Quantitative Aptitude - MCQ Practice Questions

Step 1: Discount = 15% of ₹800 = 0.15 × 800 = ₹120.

Step 2: Selling Price = Marked Price - Discount = 800 - 120 = ₹680.

Therefore, option A is correct.

Step 1: Increase in price = ₹50 - ₹40 = ₹10.

Step 2: Percentage increase = (Increase/Original Price) × 100 = ($\frac{10}{40}$) × 100 = 25%.

Therefore, option C is correct.

Step 1: Percentage = (Marks obtained/Total marks) × 100 = ($\frac{72}{120}$) × 100.

Step 2: = 0.6 × 100 = 60%.

Therefore, option B is correct.

Step 1: Profit = 20% of ₹250 = 0.20 × 250 = ₹50.

Step 2: Selling Price = Cost Price + Profit = 250 + 50 = ₹300.

Therefore, option B is correct.

Step 1: Population increase = 8% of 50,000 = 0.08 × 50,000 = 4,000.

Step 2: Current population = 50,000 + 4,000 = 54,000.

Therefore, option B is correct.

Step 1: Percentage spent on rent and food = 35% + 20% = 55%.

Step 2: Amount spent = 55% of ₹45,000 = 0.55 × 45,000 = ₹24,750.

Therefore, option A is correct.

Step 1: Profit = 25% of ₹400 = 0.25 × 400 = ₹100.

Step 2: Selling Price = Cost Price + Profit = 400 + 100 = ₹500.

Step 3: Let Marked Price = M.

After 15% discount: M × (1 - 0.15) = 500, so M × 0.85 = 500, M = $\frac{500}{0}$.85 = ₹588.24 (approximately).

Therefore, option A is correct.

Step 1: After 12% increase: Price = 500 × (1 + 0.12) = 500 × 1.12 = ₹560.

Step 2: After 10% decrease: Price = 560 × (1 - 0.10) = 560 × 0.90 = ₹504.

Therefore, option B is correct.

Step 1: Reduction in production = 18% of 5,000 = 0.18 × 5,000 = 900 units.

Step 2: Current production = 5,000 - 900 = 4,100 units per day.

Therefore, option C is correct.

To find the time saved, we need to calculate the original travel time and the new travel time after the speed increase, then find the difference.

Step 1: Calculate original travel time

Using the formula $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$:

Step 2: Find the new speed after 20% increase

A 20% increase means the new speed is:

Alternatively: ${\text{Speed}}_{\text{new}}=60\times 1.20=72\text{ km/h}$

Step 3: Calculate new travel time at increased speed

Step 4: Find time saved

Converting $\frac{1}{3}$ hour to minutes: $\frac{1}{3}\times 60=20\text{ minutes}$

Therefore, time saved = 1 hour 20 minutes

Answer: 1 hour 20 minute (Option A)

To solve this problem, we need to find the original speed, calculate the reduced speed, and then determine the time needed to cover the reduced distance.

Step 1: Calculate Original Speed

The train travels 240 km in 4 hours, so we divide distance by time to find speed.

Step 2: Calculate Reduced Speed

The speed reduces by 25%, so the new speed is 75% of the original speed.

Step 3: Calculate Time for 180 km at Reduced Speed

Using the formula Time = Distance ÷ Speed, we find how long it takes to cover 180 km.

The train will take 4 hours to cover 180 km at the reduced speed.

Original speed of the train:

Speed=

4

240

​

=60 km/h

Speed reduced by 25%:

Reduced speed=60−25% of 60

=60−15=45 km/h

Time to cover 180 km at reduced speed:

Time=

45

180

​

=4 hours

Therefore, the train will take:

Answer: (A) 4 hours

Let original price = 100. After 15% increase: 115. After 15% decrease: 115 × 0.85 = 97.75. Change = 97.75 - 100 = -2.25. Percentage change = -2.25%.

To find the total salary, we need to determine what percentage is saved, then use it to calculate the full amount.

Step 1: Find the percentage spent

The man spends money on food and rent:

Step 2: Find the percentage saved

Since total salary = 100%, the remaining amount is saved:

Step 3: Set up the equation

If 30% of his salary equals ₹3000:

Step 4: Solve for total salary

Verification: 40% of ₹10,000 = ₹4,000 (food); 30% of ₹10,000 = ₹3,000 (rent); Remaining = ₹10,000 − ₹7,000 = ₹3,000 ✓

**Answer: The total monthly salary is $₹10,000$ (Option A)**

Let original price = 100. After 20% increase: 120. After 10% increase: 120 × 1.10 = 132. Total increase = 32%.

Let MP = x. SP = x × (1 - 0.12) = 0.88x = 440. x = $\frac{440}{0}$.88 = 500.

Increase = 30,000 - 25,000 = 5,000. Percentage = (5,$\frac{000}{25}$,000) × 100 = 20%.

Let CP of 1 article = 1. CP of 8 = 8. SP of 10 = 8, so SP of 1 = 0.8. Loss = 1 - 0.8 = 0.2 = 20%.

Average = (65 + 75)/2 = $\frac{140}{2}$ = 70%.

Let number = 100. After 50% increase: 150. After 50% decrease: 150 × 0.5 = 75. Net change = (75-100)/100 × 100 = -25%.

New production = 5000 × (1 - 0.30) = 5000 × 0.70 = 3500 units.