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%

Rate = ($\frac{1728}{1296}$ - 1) × 100 = (1.333 - 1) × 100 = 33.33% p.a.

When two trains move in opposite directions, their speeds add up. The total distance to cover equals the sum of their lengths.

Step 1: Find relative speed

Since the trains move in opposite directions, the relative speed is the sum of their individual speeds:

Step 2: Convert to m/s

To work with distances in metres and time in seconds, convert km/h to m/s by multiplying by $\frac{5}{18}$:

Step 3: Find total distance to cover

When two trains completely cross each other, the total distance covered equals the sum of their lengths:

Step 4: Calculate time

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

Answer: The trains take 9 seconds to completely cross each other. (Option A)

Let amount at 5% = x. Then (x × 5 × 3)/100 + ((4000-x) × 7 × 3)/100 = 820. Solving: x = 2500

Net rate = $\frac{1}{12}$ - $\frac{1}{15}$ = (5-4)/60 = $\frac{1}{60}$. Time = 60 hours

CP per unit = $\frac{40}{12}$ = $\frac{10}{3}$. SP per unit = $\frac{1}{5}$. Loss = ($\frac{10}{3}$ - $\frac{1}{5}$)/($\frac{10}{3}$) × 100 = 33.33%

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

Let initial salary = 100. After 15% increase = 115. After 10% decrease = 115 × 0.90 = 103.5. Net change = 3.5% increase

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]

SI = 50000×12×$\frac{2}{100}$ = ₹12,000. CI = 50000(1.12)² - 50000 = ₹12,720. Difference = ₹720.

Rate = $\frac{1}{20}$ + $\frac{1}{30}$ - $\frac{1}{60}$ = $\frac{3}{60}$ + $\frac{2}{60}$ - $\frac{1}{60}$ = $\frac{4}{60}$ = $\frac{1}{15}$. Time = 15 minutes.

Distance = 150 + 250 = 400m. Speed = $\frac{400}{20}$ = 20 m/s = 20×3.6 = 72 km/h.

Upstream speed = $\frac{24}{6}$ = 4 km/h. Downstream speed = $\frac{24}{3}$ = 8 km/h. Current = (8-4)/2 = 2 km/h.

MP = 80×1.25 = ₹100. SP = 100×0.9 = ₹90. Profit = $\frac{10}{80}$ × 100 = 12.5%.

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