Quantitative Aptitude PYQs

Given:

A works twice as fast as B.

Both together can finish the work in 12 days.

Formula used:

Work done together = Total work / Time

Let B's 1 day work = x, then A's 1 day work = 2x.

Calculation:

Let the Total work = 1 unit

A's 1 day work + B's 1 day work = 1/12

⇒ 2x + x = 1/12

⇒ 3x = 1/12

⇒ x = 1/36

B's time to complete the work alone = Total work / B's 1 day work

⇒ Time = 1/(1/36) = 36 days

∴ The correct answer is option (3).

Given:

Average of 5 consecutive numbers = 24

Formula used:

If the average of n consecutive numbers is given, then the middle number is equal to the average.

For 5 consecutive numbers: Smallest number = Middle number - 2

Calculations:

Middle number = 24

⇒ Smallest number = 24 - 2

⇒ Smallest number = 22

∴ The correct answer is option (2).

Given:

Total persons in 2022 = 8,375

Reduction in persons from 2021 to 2022 = 33%

Equal number of students in Hostel P and Hostel Q in 2021

Formula used:

Total persons in 2021 = Total persons in 2022 / (1 - Reduction%)

Number of students in Hostel P in 2021 = Total persons in 2021 / 2

Calculation:

Total persons in 2021 = 8,375 / (1 - 33/100)

⇒ Total persons in 2021 = 8,375 / 0.67

⇒ Total persons in 2021 = 12,500

Number of students in Hostel P in 2021 = 12,500 / 2 = 6,250

∴ The correct answer is option (3).

Given:

Speed at which student reaches 6 min late = 20 km/h

Speed at which student reaches 6 min early = 30 km/h

Time difference = 6 min late + 6 min early = 12 min = 12/60 hours = 1/5 hours

Formula used:

Distance = Speed × Time

Calculation:

Let the distance between his college and his starting point = x km

According to question,

⇒ x/20 – x/30= 1/5

⇒ (3x – 2x)/60 = 1/5

⇒ x = [(1 × 60) / 5] km

⇒ 12 km

∴ The correct answer is option (3).

Given:

Side of triangle (base) = 17 cm

Height of triangle = 10 cm

Formula used:

Area of triangle = (1/2) × base × height

Calculation:

Area = (1/2) × 17 × 10

⇒ Area = 5 × 17

⇒ Area = 85 cm²

∴ The correct answer is option (3).

Given:

Marks in the first semester: 45, 78, 36, 65, 56

Marks in the second semester: 48, 72, 55, 46, 79

Formula Used:

Average Marks = Total Marks / Number of Subjects

Calculation:

First Semester:

Total Marks = 45 + 78 + 36 + 65 + 56

Total Marks = 280

Second Semester:

Total Marks = 48 + 72 + 55 + 46 + 79

Total Marks = 300

Total Marks in the First Year = 280 + 300

Total Marks in the First Year = 580

Number of Subjects = 5 + 5

Number of Subjects = 10

Average Marks = Total Marks / Number of Subjects

Average Marks = 580 / 10

⇒ Average Marks = 58

The average marks in the first year are 58.

Given:

A garrison of 1500 men has provisions for 11 days.

The garrison is increased by 1800 men.

Formula Used:

Provisions Days = (Initial Provisions Days × Initial Men) / Total Men

Calculation:

Initial Provisions Days = 11

Initial Men = 1500

Total Men = 1500 + 1800 = 3300

Provisions Days = (11 × 1500) / 3300

⇒ Provisions Days = 16500 / 3300

⇒ Provisions Days = 5

The provisions will last for 5 days if the garrison is increased by 1800 men.

Given:

Speed of Shanti's car = 68 km/h.

Initial distance between Shanti's car and the bus = 40 m.

After 10 seconds, Shanti's car is 60 m ahead of the bus.

Formula Used:

Relative Speed = (Distance covered by Shanti's car relative to the bus) / Time

Speed of the bus = Speed of Shanti's car - Relative Speed

Calculation:

Distance covered by Shanti's car relative to the bus = 40 m + 60 m = 100 m.

Time = 10 seconds = 10/3600 hours = 1/360 hours.

Relative Speed (in km/h) = Distance / Time

Relative Speed = 100 m / (1/360 hours)

Relative Speed = 100 m × 360 hours

Relative Speed = 36000 m/h = 36 km/h.

Speed of the bus = Speed of Shanti's car - Relative Speed

Speed of the bus = 68 km/h - 36 km/h

Speed of the bus = 32 km/h.

Shortcut Trick

 Annual Instalment = \(\frac {Due Debt \times 100}{100 \times t + r \times t \frac{(t - 1)}{2}}\)

Annual instalment = \(\frac {9048 \times 100}{100 \times 5 + 8 \times 5 \frac{(5 - 1)}{2}}\) = 1,560

∴ Annual instalment is Rs. 1,560.

Shortcut Trick - 2100P + 108P + 116P + 124P + 132P = 9048 × 100

⇒ 580P = 904800

P = 1560

∴ Annual instalment is Rs. 1,560. 

Alternate MethodGiven:

Amount = Rs. 9,048

Rate per annum = 8%

Time - 5 year

Concept used:

Increase instalment = P × (100 + r)/100 + P × (100 + 2r)/100 + ..................+ P × (100 + tr)/100

Calculation:

Let yearly instalment = a

then 1^st instalment = 100a unit

2^nd instalment = 100a × (100 + 8)/100

⇒ 108a

3^rd instalment = 100a × [100 + (2 × 8)]/100

⇒ 116a

4^th instalment = 100a × [100 + (3 × 8)]/100

⇒ 124a

5^th instalment = 100a × [100 + (4 × 8)]/100

⇒ 132a

Hence total amount = Sum of all instalment/100

⇒ 9048 = (100a + 108a + 116a + 124a + 132a)/100

⇒ 580a = 9048 × 100

⇒ a =1560

∴ Annual instalment is Rs. 1,560.

Given:

The average age of three persons = 54 years

The proportion of their ages = 2 : 3 : 4

Formula Used:

Average Age = (Sum of Ages) / (Number of Persons)

Calculation:

Let the ages of the three persons be 2x, 3x, and 4x respectively.

The sum of their ages = 2x + 3x + 4x = 9x

Average Age = (Sum of Ages) / 3

⇒ 54 = 9x / 3

⇒ 54 = 3x

⇒ x = 54 / 3

⇒ x = 18

The age of the youngest person = 2x = 2 × 18 = 36 years

The age of the youngest person is 36 years.

Given:

Principal (P) = ₹12,500

Rate of interest (R) = 15%

Time (T) = 1.5 years

Interest is compounded half-yearly.

Formula Used:

Amount (A) = P (1 + (R)/(100n))^nT

where,

P = Principal amount

R = Rate of interest

n = number of times interest is compounded per year

T = Time period in years

Calculation:

Given, n = 2 (since interest is compounded half-yearly)

R = 15%

T = 1.5 years

Substitute the values into the formula:

⇒ A = 12500 (1 + (15)/(100 × 2))^{2 × 1.5}

⇒ A = 12500 (1 + (15)/(200))³

⇒ A = 12500 (1 + 0.075)³

⇒ A = 12500 (1.075)³

⇒ A = 12500 (1.242)

⇒ A = 15528.71

The amount Asha earns in 1.5 years is ₹15528.71.

Given:

Numbers: 126, 198, 162

Formula Used:

To find the greatest number that e×actly divides the given numbers, we use the concept of the Greatest Common Divisor (GCD).

Calculation:

Prime factorization of 126: 126 = 2 × 3² × 7

Prime factorization of 198: 198 = 2 × 3² × 11

Prime factorization of 162: 162 = 2 × 3⁴

GCD is the product of the smallest power of all common prime factors.

Common prime factors: 2 and 3

Smallest power of 2: 2¹

Smallest power of 3: 3²

Therefore, GCD = 2¹ × 3²

⇒ GCD = 2 × 9

⇒ GCD = 18

The greatest number that e×actly divides 126, 198, and 162 is 18.

Given:

Cost Price (CP) = ₹225

Selling Price (SP) = ₹300

Formula Used:

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

Calculation:

Gain = SP - CP

⇒ Gain = ₹300 - ₹225

⇒ Gain = ₹75

Gain Percentage = (75)/(225) × 100

⇒ Gain Percentage = (1)/(3) × 100

⇒ Gain Percentage = 33.33%

The gain percentage is 33.33%.

Given:

Area of smaller triangle = 48 cm²

Area of bigger triangle = 156 cm²

Height of smaller triangle = 6 cm

Formula Used:

\(\frac{\text{Area of smaller triangle}}{\text{Area of bigger triangle}} = (\frac{\text{Height of smaller triangle}}{\text{Height of bigger triangle}})^2\)

Calculation:

Let the height of the bigger triangle be h cm.

\(\frac{48}{156} = (\frac{6}{h})^2\)

⇒ \(\frac{48}{156} = \frac{36}{h^2}\)

⇒ \(\frac{1}{3.25} = \frac{36}{h^2}\)

⇒ 1 × h² = 36 × 3.25

⇒ h² = 117

⇒ h = √117

Therefore, the height of the bigger triangle is √117 cm.

Given:

Q is 12% more than 156.

P is 15% less than Q.

Formula Used:

Q = 156 + 12% of 156

P = Q - 15% of Q

Calculation:

Q = 156 + (12/100) × 156

Q = 156 + 0.12 × 156

Q = 156 + 18.72

Q = 174.72

P = 174.72 - (15/100) × 174.72

P = 174.72 - 0.15 × 174.72

P = 174.72 - 26.208

⇒ P = 148.512

The value of P is 148.51 (approx).

Given:

The sum of three distinct prime numbers is 44.

Formula Used:

Sum of prime numbers = Prime₁ + Prime₂ + Prime₃

Calculation:

Let the prime numbers be p₁, p₂, and p₃.

Given that p₁ + p₂ + p₃ = 44

To find the largest possible prime number, we need to maximize one of the prime numbers while ensuring the sum remains 44 and all numbers are prime.

Consider the largest prime number, let p₃ be the largest prime:

If p₃ = 37, then:

p₁ + p₂ + 37 = 44

⇒ p₁ + p₂ = 44 - 37

⇒ p₁ + p₂ = 7

The possible prime pairs (p₁, p₂) that sum to 7 are (2, 5) and (5, 2).

So, the prime numbers are 2, 5, and 37.

Verify if there is a larger prime number:

If p₃ = 41, then:

p₁ + p₂ + 41 = 44

⇒ p₁ + p₂ = 44 - 41

⇒ p₁ + p₂ = 3

The possible prime pairs (p₁, p₂) that sum to 3 are (2, 1) and (1, 2), but 1 is not a prime number.

If p₃ = 43, then:

p₁ + p₂ + 43 = 44

⇒ p₁ + p₂ = 44 - 43

⇒ p₁ + p₂ = 1

The possible prime pairs (p₁, p₂) that sum to 1 are (0, 1) and (1, 0), but neither 0 nor 1 are prime numbers.

The largest possible prime number is 37.

Given:

A person got a new chair at 35% discount.

Had he got no discount, he would have had to pay ₹224 more.

Formula Used:

Discount = Marked Price × Discount Percentage

Discount Amount = Marked Price - Selling Price

Calculation:

Let the marked price of the chair be ₹x.

Discount percentage = 35%

Discount amount = ₹224

Discount = Marked Price × 35%

⇒ 224 = x × 0.35

⇒ × = 224 / 0.35

⇒ × = 640

The marked price (in ₹) of the chair is ₹640.

Given:

The price of sugar goes up by 25%.

Formula Used:

Percentage reduction in consumption = \(\frac{\text{Increase in price}}{(100 + \text{Increase in price})} × 100\)

Calculation:

Increase in price = 25%

Percentage reduction in consumption = \(\frac{25}{100 + 25} × 100\)

⇒ Percentage reduction in consumption = \(\frac{25}{125} × 100\)

⇒ Percentage reduction in consumption = 0.2 × 100

⇒ Percentage reduction in consumption = 20%

The family must reduce their consumption of sugar by 20% to keep the expenditure the same.

Shortcut Trick

For a set of numbers where the smallest number is a factor of all other numbers, the HCF is simply the smallest number itself.

Here, all numbers (0.006, 0.06, 0.6, and 6) are multiples of 0.0006:

0.006 = 0.0006 × 10 | 0.06 = 0.0006 × 100 | 0.6 = 0.0006 × 1000 | 6 = 0.0006 × 10000

Since 0.0006 divides every other number in the set, it is the Highest Common Factor.

∴ The correct answer is 0.0006.

Alternate Method

Given numbers: 0.0006, 0.006, 0.06, 0.6, 6

Step 1: Convert all decimals to like decimals by making the number of decimal places equal (4 decimal places).

⇒ 0.0006, 0.0060, 0.0600, 0.6000, 6.0000

Step 2: Consider these as integers by removing the decimal points.

⇒ 6, 60, 600, 6000, 60000

Step 3: Find the HCF of these integers.

⇒ HCF (6, 60, 600, 6000, 60000) = 6

Step 4: Place the decimal point back at the original position (4 places from the right).

⇒ HCF = 0.0006

∴ The correct answer is 0.0006.

Additional Information

HCF of Fractions

HCF of fractions is calculated as (HCF of Numerators) ÷ (LCM of Denominators).

LCM of Decimals

To find the LCM, convert decimals to like decimals, find the LCM of the resulting integers, and then place the decimal point back.

HCF Property

The HCF of two or more numbers is always less than or equal to the smallest number in the set.

Given:

Radius of the base of the conical tent, r = 20 feet.

Slant height of the cone, l = 35 feet.

Formula Used:

Area of the canvas required (lateral surface area of the cone) = πrl

Calculation:

Using π = \(\frac{22}{7}\)

Area of the canvas required = π × r × l

⇒ Area = \(\frac{22}{7}\) × 20 × 35

⇒ Area = \(\frac{22}{7}\) × 700

⇒ Area = 22 × 100

⇒ Area = 2200 square feet

The area of the canvas required for making the tent is 2200 square feet.