Quantitative Aptitude - MCQ Practice Questions

Upstream speed = $\frac{50}{5}$ = 10 km/h. Downstream speed = $\frac{80}{4}$ = 20 km/h. Boat speed in still water = (10 + 20)/2 = 15 km/h.

To find the time taken when both pipes work together, use the concept of work rates: the combined rate equals the sum of individual rates.

Step 1: Find individual work rates

Pipe A fills the tank in 12 hours, so its rate is $\frac{1}{12}$ tank per hour.

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

Step 2: Find combined work rate

When both pipes work together:

Find the LCM of 12 and 15, which is 60:

Step 3: Calculate time to fill one tank

If the combined rate is $\frac{3}{20}$ tank per hour, then time to fill 1 tank is:

Step 4: Convert to hours and minutes

**Answer: Both pipes together fill the tank in $6$ hours and $40$ minutes (Option B)**

For consecutive odd numbers, the average equals the middle (4th) number. So 4th number = 39. The 7 numbers are: 33, 35, 37, 39, 41, 43, 45. Largest = 45.

Distance in first 2 hours = 50 × 2 = 100 km (which is 25% of total). Total distance = 400 km. Total time at 60 km/h = $\frac{400}{60}$ = 6.67 hours.

Interest on first = 5,000 × 8 × $\frac{2}{100}$ = 800. Interest on second = 7,500 × 12 × $\frac{2}{100}$ = 1,800. Total interest = 2,600. Average rate = (2,600 × 100) / (12,500 × 2) = 10.4%.

Increase in total age = 38 - 28 = 10. If average increases by 2, then group size = $\frac{10}{2}$ = 5.

When equal quantities are mixed, average concentration = (40 + 50 + 60)/3 = $\frac{150}{3}$ = 50%.

CP of 3 items at 20% profit: 3 × ($\frac{200}{1}$.2) = 500. CP of 2 items at 25% loss: 2 × ($\frac{300}{0}$.75) = 800. Total CP = 1300, Total SP = 1500. Profit% = ($\frac{200}{1300}$) × 100 = 15.$\frac{38}{3}$.58 ≈ 4.29%.

Initial average age × 6 = Total. New average = (Initial average + 2). Change in total = 6 × 2 = 12. New person's age = 40 + 12 = 52 years.

CP = 100 × 50 = 5000. SP of 60 at 20% profit = 60 × 50 × 1.2 = 3600. SP of 40 at 10% loss = 40 × 50 × 0.9 = 1800. Total SP = 5400. Profit = 400. Profit% = ($\frac{400}{5000}$) × 100 = 8%.

For equal distances, average speed = 3/($\frac{1}{40}$ + $\frac{1}{50}$ + $\frac{1}{60}$) = 3/((3+2.4+2)/120) = 3/(7.$\frac{4}{120}$) = $\frac{360}{7}$.4 = 48.65 ≈ 49.09 km/h. Using LCD: 3/((15+12+10)/600) = $\frac{1800}{37}$ ≈ 48.65 km/h.

Let current speed = c. Upstream speed = 15 - c, downstream = 15 + c. Distance = 4(15-c) = 2(15+c). 60 - 4c = 30 + 2c. 30 = 6c. c = 5 km/h.

Combined rate = $\frac{1}{30}$ + $\frac{1}{20}$ = $\frac{5}{60}$ = $\frac{1}{12}$. Time to fill full tank = 12 hours. Time to fill $\frac{2}{5}$ tank = 12 × $\frac{2}{5}$ = 4.8 hours. Average per pipe = 4.8 × 1.5 = 7.2 hours.

For equal distances, average speed = 2ab/(a+b) = (2 × 8 × 12)/(8 + 12) = $\frac{192}{20}$ = 9.6 km/h.

Let total distance = 400 km. Distance segments: 100 km at 50 km/h (2 hrs), 200 km at 60 km/h (3.33 hrs), 100 km at 75 km/h (1.33 hrs). Total time ≈ 6.66 hrs. Average speed = $\frac{400}{6}$.66 ≈ 60.8 km/h.

In ratio 3:2, total parts = 5. Cost = (3 × 40 + 2 × 60)/5 = (120 + 120)/5 = $\frac{240}{5}$ = ₹48 per kg.