Quantitative Aptitude - MCQ Practice Questions

Combined rate = $\frac{1}{12}$ + $\frac{1}{15}$ = $\frac{5}{60}$ + $\frac{4}{60}$ = $\frac{9}{60}$ = $\frac{3}{20}$. Time = $\frac{20}{3}$ = 6.67 hours

Distance covered = 250 + 150 = 400m. Speed = $\frac{400}{20}$ = 20 m/s = 20 × $\frac{18}{5}$ = 72 km/h

Downstream speed = 12 + 3 = 15 km/h. Time = $\frac{60}{15}$ = 4 hours

Let original number be x. x × $\frac{60}{100}$ = 300. x = 300 × $\frac{100}{60}$ = 500

Selling Price = 2500 × $\frac{92}{100}$ = Rs. 2300

Number = 45 × $\frac{100}{15}$ = 300. 40% of 300 = 300 × $\frac{40}{100}$ = 120

# Solution: Net Percentage Change After Two Operations

When a quantity undergoes successive percentage changes, we apply each change sequentially to find the overall effect.

Step 1: Represent the original number

Let the original number be $x$.

Step 2: Apply the 25% increase

After a 25% increase:

Step 3: Apply the 20% decrease

The decrease is applied to the new value $1.25x$, not the original:

Step 4: Calculate the net change

The final value equals the original value. Therefore:

Assume the original number is 100.

Increase by 25%:

100×1.25=125

Then decrease by 20%:

125×0.80=100

So, the final value is again 100.

Therefore, the net percentage change is 0%.

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

---

Key Insight: A 25% increase followed by a 20% decrease does NOT cancel out to 0% by simple arithmetic (25 − 20 = 5). Instead, multiplying the multipliers $(1.25\times 0.80=1.00)$ gives the true result because the 20% decrease applies to the larger amount.

SP = CP + Profit = 500 + (35% of 500) = 500 + 175 = Rs. 675

A's work rate = $\frac{1}{20}$, B's work rate = $\frac{1}{30}$. Combined rate = $\frac{1}{20}$ + $\frac{1}{30}$ = $\frac{5}{60}$ = $\frac{1}{12}$. Time = 12 days

Combined rate = $\frac{1}{15}$ + $\frac{1}{10}$ = $\frac{2}{30}$ + $\frac{3}{30}$ = $\frac{5}{30}$ = $\frac{1}{6}$. Time = 6 hours

Time = Distance/Speed = $\frac{300}{60}$ = 5 hours

Speed of current = (Downstream speed - Upstream speed)/2 = (12 - 8)/2 = 2 km/h

SI = (5000 × 12 × 3)/100 = 1800. Amount = 5000 + 1800 = Rs. 6800

To solve percentage problems, we first find the original number, then calculate the required percentage of it.

Step 1: Find the Original Number

Since 25% of the number equals 80, we set up an equation where the unknown number is x.

Step 2: Calculate 60% of That Number

Now that we know the original number is 320, we find 60% of it.

The answer is (A) 192

Discount = 30% of 1000 = 300. Selling price = 1000 - 300 = Rs. 700

When a number undergoes multiple percentage changes, we must apply each change sequentially to find the net effect.

Step 1: Apply the 40% Increase

Starting with an original number, a 40% increase multiplies it by 1.40.

Step 2: Apply the 30% Decrease

A 30% decrease means we retain 70% of the value, so we multiply by 0.70.

Step 3: Calculate Net Change

The final value is 0.98x, which is 98% of the original value x.

The net change is a 2% decrease. Answer: (B)

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)

SI = P × R × T / 100. 1200 = P × 8 × $\frac{3}{100}$. P = (1200 × 100) / 24 = 5000.

A = P(1 + R/100)^T = 12000(1.1)^2 = 12000 × 1.21 = 14,520.

Let CP = 100. MP = 160. After 25% discount, SP = 160 × 0.75 = 120. Profit = 120 - 100 = 20. Profit% = 20%.