Quantitative Aptitude PYQs

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.

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).

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).

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.

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.

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.

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.

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.

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

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).

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.

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).

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

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).

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).

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

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).

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°.

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.

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

⇒ 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.