Quantitative Aptitude - MCQ Practice Questions

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

​

Combined rate = $\frac{1}{6}$ + $\frac{1}{8}$ + $\frac{1}{12}$ = $\frac{4}{24}$ + $\frac{3}{24}$ + $\frac{2}{24}$ = $\frac{9}{24}$ = $\frac{3}{8}$. Time = $\frac{8}{3}$ ≈ 2.67 hours ≈ 2.4 hours (approx).

We need to calculate simple interest for both borrowers using the formula $SI=\frac{P\times R\times T}{100}$, then find the difference.

Step 1: Calculate interest paid by A

A borrowed Rs. 5000 at 8% per annum for 3 years.

Step 2: Calculate interest paid by B

B borrowed Rs. 5000 at 10% per annum for 2 years.

Step 3: Find the difference in interest

Answer: The difference in interest paid is Rs. 200 (Option A)

Train speed = $\frac{280}{14}$ = 20 m/s. In 26 seconds, train + bridge = 520m. Bridge = 520 - 280 = 240m.

[To keep expenditure constant when price increases, consumption must decrease proportionally.]

Step 1: [Set Up the Expenditure Equation]

[Expenditure is the product of price and consumption. Let initial consumption be C liters.]

Step 2: [Equate Expenditures to Find New Consumption]

[Since expenditure remains the same, we set both expressions equal.]

Step 3: [Calculate Percentage Reduction in Consumption]

[The reduction in consumption is the difference between original and new consumption, expressed as a percentage of original.]

The man should reduce his consumption by 20% to keep expenditure the same.

Answer: (B) 20%

Original value = Current/(0.9)³ = $\frac{900000}{0}$.729 ≈ 1234567 ≈ 1235000.

A's work in 5 days = $\frac{5}{15}$ = $\frac{1}{3}$. Remaining = $\frac{2}{3}$. Combined rate = $\frac{1}{15}$ + $\frac{1}{20}$ = $\frac{7}{60}$. Time for remaining = ($\frac{2}{3}$)/($\frac{7}{60}$) = ($\frac{2}{3}$) × ($\frac{60}{7}$) = $\frac{40}{7}$ ≈ 5.7 ≈ 6 days.

To find the net percentage change, we track how a number changes through two successive percentage operations using the compound percentage formula.

Step 1: Apply the 25% increase

Let the original number be $x$.

After increasing by 25%:

Step 2: Apply the 20% decrease

Now decrease this new value by 20%:

Step 3: Calculate the final value

Step 4: Find the net percentage change

**Answer: The net percentage change is $0\%$ (Option A)**

---

Assume the original number is 100.

After a 25% increase:

100×1.25=125

After a 20% decrease:

125×0.80=100

Final value =100, which is the same as the original value.

Therefore, the net percentage change is:

0%

​

Key insight: When we increase by 25%, we multiply by 1.25. When we decrease by 20%, we multiply by 0.80. Since $1.25\times 0.80=1.00$, the number returns to its original value.

Boys:Girls = 3:2. If boys = 45, then 3x = 45, x = 15. Girls = 2x = 30

SP = CP + Profit = 400 + (35% of 400) = 400 + 140 = Rs. 540

Let numbers be 5x and 8x. 5x + 8x = 130, 13x = 130, x = 10. Larger number = 8 × 10 = 80

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

SI = 1440 - 1200 = 240. Rate = (SI × 100)/(P × T) = (240 × 100)/(1200 × 2) = 10%

Speed = Distance/Time = $\frac{450}{5}$ = 90 km/h

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)