Quantitative Aptitude PYQs

General solution:

Total number of sarees sold in all the shops A, B and C = 2000

Number of sarees sold in shop A = x

Number of sarees sold in shop B = x ×  (125/100) = 1.25x

Number of sarees sold in shop C = (x + 50)

(x + 1.25x + x + 50) = 2000

3.25x = 1950

x = 600

Number of sarees sold in shop A = 600

Number of sarees sold in shop B = 1.25 ×  600 = 750

Number of sarees sol in shop C = (600 + 50) = 650

Number of shirts sold in shop A = 600/3 ×  2 = 400

Number of shirts sold in shop B = (400 + 100) = 500

Total shirts sold in all the shops = 450 ×  3 = 1350

Number of shirts sold in shop C = (1350 - (400 + 500)) = 450

Solution:

Number of sarees sold in shop A = 600

Number of shirts sold in shop C = 450

Required difference = (600 - 450) = 150

General solution:

Total number of sarees sold in all the shops A, B and C = 2000

Number of sarees sold in shop A = x

Number of sarees sold in shop B = x ×  (125/100) = 1.25x

Number of sarees sold in shop C = (x + 50)

(x + 1.25x + x + 50) = 2000

3.25x = 1950

x = 600

Number of sarees sold in shop A = 600

Number of sarees sold in shop B = 1.25 ×  600 = 750

Number of sarees sol in shop C = (600 + 50) = 650

Number of shirts sold in shop A = 600/3 ×  2 = 400

Number of shirts sold in shop B = (400 + 100) = 500

Total shirts sold in all the shops = 450 ×  3 = 1350

Number of shirts sold in shop C = (1350 - (400 + 500)) = 450

Solution:

Number of sarees sold in shop B = 750

Number of sarees sold in shop C = 650

Number of shirts sold in shop B = 500

Number of shirts sold in shop C = 450

Required sum = (750 + 650 + 500 + 450) = 2350

General solution:

Total number of sarees sold in all the shops A, B and C = 2000

Number of sarees sold in shop A = x

Number of sarees sold in shop B = x ×  (125/100) = 1.25x

Number of sarees sold in shop C = (x + 50)

(x + 1.25x + x + 50) = 2000

3.25x = 1950

x = 600

Number of sarees sold in shop A = 600

Number of sarees sold in shop B = 1.25 ×  600 = 750

Number of sarees sol in shop C = (600 + 50) = 650

Number of shirts sold in shop A = 600/3 ×  2 = 400

Number of shirts sold in shop B = (400 + 100) = 500

Total shirts sold in all the shops = 450 ×  3 = 1350

Number of shirts sold in shop C = (1350 - (400 + 500)) = 450

Solution:

Number of sarees sold in shop C = 650

Number of shirts sold in shop B = 500

Required ratio = 650: 500 = 13: 10

General solution:

Total number of sarees sold in all the shops A, B and C = 2000

Number of sarees sold in shop A = x

Number of sarees sold in shop B = x ×  (125/100) = 1.25x

Number of sarees sold in shop C = (x + 50)

(x + 1.25x + x + 50) = 2000

3.25x = 1950

x = 600

Number of sarees sold in shop A = 600

Number of sarees sold in shop B = 1.25 ×  600 = 750

Number of sarees sol in shop C = (600 + 50) = 650

Number of shirts sold in shop A = 600/3 ×  2 = 400

Number of shirts sold in shop B = (400 + 100) = 500

Total shirts sold in all the shops = 450 ×  3 = 1350

Number of shirts sold in shop C = (1350 - (400 + 500)) = 450

Solution:

Number of sarees from shop B sold at discounted price = 750 ×  (20/100) = 150

Number of sarees from shop C sold at discounted price = 650 ×  (40/100) = 260

Number of sarees sold at marked price in shop B and C = (750 + 650) - (150 + 260) = 990

General solution:

Total number of sarees sold in all the shops A, B and C = 2000

Number of sarees sold in shop A = x

Number of sarees sold in shop B = x ×  (125/100) = 1.25x

Number of sarees sold in shop C = (x + 50)

(x + 1.25x + x + 50) = 2000

3.25x = 1950

x = 600

Number of sarees sold in shop A = 600

Number of sarees sold in shop B = 1.25 ×  600 = 750

Number of sarees sol in shop C = (600 + 50) = 650

Number of shirts sold in shop A = 600/3 ×  2 = 400

Number of shirts sold in shop B = (400 + 100) = 500

Total shirts sold in all the shops = 450 ×  3 = 1350

Number of shirts sold in shop C = (1350 - (400 + 500)) = 450

Solution:

Number of shirts sold in shop B = 500

Number of shirts sold in shop C = 450

Required percentage = (450/500) ×  100 = 90%

Shortcut Trick

Let speed of boat be u and speed of stream be v.

Case 1: 18/(u + v) + 18/(u − v) = 8 ⇒ 36u / (u² − v²) = 8 ⇒ u² − v² = 4.5u

Case 2: 18/(2u + v) + 18/(2u − v) = 3.2 ⇒ 72u / (4u² − v²) = 3.2 ⇒ 4u² − v² = 22.5u

Subtracting Eq 1 from Eq 2: 3u² = 18u ⇒ u = 6 km/hr.

∴ The correct answer is 6 km/hr.

Alternate Method

Given: Total Distance = 18 km each way, Total Time 1 = 8 hours, Total Time 2 (with 2u) = 3.2 hours.

Formula Used: Time = Distance ÷ Speed

⇒ 18/(u + v) + 18/(u − v) = 8

⇒ 18 × (2u / (u² − v²)) = 8

⇒ 36u = 8(u² − v²) ⇒ v² = u² − 4.5u --- (Equation 1)

⇒ 18/(2u + v) + 18/(2u − v) = 3.2

⇒ 18 × (4u / (4u² − v²)) = 3.2

⇒ 72u = 3.2(4u² − v²) ⇒ v² = 4u² − 22.5u --- (Equation 2)

⇒ Equating v²: u² − 4.5u = 4u² − 22.5u

⇒ 3u² = 18u

⇒ u = 6 km/hr

∴ The correct answer is 6 km/hr.

Additional Information

Downstream and Upstream Speed

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

Speed of Boat in Still Water

Speed of boat (u) = (D + U) ÷ 2.

Speed of Current

Speed of current (v) = (D − U) ÷ 2.

Shortcut Trick

Total Work (LCM of 18, 24, 30) = 360 units.

Efficiencies: P = 20, Q = 15, R = 12.

3-day cycle: Day 1 (P) + Day 2 (P) + Day 3 (P+Q+R) = 20 + 20 + 47 = 87 units.

In 4 cycles (12 days): 87 × 4 = 348 units. Remaining = 360 − 348 = 12 units.

Time for remaining work = 12 ÷ 20 = 3/5 days.

Total time = 12 + 3/5 = 12 3/5 days.

∴ The correct answer is 12 3/5 days.

Alternate Method

Given: Time taken by P = 18 days, Q = 24 days, R = 30 days.

⇒ Efficiency of P = 360 ÷ 18 = 20 units/day

⇒ Efficiency of Q = 360 ÷ 24 = 15 units/day

⇒ Efficiency of R = 360 ÷ 30 = 12 units/day

⇒ Work done in Day 1 (P only) = 20 units

⇒ Work done in Day 2 (P only) = 20 units

⇒ Work done in Day 3 (P+Q+R) = 20 + 15 + 12 = 47 units

⇒ Total work in 1 cycle (3 days) = 20 + 20 + 47 = 87 units

⇒ Work done in 4 cycles (12 days) = 87 × 4 = 348 units

⇒ Remaining work = 360 − 348 = 12 units

⇒ Time taken by P to do remaining work = 12 ÷ 20 = 3/5 days

⇒ Total time = 12 + 3/5 = 12 3/5 days

∴ The correct answer is 12 3/5 days.

Additional Information

LCM Method for Time and Work

Total work is assumed as the LCM of individual time periods to ensure efficiencies are whole numbers. Efficiency = Total Work ÷ Time.

Alternate Day / Periodic Help Work

In periodic help problems, identify the repeating cycle (e.g., 3 days here) and calculate the total work done in one complete cycle.

Work and Wages Relation

Wages are always distributed in the ratio of the actual work done by each person, which is Efficiency × Time.

Shortcut Trick

In 22 liters from Vessel P (Ratio 6:5), Milk = 12 liters and Water = 10 liters.

Let Vessel Q initial quantity be 7x. Milk = 4x, Water = 3x.

New ratio in Q: (4x + 12) / (3x + 10) = 22 / 17 ⇒ 68x + 204 = 66x + 220 ⇒ 2x = 16 ⇒ x = 8.

New quantity in Q = 7 × 8 + 22 = 78 liters = Initial quantity in P.

Quantity in P after taking out 22 liters = 78 − 22 = 56 liters. Milk = (6/11) × 56 = 336/11 liters.

∴ The correct answer is 336/11 liters.

Alternate Method

Given: Ratio P (6:5), Ratio Q (4:3), Transfer 22 L, New Q ratio (22:17).

⇒ In 22 L from P: Milk = 22 × (6/11) = 12 L; Water = 22 × (5/11) = 10 L.

⇒ Let initial Milk and Water in Q be 4x and 3x.

⇒ (4x + 12) / (3x + 10) = 22 / 17

⇒ 17(4x + 12) = 22(3x + 10) ⇒ 68x + 204 = 66x + 220

⇒ 2x = 16 ⇒ x = 8.

⇒ Initial quantity in Q = 7 × 8 = 56 liters.

⇒ New quantity in Q = 56 + 22 = 78 liters.

⇒ Initial quantity in P = 78 liters.

⇒ Quantity in P after removal = 78 − 22 = 56 liters.

⇒ Milk in remaining P = 56 × (6/11) = 336/11 liters.

∴ The correct answer is 336/11 liters.

Additional Information

Mixture Removal Property

When a part of a mixture is removed, the ratio of components in the remaining mixture and the removed part remains the same as the initial ratio.

Alligation Method

Can be used when two mixtures are mixed to find the resulting concentration or ratio. Quantity₁/Quantity₂ = (C₂ − C_m)/(C_m − C₁).

Replacement Formula

Final Quantity = Initial Quantity × (1 − x/V)ⁿ, where x is volume replaced and V is total volume.

Solution:

The number of cars sold by company Y in city K = 250 x 10/100 = 25

The number of cars and bikes sold by company X in city K = 400 – 250 – 25 = 125

The number of cars sold by company X in city K = 125 x 3/5 = 75

Required percentage = 75/200 x 100 = 37.5%

Solution:

The number of bikes sold by company Y in city L = 150

The number of bikes sold by company X in city L = 150 x 33.33/100 = 50

The number of cars sold by company X in city L = 250 x 50/100 = 125

The number of cars sold by company Y in city L = 350 – 150 – 50 – 125 = 25

Solution:

The number of bikes sold by company X in city M = 20

The total number of cars sold by companies X and Y in city M = 300 – 200 – 20 = 80

The number of bikes sold by company Y in city M = (80 - 20)/2 = 30

Solution:

The number of bikes sold by company X in city N = 250 x 30/100 = 75

The number of cars sold by company Y in city N = (450 – 250 - 75) x 9/25 = 125 x 9/25

The number of cars sold by company Y in city N = 45

Solution:

The number of vehicles (car + bike) sold by companies X and Y in city O = 350 x 3 – 300 – 450 = 300

The number of bikes sold by company Y in city O = 150 x 60/100 = 90

The number of cars sold by company Y in city O = 90 – 30 = 60

Required total = 300 – 90 – 60 = 150

Given:

1000 , 1001 , 993 , 1002 , 936 , 963

Calculation:

Here the logic follows as,

Hence, 936 is the wrong term

\(\therefore\) The correct answer is 936.

Given:

5 , 14 , 53 , 128 , 240, 390

Calculation:

Here the logic follows as,

14 - 4 = 10 

Hence, 5 is the wrong term

\(\therefore\) The correct answer is 5.