Quantitative Aptitude PYQs

Given:

a = -5%

b = 5%

R = Rs. 80

Formula used:

Profit% = {R/(b - a) × 100}

Calculation:

Here, a = -5%, b = 5%, R = Rs. 80

⇒ According to the formula,

⇒ CP of scooter = {R/(b - a) × 100}

⇒ 80/(5 + 5) × 100 = Rs. 800

∴ The cost price of the scooter is Rs. 800.

Alternate Method

Let CP = x

⇒ Then, according to the Q.,

⇒ 105x/100 – 95x/100 = 80

⇒ 21x/20 – 19x/20 = 80

⇒ 2x/20 = 80

⇒  x = 800

∴ The cost price of the scooter is Rs. 800.

The series follows the following pattern:

9 × 2 - 2 = 16

16 × 2 - 2 = 30

30 × 2 - 2 = 58

58 × 2 - 2 = 114

114 × 2 - 2 = 226

The series follows the following pattern:

8 × 2 = 16 

16 + 3 = 19 

19 × 4 = 76

76 + 5 = 81

81 × 6 = 486

The series follows the following pattern:

13 + 4 = 17

17 + 9 = 26

26 + 16 = 42

42 + 25 = 67

67 + 36 = 103

The series follows the following pattern:

450 – 23 = 442 

442 – 33 = 415 

415 – 43 = 351 

351 – 53 = 226 

226 – 63 = 10

The series follows the following pattern:

13 + 2 = 15

15 + 3 = 18

18 + 5 = 23

23 + 7 = 30

30 + 11 = 41

Here, we see the difference of consecutive prime number

Given:

First distance = 40 km

Time to cover the distance of 40 km by car = 20 min

Second distance = 20 km

Time to cover the distance of 20 km by bus = 30 min

Concept used:

Average speed = Total distance/Total time

Calculation:

Total distance cover by the lady during her whole journey = (40 + 20) km = 60 km

Total time to cover the Whole distance = (20 + 30)/60 = 5/6 hr

According to the question,

⇒ 60/(5/6) = (60 × 6)/5 = 72

∴ Required average speed = 72 km/hr

Given:

Sid started a business with Rs. 15000.

Profit is divided between them in the ratio 5 ∶ 4

Calculation:

Let, investment of Mohan = Rs. 100x

Sid invested Rs. 15000 for 12 months.

Mohan invested Rs. 100x for (12 – 4) = 8 months.

∴ Ratio of profit of Sid and Mohan = (15000 × 12) ∶ (100x × 8) = 225 ∶ x

According to the question,

⇒ 225 ∶ x = 5 ∶ 4

⇒ x = 225 × (4/5)

⇒ x = 180

∴ Investment of B = Rs. (100 × 180) = Rs. 18000

Calculation

Bus sold in city B = x + 30

→ Given: Bus sold in city B = 80

So,

x + 30 = 80

⇒ x = 50

Buses sold in city F = 5x = 5 × 50 = 250

Answer: 250 buses in city F

Calculation

Total Buses = 120 + 80 + 150 + 180 + 160 = 690

Total Cars = 320 + 240 + 300 + 420 + 350 = 1630

Number of cities = 5

Average Buses = 690 / 5 = 138

Average Cars = 1630 / 5 = 326

Difference = 326 - 138 = 188

Answer: 188

Calculation

Ratio of AC to Non-AC buses in city A = 5:3. Total buses in city A = 120

Let common multiple be x. So,

AC buses = 5x

Non-AC buses = 3x

Total = 5x + 3x = 8x = 120

⇒ x = 15

AC = 5x = 75

Non-AC = 3x = 45

Difference = 75 - 45 = 30

Answer: 30 buses

Calculation

Cars in city B = 240

25% of 240 = (25/100) × 240 = 60

So, tractors sold = 240 - 60 = 180

Answer: 180 tractors

Calculation

Buses in D + E = 180 + 160 = 340

Cars in C + D = 300 + 420 = 720

Ratio = 340 : 720 = 17 : 36

Answer: 17 : 36

The concentration of milk has reduced from 24% to 20% i.e. to 5/6^th

The concentration of water now has become 6^th of total = 6/5 × 20 = 24 litres

The amount of water added = 24 - 20 = 4 litres

Given:

The initial amount of solution is 20 litres

Calculation:

The initial amount of milk = 24/100 × 20 = 24/5

The initial amount of water = (100 - 24)/100 × 20 = 76/5

Let the amount of water added be x

According to the question;

⇒ 24/5/(76/5 + x) = 20/80

⇒ 24/5/(76/5 + x) = 1/4

⇒ 20/5 = x

⇒ x = 4

The amount of water added is 4 litres.

Given:

The initial ratio between the two numbers = 2 : 3

Each number reduced by 5

The final ratio between the two numbers = 1 : 2

Calculation:

Let the first and second number be 2x and 3x respectively.

According to the question,

(2x - 5)/(3x - 5) = 1/2

⇒ 2(2x - 5) = 3x - 5

⇒ 4x - 10 = 3x - 5

⇒ 4x - 3x = 10 - 5

⇒ x = 5

The first number = 2x

⇒ 2 × 5

⇒ 10

∴ The required first number will be 10.

Shortcut Trick

Calculation

A = 1.2B (20% more)

B = 0.75C (25% less)

Substitute:

A = 1.2 × 0.75C = 0.9C

Thus,

A : C=0.9

Or, A : C = 9  : 10

Given:

Average weight of 5 persons = 64 kg

Average weight of 3 girls = 58 kg

Calculations:

Total weight of 5 persons = 5 × 64 = 320 kg

Total weight of 3 girls = 3 × 58 = 174 kg

Total weight of the 2 boys = 320 - 174 = 146 kg

Average weight of the 2 boys = 146 / 2 = 73 kg

∴ The average weight of the two boys is 73 kg.

GIVEN :

Total Students = 1200

Boys : Girls = 5 : 3

Required ratio = 1 : 1

 

CONCEPT :

Ratio and its types

 

ASSUMPTION :

Let the number of original number of boys and girls in the school be 5x and 3x

 

CALCULATION :

According to the question

 

5x + 3x = 1200

8x = 1200

x = 150

5x = 5 × 150 = 750

3x = 3 × 150 = 450

Let the number of girls joined be y

Now,

\(\frac{{750}}{{450; + ;y}}; = ;\frac{1}{1}\)

750 = 450 + y

y = 300

∴, number of girls added was 300

Calculation

People on Tuesday + Wednesday = 320 + 280 = 600

People on Thursday + Friday = 360 + 400 = 760

Required ratio:

600 : 760 = 15 : 19

Answer: 15 : 19

Calculation

People attend on Friday = 400

Saturday = 20% more than Friday

Thus:

[ 400 + [(20/100) ×400] = [400 + 80] = 480

480 people on Saturday