Test: Time & Work (March 5) - CAT MCQ

To solve this problem, let's first find out the total time Anoop takes to read all four passages and answer all the questions.

Let the time he takes to read one passage be T. Since there are four passages, he takes 4T time to read all the passages. It is given that he can answer 12 questions in the time he takes to read one passage. So, the time he takes to answer one question is T/12.

There are a total of 5+8+8+6 = 27 questions. The time he takes to answer all the questions is 27 * (T/12) = 27T/12 = 9T/4.

Now, the total time spent on the section is the sum of the time spent on reading all the passages and answering all the questions: 4T + 9T/4 = 25T/4.

To cut down on his total time spent on the section by 20%, the new total time should be 80% of the original time, which is 0.8 * (25T/4) = 5T.

Since the time spent on answering the questions remains constant, the time spent on reading should reduce to 5T - 9T/4 = 11T/4. The new time he takes to read one passage is (11T/4) / 4 = 11T/16.

Now, let's find out the percentage increase in reading speed. The original time to read one passage is T, and the new time is 11T/16. Since speed is inversely proportional to time, the new speed will be 16/11 times the original speed.

The percentage increase in speed is [(16/11 - 1) * 100] = [(5/11) * 100] = 45.45%.

So, Anoop should increase his reading speed by 45.45% to cut down on his total time spent on the section by 20%.

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Tap A can fill the tank in 25 hrs.

Tap B cab fill the tank in 20 hrs.

Tap C can fill the tank in 10 hrs.

Total volume of the tank filled by 3 pipes = LCM of (25, 20, 10) = 100 units

⇒ Pipe A can fill 4 units of water in 1 hr.

⇒ Pipe B can fill 5 units of water in 1 hr.

⇒ Pipe C can fill 10 units of water in 1 hr.

⇒ Pipe (A + B + C) can fill 19 units of water in 1 hr.

According to question,

Pipe (A + B + C) are opened for 1 hr, then tap C is closed.

⇒ 19 units of water are filled in the tank.

After 4^th hr, tap B is also closed

⇒ for 3 hrs, pipe A and B are open

⇒ (4 + 5) × 3 = 27 units of water are filled.

⇒ 100 - (19 + 27) = 54 units of water are filled by tap A alone.

Total units of water filled by tap A is 4 × 1 + 4 × 3 + 54 = 4 + 12 + 54 = 70 units

∴ Percentage of work done by tap A = 70%

Let us assume that digging one well = 40 unit

Efficiency of Anup = 40/10 = 4 unit/day

Now efficiency is falling by 10%

So, this is case of GP

⇒ 4,18/5,81/25,…… where common ratio = 9/10

Now, sum of infinite GP = a / (1 – r)

⇒ 4/ (1 – 9/10) = 40

Hence, it is clear that Anup will takes infinite time, so he will never finish the work.

Since the amount of water flowing through each pipe is proportional to square of its diameter so if efficiency of longest pipe (3 cm) = 1/49

Then efficiency of pipe (2 cm) = 4/(49 x 9)

and efficiency of pipe (1 cm) = 1/ (49 x 9) 

Now let cistern is filled by all three pipes in x minutes.

(A+B)'s one day's work = 1/3 part
(A+B) works 2 days together = 2/3 part
Remaining work = 1−2/3 = 1/3 part

1/3 part of work is completed by A in two days
Hence, one day's work of A = 1/6
Then, one day's work of B = 1/3−1/6 = 1/6
So, B alone can complete the whole work in 6 days.

A, B and C complete a job in 30 days working together,

⇒ 1/A + 1/B + 1/C = 1/30

A can work twice as fast as B,

⇒ 1/B = 1/2A

A and C together can work thrice as fast as B,

⇒ 1/B = 1/3(1/A + 1/C)

Solving,

⇒ 3/B = 1/A + 1/C

⇒ 3/B = 1/30 – 1/B

⇒ 1/B = 120

Then,

⇒ 1/A = 1/60

⇒ 1/C = 1/120

∴ A, B and C can complete work in 60, 120 and 120 days respectively. 

Let us assume that, first three pumps fills the tank in x hours .

so,

→ Efficiency of each pump = (1/x) m³ / hour .

then,

→ Efficiency of three pump = (3/x) m³ / hour .

now,

→ First three pumps works for = 2.5h + 1h + 1h = 4.5 hours.

so,

→ Water filled by 3 pumps in 4.5 hours = 4.5 * (3/x) = (13.5/x) m³ .

now, given that,

→ Time taken by additional pump to fill the tank = 40 hours.

so,

→ Efficiency of 2 additional tanks = 2 * (1/40) = (1/20) m³ / h .

and,

→ Additional pumps work for = 1 + 1 = 2 hours.

so,

→ Water filled by additional pumps in 2 hours = 2 * (1/20) = (1/10) m³ .

therefore,

→ (13.5/x) + (1/10) = 1

→ (13.5/x) = 1 - (1/10)

→ (13.5/x) = (9/10)

→ x = 135/9 = 15 hours.

since given that, in last 1 hour they filled 15 m³ .

hence,

→ 3 * (1/15) + (1/20) = 15 m³

→ (1/5) + (1/20) = 15

→ (4 + 1)/20 = 15

→ (5/20) = 15

→ (1/4) = 15

→ 1 = 60 m³ (Ans.) (Option A)

Total time required to fill the tank: 1/10 + 1/20 + 1/25 - 1/100 = 18/100
Time required to fill the tank in 1hr = capacity/total time
= 50000*18/100
= 9000litre