Quiz 4

Let the capacity of the tank be C.
Rate of Pipe A = CACA
Rate of Pipe B = CBCB
Rate of Pipe A and Pipe B together = CACA + CBCB = C8C8
C12C12 + CBCB = C8C8
CBCB = C8C8 - C12C12 = C24C24
Pipe B independently will take 24 hours to fill the tank.

The question is In how many hours can pipe B fill the same tank independently?

Hence, the answer is 24 hrs.

G α No of workers²
G α √(No. of Machines)
G α No of hours
200 α 4² * √4 * 8
200 α 16 * 2 * 8
200 α 256
200 = k x 256
K = 200256200256
K = 25322532

The question is When 3 people work for 12 hours each with 9 machines, how many goods will be produced?

Hence, the answer is K = 25/32.

A can complete a task in 12 days and B can complete the task in 18 days. So,
In one day, A can do 112112, in one day B can do 118118.
In two days, they can complete 112112 + 118118 = which is 536536.
In 4 days, they can finish 10361036, and so on.
We can have 7 such sets of 2days each.
So, at the end of 14 days, they would have done 35363536 of task.
On the 15th day A would begin work with 136136of the task to finish. He can finish 112112 in a day. So, he would take one-third of a day. So, they can finish the whole task in 14.33 days.

The question is how many days do they finish the entire task?

Hence, the answer is "14.33 days".

Choice B is the correct answer.

Let time taken by A be 'a' hours and time taken by B be 'a+4' hours
Then A does 1a1a of the work in an hour.
B does 1a+41a+4 of the work in an hour.
Together they take 288 minutes to finish the job , 288 minutes = 2886028860 = 245245 hours.
Therefore, both A and B together complete 524524 every hour.
1a1a + 1a+41a+4 = 524524
2a+4a(a+4)2a+4a(a+4) = 524524
We get, 48a + 96 = 5(a² + 4a
=> 5a² - 28a - 96 = 0
=> 5a² - 40a + 12a - 96 = 0
5a (a - 8) + 12(a - 8) = 0
(5a + 12)(a - 8) = 0 . Therefore, Since a cannot be negative, a = 8 hours.
Hence, a + 4 = 12 hours. Therefore, Time taken by B to complete the work on his own is 12 hours.

The question is  how long does B alone take to complete the task?

Hence, the answer is "12 hrs".

Time taken to fill the tank completely without leakage = 1(1/12)+(1/18)1(1/12)+(1/18) hrs
= 1(9+6)/1081(9+6)/108 = 1081510815 = 7 hours and 12 minutes
Given, it takes 48 minutes extra with the leakage. So, total time taken = 8 hours
= 1(1/12)+(1/18)−(1/x)1(1/12)+(1/18)−(1/x) = 8 (Where, x is the time taken by the leakage to completely empty the tank)
= 112112 + 118118 - 1818 = 1x1x
= (9+6−13.5)108(9+6−13.5)108 = 1x1x
= x = 1081.51081.5 = 72 hours

The question is what time will it take to completely empty due to the leakage.

Hence, the answer is "72 hrs".

Initially, net flow in the tank = 40 litres/ 8 minutes = 5 litres/minute
Now, the increase in outflow should be greater than 5 litres/min for the tank to fill up.
Increase in outflow = 1.5x – x = 0.5x
Only in option C, 0.5 * 12 = 6 litres/min > 5 litres/min

The question is  Which of the following can be the outflow rate?

Hence, the answer is "12 litres/minute".

The 475 litres is filled in 19 hours by the two pipes when they are opened simultaneously.
Therefore, rate of water flow by A and B,
A + B = 4751947519 = 25 litres/hour………….(1)
Let pipe B was opened for t hours
Then, pipe A was opened for t + 5 hours
A x (t+5) + B x t = 475
(A + B) x t + 5A = 475
25t + 5A = 475
=> 5t + A = 95
Now, using option analysis
a) t = 10 => 5 x 10 + A = 95 => A = 45 (not possible from (1))
b) t = 14 => 5 x 14 + A = 95 => A = 25 (not possible from (1) as B can’t be zero)
c) t = 16 => 5 x 16 + A = 95 => A = 15

The question is  if the two pipes were open simultaneously the tank would fill in 19 hours.

Hence, the answer is "16 hrs".

Given, 295 m was mined out of which first 5 days – planned. Using option analysis:
a) 30 * 5 + (30 + 5) * (3003030030 – 5 – 4) = 150 + 35 * 1 = 185 ≠ 295
b) 12.5 * 5 + 17.5 * (30012.530012.5 – 5 – 4) = 62.5 + 17.5 * 15 = 325 ≠ 295
c) 15 * 5 + 20 * (3001530015 – 5 – 4)1 = 75 + 20 * 11 = 295

The question is How many meters of mining was initially planned for each day?

Hence, the answer is "15 m".

Suraj took 23 minutes to finish the remaining work = 23402340
Work finished by Ajay and Suraj together = 17401740
Time taken = 17/40(1/45)+(1/40)17/40(1/45)+(1/40)= 178/9+1178/9+1 = 1717/91717/9 = 9 mins

The question is how many minutes had Ajay left?

Hence, the answer is "9 min".

Let the work done by Ahmed be 60 (LCM of 2,3,4,5)
So, work done per hour by:
Ahmed – 602602 = 30
Bibhor – 603603 = 20
Bibhor in 4 hours – 20 x 4 = 80
Deepak – 805805 = 16
Therefore, total work done in an hour = 30 + 20 + 16 = 66
If all 3 work for 20 hours, total work done = 66 * 20
If Bibhor alone has to work, time taken to do 66 * 20 of work = 66 * 20202020 (work done by Bibhor in 1 hour)
= 66 hours

The question is What time would Bibhor take to fill the tank alone?

Hence, the answer is "66 hrs".