Quantitative Aptitude PYQs

Given:

Selling Price (SP) = ₹8,000

Increase = 60%

Formula used:

SP = CP + (Increase% of CP)

Where, CP = Cost Price

Calculation:

8000 = CP + (60% of CP)

⇒ 8000 = CP + 0.60 × CP

⇒ 8000 = 1.60 × CP

⇒ CP = 8000 / 1.60

⇒ CP = 5000

∴ The correct answer is option 2.

Given:

Prime numbers between 15 and 30

Formula used:

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

Calculation:

Prime numbers between 15 and 30 are: 17, 19, 23, 29

Sum of prime numbers = 17 + 19 + 23 + 29

⇒ Sum = 88

Number of prime numbers = 4

Average = 88 / 4

⇒ Average = 22

∴ The correct answer is option (4).

Given:

Principal (P) = ₹3,00,000

Time (t) = 2 years

Rate for Bank A (r_A) = 5.5% per annum

Rate for Bank B (r_B) = 6.5% per annum

Formula used:

Simple Interest (SI) = \( \frac{P \times r \times t}{100} \)

Calculations:

SI_A = \( \frac{3,00,000 \times 5.5 \times 2}{100} \)

⇒ SI_A = \( \frac{3,00,000 \times 11}{100} \)

⇒ SI_A = 33,000

SI_B = \( \frac{3,00,000 \times 6.5 \times 2}{100} \)

⇒ SI_B = \( \frac{3,00,000 \times 13}{100} \)

⇒ SI_B = 39,000

Positive difference between SI_B and SI_A = 39,000 - 33,000 = 6,000

∴ The correct answer is option 4.

Calculation:

We have,

[2\(\frac{5}{13}\) % of 9100 + 1\(\frac{1}{17}\)% of 10200 + \(\frac{3}{5}\)(17 + 13)]1/3

⇒ [\(\frac{31}{13}\)% of 9100 + \(\frac{18}{17}\)% of 10200 + \(\frac{3}{5}\)(30)]^1/3

⇒ [\(\frac{31}{13}\) × 9100/100 + \(\frac{18}{17}\) × 10200/100 + 3 × 6]1/3

⇒ [\(\frac{31}{13}\) × 91 + \(\frac{18}{17}\) × 102 + 18]1/3

⇒ [31 × 7 + 18 × 6 + 18]^1/3

⇒ [217 + 108 + 18]1/3

⇒ [343]1/3 = 7

Hence, 7 is the correct answer.

Given:

Height (h) = 48 cm

Base (b) = (1/6) × height

Formula used:

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

Calculation:

Base (b) = (1/6) × 48

⇒ b = 8 cm

Area = (1/2) × 8 × 48

⇒ Area = 4 × 48

⇒ Area = 192 cm²

∴ The correct answer is option 3.

Given:

Principal (P) = ₹1,500

Rate of Interest (R) = 5% p.a.

Time (T) = 3 years

Formula Used:

Simple Interest (SI) = \(\frac{P \times R \times T}{100}\)

Calculation:

SI = \(\frac{1,500 \times 5 \times 3}{100} \)

⇒ SI = \(\frac{22,500}{100}\)

⇒ SI = ₹225

The simple interest on a sum of ₹1,500 at 5% p.a. for 3 years is ₹225.

Given:

Cost Price (CP) = 100 (assume)

Marked Price (MP) = CP + 35% of CP

Discount = 20% on MP

Formula used:

Profit Percentage = ((Selling Price - Cost Price) / Cost Price) × 100

Calculation:

Marked Price = 100 + 35% of 100

Marked Price = 100 + 35

Marked Price = 135

Discount = 20% of 135

Discount = (20/100) × 135

Discount = 27

Selling Price (SP) = Marked Price - Discount

Selling Price = 135 - 27

Selling Price = 108

Profit = Selling Price - Cost Price

Profit = 108 - 100

Profit = 8

Profit Percentage = (8 / 100) × 100

Profit Percentage = 8%

The correct answer is option 1.

Given:

We need to find the unit digit of the product of all prime numbers between 51 and 78.

Formula used:

To determine the unit digit of the product, focus on the unit digits of each prime number and multiply them together.

Calculation:

The prime numbers between 51 and 78 are: 53, 59, 61, 67, 71, 73.

The unit digits of these primes are: 3, 9, 1, 7, 1, 3.

Now, calculate the product of these unit digits:

⇒ (3 × 9 × 1 × 7 × 1 × 3)

⇒ 3 × 9 = 27 (unit digit is 7)

⇒ 7 × 1 = 7 (unit digit remains 7)

⇒ 7 × 7 = 49 (unit digit is 9)

⇒ 9 × 1 = 9 (unit digit remains 9)

⇒ 9 × 3 = 27 (unit digit is 7)

∴ The unit digit of the product of all prime numbers between 51 and 78 is 7.

Given:

Selling Price (SP) = Rs. 4,650

Loss = Rs. 350

Formula Used:

Cost Price (CP) = SP + Loss

Percentage Loss = (Loss / CP) × 100

Calculation:

CP = SP + Loss

CP = 4,650 + 350

CP = 5,000

Percentage Loss = (Loss / CP) × 100

⇒ Percentage Loss = (350 / 5,000) × 100

⇒ Percentage Loss = 0.07 × 100

⇒ Percentage Loss = 7%

The percentage loss is 7%.

Given:

The product of the LCM and HCF of two numbers is 48.

The difference between the two numbers is 13.

Formula Used:

Product of two numbers = LCM × HCF

Calculation:

Let the two numbers be x and y.

Given: LCM × HCF = 48

Also, given: |x - y| = 13

We need to find two numbers such that their product is 48 and their difference is 13.

Let's start with the given options:

Option 1: 9 and 22

Product = 9 × 22 = 198

Difference = |9 - 22| = 13

Option 2: 6 and 19

Product = 6 × 19 = 114

Difference = |6 - 19| = 13

Option 3: 3 and 16

Product = 3 × 16 = 48

Difference = |3 - 16| = 13

Option 4: 10 and 23

Product = 10 × 23 = 230

Difference = |10 - 23| = 13

From the options, only Option 3 satisfies both conditions.

The correct answer is option 3: 3 and 16.

Given:

Slant height of the cone = 5 cm

Diameter of the cone = 8 cm

Radius of the cone = 8/2 = 4 cm

Formula Used:

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

Using the Pythagorean theorem to find the height (h):

h = √(l² - r²)

where l is the slant height and r is the radius.

Calculation:

h = √(5² - 4²)

⇒ h = √(25 - 16)

⇒ h = √9

⇒ h = 3 cm

Volume of the cone:

Volume = (1/3) × π × 4² × 3

⇒ Volume = (1/3) × (22/7) × 16 × 3

⇒ Volume = (1/3) × (22/7) × 48

⇒ Volume = (22 × 48) / 21

⇒ Volume = 1056 / 21

⇒ Volume ≈ 50.28 cm³

The volume of the cone is approximately 50.28 cm³.

Given:

After a 15% decline, the income of Ajay stands at Rs. 2,12,500.

Formula Used:

Original Income = Declined Income / (1 - Decline Percentage)

Calculation:

Let the original income be X.

After a 15% decline, income becomes 85% of the original income.

85% of X = Rs. 2,12,500

0.85 × X = 2,12,500

⇒ X = 2,12,500 / 0.85

⇒ X = 2,50,000

Ajay's original income was Rs. 2,50,000.

Given:

In a 600 m race, A gave B a head start of 5 seconds and still won the race by 20 seconds.

The ratio of the speed of A to the speed of B is 2 ∶ 1.

Formula Used:

Speed = Distance / Time

Calculation:

Let the speed of B be x m/sec.

Then, the speed of A = 2x m/sec.

Time taken by A to cover 600 m = 600 / 2x

Time taken by B to cover 600 m = 600 / x

According to the question,

Time taken by B - Time taken by A = 25 sec (B had a head start of 5 seconds and lost by 20 seconds)

⇒ (600 / x) - (600 / 2x) = 25

⇒ 600(2 - 1) / 2x = 25

⇒ 600 / 2x = 25

⇒ 600 = 50x

⇒ x = 600 / 50

⇒ x = 12 m/sec

The speed of B is 12 m/sec.

Given:

Number of students in the class = 50

Average marks of 20 students = 70

Average marks of remaining 30 students = 80

Formula Used:

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

Calculation:

Total marks of 20 students = 20 × 70

Total marks of 20 students = 1400

Total marks of 30 students = 30 × 80

Total marks of 30 students = 2400

Total marks of all students = 1400 + 2400

Total marks of all students = 3800

Total number of students = 50

Average marks = 3800 / 50

Average marks = 76

The average marks of the whole class is 76.

Given:

The man agreed to pay:

Rs. 4,200 at the end of the first year.

Rs. 4,410 at the end of the second year.

Compound interest rate = 5% per annum.

We need to find the sum borrowed.

Formula used:

Compound Interest Formula: \(A = P(1 + \frac{r}{100})^t\)

Where:

A = Amount after t years.

P = Principal.

r = Rate of interest per annum.

t = Time (in years).

Calculations:

The payment of Rs. 4,200 at the end of the first year and Rs. 4,410 at the end of the second year cover the borrowed amount and interest.

Total borrowed amount = Present value of Rs. 4,200 (first year) + Present value of Rs. 4,410 (second year).

Present value = Future value / (1 + r/100)^t

Present value of Rs. 4,200:

\(PV_1 = 4,200 / (1 + 5/100)^1\) = 4,200 / 1.05 = Rs. 4,000

Present value of Rs. 4,410:

\(PV_2 = 4,410 / (1 + 5/100)^2 = 4,410 / (1.05)^2\) = 4,410 / 1.1025 = Rs. 4,000

Total borrowed amount = PV₁ + PV₂ = 4,000 + 4,000 = Rs. 8,000

The sum borrowed is Rs. 8,000.

Given:

Time reduction = 80%

Formula Used:

Speed = Distance / Time

New Speed = (Initial Speed × 100) / (100 - Time Reduction Percentage)

Percentage Increase in Speed = (New Speed - Initial Speed) / Initial Speed × 100

Calculation:

Let the initial speed be S and initial time be T.

Since the time is reduced by 80%, the new time will be 20% of the initial time.

New Time = 0.2T

Since the distance is fixed,

Initial Distance = S × T

New Distance = New Speed × New Time

Since the distance remains the same,

S × T = New Speed × 0.2T

⇒ New Speed = S / 0.2

⇒ New Speed = 5S

Percentage Increase in Speed = (5S - S) / S × 100

⇒ Percentage Increase = 4S / S × 100

⇒ Percentage Increase = 400%

The woman should increase the speed of her car by 400% to reduce the time by 80% to cover a fixed distance.

Given:

A can finish the work in 10 days.

B can finish the work in 16 days.

C can finish the work in 12 days.

A, B, and C work one after the other in that order.

They complete 3 cycles of ABC before A and C leave.

Formula Used:

Work completed per day = 1 / Number of days to complete the work individually

Work completed in 1 cycle (ABC) = Work completed by A + Work completed by B + Work completed by C

Total work completed in 3 cycles = 3 × Work completed in 1 cycle

Remaining work = Total work - Work completed in 3 cycles

Time taken by B to complete the remaining work = Remaining work / Work completed by B per day

Calculation:

Work completed by A in 1 day = 1 / 10

Work completed by B in 1 day = 1 / 16

Work completed by C in 1 day = 1 / 12

Work completed in 1 cycle (ABC) = (1 / 10) + (1 / 16) + (1 / 12)

\( \Rightarrow \frac{1}{10} + \frac{1}{16} + \frac{1}{12} \)

\( \Rightarrow \frac{24 + 15 + 20}{240} \)

\( \Rightarrow \frac{59}{240} \)

Total work completed in 3 cycles = 3 × (59 / 240)

\( \Rightarrow \frac{177}{240} \)

Remaining work = 1 - (177 / 240)

\( \Rightarrow \frac{240 - 177}{240} \)

\( \Rightarrow \frac{63}{240} \)

Time taken by B to complete the remaining work = (63 / 240) / (1 / 16)

\( \Rightarrow \frac{63}{240} \times 16 \)

\( \Rightarrow \frac{1008}{240} \)

\( \Rightarrow 4 \frac{1}{5}\) days

The time B alone will complete the rest of the work in 4¹/₅ days.

Given:

Volume of the cone = \(\frac{128}{3} \pi\)

Height of the cone (h) = 2 × radius of the cone (r)

Formula Used:

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

Slant height (l) of the cone = \( √{r^2 + h^2} \)

Calculation:

We know, Volume of cone = \(\frac{1}{3} \pi r^2 h\)

Given volume is \(\frac{128}{3} \pi\), so:

\(\frac{1}{3} \pi r^2 h = \frac{128}{3} \pi\)

⇒ \(\pi r^2 h = 128 \pi\)

⇒ \(r^2 h = 128\)

Since h = 2r:

⇒ \(r^2 (2r) = 128\)

⇒ \(2r^3 = 128\)

⇒ \(r^3 = 64\)

⇒ \(r = \sqrt[3]{64}\)

⇒ \(r = 4\)

Now, h = 2r = 2 × 4 = 8

Slant height (l) = \( \sqrt{r^2 + h^2} \)

⇒ \( \sqrt{4^2 + 8^2} \)

⇒ \( \sqrt{16 + 64} \)

⇒ \( \sqrt{80} \)

The slant height of the cone is √80

Formula Used:

Two numbers are co-prime if their greatest common divisor (GCD) is 1.

Calculation:

Option 1: (7, 35)

GCD(7, 35) = 7

Since GCD ≠ 1, they are not co-primes.

Option 2: (217, 651)

GCD(217, 651) = 217

Since GCD ≠ 1, they are not co-primes.

Option 3: (32, 62)

GCD(32, 62) = 2

Since GCD ≠ 1, they are not co-primes.

Option 4: (198, 175)

GCD(198, 175)

⇒ 198 = 2 × 3² × 11

⇒ 175 = 5² × 7

Since there are no common factors other than 1, GCD(198, 175) = 1

Since GCD = 1, they are co-primes.

The correct answer is option 4.

Given:

Let x = 5 and y = 8. We need to arrange the following fractions in ascending order:

\(\frac{x+1}{y-1}, \frac{2 x}{2 y-3}, \frac{y+1}{2 x-1}, \frac{y+1}{3 x-y}\)

Calculation:

Substitute x = 5 and y = 8 into each fraction:

1. (x+1)/(y-1) = (5+1)/(8-1) = 6/7
2. (2x)/(2y-3) = (2×5)/(2×8-3) = 10/13
3. (y+1)/(2x-1) = (8+1)/(2×5-1) = 9/9 = 1
4. (y+1)/(3x-y) = (8+1)/(3×5-8) = 9/7

Convert the fractions to decimals:

6/7 ≈ 0.857, 10/13 ≈ 0.769, 1 = 1, 9/7 ≈ 1.286

Arrange in ascending order:

10/13 (≈ 0.769), 6/7 (≈ 0.857), 1, 9/7 (≈ 1.286)

The numbers in ascending order are: \(\frac{2 x}{2 y-3}, \frac{x+1}{y-1}, \frac{y+1}{2 x-1}, \frac{y+1}{3 x-y}\)