Quantitative Aptitude PYQs

Given:

A's money is 20% less than B's money.

B gives 16% of her money to her brother.

Formula used:

Percentage = (Part / Whole) × 100

Calculation:

Let B's initial money be 100x.

A's initial money is 20% less than B's:

A's money = 100x - (20/100) × 100x = 100x - 20x = 80x

B gives 16% of her money to her brother:

Money given = (16/100) × 100x = 16x

B's remaining money = 100x - 16x = 84x

For A and B to have identical amounts, A's money should be 84x.

Increase in A's money:

Increase = 84x - 80x = 4x

Percentage increase in A's money = (Increase / A's initial money) × 100

Percentage increase = (4x / 80x) × 100

Percentage increase = (1/20) × 100

Percentage increase = 5%

∴ A's sum should increase by 5%.

Given:

A can do the work in 10 days.

B can do the work in 20 days.

C can do the work in 40 days.

First day: A and B work together.

Second day: A and C work together.

Formula used:

Total work = LCM of the number of days

Work done per day = Total work / Number of days

Calculation:

LCM(10, 20, 40) = 40

Let the total work be 40 units.

A's work per day = 40 / 10 = 4 units

B's work per day = 40 / 20 = 2 units

C's work per day = 40 / 40 = 1 unit

Work done on the first day (A and B):

Work done (A + B) = 4 + 2 = 6 units

Work done on the second day (A and C):

Work done (A + C) = 4 + 1 = 5 units

Work done in the 2-day cycle:

Work done (2 days) = 6 + 5 = 11 units

2-day cycles are needed:

Number of cycles = 40 / 11 = 3.63...

So, 3 full cycles are completed.

Work done in 3 cycles (6 days):

Work done (6 days) = 3 × 11 = 33 units

Remaining work = 40 - 33 = 7 units

Work done on the 7^th day (A and B):

Work done (A + B) = 6 units

Remaining work = 7 - 6 = 1 unit

Work done on the 8^th day (A and C):

Work done (A + C) = 5 units

Since only 1 unit of work remains, A and C will complete it.

Fraction of 8^th day needed = 1/5

Total days to complete the work:

6 days + 1 day + 1/5 day = 7 and 1/5 days.

Since we need the whole number of days, the work will be completed in 8 days.

It would take 8 days to complete the work.

Given:

Principal (P) = ₹24,000

Rate (r) = 10% per annum

Time (t) = 3 years

Formula used:

Amount (A) = P × (1 + r/100)^t

Compound Interest (CI) = Amount (A) - Principal (P)

Calculation:

A = 24000 × (1 + 10/100)³

A = 24000 × (1 + 0.1)³

A = 24000 × (1.1)³

A = 24000 × 1.331 = 31944

Now,

CI = A - P

CI = 31944 - 24000

CI = 7944

∴ The compound interest is ₹7944.

Given:

Vertical height of conical tent (h) = 42 decimeters

Diameter of base (d) = 5.4 meters

Space per person = 2673 cubic decimeters

Formula used:

Volume of cone (V) = (1/3) × π × r² × h

Number of persons = Volume of cone / Space per person

Calculation:

1 meter = 10 decimeters

5.4 meters = 5.4 × 10 = 54 decimeters

r = d / 2 = 54 / 2 = 27 decimeters

Volume of the cone:

V = (1/3) × (22/7) × (27)² × 42

V = (1/3) × (22/7) × 729 × 42

V = 22 × 243 × 6 = 32076 cubic decimeters

Now, Number of persons = V / Space per person

Number of persons = 32076 / 2673 = 12

∴ The conical tent can accommodate 12 persons.

Given:

Area of right triangle = 48 m²

One side containing right angle = 8 m

Formula used:

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

Calculation:

Let the other side containing the right angle be 'x' meters.

Substitute the given values into the formula:

48 = (1/2) × 8 × x

48 = 4 × x

x = 48 / 4 = 12

∴ The length of the other side containing the right angle is 12 m.

Given:

Cost Price (CP) = ₹1350

Discount = 10%

Gain = 15%

Formula used:

Selling Price (SP) = CP × (1 + Gain/100)

Selling Price (SP) = Marked Price (MP) × (1 - Discount/100)

Calculation:

Selling price (SP):

SP = 1350 × (1 + 15/100)

SP = 1350 × (1 + 0.15)

SP = 1350 × 1.15 = ₹1552.50

Let the marked price (MP) be x.

SP = MP × (1 - Discount/100)

1552.50 = x × (1 - 10/100)

1552.50 = x × (1 - 0.1)

1552.50 = x × 0.9

x = 1552.50 / 0.9

x = 1725

∴ The marked price of the article is ₹1725.

Given:

P's income is 25% less than Q's income.

Formula used:

Percentage increase = (Difference / Original value) × 100

Calculation:

Let Q's income be 100x.

P's income = 100x - (25/100) × 100x = 100x - 25x = 75x

Combined income of P and Q:

Combined income = 75x + 100x = 175x

Difference between the combined income and P's income:

Difference = 175x - 75x = 100x

Percentage more than P = (Difference / P's income) × 100

⇒ (100x / 75x) × 100 = (4/3) × 100 = 400/3 %

Percentage more than P = 133.33%

∴ The combined income of P and Q is 133.33% more than P.

Given:

Numbers: 24, 39, y

Average = 33

Formula used:

Average = (Sum of numbers) / (Number of numbers)

Calculation:

33 = (24 + 39 + y) / 3

⇒ 33 × 3 = 24 + 39 + y

⇒ 99 = 63 + y

⇒ y = 99 - 63 = 36

∴ The value of y is 36.

Given:

Walk + Ride = 1 hour 20 minutes = 80 minutes

Walk + Walk = 2 hours 30 minutes = 150 minutes

Formula used:

Ride + Ride = 2 x (Walk + Ride) - (Walk + Walk)

Calculation:

Walk + Ride = 1 hour 20 minutes = 80 minutes

2 x (Walk + Ride) = 2 x 80 minutes = 160 minutes

Walk + Walk = 2 hours 30 minutes = 150 minutes

Ride + Ride = 2 x (Walk + Ride) - (Walk + Walk)

Ride + Ride = 160 minutes - 150 minutes

Ride + Ride = 10 minutes

∴ The time taken by him to ride back both ways is 10 minutes.

Given:

Amount after 3 years = ₹16,200

Amount after 4 years = ₹17,400

Formula used:

Simple Interest (SI) = (P × r × t) / 100

Amount (A) = Principal (P) + Simple Interest (SI)

Calculation:

SI for 1 year = ₹17,400 - ₹16,200 = ₹1,200

SI for 3 years = 3 × ₹1,200 = ₹3,600

Principal (P) = Amount after 3 years - SI for 3 years

P = ₹16,200 - ₹3,600

P = ₹12,600

∴ The sum she invested is ₹12,600.

Given:

Total number of votes = 30,000

Invalid votes = 6% of total votes

Votes received by the candidate = 55% of valid votes

Formula Used:

Valid votes = Total votes - Invalid votes

Votes in favor of the candidate = 55% of valid votes

Calculation:

Invalid votes = 6% of 30,000

Invalid votes = 0.06 × 30,000

Invalid votes = 1,800

Valid votes = 30,000 - 1,800

Valid votes = 28,200

Votes in favor of the candidate = 55% of 28,200

Votes in favor of the candidate = 0.55 × 28,200

Votes in favor of the candidate = 15,510

The number of valid votes polled in favour of the candidate is 15,510.

Given:

Principal (P) = ₹9,000

Amount (A) = ₹10,200

Time (T) = 3 years

Simple Interest (SI) = A - P = ₹10,200 - ₹9,000 = ₹1,200

Formula Used:

Simple Interest = (P × R × T) / 100

New Simple Interest Rate = Old Rate + 2%

New Amount = Principal + New Simple Interest

Calculation:

Let's find the old interest rate first:

⇒ ₹1,200 = (₹9,000 × R × 3) / 100

⇒ 120,000 = 27,000R

⇒ R = 120,000 / 27,000

⇒ R = 40/9%

New Interest Rate = 40/9% + 2% = 58/9%

New Simple Interest:

⇒ SI = (₹9,000 × 58/9 × 3) / 100

⇒ SI = ₹1,740

New Amount:

⇒ A = ₹9,000 + ₹1,740

⇒ A = ₹10,740

The amount Namita would get if the interest rate had been 2% more is ₹10,740.

Given:

Anil's income is 18% more than Ramesh's income.

Formula Used:

Percentage decrease = \(\frac{\text{Difference}}{\text{Original}} \times 100 \)

Calculation:

Let Ramesh's income be R .

Anil's income = R + 0.18R = 1.18R .

Difference = Anil's income - Ramesh's income

Difference = 1.18R - R = 0.18R

Percentage decrease in Ramesh's income = \(\frac{0.18R}{1.18R} \times 100\)

⇒ Percentage decrease in Ramesh's income = \(\frac{0.18}{1.18} \times 100\)

⇒ Percentage decrease in Ramesh's income ≈ 15.25%

The correct answer is option 1.

Given:

P can finish the work in = 30 days

Q can finish the work in = 45 days

P works alone for 5 days

Then P and Q work together for 6 days

Then P leaves

Formula used:

Use LCM of 30 and 45 to assume total work

Calculation:

LCM(30, 45) = 90 units (Assume total work = 90 units)

P's 1 day work = \( \frac{90}{30} = 3 \) units/day

Q's 1 day work = \( \frac{90}{45} = 2 \) units/day

P works for 5 days:  5 × 3 = 15  units

P + Q work together for 6 days:  6 × (3 + 2) = 6 ×5 = 30  units

Total work done = \( 15 + 30 = 45 \) units

Work left =  90 - 45 = 45  units

Fraction of work left = \( \frac{45}{90} = \frac{1}{2} \) = 0.5

∴ The part of the work left is 0.5.

Given:

Sides of the triangle are 9 cm, 40 cm, and 41 cm.

Formula Used:

Area of a right-angled triangle = (1/2) × base × height

Calculation:

In the given triangle, the sides 9 cm and 40 cm can be considered as the base and height respectively, because the hypotenuse (longest side) is 41 cm.

Area = (1/2) × base × height

⇒ Area = (1/2) × 9 cm × 40 cm

⇒ Area = (1/2) × 360 cm²

⇒ Area = 180 cm²

The area of the triangle is 180 cm².

Given:

Volume of silver, V = 66 cm³

Diameter of the wire, d = 2 mm = 0.2 cm

Radius of the wire, r = d/2 = 0.1 cm

Formula Used:

Volume of a cylinder, V = πr²h

Where,

π = 22/7

r = radius

h = height (or length of the wire)

Calculation:

Using the formula for the volume of a cylinder, we have:

66 = (22/7) × (0.1)² × h

Simplifying the equation:

⇒ 66 = (22/7) × 0.01 × h

⇒ 66 = (22/700) × h

⇒ 3 = (1/700) × h

⇒ h = 2100

Convert the length from cm to m:

⇒ h = 2100 / 100

⇒ h = 21 m

The length of the wire is approximately 21 m.

Given:

Time taken to cross a pole = 3 sec

Time taken to cross a 700 m bridge = 10 sec

Formula used:

Speed = \( \frac{\text{Distance}}{\text{Time}} \)

When crossing a bridge: Distance = Length of train + Length of bridge

Calculation:

Let length of train = L meters

Train crosses pole in 3 sec ⇒ L = Speed × 3

Train crosses 700 m bridge in 10 sec ⇒ L + 700 = Speed × 10

From first equation: L = 3 × Speed

Substitute in second:

⇒ 3 × Speed + 700 = 10 × Speed

⇒ 700 = 10 × Speed - 3 × Speed = 7 × Speed

⇒ Speed = \( \frac{700}{7} = 100 \) m/s

∴ Speed of the train = 100 m/s

Shortcut Trick

The given expression follows the algebraic pattern: (a³ − b³) ÷ (a² + ab + b²) = a − b.

Here, replacing a with 6.1 and b with 1 perfectly matches the terms in the question.

Directly calculating a − b gives the final result.

Result = 6.1 − 1 = 5.1

∴ The correct answer is 5.1.

Alternate Method

Given:

Expression = (6.1 × 6.1 × 6.1 − 1) ÷ (6.1 × 6.1 + 6.1 + 1)

Formula Used:

a³ − b³ = (a − b)(a² + ab + b²)

Calculation:

⇒ Numerator = (6.1)³ − 1³

⇒ Denominator = (6.1)² + 6.1 × 1 + 1²

⇒ Expression = [(6.1)³ − 1³] ÷ [(6.1)² + 6.1 + 1]

⇒ Applying the formula for the difference of cubes in the numerator:

⇒ Expression = [(6.1 − 1)((6.1)² + 6.1 × 1 + 1²)] ÷ [(6.1)² + 6.1 + 1]

⇒ The term ((6.1)² + 6.1 + 1) is common in both numerator and denominator.

⇒ Canceling the common term, we are left with:

⇒ Expression = 6.1 − 1

⇒ Expression = 5.1

∴ The correct answer is 5.1.

Additional Information

Sum of Cubes Formula

a³ + b³ = (a + b)(a² − ab + b²). This identity is frequently used to rapidly simplify fraction expressions containing positive cubic terms.

Difference of Squares Formula

a² − b² = (a − b)(a + b). It helps in instantly factoring and reducing fractions containing squared differences.

Perfect Square Formulas

(a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². These are primary tools for algebraic expansion and solving polynomial equations.

Given:

We need to find the average of the prime numbers between 57 and 99 (both inclusive).

Formula Used:

Average of prime numbers = (Sum of prime numbers) / (Number of prime numbers)

Calculation:

Prime numbers between 57 and 99 are: 59, 61, 67, 71, 73, 79, 83, 89, 97

Sum of these prime numbers = 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 97

Sum of prime numbers = 679

Number of prime numbers = 9

Average of prime numbers = 679 / 9

Average of prime numbers = 75.4

The average of the prime numbers between 57 and 99 (both inclusive) is 75.4.

Given:

N is a 3-digit number

Unit digit of N is prime

Product of all digits of N is also prime

Concept used:

Prime digits = 2, 3, 5, 7 (for unit place)

A product of digits is prime ⇒ only one digit is prime, rest digits must be 1 (since 1 × prime = prime)

Calculation:

Let N = xyz, where x, y, z are digits (x ≠ 0)

Product x × y × z = prime ⇒ only one digit among x, y, z is a prime number (2, 3, 5, 7), others must be 1

Try placing prime digit at units place (z), as given:

If z = 2 → x, y = 1 ⇒ N = 112

If z = 3 → x, y = 1 ⇒ N = 113

If z = 5 → x, y = 1 ⇒ N = 115

If z = 7 → x, y = 1 ⇒ N = 117

Total 4 valid numbers if prime digit is at units place

Try placing prime digit at tens place (y): But unit digit must also be prime → contradiction unless y = prime & z = prime → product will not be prime (e.g., 3 × 3 = 9)

Only valid case: one digit is prime, others = 1, and prime is at unit place

∴ The number of possible values of N is 4.