Practice Test: Time & Work- 2 - UPSC MCQ

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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

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.

(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.

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%.

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. 

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.

To maximize the production, four men will work in each shift. 2 men will work with machines and 2 men work alone.
Total cost incurred in one hour

Assume total amount of work = 100 units.
A does = 10 units/day, B does = 5 units/day, and C does = 4 units/day
Possible Pairs:
A+B = 15 units/day, A+C = 14 units/day, B+C = 9 units/day
To minimize time, we will use the first two pairs.
So, 15 +14 +15 + 14 + 15 +14 + 15 = 102 units.
So, little less than 7 days are required

Hence, the correct option is 'B'.

Let Anuj do the work in x days.

 So, Bhanu = 3x days, Chandu = 9x days and Dodo = 27x days

Now, using options,

Anuj + Chandu = 1/x + l/9x = 10/9x,

So, they take 9x/10 days

Bhanu + Dodo = l/3x + 1/27x = 10/27x, So, they take 27x/10 days

Now, 1/3 of 27x/ 10 = 9x /10.

 So, Anuj and Chandu is first pair.

Answer: Option D

Explanation :Time to completely fill the tank by the two pipe:

1/20 + 1/30= 1/n 

⇒ n = 12 hours

So, 3/4^th of the tank will be filled in 3/4 × 12 = 9 hours.

Remaining time = 12 – 9 = 3 hours.

But, for the remaining 1/4^th of the tank, the combined efficiency drops to 3/4^th (1/4^th is getting leaked), 

∴ Time required will be come 4/3 times, i.e. 4/3 × 3 = 4 hours.

Hence, total time taken to fill the tank = 9 + 4 = 13 hours.

Hence, option (d).

Answer: Option A

Explanation :We know, 

Time ∝ 1/no. of pumps

∴ Time × (Number of pumps) = constant.

⇒ 20 × 30 = 15 × x

⇒ x = 20 × 30/15 = 40 mins

Hence, option (a).

Let eating capacity of each soldier initially = 1 unit/day.

Initial food supply = 200 × 1 × 40 = 8000 units.

Supply left after 10 days = 200 × 1 × 30 units.

Let ‘s’ numbers of soldiers joined the camp.

New capacity of each soldier = 1.5 units/day

∴ The remaining supply is consumed by (200 + s) solders in another 10 days.

∴ 200 × 1 × 30 = (200 + s) × 1.5 × 10

⇒ 400 = 200 + s

⇒ s = 200

Hence, option D.

Explanation :Let the total work to be done = 60 units

∴ Efficiency of a man = 60/20 = 3 units/day
∴ Efficiency of a woman = 60/30 = 2 units/day
∴ Efficiency of a boy = 60/60 = 1 units/day

Let x boys assist 4 men and 5 women.

∴ (4m + 5w + xb) × 2 = 60

⇒ (12 + 10 + x) = 30

⇒ x = 8

Hence, option (D).

Let the time taken by Anu, Tanu and Manu be 5x, 8x and 10x hours.

Total work = LCM(5x, 8x, 10x) = 40x

Anu can complete 8 units in one hour

Tanu can complete 5 units in one hour

Manu can complete 4 units in one hour

It is given, three of them together can complete in 32 hours.

32(8 + 5 + 4) = 40x

x = 685568​

It is given,

Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours

40(8 + 5) + y(4) = 40x

4y = 24

y = 6

Manu alone will complete the remaining work in 6 hours.
Option C