Quantitative Aptitude - MCQ Practice Questions

Total distance = 200 + 100 = 300 m. Speed = 54 km/h = 15 m/s. Time = $\frac{300}{15}$ = 20 seconds

Total distance = 5×3 + 4×2 = 15 + 8 = 23 km. Total time = 5 hours. Avg speed = $\frac{23}{5}$ = 4.6 km/h

After 20% discount = 1000 × 0.80 = ₹800. After 10% discount = 800 × 0.90 = ₹720

SI₁ = (1000 × 10 × 3)/100 = 300. SI₂ = (2000 × 12 × 2)/100 = 480. Total = 300 + 480 = ₹780 (closest: should verify - 300+480=780, option A)

A's rate = $\frac{1}{30}$, B's rate = $\frac{1}{40}$. Combined rate = $\frac{7}{120}$. Work in 10 days = $\frac{70}{120}$. Remaining = $\frac{50}{120}$. B alone = ($\frac{50}{120}$)/($\frac{1}{40}$) = 16.67 days. (Re-check: Remaining work = 1 - 10($\frac{1}{30}$ + $\frac{1}{40}$) = 1 - 10×$\frac{7}{120}$ = $\frac{50}{120}$. Time for B = ($\frac{50}{120}$)÷($\frac{1}{40}$) = $\frac{50}{120}$ × 40 = 16.67 ≈ 16-17 days, check options)

SP = CP × 1.20. 480 = CP × 1.20. CP = $\frac{480}{1}$.20 = ₹400

Fill rate = $\frac{1}{10}$, Empty rate = $\frac{1}{15}$. Net = $\frac{1}{10}$ - $\frac{1}{15}$ = $\frac{1}{30}$. Time = 30 hours

Amount = P(1.10)³ = P × 1.331 = 1331. P = ₹1000

Let CP = 100. MP = 140. For 12% profit, SP = 112. Discount = 140 - 112 = 28. Discount % = $\frac{28}{140}$ × 100 = 20%

Let A, B, C complete work in a, b, c days respectively. 1/a + 1/b = $\frac{1}{12}$, 1/b + 1/c = $\frac{1}{15}$, 1/a + 1/c = $\frac{1}{20}$. Adding all three: 2(1/a + 1/b + 1/c) = $\frac{1}{12}$ + $\frac{1}{15}$ + $\frac{1}{20}$ = $\frac{12}{60}$ = $\frac{1}{5}$. So 1/a + 1/b + 1/c = $\frac{1}{10}$. Therefore, 1/a = $\frac{1}{10}$ - $\frac{1}{15}$ = $\frac{1}{30}$. A completes work in 30 days.

# 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