Quantitative Aptitude PYQs

Given:

Two stations A and B are 180 km apart

Two trains meet after 5 hours

The local train is 4 km/hr faster than the goods train

Concept Used:

Distance = speed × time

Calculation:

Let, the speed of the goods train = A km/hr

And, speed of local train = (A + 4) km/hr

According to the question,

⇒ 180/(A + A + 4) = 5

⇒ 2A = 32

⇒ A = 16

The speed of the local train = (16 + 4) = 20 km/hr

∴ The correct answer is option 1

Calculation

A + B can complete work in 40 days

B + C can complete work in 60 days

A + C can complete work in 30 days

A + B + C complete 3/2 work together

Total wage = Rs. 5400

We are to find C's share.

Take total work as LCM of 40, 60, 30

LCM of 40, 60, and 30 = 120 units

Now we compute the combined efficiencies:

A + B = 120 / 40 = 3 units/day

B + C = 120 / 60 = 2 units/day

A + C = 120 / 30 = 4 units/day

Let’s add:

(A+B) + (B+C) + (A+C) = 3+2+4=9 units/day

A+B+C = 9/2 = 4.5 units/day

Total work done by A + B + C

They complete 3/2 of total work.

Total work = 120 units

⇒ 3/2 × 120 = 180 units

Since A + B + C = 4.5 units/day:

Days = 180 / 4.5 = 40 days

C’s efficiency = 4.5 – 3 = 1.5

In 40 days C did 40  × 1.5 = 60 unit work

Wage for per unit work is 5400 / 180 = 30

So, C’s wage = 60 × 30 = 1800

Given Series:

3375, 289, 6859, 441, 12167, (?)

The logic followed here is as follows:

⇒ (15)³ = 3375

⇒ (17)² = 289

⇒ (19)³ = 6859

⇒ (21)² = 441

⇒ (23)³ = 12167

⇒ (25)² = 625

Hence, 625 will come in the place of (?).

Given Series:

1220, 1617, 2086, 2633, 3264, (?)

The logic followed here is as follows:

⇒ (11)³ - 111 = 1220

⇒ (12)³ - 111 = 1617

⇒  (13)3 - 111 = 2086

⇒  (14)3 - 111 = 2633

⇒  (15)3 - 111 = 3264

⇒  (16)3 - 111 = 3985

Hence, 3985 will come in the place of (?).

Given Series:

2, 12, 130, 56, (?), 132

The logic followed here is as follows:

⇒ 1 + 1³ = 2

⇒ 3 + 3² = 12

⇒ 5 + 5³ = 130

⇒ 7 + 7² = 56

⇒ 9 + 9³ = 738

⇒ 11 + 11² = 132

Hence, 738 will come in the place of (?).

 

Original candidates in centre A,

Male = 250 and Female = 200

Seventy candidates are transferred from E to A in the ratio = 4:3

Total parts = 4 + 3 = 7

Male candidates transferred to A = 70 x 4/7 = 40

Female candidates transferred to A = 70 x 3/7 = 30

Also, 20% of the original female candidates of A are transferred to E.

20% x 200 = 40

Therefore, the new number of male candidates in A = 250 + 40 = 290

The new number of female candidates in A = 200 + 30 – 40 = 190

Required ratio = 290:190 = 29:19

 

New males in D = 120% x 210 = 252

New females = 90% x 150 = 135

New total = 252 + 135 = 387

Required Difference = 550 – 387 = 163

Common Calculation

Given:

Formula:

Girls = Boys − Difference (when Boys are more)

Girls = Boys + Difference (when Girls are more)

Calculation:

⇒ Girls in School A = 360 – 40 = 320

⇒ Girls in School B = 420 + 84 = 504

⇒ Girls in School C = 300 + 60 = 360

⇒ Girls in School D = 480 – 120 = 360

Given:

Girls in School A = 320

Boys in School B = 420

Formula:

New Value

= Original Value × (100 ± Percentage)/100

Required Ratio

= New Girls : New Boys

Calculation:

⇒ New Girls in School A = 320 × 125/100 = 400

⇒ New Boys in School B = 420 × 80/100 = 336

⇒ Required Ratio = 400 : 336 = 25 : 21

∴ Answer = 25 : 21

Common Calculation

Given:

Formula:

Girls = Boys − Difference (when Boys are more)

Girls = Boys + Difference (when Girls are more)

Calculation:

⇒ Girls in School A = 360 – 40 = 320

⇒ Girls in School B = 420 + 84 = 504

⇒ Girls in School C = 300 + 60 = 360

⇒ Girls in School D = 480 – 120 = 360

Given:

Girls in School A = 320

Girls in School B = 504

Formula:

Required Percentage

= (Difference/Base Quantity) × 100

Calculation:

⇒ Difference = 504 – 320 = 184

⇒ Required Percentage = (184/320) × 100 = 57.5%

∴ Answer = 57.5%

Common Calculation

Given:

Formula:

Girls = Boys − Difference (when Boys are more)

Girls = Boys + Difference (when Girls are more)

Calculation:

⇒ Girls in School A = 360 – 40 = 320

⇒ Girls in School B = 420 + 84 = 504

⇒ Girls in School C = 300 + 60 = 360

⇒ Girls in School D = 480 – 120 = 360

Given:

Boys in School C = 300

Boys in School D = 480

Transferred Boys = 40

Formula:

New Number of Boys

= Original Number of Boys ± Transferred Boys

Required Ratio

= New Boys in School C : New Boys in School D

Calculation:

⇒ New Boys in School C = 300 + 40 = 340

⇒ New Boys in School D = 480 – 40 = 440

⇒ Required Ratio = 340 : 440 = 17 : 22

∴ Answer = 17 : 22

Common Calculation

Given:

Formula:

Girls = Boys − Difference (when Boys are more)

Girls = Boys + Difference (when Girls are more)

Calculation:

⇒ Girls in School A = 360 – 40 = 320

⇒ Girls in School B = 420 + 84 = 504

⇒ Girls in School C = 300 + 60 = 360

⇒ Girls in School D = 480 – 120 = 360

Given:

Girls in Schools B and C

Boys in Schools A and D

Formula:

Required Percentage

= (Required Quantity/Base Quantity) × 100

Calculation:

⇒ Total Girls in Schools B and C = 504 + 360 = 864

⇒ Total Boys in Schools A and D = 360 + 480 = 840

⇒ Required Percentage = (864/840) × 100 = 102.86% ≈ 103%

∴ Answer = 103%

Common Calculation

Given:

Formula:

Girls = Boys − Difference (when Boys are more)

Girls = Boys + Difference (when Girls are more)

Calculation:

⇒ Girls in School A = 360 – 40 = 320

⇒ Girls in School B = 420 + 84 = 504

⇒ Girls in School C = 300 + 60 = 360

⇒ Girls in School D = 480 – 120 = 360

Given:

Students in Schools B and D

Students in Schools A and C

Formula:

Average

= Sum of Students/Number of Schools

Required Difference

= Average of (B & D) − Average of (A & C)

Calculation:

⇒ Average Students in Schools B and D = (924 + 840)/2 = 1764/2 = 882

⇒ Average Students in Schools A and C = (680 + 660)/2 = 1340/2 = 670

⇒ Required Difference = 882 – 670 = 212

∴ Answer = 212

Common Calculation

Given:

Formula:

Girls = Boys − Difference (when Boys are more)

Girls = Boys + Difference (when Girls are more)

Calculation:

⇒ Girls in School A = 360 – 40 = 320

⇒ Girls in School B = 420 + 84 = 504

⇒ Girls in School C = 300 + 60 = 360

⇒ Girls in School D = 480 – 120 = 360

Given:

Total Students in Schools A and C

Total Students in Schools B and D

Formula:

Required Ratio

= Total Students in Schools A and C : Total Students in Schools B and D

Calculation:

⇒ Total Students in Schools A and C = 680 + 660 = 1340

⇒ Total Students in Schools B and D = 924 + 840 = 1764

⇒ Required Ratio = 1340 : 1764 = 335 : 441

Initial milk = 140 x 4/7 = 80 litres

Initial water = 140 x 3/7 = 60 litres

Water removed = 25% of 60 = 15 litres

Therefore, remaining water = 60 – 15 = 45 litres

An equal quantity, that is 15 litres of milk, is added.

Final milk = 80 + 15 = 95 litres

Required ratio = 19:9

Cost Price = ₹1,500

Marked Price = 1500 x 140% = ₹ 2100

Selling Price after 15% discount = 2100 x 85% = ₹1785

New Selling Price = 1785 – 90 = ₹ 1695

Profit = ₹1695 – 1500 = ₹195

Profit percentage = (195/1500) x 100 = 13%