Quantitative Aptitude PYQs

Given:

A boy goes to his school from his house at a speed of 5 km/h and returns at a speed of 10 km/h.

Formula used:

Average Speed = \(\dfrac{2ab}{a+b}\)

Where, a = speed while going, b = speed while returning

Calculation:

Average Speed = \(\dfrac{2 \times 5 \times 10}{5 + 10}\)

⇒ Average Speed = \(\dfrac{100}{15}\)

⇒ Average Speed = 6(2/3) km/h

∴ The correct answer is option (4).

Given:

Radius (r) = 3 units

Height (h) = 8 units

Formula used:

For the largest straight rod (diagonal of the cylinder), we use the formula:

Length = \(\sqrt{d^2 + h^2}\)

Where d is the diameter of the base.

Calculation:

Diameter (d) = 2 × radius = 2 × 3 = 6 units

Length = \(\sqrt{6^2 + 8^2} \)

⇒ Length = \(\sqrt{36 + 64}\)

⇒ Length = \(\sqrt{100}\)

⇒ Length = 10 units

∴ The correct answer is option (1).

Given:

The income of a person first increased by 10% and then it is decreased by 10%.

Formula used:

New Income = Original Income × (1 + Increase Rate) × (1 - Decrease Rate)

Calculations:

Let the original income be ₹100.

After a 10% increase:

New Income = 100 × (1 + 10/100)

⇒ New Income = 100 × 1.1

⇒ New Income = ₹110

After a 10% decrease on ₹110:

New Income = 110 × (1 - 10/100)

⇒ New Income = 110 × 0.9

⇒ New Income = ₹99

Change in income = New Income - Original Income

⇒ Change in income = 99 - 100

⇒ Change in income = -₹1

Percentage change = (Change in income / Original Income) × 100

⇒ Percentage change = (-1 / 100) × 100

⇒ Percentage change = -1%

∴ The correct answer is option (2).

Given:

The HCF of 408 and 1309

Formula used:

Euclidean Algorithm

Calculations:

Using the Euclidean Algorithm:

1309 ÷ 408 = 3 remainder 85

408 ÷ 85 = 4 remainder 68

85 ÷ 68 = 1 remainder 17

68 ÷ 17 = 4 remainder 0

⇒ The HCF is 17

∴ The correct answer is option (3).

Given:

Scooty price = Rs. 1,00,000

Down payment = Rs. 20,800

Number of installments = 4

Rate of interest = 25% 

Concept used : 

A = P + SI 

Where,

\(SI = \frac{{P}×{R}×{T}}{100}\)

Calculation:

Since Rs.20,800 was given for down payments.

∴ Amount to be paid  = Rs. 79200

According to the question,

⇒ 79200 + 79200 × 25/100 × 4 =

\([x + (x + \frac{25x}{100}) + (x + \frac{{25}×{2x}}{100}) + (x +\frac{{25}×{3x}}{100}) ]\)

⇒ 79200 + 79200 × 25/100 × 4 = \([x + \frac{5x}{4} + \frac{3x}{2} + \frac{7x}{4}]\)

⇒ 158400 = \(\frac{22x}{4}\)

⇒ 14400 = 2x/4

⇒ x = 28800

∴ The correct answer is Rs.28800.

Given:

The ratio of A and B is 4 : 3

The ratio of B and C is 7 : 2

Formula used:

To find the ratio of A : B : C, we need to make the B term the same in both ratios.

Calculation:

The ratio of A and B is 4 : 3

The ratio of B and C is 7 : 2

To make the B term the same in both ratios:

A : B = 4 : 3 = 4 × 7 : 3 × 7 = 28 : 21

B : C = 7 : 2 = 21 : 6

Therefore, A : B : C = 28 : 21 : 6

∴ The correct answer is option 1.

Given:

The concentration of three acids, P, Q and R, is given as 15%, 25% and 35%, respectively.

They are mixed in the ratio of x : 3 : 5, resulting in a 28% concentration solution.

Formula used:

Concentration of mixture = (Concentration_P ×  x + Concentration_Q × 3 + Concentration_R × 5) / (x + 3 + 5)

Calculation:

28 = (15x + 25 ×  3 + 35 × 5) / (x + 3 + 5)

⇒ 28(x + 8) = 15x + 75 + 175

⇒ 28x + 224 = 15x + 250

⇒ 28x - 15x = 250 - 224

⇒ 13x = 26

⇒ x = 2

∴ The correct answer is option 1.

Given:

The sum of the smallest three-digit prime number and the largest three-digit prime number is:

Formula used:

Smallest three-digit prime number = 101

Largest three-digit prime number = 997

Calculation:

Sum = Smallest three-digit prime number + Largest three-digit prime number

⇒ Sum = 101 + 997

⇒ Sum = 1098

∴ The correct answer is option (1).

Given:

Radius (r) = 12 cm

Cost of painting per cm² = ₹7

Formula used:

Surface area of a sphere = 4πr²

Calculations:

Surface area = 4 × (22/7) × 12²

⇒ Surface area = 4 × (22/7) × 144

⇒ Surface area = 4 × 22 × 144 / 7

⇒ Surface area = 12672 / 7

⇒ Surface area ≈ 1810 cm²

Cost = 1810 × 7

⇒ Cost ≈ ₹12,672

∴ The correct answer is option 1.

Given:

Original Salary = ₹40,000

New Salary = ₹45,000

Formula Used:

Percentage Increment = [(New Salary - Original Salary) ÷ Original Salary] × 100

Calculations:

Increment = ₹45,000 - ₹40,000 = ₹5,000

Percentage Increment = (₹5,000 ÷ ₹40,000) × 100

⇒ Percentage Increment = 12.5%

Final Answer:

The percentage increment in Rohan’s monthly salary is 12.5%.

Given:

The tag price of an item is ₹250.

After a discount of x% of tag price, the shopkeeper earns 1.25x% profit.

x = 10

Formula used:

Let the cost price be C.

Selling Price (SP) = Tag Price - Discount

Profit Percentage = ((SP - C) / C) × 100

Calculation:

Tag Price = ₹250

Discount = x% of ₹250

⇒ Discount = 10% of ₹250

⇒ Discount = ₹250 × 10/100

⇒ Discount = ₹25

SP = ₹250 - ₹25

⇒ SP = ₹225

Profit = 1.25x% of C

⇒ Profit = 1.25 × 10% of C

⇒ Profit = 12.5% of C

⇒ Profit = 0.125C

SP = C + Profit

⇒ ₹225 = C + 0.125C

⇒ ₹225 = 1.125C

⇒ C = ₹225 / 1.125

⇒ C = ₹200

∴ The correct answer is option (4).

Alternate Method 

Formula Used:

Cost Price = Selling Price ÷ (1 + Profit% ÷ 100)

Selling Price = Tag Price × (1 - Discount% ÷ 100)

Calculations:

Selling Price = ₹250 × (1 - 10 ÷ 100) = ₹250 × 0.9 = ₹225

Cost Price = ₹225 ÷ (1 + 12.5 ÷ 100) = ₹225 ÷ 1.125 = ₹200

Final Answer:

The cost of the item is ₹200.

Given:

Cost Price (CP) = ₹9,200

Repair Cost = ₹1,200

Selling Price (SP) = ₹12,000

Formula used:

Profit Percentage = \(\frac{Profit}{Cost Price} \times 100\)

Where, Profit = SP - (CP + Repair Cost)

Calculations:

Profit = 12000 - (9200 + 1200)

⇒ Profit = 12000 - 10400

⇒ Profit = 1600

Profit Percentage = \(\frac{1600}{10400} \times 100\)

⇒ Profit Percentage = \(\frac{1600}{104}\)

⇒ Profit Percentage = 15.38%

∴ The correct answer is option (1).

Given:

98, 145, 505

Formula used:

LCM of a set of numbers is found by using their prime factorization and taking the highest power of each prime factor.

Calculation:

Prime factorization of 98 = 2 × 7²

Prime factorization of 145 = 5 × 29

Prime factorization of 505 = 5 × 101

LCM = 2 × 7² × 5 × 29 × 101

⇒ LCM = 2 × 49 × 5 × 29 × 101

⇒ LCM = 2 × 49 × 5 × 2929

⇒ LCM = 2 × 49 × 14645

⇒ LCM = 2 × 717605

⇒ LCM = 1435210

∴ The correct answer is option 4.

Given:

Prakash can do a piece of work in 18 days.

Mukesh works twice as fast as Prakash.

Formula used:

Work rate of Prakash = 1 work / 18 days

Work rate of Mukesh = 2 × (1 work / 18 days)

Total work rate when working together = Work rate of Prakash + Work rate of Mukesh

Calculation:

Work rate of Prakash = 1/18

Work rate of Mukesh = 2 × (1/18) = 2/18 = 1/9

Total work rate when working together = 1/18 + 1/9

⇒ Total work rate = 1/18 + 2/18

⇒ Total work rate = 3/18

⇒ Total work rate = 1/6

⇒ Time to complete the work when working together = 1 / (Total work rate)

⇒ Time = 1 / (1/6)

⇒ Time = 6 days

∴ The correct answer is option 1.

Given:

A sum becomes 1331/729 times of itself in 3 years.

Formula used:

A = P × (1 + r/100)^t

Where, r = rate

Calculations:

1331/729 = (1 + r/100)³

⇒ (1331/729)^1/3 = 1 + r/100

⇒ 11/9 = 1 + r/100

⇒ 11/9 - 1 = r/100

⇒ 2/9 = r/100

⇒ r = (2/9) × 100

⇒ r = 200/9

⇒ r ≈ 22.22% or \(22\frac{2}{9}%\)

∴ The correct answer is option (3).

Given:

Radius (r) = 7 cm

Height (h) = 12 cm

π = 22/7

Formula used:

Total Surface Area (TSA) of a cylinder = 2πr² + 2πrh

Calculation:

TSA = 2πr² + 2πrh

Substitute the given values:

TSA = 2 × (22/7) × (7)² + 2 × (22/7) × 7 × 12

TSA = 2 × (22/7) × 49 + 2 × (22/7) × 7 × 12

TSA = (22/7) × 98 + (22/7) × 168

TSA = (22 × 98)/7 + (22 × 168)/7

TSA = (2156)/7 + (3696)/7

TSA = (2156 + 3696)/7

TSA = 5852/7

TSA = 835.71 cm²

∴ The total surface area of the cylinder is 836 cm².

Given:

Principal (P) = ₹3,90,625

Rate (r) = 8% per annum

Time (t) = 2 years

Interest is compounded half-yearly, so the number of compounding periods (n) = 2 × 2 = 4 compounding periods.

Step 1: Determine the number of compounding periods

Since interest is compounded half-yearly, there are 2 compounding periods per year. So, for 2 years, the total number of compounding periods will be:

n = 2 × 2 = 4 compounding periods.

Step 2: Calculate the interest rate per period

The annual interest rate is 8%, so the interest rate per half-year will be:

Rate per period = 8% / 2 = 4%.

Step 3: Calculate the amount after 2 years

Using the compound interest formula:

Amount = Principal × (1 + (rate / 100))ⁿ

Substituting the values:

Amount = ₹3,90,625 × (1 + (4 / 100))⁴

Amount = ₹3,90,625 × (1.04)⁴

Amount = ₹3,90,625 × 1.16985856

Amount = ₹4,56,976

Step 4: Calculate the compound interest

Compound Interest (CI) = Amount - Principal

CI = ₹4,56,976 - ₹3,90,625

CI = ₹66,351

∴ The compound interest on ₹3,90,625 in 2 years at 8% per annum compounded half-yearly is ₹66,351.

Given:

Percentage increase = 30%

To find:

Percentage decrease that cancels out the 30% increase.

Concept:

If the salary is increased by a percentage, it becomes (100 + increase)% of the original salary.

If the salary is then decreased by a percentage, the new salary becomes (100 - decrease)% of the new increased salary.

To cancel out the increase, the product of the increase and the decrease should be 100%. In other words:

\( (1 + \frac{30}{100}) \times (1 - \frac{d}{100}) = 1 \)

Step 1: Solve for the percentage decrease (d):

\( 1.30 \times (1 - \frac{d}{100}) = 1 \)

\( 1 - \frac{d}{100} = \frac{1}{1.30} \)

\( 1 - \frac{d}{100} = \frac{10}{13} \)

\( \frac{d}{100} = 1 - \frac{10}{13} = \frac{3}{13} \)

\( d = \frac{3}{13} \times 100 \)

\( d = \frac{300}{13} \approx 23 \frac{1}{3} \)%

Conclusion:

The percentage decrease required to exactly cancel out a 30% increase is 23 (1/3)%.

Given:

1. The ratio of marks of Prabhu and Raghu is 7 : 8.
2. The ratio of marks of Raghu and Raghav is 5 : 4.

To find:

The ratio of marks of Prabhu and Raghav.

Step 1: Let the marks of Prabhu, Raghu, and Raghav be:

Marks of Prabhu = 7x

Marks of Raghu = 8x

Marks of Raghav = 4y

Step 2: Use the second ratio to relate Raghu's marks:

From the second ratio (Raghu : Raghav = 5 : 4), we can write:

8x = 5k and 4y = 4k (where k is a constant).

Step 3: Solve for x and y:

From 8x = 5k, we get:

x = 5k / 8

From 4y = 4k, we get:

y = k

Step 4: Substitute the values of x and y in terms of k into the Prabhu to Raghav ratio:

The ratio of marks of Prabhu to Raghav is:

Prabhu's marks : Raghav's marks = (7x) : (4y)

= (7 × 5k / 8) : (4 × k)

= (35k / 8) : (4k)

Canceling k from both terms:

Step 5: Simplify the ratio:

35 / 8 : 4 = 35 : 32

Therefore, the ratio of marks obtained by Prabhu and Raghav is 35 : 32.

Given:

The e×pression to be simplified is:

(231 + 19)² + (231 - 19)² / (231 × 231 + 19 × 19)

Step 1: Expand the numerator:

Numerator = (231 + 19)² + (231 - 19)²

Expanding each square:

(231 + 19)² = 231² + 2 × 231 × 19 + 19²

(231 - 19)² = 231² - 2 × 231 × 19 + 19²

Adding both expressions together:

Numerator = (231² + 2 × 231 × 19 + 19²) + (231² - 2 × 231 × 19 + 19²)

Numerator = 231² + 231² + 19² + 19²

Numerator = 2 × 231² + 2 × 19²

Step 2: Simplify the denominator:

Denominator = 231 × 231 + 19 × 19

Denominator = 231² + 19²

Step 3: Simplify the entire expression:

Now, substitute the simplified e×pressions for the numerator and denominator:

(2 × 231² + 2 × 19²) / (231² + 19²)

Factor out the common term "2" from the numerator:

2 × (231² + 19²) / (231² + 19²)

Step 4: Cancel common terms:

The (231² + 19²) terms cancel out, leaving:

2

Therefore, the value of the given e×pression is 2.