Quantitative Aptitude PYQs

Given:

Ratio 1 = 5 : 8

Ratio 2 = 4 : 7

Ratio 3 = 3 : 5

Ratio 4 = 1 : 3

Formula Used:

To compare ratios, convert them to fractions and compare the values.

Calculation:

Convert the ratios to fractions:

Ratio 1 = 5 / 8

Ratio 2 = 4 / 7

Ratio 3 = 3 / 5

Ratio 4 = 1 / 3

Compare the fractions by finding a common denominator or converting to decimals:

\( \frac{5}{8} = 0.625 \)

\( \frac{4}{7} \approx 0.571 \)

\( \frac{3}{5} = 0.6 \)

\( \frac{1}{3} \approx 0.333 \)

Compare the decimal values:

0.625 > 0.6 > 0.571 > 0.333

The greatest ratio is 5 : 8.

Given:

Principal (P) = ₹2,400

Simple Interest (SI) = ₹460

Time (T) = 2 years and 4 months = 2 + 4/12 = 2 + 1/3 = 7/3 years

Formula Used:

Simple Interest (SI) = (P × R × T) / 100

Calculation:

Using the formula for simple interest:

₹460 = (₹2,400 × R × 7/3) / 100

First, we simplify the equation:

₹460 = (₹2,400 × R × 7) / 300

⇒ ₹460 = ₹16,800R / 300

⇒ ₹460 = ₹56R

⇒ R = ₹460 / ₹56 = \(8\frac{3}{14}%\)

The rate of interest per annum is \(8\frac{3}{14}%\).

The correct answer is option 4.

Given:

Anjali and Manju together can do a piece of work in 5 days.

Anjali does twice as much work as Manju in a given time.

Formula Used:

Work done by Anjali and Manju together in 1 day = 1 / Total days

Work done by Anjali alone in 1 day = 2 × Work done by Manju alone in 1 day

Let the work done by Manju alone in 1 day be M.

Then, the work done by Anjali alone in 1 day = 2M.

Work done by Anjali and Manju together in 1 day = 2M + M = 3M .

Since they can complete the work together in 5 days,

Work done by Anjali and Manju together in 1 day = 1/5

Calculation:

3M = 1/5

⇒ M = 1/(5 × 3)

⇒ M = 1/15

Therefore, the work done by Anjali alone in 1 day = 2M = 2 × 1/15 = 2/15

Number of days required by Anjali alone to complete the work = 1 / (2/15)

⇒ Number of days = 15 / 2

⇒ Number of days = 7.5

Anjali alone would take 7.5 days to do the work.

Given:

Loss = (1/6) of the Selling Price

Formula Used:

Loss Percentage = (Loss / Cost Price) × 100

Calculation:

Let the Selling Price (SP) be 6 units.

Loss = 1/6 of SP

Loss = 1 unit

Therefore, Cost Price (CP) = SP + Loss

CP = 6 + 1 = 7 units

Loss Percentage = (Loss / CP) × 100

Loss Percentage = (1 / 7) × 100

Loss Percentage = \(\frac{100}{7}\)

Loss Percentage = \(14\frac{2}{7}%\)

The loss percentage is \(14\frac{2}{7}%\).

Given:

The price of a car increased by 25%.

Formula Used:

Percentage reduction required = \(\left(\frac{\text{Increase percentage}}{100 + \text{Increase percentage}}\right) \times 100\)

Calculation:

Let the original price be 100.

New price after 25% increase = 100 + 25 = 125

To find the percentage reduction required to bring the price back to 100:

Percentage reduction required = \(\left(\frac{25}{125}\right) \times 100\)

Percentage reduction required = \(\left(\frac{25}{125}\right) \times 100\)

⇒ \(\left(\frac{25}{125}\right) \times 100\) = 20%

The correct answer is option 1.

Given:

Distance = 10 km

Initial time = 10 minutes = 10/60 hours = 1/6 hours

Formula Used:

Speed = Distance / Time

Time = Distance / Speed

Calculation:

Initial Speed = 10 km / (1/6) hours

Initial Speed = 10 × 6 = 60 km/h

Decreased Speed = 60 km/h - 20 km/h = 40 km/h

Time taken to cover the same distance with decreased speed

Time = Distance / Speed

Time = 10 km / 40 km/h

Time = 1/4 hours

Time = (1/4) × 60 minutes

Time = 15 minutes

The time taken by the train to cover the same distance with decreased speed is 15 minutes.

Given:

Marked price (MP) = ₹2,100

Discount = 15%

Profit = 25%

Formula Used:

Selling Price (SP) = MP × (1 - Discount%)

Cost Price (CP) = SP / (1 + Profit%)

Calculation:

Selling Price (SP) = 2,100 × (1 - 0.15)

SP = 2,100 × 0.85

SP = ₹1,785

Cost Price (CP) = SP / (1 + 0.25)

CP = 1,785 / 1.25

CP = ₹1,428

The cost price of the article is ₹1,428.

Given:

Interest rate = 6% per annum, compounded annually

Loan repayment amount after 5 years = ₹20,000

Formula Used:

Compound Interest Formula:

\(A = P(1 + \frac{r}{100})^n\)

Where:

A = Final amount (₹20,000)

P = Principal amount (Loan amount)

r = Annual interest rate (6%)

n = Number of years (5)

Calculation:

Using the compound interest formula:

\(20000 = P(1 + \frac{6}{100})^5\)

20000 = P(1.06)⁵

20000 = P × 1.338225

⇒ \(P = \frac{20000}{1.338225}\)

⇒ P ≈ 14946.94

Rounding to the nearest thousand:

Loan amount ≈ ₹15,000

The amount of loan that he had taken was ₹15,000.

Given:

The ratio of the salaries of Ram and Shyam is 8:5.

After an increment of ₹2,600 in each of their salaries, the ratio becomes 15:11.

Formula Used:

Let Ram's original salary be 8x and Shyam's original salary be 5x.

After the increment, Ram's salary = 8x + 2600

Shyam's salary = 5x + 2600

According to the given condition:

\(\frac{8x + 2600}{5x + 2600} = \frac{15}{11}\)

Calculation:

We solve the equation:

⇒ \(\frac{8x + 2600}{5x + 2600} = \frac{15}{11}\)

⇒ 11 × (8x + 2600) = 15 × (5x + 2600)

⇒ 88x + 28600 = 75x + 39000

⇒ 88x - 75x = 39000 - 28600

⇒ 13x = 10400

⇒ x = 800

Shyam's original salary = 5x = 5 × 800 = 4000

The salary of Shyam is ₹4,000.

Given:

We need to find the highest common factor (HCF) of 161 and 207.

Formula Used:

To find the HCF of two numbers, we use the Euclidean algorithm.

Calculation:

Using the Euclidean algorithm:

⇒ 207 ÷ 161 = 1 remainder 46

⇒ 161 ÷ 46 = 3 remainder 23

⇒ 46 ÷ 23 = 2 remainder 0

The remainder is 0, so the divisor at this step is the HCF.

The highest common factor of 161 and 207 is 23.

Given:

\(\frac{844}{0.844} = \frac{84.4}{K}\)

Formula Used:

Cross-multiplication to solve for K.

Calculation:

Using cross-multiplication:

⇒ \(844 \times K = 84.4 \times 0.844\)

⇒ 844K = 71.2336

⇒ K = \(\frac{71.2336}{844}\)

⇒ K = 0.0844

The value of K is 0.0844.

Given:

Side of the equilateral triangle = 8√3 cm

Formula Used:

Area of an equilateral triangle = \(\frac{\sqrt{3}}{4} \times \text{side}^2\)

Calculation:

Side = 8√3 cm

Area = \(\frac{\sqrt{3}}{4} \times (8√3)^2\)

⇒ Area = \(\frac{\sqrt{3}}{4} \times 64 \times 3\)

⇒ Area = \(\frac{\sqrt{3}}{4} \times 192\)

⇒ Area = \(48\sqrt{3}\) cm²

The area of the equilateral triangle is 48√3 cm².

Given:

The average of n natural odd numbers, where n is odd.

Formula Used:

Average of n natural odd numbers = n

Calculation:

Let the n natural odd numbers be:

1, 3, 5, ..., (2n-1)

Sum of these n odd numbers = n²

Average = Sum of numbers / n

Average = n² / n

Average = n

The average of n natural odd numbers is n.

Given:

Area of great circle = 36π m²

π = 3.14

Formula Used:

Area of great circle = πr²

Volume of a hemisphere = (2/3) π r³

Calculation:

Area of great circle = πr²

36π = 3.14r²

r² = 36

r = 6 m

Volume of a hemisphere = (2/3) π r³

Volume of a hemisphere = (2/3) × 3.14 × 6³

Volume of a hemisphere = (2/3) × 3.14 × 216

Volume of a hemisphere = (2/3) × 678.24

Volume of a hemisphere = 452.16 m³

The correct answer is option 3.

Given:

\(\frac{17}{250}\) is to be converted to its decimal equivalent.

Formula Used:

Division Method: To convert a fraction to a decimal, divide the numerator by the denominator.

Calculation:

To find the decimal equivalent of \(\frac{17}{250}\) :

⇒ 17 ÷ 250

⇒ 0.068

The correct answer is option 2.

Given:

Distance between city A and city B = 480 km

A car starts from A at (3/5) of its original speed and reaches B late by 80 minutes

Formula Used:

Time = Distance / Speed

Calculation:

Let the original speed of the car be x km/h

Original time taken = 480 / x hours

New speed = (3/5) × x = 3x/5 km/h

New time taken = 480 / (3x/5) hours

New time taken = 480 × (5/3x) hours

New time taken = 800 / x hours

New time taken is 80 minutes (or 80/60 = 4/3 hours) more than the original time taken

Therefore, 800 / x = 480 / x + 4/3

⇒ 800 / x - 480 / x = 4/3

⇒ (800 - 480) / x = 4/3

⇒ 320 / x = 4/3

⇒ 320 × 3 = 4x

⇒ 960 = 4x

⇒ x = 960 / 4

⇒ x = 240 km/h

The original speed of the car is 240 km/h.

Given:

Number leaves a remainder of 8 when divided by 15, 30, and 45.

Formula Used:

Required number = LCM(15, 30, 45) × k + Remainder

Calculation:

LCM of 15, 30, and 45:

15 = 3 × 5

30 = 2 × 3 × 5

45 = 3² × 5

LCM = 2 × 3² × 5 = 90

Required number = 90k + 8

For k = 0,

⇒ Required number = 90 × 0 + 8 = 8

For k = 1,

⇒ Required number = 90 × 1 + 8 = 98

The least number which, when divided by 15, 30, and 45, leaves a remainder of 8 is 98.

Given:

Number 1 = 216, remainder = 9

Number 2 = 358, remainder = 13

Number 3 = 454, remainder = 17

Formula Used:

We need to find the largest positive integer that divides the differences of the given numbers and their remainders.

Calculation:

Let the required number be x.

216 - 9 = 207

358 - 13 = 345

454 - 17 = 437

We need to find the largest positive integer that will divide 207, 345, and 437.

First, find the GCD of 207 and 345:

207 = 3 × 3 × 23

345 = 3 × 5 × 23

GCD(207, 345) = 3 × 23 = 69

Next, find the GCD of 69 and 437:

69 = 3 × 23

437 = 19 × 23

GCD(69, 437) = 23

The largest positive integer that will divide 216, 358, and 454 leaving remainders 9, 13, and 17 respectively is 23.

Given:

The ratio of the incomes of Ram and Ganga is 3 ∶ 4.

The ratio of their expenditures is 2 ∶ 3.

Each person saves Rs. 2,000.

Formula Used:

Total income together = Income of Ram + Income of Ganga

Total expenditure together = Expenditure of Ram + Expenditure of Ganga

Income - Expenditure = Savings

Calculation:

Let the income of Ram be 3x and the income of Ganga be 4x.

Let the expenditure of Ram be 2y and the expenditure of Ganga be 3y.

Savings of Ram: 3x - 2y = 2000

Savings of Ganga: 4x - 3y = 2000

We have two equations:

1. 3x - 2y = 2000

2. 4x - 3y = 2000

Multiply equation 1 by 3 and equation 2 by 2:

9x - 6y = 6000

8x - 6y = 4000

Subtract the second equation from the first:

⇒ (9x - 6y) - (8x - 6y) = 6000 - 4000

⇒ x = 2000

Substitute x = 2000 into equation 1:

⇒ 3(2000) - 2y = 2000

⇒ 6000 - 2y = 2000

⇒ 2y = 4000

⇒ y = 2000

Income of Ram = 3x = 3 × 2000 = 6000

Income of Ganga = 4x = 4 × 2000 = 8000

Expenditure of Ram = 2y = 2 × 2000 = 4000

Expenditure of Ganga = 3y = 3 × 2000 = 6000

Total income together = 6000 + 8000 = 14000

Total expenditure together = 4000 + 6000 = 10000

Ratio of their total income together to their total expenditure together:

⇒ 14000 : 10000 = 14 : 10 = 7 : 5

The correct answer is option 2 (7 ∶ 5).

Given:

Investment amount = Rs. x

Return = 12%

Extra amount received = Rs. 12,000

Formula Used:

Return = Principal × Rate × Time / 100

Calculation:

Let the investment amount be Rs.

x. The return on investment is 12%.

Extra amount received = 12% of x

12% of x = 12,000

12/100 × x = 12,000

⇒ x = 12,000 × 100 / 12

⇒ x = 1,200,000 / 12

⇒ x = 100,000

The investment amount is Rs. 100,000.