Quantitative Aptitude PYQs

Given:

Weight of the rice in the bag = 4800 g

Formula Used:

New Weight = Original Weight + Added Weight - Removed Weight

Calculation:

20% of 4800 g is added:

Added Weight = 4800 × 20 / 100 = 960 g

New Weight = 4800 + 960 = 5760 g

10% of 5760 g is taken out:

Removed Weight = 5760 × 10 / 100 = 576 g

New Weight = 5760 - 576 = 5184 g

15% of 5184 g is added back:

Added Weight = 5184 × 15 / 100 = 777.6 g

New Weight = 5184 + 777.6 = 5961.6 g

25% of 5961.6 g is taken out:

Removed Weight = 5961.6 × 25 / 100 = 1490.4 g

New Weight = 5961.6 - 1490.4 = 4471.2 g

The weight of the bag now is 4471.2 grams.

Given:

Cost of one packet of pencils = Rs. 15

Cost of 2 erasers = 2 × Rs. 2 = Rs. 4

Formula Used:

Discount = (Total free items value / Total cost paid) × 100

Calculation:

Total cost = Cost of erasers + Cost of pencils

⇒ Total cost = 4 + 15 = Rs. 19

Total free items value = Cost of 2 erasers = Rs. 4

Now, 

Discount%  = (4 / 19) × 100

⇒ Discount% = 400 / 19

⇒ Discount% = \(21\frac{1}{19}\)%

∴ The correct answer is option (1).

Given:

Total distance covered by Ranjan = 108 km

Average speed = 12 km/h

He covers (1/6)^th of total distance in 2 hours

He covers (1/4)^th of the total distance at a speed of 9 km/h

Formula Used:

Time = Distance / Speed

Calculation:

Distance covered in the first part = (1/6) × 108 km = 18 km

Time taken for the first part = 2 hours

Distance covered in the second part = (1/4) × 108 km = 27 km

Speed for the second part = 9 km/h

Time taken for the second part ⇒ Time = Distance / Speed

Time taken for the second part ⇒ Time = 27 / 9 = 3 hours

Let time taken to cover the remaining distance = x

Average speed = 12 km/h = Total distance/ Total time

⇒ 12 = 108/(2 + 3 + x)

⇒ 12(5 + x) = 108

⇒ x = (108 - 60)/12 = 4

The correct answer is option 3: 4 hours

Given:

Total deposit = ₹5,00,000

Rate of interest (post office) = 12% p.a.

Rate of interest (bank) = 10% p.a.

Monthly interest income = ₹4,500

Formula used:

Total annual interest = \(\frac{12}{100}x + \frac{10}{100}y\)

Where, x = amount deposited in the post office and y = amount deposited in the bank.

Total deposit = x + y = 5,00,000

Calculation:

Total annual interest = ₹4,500 × 12 = ₹54,000

\(\frac{12}{100}x + \frac{10}{100}y = 54,000\)

⇒ 12x + 10y = 5,40,000 

Total deposit: x + y = 5,00,000

⇒ y = 5,00,000 - x

Substitute y in the interest equation:

12x + 10(5,00,000 - x) = 5,40,000

⇒ 12x + 50,00,000 - 10x = 5,40,000

⇒ 2x = 4,00,000

⇒ x = 2,00,000

From x + y = 5,00,000,

⇒ y = 5,00,000 - 2,00,000 = 3,00,000

Difference between the deposits: y - x = 3,00,000 - 2,00,000 = 1,00,000

∴ The difference between the deposit in the post office and in the bank is ₹1,00,000.

Given:

Cost Price (C.P.) = 7/8 of Selling Price (S.P.)

Selling Price (S.P.) = 100 (assumed value)

New Selling Price (N.S.P.) = 120% of the original S.P.

Formula used:

Profit % = (Profit / C.P.) × 100

Profit = New Selling Price - Cost Price

Calculation:

Let Selling Price (S.P.) = 100

Cost Price (C.P.) = (7/8) × 100 = 87.5

New Selling Price (N.S.P.) = (120/100) × 100 = 120

Profit = N.S.P. - C.P. = 120 - 87.5 = 32.5

Profit % = (32.5 / 87.5) × 100 = \(37\frac{1}{7}\)%

∴ The profit percentage is \(37\frac{1}{7}\)%.

Concept Used:

A prime number is a number greater than 1 that has no positive divisors other than 1 and itself.

Calculation:

Check if each number is prime:

For 144765:

144765 is divisible by 5 (ends in 5), so it is not prime.

For 144723:

144723 is divisible by 3 (sum of digits 1+4+4+7+2+3=21, which is divisible by 3), so it is not prime.

For 144757:

Check divisibility by prime numbers up to √144757 ≈ 380.5:

Divisibility by 2: 144757 is odd, not divisible.

Divisibility by 3: Sum of digits 1+4+4+7+5+7=28, not divisible.

Divisibility by 5: Does not end in 0 or 5.

Divisibility by 7: 144757 ÷ 7 = 20679.5714, not divisible.

Divisibility by 11: Alternating sum of digits 1-4+4-7+5-7= -8, not divisible.

Continue checking up to 380 and find no divisors.

144757 is prime.

For 144393:

144393 is divisible by 3 (sum of digits 1+4+4+3+9+3=24, which is divisible by 3), so it is not prime.

The correct answer is option 3, 144757.

Shortcut Trick

Let the speed of the goods train be V km/h.

Distance (Length of train) is constant: L = (V − 5.4) × 10.6 = (V − 6.3) × 11

⇒ 10.6V − 57.24 = 11V − 69.3

⇒ 0.4V = 12.06 → V = 30.15 km/h

Relative speed 1 = 30.15 − 5.4 = 24.75 km/h = 24.75 × (5/18) m/s

Length = 24.75 × (5/18) × 10.6 = 72.875 m

∴ The correct answer is 72.875 m.

Alternate Method

Given:

Speed of Person 1 (S₁) = 5.4 km/h | Time (t₁) = 10.6 s

Speed of Person 2 (S₂) = 6.3 km/h | Time (t₂) = 11 s

Formula Used:

Distance = Relative Speed × Time

Relative Speed (Same direction) = V_train − V_person

Calculations:

⇒ Let Train Speed = V km/h and Length = L meters

⇒ L = (V − 5.4) × (5/18) × 10.6 --- (i)

⇒ L = (V − 6.3) × (5/18) × 11 --- (ii)

⇒ (V − 5.4) × 10.6 = (V − 6.3) × 11

⇒ 10.6V − 57.24 = 11V − 69.3

⇒ 0.4V = 12.06 ⇒ V = 30.15 km/h

⇒ L = (30.15 − 5.4) × (5/18) × 10.6

⇒ L = 24.75 × (5/18) × 10.6 = 1.375 × 5 × 10.6

⇒ L = 6.875 × 10.6 = 72.875 m

∴ The correct answer is 72.875 m.

Additional Information

Relative Speed

When two objects move in the same direction, relative speed is the difference (u − v). In opposite directions, it is the sum (u + v).

Unit Conversion

To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5.

Train Crossing Objects

When a train crosses a moving person or a pole, the distance covered is equal to the length of the train itself.

Given:

a : b = 3 : 4

Formula used:

To find the ratio of two expressions, substitute the values of a and b based on their ratio.

Calculation:

Let a = 3x and b = 4x (using the given ratio).

Expression 1: 5a + 7b = 5(3x) + 7(4x) = 15x + 28x = 43x

Expression 2: 7a + 5b = 7(3x) + 5(4x) = 21x + 20x = 41x

Ratio: (5a + 7b) : (7a + 5b) = 43x : 41x

⇒ 43 : 41

∴ The ratio (5a + 7b) : (7a + 5b) is 43 : 41.

Given:

A can complete the work in 16 days.

B can complete the same work in 24 days.

Formula used:

Work done in 1 day by A = 1/16

Work done in 1 day by B = 1/24

Total work = 1 (whole work).

Least Common Multiple (LCM) of the time taken is used to calculate work capacity per day.

Calculation:

LCM of 16 and 24 = 48 (smallest number divisible by both).

Total work = 48 units (assume the total work is 48 units for simplicity).

Work done by A in 1 day = 48 ÷ 16 = 3 units.

Work done by B in 1 day = 48 ÷ 24 = 2 units.

In 2 days, A and B together complete:

Work = 3 + 2 = 5 units.

Total work = 48 units.

Number of complete 2-day cycles = 48 ÷ 5 = 9 cycles (9 × 5 = 45 units).

Remaining work = 48 - 45 = 3 units.

On the 19^th day, A works (as it is A's turn):

A completes 3 units in 1 day.

∴ The least whole number of days to complete the work is 19 days.

Given:

0.2x = 0.04y

Formula Used:

Solve for x/y and substitute in (x + y)/(y - x)

Calculation:

⇒ 0.2x = 0.04y

⇒ x/y = 0.04 / 0.2

⇒ x/y = 1/5

⇒ x = y/5      [Substitute in (x + y)/(y - x)]

⇒ (y/5 + y) / (y - y/5)

⇒ (6y/5) / (4y/5)

⇒ 6y/5 × 5/4y

⇒ 6/4

⇒ 3/2

The correct answer is option 1:

Given:

5.5x = 0.05y

Formula used:

\(\dfrac{y - 3x}{y + 3x}\)

Calculation:

5.5x = 0.05y

⇒ y = 5.5x/0.05

⇒ y = 110x

\(\dfrac{y - 3x}{y + 3x} = \dfrac{110x - 3x}{110x + 3x}\)

⇒ \(\dfrac{107x}{113x}\)

⇒ \(\dfrac{107}{113}\)

∴ The correct answer is option (4).

Given:

The diameter of a right circular cone is 14 cm

The slant height of the cone is 11 cm

Formula used:

Volume of cone, \(V = \dfrac{1}{3}\pi r^2 h\)

We need to find the height (h) using Pythagorean theorem: \(l^2 = r^2 + h^2\)

Calculations:

Radius, \(r = \dfrac{diameter}{2} = \dfrac{14}{2} = 7 \text{ cm}\)

Slant height, l = 11 cm

⇒ \(l^2 = r^2 + h^2\)

⇒ \(11^2 = 7^2 + h^2\)

⇒ \(121 = 49 + h^2\)

⇒ \(h^2 = 72 \)

⇒ \(h = \sqrt{72} = 6\sqrt{2} \text{ cm}\)

Now, calculate the volume:

⇒ \(V = \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 6\sqrt{2}\)

⇒ \(V = \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 6\sqrt{2}\)

⇒ \( 308 \sqrt{2} \mathrm{~cm}^3 \)

∴ The correct answer is option 2.

Given:

Cost of 16 articles at ₹85 each

Cost of 15 articles at ₹65 each

Cost of 12 articles at ₹75 each

Formula used:

Average Price = Total Cost / Total Number of Articles

Calculation:

Total Cost = (16 × 85) + (15 × 65) + (12 × 75)

Total Number of Articles = 16 + 15 + 12

⇒ Total Cost = 1360 + 975 + 900 = 3235

⇒ Total Number of Articles = 43

⇒ Average Price = 3235 / 43

⇒ Average Price = 75.23

∴ The correct answer is option 1.

Given:

Principal (P) = ₹3,660

Rate of Interest (R) = 10% p.a.

Time (T) = 1.5 years

Formula used:

Simple Interest (S.I.) = \(\dfrac{P \times R \times T}{100}\)

Compound Interest (C.I.) = \(P(1+\dfrac{r}{100})^n - P\)

Calculation:

S.I. = \(\dfrac{3660 \times 10 \times 1.5}{100}\)

⇒ S.I. = ₹549

For C.I. compounded half-yearly:

Rate per half year = \( \dfrac{10}{2} = 5%\), Number of half-years (n) = \(1.5 \times 2 = 3\)

C.I. = \(3660 \left(1+\dfrac{5}{100}\right)^3 - 3660\)

⇒ C.I. = \(3660 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} - 3660\)

⇒ C.I. = ₹576.9 \(\approx\) 577

Required Difference = \(577 - 549\) = ₹28

∴ The correct answer is ₹28.

Given:

Rama's marks in two subjects, mathematics and physics, are in the ratio 8 ∶ 13.

He got 25 marks more in physics than in mathematics.

Formula used:

Let marks in mathematics be 8x.

Let marks in physics be 13x.

Given: 13x - 8x = 25

Calculations:

13x - 8x = 25

⇒ 5x = 25

⇒ x = 5

Marks in mathematics = 8x = 8 × 5 = 40

∴ The correct answer is option (2).

Given:

The product of two numbers is 2028 and their HCF is 13.

Formula used:

LCM × HCF = Product of the numbers

Calculation:

Let the LCM be x.

LCM × HCF = Product of the numbers

⇒ x × 13 = 2028

⇒ x = \(\dfrac{2028}{13}\)

⇒ x = 156

∴ The correct answer is option (3).

Given:

A garrison of 1200 men has provisions for 15 days.

Formula used:

Provisions last for = \(\dfrac{\text{Initial Provisions} \times \text{Initial Men}}{\text{Increased Men}}\)

Calculations:

Initial Provisions = 1200 men × 15 days

Increased Men = 1200 men + 600 men

⇒ Initial Provisions = 18000 men-days

⇒ Increased Men = 1800 men

Provisions last for = \(\dfrac{18000}{1800}\)

⇒ Provisions last for = 10 days

∴ The correct answer is option (1).

Given:

Let the original dimensions of the cuboid be:

Length = L, Breadth = B, Height = H

Percentage increase in Length = 24%, Percentage increase in Breadth = 45%

Percentage decrease in Height = 55%

Formula used:

Volume of cuboid: \( V = L \times B \times H \)

Percentage change in volume:

\( \text{Total Change} = (1 + \dfrac{x}{100}) (1 + \dfrac{y}{100}) (1 + \dfrac{z}{100}) \)

where: x = 24, y = 45 and z = 55

Calculation:

New volume factor:

\( (1 + \dfrac{24}{100}) (1 + \dfrac{45}{100}) (1 - \dfrac{55}{100}) \)

⇒ \( \left( \dfrac{124}{100} \right) \times \left( \dfrac{145}{100} \right) \times \left( \dfrac{45}{100} \right) \)

⇒ \( \dfrac{124 \times 145 \times 45}{100^3} \)

⇒ \( \dfrac{809100}{1000000} \)

⇒ 0.8091

Percentage change in volume:

⇒ \( (1 - 0.8091 ) \times 100 \)

⇒ \(- 19.1% \)

∴ The total percentage change in volume is 19.1%.

Given:

Kumar invested a total of ₹16,900 partly in two schemes, A and B.

Rate of simple interest in Scheme A = 14% p.a.

Rate of simple interest in Scheme B = 11% p.a.

Total interest earned in 2 years = ₹4,498

Formula used:

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

Calculations:

Let the amount invested in Scheme A be ₹x.

Therefore, the amount invested in Scheme B = ₹(16900 - x).

Interest from Scheme A in 2 years = \(\frac{x × 14 × 2}{100}\) = \(\frac{28x}{100}\)

Interest from Scheme B in 2 years = \(\frac{(16900 - x) × 11 × 2}{100}\) = \(\frac{22(16900 - x)}{100}\)

Total interest = ₹4498

⇒ \(\frac{28x}{100} + \frac{22(16900 - x)}{100} = 4498\)

⇒ \(28x + 22(16900 - x) = 449800\)

⇒ \(6x = 78000\)

⇒ \(x = \frac{78000}{6}\)

⇒ x = ₹13000

Amount invested in Scheme B = ₹(16900 - 13000) = ₹3900

∴ The correct answer is option 2.

Given:

Let the actual cost price be ₹x.

Profit percentage = 23.75%

New cost price = ₹(x + 1000)

New profit percentage = 10%

Formula used:

Selling Price (S.P.) = \( C.P. \times \left(1 + \dfrac{\text{Profit %}}{100}\right) \)

Calculation:

Selling Price with actual cost price:

\( S.P. = x \times \left(1 + \dfrac{23.75}{100}\right) \)

⇒ \( S.P. = x \times \dfrac{123.75}{100} \)

New Selling Price with increased cost price:

\( S.P. = (x + 1000) \times \left(1 + \dfrac{10}{100}\right) \)

⇒ \( S.P. = (x + 1000) \times \dfrac{110}{100} \)

Since the selling price remains the same:

\( x \times \dfrac{123.75}{100} = (x + 1000) \times \dfrac{110}{100} \)

⇒ \( 123.75x = 110x + 110000 \)

⇒ \( 13.75x = 11000 \)

⇒ \( x = \dfrac{11000}{13.75} \)

⇒ \( x = 800 \)

∴ The actual cost price of the item is ₹800.