Quantitative Aptitude PYQs

Given:

A sum of money on compound interest becomes double of itself in 3 years.

Formula Used:

If a sum of money becomes n times in T years at compound interest, then to find the number of years it will take to become

\(T_m = T \times \frac{\log m}{\log n}\)

Calculation:

Here, n = 2 , T = 3 years, and m = 8 .

Using the formula:

\(T_m = 3 \times \frac{\log 8}{\log 2}\)

We know that \(\log 8 = \log (2^3) = 3 \log 2 . So,\)

\(\frac{\log 8}{\log 2} = \frac{3 \log 2}{\log 2} = 3 .\)

Therefore,

\(T_m = 3 \times 3 = 9 years\).

The correct answer is option 2.

Given:

The ratio of the two numbers = 5 : 7

The product of their HCF and LCM = 2835

Formula Used:

If two numbers are in the ratio a:b, let the numbers be 5x and 7x.

HCF × LCM = Product of the numbers

Calculation:

Let the two numbers be 5x and 7x.

HCF × LCM = 5x × 7x

Product of the numbers = 35x²

Given HCF × LCM = 2835

⇒ 35x² = 2835

⇒ x² = 2835 / 35

⇒ x² = 81

⇒ x = 9

The numbers are 5x and 7x.

5 × 9 = 45

7 × 9 = 63

The sum of the numbers = 45 + 63

⇒ 108

The sum of the numbers is 108.

Given:

Total amount = Rs. 10,400

Proportions for P, Q, and R = \(\frac{1}{2} : \frac{1}{5} : \frac{1}{6}\)

Formula Used:

Q's share = Total amount × (Q's proportion / Sum of proportions)

Calculation:

Sum of proportions = \(\frac{1}{2} + \frac{1}{5} + \frac{1}{6}\)

Sum of proportions = \(\frac{15 + 6 + 5}{30} = \frac{26}{30} = \frac{13}{15}\)

Q's proportion = \(\frac{1}{5}\)

Q's share = Rs. 10,400 × \(\frac{1}{5} / \frac{13}{15}\)

Q's share = Rs. 10,400 × \(\frac{1}{5} × \frac{15}{13}\)

Q's share = Rs. 10,400 × \(\frac{15}{65}\)

Q's share = Rs. 10,400 × \( \frac{3}{13}\)

Q's share = Rs. 2,400

Q's share is Rs. 2,400.

Given:

Interest rate at 12% = 12% per annum

Interest rate at 15% = 15% per annum

Time period = 2 years

Total interest received = ₹675

Formula Used:

Simple Interest = (Principal × Rate × Time) / 100

Calculation:

Let the amount lent at 12% be ₹x and the amount lent at 15% also be ₹x.

Interest from amount lent at 12%:

Interest_12% = (x × 12 × 2) / 100

Interest_12% = (24x) / 100

Interest from amount lent at 15%:

Interest_15% = (x × 15 × 2) / 100

Interest_15% = (30x) / 100

Total interest:

Interest_total = Interest_12% + Interest_15%

⇒ 675 = (24x) / 100 + (30x) / 100

⇒ 675 = (54x) / 100

⇒ 675 × 100 = 54x

⇒ 67500 = 54x

⇒ x = 67500 / 54

⇒ x = 1250

The amount lent at 12% was ₹1,250.

Given:

K is a prime number strictly greater than 5.

Formula Used:

To find the remainder when (K² + 19) is divided by 12.

Calculation:

Since K is a prime number greater than 5, K must be one of 6k ± 1 (as all primes greater than 3 are of this form).

Let's consider both cases:

Case 1: K = 6k + 1

⇒ (K² = (6k + 1)²

⇒ (K² = 36k² + 12k + 1)

⇒ (K² ≡ 1 (mod 12))

⇒ K² + 19 ≡ 1 + 19 ≡ 20 ≡ 8 (mod 12)

Case 2: K = 6k - 1

⇒ K² = (6k - 1)²

⇒ K² = 36k² - 12k + 1

⇒ K² ≡ 1 (mod 12)

⇒ K² + 19 ≡ 1 + 19 ≡ 20 ≡ 8 (mod 12)

The remainder when (K² + 19) is divided by 12 is 8.

Given:

Percentage of female staff = 45%

Percentage of male staff = 55%

Percentage of postgraduate female staff = 80%

Percentage of postgraduate male staff = 60%

Formula Used:

Percentage of non-postgraduate staff = 100% - Percentage of postgraduate staff

Calculation:

Percentage of postgraduate female staff = 45% × 80%

⇒ Percentage of postgraduate female staff = 0.45 × 0.80

⇒ Percentage of postgraduate female staff = 0.36 (36%)

Percentage of postgraduate male staff = 55% × 60%

⇒ Percentage of postgraduate male staff = 0.55 × 0.60

⇒ Percentage of postgraduate male staff = 0.33 (33%)

Total percentage of postgraduate staff = 36% + 33%

⇒ Total percentage of postgraduate staff = 69%

Percentage of non-postgraduate staff = 100% - 69%

⇒ Percentage of non-postgraduate staff = 31%

The percentage of the non-postgraduate staff in the school is 31%.

Given:

Selling price of each device = ₹4,982

Profit on the first device = 6%

Loss on the second device = 6%

Formula Used:

Cost Price (CP) = Selling Price (SP) / (1 ± Profit/Loss %)

Calculation:

Let the cost price of the first device be CP₁ and the second device be CP₂ .

Selling price of the first device = ₹4,982

Profit% = 6%

⇒ \(CP_1 = \frac{4,982}{1 + \frac{6}{100}}\)

⇒ \( CP_1 = \frac{4,982}{1.06}\)

⇒ \(CP_1 ≈ ₹4,700\)

Selling price of the second device = ₹4,982

Loss% = 6%

⇒ \(CP_2 = \frac{4,982}{1 - \frac{6}{100}}\)

⇒ \(CP_2 = \frac{4,982}{0.94}\)

⇒ \( CP_2 ≈ ₹5,300\)

Total cost of both devices = \(CP_1 + CP_2\)

⇒ Total cost = ₹4,700 + ₹5,300

⇒ Total cost = ₹10,000

The total cost of both the devices is ₹10,000.

Given:

A is thrice as good a workman as B.

A can finish the job 60 days less than B.

Formula Used:

If A can do a piece of work in n days, then B will take 3n days to complete the same work.

Work Rate of A = 1/n

Work Rate of B = 1/(3n)

Combined Work Rate = Work Rate of A + Work Rate of B

Time taken working together = 1 / (Combined Work Rate)

Calculation:

Let B take n days to finish the job.

Then, A will take (n - 60) days.

Since A is thrice as good as B:

n - 60 = n/3

⇒ 3(n - 60) = n

⇒ 3n - 180 = n

⇒ 2n = 180

⇒ n = 90

So, B takes 90 days to finish the job.

A takes 30 days to finish the job.

Combined Work Rate = 1/30 + 1/90

⇒ (3 + 1) / 90

⇒ 4/90

⇒ 2/45

Time taken working together = 1 / (2/45)

⇒ 45/2

⇒ 22.5 days

The correct answer is option 2 (22.5 days).

Given:

Simplify the following expression:

\(12 - [2.06 + {2.57 + (9.91 - \overline{5.09 - 0.76} }]\)

Formula Used:

Use the BODMAS/BIDMAS rule (Brackets, Orders, Division/Multiplication, Addition/Subtraction).

Calculation:

Let's start simplifying the innermost brackets:

\(\overline{5.09 - 0.76}\)

⇒ 5.09 - 0.76 = 4.33

Now substitute this back into the expression:

12 - [2.06 + {2.57 + (9.91 - 4.33)}]

Simplify inside the parenthesis:

9.91 - 4.33

⇒ 5.58

Substitute back:

12 - [2.06 + {2.57 + 5.58}]

Simplify inside the curly brackets:

2.57 + 5.58

⇒ 8.15

Substitute back:

12 - [2.06 + 8.15]

Simplify inside the square brackets:

2.06 + 8.15

⇒ 10.21

Finally, subtract this from 12:

12 - 10.21

⇒ 1.79

The correct answer is option 1.

Given:

Numbers: 144, 169, 196, 225, 256

Formula Used:

Average = Sum of numbers / Total count of numbers

Calculation:

Sum of numbers = 144 + 169 + 196 + 225 + 256

Sum of numbers = 990

Total count of numbers = 5

Average = 990 / 5

⇒ Average = 198

The average of 144, 169, 196, 225, and 256 is 198.

Given:

A number is first decreased by 30% and then increased by 40%. The resulting number is again increased by 5%.

Formula Used:

Net Increase/Decrease Percentage Calculation

Calculation:

Let the original number be 100.

After decreasing by 30%:

⇒ 100 × (1 - 30/100) = 100 × 0.7 = 70

After increasing by 40%:

⇒ 70 × (1 + 40/100) = 70 × 1.4 = 98

After increasing by 5%:

⇒ 98 × (1 + 5/100) = 98 × 1.05 = 102.9

Net change = 102.9 - 100 = 2.9

Net increase percentage = (2.9 / 100) × 100% = 2.9%

The net increase percentage is 2.9%.

Given:

Capacity of cylindrical tank (V) = 14,168 m³

Diameter of base (d) = 28 m

Radius of base (r) = d/2 = 28/2 = 14 m

π = 22/7

Formula Used:

Volume of cylinder (V) = π × r² × h

where,

V = Volume

r = Radius of the base

h = Depth of the tank

Calculation:

Using the formula for the volume of a cylinder:

V = π × r² × h

Substitute the given values:

14,168 = (22/7) × 14² × h

⇒ 14,168 = (22/7) × 196 × h

⇒ 14,168 = 22 × 28 × h

⇒ 14,168 = 616 × h

⇒ h = 14,168 / 616

⇒ h = 23 m

The depth of the tank is 23 m.

Given:

Speed for the first part = 75 km/h

Time for the first part = 40 minutes

Speed for the second part = 60 km/h

Time for the second part = 20 minutes

Formula Used:

Average speed = Total distance travelled / Total time taken

Calculation:

Convert time from minutes to hours:

Time for the first part = 40 minutes = 40 / 60 = 2/3 hours

Time for the second part = 20 minutes = 20 / 60 = 1/3 hours

Distance for the first part = Speed × Time

Distance for the first part = 75 km/h × 2/3 hours = 50 km

Distance for the second part = Speed × Time

Distance for the second part = 60 km/h × 1/3 hours = 20 km

Total distance travelled = 50 km + 20 km = 70 km

Total time taken = (2/3) hours + (1/3) hours = 1 hour

Average speed = Total distance / Total time

Average speed = 70 km / 1 hour

Average speed = 70 km/h

Given:

Selling Price (SP) = ₹1,190

Discount = 30%

Formula Used:

Marked Price (MP) = SP / (1 - Discount)

Calculation:

Discount = 30% = 0.30

MP = SP / (1 - Discount)

MP = 1190 / (1 - 0.30)

MP = 1190 / 0.70

MP = 1190 × (10/7)

MP = 11900 / 7

MP = 1700

The correct answer is option 3.

Shortcut Trick

Surface area of the dome (3πr²) = 363π sq.ft.

r² = 363 / 3 = 121 ⇒ r = 11 ft.

Volume of the dome = (2/3)πr³ = (2/3) × π × 11³.

Volume = (2/3) × 1331 × π = 887.33π ≈ 887.3π cubic ft.

∴ The correct answer is 887.3π cubic ft.

Alternate Method

Given:

Surface area of the hemispherical dome = 363π sq.ft.

Formula Used:

Total Surface Area of Hemisphere = 3πr²

Volume of Hemisphere = (2/3)πr³

r = 11 ft
Hemispherical Dome
Area = 3πr² = 363π
Volume = (2/3)πr³

Calculations:

⇒ 3πr² = 363π

⇒ r² = 363 / 3

⇒ r² = 121

⇒ r = √121 = 11 ft

⇒ Volume = (2/3) × π × (11)³

⇒ Volume = (2/3) × π × 1331

⇒ Volume = 2662π / 3

⇒ Volume ≈ 887.33π cubic ft

∴ The correct answer is 887.3π cubic ft.

Additional Information

Surface Area of Sphere

A sphere has a total surface area of 4πr² and a volume of (4/3)πr³.

Hemisphere Variations

Curved Surface Area (CSA) is 2πr², while Total Surface Area (TSA) including the base is 3πr².

Volume Relationship

The volume of a hemisphere is exactly half the volume of a sphere with the same radius: V = (2/3)πr³.

Given:

y is 20% more than x.

When the same number is added to x and y, x becomes (3/2) times its previous value.

Formula Used:

y = x + 0.2x

New x = (3/2) × x

New y = (x + 0.2x + k)

Calculation:

Let y = 1.2x

New x = (3/2) × x

Let the same number added be k.

x + k = (3/2) × x

⇒ k = (3/2) × x - x

⇒ k = (3x - 2x) / 2

⇒ k = x / 2

Now, New y = 1.2x + k

⇒ New y = 1.2x + x / 2

⇒ New y = 1.2x + 0.5x

⇒ New y = 1.7x

Previous y = 1.2x

New value of y is 1.7x / 1.2x times its previous value.

⇒ New value of y = (17 / 12)

The correct answer is option 4.

Shortcut Trick

Let the Marked Price (MP) be 100 units.

Cost Price (CP) after 25% discount = 100 − 25 = 75 units.

Selling Price (SP) at 20% higher than MP = 100 + 20 = 120 units.

Actual Profit = SP − CP = 120 − 75 = 45 units.

Profit % = (45 ÷ 75) × 100 = 60%.

∴ The correct answer is 60%.

Alternate Method

Given:

Discount given on Marked Price = 25%

Selling Price is 20% higher than the Marked Price.

Formula Used:

Profit = SP − CP

Profit % = (Profit ÷ CP) × 100

MP = 100

CP = 75

SP = 120

−25%

+20%

Profit = 45 → Profit % = 60%

Calculation:

⇒ Let the Marked Price (MP) of the car be 100 units.

⇒ The person buys the car at 25% discount.

⇒ Cost Price (CP) for the person = 100 − (25 ÷ 100) × 100 = 75 units.

⇒ The person sells the car at 20% higher than the marked price.

⇒ Selling Price (SP) = 100 + (20 ÷ 100) × 100 = 120 units.

⇒ Actual Gain = SP − CP = 120 − 75 = 45 units.

⇒ Actual Gain Percentage = (Actual Gain ÷ CP) × 100

⇒ Actual Gain Percentage = (45 ÷ 75) × 100

⇒ Actual Gain Percentage = (3 ÷ 5) × 100 = 60%

∴ The correct answer is 60%.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP) of the article. Selling Price (SP) = Marked Price − Discount.

Profit & Loss Percentage

Actual profit or loss is always calculated based on the Cost Price (CP), which is the price originally paid by the seller. Profit % = (Profit ÷ CP) × 100.

Markup Percentage

Markup percentage is the percentage by which the Marked Price is set above the original Cost Price. Markup % = ((MP − CP) ÷ CP) × 100.

Given:

Principal (P) = Rs. 9,500

Amount (A) = Rs. 11,704.95

Time (T) = 2 years

Formula Used:

Compound Interest formula: \(A = P \left(1 + \frac{R}{100}\right)^{T}\)

Calculation:

We need to find the rate of interest (R).

A = \(P \left(1 + \frac{R}{100}\right)^{T}\)

11,704.95 = 9,500 \(\left(1 + \frac{R}{100}\right)^{2}\)

⇒ \(\frac{11,704.95}{9,500} = \left(1 + \frac{R}{100}\right)^{2}\)

⇒ \(1.23105 = \left(1 + \frac{R}{100}\right)^{2}\)

⇒ \(\sqrt{1.23105} = 1 + \frac{R}{100}\)

⇒ 1.11 = 1 + \(\frac{R}{100}\)

⇒ 1.11 - 1 = R/100

⇒ 0.11 = R/100

⇒ R = 0.11 × 100

⇒ R = 11%

The rate of interest is 11%, which corresponds to option 1.

Given:

3x ∶ y ∶ 2z = 6 ∶ 5 ∶ 4

Equation: 5x - 3y + 4z = 48

Formula Used:

Proportion formula and solving linear equations.

Calculation:

From the given ratio, we can write:

\(\frac{3x}{6} = \frac{y}{5} = \frac{2z}{4}\)

Let \(\frac{3x}{6} = \frac{y}{5} = \frac{2z}{4} = k\)

⇒ 3x = 6k

⇒ x = 2k

⇒ y = 5k

⇒ 2z = 4k

⇒ z = 2k

Substitute these values into the equation 5x - 3y + 4z = 48

⇒ 5(2k) - 3(5k) + 4(2k) = 48

⇒ 10k - 15k + 8k = 48

⇒ 3k = 48

⇒ k = 16

Therefore, 2z = 4k = 4(16) = 64

The correct answer is option 1.

Given:

The simple interest becomes \( \frac{3^{\text{th }}}{4} \) of the sum invested in 15 years.

Sum invested (P) = Rs. 20,000

Formula Used:

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

Calculation:

Simple Interest = \( \frac{3}{4} \times P \)

⇒ SI = \( \frac{3}{4} \times 20000 \)

⇒ SI = 15000

Using the formula for Simple Interest:

SI = P × R × T / 100

⇒ 15000 = 20000 × R × 15 / 100

⇒ 15000 = 300000 × R / 100

⇒ 15000 = 3000 × R

⇒ R = 15000 / 3000

⇒ R = 5%

The rate of interest is 5%.