Quantitative Aptitude PYQs

Given:

\(\dfrac{\dfrac{6}{7}+\dfrac{2}{3}}{\dfrac{1}{6}-\dfrac{3}{20}}\)

Formula used:

\(\dfrac{\dfrac{a}{b}+\dfrac{c}{d}}{\dfrac{e}{f}-\dfrac{g}{h}}\)

Calculation:

\(\dfrac{\dfrac{6}{7}+\dfrac{2}{3}}{\dfrac{1}{6}-\dfrac{3}{20}}\)

⇒ \(\dfrac{\dfrac{6 \times 3 + 2 \times 7}{7 \times 3}}{\dfrac{1 \times 20 - 3 \times 6}{6 \times 20}}\)

⇒ \(\dfrac{\dfrac{18 + 14}{21}}{\dfrac{20 - 18}{120}}\)

⇒ \(\dfrac{\dfrac{32}{21}}{\dfrac{2}{120}}\)

⇒ \(\dfrac{32}{21} \times \dfrac{120}{2}\)

⇒ \(\dfrac{32 \times 120}{21 \times 2}\)

⇒ \(\dfrac{3840}{42}\)

⇒ \(\dfrac{640}{7}\)

∴ The correct answer is option 1

Given:

Let votes polled for S = x

Votes polled for R = x + ¹/₄x

Total votes = R + S = x + x + ¹/₄x

Formula used:

Percentage votes for R = \(\frac{\text{Votes polled for R}}{\text{Total votes}} \times 100\)

Calculation:

Total votes = x + x + ¹/₄x = ⁹/₄x

Votes for R = x + ¹/₄x = ⁵/₄x

⇒ Percentage votes for R = \(\frac{\frac{5}{4}x}{\frac{9}{4}x} \times 100\)

⇒ \(\frac{5}{9} \times 100\)

⇒ 55.56%

∴ The percentage of votes polled in favour of R is 55.56%.

Given:

There are four prime numbers written in ascending order. The product of the first three is 30 and that of the last three is 165.

Formula used:

Let the four prime numbers be p₁, p₂, p₃, and p₄ in ascending order.

Given:

p₁ × p₂ × p₃ = 30

p₂ × p₃ × p₄ = 165

Calculation:

Prime factors of 30: 2, 3, 5

Prime factors of 165: 3, 5, 11

From the given product equations:

p₁ = 2

p₂ = 3

p₃ = 5

p₄ = 11

⇒ The last number is 11

∴ The correct answer is option (1).

Given:

The cost of an item increased to ₹250 from ₹225.

Formula used:

Percentage Increase = \(\dfrac{(New Value - Old Value)}{Old Value} \times 100\)

Calculation:

Percentage Increase = \(\dfrac{(250 - 225)}{225} \times 100\)

⇒ Percentage Increase = \(\dfrac{25}{225} \times 100\)

⇒ Percentage Increase = \(\dfrac{25 \times 100}{225}\)

⇒ Percentage Increase = 11.11%

∴ The correct answer is option 1.

Given:

Discount = 10%

Selling Price (SP) = ₹6,750

Formula used:

SP = Marked Price (MP) - Discount

Discount = MP × (Discount%)

Calculation:

Let the Marked Price be MP.

Since the selling price is 90% of the marked price (100% - 10%):

⇒ 6750 = MP × 90/100

⇒ MP = 6750 × 100/90

⇒ MP = 7500

Discount = MP - SP

⇒ Discount = 7500 - 6750

⇒ Discount = ₹750

∴ The correct answer is option (1).

Given:

A can complete the work in 16 days.

B can complete the work in 26 days.

Formula used:

Work done by A and B together in 1 day = (1/A's days) + (1/B's days)

Total days required = 1 / (Work done in 1 day)

Calculation:

Work done by A in 1 day = 1/16

Work done by B in 1 day = 1/26

Work done by A and B together in 1 day = (1/16) + (1/26)

⇒ (26 + 16) / (16 × 26)

⇒  42 / 416

⇒ 21 / 208

Total days required = 1 / (21 / 208) = 208 / 21

⇒  9 (19/21) days

Therefore, the correct answer is option 3: 9 \(\frac{19}{21}\)

Given:

If the average of 10 numbers is 625 and the average of 25 numbers is 100

Formula used:

Average = Total Sum / Total Count

Calculation:

Total Sum of 10 numbers = 10 × 625

Total Sum of 25 numbers = 25 × 100

Total Sum of 35 numbers = (10 × 625) + (25 × 100)

⇒ Total Sum of 35 numbers = 6250 + 2500

⇒ Total Sum of 35 numbers = 8750

Average of 35 numbers = Total Sum of 35 numbers / 35

⇒ Average of 35 numbers = 8750 / 35

⇒ Average of 35 numbers = 250

∴ The correct answer is option (4).

Given:

Principal (P): Unknown

Amount after 3 years: ₹1,725

Amount after 5 years: ₹1,875

Time difference: 5 - 3 = 2 years

Calculation:

Step 1: Find the simple interest for 2 years

Simple Interest for 2 years = 1,875 - 1,725 = ₹150

Step 2: Find the simple interest for 1 year

Simple Interest for 1 year = 150 ÷ 2 = ₹75

Step 3: Find the total interest for 3 years

Simple Interest for 3 years = 75 × 3 = ₹225

Step 4: Find the principal amount

Principal (P) = 1,725 - 225 = ₹1,500

Step 5: Calculate the rate of interest

Using the formula for simple interest:

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

Substitute the values:

225 = (1,500 × R × 3) ÷ 100

225 = 45R

R = 225 ÷ 45 = 5%

Final Answer:

The rate of interest per annum is 5%.

Given:

Compound amount for 2 years = ₹12,260

Compound amount for 3 years = ₹12,873

Formula used:

Compound interest: A = P(1 + r/100)ⁿ, where:

A = Amount after time n years

P = Principal

r = Rate of interest

Calculations:

Calculate the ratio of amounts for consecutive years:

A₃ / A₂ = 12,873 / 12,260

Simplify:

A₃ / A₂ = 1 + r/100

12,873 / 12,260 ≈ 1.05

Thus, 1 + r/100 = 1.05.

Find the rate of interest:

r/100 = 1.05 - 1 = 0.05

r = 0.05 × 100 = 5%

Final Answer:

The rate of interest is 5% per annum.

Given:

A football can occupy 36π cm³ of air.

Formula used:

Volume of a sphere (V) = (4/3)πr³

Where, r = radius

Calculation:

36π = (4/3)πr³

⇒ 36 = (4/3)r³

⇒ 36 × (3/4) = r³

⇒ 27 = r³

⇒ r = 3

∴ The correct answer is option (4).

Given:

Distance between A and B = 100 km

Speed of Train A = 20 km/h

Speed of Train B = 25 km/h

Train A starts at 7 A.M.

Train B starts at 8 A.M.

Formula used:

Relative speed when two trains are moving towards each other = Speed of Train A + Speed of Train B

Time taken to meet = Distance / Relative Speed

Calculation:

Relative speed = 20 km/h + 25 km/h = 45 km/h

Train A travels 1 hour before Train B starts, so distance covered by Train A in 1 hour = 20 km

Remaining distance = 100 km - 20 km = 80 km

Time taken to cover remaining distance at relative speed = 80 km / 45 km/h

⇒ Time = 80 / 45 hours

⇒ Time = 1.78 hours

Convert 1.78 hours into hours and minutes:

1.78 hours = 1 hour + (0.78 × 60) minutes ≈ 1 hour 47 minutes

⇒ Time ≈ 1 hour 47 minutes

Train B starts at 8 A.M., so they will meet at 8 A.M. + 1 hour 47 minutes

⇒ Meeting time = 9:47 A.M.

∴ The correct answer is option 4.

Shortcut Trick

Identify the triplet: The sides 6, 8, and 10 follow the ratio 3 : 4 : 5.

Since 3² + 4² = 5², the triangle is right-angled.

Area = (1/2) × Base × Height

Area = (1/2) × 6 × 8 = 24 cm².

∴ The correct answer is 24 cm².

Alternate Method

Given: Sides of the triangle are a = 6 cm, b = 8 cm, c = 10 cm.

Formula Used: Heron's Formula, Area = √(s(s−a)(s−b)(s−c)), where s = (a + b + c) ÷ 2.

6 cm 8 cm 10 cm Right-Angled Triangle

⇒ Semi-perimeter (s) = (6 + 8 + 10) ÷ 2

⇒ s = 24 ÷ 2 = 12 cm

⇒ Area = √(12 × (12−6) × (12−8) × (12−10))

⇒ Area = √(12 × 6 × 4 × 2)

⇒ Area = √576

⇒ Area = 24 cm²

∴ The correct answer is 24 cm².

Additional Information

Pythagorean Triplets

A set of three positive integers a, b, and c, such that a² + b² = c². Common examples include (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Equilateral Triangle Area

The area of a triangle with all sides equal (a) is calculated as (√3 ÷ 4) × a².

Semi-perimeter Property

The semi-perimeter (s) is always greater than any single side of the triangle (s > a, s > b, s > c).

Shortcut Trick
Observe the antecedent (first term) and consequent (second term) of each given ratio.

Only in the ratio 6 : 5 is the antecedent strictly greater than the consequent.
This means 6 : 5 evaluates to a value greater than 1, while all other given ratios (3 : 5, 4 : 5, 7 : 10) are proper fractions less than 1.
Thus, without any complex calculation, we can determine the largest value.
∴ The correct answer is 6 : 5.

Alternate Method
Given:
Four ratios: 3 : 5, 6 : 5, 4 : 5, and 7 : 10

Method Used:
Convert each ratio into its equivalent decimal value by dividing the first term by the second term.

Calculations:
⇒ 3 : 5 = 3 ÷ 5 = 0.6
⇒ 6 : 5 = 6 ÷ 5 = 1.2
⇒ 4 : 5 = 4 ÷ 5 = 0.8
⇒ 7 : 10 = 7 ÷ 10 = 0.7
⇒ Comparing the calculated decimal values: 1.2 > 0.8 > 0.7 > 0.6
⇒ The largest decimal value is 1.2, which directly corresponds to the ratio 6 : 5.
∴ The correct answer is 6 : 5.

Additional Information
Comparing by Denominator Equalization (LCM Method)
Fractions can be rapidly compared by equating their denominators. The LCM of 5 and 10 is 10. Expanding the given ratios yields 6/10, 12/10, 8/10, and 7/10. Comparing numerators: 12 > 8 > 7 > 6, confirming 6 : 5 is the maximum.

Proper vs Improper Ratios
A ratio a : b where a < b behaves like a proper fraction (less than 1). When a > b, it forms an improper fraction (greater than 1). Spotting this visually saves valuable examination time.

Percentage Conversion Technique
Converting ratios to percentages using Base × 100 provides a simple uniform scale. Applying this gives 60%, 120%, 80%, and 70%. Comparing whole integers helps to eliminate decimal point calculation errors.

Given:

Selling Price of each article = ₹840

Profit on first article = 20%

Loss on second article = 20%

Formula used:

Cost Price (CP) = \(\dfrac{Selling Price}{1 + Profit%/100}\) for profit

Cost Price (CP) = \(\dfrac{Selling Price}{1 - Loss%/100}\) for loss

Calculations:

CP of first article = \(\dfrac{840}{1 + 20/100}\)

⇒ CP of first article = \(\dfrac{840}{1.2}\)

⇒ CP of first article = ₹700

CP of second article = \(\dfrac{840}{1 - 20/100}\)

⇒ CP of second article = \(\dfrac{840}{0.8}\)

⇒ CP of second article = ₹1050

Total CP = ₹700 + ₹1050 = ₹1750

Total SP = ₹840 + ₹840 = ₹1680

Loss = Total CP - Total SP

⇒ Loss = ₹1750 - ₹1680 = ₹70

Loss% = \(\dfrac{Loss}{Total CP} \times 100\)

⇒ Loss% = \(\dfrac{70}{1750} \times 100\)

⇒ Loss% = 4%

∴ The correct answer is option 3.

Given:

If the product of two co-prime numbers is 306

Formula used:

LCM × GCD = Product of the numbers

Calculation:

Let the two co-prime numbers be a and b.

Given, a × b = 306 and GCD(a, b) = 1 (since they are co-prime)

⇒ LCM(a, b) × GCD(a, b) = a × b

⇒ LCM(a, b) × 1 = 306

⇒ LCM(a, b) = 306

∴ The correct answer is option 1.

Given:

If a man walks at five-sixth of his usual speed from his home towards his office, he is 20 minutes late.

Formula used:

Let the usual time be T minutes. If the speed is reduced to five-sixth, the time taken will be \(\dfrac{6}{5}T\) .

Difference in time = \(\dfrac{6}{5}T - T\) = 20 minutes

Calculations:

\(\dfrac{6}{5}T - T\) = 20

⇒ \(\dfrac{6T - 5T}{5}\) = 20

⇒ \(\dfrac{T}{5}\) = 20

⇒ T = 20 × 5

⇒ T = 100 minutes

∴ The correct answer is option 3.

Given:

\(47\frac{37}{50}\times 4\frac{1}{6}\times 24\)

Formula used:

\(\text{Convert mixed fractions to improper fractions and perform multiplication.}\)

Calculation:

\(\text{Convert mixed fractions: } 47\frac{37}{50} = \frac{2387}{50}, 4\frac{1}{6} = \frac{25}{6}\)

⇒ \(\frac{2387}{50} \times \frac{25}{6} \times 24\)

⇒ \(\frac{2387 \times 25 \times 24}{50 \times 6}\)

⇒ \(4774\)

∴ The final answer is 4774.

Given:

A alone can build a house in 16 days

B alone can build a house in 12 days

A and B work on alternate days

A works on the first day

Formula used:

Work rate of A = 1/16 per day

Work rate of B = 1/12 per day

Combined work in 2 days = (Work rate of A) + (Work rate of B)

Calculations:

Work rate of A = 1/16

Work rate of B = 1/12

Combined work in 2 days = (1/16) + (1/12)

⇒ Combined work in 2 days = (3 + 4) / 48

⇒ Combined work in 2 days = 7/48

To find the number of days required to complete 1 unit of work:

Let x be the number of days required

Since they work on alternate days, we can calculate the work done in (2n) days and (2n+1) days:

Work done in 14 days (7 cycles) = 7 × (7/48)

⇒ Work done in 14 days = 49/48

This is slightly more than 1, hence we need to consider up to the exact day:

Work done in 13 days (6 full cycles + 1 day of A) = 6 × (7/48) + (1/16)

⇒ Work done in 13 days = 42/48 + 3/48

⇒ Work done in 13 days = 45/48

Remaining work = 1 - 45/48

⇒ Remaining work = 3/48

⇒ Remaining work = 1/16

This remaining work will be completed by A on the 14^th day:

⇒ The house will be built in \(13\frac{3}{4}\) days

∴ The correct answer is option 1.

Given:

Cost price (CP) of 12 dozen eggs = ₹60 per dozen

Selling price (SP) of 1 egg = ₹6 per piece

Formula used:

Profit percentage = \(\dfrac{Profit}{CP} \times 100\)

Profit = SP - CP

Calculation:

CP of 12 dozen eggs = 12 × ₹60 = ₹720

Number of eggs = 12 × 12 = 144

SP of 144 eggs = 144 × ₹6 = ₹864

Profit = SP - CP

⇒ Profit = ₹864 - ₹720 = ₹144

Profit percentage = \(\dfrac{144}{720} \times 100\)

⇒ Profit percentage = 0.2 × 100 = 20%

∴ The correct answer is option (3).

Given:

An isosceles right-angled triangle with hypotenuse length = 10 units.

Formula used:

Area of an isosceles right-angled triangle = 1/2 × (side²)

In an isosceles right-angled triangle, if hypotenuse = c, then each of the other two sides = c/√2

Calculation:

Hypotenuse (c) = 10

Side = 10/√2

⇒ Side = 10/√2 = 10√2/2 = 5√2

⇒ Area = 1/2 × (5√2)²

⇒ Area = 1/2 × (25 × 2)

⇒ Area = 25 square units

∴ The correct answer is option (2).