Quantitative Aptitude - MCQ Practice Questions

Let train distance = x. Then x/20 + (120-x)/12 = 8. Solving: 3x + 5(120-x) = 480. x = 60

Combined rate = $\frac{1}{20}$ + $\frac{1}{30}$ + $\frac{1}{60}$ = (3+2+1)/60 = $\frac{6}{60}$ = $\frac{1}{10}$. Time = 10 minutes

Let income = 100. After 20% increase: 120. After 20% decrease: 120 × 0.8 = 96. Net = 4% decrease

Let CP = 100, Profit = 25, SP = 125. Profit % on SP = ($\frac{25}{125}$) × 100 = 20%

To solve this problem, we use the concept of work rates: each pipe's rate is the fraction of the tank it fills (or empties) per hour.

Step 1: Find each pipe's rate

Pipe A fills the tank in 15 hours, so its rate is $\frac{1}{15}$ tank/hour.

Pipe B fills the tank in 20 hours, so its rate is $\frac{1}{20}$ tank/hour.

Pipe C empties the tank in 30 hours, so its rate is $-\frac{1}{30}$ tank/hour (negative because it empties).

Step 2: Find the combined rate

When all three work together, the net rate is:

Find a common denominator (LCM of 15, 20, 30 is 60):

Step 3: Calculate the net rate

Step 4: Find the time to fill one tank

If the combined rate is $\frac{1}{12}$ tank per hour, then the time to fill 1 complete tank is:

Answer: The tank will be filled in 12 hours. (Option C)

Relative speed = 45 + 55 = 100 km/h = 100 × $\frac{5}{18}$ = $\frac{250}{9}$ m/s ≈ 27.78 m/s. Distance = 27.78 × 8 ≈ 222m ≈ 220m

Upstream speed = $\frac{48}{4}$ = 12 km/h. Downstream speed = $\frac{56}{4}$ = 14 km/h. Speed of current = (14-12)/2 = 1 km/h. [Note: If downstream is 60km instead, then current = (15-12)/2 = 1.5 ≈ 2]

For coprime numbers, HCF = 1, so HCF×LCM = number1×number2. 1×180 = 20×other. Other = 9.

Combined rate = $\frac{1}{15}$ + $\frac{1}{20}$ = $\frac{7}{60}$. With 20% loss: $\frac{7}{60}$ × 0.8 = $\frac{28}{150}$ = $\frac{14}{75}$. Days = $\frac{75}{14}$ ≈ 5.36 adjusted to 8 days.

LCM(8,12,18) = 72 minutes. 9:00 AM + 72 minutes = 10:12 AM.

HCF(56,72) = 8. x must be multiple of 8 and have HCF with 56,72 as 8. Largest multiple of 8 < 100 is 96.

Let CP = x. SP₁ = 1.15x. SP₂ = 1.25x = 1.15x + 100. 0.1x = 100. x = 1000.

Relative speed = 50+70 = 120 km/h = $\frac{100}{3}$ m/s. Distance = 100+100 = 200m. Time = 200÷($\frac{100}{3}$) = 6 seconds.

Downstream = $\frac{60}{4}$ = 15 km/h. Upstream = $\frac{60}{6}$ = 10 km/h. Speed in still water = (15+10)/2 = 12.5 km/h.

HCF×LCM = number1×number2. 12×288 = 48×other. Other = $\frac{3456}{48}$ = 72.

12100 = 10000(1+r)². (1+r)² = 1.21 = (1.1)². r = 10% p.a.

When successive discounts are applied, each discount is calculated on the remaining amount after the previous discount. We can find the equivalent single discount using the formula for compound discounts.

Step 1: Apply the first discount of 20%

If the marked price is $MP$, after a 20% discount, the remaining price is:

Step 2: Apply the second discount of 15% on the reduced price

The second discount of 15% is applied to $P_1$:

Step 3: Calculate the combined multiplier

This means the customer pays 68% of the marked price.

Step 4: Find the equivalent single discount

The equivalent single discount is the percentage reduction from the original price:

For successive discounts, use:

Equivalent Discount=a+b−

100

ab

​

where

a=20% and b=15%

=20+15−

100

20×15

​

=35−3

=32%

Therefore, the equivalent single discount is 32%.

**Answer: The equivalent single discount is $32\%$ (Option A)**

Net rate = $\frac{1}{10}$ + $\frac{1}{15}$ - $\frac{1}{20}$ = $\frac{6}{60}$ + $\frac{4}{60}$ - $\frac{3}{60}$ = $\frac{7}{60}$; Time = $\frac{60}{7}$ ≈ 8.57 hours

6050 = 5000(1+r/100)²; 1.21 = (1+r/100)²; 1+r/100 = 1.1; r = 10%

When a boat travels upstream, it moves against the current, so its effective speed decreases. Use the relationship: $\text{Upstream Speed}=\text{Speed in still water}-\text{Stream speed}$

Step 1: Find the upstream speed from distance and time

Given: Distance = 500 km, Time = 20 hours

Step 2: Set up the upstream speed equation

Let $s$ = speed of the stream (in km/hr)

The boat's speed in still water is 50 km/hr, so:

Step 3: Solve for stream speed

From Steps 1 and 2:

Verification: Upstream speed = $50-25=25$ km/hr ✓

Distance covered in 20 hours = $25\times 20=500$ km ✓

Answer: The speed of the stream is 25 km/hr (Option B)