Quantitative Aptitude PYQs

**Given:**​

\(17{1\over3} +{21{1\over2}} = 33{1\over3}%\)% of ?

Concept Used:

​

Calculation:

\(17{1\over3} +{21{1\over2}} = 33{1\over3}% \)% of ?

⇒ (52/3) + (43/2) = (1/3) of ?

⇒ (1/3) of (?) = (104 + 129)/6

⇒ ? = 233/2

⇒ ? = 116.5

∴ The value of (?) is 116.5

Given:

(8)- 3 × (2)5 × (4)5 ÷ {(2)4 × (4)4 } = (8)?/2

Concept Used:

​Formula used:

x^m × xⁿ = x^{m + n}

xm ÷ xn = xm - n

Calculation:

(8)- 3 × (2)5 × (4)5 ÷ {(2)4 × (4)4 } = (8)?/2

⇒ (8)- 3 × (8)5 ÷ (8)4 = (8)?/2

⇒ 8^{(-3 + 5 - 4)} = (8)?/2

⇒ 8^-2 = (8)?/2

Comparing powers from both sides

⇒ ?/2 = - 2

⇒ ? = - 4

∴ The value of (?) is - 4.

 

When degrees are the same and base are different.

Then, you can multiply both bases and the degree remains the same.

for example,  (2)5 × (4)⁵ = (8)⁵.

Given:

√(113 - 31 + 122) ÷ √(62 × 10 + 52 - 23 × 3) = ?

Concept Used:

Calculation:

√(113 - 31 + 122) ÷ √(62 × 10 + 52 - 23 × 3) = ?

⇒ √1444 ÷ √361 = ?

⇒ 38 ÷ 19 = ?

⇒ ? = 2

∴ The value of ? is 2.

Given:

(256)1/2 × (4)3 × (216)1/3 ÷ 48 = 2?

Concept Used:

Follow the BODMAS rule according to the table given below:

 

Calculation:

(256)1/2 × (4)3 × (216)1/3 ÷ 48 = 2?

⇒ 16 × 64 × 6 ÷ 48 = 2^?

⇒ 16 × 64 × (1/8) = 2^?

⇒ 16 × 8 = 2^?

⇒ 2⁷ = 2^?

⇒ ? = 7

∴ 7 will come in place of the question mark ('?').

Important Points

If (x)^a = (x)^b then a = b.

Given:

∛2197 + ∛1728 + ∛3375 = 8 × ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

∛2197 + ∛1728 + ∛3375 = 8 × ?

⇒ 13 + 12 + 15 = 8 × ?

⇒ 40 = 8 × ?

⇒ ? = 40/8 = 5

⇒ ? = 5

∴ The value of ‘?’ is 5.

Solution:

Cost of each magazine in 2010 = x

Cost of each magazine in 2012 = 150/100 × x = 1.5x

Total revenue generated by selling magazine by person A = (20 × x + 1.5x × 12) = 380

(20x + 18x) = 380

38x = 380

x = 10

Solution:

Number of magazine sold by person B in 2010 = 16

1/4 th of the magazine sold by person B in 2010 = 16/4 = 4

Number of magazine sold by person C in 2012 = 21

1/7 th of the number of magazine sold by person C = 21/7 = 3

Number of magazines sold by person C in all the three years = (30 + 14 + 21) = 65

Required percentage = (4 + 3)/65 × 100 = 10.76%

Solution:

Number of magazines sold by the person C in 2011 = 14

1/7 th of the number of magazines sold by the person C in 2011 = 14/7 = 2

Number of magazine sold by the person D in 2010 = 25

1/5 th of the number of magazines sold by person D in 2010 = 25/5 = 5

Required sum = (2 + 5) = 7

Solution:

Number of magazines sold by the person C in 2010 = 30

Number of magazines sold by the person E in 2010 = 30 × (120/100) = 36

If the number of magazines sold by person E in 2011 = 36 × (125/100) = 45

Number of magazines sold by person E in 2012 = 45 × (120/100) = 54

Required sum = (36 + 45 + 54) = 135

Solution:

Total number of magazines sold by person A initially in all the years = (20 + 18 + 12) = 50

Number of increased magazines =(55- 50) = 5

To find the value of x

New increased number of sold magazine = (20 + 5) = 25

Value of x

(25 - 20)/20 ×  100 = 100/4 = 25%

The given series is cubes of prime numbers added with 1

2³ + 1 = 9

3³ + 1 = 28

5³ + 1 = 126

7³ + 1 = 344

11³ + 1 = 1332 = ?

The given series is the squares of all the prime numbers

2² = 4

3² = 9

5² = 25

7² = 49

11² = 121

13² = 169

14 × 2 – 2 = 26

26 × 2 – 2 = 50

50 × 2 – 2 = 98

98 × 2 – 2 = 194

194 × 2 – 2 = 386

The first element in the series is 18.

And,

18 × 1 = 18

18 × 2 = 36

36 × 3 = 108

108 × 4 = 432

432 × 5 = 2160

Given that, mixture of 90 litres of the spirit and water contains 20% water.

In the first mixture

Water = (20/100) × 90 = 18 litres

∴ Spirit = 90 – 18 = 72 litres

After adding some water to the mixture, water became 50% of the new mixture

∴ 72 litres of water in new mixture = 50% of new mixture

∴ Total litres of new mixture = \(\left( {\frac{{100}}{{50}} \times 72} \right) = 144{\rm{;litres}}\)

So, water to be added = 144 – 90 = 54 litres

Given: average age of 28 students in class = 15

Total age of 28 students in class = 28 × 15 = 420

Let the age of teacher be x years.

If the teacher’s age is included, average age of class = 16

Thus, (420 + x)/29 = 16

⇒ 420 + x = 16 × 29

⇒ 420 + x = 464

∴ x = 44

∴ The age of teacher = 44 years

Given:

Amount = Rs. 15,950

Time = 2 years

New Amount = Rs. 16,820

Assumption:

Let the rate of interest = x%

Formula used:

Amount = Principal + Interest

Calculation:

Case 1:

P + I = 15950 (where, I = PRN/100)

⇒ P + [P × (x/100) × 2] = 15950

⇒ P + (2Px/100) = 15950

⇒ 2Px/100 = 15950 - P      ----(1)

Case 2:

New interest rate = x% + (60% of x) = x + 0.6x = 1.6x

⇒ P + I = 16820

⇒ P + [P × (160x/100)/100 × 2] = 16820

⇒ P + [2Px/100 × (160/100)] = 16820

⇒ P + [160/100(15950 - P)] = 16820

⇒ 100P + 2552000 - 160P = 1682000

⇒ 60P = 870000

⇒ P = Rs. 14500

Put the value in equation 1,

⇒ 14500 × (x/100) × 2 = 15950 - 14500

⇒ 29000 × (x/100) = 1450

⇒ 290x = 1450

⇒ Rate of Interest = 5%

∴ Required value = 5%

Ratio (Vessle1 : Vessel2)

⇒ (5/8 − 3/7) : (3/7 − 3/8)

⇒ 11 : 3

Detailed solution:

Given

Milk and water ration in Vessels₁ = 3 : 5

Milk and water ration in Vessels2 = 5 : 3

Calculation

In the first vessel milk = 3/8 of the mixture

In the second vessel milk = 5/8 of the mixture

It is required to have a mixture in which milk is 3/7 of the new mixture. 

Let 1^st mixture : 2^nd mixture be 1 : x

So the vol. of resulting mixture = 1 + x

(milk in the 1^st mix.) + (milk in the 2^nd mix.) = (milk in the resulting mix.)

⇒ (3/8) × 1 + (5/8) × x = (1 + x) × (3/7)

⇒ (3/8) + (5x/8) = (3/7) + (3x/7)

⇒ (5x/8 − 3x/7) = (3/7 − 3/8)

⇒ (35x − 24x)/56 =  (24 − 21)/56

⇒ 11x = 3

∴ x = 3/11

∴ These mixtures are to be mixed in 1 : 3/11 i.e. 11 : 3

Given:

Amount invested by Bani = Rs. 55,000

Share of profit of Bani and Sandeep = 11 ∶ 14

Formula used:

(Capital of Bani × Time period of Bani) ∶ (Capital of Sandeep × Time period of Sandeep) = Profit of Bani : Profit of Sandeep

Calculation:

Let the amount invested by Sandeep be Rs. ‘x’.

(Capital of Bani × Time period of Bani) : (Capital of Sandeep × Time period of Sandeep) = Profit of Bani : Profit of Sandeep

⇒ (55000 × 12) ∶ (x × 7) = 11 ∶ 14

⇒ (55000 × 12)/7x = 11/14

⇒ (5000 × 12)/x = 1/2

⇒ x = 1,20,000

∴ The amount invested by Sandeep is Rs. 1,20,000.

1. If two partners A and B are investing their money for different period of time to run a business then:

(Capital of A × Time period of A) ∶ (Capital of B × Time period of B) = (Profit or Loss of A) ∶ (Profit or Loss of B)

2. If no information has been provided about the time of investment of a partner, it means that he invests for the entire time.

Given:

MRP = 20% more than C.P

Discount = 30%

Concept Used:

Cost price × (100 + Profit %)/100 = Selling price

Discount % = {(MRP - SP)/MRP} × 100

Calculation:

Cost price = CP,  Selling price = SP,  Marked price = MRP

Let CP be 100.

MRP = 20% more than CP

⇒ MRP = 120

Discount = 120 × 30% 

⇒ 36

SP = 120 - 36

⇒ 84 

⇒ Loss = {(100 - 84)/100} × 100

∴ Loss percent is 16%