c)(c − a)

So: x + y + z = $\frac{3(a - b)(b - c)(c - a)}{(a - b)(b - c)(c - a)} = 3$

Final Expression = x + y + z − 3 = 3 − 3 = 0

∴ The correct answer is: 0

Given that $\frac{100 \times 99 \times 98 \times \ldots \times 3 \times 2 \times 1}{100^n}$ is an integer. What is the largest value of n for which this is true?

((a))
20
((b))
21
((c))
24
((d))
None of these

Show Answer

Answer: ((d))

None of these

Given:

Expression: $\dfrac{100!}{100^n}$ is an integer

Formula used:

We need to find the largest n such that 100ⁿ divides 100!

100 = 2² × 5² ⇒ 100ⁿ = 2²ⁿ × 5²ⁿ

So, we must find maximum n such that both 2²ⁿ and 5²ⁿ divide 100!

Calculations:

Powers of 2 in 100!

⇒ [100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64]

⇒ 50 + 25 + 12 + 6 + 3 + 1 = 97

Powers of 5 in 100!

⇒ [100/5] + [100/25] + [100/125]

⇒ 20 + 4 + 0 = 24

Now: 2²ⁿ must ≤ 2⁹⁷ ⇒ 2n ≤ 97 ⇒ n ≤ 48.5

And: 5²ⁿ must ≤ 5²⁴ ⇒ 2n ≤ 24 ⇒ n ≤ 12

So, the limiting factor is powers of 5 ⇒ n ≤ 12

The largest value of n is 12.

∴ The correct answer is option 4.

A man starting from a place P went x metre (x > 120 m) East before turning South. He went 40 m straight before turning to West. He went 60 m to reach a place Q. From Q he went 200 m North and reached a place R. If PR = 200 m, then what is x equal to?

((a))
150 m
((b))
180 m
((c))
200 m
((d))
240 m

Show Answer

Answer: ((b))

180 m

Given:

Distance traveled East = x meters

Distance traveled South = 40 meters

Distance traveled West = 60 meters

Distance traveled North = 200 meters

PR = 200 meters

Formula used:

Pythagoras theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Calculation:

PR = 200 meters

Horizontal distance between P and R = x - 60

Vertical distance between P and R = 200 - 40 = 160

Using Pythagoras theorem:

⇒ (PR)² = (Horizontal distance)² + (Vertical distance)²

⇒ 200² = (x - 60)² + 160²

⇒ 40000 = (x - 60)² + 25600

⇒ (x - 60)² = 40000 - 25600

⇒ (x - 60)² = 14400

⇒ x - 60 = √14400

⇒ x - 60 = 120

⇒ x = 120 + 60

⇒ x = 180

∴ The correct answer is option (2).

If x²+ y²+ z² = 3, where x, y and z are integers, then how many values can (xy + yz + zx) have?

((a))
One
((b))
Two
((c))
Three
((d))
Four

Show Answer

Answer: ((b))

Two

Given:

Equation: x² + y² + z² = 3

x, y, and z are integers.

Formula used:

The algebraic identity: (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

Calculations:

We are given x² + y² + z² = 3, where x, y, and z are integers.

Since x, y, and z are integers, their squares (x², y², z²) must be non-negative integers (0, 1, 4, 9, ...).

The only way to sum three non-negative integer squares to get 3 is if each square is 1

⇒ x² = 1, y² = 1, z² = 1

This implies that x, y, and z can only be either 1 or -1.

Now, let P = (xy + yz + zx) and S = (x + y + z).

Using the algebraic identity:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

⇒ S² = 3 + 2P

Rearrange the equation to solve for P:

⇒ 2P = S² − 3

⇒ P = $(\dfrac{S^2 - 3}{2})$

Now, we will consider all possible combinations of x, y, and z from {1, -1} and calculate S and P for each case:

Case 1: All three integers are 1.

(x, y, z) = (1, 1, 1)

S = 1 + 1 + 1 = 3

P = $(\dfrac{3^2 - 3}{2})$ = $(\dfrac{9 - 3}{2})$ = $(\dfrac{6}{2})$ = 3

Case 2: Two integers are 1, and one is -1.

Possible combinations: (1, 1, -1), (1, -1, 1), (-1, 1, 1)

For any of these combinations, S = 1 + 1 − 1 = 1

P = $(\dfrac{1^2 - 3}{2})$ = $(\dfrac{1 - 3}{2})$ = $(\dfrac{-2}{2})$ = -1

Case 3: One integer is 1, and two are -1.

Possible combinations: (1, -1, -1), (-1, 1, -1), (-1, -1, 1)

For any of these combinations, S = 1 − 1 − 1 = -1

P = $(\dfrac{(-1)^2 - 3}{2})$ = $(\dfrac{1 - 3}{2})$ = $(\dfrac{-2}{2})$ = -1

Case 4: All three integers are -1.

(x, y, z) = (-1, -1, -1)

S = -1 − 1 − 1 = -3

P = $(\dfrac{(-3)^2 - 3}{2})$ = $(\dfrac{9 - 3}{2})$ = $(\dfrac{6}{2})$ = 3

The distinct values that (xy + yz + zx) can have are 3 and -1.

Therefore, there are 2 possible values for (xy + yz + zx).

If x, y, z are real numbers such that x + y + z=10 and xy + yz + zx = 18, then what is the value of x³+y³+z³− 3xyz?

((a))
400
((b))
440
((c))
460
((d))
500

Show Answer

Answer: ((c))

460

Given:

x + y + z = 10

xy + yz + zx = 18

Formula used:

The expression for the sum of cubes of three numbers can be expressed as:

x³ + y³ + z³ - 3xyz = (x + y + z) [(x + y + z)² - 3(xy + yz + zx)]

Calculation:

Substituting the given values:

x³ + y³ + z³ - 3xyz = (10) [(10)² - 3(18)]

⇒ x³ + y³ + z³ - 3xyz = 10 [100 - 54]

⇒ x³ + y³ + z³ - 3xyz = 10 × 46

⇒ x³ + y³ + z³ - 3xyz = 460

∴ The correct answer is option (3).

What is $\sqrt{17 - 4\sqrt{15}} + \sqrt{8 - 2\sqrt{15}}$ equal to?

((a))
$\sqrt{3}$
((b))
$2\sqrt{3}$
((c))
$2\left(\sqrt{5} - \sqrt{3}\right)$
((d))
$2\left(\sqrt{5} + \sqrt{3}\right)$

Show Answer

Answer: ((a))

$\sqrt{3}$

Given:

The required expression: $\sqrt{17 - 4\sqrt{15}} + \sqrt{8 - 2\sqrt{15}}$

Calculation:

$\sqrt{17 - 4\sqrt{15}}$ = $\sqrt{(2\sqrt3)^2 + (\sqrt5)^2 - 2 \times 2\sqrt{3}\times \sqrt5}$ = $\sqrt{(2\sqrt3 -\sqrt 5)^2}$ = $2\sqrt3 - \sqrt5$

$\sqrt{8 - 2\sqrt{15}}$ = $\sqrt{(\sqrt{3})^2 + (\sqrt{5})^2 - 2 \times \sqrt{3} \times \sqrt{5}}$ = $\sqrt{(\sqrt{5} -\sqrt{3})^2 }$ = $\sqrt{5} - \sqrt{3}$

Now, adding both terms:

$2\sqrt3 - \sqrt5$ + $\sqrt{5} - \sqrt{3}$ = $\sqrt3$

Therefore, the final answer is option 1.

What is the maximum value of the sum of the numbers 36, 33, 30, 27, 24, ...?

((a))
240
((b))
237
((c))
234
((d))
231

Show Answer

Answer: ((c))

234

Given:

The sequence is: 36, 33, 30, 27, 24, ...

We need to find the maximum value of the sum of the numbers.

This is an arithmetic progression (AP) where:

First term (a) = 36

Common difference (d) = 33 - 36 = -3

The sum of n terms of an AP is given by:

Formula used:

Sum (S_n) = $\dfrac{n}{2}[2a + (n-1)d]$

Calculation:

We need to find the maximum value of the sum. The AP ends when the last term becomes ≥ 0.

Last term (l) = a + (n-1)d

For the last term to be ≥ 0:

⇒ 36 + (n-1)(-3) ≥ 0

⇒ 36 - 3(n-1) ≥ 0

⇒ 36 - 3n + 3 ≥ 0

⇒ 39 - 3n ≥ 0

⇒ 3n ≤ 39

⇒ n ≤ 13

Thus, the sequence has a maximum of n = 13 terms.

Now, calculate the sum of these 13 terms:

S₁₃ = $\dfrac{13}{2}[2(36) + (13-1)(-3)]$

⇒ S₁₃ = $\dfrac{13}{2}[72 + 12(-3)]$

⇒ S₁₃ = $\dfrac{13}{2}[72 - 36]$

⇒ S₁₃ = $\dfrac{13}{2}[36]$

⇒ S₁₃ = 13 × 18

⇒ S₁₃ = 234

∴ The correct answer is option (3).

There are two natural numbers m and n (m > n). When m is divided by 12, it leaves a remainder 4. When n is divided by 12, it leaves a remainder 6. Which of the following statements is/are correct?

I. The remainder when (m + n) is divided by 12 is 10.

II. The remainder when (m - n) is divided by 12 is 10.

Select the correct answer using the code given below :

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((c))

Both I and II

Given:

m ≡ 4 (mod 12)

n ≡ 6 (mod 12)

Formula used:

If a ≡ r₁ (mod m) and b ≡ r₂ (mod m), then:

a ± b ≡ r₁ ± r₂ (mod m)

Calculations:

Statement I: (m + n) mod 12

⇒ m + n ≡ 4 + 6 ≡ 10 (mod 12)

The statement I is correct.

Statement II: (m − n) mod 12

⇒ m − n ≡ 4 − 6 ≡ −2 ≡ 10 (mod 12)

The statement II is correct.

∴ The correct answer is: Both I and II

If (x + y) : (y + z) : (z + x) = 3 : 5 : 6 and x + y + z = 14, then what is x² + y² + z² equal to ?

((a))
81
((b))
84
((c))
87
((d))
90

Show Answer

Answer: ((b))

Given:

(x + y) : (y + z) : (z + x) = 3 : 5 : 6

x + y + z = 14

Formula used:

x² + y² + z² = (x + y + z)² − 2(xy + yz + zx)

Calculations:

Let (x + y) = 3k, (y + z) = 5k, (z + x) = 6k

⇒ (x + y) + (y + z) + (z + x) = 3k + 5k + 6k = 14k

But (x + y) + (y + z) + (z + x) = 2(x + y + z) = 2 × 14 = 28

⇒ 14k = 28 ⇒ k = 2

So, x + y = 6, y + z = 10, z + x = 12

Adding all: 2(x + y + z) = 28 ⇒ x + y + z = 14 ✔

Now, solve individually:

x + y = 6, x + y + z = 14 ⇒ z = 8

y + z = 10 ⇒ y + 8 = 10 ⇒ y = 2

x + z = 12 ⇒ x + 8 = 12 ⇒ x = 4

Now, x² + y² + z² = 4² + 2² + 8² = 16 + 4 + 64 = 84

∴ x² + y² + z² = 84

The ratio of sum of two numbers to their difference is 5 : 1. What is the ratio of the sum of their squares to the difference of their squares?

((a))
13 : 5
((b))
25 : 1
((c))
9 : 4
((d))
16 : 1

Show Answer

Answer: ((a))

13 : 5

Given:

The ratio of the sum of two numbers to their difference is 5 : 1.

Formula used:

If the ratio of the sum of two numbers (a + b) to their difference (a - b) is given as 5 : 1, then:

(a + b)/(a - b) = 5/1

The ratio of the sum of their squares to the difference of their squares can be calculated using:

(a² + b²) / (a² - b²) = [(a + b)² - 2ab] / [(a - b)² + 2ab]

Calculation:

From the ratio (a + b)/(a - b) = 5/1:

⇒ a + b = 5k and a - b = k

⇒ a = (5k + k)/2 = 3k, b = (5k - k)/2 = 2k

Now calculate the ratio (a² + b²) / (a² - b²):

⇒ a² + b² = (3k)² + (2k)² = 9k² + 4k² = 13k²

⇒ a² - b² = (3k)² - (2k)² = 9k² - 4k² = 5k²

⇒ (a² + b²) / (a² - b²) = 13k² / 5k²

⇒ Ratio = 13 : 5

∴ The correct answer is option (1).

Travelling at

Code

3/5<sup>th</sup> of his usual speed, a man is late by 20 minutes. What is the usual time if he travels with his usual speed ?

((a))
25 minutes
((b))
30 minutes
((c))
32 minutes
((d))
35 minutes

Show Answer

Answer: ((b))

30 minutes

Given:

A man is traveling at 3/5th of his usual speed and is late by 20 minutes.

Formula used:

Time = Distance ÷ Speed

Let the usual time taken by the man be T minutes.

When traveling at 3/5th of the usual speed, time taken = T × (5/3).

Difference in time: T × (5/3) - T = 20 minutes.

Calculation:

⇒ T × (5/3) - T = 20

⇒ (5T/3) - T = 20

⇒ (5T - 3T)/3 = 20

⇒ 2T/3 = 20

⇒ T = 20 × (3/2)

⇒ T = 30 minutes

∴ The correct answer is option (2).

What is the remainder when 2^p−1 is divided by p, where p > 5 is a prime number?

((a))
1
((b))
2
((c))
3
((d))
4

Show Answer

Answer: ((a))

Given:

Expression: 2^p−1

Divisor: p

Calculations:

We need to find the remainder when 2^p−1 is divided by p.

Since p is a prime number greater than 5: Direct putting value p = 7.

Substitute p = 7 into the expression:

Expression = 2⁷ − 1

2⁷ = 128

Expression = 128 − 1

⇒ Expression = 127

Now, we need to divide 127 by p = 7 and find the remainder.

127 ÷ 7:

127 = 7 × 18 + 1

The remainder when 2⁷−1 is divided by 7 is 1.

∴ The correct answer is option 1.

What is the number of factors of 24³- 16³ - 8³ ?

((a))
33
((b))
30
((c))
28
((d))
24

Show Answer

Answer: ((a))

Given:

Expression: 24³ − 16³ − 8³

Calculations:

⇒ 8³(27 − 8 − 1)

⇒ 8³ × 18

⇒ 512 × 18 = 9216

Prime factorization of 9216:

9216 = 2¹⁰ × 3²

Number of factors = (10 + 1)(2 + 1) = 11 × 3 = 33

∴ The correct answer is: 33

What is the least number of complete years in which a sum of money put out at 20% compound interest (compounded annually) will be more than doubled?

((a))
2
((b))
3
((c))
4
((d))
5

Show Answer

Answer: ((c))

Given:

Rate of Interest (r) = 20%

Amount (A) = 2 × Principal (P)

Formula used:

A = P(1 + r/100)^t

Calculation:

2P = P(1 + 20/100)^t

⇒ 2 = (1 + 0.2)^t

⇒ 2 = (1.2)^t

Now, we calculate t for different values:

For t = 2, (1.2)² = 1.44

For t = 3, (1.2)³ = 1.728

For t = 4, (1.2)⁴ = 2.0736

⇒ At t = 4, the amount exceeds double the principal.

∴ The least number of complete years is 4, and the correct answer is option (3).

A train of certain length takes time t to pass completely through a station of length x. The same train with same speed takes time 2t to pass completely through another station of length y. What is the time taken by the train to pass completely through a station of length (x + y)?

((a))
(2yt + xt) / (y - x)
((b))
(yt + xt) / (y - x)
((c))
(3yt - xt) / (2y - x)
((d))
(2yt - xt) / (y - x)

Show Answer

Answer: ((d))

(2yt - xt) / (y - x)

Given:

A train of length L passes a station of length x in time t.

The same train with same speed passes another station of length y in time 2t.

We need to find the time taken to pass a station of length (x + y).

Formula used:

Time = (Distance covered) / (Speed)

When a train passes a station, distance covered = (Length of train + Length of station).

Calculation:

Case 1: Through station of length x

(L + x) / v = t … (i)

Case 2: Through station of length y

(L + y) / v = 2t … (ii)

From (i): v = (L + x)/t

From (ii): v = (L + y)/(2t)

Equating both values of v:

(L + x)/t = (L + y)/(2t)

2(L + x) = (L + y)

2L + 2x = L + y

L = y − 2x … (iii)

Now, time to pass station of length (x + y):

Required time = (L + (x + y)) / v

Using (i): v = (L + x)/t

So, Required time = (L + x + y) / ((L + x)/t)

= [(L + x + y) × t] / (L + x)

Substitute L = y − 2x:

= [(y − 2x + x + y) × t] / (y − 2x + x)

= [(2y − x) × t] / (y − x)

∴ The correct answer is (2yt - xt) / (y - x).

A frustum of a right cone has a top of diameter 2k, bottom of diameter 2·5k and height k.

What is the whole surface area of the frustum?

((a))
39 $\pi$ k²/8
((b))
41 $\pi$ k²/8
((c))
43 $\pi$ k²/8
((d))
45 $\pi$ k²/8

Show Answer

Answer: ((a))

39 $\pi$ k²/8

Given:

Top diameter of the frustum (d₁) = 2k

Bottom diameter of the frustum (d₂) = 2.5k

Height of the frustum (h) = k

Formula used:

Radius of top (r₁) = d₁ / 2

Radius of bottom (r₂) = d₂ / 2

Slant height of frustum (l) = $\sqrt{h^2 + (r_2 - r_1)^2}$

Whole Surface Area of Frustum = Area of top base + Area of bottom base + Lateral Surface Area

Area of top base = $\pi r_1^2$

Area of bottom base = $\pi r_2^2$

Lateral Surface Area = $\pi(r_1 + r_2)l$

Calculations:

Calculate radii:

r₁ = 2k / 2 = k

r₂ = 2.5k / 2 = 1.25k

Calculate the difference in radii:

r₂ - r₁ = 1.25k - k = 0.25k

Calculate the slant height (l):

l = $\sqrt{k^2 + (0.25k)^2}$

⇒ l = $\sqrt{k^2 + 0.0625k^2}$

⇒ l = $\sqrt{1.0625k^2}$

⇒ l = k $\sqrt{1.0625}$

Now, calculate the surface areas:

Area of top base = $\pi r_1^2$ = $\pi (k)^2$ = $\pi k^2$

Area of bottom base = $\pi r_2^2$ = $\pi (1.25k)^2$ = $\pi (1.5625k^2)$ = 1.5625 $\pi k^2$

Lateral Surface Area = $\pi(r_1 + r_2)l$

⇒ Lateral Surface Area = $\pi(k + 1.25k)k\sqrt{1.0625}$

⇒ Lateral Surface Area = $\pi(2.25k)k\sqrt{1.0625}$

⇒ Lateral Surface Area = 2.25 $\pi k^2\sqrt{1.0625}$

Whole Surface Area = $\pi k^2$ + 1.5625 $\pi k^2$ + 2.25 $\pi k^2\sqrt{1.0625}$

⇒ Whole Surface Area = $\pi k^2 (1 + 1.5625 + 2.25\sqrt{1.0625})$

⇒ Whole Surface Area = $\pi k^2 (2.5625 + 2.25 \times 1.030776...)$

⇒ Whole Surface Area = $\pi k^2 (2.5625 + 2.319246...)$

⇒ Whole Surface Area = $\pi k^2 (4.881746...)$

Let's use the fraction form for sqrt(1.0625) for precision:

1.0625 = 10625 / 10000 = 425 / 400 = 85 / 80 = 17/16 (Error in calculation here, 10625/10000 = 17/16 was for 1.0625, but 0.25k is (1/4)k)

0.25k = (1/4)k

(0.25k)² = (1/16)k²

l = $\sqrt{k^2 + \frac{1}{16}k^2}$ = $\sqrt{\frac{16k^2 + k^2}{16}}$ = $\sqrt{\frac{17k^2}{16}}$ = $\frac{k\sqrt{17}}{4}$

Lateral Surface Area = $\pi(k + 1.25k)\frac{k\sqrt{17}}{4}$

⇒ Lateral Surface Area = $\pi(2.25k)\frac{k\sqrt{17}}{4}$

⇒ Lateral Surface Area = $\pi(\frac{9}{4}k)\frac{k\sqrt{17}}{4}$

⇒ Lateral Surface Area = $\frac{9\pi k^2\sqrt{17}}{16}$

Whole Surface Area = $\pi k^2$ + $\pi (\frac{5}{4}k)^2$ + $\frac{9\pi k^2\sqrt{17}}{16}$

⇒ Whole Surface Area = $\pi k^2$ + $\frac{25\pi k^2}{16}$ + $\frac{9\pi k^2\sqrt{17}}{16}$

⇒ Whole Surface Area = $\pi k^2 \left(1 + \frac{25}{16} + \frac{9\sqrt{17}}{16}\right)$

⇒ Whole Surface Area = $\pi k^2 \left(\frac{16}{16} + \frac{25}{16} + \frac{9\sqrt{17}}{16}\right)$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (16 + 25 + 9\sqrt{17})$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (41 + 9\sqrt{17})$

Using $\sqrt{17} \approx 4.123$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (41 + 9 \times 4.123)$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (41 + 37.107)$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (78.107)$

⇒ Whole Surface Area ≈ 4.8816 $\pi k^2$

The exact expression is $\frac{\pi k^2}{16} (41 + 9\sqrt{17})$

∴ The whole surface area of the frustum is $\frac{\pi k^2}{16} (41 + 9\sqrt{17})$ .

What is the volume of the frustum?

((a))
61 $\pi$ k³/48
((b))
59 $\pi$ k³/48
((c))
57 $\pi$ k³/48
((d))
53 $\pi$ k³/48

Show Answer

Answer: ((a))

61 $\pi$ k³/48

Given:

Top diameter of the frustum (d₁) = 2k

Bottom diameter of the frustum (d₂) = 2.5k

Height of the frustum (h) = k

Formula used:

Radius of top (r₁) = d₁ / 2

Radius of bottom (r₂) = d₂ / 2

Volume of frustum (V) = $\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)$

Calculations:

Calculate radii:

r₁ = 2k / 2 = k

r₂ = 2.5k / 2 = 1.25k

Substitute the values into the volume formula:

V = $\frac{1}{3}\pi (k) (k^2 + (1.25k)^2 + k \times 1.25k)$

⇒ V = $\frac{1}{3}\pi k (k^2 + 1.5625k^2 + 1.25k^2)$

⇒ V = $\frac{1}{3}\pi k (1k^2 + 1.5625k^2 + 1.25k^2)$

⇒ V = $\frac{1}{3}\pi k (k^2(1 + 1.5625 + 1.25))$

⇒ V = $\frac{1}{3}\pi k (3.8125k^2)$

⇒ V = $\frac{3.8125}{3}\pi k^3$

To express 3.8125 as a fraction:

3.8125 = 38125 / 10000 = 1525 / 400 = 305 / 80 = 61 / 16

So, V = $\frac{1}{3}\pi k \left(\frac{61}{16}k^2\right)$

⇒ V = $\frac{61}{48}\pi k^3$

∴ The volume of the frustum is $\frac{61}{48}\pi k^3$ .

ABC is a triangle right-angled at B. The perimeter of the triangle is 24 cm and the difference between the sum of the perpendicular sides and the hypotenuse is 4 cm.

What is the area of the triangle ABC?

((a))
18 square cm
((b))
24 square cm
((c))
36 square cm
((d))
48 square cm

Show Answer

Answer: ((b))

24 square cm

Given:

p + b + h = 24

p + b - h = 4

Formula used:

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{p + b - h}{2}$

Where: p and b are the perpendicular sides of the triangle

Area of triangle = radius (r) of the inscribed circle × semi - perimeter.

c is the hypotenuse of the triangle

Calculation:

p + b + h = 24 cm

p + b - h = 4 cm

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{p + b - h}{2}$

r = 4/2 = 2 cm

Area of triangle = 2 × 24/2 = 24 cm²

∴ The correct anser is option 2.

A circle is inscribed in the triangle. What is its radius?

((a))
1 cm
((b))
1·5 cm
((c))
2 cm
((d))
2·5 cm

Show Answer

Answer: ((c))

2 cm

Given:

p + b + h = 24

p + b - h = 4

Formula used:

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{a + b - c}{2}$

Where: a and b are the perpendicular sides of the triangle

c is the hypotenuse of the triangle

Calculation:

p + b + h = 24 cm

p + b - h = 4 cm

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{p + b - h}{2}$

r = 4/2 = 2 cm

Therefore, the radius of the inscribed circle is: 2 cm.

A circle M of radius 8 cm touches externally with another circle N of radius 16 cm. Let P, Q be the points where the common tangent touches the circles M and N respectively.

What is the length of the common tangent PQ?

((a))
16 cm
((b))
$16\sqrt{2}$ cm
((c))
24 cm
((d))
$24\sqrt{2}$ cm

Show Answer

Answer: ((b))

$16\sqrt{2}$ cm

Given:

Radius of circle M = 8 cm

Radius of circle N = 16 cm

Distance between centers = 8 + 16 = 24 cm

Formula used:

Length of direct common tangent = $\sqrt{d^2 - (r_1 - r_2)^2}$

Calculation:

⇒ d = 24, r₁ = 16, r₂ = 8

⇒ PQ = √(24² - (16 - 8)²)

⇒ PQ = √(576 - 64)

⇒ PQ = √512

⇒ PQ = 16√2 cm

∴ The length of common tangent PQ = 16√2 cm.

If U, V are the centres of the circles M and N respectively, then what is the area of the quadrilateral formed by the points P, Q, V and U?

((a))
$192\sqrt{2}$ square cm
((b))
192 square cm
((c))
$96\sqrt{2}$ square cm
((d))
96 square cm

Show Answer

Answer: ((a))

$192\sqrt{2}$ square cm

Given:

Circle M radius = 8 cm (centre U).

Circle N radius = 16 cm (centre V).

They touch externally ⇒ UV = 8 + 16 = 24 cm.

PQ = length of common external tangent = 16√2 cm.

Formula used:

If two radii to tangent points are perpendicular to the tangent, those two radii are parallel.

Area of a trapezoid with parallel sides a and b separated by distance d = (a + b)/2 × d.

Calculations:

⇒ UP = radius of M = 8 cm (UP ⟂ PQ).

⇒ VQ = radius of N = 16 cm (VQ ⟂ PQ).

⇒ UP ∥ VQ and the distance between these parallel segments = length PQ = 16√2 cm.

⇒ Area of quadrilateral P – Q – V – U = (UP + VQ)/2 × PQ

⇒ Area = (8 + 16)/2 × 16√2

⇒ Area = 24/2 × 16√2 = 12 × 16√2

⇒ Area = 192√2 cm²

∴ Area of quadrilateral P Q V U = 192√2 cm².

The perimeter of a triangle ABC is 105 cm. The altitudes AD, BE and CF are in the ratio 3 : 5 : 6.

What is AB : BC : CA equal to?

((a))
10 : 6 : 5
((b))
5 : 10 : 6
((c))
6 : 5 : 3
((d))
3 : 5 : 6

Show Answer

Answer: ((b))

5 : 10 : 6

Given:

Perimeter of triangle ABC = 105 cm

Ratio of altitudes AD : BE : CF = 3 : 5 : 6

Formula used:

The area of a triangle (Area) can be expressed as: Area = $(\dfrac{1}{2}) × base × height.$

For a triangle with sides a, b, c and corresponding altitudes ha, hb, hc, we have:

2 × Area = a × ha = b × hb = c × hc

This implies that the sides of a triangle are inversely proportional to their corresponding altitudes. That is, $a : b : c = (\dfrac{1}{h_a}) : (\dfrac{1}{h_b}) : (\dfrac{1}{h_c})$

Calculations:

Let the sides of the triangle be a, b, c, where:

a = BC (side opposite to vertex A)

b = CA (side opposite to vertex B)

c = AB (side opposite to vertex C)

Let the altitudes be ha, hb, hc, where:

ha = AD

hb = BE

hc = CF

Given the ratio of altitudes: ha : hb : hc = 3 : 5 : 6.

Using the property that sides are inversely proportional to altitudes:

$a : b : c = (\dfrac{1}{3}) : (\dfrac{1}{5}) : (\dfrac{1}{6})$

⇒ $a : b : c = (\dfrac{1}{3}) × 30 : (\dfrac{1}{5}) × 30 : (\dfrac{1}{6}) × 30$

⇒ a : b : c = 10 : 6 : 5

This means the ratio of the sides BC : CA : AB is 10 : 6 : 5.

Therefore, AB : BC : CA = c : a : b = 5 : 10 : 6.

∴ The correct answer is option 2.

What is the approximate area of the triangle ABC?

((a))
175 square cm
((b))
190 square cm
((c))
205 square cm
((d))
285 square cm

Show Answer

Answer: ((d))

285 square cm

Given:

Perimeter of triangle ABC = 105 cm

Ratio of sides AB : BC : CA = 5 : 10 : 6

Formula used:

Heron's Formula for the area of a triangle:

Area = $(\sqrt{s(s-a)(s-b)(s-c)})$

Where a, b, c are the lengths of the sides of the triangle, and

s is the semi-perimeter s = $(\dfrac{\text{a + b + c}}{2})$

Calculations:

From the previous problem, we established

The ratio of the sides AB : BC : CA = 5 : 10 : 6.

Let the sides be AB = 5x, BC = 10x, and CA = 6x.

The perimeter is the sum of the sides:

Perimeter = AB + BC + CA = 5x + 10x + 6x = 21x

Given Perimeter = 105 cm.

⇒ 21x = 105

⇒ x = $(\dfrac{105}{21})$

⇒ x = 5

a (BC) = 10x = 10 × 5 = 50 cm

b (CA) = 6x = 6 × 5 = 30 cm

c (AB) = 5x = 5 × 5 = 25 cm

The semi-perimeter (s): s = $(\dfrac{\text{Perimeter}}{2})$ = $(\dfrac{105}{2})$ = 52.5 cm

Area = $(\sqrt{s(s-a)(s-b)(s-c)})$

Area = $(\sqrt{52.5 \times (52.5 - 50) \times (52.5 - 30) \times (52.5 - 25)})$

Area = $(\sqrt{52.5 \times 2.5 \times 22.5 \times 27.5})$

Area = $(\sqrt{99738.28125})$

Area ≈ 284.975 cm²

Area ≈ 285 cm²

∴ The correct answer is option 4.

A pot is made from a hollow sphere of inner radius 20 cm by cutting its upper portion horizontally. The height of the pot is 30 cm.

What is the inner radius of the circular opening of the pot so formed?

((a))
$10\sqrt{2}$ cm
((b))
15 cm
((c))
$10\sqrt{3}$ cm
((d))
12 cm

Show Answer

Answer: ((c))

$10\sqrt{3}$ cm

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

Calculations:

Let 'r' be the inner radius of the circular opening of the pot.

Draw a vertical line from the center of the sphere to the center of the circular opening. This line segment will be perpendicular to the plane of the circular opening.

Let the center of the sphere be O. Let the center of the circular opening be C'.

The radius of the sphere (R) goes from O to any point on the surface of the sphere, including the edge of the circular opening.

Consider a right-angled triangle formed by:

The radius of the sphere (R) as the hypotenuse, from the center of the sphere to any point on the edge of the circular opening.

d = |R - h| = |20 - 30| = |-10| = 10 cm.

Now, substitute the values into the Pythagorean theorem:

R² = r² + d²

20² = r² + 10²

400 = r² + 100

r² = 400 - 100

r² = 300

r = $\sqrt{300}$

r = $\sqrt{100 \times 3}$

r = 10 $\sqrt{3}$ cm

∴ The inner radius of the circular opening of the pot is 10 $\sqrt{3}$ cm.

What is the angle made by the line joining the centre of the sphere and any point on the rim of the circular opening with a vertical line passing through the centre?

((a))
$\pi/3$
((b))
$\pi/4$
((c))
$\pi/6$
((d))
$\pi/12$

Show Answer

Answer: ((a))

$\pi/3$

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

From the previous calculation, the radius of the circular opening (r) = 10 $\sqrt{3}$ cm

From the previous calculation, the perpendicular distance from the center of the sphere to the plane of the circular opening (d) = 10 cm

Calculations:

The perpendicular distance from the center of the sphere to the plane of the circular opening (d) as the adjacent side to the angle θ (this is the vertical line segment from the center to the center of the opening).

The inner radius of the circular opening (r) as the opposite side to the angle θ.

We can use trigonometric ratios:

cos(θ) = Adjacent / Hypotenuse = d / R

cos(θ) = d / R = 10 cm / 20 cm = 1/2

The angle whose cosine is 1/2 is 60°.

⇒ θ = 60° = π/3

∴ The angle made by the line joining the center of the sphere and any point on the rim of the circular opening with a vertical line passing through the center is 60°.

A hall is of length l, breadth b and height h. The maximum distance between any two points (say P and Q) inside the hall is 14 m, whereas the maximum distance between two points (say P and R) on the floor is 6√5 m.

What is h equal to?

((a))
3·5 m
((b))
4 m
((c))
4·5 m
((d))
5 m

Show Answer

Answer: ((b))

4 m

Given:

Hall dimensions = l (length), b (breadth), h (height)

Maximum distance inside (space diagonal) = 14 m

Maximum distance on floor (floor diagonal) = 6√5 m

Formula used:

Space diagonal = √(l² + b² + h²)

Floor diagonal = √(l² + b²)

Calculations:

⇒ l² + b² + h² = 14² = 196

⇒ l² + b² = (6√5)² = 36 × 5 = 180

⇒ h² = 196 − 180 = 16

⇒ h = √16 = 4

∴ h = 4 m

If α is the angle between the line joining P and Q, and the line joining P and R, then what is cos α equal to?

((a))
$\frac{2\sqrt{5}}{7}$
((b))
$\frac{3\sqrt{5}}{7}$
((c))
$\frac{1}{3}$
((d))
$\frac{2}{3}$

Show Answer

Answer: ((b))

$\frac{3\sqrt{5}}{7}$

Given:

PQ = 14, PR = 6√5, h = 4 (from previous)

Concept Used:

PQ is the space diagonal; its horizontal projection on the floor is PR and the vertical component is h. So the angle α between PQ and PR has adjacent = PR and hypotenuse = PQ. Hence

cos α = adjacent/hypotenuse = PR / PQ

Calculation:

⇒ cos α = (6√5) / 14

⇒ cos α = (3√5) / 7

∴ cos α = (3√5) / 7.

Alternate Method

Given:

Hall dimensions = l (length), b (breadth), h (height)

Maximum distance inside (space diagonal) = 14 m

Maximum distance on floor (floor diagonal) = 6√5 m

Formula used:

Space diagonal = √(l2 + b2 + h2)

Floor diagonal = √(l2 + b2)

Calculations:

⇒ l2 + b2 + h2 = 142 = 196

⇒ l2 + b2 = (6√5)2 = 36 × 5 = 180

⇒ h2 = 196 − 180 = 16

⇒ h = √16 = 4

Using dot product to find cos α

PQ = (l, b, h), PR = (l, b, 0)

PQ · PR = l² + b² + 0 = 180

|PQ| = 14, |PR| = 6√5

cos α = (PQ · PR) / (|PQ| × |PR|)

cos α = 180 / (14 × 6√5)

cos α = 180 / (84√5)

cos α = 15 / (7√5)

∴ cos α = 15 / (7√5)

The sides of an open box are 0·5 cm thick and bottom is 1 cm thick. The internal length, breadth and depth are respectively 14 cm, 10 cm and 8 cm. It is completely filled with water.

If the material weighs 2000 kg per cubic metre, then what is the weight of the material used in the construction of the box?

((a))
360 gm
((b))
365 gm
((c))
720 gm
((d))
730 gm

Show Answer

Answer: ((d))

730 gm

Given:

Side thickness = 0.5 cm, bottom thickness = 1 cm

Internal L × B × D = 14 cm × 10 cm × 8 cm

Material density = 2000 kg/m³

Formula used:

Outer dimension = internal + 2×side thickness (for L and B), depth adds bottom thickness

Volume of material = Volume_outer − Volume_inner

1 cm³ = 10⁻⁶ m³; Mass = Density × Volume

Calculations:

⇒ Outer L = 14 + 2 × 0.5 = 15 cm

⇒ Outer B = 10 + 2 × 0.5 = 11 cm

⇒ Outer D = 8 + 1 = 9 cm

⇒ Volume_outer = 15 × 11 × 9 = 1485 cm³

⇒ Volume_inner = 14 × 10 × 8 = 1120 cm³

⇒ Volume_material = 1485 − 1120 = 365 cm³

⇒ Volume (m³) = 365 × 10⁻⁶ = 3.65 × 10⁻⁴ m³

⇒ Mass = 2000 × 3.65 × 10⁻⁴ = 0.73 kg = 730 gm

∴ Weight of material = 730 gm.

If water weighs 1000 kg per cubic metre, then what is the weight of the box with water?

((a))
1·850 kg
((b))
1·900 kg
((c))
2·050 kg
((d))
2·100 kg

Show Answer

Answer: ((a))

1·850 kg

Given:

Side thickness = 0.5 cm, bottom thickness = 1 cm

Internal L × B × D = 14 cm × 10 cm × 8 cm

Material density = 2000 kg/m³

Formula used:

Outer dimension = internal + 2×side thickness (for L and B), depth adds bottom thickness

Volume of material = Volume_outer − Volume_inner

1 cm³ = 10⁻⁶ m³; Mass = Density × Volume

Calculations:

⇒ Outer L = 14 + 2 × 0.5 = 15 cm

⇒ Outer B = 10 + 2 × 0.5 = 11 cm

⇒ Outer D = 8 + 1 = 9 cm

⇒ Volume_outer = 15 × 11 × 9 = 1485 cm³

⇒ Volume_inner = 14 × 10 × 8 = 1120 cm³

⇒ Volume_material = 1485 − 1120 = 365 cm³

⇒ Volume (m³) = 365 × 10⁻⁶ = 3.65 × 10⁻⁴ m³

⇒ Mass = 2000 × 3.65 × 10⁻⁴ = 0.73 kg = 730 gm

Weight of material = 730 gm.

Volume of water = Volumeinner = 14 × 10 × 8 = 1120 cm3 = 1120 × 10^-6 m³

Weight of water = 1000 × 1120 × 10-6 kg = 1.12 kg

Total weight of the box with water = 0.73 kg + 1.12 kg = 1.850 kg

∴ The correct answer is option 1.

ABC is a triangle right-angled at A. Further,

AB = 8 cm, BC = 10 cm. D is the point on BC such that AD is perpendicular to BC.

What is AD equal to?

((a))
4·8 cm
((b))
5·0 cm
((c))
5·2 cm
((d))
5·4 cm

Show Answer

Answer: ((a))

4·8 cm

Given:

The triangle ABC is right-angled at A.

AB = 8 cm, BC = 10 cm,

D is perpendicular to BC.

Calculation:

AC² = BC² - AB²

AC = √(100 - 64)

AC = √36 = 6 cm

AD = (AB × AC)/(BC)

AD = (6 × 8)/10 = 4.8 cm

∴ The correct answer is option 1.

What is ratio of area of triangle ADC to area of triangle ADB?

((a))
7 : 15
((b))
9 : 16
((c))
2 : 3
((d))
3 : 4

Show Answer

Answer: ((b))

9 : 16

Given:

The triangle ABC is right-angled at A.

AB = 8 cm, BC = 10 cm,

D is perpendicular to BC.

Calculation:

AC2 = BC2 - AB2

AC = √(100 - 64)

AC = √36 = 6 cm

ΔADB ~ ΔADB

So, $\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{6^2}{8^2}$

⇒ $\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{36}{64}$

⇒ $\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{9}{16}$

∴ The correct answer is option 2.

The annual rainfall at a place is 40 cm. The weight of water is 1 metric tonne per cubic meter.

What is the volume of rainfall in cubic meter per hectare?

((a))
40
((b))
400
((c))
4000
((d))
40000

Show Answer

Answer: ((c))

4000

Given:

Annual rainfall = 40 cm = 0.40 m

Area = 1 hectare = 10,000 m²

Formula used:

Volume = Area × Rainfall depth

Calculations:

⇒ Volume = 10,000 × 0.40

⇒ Volume = 4,000 m³

∴ Volume of rainfall = 4,000 m³ per hectare (annually).

What is the weight of water (in metric tonnes) of annual rainfall falling there on a hectare of land?

((a))
40
((b))
400
((c))
4000
((d))
40000

Show Answer

Answer: ((c))

4000

Given:

Weight of water = 1 metric tonne per m³

Formula used:

Volume = Area × Rainfall depth

Weight = Volume × Density (tonnes/m3)

Calculations:

⇒ Volume = 10,000 × 0.40

⇒ Volume = 4,000 m3

⇒ Weight = 4,000 × 1

⇒ Weight = 4,000 metric tonnes

∴ Weight of water = 4,000 metric tonnes per hectare (annually).

The angle at the vertex of a conical body is 120°.

What is the ratio of the radius of the conical body to its slant height?

((a))
1 : 2
((b))
$\sqrt{3}$ : 1
((c))
$\sqrt{3}$ : 2
((d))
$\sqrt{2}$ : 1

Show Answer

Answer: ((c))

$\sqrt{3}$ : 2

Calculation:

In Δ AOB,

sin 60° = p/h = OB / AB

⇒ √3/2 = r / l

⇒ r : l = √3 : 2

Hence, the correct answer is option (3).

If the sum of slant height, height and radius is ( 9 + $3\sqrt{3}$ ) cm, then what is the volume of the cone?

((a))
27 $\pi$ cubic cm
((b))
18 $\sqrt{3}$ cubic cm
((c))
24 $\pi$ cubic cm
((d))
$27\sqrt{3}$ cubic cm

Show Answer

Answer: ((a))

27 $\pi$ cubic cm

Given:

Angle at the vertex of the conical body = 120°

Sum of slant height, height, and radius = 9 + 3√3 cm

Formula used:

Volume of the cone = (1/3)πr²h

Relationship between slant height, radius, and height = l² = r² + h²

Calculations:

Let the height of the cone be h, radius be r, and slant height be l.

We know that angle at the vertex is 120° and tan(60°) = r/h, so r = √3 × h.

We are given that r + h + l = 9 + 3√3, substituting r = √3 × h and l = √(r² + h²) = 2h:

(√3h) + h + 2h = 9 + 3√3

⇒ (√3 + 3)h = 9 + 3√3

⇒ h = (9 + 3√3) / (√3 + 3)

Rationalize the denominator:

h = (9 + 3√3)(√3 - 3) / (3 - 9)

⇒ h = -18 / -6

⇒ h = 3 cm

Now, r = √3 × h = √3 × 3 = 3√3 cm.

Now, volume of the cone = (1/3)πr²h = (1/3)π(3√3)² × 3

⇒ Volume = (1/3)π × 27 × 3 = 27π cubic cm.

∴ The volume of the cone is 27π cubic cm.

A person sells article X for ₹34,500 and makes a profit of 15%. He sells article Y at a loss of 10%. He neither loses nor gains on the whole because of these two transactions. What is the selling price of article Y?

((a))
₹40,000
((b))
₹40,500
((c))
₹41,000
((d))
₹51,500

Show Answer

Answer: ((b))

₹40,500

Given:

Profit on article X = 15%

Selling price of article X = ₹34,500

Loss on article Y = 10%

No overall gain or loss (Net profit = ₹0)

Formula used:

Cost price (CP) = $\frac{\text{Selling Price (SP)}}{1+\frac{\text{Profit%}}{100}}$

Cost price (CP) = $\frac{\text{Selling Price (SP)}}{1-\frac{\text{Loss%}}{100}}$

Calculations:

For article X:

CP of article X = $\frac{34500}{1+\frac{15}{100}}$

⇒ CP of article X = $\frac{34500}{1.15}$

⇒ CP of article X = ₹30,000

For article Y:

Total cost price of X and Y = Total selling price of X and Y

⇒ CP of X + CP of Y = SP of X + SP of Y

⇒ ₹30,000 + CP of Y = ₹34,500 + SP of Y

⇒ CP of Y = SP of Y - ₹4,500

CP of article Y = $\frac{\text{SP of Y}}{1-\frac{10}{100}}$

⇒ SP of Y - ₹4,500 = $\frac{\text{SP of Y}}{0.9}$

⇒ 0.9(SP of Y - ₹4,500) = SP of Y

⇒ 0.9SP of Y - ₹4,050 = SP of Y

⇒ SP of Y - 0.9SP of Y = ₹4,050

⇒ 0.1SP of Y = ₹4,050

⇒ SP of Y = ₹4,050 × 10

⇒ SP of Y = ₹40,500

∴ The correct answer is option (2).

100 quintals is what percent of 10 metric tonnes?

((a))
1%
((b))
10%
((c))
100%
((d))
1000%

Show Answer

Answer: ((c))

100%

Given:

Quantity = 100 quintals

Total = 10 metric tonnes

Conversion:

1 quintal = 100 kg

1 metric tonne = 1000 kg

So,

100 quintals = 100 × 100 = 10,000 kg

10 metric tonnes = 10 × 1000 = 10,000 kg

Calculations:

⇒ (10000 / 10000) × 100 = 100%

∴ The correct answer is: 100%

A circle is inscribed in an equilateral triangle. The radius of the circle is 2 cm. What is the area of the triangle?

((a))
$12\sqrt{3}$ square cm
((b))
12 square cm
((c))
$9\sqrt{3}$ square cm
((d))
9 square cm

Show Answer

Answer: ((a))

$12\sqrt{3}$ square cm

Given:

Radius of the circle (r) = 2 cm

The circle is inscribed in an equilateral triangle.

Formula used:

Area of an equilateral triangle = $\dfrac{\sqrt{3}}{4}a^2$

Where, a = side of the equilateral triangle

Relation between radius (r) and side (a) of an equilateral triangle:

$\text{r} = \dfrac{\sqrt{3}}{6}a$

Calculation:

$\text{r} = \dfrac{\sqrt{3}}{6}a$

⇒ 2 = $\dfrac{\sqrt{3}}{6}a$

⇒ $\text{a} = \dfrac{2 \times 6}{\sqrt{3}} = \dfrac{12}{\sqrt{3}} = 4\sqrt{3}$

Area of the triangle:

$\text{Area} = \dfrac{\sqrt{3}}{4}a^2$

⇒ $\text{Area} = \dfrac{\sqrt{3}}{4}(4\sqrt{3})^2$

⇒ $\text{Area} = \dfrac{\sqrt{3}}{4} \times 16 \times 3$

⇒ $\text{Area} = 12\sqrt{3}$

∴ The correct answer is option (1).

The sides of a triangle are k, 1·5k and 2·25k. What is the sum of the squares of its medians?

((a))
359k²/64
((b))
379k²/64
((c))
389k²/64
((d))
399k²/64

Show Answer

Answer: ((d))

399k²/64

Given:

The sides of a triangle are k, 1.5k, and 2.25k.

Formula used:

The sum of the squares of medians of a triangle is given by:

$\text{Sum of squares of medians} = \frac{3}{4} (a^2 + b^2 + c^2)$

Where, a, b, and c are the sides of the triangle.

Calculation:

Let a = k, b = 1.5k = 3k/2, c = 2.25k = 9k/4.

$a^2 + b^2 + c^2 = k^2 + \frac{9}{4}k^2 + \frac{81}{16}k^2$

$a^2 + b^2 + c^2 = \frac{16}{16}k^2 + \frac{36}{16}k^2 + \frac{81}{16}k^2 = \frac{133}{16}k^2$

Sum of squares of medians = $\frac{3}{4} \times \frac{133}{16}k^2$

Sum of squares of medians = $\frac{3 \times 133}{4 \times 16}k^2 = \frac{399}{64}k^2$

Therefore, the sum of the squares of the medians is: $\frac{399}{64}k^2$

if 2s = a + b + c then what is:

s(s - a)(s - b)(s - c) $\left[ \frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s} \right]$

((a))
abc
((b))
2abc
((c))
4abc
((d))
ab + bc + ca

Show Answer

Answer: ((a))

abc

Given:

2s = a + b + c ⇒ s = (a + b + c)/2

s(s − a)(s − b)(s − c) × $\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]$

Let us take:

a = b = c = 2 ⇒ a + b + c = 6 ⇒ 2s = 6 ⇒ s = 3

Then: s − a = s − b = s − c = 1

Now evaluate:

s(s − a)(s − b)(s − c) × $\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]$

= 3 × 1 × 1 × 1 × [1/1 + 1/1 + 1/1 − 1/3]

= 3 × (3 − 1/3) = 3 × (8/3) = 8

Check which option gives 8:

abc = 2 × 2 × 2 = 8 (Match)

2abc = 16 (Not matching)

4abc = 32 (Not matching)

ab + bc + ca = 12 (Not matching)

Therefore, the correct answer is option 1.

How much will ₹10,000 amount to in one year's time at 4% rate of interest per annum if the interest is compounded once in every three months? (take approximate value)

((a))
₹10,406
((b))
₹10,416
((c))
₹10,426
((d))
₹10,436

Show Answer

Answer: ((a))

₹10,406

Given:

Principal (P) = ₹10,000

Rate of Interest (r) = 4% per annum

Time (t) = 1 year

Compounding frequency = Quarterly (every 3 months), so n = 4

Formula used:

A = P $(1+\frac{r}{n×100})^{n×t}$

Calculation:

A = 10,000 $(1+\frac{4}{4×100})^{4×1}$

⇒ A = 10,000 $(1+\frac{1}{100})^4$

⇒ A = 10,000 $(1+0.01)^4$

⇒ A = 10,000 $(1.01)^4$

⇒ A = 10,000 × 1.04060401

⇒ A ≈ ₹10,406

∴ The correct answer is option (1).

If p = $0.0{\overline9}$ , then what is the value of 70p² + 43p - 5:

((a))
-1
((b))
0
((c))
1
((d))
10

Show Answer

Answer: ((b))

Given:

p = 0.099999...

Calculation:

Let p = $0.0{\overline9}$ = x. Multiply both sides by 10:

10x = 0.9999

100x = 9.99999...

100x − 10x = 9.99999... − 0.99999...

90x = 9

x = 0.1

So, x = 1. Therefore, p = 1.

p = 0.1 into the expression 70p² + 43p − 5:

70p² + 43p − 5 = 70(0.1)² + 43(0.1) − 5

= 0.70 + 4.3 − 5

= 0

∴ The value of 70p² + 43p − 5 is 0.

What is the remainder when 2¹⁰¹ is divided by 101 ?

((a))
1
((b))
2
((c))
5
((d))
7

Show Answer

Answer: ((b))

Given:

Find the remainder when 2¹⁰¹ is divided by 101

Formula used:

By Fermat's Little Theorem:

If p is a prime and a is not divisible by p, then:

a^p−1 ≡ 1 (mod p)

So, 2¹⁰⁰ ≡ 1 (mod 101)

Calculations:

2¹⁰¹ = 2 × 2¹⁰⁰

⇒ 2¹⁰¹ ≡ 2 × 1 ≡ 2 (mod 101)

∴ The remainder is 2.

If p( ≠ 0 ) and q( ≠ 0 ) are the roots of the equation x ²+ px + q = 0, then what is p²+ q²?

((a))
2
((b))
3
((c))
4
((d))
5

Show Answer

Answer: ((d))

Given:

The quadratic equation is: x² + p·x + q = 0

Roots of the equation are: p ≠ 0 and q ≠ 0 (same as the coefficients)

Formula used:

Let α and β be the roots of the equation:

α + β = −p

α · β = q

But it's given that roots are p and q themselve

p + q = −p ...(i)

p × q = q ...(ii)

Calculations:

From (ii): p × q = q

Divide both sides by q (since q ≠ 0):

⇒ p = 1

Substitute p = 1 into (i):

⇒ 1 + q = −1 ⇒ q = −2

Now find: p² + q²

= 1² + (−2)² = 1 + 4 = 5

∴ The correct answer is option 4.

The equation x² + px +q = 0 and x² + qx +p = 0 ( p ≠ q ) have a common root. what is the value of (p + q) ?

((a))
-1
((b))
0
((c))
1
((d))
2

Show Answer

Answer: ((a))

-1

Given:

The equations: x² + px + q = 0 and x² + qx + p = 0 (p ≠ q) have a common root.

Formula used:

If the equations have a common root, let the common root be α. Substituting α in both equations, we have:

α² + pα + q = 0

α² + qα + p = 0

By subtracting these equations, we eliminate α² and solve for p and q.

Calculation:

From the two equations:

α² + pα + q = 0

α² + qα + p = 0

Subtract the second equation from the first:

⇒ (pα + q) - (qα + p) = 0

⇒ pα - qα + q - p = 0

⇒ α(p - q) = p - q

If p ≠ q, we can cancel (p - q):

⇒ α = 1

Substitute α = 1 in the first equation:

1² + p(1) + q = 0

⇒ 1 + p + q = 0

⇒ p + q = -1

∴ The correct answer is option (1).

"If x² - 5x + 4 is a factor of x⁴ - px² + q, then what are the values of p and q respectively?"

((a))
17, 16
((b))
16, 17
((c))
15, 16
((d))
16, 15

Show Answer

Answer: ((a))

17, 16

Given:

If x² - 5x + 4 is a factor of x⁴ - px² + q.

Formula used:

If (x² - 5x + 4) is a factor of (x⁴ - px² + q), then the roots of x² - 5x + 4 (i.e., 1 and 4) must satisfy x⁴ - px² + q = 0.

Calculation:

Roots of x² - 5x + 4 are x = 1 and x = 4.

Substitute x = 1 in x⁴ - px² + q = 0

⇒ 1⁴ - p(1²) + q = 0

⇒ 1 - p + q = 0

⇒ q = p - 1 ----- (1)

Substitute x = 4 in x⁴ - px² + q = 0

⇒ 4⁴ - p(4²) + q = 0

⇒ 256 - 16p + q = 0

⇒ q = 16p - 256 ------ (2)

Equate q values from 1 and 2

⇒ p - 1 = 16p - 256

⇒ 255 = 15p

⇒ p = 17

Substitute p = 17 into q = p - 1

⇒ q = 17 - 1

⇒ q = 16

∴ The values of p and q are 17 and 16 respectively.

The correct answer is option (1).

"If two quadratic equations px² + px + 4 = 0 and x² + qx + q = 0 have a common root 2, then what is p + q equal to?"

((a))
-3
((b))
-2
((c))
0
((d))
3

Show Answer

Answer: ((b))

-2

Given:

Two quadratic equations: $px^2 + px + 4 = 0$ and $x^2 + qx + q = 0$

Common root = 2

Formula used:

For a quadratic equation $ax^2 + bx + c = 0$ , if a root is given, substitute it into the equation to find the unknowns.

Calculation:

Substituting the root 2 into the first equation:

$p(2)^2 + p(2) + 4 = 0$

⇒ $4p + 2p + 4 = 0$

⇒ $6p + 4 = 0$

⇒ $6p = -4$

⇒ $p = -\frac{2}{3}$

Substituting the root 2 into the second equation:

$(2)^2 + q(2) + q = 0$

⇒ $4 + 2q + q = 0$

⇒ $4 + 3q = 0$

⇒ $3q = -4$

⇒ $q = -\frac{4}{3}$

Now, calculate $p + q$ :

$p + q = -\frac{2}{3} - \frac{4}{3}$

⇒ $p + q = -\frac{6}{3}$

⇒ $p + q = -2$

∴ The correct answer is option (2).

what is the Hcf of the polynomials x⁸ + x⁴ + 1 and x⁴ + x² + 1 ?

((a))
1
((b))
x⁴− x²+ 1
((c))
x⁴ + x² + 1
((d))
x⁴− x²− 1

Show Answer

Answer: ((c))

x⁴ + x² + 1

Given:

Find the HCF of: f(x) = x⁸ + x⁴ + 1 and g(x) = x⁴ + x² + 1

Formula used:

a² + ab + b² = (a + b)² − ab

f(x) = x⁸ + x⁴ + 1 = (x⁴)² + x⁴ × 1 + 1²

So f(x) = x⁸ + x⁴ + 1 = a² + ab + b² where a = x⁴, b = 1

⇒ f(x) = (x⁴)² + x⁴ × 1 + 1² = x⁸ + x⁴ + 1

Similarly, g(x) = x⁴ + x² + 1 = a² + ab + b² where a = x², b = 1

So: f(x) = a² + ab + b² where a = x⁴, b = 1

g(x) = a² + ab + b² where a = x², b = 1

Now, we know that: a² + ab + b² = (a + b)² − ab

So: f(x) = (x⁴ + 1)² − x⁴

f(x) = (x⁴ + x² + 1)(x⁴ − x² + 1)

f(x) = x⁸ + x⁴ + 1 = (x⁴ + x² + 1)(x⁴ − x² + 1)

g(x) = x⁴ + x² + 1

Since x4 + x2 + 1 is common in both terms.

∴ HCF is: x⁴ + x² + 1

An arc AB of a circle subtends an angle x radians at the centre O of the circle. If the area of sector AOB is equal to half the square of the length of arc AB, then what is x equal to?

((a))
1/4
((b))
1/2
((c))
1
((d))
2

Show Answer

Answer: ((c))

Given:

An arc AB of a circle subtends an angle x radian at the center O.

The area of sector AOB = Half of the square of the length of arc AB.

Formula used:

Area of sector = (1/2) × r² × x

Length of arc = r × x

Given condition: (1/2) × r² × x = (1/2) × (r × x)²

Calculations:

(1/2) × r² × x = (1/2) × r² × x²

⇒ r² × x = r² × x²

⇒ x = x²

⇒ x² - x = 0

⇒ x(x - 1) = 0

⇒ x = 0 or x = 1

Since x cannot be 0 (as it represents an angle),

∴ x = 1 radian.

Correct answer is option (3).

Consider the following statements in respect of prime numbers p and q:

I. Their LCM is always an odd number.

II. Sum of their LCM and HCF is always an even number.

Which of the statements given above is/are correct?

((a))
I only
((b))
II only
((c))
Both I and II
((d))
Neither I nor II

Show Answer

Answer: ((d))

Neither I nor II

Given:

p and q are prime numbers

Statement I: Their LCM is always an odd number

Formula used:

LCM of two numbers is odd only if both numbers are odd

Among prime numbers, 2 is the only even prime

Calculation:

If p = 2, q = 3 ⇒ LCM = 6 (even)

So, LCM is not always odd

Statement I is incorrect

Statement II: Sum of their LCM and HCF is always even

HCF of two distinct primes = 1

LCM of p and q = p × q

Sum = p × q + 1

Check:

p = 2, q = 3 ⇒ LCM = 6, HCF = 1 ⇒ Sum = 6 + 1 = 7 odd

p = 3, q = 5 ⇒ LCM = 15, HCF = 1 ⇒ Sum = 15 + 1 = 16 even

p = 3, q = 7 ⇒ 21 + 1 = 22 even

But counter example: p = 2, q = 3 ⇒ 6 + 1 = 7

Statement II is incorrect

∴ The correct answer is: Neither I nor II.

The frequency distribution of 205 observations on X is given below:

x	3	5	6	7
Frequency	f	f + 2	f − 3	f + 6

What is the value of f?

((a))
50
((b))
60
((c))
70
((d))
80

Show Answer

Answer: ((a))

Given:

Frequency distribution of X with total observations = 205

x = 3, 5, 6, 7

Frequencies = f, f + 2, f - 3, f + 6

Formula used:

Sum of all frequencies = Total observations

Calculations:

f + (f + 2) + (f - 3) + (f + 6) = 205

⇒ 4f + 5 = 205

⇒ 4f = 200

⇒ f = 50

∴ The value of f is 50.

What is the median of the frequency distribution?

((a))
3
((b))
5
((c))
6
((d))
It cannot be determined from the given data

Show Answer

Answer: ((c))

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56 (from previous calculation f = 50)

Total observations = 205

Formula used:

Median for grouped data = L + [(N/2 - CF)/f] × h

Where, L = lower boundary of median class, N = total observations, CF = cumulative frequency before median class, f = frequency of median class, h = class width

Calculations:

Cumulative frequencies:

3 → 50

5 → 50 + 52 = 102

6 → 102 + 47 = 149

7 → 149 + 56 = 205

N/2 = 205/2 = 102.5 → lies in class x = 6 (since CF before = 102)

Assuming class width h = 1, L = 6

Median = 6 + [(102.5 - 102)/47] × 1

⇒ Median = 6 + [0.5/47]

⇒ Median ≈ 6 + 0.0106 ≈ 6

∴ The median of the frequency distribution is approximately 6.

What is the mode of the frequency distribution?

((a))
5
((b))
6
((c))
7
((d))
It cannot be determined from the given data

Show Answer

Answer: ((c))

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Total observations = 205

Formula used:

Mode for grouped data = L + [(f1 - f0) / (2f1 - f0 - f2)] × h

Where, L = lower boundary of modal class, f1 = frequency of modal class, f0 = frequency of previous class, f2 = frequency of next class, h = class width

Calculations:

Highest frequency = 56 → modal class x = 7

f1 = 56, f0 = 47 (previous), f2 = 0 (no class after 7), h = 1, L = 7

Mode = 7 + [(56 - 47)/(2×56 - 47 - 0)] × 1

⇒ Mode = 7 + [9 / (112 - 47)]

⇒ Mode = 7 + 9 / 65 ≈ 7 + 0.138 ≈ 7

∴ The mode of the frequency distribution is approximately 7.

What is the most appropriate graphical representation of the given frequency distribution of X?

((a))
Bar diagram
((b))
Histogram
((c))
Frequency polygon
((d))
Pie Chart

Show Answer

Answer: ((a))

Bar diagram

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Formula used:

For discrete data, suitable graphs are bar graph or histogram (for continuous intervals); here X is discrete.

Calculations:

Plot x-values on x-axis (3, 5, 6, 7)

Plot frequencies on y-axis (50, 52, 47, 56)

Draw bars corresponding to each x-value with height equal to its frequency

∴ The most appropriate graphical representation is a bar graph.

What is the mean of the frequency distribution?

((a))
3·29
((b))
4·29
((c))
5·29
((d))
6·29

Show Answer

Answer: ((c))

5·29

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Total observations = 205

Formula used:

Mean = Σ(f × x) ÷ Σf

Calculations:

f × x:

3 × 50 = 150

5 × 52 = 260

6 × 47 = 282

7 × 56 = 392

Σ(f × x) = 150 + 260 + 282 + 392 = 1084

Mean = 1084 ÷ 205 ≈ 5.29

∴ The mean of the frequency distribution is approximately 5.29.

The frequency distribution of marks obtained by students in an English examination is given below:

Marks obtained	Number of Students
Below 40	50
Below 50	125
Below 60	210
Below 70	315
Below 80	350

<br>

What is the number of students who scored between 60 and 70 marks?

((a))
105
((b))
110
((c))
205
((d))
210

Show Answer

Answer: ((a))

105

Given:

Cumulative frequency distribution of marks:

Below 40 = 50

Below 50 = 125

Below 60 = 210

Below 70 = 315

Below 80 = 350

Formula used:

Number of students in a class = Cumulative frequency of upper limit - Cumulative frequency of lower limit

Calculations:

Students between 60 and 70 = 315 - 210 = 105

∴ The number of students who scored between 60 and 70 marks is 105.

What is the number of students who scored more than 50 marks?

((a))
100
((b))
125
((c))
200
((d))
225

Show Answer

Answer: ((d))

225

Given:

Cumulative frequency distribution of marks:

Below 40 = 50

Below 50 = 125

Below 60 = 210

Below 70 = 315

Below 80 = 350

Formula used:

Number of students scoring more than a mark = Total students - Cumulative frequency of that mark

Calculations:

Total students = 350 (Below 80)

Students scoring more than 50 = 350 - 125 = 225

∴ The number of students who scored more than 50 marks is 225.

Circulation figures (in thousands) of different newspapers (A, B, C, D, E) for five years are given below:

Year	A	B	C	D	E
2019	20	10	15	8	20
2020	12	12	18	12	12
2021	24	14	17	14	15
2022	26	10	16	15	9
2023	22	16	14	16	11

In which of the given years was the circulation of the newspaper D close to its average circulation over all the years?

((a))
2020 and 2021
((b))
2022 and 2023
((c))
2022 only
((d))
2020 only

Show Answer

Answer: ((a))

2020 and 2021

Calculations:

Sum of D = 8 + 12 + 14 + 15 + 16 = 65 (thousands)

Average of D = 65 ÷ 5 = 13 (thousands)

Differences from average:

2019: |8 − 13| = 5

2020: |12 − 13| = 1

2021: |14 − 13| = 1

2022: |15 − 13| = 2

2023: |16 − 13| = 3

∴ The circulation of D was closest to its average (13 thousands) in the years 2020 and 2021 (difference = 1 thousand).

In which of the years from 2019 to 2022 was the circulation of the newspaper D close to the average circulation of all the newspapers in that year?

((a))
2019
((b))
2020
((c))
2021
((d))
2022

Show Answer

Answer: ((d))

2022

Given:

Circulation figures (in thousands) for newspapers A, B, C, D, E from 2019 to 2022

Formula used:

Average circulation = (A + B + C + D + E) ÷ 5

Calculations:

2019: Total = 20 + 10 + 15 + 8 + 20 = 73

Average = 73 ÷ 5 = 14.6; D = 8 → Difference = 6.6

2020: Total = 12 + 12 + 18 + 12 + 12 = 66

Average = 66 ÷ 5 = 13.2; D = 12 → Difference = 1.2

2021: Total = 24 + 14 + 17 + 14 + 15 = 84

Average = 84 ÷ 5 = 16.8; D = 14 → Difference = 2.8

2022: Total = 26 + 10 + 16 + 15 + 9 = 76

Average = 76 ÷ 5 = 15.2; D = 15 → Difference = 0.2

∴ The circulation of newspaper D was closest to the average in 2022.

How many cases are there in which average of the circulation for an individual newspaper was more than the average of the circulation of all the newspapers?

((a))
One
((b))
Two
((c))
Three
((d))
Four

Show Answer

Answer: ((b))

Two

Given:

Circulation figures (in thousands) of newspapers A, B, C, D, E for years 2019–2023

Formula used:

Average for individual newspaper = Sum of its circulation ÷ 5

Average of all newspapers for all years = Total circulation of all newspapers ÷ (5 × 5)

Calculations:

Total circulation for each newspaper:

A = 20 + 12 + 24 + 26 + 22 = 104 → Average = 104 ÷ 5 = 20.8

B = 10 + 12 + 14 + 10 + 16 = 62 → Average = 62 ÷ 5 = 12.4

C = 15 + 18 + 17 + 16 + 14 = 80 → Average = 80 ÷ 5 = 16

D = 8 + 12 + 14 + 15 + 16 = 65 → Average = 65 ÷ 5 = 13

E = 20 + 12 + 15 + 9 + 11 = 67 → Average = 67 ÷ 5 = 13.4

Total circulation of all newspapers = 104 + 62 + 80 + 65 + 67 = 378

Average circulation of all newspapers = 378 ÷ 25 = 15.12

Compare individual averages with 15.12:

A = 20.8 > 15.12 ✅

B = 12.4 < 15.12 ❌

C = 16 > 15.12 ✅

D = 13 < 15.12 ❌

E = 13.4 < 15.12 ❌

∴ There are 2 newspapers (A and C) whose average circulation is more than the average of all newspapers.

Study the following Pie Charts:

On an average, how many persons live in every pucca house?

((a))
8
((b))
8·54
((c))
9·54
((d))
7·54

Show Answer

Answer: ((c))

9·54

Given:

From Chart-I: Number of houses = 1,20,000

Angle for Pucca houses = 110°

From Chart-II: Population = 21,00,000

Angle for population living in Pucca houses = 60°

Formula used:

Part of total = (Angle / 360) × Total Value

Average = Total Value / Number of units

Calculations:

Number of Pucca Houses = (110 / 360) × 1,20,000

⇒ Number of Pucca Houses = (11 / 36) × 1,20,000

⇒ Number of Pucca Houses = 11 × (120000 / 36)

⇒ Number of Pucca Houses = 110000 / 3

Population Living in Pucca Houses = (60 / 360) × 21,00,000

⇒ Population Living in Pucca Houses = (1 / 6) × 21,00,000

⇒ Population Living in Pucca Houses = 3,50,000

Average Persons per Pucca House = Population Living in Pucca Houses / Number of Pucca Houses

⇒ Average Persons per Pucca House = 350000 / (110000 / 3)

⇒ Average Persons per Pucca House = (350000 × 3) / 110000

⇒ Average Persons per Pucca House = (35 × 3) / 11

⇒ Average Persons per Pucca House = 105 / 11

⇒ Average Persons per Pucca House ≈ 9.5454...

∴ The correct answer is option 3.

If 5000 more Kutcha houses are built, then what will be approximate change in angle for Kutcha houses in Pie Chart-I?

((a))
11°
((b))
13°
((c))
15°
((d))
23°

Show Answer

Answer: ((b))

13°

Given:

From Chart-I: Initial total number of houses = 1,20,000

Initial angle for Kutcha houses = 40°

Increase in Kutcha houses = 5000

Formula used:

Number of houses for a category = (Angle for category / 360°) × Total number of houses

Angle for a category = (Number of houses for category / Total number of houses) × 360°

Calculations:

Initial number of Kutcha houses = (40 / 360) × 1,20,000

⇒ Initial number of Kutcha houses = (1 / 9) × 1,20,000

⇒ Initial number of Kutcha houses = 120000 / 9

New total number of houses = Initial total number of houses + 5000

⇒ New total number of houses = 1,20,000 + 5000

⇒ New total number of houses = 1,25,000

New number of Kutcha houses = Initial number of Kutcha houses + 5000

⇒ New number of Kutcha houses = (120000 / 9) + 5000

⇒ New number of Kutcha houses = (120000 + 45000) / 9

⇒ New number of Kutcha houses = 165000 / 9

New angle for Kutcha houses = (New number of Kutcha houses / New total number of houses) × 360°

⇒ New angle for Kutcha houses = ((165000 / 9) / 125000) × 360°

⇒ (165000 / (9 × 125000)) × 360°

⇒ (165 / (9 × 125)) × 360°

⇒ (11 / 75) × 360° = 11 × 4.8 = 52.8°

Change in angle = New angle for Kutcha houses - Initial angle for Kutcha houses

⇒ Change in angle = 52.8° - 40°

⇒ Change in angle = 12.8° ≈ 13°

∴ The approximate change in angle for Kutcha houses in Pie Chart-I will be 12.8°.

If 300 families from the "Houseless" category shift into Kutcha houses, what will be the average number of families in every Kutcha house?

((a))
15
((b))
13.9
((c))
12.9
((d))
10.5

Show Answer

Answer: ((b))

13.9

Given:

From Chart-III: Total number of families = 3,60,000

Angle for families in "Houseless" category = 20°

Angle for families "Living in Kutcha houses" = 185°

Number of families from "Houseless" category shifting to Kutcha houses = 300

Formula used:

Number of families for a category = (Angle for category / 360°) × Total number of families

Average families per house = Total families in houses / Total number of houses of that type

Calculations:

Initial Families in Kutcha Houses = (185 / 360) × 3,60,000

⇒ Initial Families in Kutcha Houses = 185 × (3,60,000 / 360) = 1,85,000

Initial Houseless Families = (20 / 360) × 3,60,000

⇒ Initial Houseless Families = 20 × (3,60,000 / 360) = 20,000

New Families in Kutcha Houses = Initial Families in Kutcha Houses + 300

⇒ New Families in Kutcha Houses = 1,85,000 + 300 = 1,85,300

From previous calculations or Chart-I:

Number of Kutcha Houses = (40 / 360) × 1,20,000

⇒ Number of Kutcha Houses = 120000 / 9

Average families per Kutcha house = New Families in Kutcha Houses / Number of Kutcha Houses

⇒ Average families per Kutcha house = 1,85,300 / (120000 / 9)

⇒ Average families per Kutcha house = (1,85,300 × 9) / 120000

⇒ Average families per Kutcha house = 1667700 / 120000

⇒ Average families per Kutcha house = 16677 / 1200

⇒ Average families per Kutcha house ≈ 13.8975 ≈ 13.9

∴ The average number of families in every Kutcha house will be 13.90.

Marks obtained by 60 students in a test (Maximum Marks-100) are given below :

60, 82, 63, 40, 88, 80, 24, 60, 09, 58, 44, 74 57, 60, 63, 30, 31, 64, 00, 62, 37, 54, 10, 99 35, 83, 58, 33, 92, 43, 66, 41, 78, 72, 32, 33 65, 11, 60, 17, 53, 93, 57, 03, 29, 31, 82, 50 63, 15, 41, 55, 72, 42, 32, 23, 30, 48, 67, 34

What is the percentage (approximate) of students getting marks higher than or equal to 60 but less than 80 ?

((a))
25.0
((b))
26.7
((c))
28.3
((d))
30.0

Show Answer

Answer: ((b))

26.7

Given:

Total students = 60

Marks range considered = ≥ 60 and < 80

Calculations:

Number of students in the range = 16

Percentage = (16 ÷ 60) × 100

⇒ Percentage = 26.67% ≈ 26.7%

∴ Approximately 26.7% of students scored between 60 and 79 marks.

100

The arithmetic mean of 100 observations was 50. Later on, it was found that two observations were misread as 82 and 6 instead of 182 and 56. What is the value of the corrected mean?

((a))
50.1
((b))
51.5
((c))
52.3
((d))
53.4

Show Answer

Answer: ((b))

51.5

Given:

Initial arithmetic mean of 100 observations = 50

Misread observations = 82 and 6

Correct observations = 182 and 56

Formula used:

Mean = (Sum of all observations) ÷ (Number of observations)

Calculations:

Initial sum of observations = Mean × Number of observations = 50 × 100 = 5000

Difference in sum due to misread observations = (182 + 56) - (82 + 6) = 238 - 88 = 150

Corrected sum of observations = 5000 + 150 = 5150

Corrected mean = Corrected sum ÷ Number of observations = 5150 ÷ 100 = 51.5

∴ The corrected mean is 51.5

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