Quantitative Aptitude PYQs

Given:

The average weight of A, B, and C = 65 kg

The average weight of A and B = 62 kg

The average weight of B and C = 57 kg

Formula Used:

Average weight = (Sum of individual weights) / (Number of individuals)

Calculation:

Let the weights of A, B, and C be a, b, and c respectively.

According to the given data:

(a + b + c) / 3 = 65

⇒ a + b + c = 195

(a + b) / 2 = 62

⇒ a + b = 124

(b + c) / 2 = 57

⇒ b + c = 114

Now, we have two equations:

a + b + c = 195

a + b = 124

b + c = 114

Subtracting the second equation from the first:

⇒ (a + b + c) - (a + b) = 195 - 124

⇒ c = 71

Subtracting the third equation from the first:

⇒ (a + b + c) - (b + c) = 195 - 114

⇒ a = 81

Now, using the value of a in the second equation:

a + b = 124

⇒ 81 + b = 124

⇒ b = 43

The weight of B is 43 kg.

Given:

The train increases its speed by 16% and reaches its destination 30 minutes (0.5 hours) earlier. We need to find the normal time taken by the train for the journey.

Formula Used:

Let the normal time be T hours and the normal speed be S. Then:

New speed = 1.16S

Distance (D) = Speed × Time

At normal speed: D = S × T

At increased speed: D = 1.16S × (T – 0.5)

Equating the distances:

S × T = 1.16S × (T – 0.5)

Calculation:

From S × T = 1.16S × (T – 0.5):

⇒ T = 1.16(T – 0.5)

⇒ T = 1.16T – 0.58

⇒ 0.58 = 1.16T – T

⇒ 0.58 = 0.16T

⇒ T = 0.58 / 0.16

⇒ T = 58/16 hours = 29/8 = \(3\frac{5}{8}\) hours

∴ The normal time taken by the train is \(3\frac{5}{8}\) hours (or 3 hours and 37.5 minutes).

Given:

Discount given = 25%

Gain = 20%

Formula Used:

MP / CP = (100 + Gain%) / (100 - discount%)

Calculation:

MP / CP = (100 + Gain%) / (100 - discount%)

⇒ MP / CP = (100 + 20) / (100 - 25)

⇒ MP / CP = 120 / 75 = 8 / 5

So, Required percentage = (8 - 5) / 5 × 100

⇒ 3/5 × 100 = 3 × 20 = 60%

Hence, the correct answer is option 2.

Given:

Radius of the sphere = 4 cm

Radius of the cone = 2 cm

Formula Used:

Surface area of a sphere = 4πr²

Curved surface area of a cone = πrl

Volume of a cone = (1/3)πr²h

Calculation:

Surface area of the sphere = 4π(4)²

⇒ Surface area of the sphere = 4π(16)

⇒ Surface area of the sphere = 64π cm²

Let the curved surface area of the cone be A.

Given, 64π = 8A

⇒ A = 64π / 8

⇒ A = 8π cm²

Curved surface area of the cone = πrl

8π = π × 2 × l

⇒ 8 = 2l

⇒ l = 4 cm

Using Pythagoras theorem in the cone:

l² = r² + h²

4² = 2² + h²

⇒ 16 = 4 + h²

⇒ h² = 12

⇒ h = √12

⇒ h = 2√3 cm

Volume of the cone = (1/3)πr²h

⇒ Volume of the cone = (1/3)π(2)²(2√3)

⇒ Volume of the cone = (1/3)π(4)(2√3)

⇒ Volume of the cone = (8π√3)/3

⇒ Volume of the cone = (8π)/(√3) cm³

The correct answer is option 3.

Shortcut Trick

Range 51−60: 53, 59 (2 primes)

Range 61−70: 61, 67 (2 primes)

Range 71−80: 71, 73, 79 (3 primes)

Range 81−90: 83, 89 (2 primes)

Range 91−100: 97 (1 prime)

Total count = 2 + 2 + 3 + 2 + 1 = 10.

∴ The correct answer is 10.

Alternate Method

Given: Range of numbers between 50 and 100.

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

⇒ Step 1: List potential odd numbers: 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99.

⇒ Step 2: Remove multiples of 3: 51, 57, 63, 69, 75, 81, 87, 93, 99.

⇒ Step 3: Remove multiples of 5: 55, 65, 85, 95.

⇒ Step 4: Remove multiples of 7 and 11: 77 (7×11), 91 (7×13).

⇒ Step 5: Remaining prime numbers: 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

⇒ Step 6: Count the remaining numbers = 10.

∴ The correct answer is 10.

Additional Information

Prime Numbers from 1 to 50

There are 15 prime numbers between 1 and 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

Total Primes from 1 to 100

The total number of prime numbers between 1 and 100 is 15 (from 1−50) + 10 (from 51−100) = 25.

Twin Primes

Pairs of prime numbers with a difference of 2 are called twin primes. Examples between 50 and 100 include (59, 61) and (71, 73).

Sieve of Eratosthenes

A systematic method to find all primes up to a limit by repeatedly marking the multiples of each prime starting from 2.

Given:

1^st prime number = 2

5^th prime number = 11

3^rd prime number = 5

9^th prime number = 23

Formula Used:

Sum of the product of the 1^st & 5^th prime numbers and the 3^rd & 9^th prime numbers

Calculation:

Product of 1^st and 5^th prime numbers = 2 × 11

⇒ 2 × 11 = 22

Product of 3^rd and 9^th prime numbers = 5 × 23

⇒ 5 × 23 = 115

Sum of the products = 22 + 115

⇒ 22 + 115 = 137

The correct answer is option 4: 137

Shortcut Trick

For 1 year, Yearly CI = 6%.

For Half-yearly CI, rate = 6% ÷ 2 = 3% and time = 2 periods.

Effective rate for 2 periods = 3 + 3 + (3 × 3) ÷ 100 = 6.09%.

Difference in rate = 6.09% − 6% = 0.09%.

Difference in amount = 0.09% of ₹8,000 = 0.0009 × 8000 = ₹7.2.

∴ The correct answer is Rs. 7.2.

Alternate Method

Given: Principal (P) = ₹8,000, Rate (R) = 6% p.a., Time (n) = 1 year.

Formula Used: CI = P[(1 + r/100)ⁿ − 1]

Case 1: Compounded Yearly

⇒ CI₁ = 8000 × (6/100) × 1 = ₹480.

Case 2: Compounded Half-Yearly

⇒ New Rate = 6% ÷ 2 = 3% per half-year.

⇒ Time periods = 1 × 2 = 2.

⇒ Amount = 8000 × (1 + 3/100)²

⇒ Amount = 8000 × (1.03) × (1.03) = 8000 × 1.0609 = ₹8487.2.

⇒ CI₂ = 8487.2 − 8000 = ₹487.2.

Difference:

⇒ Difference = CI₂ − CI₁ = 487.2 − 480 = ₹7.2.

∴ The correct answer is Rs. 7.2.

Additional Information

Half-Yearly Compounding Rule

When interest is compounded half-yearly, the annual rate is halved (R/2) and the time period is doubled (2n).

CI vs SI for 1 Year

For the first year, if interest is compounded annually, Simple Interest and Compound Interest are equal.

Effective Rate Formula

For two successive growth rates of a%, the effective rate is given by [a + a + (a × a)/100]%.

Given:

The radius of the pond is 14 m.

The diameter of the fountain is 14 m, so its radius is 7 m.

Formula Used:

Area of a circle = π × r²

Area of the pond excluding the fountain = Area of the pond - Area of the fountain

Calculation:

Step 1: Calculate the area of the pond:

Area of the pond = π × (14)²

⇒ Area of the pond = (22/7) × 196 = 616 m²

Step 2: Calculate the area of the fountain:

Area of the fountain = π × (7)²

⇒ Area of the fountain = (22/7) × 49 = 154 m²

Step 3: Subtract the area of the fountain from the area of the pond:

Excluding area = 616 - 154

⇒ Excluding area = 462 m²

∴ The area of the pond excluding the fountain is 462 m².

Given:

Current population = 1,00,000

Increase by 12% next year

Decrease by 10% the year after

Formula Used:

Population after increase = Initial population × (1 + Rate of increase)

Population after decrease = Population after increase × (1 - Rate of decrease)

Calculation:

Population after first year = 1,00,000 × (1 + 12/100)

⇒ Population after first year = 1,00,000 × 1.12

⇒ Population after first year = 1,12,000

Population after second year = 1,12,000 × (1 - 10/100)

⇒ Population after second year = 1,12,000 × 0.90

⇒ Population after second year = 1,00,800

The new population would be 1,00,800.

Given:

Initial number of men = 400

Initial duration of provisions = 23 weeks

Additional men joined = 60

Total men = 400 + 60 = 460

Formula Used:

Men × Weeks = Constant

Calculation:

Initial Men × Initial Weeks = Total Men × New Weeks

400 × 23 = 460 × New Weeks

⇒ New Weeks = (400 × 23) / 460

⇒ New Weeks = 9200 / 460

⇒ New Weeks = 20 weeks

The provisions will last for 20 weeks.

Given:

The ratio of two sums of money = 5:9

The first sum = 55

Formula Used:

If two sums of money are in the ratio a:b, and the first sum is A, then the second sum B is given by:

B = (b/a) × A

Calculation:

Given ratio a:b = 5:9 and A = 55

Second sum B = (9/5) × 55

⇒ B = (9 × 55) / 5

⇒ B = 495 / 5

⇒ B = 99

The second sum is 99.

Given:

A vendor bought buttons at 22 for a rupee.

Formula Used:

Selling Price per button = Cost Price per button × (1 + Gain%)

Gain% = 10%

Calculation:

Cost Price per button = 1/22 rupees

Selling Price per button = (1/22) × (1 + 10/100)

Selling Price per button = (1/22) × (1 + 0.1)

Selling Price per button = (1/22) × 1.1

Selling Price per button = 1.1/22 rupees

To find how many buttons he must sell for 1 rupee:

Number of buttons = 1 / (1.1/22)

Number of buttons = 22 / 1.1

Number of buttons = 20

The correct answer is option 1.

Given:

Numbers: 44, 108, 135, 162, 198

Formula Used:

Least Common Multiple (LCM) is the smallest number that is a multiple of each of the given numbers.

Calculation:

Prime factorization of each number:

44 = 2² × 11

108 = 2² × 3³

135 = 3³ × 5

162 = 2 × 3⁴

198 = 2 × 3² × 11

LCM is calculated by taking the highest power of each prime number that appears in the factorization of any of the numbers:

LCM = 2² × 3⁴ × 5 × 11

Calculating the LCM step-by-step:

2² = 4

3⁴ = 81

5 = 5

11 = 11

⇒ LCM = 4 × 81 × 5 × 11

⇒ LCM = 4 × 81 = 324

⇒ LCM = 324 × 5 = 1620

⇒ LCM = 1620 × 11 = 17820

The least common multiple of 44, 108, 135, 162, and 198 is 17820.

Given:

A sum of money doubles itself in 8 years.

Formula Used:

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

Calculation:

Let the principal amount be P.

Since the sum doubles itself, the amount becomes 2P.

Simple Interest (SI) = 2P - P = P

Time (T) = 8 years

Rate (R) = ?

Using the formula for Simple Interest:

SI = P × R × T / 100

⇒ P = P × R × 8 / 100

⇒ 1 = R × 8 / 100

⇒ R = 100 / 8

⇒ R = 12.5

The rate per annum on simple interest is 12.5%.

Given:

Net profit after paying taxes = Rs. 2,340

Net profit = 18% of expenditure (other than taxes)

Tax rate on profit = 22%

Formula Used:

Net profit = Profit after tax

Profit after tax = Profit before tax × (1 - Tax rate)

Income = Expenditure + Profit before tax

Calculation:

Step 1: Let expenditure = E and profit before tax = P.

Net profit = P × (1 - 0.22) = Rs. 2,340

⇒ P × 0.78 = 2,340

⇒ P = 2,340 / 0.78

⇒ P = Rs. 3,000

Step 2: Expenditure is 18% of the profit after tax:

0.18E = Rs. 2,340

⇒ E = 2,340 / 0.18

⇒ E = Rs. 13,000

Step 3: Income = Expenditure + Profit before tax:

Income = Rs. 13,000 + Rs. 3,000

⇒ Income = Rs. 16,000

∴ The income of the company is Rs. 16,000.

Given:

Side of square field = 25 m 

Length of rope = 14 m

Formula Used:

Area of sector = (θ/360) × πr²

Calculation:

Let ABCD be a square field

The area cow can grass will be the sector of the square field

The length of rope = radius of the sector = 14 m

 ∠QBP = 90°  (In squares all angle are 90°)

The area of a sector in which cow can graze =  (θ/360) × πr2  

⇒ (θ/360) × πr2 = (90°/360°) × π(14)² ⇒ 1/4 × 22/7 × 14 × 14

⇒ Area of sector = 1/4 × 22/7 × 14 × 14 = 154 m² 

∴ The area of that part of the field in which the cow can graze is 154 m2

Given:

The average monthly salary of five friends is Rs. 62,000.

The new average of their salaries after the promotion is Rs. 64, 250

Formula used:

Average = Sum of observation/Total number of observation

Calculation:

The total salary of five friends is

⇒ 62000 × 5 = Rs. 310000

The total salary after the promotion is

⇒ 64250 × 5 = Rs. 321250

So, the increment the person gets is

⇒ 321250 - 310000 = Rs.11250

∴ The increment he got is Rs. 11250.

Shortcut TrickTotal change value = change in average × number of observations.

So, Increment in salary is

⇒ (64250 - 62000) × 5

⇒ 2250 × 5 = Rs. 11250

∴ The increment he got is Rs. 11250.

Given:

A sum invested at 10% per annum amounts to Rs. 20,947.50 in one year, interest is compounded half-yearly

Formula used:

Simple Interest = (P × r × t)/100

Compound interest

⇒ A = P × {1 + (r/100)}^t

P = Principal, A = Amount

r = Rate of Interest, t = Time

Calculation:

The rate of interest is 10% per annum

But the interest is compounded half-yearly

So, rate of interest will become 10/2 = 5%

⇒ 20947.50 = P × {1 + (5/100)}²

⇒ 20947.50 = P × (21/20)²

⇒ P = 20947.50 × (20/21)²

⇒ P = 19000

According to the question, the rate of interest is tripled.

⇒ r = 10 × 3 = 30%

The simple interest for 22/5 years is

⇒ {19000 × 30 × (22/5)}/100

⇒ (19000 × 6 × 22)/100

⇒ 2508000/100 = Rs. 25080

∴ The simple interest for \(4\frac{2}{5}\) is Rs. 25080.

Given:

A can do a work in 40 days and B can do the same work in 50 days

They worked together for 5 days and then B left the work

Concept used:

Efficiency ∝ 1/Time

Calculation:

Let A complete the remaining work in x days.

The efficiency of A is 1/40

And the efficiency of B is 1/50

So, the efficiency of A and B together is

⇒ (1/40) + (1/50)

⇒ (5 + 4)/200 = 9/200

A and B worked for 5 days, rest completed by A

⇒ (9/200) × 5 + (1/40) × x = 1

⇒ 9/40 + x/40 = 1

⇒ 9 + x = 40

⇒ x = 31 days

∴ A will complete the rest of the work in 31 days.

Alternate Method

The total efficiency of A and B is

⇒ 5 + 4 = 9

They worked for 5 days

⇒ 9 × 5 = 45

Rest of the work completed by A

⇒ (200 - 45)/5 = 31 days.

∴ A will complete the rest of the work in 31 days.

Given:

A person spends 22% of the monthly income on house rent,

40% of the income on house expenses and groceries,

50% of the remaining on children's education and other heads 

And his monthly saving is Rs, 3,800

Formula used:

Percentage = (Part value/Whole value) × 100

Calculation:

Let the monthly income of the person be 100x

The amount he spends on rent

⇒ (22/100) × 100x = 22x

The amount he spends on house expenses and groceries

⇒ (40/100) × 100x = 40x

Remaining amount is

⇒ 100x - 22x - 40x = 38x

50% of the remaining on children's education and other heads 

⇒ (50/100) × 38x = 19x

The person saves 19x of his income

⇒ 19x = 3800

⇒ x = Rs. 200

The amount he spends on children's education and other heads 

⇒ 19 × 200 = Rs. 3800

∴ The person spends Rs. 3800 on children's education and other heads.

Shortcut TrickLet total income be 100%

He spend 22% + 40% = 62%

⇒ Education = (50/100) × 38 = 19%

19% is savings of the person

⇒ savings = education = 19% = 3800

∴ The person spends Rs. 3800 on children's education and other heads.