Quantitative Aptitude PYQs

Given:

The radii of the bases of two cylinders are in the ratio 3 ∶ 7.

The heights of the cylinders are in the ratio 7 ∶ 9.

(Тake π = 22/7)

Formula used:

Curved surface area of a cylinder = 2πrh

Calculation:

Let the radius of the first cylinder be 3x and the height be 7y.

Let the radius of the second cylinder be 7x and the height be 9y.

Curved surface area of the first cylinder = 2π(3x)(7y)

Curved surface area of the second cylinder = 2π(7x)(9y)

Ratio of curved surface areas:

⇒ (2π(3x)(7y)) / (2π(7x)(9y))

⇒ (3×7) / (7×9)

⇒ 21 / 63

⇒ 1 / 3

∴ The correct answer is option (3).

Given:

A fruit seller sells (2/5) part of the fruits at a profit of 15%, and remaining at a loss 5%

Calculation:

Let the cost price be 500 

A fruit seller sells (2/5) part of the fruits at a profit of 15% = (2/5)× 500 

⇒ 200 × (115/100)

⇒ 230

Remaaining part of the fruits at a loss 5% = (1 - 2/5) × 500 

⇒ 300 × (100 - 5)/100

⇒ 285 

CP = 500 

SP = 230 + 285 = 515

Total profit = 515 - 500 = 15 

15 → 1800 

1 → 120 

500 → 120 × 500 = Rs. 60,000

∴ The total cost price of the fruits is Rs. 60,000

Given:

Speed of the first train = 54 km/h

Distance = 70 m

Time = 14 seconds

Formula used:

Relative speed = Distance / Time

Speed of the second train = Speed of the first train - Relative speed

Calculation:

Relative speed = 70 m / 14 s

⇒ Relative speed = 5 m/s

Convert Relative speed to km/h:

Relative speed = 5 m/s × 18/5

⇒ Relative speed = 18 km/h

Speed of the second train = 54 km/h - 18 km/h

⇒ Speed of the second train = 36 km/h

∴ The correct answer is option (2).

Given:

Installments X = ₹750

Number of years N = 5

Rate of simple interest R = 5%

Formula used:

X = P × 100 / [(N × 100) + {(N-1) + (N-2) + (N-3) + …. + 1} × R)]

Calculation:

Sum of series = (5-1) + (5-2) + (5-3) + (5-4) + 1 = 4 + 3 + 2 + 1 = 10

X = P × 100 / [(N × 100) + {(N-1) + (N-2) + (N-3) + …. + 1} × R)]

750 = P × 100 / [(5 × 100) + {10} × 5]

⇒ 750 = P × 100 / [500 + 50]

⇒ 750 = P × 100 / 550

⇒ 750 = P × 10 / 55

⇒ P = 4125

The debt is ₹4125.

Given:

Marked Price (MP) = ₹2875

Discount = 8%

Profit = 15%

Formula used:

Selling Price (SP) = MP - Discount

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

Calculation:

Discount Amount = MP × (Discount / 100)

⇒ ₹2875 × 0.08 = ₹230

Selling Price (SP) = MP - Discount Amount

⇒ ₹2875 - ₹230 = ₹2645

Cost Price (CP) = SP / (1 + Profit / 100)

⇒ ₹2645 / 1.15

⇒ ₹2645 / 1.15 = ₹2300

∴ The correct answer is option (4).

Given:

Radius (r) = 1.65 cm

Formula used:

Surface Area of Sphere = 4\(\pi r^2 \)

Calculation:

Surface Area = 4\(\pi (1.65)^2 \)

⇒ Surface Area = 4 × \(\pi (2.7225) \)

⇒ Surface Area = 10.89\(\pi\)

∴ The correct answer is option (3).

Given:

The ratio of boys to girls = 9 ∶ 13

Total number of boys and girls = 2134

Formula used:

Total = Boys + Girls

Boys = \(\frac{9}{9+13} \times \text{Total}\)

Calculations:

Boys = \(\frac{9}{9+13} \times 2134\)

⇒ Boys = \(\frac{9}{22} \times 2134\)

⇒ Boys = \(\frac{19206}{22}\)

⇒ Boys = 873

∴ The correct answer is option (4).

Given:

Price of the computer (Original Price) = ₹34,000

Reduction Percentage = 17%

Formula used:

Reduced Price = Original Price × (1 - Reduction Percentage/100)

Calculation:

Reduced Price = 34,000 × (1 - 17/100)

⇒ Reduced Price = 34,000 × (1 - 0.17)

⇒ Reduced Price = 34,000 × 0.83

⇒ Reduced Price = ₹28,220

∴ The correct answer is option (3).

Given:

Average height of 16 students = 160 cm

Average height of first 6 students = 150 cm

Average height of last 8 students = 165 cm

Formula used:

Total height = Average × Number of students

Remaining students' average height = Remaining students' total height ÷ Number of remaining students

Calculation:

Total height of 16 students = 160 × 16

⇒ Total height = 2560 cm

Total height of first 6 students = 150 × 6

⇒ First 6 students' height = 900 cm

Total height of last 8 students = 165 × 8

⇒ Last 8 students' height = 1320 cm

Remaining students' total height = 2560 - (900 + 1320)

⇒ Remaining students' total height = 2560 - 2220

⇒ Remaining students' total height = 340 cm

Remaining students' average height = 340 ÷ 2

⇒ Remaining students' average height = 170 cm

∴ The correct answer is option (3).

Given:

The salaries of P and Q together amount to ₹4,32,000.

P spends 75% of his salary, and Q spends 61% of his salary.

P's savings = Q's savings.

Formula used:

Savings = Salary - Expenses

Total salaries of P and Q = ₹4,32,000.

Calculations:

Let P's salary be ₹x, then Q's salary = ₹(4,32,000 - x).

P's savings = Q's savings

⇒ x - 0.75x = (4,32,000 - x) - 0.61(4,32,000 - x)

⇒ 0.25x = 4,32,000 - x - 0.61 × 4,32,000 + 0.61x

⇒ 0.25x = 4,32,000 - 0.61 × 4,32,000 - x + 0.61x

⇒ 0.25x = 4,32,000 - 2,63,520 + 0.61x - x

⇒ 0.25x = 1,68,480 - 0.39x

⇒ 0.25x + 0.39x = 1,68,480

⇒ 0.64x = 1,68,480

⇒ x = 1,68,480 ÷ 0.64

⇒ x = 2,63,250

∴ The correct answer is option (3): ₹2,63,250.

Given:

Principal (P) = ₹24,000

Simple Interest (SI) = ₹7,200

Time (t) = 3 years

Formula used:

SI = \(\frac{P \times r \times t}{100}\)

CI = P\((1+\frac{r}{100})^t - 1\)

Calculation:

For Simple Interest:

7200 = \(\frac{24000 \times r \times 3}{100}\)

⇒ r = \(\frac{7200 \times 100}{24000 \times 3}\)

⇒ r = 10%

For Compound Interest:

CI = 24000\((1+\frac{10}{100})^3 - 1\)

⇒ CI = 24000\((1.1)^3 - 1\)

⇒ CI = 24000\((1.331 - 1)\)

⇒ CI = 24000 × 0.331

⇒ CI = ₹7,944

∴ The correct answer is option (4).

Given:

Diameter of hemisphere = 14 cm

Radius of hemisphere (r) = Diameter/2 = 7 cm

\(\pi = \frac{22}{7}\)

Formula used:

Total Surface Area (TSA) of a hemisphere = 3\(\pi r^2\)

Curved Surface Area (CSA) of a hemisphere = 2\(\pi r^2\)

Difference = TSA - CSA

Calculation:

Total Surface Area = 3\(\pi r^2\)

⇒ TSA = 3 × \(\frac{22}{7} × 7^2\)

⇒ TSA = 3 × \(\frac{22}{7} × 49\)

⇒ TSA = 3 × 154 = 462 cm²

Curved Surface Area = 2\(\pi r^2\)

⇒ CSA = 2 × \(\frac{22}{7} × 7^2\)

⇒ CSA = 2 × \(\frac{22}{7} × 49\)

⇒ CSA = 2 × 154 = 308 cm²

Difference = TSA - CSA

⇒ Difference = 462 - 308

⇒ Difference = 154 cm²

∴ The correct answer is option (3).

Shortcut TrickDifference = TSA - CSA

⇒ Difference =  3\(\pi r^2\)  - 2\(\pi r^2\)

⇒ Difference = \(\pi r^2\)

\(\pi = \frac{22}{7}\)

⇒ Difference = 22/7 × 7²

Difference = 22/7 × 49 = 154 cm2

Given:

A garrison of 1500 men has provisions for 12 days.

The garrison is increased to 2000 men.

Formula used:

Provisions are inversely proportional to the number of men.

Men₁ × Days₁ = Men₂ × Days₂

Calculation:

1500 × 12 = 2000 × Days₂

⇒ Days₂ = (1500 × 12) ÷ 2000

⇒ Days₂ = 18000 ÷ 2000

⇒ Days₂ = 9

∴ The provisions will last for 9 days, and the correct answer is option (2).

Given:

Number = 1560

Formula used:

Prime Factorization of a number = Find all prime factors of the number and sum them.

Calculation:

Prime factorization of 1560 = 2 × 2 × 2 × 3 × 5 × 13

⇒ Prime factors = 2, 3, 5, 13

Sum of prime factors = 2 + 3 + 5 + 13

⇒ Sum of prime factors = 23

Given options:

Option 1: 21

Option 2: 22

Option 3: 23

Option 4: 41

From the options, the number greater than the sum of prime factors (23) is:

⇒ Option 4: 41

∴ The correct answer is option (4).

Calculating the Total Interest

Since the interest is calculated on the original principal for the entire duration, the total simple interest for 4 years will be:

Total Interest = Principal × Rate × Time

Total Interest = ₹30,000 × (20/100) × 4

Total Interest = ₹30,000 × 0.20 × 4

Total Interest = ₹6,000 × 4

Total Interest = ₹24,000

Calculating the Total Amount to be Repaid

The total amount Ram needs to repay is the principal plus the total interest:

Total Amount = Principal + Total Interest

Total Amount = ₹30,000 + ₹24,000

Total Amount = ₹54,000

Calculating the Amount of Each Instalment

Ram agrees to repay this total amount in four equal annual installments:

Amount of Each Instalment = Total Amount / Number of Instalments

Amount of Each Instalment = ₹54,000 / 4

Amount of Each Instalment = ₹13,500

Therefore, the amount of each instalment will be ₹13,500.

Given:

Simplify (0.85 × 3.4 × 2.4) ÷ 0.3

Formula used:

To simplify: Perform multiplication first, then divide.

Calculations:

(0.85 × 3.4 × 2.4) ÷ 0.3

⇒ (2.89 × 2.4) ÷ 0.3

⇒ 6.936 ÷ 0.3

⇒ 23.12

∴ The correct answer is option (4).

Given:

Marked Price (MP) = ₹1,500

Discount = 10%

Cost Price (CP) = MP / 1.2 (as MP is 20% above CP)

Formula used:

Selling Price (SP) = MP × (1 - Discount/100)

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

Calculation:

Marked price is 20% above the cost price

CP = ₹1,500 / 1.2

⇒ CP = ₹1,250

SP = ₹1,500 × (1 - 10/100)

⇒ SP = ₹1,500 × 0.9

⇒ SP = ₹1,350

Profit Percentage = ((₹1,350 - ₹1,250) / ₹1,250) × 100

⇒ Profit Percentage = (₹100 / ₹1,250) × 100

⇒ Profit Percentage = 8%

∴ The correct answer is option (3).

Given:

Total cost price = ₹1800

First case: Profit% = 12% and 20%

Second case: Profit% = 20% and 12%

Difference in total amount = ₹36

Formula used:

Profit = (Profit Percentage × Cost Price)/100

Calculation:

Let: Cost price of first chair = ₹x

Cost price of second chair = ₹(1800 − x)

Case 1 – Original profit:

Selling price = x + 12% of x + (1800 − x) + 20% of (1800 − x)

Selling price = x + 0.12x + (1800 − x) + 0.20(1800 − x)

Case 2 – Swapped profit:

Selling price = x + 20% of x + (1800 − x) + 12% of (1800 − x)

Selling price = x + 0.20x + (1800 − x) + 0.12(1800 − x)

Difference between second and first case = ₹36

⇒ (x + 0.20x + 1800 − x + 0.12(1800 − x)) − (x + 0.12x + 1800 − x + 0.20(1800 − x)) = 36

⇒ (1.20x + 1800 − x + 216 − 0.12x) − (1.12x + 1800 − x + 360 − 0.20x) = 36

⇒ (0.08x + 2016) − (−0.08x + 2160) = 36

⇒ 0.08x + 2016 + 0.08x − 2160 = 36

⇒ 0.16x − 144 = 36

⇒ 0.16x = 180

⇒ x = 180 ÷ 0.16 = 1125

Cost price of first chair = ₹1125

Cost price of second chair = ₹1800 − ₹1125 = ₹675

∴ Required difference = ₹1125 − ₹675 = ₹450

∴ The correct answer is option (1): ₹450

Given:

LCM = 18 × HCF

LCM + HCF = 855

One number = 81

Formula used:

For two numbers, LCM × HCF = Product of the numbers

Calculation:

Let HCF = x

⇒ LCM = 18 × x

LCM + HCF = 855

⇒ 18x + x = 855

⇒ 19x = 855

⇒ x = 855 ÷ 19

⇒ x = 45

HCF = 45, LCM = 18 × 45 = 810

Let the other number = y

Using the formula: LCM × HCF = Product of the numbers

⇒ 810 × 45 = 81 × y

⇒ y = (810 × 45) ÷ 81

⇒ y = 450

∴ The correct answer is option (2).

Formula used:

Distance (D) = Speed (S) × Time (T)

Let the normal speed = S km/h

Let the distance = D km

Calculations:

Normal time taken = D/S hours

Condition 1: If speed = (S + 4) km/h, time taken = D/(S + 4) hours

Condition 2: If speed = (S - 2) km/h, time taken = D/(S - 2) hours

From the given conditions:

⇒ Normal time - Time with increased speed = 3 minutes

⇒ D/S - D/(S + 4) = 3/60

⇒ (D × (S + 4) - D × S) / (S × (S + 4)) = 1/20

⇒ 4D / (S × (S + 4)) = 1/20

⇒ 80D = S × (S + 4) ...(i)

From the second condition:

⇒ Time with reduced speed - Normal time = 2 minutes

⇒ D/(S - 2) - D/S = 2/60

⇒ (D × S - D × (S - 2)) / (S × (S - 2)) = 1/30

⇒ (2D) / (S × (S - 2)) = 1/30

⇒ 60D = S × (S - 2) ...(ii)

From equations (i) and (ii):

⇒ 80D = S² + 4S

⇒ 60D = S² - 2S

Subtract equation (ii) from (i):

⇒ 20D = 6S

⇒ D = 6S / 20 = 3S / 10

Substitute D = 3S/10 in equation (ii):

⇒ 60 × (3S/10) = S² - 2S

⇒ 18S = S² - 2S

⇒ S² - 20S = 0

⇒ S(S - 20) = 0

⇒ S = 20 (as S = 0 is not valid)

Substitute S = 20 in D = 3S/10:

⇒ D = 3 × 20 / 10 = 6 km

∴ The correct answer is option (4).