Quantitative Aptitude PYQs

Simplify the right-hand side:

⇒ 428 - 195 = 233

Simplify known values on the left-hand side:

⇒ 23² = 529

⇒ √624 ≈ 24.98 ≈ 25 (approx)

Plug into the equation:

⇒ (?) × 17.93 - 529 + 25 = 233

⇒ (?) × 17.93 - 504 = 233

⇒ (?) × 17.93 = 233 + 504 = 737

⇒ ? = 737 ÷ 17.93 ≈ 737 ÷ 18 ≈ 41

Thus, the correct answer is 41.

Approximate the numbers:

189.84 ≈ 190

327.11 ≈ 327

215.96 ≈ 216

4.98 ≈ 5

Perform the left-side calculation:

190 + 327 = 517

517 - 216 = 301

301 + 5 = 306

Set up the equation:

306 = 9 × ?

Solve for ?:

? = 306 ÷ 9 = 34

Thus, the correct answer is 34.

Given Expression:

18.45² + ∛(729.08) + 89.32 = ?² - 42.99

Approximate each term:

18.45² ≈ 340.2 (since 18.45 × 18.45 ≈ 340.2025)

∛729.08 ≈ ∛729 = 9

89.32 stays as is

LHS: 340.2 + 9 + 89.32 = 438.52

⇒ 438.52 = ?² - 42.99

⇒ ?² = 438.52 + 42.99 = 481.51

√481.51 ≈ 21.93 ≈ 22

Thus, the correct answer is 22.

Approximate each term by rounding to the nearest whole number:

49.95 ≈ 50

12.12 ≈ 12

19.98 ≈ 20

7.95 ≈ 8

1.98 ≈ 2

60.05 ≈ 60

Calculate each component:

First term: 50 × 12 = 600

Second term: 20 × 8 = 160

Third term: 2 × 60 = 120

Combine the terms:

600 - 160 - 120 = 320

Thus, the correct answer is 320.

⇒ (1845.89 + 3126.47 - 1280.35)

⇒ (4972.36 - 1280.35) = 3692.01

⇒ 3692.01 ÷ 5.25 ≈ 703.24

⇒ 703.24 × 3 ≈ 2109.72

⇒ 2109.72 = ? + 187.263

⇒ ? = 2109.72 - 187.263 ≈ 1922.46

Thus, the correct answer is 1922.5.

Shortcut Trick

Speed of Current (v) = 14.28% of Boat Speed (u) ⇒ v = (1/7) × u.

Ratio of u : v = 7 :

1. Let u = 7x and v = x.

Upstream speed (u−v) = 6x and Downstream speed (u+v) = 8x.

Time difference: 300/(6x) − 280/(8x) = 3 ⇒ 50/x − 35/x = 3 ⇒ 15/x = 3 ⇒ x = 5.

Boat speed in still water = 7x = 7 × 5 = 35 km/hr.

∴ The correct answer is 35 km/hr.

Alternate Method

Given: Upstream distance = 300 km, Downstream distance = 280 km, Time diff = 3 hrs.

Formula: Speed Upstream = u − v; Speed Downstream = u + v; Time = Distance ÷ Speed.

⇒ v = 14.28% of u = (1/7)u

⇒ Let u = 7x, then v = x

⇒ Upstream speed (u−v) = 6x

⇒ Downstream speed (u+v) = 8x

⇒ (300 / 6x) − (280 / 8x) = 3

⇒ 50/x − 35/x = 3

⇒ 15/x = 3 ⇒ x = 5

⇒ Speed of boat in still water = 7 × 5 = 35 km/hr

∴ The correct answer is 35 km/hr.

Additional Information

Relative Speed in Water

Downstream speed (D) = u + v and Upstream speed (U) = u − v, where u is boat speed and v is stream speed.

Calculating Component Speeds

Speed of boat in still water u = (D + U) ÷ 2 and Speed of current v = (D − U) ÷ 2.

Percentage to Fraction Conversion

14.28% is approximately equal to 1/7. This is a common fraction used in competitive exams for 1/7, 2/7 (28.56%), etc.

Shortcut Trick

Profit ratio is the product of investment and time period.

Total profit = 3800 and B's share = 500. Ratio of B's profit to total profit = 500 ÷ 3800 = 5 ÷ 38.

B's weighted investment = (x + 1000) × 5. Total weighted investment = 5x + 5(x + 1000) + 8000 × 12.

Using Ratio-Unitary: [5(x + 1000)] ÷ [10x + 5000 + 96000] = 5 ÷ 38.

Solving: 38(x + 1000) = 10x + 101000 ⇒ 38x + 38000 = 10x + 101000 ⇒ 28x = 63000.

∴ The correct answer is Rs. 2250.

Alternate Method

Given → Investment of A = x, Investment of B = x + 1000, Investment of C = 8000.
Time → t_A = 5, t_B = 5, t_C = 12.
Formulas: Profit Ratio = (I₁ × T₁) : (I₂ × T₂) : (I₃ × T₃).

Steps:
⇒ Profit Share Ratio = (x × 5) : ((x + 1000) × 5) : (8000 × 12)
⇒ Profit Share Ratio = 5x : (5x + 5000) : 96000
⇒ Total Profit Units = 5x + 5x + 5000 + 96000 = 10x + 101000
⇒ (B's Profit) / (Total Profit) = (5x + 5000) / (10x + 101000) = 500 / 3800
⇒ (5x + 5000) / (10x + 101000) = 5 / 38
⇒ Divide numerator by 5: (x + 1000) / (10x + 101000) = 1 / 38
⇒ 38x + 38000 = 10x + 101000
⇒ 28x = 63000
⇒ x = 63000 ÷ 28 = 2250

∴ The correct answer is Rs. 2250.

Additional Information

Partnership Basics

When time periods are different, the profit is distributed in the ratio of capital × time. This is known as compound partnership.

Ratio Proportionality

If Profit Ratio is P₁:P₂ and Capital Ratio is C₁:C₂, then the Time Ratio T₁:T₂ = (P₁/C₁) : (P₂/C₂).

General Solution:

 

Calculations:

We are given:

Total passengers in Train B on Thursday = 90 (from general solution)

Female passengers are 25% more than male passengers in Train B on Thursday

Let male passengers = x

⇒ Female passengers = 1.25x

⇒ Total in Train B on Thursday = x + 1.25x = 2.25x

⇒ 2.25x = 90

⇒ x = 90 ÷ 2.25 = 40

Thus, 40 male passengers travelled in train B on Thursday.

General Solution:

 

Calculations:

Passengers in Train B on Monday = 60 (from general solution)

Passengers in Train B on Wednesday = 84 (from general solution)

Train C (Monday): 8/5 × 60 = (8 × 60) ÷ 5 = 480 ÷ 5 = 96

Required Ratio:

Train C (Monday) : Train B (Wednesday) = 96 : 84

Simplify: Divide both by 12 → 96 ÷ 12 = 8, 84 ÷ 12 = 7

Thus, the correct answer is 8:7.

General Solution:

 

Calculations:

Passengers in Train B each day:

Monday: 60

Tuesday: 81

Wednesday: 84

Thursday: 90

Total = 60 + 81 + 84 + 90 = 315

Thus, the correct answer is 315.

General Solution:

 

Calculations:

Let’s calculate % difference between Train B and Train A for each day:

Monday:

A = 105 → 105/165 × 100 = 63.64%

B = 60 → 60/165 × 100 = 36.36%

Diff = 36.36% – 63.4% = -27.28%

Tuesday:

A = 54 → 40%

B = 81 → 60%

Diff = 20%

Wednesday:

A = 96 → 53.33%

B = 84 → 46.67%

Diff = 6.66%

Thursday:

A = 54 → 37.5%

B = 90 → 62.5%

Diff = 25%

Thursday: 62.5 – 37.5 = 25% → greatest margin

Thus, the correct answer is Thursday.

General Solution:

 

Calculations:

Answer Key:

​Sum A (Mon + Wed) = 105 + 96 = 201

Sum B (Tue + Thu) = 81 + 90 = 171

Ratio = 201 : 171 ⇒ 67 : 57

⇒ p + q = 67 + 57 = 124

Thus, the correct answer is 124.

Shortcut Trick

Net multiplier after 3 years depreciation at 10% = (0.9)³ = 0.729.

Value after 3 years = 10,000 × 0.729 = Rs. 7,290.

Final Selling Price (SP) = 7,290 − 270 = Rs. 7,020.

Total Loss = CP − SP = 10,000 − 7,020 = Rs. 2,980.

Total Loss % = (2,980 ÷ 10,000) × 100 = 29.8%.

∴ The correct answer is 29.8%.

Alternate Method

Given → Initial Cost Price (CP) = Rs. 10,000; Depreciation Rate = 10% per year; Time = 3 years; Additional Discount = Rs. 270. Formulas: Final Value = P × (1 − R÷100)ⁿ | Total Loss = CP − Final SP | Loss% = (Total Loss ÷ CP) × 100.

⇒ Value after 1^st year = 10,000 − 10% of 10,000 = Rs. 9,000.
⇒ Value after 2^nd year = 9,000 − 10% of 9,000 = Rs. 8,100.
⇒ Value after 3^rd year = 8,100 − 10% of 8,100 = Rs. 7,290.
⇒ Selling Price (SP) after Rs. 270 discount = 7,290 − 270 = Rs. 7,020.
⇒ Total Loss Amount = 10,000 − 7,020 = Rs. 2,980.
⇒ Total Loss % = (2,980 ÷ 10,000) × 100 = 29.8%.

∴ The correct answer is 29.8%.

Additional Information

Depreciation

Depreciation is the reduction in the value of an asset over time. Formula: A = P(1 − R/100)ⁿ.

Successive Percentage Decrease

Three successive decreases of 10% result in a net decrease of [1 − (0.9 × 0.9 × 0.9)] × 100 = 27.1%.

Shortcut Trick

Total present age of P, Q, and R = 61 + (3 × 12) = 97 years.

Let P = 2x, Q = 3x, then R = 2x − 8.

Sum: 2x + 3x + (2x − 8) = 97 → 7x = 105 → x = 15.

P's age = 2 × 15 = 30 years.

Ratio P : S = 5 : 7 → S = (30 ÷ 5) × 7 = 42 years.

∴ The correct answer is 42 years.

Alternate Method

Given: Sum of ages (P+Q+R) 12 years ago = 61 years; P:Q = 2:3; P = R + 8; P:S = 5:7.

⇒ Present sum of P, Q, and R = 61 + (12 × 3) = 97 years.

⇒ Let Present age of P = 2x and Q = 3x.

⇒ Since P is 8 years older than R, R = P − 8 = 2x − 8.

⇒ 2x + 3x + (2x − 8) = 97

⇒ 7x − 8 = 97

⇒ 7x = 105

⇒ x = 15

⇒ Present age of P = 2 × 15 = 30 years.

⇒ Given Ratio P : S = 5 : 7

⇒ 5 units = 30 years

⇒ 1 unit = 6 years

⇒ Present age of S = 7 × 6 = 42 years.

∴ The correct answer is 42 years.

Additional Information

Age Sum Calculation

When moving "n" years forward or backward for a group of "k" people, the total sum changes by (n × k) years.

Ratio Balancing

To combine ratios P:Q and P:S, make the value of the common term (P) equal using the LCM of its ratio parts.

Shortcut Trick

Ratio of White balls to Total balls = 1 : 4.

Total balls = 6 + y + (y + 4) = 10 + 2y. Since y / (10 + 2y) = 1/4 ⇒ 4y = 10 + 2y ⇒ 2y = 10 ⇒ y = 5.

Black balls = y + 4 = 9; Total balls = 10 + 2(5) = 20.

Probability (Both Black) = (9 × 8) / (20 × 19) = 18 / 95.

∴ The correct answer is 18/95.

Alternate Method

Given: Blue = 6, White = y, Black = y + 4.

Formula Used: Probability = (Favorable Outcomes) ÷ (Total Outcomes)

⇒ Total balls = 6 + y + y + 4 = 10 + 2y

⇒ P(White) = y / (10 + 2y) = 1/4

⇒ 4y = 10 + 2y ⇒ 2y = 10 ⇒ y = 5

⇒ Number of Black balls = 5 + 4 = 9

⇒ Total number of balls = 10 + 2(5) = 20

⇒ P(Both Black) = ⁹C₂ / ²⁰C₂

⇒ P(Both Black) = (9 × 8 / 2) / (20 × 19 / 2) = 36 / 190

⇒ P(Both Black) = 18 / 95

∴ The correct answer is 18/95.

Additional Information

Combinations Formula

For selecting r items from n, use ⁿC_r = n! / [r!(n−r)!]. For r=2, it simplifies to n(n−1)/2.

Simultaneous vs. Sequential Draw

Drawing two balls simultaneously is equivalent to drawing them without replacement one after another.

Complementary Probability

P(A) + P(not A) = 1. This is useful for 'at least' or 'at most' probability questions.

General Solution:

Let the runs scored by A in Test matches = x

Then:

ODIs = 55% of x = 0.55x

T20s = Total of A’s average = 230 × 3 = 690

So, A's T20 runs = 690 − (x + 0.55x) = 690 − 1.55x

C's T20 runs = 40 more than one-third of A’s Test runs:

C's T20 = (1/3) × x + 40

We're told:

C and D scored 180 and 95 runs respectively in T20

So:

C's T20 = 180 = (1/3)x + 40

⇒ (1/3)x = 140 ⇒ x = 420

Now that we know x = 420, substitute values for A:

A's Test = 420

A's ODI = 0.55 × 420 = 231

A's T20 = 690 − (420 + 231) = 39

ODI data:

C's ODI = c

D's ODI = c − 42

B's ODI = (1/3) × A ODI = (1/3) × 231 = 77

Total ODI by B, C, D = 471

⇒ 77 + c + (c − 42) = 471

⇒ 2c + 35 = 471 ⇒ 2c = 436 ⇒ c = 218

So:

C's ODI = 218

D's ODI = 176

B's ODI = 77

Test Match Data:

D's Test = 4 × D ODI = 4 × 176 = 704

C's Test = D's Test + 8 = 712

Also:

C's Test = 2 × B's Test ⇒ B's Test = 712 / 2 = 356

T20 Data:

C's T20 = 180 (already given)

D's T20 = 95

B's T20 = (1/7) × B ODI = (1/7) × 77 = 11

Thus,

 

Calculations:

From the general solution:

A's Test runs = 420

B's T20 runs = 11

D's T20 runs = 95

B + D T20 runs = 11 + 95 = 106

Required Ratio:

Ratio = A's Test : (B + D)'s T20 = 420 : 106 = 210:53

Thus, the correct answer is 210:53.

General Solution:

Let the runs scored by A in Test matches = x

Then:

ODIs = 55% of x = 0.55x

T20s = Total of A’s average = 230 × 3 = 690

So, A's T20 runs = 690 − (x + 0.55x) = 690 − 1.55x

C's T20 runs = 40 more than one-third of A’s Test runs:

C's T20 = (1/3) × x + 40

We're told:

C and D scored 180 and 95 runs respectively in T20

So:

C's T20 = 180 = (1/3)x + 40

⇒ (1/3)x = 140 ⇒ x = 420

Now that we know x = 420, substitute values for A:

A's Test = 420

A's ODI = 0.55 × 420 = 231

A's T20 = 690 − (420 + 231) = 39

ODI data:

C's ODI = c

D's ODI = c − 42

B's ODI = (1/3) × A ODI = (1/3) × 231 = 77

Total ODI by B, C, D = 471

⇒ 77 + c + (c − 42) = 471

⇒ 2c + 35 = 471 ⇒ 2c = 436 ⇒ c = 218

So:

C's ODI = 218

D's ODI = 176

B's ODI = 77

Test Match Data:

D's Test = 4 × D ODI = 4 × 176 = 704

C's Test = D's Test + 8 = 712

Also:

C's Test = 2 × B's Test ⇒ B's Test = 712 / 2 = 356

T20 Data:

C's T20 = 180 (already given)

D's T20 = 95

B's T20 = (1/7) × B ODI = (1/7) × 77 = 11

Thus,

 

Calculations:

Original Runs Scored by A:

Test = 420

ODI = 231

T20 = 39

New T20 Runs (if doubled):

New T20 = 39 × 2 = 78

New Total Runs:

Total = 420 (Test) + 231 (ODI) + 78 (New T20) = 729

New average = 729 ÷ 3 = 243

 

Thus, the correct answer is 243.

General Solution:

Let the runs scored by A in Test matches = x

Then:

ODIs = 55% of x = 0.55x

T20s = Total of A’s average = 230 × 3 = 690

So, A's T20 runs = 690 − (x + 0.55x) = 690 − 1.55x

C's T20 runs = 40 more than one-third of A’s Test runs:

C's T20 = (1/3) × x + 40

We're told:

C and D scored 180 and 95 runs respectively in T20

So:

C's T20 = 180 = (1/3)x + 40

⇒ (1/3)x = 140 ⇒ x = 420

Now that we know x = 420, substitute values for A:

A's Test = 420

A's ODI = 0.55 × 420 = 231

A's T20 = 690 − (420 + 231) = 39

ODI data:

C's ODI = c

D's ODI = c − 42

B's ODI = (1/3) × A ODI = (1/3) × 231 = 77

Total ODI by B, C, D = 471

⇒ 77 + c + (c − 42) = 471

⇒ 2c + 35 = 471 ⇒ 2c = 436 ⇒ c = 218

So:

C's ODI = 218

D's ODI = 176

B's ODI = 77

Test Match Data:

D's Test = 4 × D ODI = 4 × 176 = 704

C's Test = D's Test + 8 = 712

Also:

C's Test = 2 × B's Test ⇒ B's Test = 712 / 2 = 356

T20 Data:

C's T20 = 180 (already given)

D's T20 = 95

B's T20 = (1/7) × B ODI = (1/7) × 77 = 11

Thus,

 

Calculations:

Calculate the difference between C's Test runs and (A + B) T20 runs:

C's Test runs = 712

A's T20 runs = 39

B's T20 runs = 11

Sum of A and B T20 runs = 39 + 11 = 50

Difference = 712 − 50 = 662

Calculate total runs by D:

D's Test runs = 704

D's ODI runs = 176

D's T20 runs = 95

Total runs by D = 704 + 176 + 95 = 975

Calculate percentage:

(Difference ÷ D's total runs) × 100 = (662 ÷ 975) × 100 ≈ 67.9%

Thus, the correct answer is 67.9%.

General Solution:

Let the runs scored by A in Test matches = x

Then:

ODIs = 55% of x = 0.55x

T20s = Total of A’s average = 230 × 3 = 690

So, A's T20 runs = 690 − (x + 0.55x) = 690 − 1.55x

C's T20 runs = 40 more than one-third of A’s Test runs:

C's T20 = (1/3) × x + 40

We're told:

C and D scored 180 and 95 runs respectively in T20

So:

C's T20 = 180 = (1/3)x + 40

⇒ (1/3)x = 140 ⇒ x = 420

Now that we know x = 420, substitute values for A:

A's Test = 420

A's ODI = 0.55 × 420 = 231

A's T20 = 690 − (420 + 231) = 39

ODI data:

C's ODI = c

D's ODI = c − 42

B's ODI = (1/3) × A ODI = (1/3) × 231 = 77

Total ODI by B, C, D = 471

⇒ 77 + c + (c − 42) = 471

⇒ 2c + 35 = 471 ⇒ 2c = 436 ⇒ c = 218

So:

C's ODI = 218

D's ODI = 176

B's ODI = 77

Test Match Data:

D's Test = 4 × D ODI = 4 × 176 = 704

C's Test = D's Test + 8 = 712

Also:

C's Test = 2 × B's Test ⇒ B's Test = 712 / 2 = 356

T20 Data:

C's T20 = 180 (already given)

D's T20 = 95

B's T20 = (1/7) × B ODI = (1/7) × 77 = 11

Thus,

 

Calculations:

Identify top 2 highest total run-scorers from total runs:

C: 1110

D: 975

Among these two, who satisfies:

Test runs > 700 and ODI runs ≥ 200

C:

Test = 712 (> 700)

ODI = 218 (≥ 200)

C qualifies

D:

Test = 704 (> 700)

ODI = 176 (< 200)

D does not qualify

Probability:

Only 1 player (C) out of 2 satisfies both conditions.

So, Required probability = 1/2

Thus, the correct answer is 1/2.

General Solution:

Let the runs scored by A in Test matches = x

Then:

ODIs = 55% of x = 0.55x

T20s = Total of A’s average = 230 × 3 = 690

So, A's T20 runs = 690 − (x + 0.55x) = 690 − 1.55x

C's T20 runs = 40 more than one-third of A’s Test runs:

C's T20 = (1/3) × x + 40

We're told:

C and D scored 180 and 95 runs respectively in T20

So:

C's T20 = 180 = (1/3)x + 40

⇒ (1/3)x = 140 ⇒ x = 420

Now that we know x = 420, substitute values for A:

A's Test = 420

A's ODI = 0.55 × 420 = 231

A's T20 = 690 − (420 + 231) = 39

ODI data:

C's ODI = c

D's ODI = c − 42

B's ODI = (1/3) × A ODI = (1/3) × 231 = 77

Total ODI by B, C, D = 471

⇒ 77 + c + (c − 42) = 471

⇒ 2c + 35 = 471 ⇒ 2c = 436 ⇒ c = 218

So:

C's ODI = 218

D's ODI = 176

B's ODI = 77

Test Match Data:

D's Test = 4 × D ODI = 4 × 176 = 704

C's Test = D's Test + 8 = 712

Also:

C's Test = 2 × B's Test ⇒ B's Test = 712 / 2 = 356

T20 Data:

C's T20 = 180 (already given)

D's T20 = 95

B's T20 = (1/7) × B ODI = (1/7) × 77 = 11

Thus,

 

Calculations:

Redistribution of A's ODI runs

A's ODI = 231 runs redistributed to B and C in the ratio 3:4.

Total parts = 3 + 4 = 7

B gets = (3/7) × 231 = 99

C gets = (4/7) × 231 = 132

So, update B's and C’s ODI:

B new ODI = 77 + 99 = 176

C new ODI = 218 + 132 = 350

A ODI = 0 (since it’s fully redistributed)

Double D’s T20 runs

D original T20 = 95

New T20 = 95 × 2 = 190

New Total Runs:

 

Difference between the highest and lowest scorers

Highest = C = 1242

Lowest = A = 459

Difference = 1242 − 459 = 783

Thus, the correct answer is 783.