Quantitative Aptitude - MCQ Practice Questions

For SI, Amount = P(1 + RT/100). For doubling: 2P = P(1 + 8T/100). T = 12.5 years

Total distance = 200 + 300 = 500m. Speed = $\frac{500}{25}$ = 20 m/s = 20 × 3.6 = 72 km/h

Final price = 100 × 0.9 × 0.8 × 0.7 = 50.4. Discount = 49.6%.

A = 6000(1.08)³ = 6000 × 1.2597 ≈ 7558.27. CI ≈ 1558.27 ≈ 1548.

A = P(1.1)³ = P × 1.331. 13310 = P × 1.331. P = 10000.

When a man sells at a loss, the selling price is less than the cost price. Here we use the loss percentage formula: $\text{Selling Price}=\text{Cost Price}\times \left(1-\frac{\text{Loss\%}}{100}\right)$

Step 1: Identify given information

Step 2: Apply the loss formula

When there is a loss of 15%, the selling price is 85% of the cost price:

Step 3: Calculate the selling price

Answer: The selling price is Rs. 15300 (Option D)

Downstream speed = 15 km/h. Upstream speed = 10 km/h. Stream speed = (15-10)/2 = 2.5 km/h.

Total work = 15 × 20 = 300 worker-days. Workers needed = $\frac{300}{12}$ = 25.

Let CP = 100. MP = 150. For 25% profit, SP = 125. Discount = (150-125)/150 × 100 = 16.67%.

SI = (20000 × 10 × 2)/100 = 4000. CI = 20000(1.1)² - 20000 = 24200 - 20000 = 4200. Difference = 200.

Combined speed = 50 + 70 = 120 km/h. Time = $\frac{360}{120}$ = 3 hours.

Let salary = 100. After increase: 120. After decrease: 120 × 0.9 = 108. Net change = 8%.

This problem involves using the relationship between distance, speed, and time where the same distance is covered in different times at different speeds.

Step 1: Set Up Equations Using Distance Formula

Let the distance be d km and the required time be t hours. Since distance = speed × time, we can write two equations based on the given conditions.

Step 2: Solve for Time

Since both expressions equal d, we set them equal and solve for t.

Step 3: Calculate Distance

Substitute t back into either equation to find d.

The distance to the office is 7.5 km.

Let the distance to the office be d km.

Time taken at 5 km/h:

5

d

​

hours

Time taken at 6 km/h:

6

d

​

hours

The difference in arrival times is:

10 minutes late at 5 km/h

5 minutes early at 6 km/h

So, total difference =15 minutes =

60

15

​

=

4

1

​

hour.

Thus,

5

d

​

−

6

d

​

=

4

1

​

d(

5

1

​

−

6

1

​

)=

4

1

​

d(

30

6−5

​

)=

4

1

​

30

d

​

=

4

1

​

d=30×

4

1

​

=7.5

Therefore, the distance to the office is:

7.5 km

​

Let CP = 100. MP = 150. SP = 150 × 0.8 = 120. Profit = 20%

Work rate combined = $\frac{1}{12}$ + $\frac{1}{15}$ = $\frac{5}{60}$ + $\frac{4}{60}$ = $\frac{9}{60}$ = $\frac{3}{20}$. Time = $\frac{20}{3}$ ≈ 6.67 days

Upstream speed = $\frac{30}{5}$ = 6 km/h. Boat speed = Upstream speed + Stream speed = 6 + 2 = 8 km/h

To solve this problem, use the work rate method: find how much of the tank each pipe fills/empties per hour, then combine the rates.

Step 1: Find individual rates (fraction of tank per hour)

Pipe 1 fills the tank in 20 hours:

Pipe 2 fills the tank in 30 hours:

Drain empties the tank in 60 hours (negative rate):

Step 2: Find combined net rate

When all three are open, add the rates:

Step 3: Find common denominator and simplify

The LCD of 20, 30, and 60 is 60:

Step 4: Calculate time to fill one tank

If the net rate is $\frac{1}{15}$ tank per hour, then:

Answer: The tank fills in 15 hours (Option C)

(3x/4) ÷ ($\frac{2}{5}$) = (3x/4) × ($\frac{5}{2}$) = 15x/8

Total increase = (1.2 × 1.1 - 1) × 100 = (1.32 - 1) × 100 = 32%

Let stock = 100. 40 units at 10% profit = 44. 60 units at 5% loss = 57. Total SP = 101. Profit = 1%