Quantitative Aptitude PYQs

Given:

We need to find the smallest number which leaves remainder 3 when divided by 5, 8, and 9, but leaves no remainder when it is divided by 11.

Formula used:

Number = LCM(5, 8, 9) + 3

The number must also be divisible by 11.

Calculation:

LCM of 5, 8, and 9 = 360

Number = 360k + 3 (where k is an integer)

The number must be divisible by 11

⇒ 360k + 3 ≡ 0 (mod 11)

⇒ 360k ≡ -3 (mod 11)

⇒ 360 ≡ 8 (mod 11) (since 360 mod 11 = 8)

⇒ 8k ≡ -3 (mod 11)

⇒ 8k ≡ 8 (mod 11)

⇒ k ≡ 1 (mod 11)

⇒ k = 1

⇒ Number = 360 × 1 + 3 = 363

∴ The correct answer is option 1.

Given:

Amit's expenditure to savings ratio = 3:2

Formula used:

Total salary = Expenditure + Savings

Let the total salary be 5 units (since 3 + 2 = 5)

Expenditure = 3 units

Savings = 2 units

Calculation:

Percentage of expenditure = \(\dfrac{3}{5} \times 100\)

⇒ \(\dfrac{3}{5} \times 100 = 60%\)

Percentage of savings = \(\dfrac{2}{5} \times 100\)

⇒ \(\dfrac{2}{5} \times 100 = 40%\)

∴ The correct answer is option (1).

Solution:

We need to find the number of ways to express 960 as a product of two numbers that are co-prime to each other. Two numbers are co-prime if their greatest common divisor (GCD) is 1.

Step 1: Prime factorization of 960

960 = 26 × 3 × 5

Step 2: Concept of co-prime pairs

For two numbers to be co-prime, all prime factors of 960 must be assigned to exactly one of the two numbers. This ensures that there are no common factors between them.

Step 3: Distribute the prime factors

There are three distinct prime factors of 960: 2, 3, and 5.

Each prime factor can either go to the first number or the second number. Thus, there are 2³ = 8 ways to assign the factors to the two numbers.

However, since swapping the two numbers does not create a new pair, we divide the result by 2.

Number of unique ways = 8 / 2 = 4.

Step 4: Verify the pairs

The four ways to divide the prime factors are:

First number: 2⁶, Second number: 3 × 5

First number: 26 × 3, Second number: 5

First number: 26 × 5, Second number: 3

First number: 26 × 3 × 5, Second number: 1

∴ The number of ways to write 960 as the product of two co-prime numbers is 4.

Given:

A person has some amount initially.

He gives 30% of the amount to his wife.

From the remaining amount, he gives 20%, 20%, and 15% respectively to his two sons and one daughter.

From the amount left, he pays 1/3 as house rent.

Remaining amount = ₹33,600

Formula used:

Initial amount = Final remaining amount ÷ Remaining fraction after deductions

Calculation:

Let the initial amount be x

Amount given to wife = 0.3x

Remaining = 0.7x

Amount given to two sons and one daughter = 0.2 × 0.7x + 0.2 × 0.7x + 0.15 × 0.7x

⇒ 0.14x + 0.14x + 0.105x = 0.385x

Remaining after giving to children = 0.7x - 0.385x = 0.315x

House rent = \(\dfrac{1}{3} \times 0.315x\) = 0.105x

Remaining after rent = 0.315x - 0.105x = 0.21x

Given remaining amount = ₹33,600

⇒ 0.21x = 33,600

⇒ x =\(\dfrac{33,600}{0.21}\)

⇒ x = 1,60,000

∴ The correct answer is option (3).

Given:

The selling price of 61 items is equal to the cost price of 53 items.

Formula used:

Profit or Loss Percentage = \(\dfrac{(Selling Price - Cost Price)}{Cost Price} \times 100\)

Calculation:

Let Cost Price (CP) of 1 item be ₹1.

Then, Cost Price (CP) of 53 items = ₹53.

Selling Price (SP) of 61 items = ₹53.

Cost Price (CP) of 61 items = ₹61.

⇒ Loss = CP - SP

⇒ Loss = ₹61 - ₹53

⇒ Loss = ₹8

Loss Percentage = \(\dfrac{8}{61} \times 100\)

⇒ Loss Percentage = \(\dfrac{800}{61}\)

⇒ Loss Percentage ≈ 13.11%

∴ The correct answer is option (3).

Given:

Original breadth of the rectangle = b

Original length of the rectangle = l

Breadth decreased by 40% ⇒ New breadth = 0.6b

Length increased by 25% ⇒ New length = 1.25l

Formula Used:

Area of a rectangle = length × breadth

Percentage change in area = \(\frac{New Area - Original Area}{Original Area} \times 100\)

Calculation:

Original Area = l × b

New Area = (1.25l) × (0.6b)

New Area = 0.75lb

Percentage change in area = \(\frac{0.75lb - lb}{lb} \times 100\)

Percentage change in area = \(\frac{-0.25lb}{lb} \times 100\)

Percentage change in area = -25%

The percentage change in its area is 25% decrease.

Given:

When 24 is subtracted from a number, it reduces to 60%.

Formula Used:

Let the number be x .

Equation: x - 24 = 0.6x

Solve for x and then find two-fifth of that number: $\frac{2}{5}x$

Calculation:

x - 24 = 0.6x

⇒ x - 0.6x = 24

⇒ 0.4x = 24

⇒ x = \(\frac{24}{0.4}\)

⇒ x = 60

Two-fifth of 60:

⇒ \(\frac{2}{5} \times 60\) = 24

The two-fifth of that number is 24.

Given:

Speed of the bus = 40 km/h

Speed of the car = 75 km/h

Time taken for the car to catch up = 15 minutes = 15/60 hours = 0.25 hours

Formula Used:

Distance = Relative Speed × Time

Calculation:

Relative Speed = Speed of the car - Speed of the bus

Relative Speed = 75 - 40

Relative Speed = 35 km/h

Distance = Relative Speed × Time

⇒ Distance = 35 × 0.25

⇒ Distance = 8.75 km

The distance between the bus and the car is 8.75 km.

Given:

\(\frac{1}{2-\frac{1}{5+\frac{1}{6-\frac{2}{5}}}}\)

Calculation:

\(\frac{1}{2-\frac{1}{5+\frac{1}{6-\frac{2}{5}}}}\)

⇒ \(\frac{1}{2-\frac{1}{5+\frac{1}{\frac{30-2}{5}}}} \)

⇒ \(\frac{1}{2-\frac{1}{5+\frac{5}{28}}}\)

⇒ \( \frac{1}{2-\frac{1}{\frac{140+5}{28}}} \)

⇒ \( \frac{1}{2-\frac{28}{145}} \)

⇒ \(\frac{1}{\frac{290 - 28}{145}} \)

⇒ \(\frac{145}{262}\)

∴ The correct answer is option 3.

Given:

Marked price of the machine = Rs. 30,000

Trade discount = 16%

Selling price after cash discount = Rs. 23,184

Formula Used:

Trade Discount Price = Marked Price × (1 - Trade Discount)

Cash Discount Price = Trade Discount Price × (1 - Cash Discount)

Calculation:

Trade Discount Price = 30,000 × (1 - 0.16)

Trade Discount Price = 30,000 × 0.84

Trade Discount Price = 25,200

Let the cash discount be x.

Cash Discount Price = 25,200 × (1 - x)

Given, Cash Discount Price = 23,184

⇒ 25,200 × (1 - x) = 23,184

⇒ 1 - x = 23,184 / 25,200

⇒ 1 - x = 0.92

⇒ x = 1 - 0.92

⇒ x = 0.08 = 8%

The cash discount is 8%.

Given:

Provisions for 60 persons for 18 days.

Strength increased to 90 persons.

Formula Used:

Total provisions = Number of persons × Number of days

New number of days = Total provisions / New number of persons

Calculation:

Total provisions = 60 × 18

Total provisions = 1080 person-days

New number of days = 1080 / 90

New number of days = 12 days

The provisions will finish off in 12 days if the strength of the camp is increased to 90 persons.

Given:

Number of students in Class 1 = 50

Average marks of Class 1 = 75

Number of students in Class 2 = 65

Average marks of Class 2 = 80

Number of students in Class 3 = 45

Average marks of Class 3 = 90

Formula Used:

Average marks = Total marks of all students / Total number of students

Calculation:

Total marks of Class 1 = 50 × 75 = 3750

Total marks of Class 2 = 65 × 80 = 5200

Total marks of Class 3 = 45 × 90 = 4050

Total marks of all students = 3750 + 5200 + 4050

Total marks of all students = 13000

Total number of students = 50 + 65 + 45

Total number of students = 160

Average marks = 13000 / 160

Average marks = 81.25

The average marks of all the students is 81.25.

Given:

After down payment, remaining amount = ₹8,000 - ₹500 = ₹7,500

Rate of interest (R) = 25% per annum

Total instalments (n) = 4

Formula used:

Instalment =  \(\frac{100 \times \text{Total Amount Payable}}{100 \times n + R \times \frac{n(n-1)}{2 \times 12}}\)

Calculation:

Step 1: Total amount payable:

Total Interest = \(\frac{7500 \times 25 \times 4}{12 \times 100} = ₹625\)

Total Amount Payable = ₹7,500 + ₹625 = ₹8,125

Step 2: Calculate the instalment:

Instalment = \(\frac{100 \times 8125}{100 \times 4 + 25 \times \frac{4(4-1)}{2 \times 12}}\)

Instalment = \(\frac{100 \times 8125}{400 + 25 \times \frac{4 \times 3}{24}}\)

Instalment = \(\frac{100 \times 8125}{400 + 25 \times 0.5}\)

Instalment = \(\frac{812500}{412.5} ≈ 1969.69\)

∴ The monthly instalment is ₹1,970.

Shortcut Trick

To check if a number N is a prime number, test its divisibility by primes up to √N.

For 1271, √1271 ≈ 35.6. Primes to check: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31.

We find that 1271 ÷ 31 = 41.

Since 1271 has factors other than 1 and itself, it is a composite number.

∴ The correct answer is 1271.

Alternate Method

Given: A set of four-digit numbers: 1571, 1471, 1271, 1171.

Method: A prime number is a natural number greater than 1 that has only two divisors: 1 and itself. A number that has more than two divisors is composite.

⇒ Let us test 1271 for factors:

⇒ 1271 ÷ 31 = 41

⇒ This means 1271 = 31 × 41.

⇒ The factors of 1271 are 1, 31, 41, and 1271.

⇒ Since it has more than two factors, 1271 is NOT a prime number.

⇒ For the other options: 1571, 1471, and 1171 have no factors other than 1 and themselves, so they are prime numbers.

∴ The correct answer is 1271.

Additional Information

Prime Number

A natural number greater than 1 that has no positive divisors other than 1 and itself. 2 is the only even prime number.

Composite Number

A positive integer greater than 1 that has at least one divisor other than 1 and itself. 1 is neither prime nor composite.

Co-prime Numbers

Two numbers are said to be co-prime if their Highest Common Factor (HCF) is 1. Example: (8, 15).

Primality Test

To check if n is prime, verify divisibility by all prime numbers p such that p ≤ √n.

Given:

The cost price of an item is one-fourth of its selling price.

Formula Used:

Gain percentage = \(\left(\frac{{\text{{Selling Price}} - \text{{Cost Price}}}}{{\text{{Cost Price}}}}\right) \times 100\)

Calculation:

Let the selling price (SP) be 4x.

Cost price (CP) = SP / 4

CP = 4x / 4

CP = x

Gain = SP - CP

Gain = 4x - x

Gain = 3x

Gain percentage = \(\left(\frac{{3x}}{{x}}\right) \times 100\)

Gain percentage = 3 × 100

Gain percentage = 300%

The correct answer is option 1 (Gain of 300%).

Given:

The sum of five consecutive integers is 235.

Formula Used:

Sum of n consecutive integers = n × average of the integers

Calculation:

Let the five consecutive integers be a - 2, a - 1, a, a + 1, a + 2.

Sum of these integers = (a - 2) + (a - 1) + a + (a + 1) + (a + 2)

Sum of five consecutive integers = 5a

Given:

5a = 235

⇒ a = 235 / 5

⇒ a = 47

So, the five integers are 45, 46, 47, 48, 49.

Among these, 47 is the only prime number.

The correct answer is option 1: 47

Given:

The ratio of the number of boys to that of the girls in a school is 5:7.

Number of girls in the school = 1148

Formula Used:

Let the number of boys be 5x and the number of girls be 7x.

Ratio of boys to girls = 5x / 7x

Calculation:

We know, Number of girls = 7x

⇒ 7x = 1148

⇒ x = 1148 / 7

⇒ x = 164

Number of boys = 5x

⇒ Number of boys = 5 × 164

⇒ Number of boys = 820

The number of boys in the school is 820.

Given:

Internal radius of the pipe (r₁) = 2 cm.

External radius of the pipe (r₂) = 2.7 cm.

Length of the pipe (h) = 10 cm.

Formula Used:

Volume of the metal used = Volume of the external cylinder - Volume of the internal cylinder

Volume of the cylinder = πr²h

Calculation:

Volume of the external cylinder = πr₂²h

Volume of the internal cylinder = πr₁²h

Volume of the metal used = πr₂²h - πr₁²h

Volume of the metal used = πh(r₂² - r₁²)

\(r_{2}=2.7 , \text{cm}, , r_{1}=2 , \text{cm}, , h=10 , \text{cm}\)

Volume of the metal used = (22/7) × 10 × (2.7² - 2²)

⇒ Volume of the metal used = (22/7) × 10 × (7.29 - 4)

⇒ Volume of the metal used = (22/7) × 10 × 3.29

⇒ Volume of the metal used = 22 × 4.7

⇒ Volume of the metal used = 103.4 cm³

The correct answer is option 4.

Given:

Principal (P) = Rs. 32,760

Rate of interest (R) = 20% compounded annually

Number of instalments (n) = 3

Formula Used:

The formula for calculating the equal annual instalment (A) for a loan with compound interest is:

\( A = \frac{P \times R}{1 - (1 + R)^{-n}} \)

Where,

P = Principal amount

R = Rate of interest per period

n = Number of instalments

Calculation:

Rate of interest per period (R) = 20% = 0.20

Substitute the values into the formula:

\( A = \frac{32760 \times 0.20}{1 - (1 + 0.20)^{-3}} \)

⇒ \( A = \frac{6552}{1 - (1.20)^{-3}} \)

⇒ \( A = \frac{6552}{1 - \frac{1}{1.20^3}} \)

\( 1.20^3 \approx 1.728 \)

⇒ \( A = \frac{6552}{1 - \frac{1}{1.728}} \)

⇒ \( A = \frac{6552}{1 - 0.5787} \)

⇒ \( A = \frac{6552}{0.4213} \)

⇒ \( A \approx 15552 \)

The value of each instalment is Rs. 15,552.

Shortcut Trick

Volume is proportional to the cube of the radius (r³).

Let R be the radius of the big sphere and r be the radius of the small sphere.

n × r³ = R³ ⇒ 27 × r³ = (6 ÷ 2)³

27 × r³ = 3³ ⇒ 27 × r³ = 27

r³ = 1 ⇒ r = 1 cm

∴ The correct answer is 1 cm.

Alternate Method

Given:

Diameter of big sphere = 6 cm ⇒ Radius (R) = 3 cm

Number of small spheres (n) = 27

Formula Used:

Volume of a sphere = (4/3) × π × r³

Calculations:

⇒ Volume of big sphere = 27 × Volume of one small sphere

⇒ (4/3) × π × R³ = 27 × (4/3) × π × r³

⇒ R³ = 27 × r³

⇒ 3³ = 27 × r³

⇒ 27 = 27 × r³

⇒ r³ = 1

⇒ r = 1 cm

∴ The correct answer is 1 cm.

Additional Information

Surface Area of Sphere

The total surface area of a sphere is given by the formula 4 × π × r².

Volume Conservation

When a solid is melted and recast into other shapes, the total volume remains constant.

Hemisphere Properties

Volume of hemisphere = (2/3) × π × r³ and Total Surface Area = 3 × π × r².