Quantitative Aptitude - MCQ Practice Questions

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%.

Using compound growth: 100(1+r)³ = 160. (1+r)³ = 1.6. 1+r = 1.6^($\frac{1}{3}$) ≈ 1.1696. r ≈ 16.96% ≈ 16.5%.

Let n = previous innings. 40n + 65 = 45(n+1). 40n + 65 = 45n + 45. 20 = 5n. n = 4.

Total distance = 40 + 60 + 100 = 200 km. Time for first = $\frac{40}{20}$ = 2 hours. Time for second = $\frac{60}{30}$ = 2 hours. Time for third = $\frac{100}{50}$ = 2 hours. Total time = 6 hours. Average speed = $\frac{200}{6}$ = 33.33 km/h.

Let boys = 3x, girls = 2x. Total weight = (3x × 60) + (2x × 55) = 180x + 110x = 290x. Average = 290x / 5x = 58 kg.

To find the new average, we must first calculate the total score change when a student leaves and a new student joins.

Step 1: Find the Total Score of Original 5 Students

The average of 5 students is 78, so we multiply the average by the number of students to get the total.

Step 2: Calculate the New Total Score

When a student scoring 68 leaves, we subtract 68. When a new student scoring 88 joins, we add 88.

Step 3: Find the New Average

The number of students remains 5. Divide the new total by 5 to get the new average.

The new average is 82.

Answer: (B) 82

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%.

Sum of 4 numbers = 35 × 4 = 140. Sum of 3 numbers = 28 + 36 + 42 = 106. Fourth number = 140 - 106 = 34.

Average speed is the total distance traveled divided by the total time taken, calculated as $\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}$.

Step 1: Calculate total distance

The train travels in two segments:

Step 2: Calculate total time

The train travels for two different periods:

Step 3: Apply the average speed formula

Step 4: Simplify the division

Total distance traveled:

200+150=350 km

Total time taken:

4+2.5=6.5 hours

Average speed:

Average Speed=

Total Time

Total Distance

​

=

6.5

350

​

=

13

700

​

≈53.85 km/h

Therefore, the average speed is approximately 53.85 km/h.

**Answer: The average speed is approximately $53.85\text{ km/h}$ (Option C)**

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

Rate of worker 1 = $\frac{1}{8}$, worker 2 = $\frac{1}{12}$. Combined rate = $\frac{1}{8}$ + $\frac{1}{12}$ = (3+2)/24 = $\frac{5}{24}$. Time = $\frac{24}{5}$ = 4.8 days.

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%.

Total height of 5 = 165 × 5 = 825 cm. After adding 180 cm: 825 + 180 = 1005 cm. New average = $\frac{1005}{6}$ = 167.5 cm.

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.

Total income of 4 = 45000 × 4 = 180000. After 5th employee: 180000 + 55000 = 235000. New average = $\frac{235000}{5}$ = 47000.

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.

Sum of 5 numbers = 48 × 5 = 240. Sum of 4 numbers = 40 + 45 + 50 + 55 = 190. Fifth number = 240 - 190 = 50.

# Work Rate Problem Solution

[To solve tank filling problems, we must find each pipe's work rate (fraction of tank filled per hour), then track cumulative work.]

Step 1: Find Individual Work Rates

[Pipe A completes the tank in 20 hours, so its hourly rate is $\frac{1}{20}$ of the tank. Pipe B completes it in 30 hours, so its rate is $\frac{1}{30}$ of the tank.]

Step 2: Calculate Work Done in First 6 Hours (Both Pipes Working)

[When both pipes work together for 6 hours, we add their rates and multiply by time:]

Step 3: Find Remaining Work

[Half the tank is already filled, so the remaining work is:]

Step 4: Calculate Time for Pipe A Alone to Finish

[Only Pipe A continues at rate $\frac{1}{20}$ tank/hour. Using Time = Work ÷ Rate:]

Pipe A needs 10 hours to finish filling the tank.

Pipe A fills the tank in 20 hours.

So, work done by A in 1 hour:

20

1

​

Pipe B fills the tank in 30 hours.

So, work done by B in 1 hour:

30

1

​

Together, in 1 hour they fill:

20

1

​

+

30

1

​

LCM of 20 and 30 is 60:

=

60

3+2

​

=

60

5

​

=

12

1

​

So together they fill

12

1

​

of the tank per hour.

In 6 hours, they fill:

6×

12

1

​

=

12

6

​

=

2

1

​

So, half the tank remains.

Now only Pipe A works.

Pipe A fills

20

1

​

of the tank per hour.

Time to fill remaining

2

1

​

tank:

20

1

​

2

1

​

​

=

2

1

​

×20=10

Therefore, Pipe A will take:

10 hours

​

to finish filling the tank.

Answer: (D) 10 hours