Practice Test: Time & Work- 2 - CAT MCQ

 .

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%

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. 

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