DevSquare Quantitative Aptitude Time and Work Questions

sita 1 day sepend time =40 min
sita 20 days spend time =20*40min
=800 min
so, convert it into hours= 800/60 hrs
13.33hrs

Time taken by both pipes together to fill the tank = 1/(1/40 +1/30) = 120/7 min

3men 8days -->1 man=1/24 days(1/3*1/8)
4boys 7days-->1 boy=1/28 days
2men+2boys==2/24+2/28-->1/12+1/14
2 such walls-->1/24+1/28
take LCM, 13/168
reciprocal of it, 168/13
i.e,12 12/13

10 days

1/A + 1/B = 2/15 or (A+B)/AB= 2/15 ----(i)

(1/2)/(1/A) + (1/2)/(1/B) = 20 or A/2 + B/2 = 20 or A+B= 40 ----(ii)
Substituting A+B= 40 from (ii) in (i), we get AB= 300
Now finding the factors of 300, whose sum is 40, we have 10 and 30
As B is more efficient than A, B can do the work in 10 days.

Inlet pipe fills the tank in 5 hr and outlet pipe empties the tank
in 36 hr.
So inlet pipe is 36/5 = 7.2 times efficient than , ensure that the
an outlet pipe.
Therefore , in order to tank never overflows , we will need total
8 outlet pipes.
Already we have 1 outlet pipe , thus we need only (8-1) = 7 outlet pipes

Consider this way....

Shanti walked and as a result she saved the total driving hours= 1
Which means her mom saves half an hour each going up and
down since Santi met her on the way.
That means both
met at 3.30 pm. So Shanti walked what the distance in 2.5 hrs
(from 1 pm to 3.30 pm) what her mom would have driven for half an hour.

So the ratio of their speeds are 0.5: 2.5 i.e. 1:5.

The clue lies in considering the to and fro distance.

Hope it is clear now.
🙂

80

Here if number of cows decrease by 10 has increasing the
number of days by 20

therefore,
40cows for 40days
30cows for 60days
20cows for 80days

Ans: 80

or you can solve by ratio
40C : 40D --> 1:1
30C : 60D --> 1:2
20C : 80D --> 1:4

ans:80

Ans : 15days
For Skilled
5 workers --------20 days
5 workers 1 day work= 1/20
1 worker's 1 day work= 1/(5*20)

Similarly,
For Semi-Skilled------
1 worker's 1 day work= 1/(8*25)

For Unskilled------
1 worker's 1 day work= 1/(10*30)

For 2 skilled,6 semi-skilled and 5 unskilled
workers
One day work= 2*[1/(5*20)] +6*[1/(8*25)] + 5*[ 1/(10*30)] = 1/15
Therefore no. of days taken= 15

Starting homework at 12 PM and Taking a break at 3:30 PM,
we came to know that each of them have already spent
3 hrs and 30 minute i.e 210 min
After taking break, they started to do their homework at 4 PM
1. Rajesh can finish 1/5 of his work in 1 hr. i.e in 60 min
He can finish 1 work in 60* 5 = 300 min
Remaining time to complete his hm work = 300-210= 90 min
Starting again at 4 PM,he will finish his homework at 5:30 PM
2. Sema can finish 3/7 of her homework in 90 min
She will finish 1 work in 90*(7/3) = 210 min
Therefore she finished her work at 3:30 PM
3.
Ramya can finish 3/4 of her work in 210 min

She will finish 1 work in 210*(4/3 ) min= 280 min

Remaining time left for Ramya to complete her work =280 - 210 = 70 min
Starting to do homework again at 4 PM,she will complete her work at
5:10 PM
Since Rajesh takes longer time to finish the work
(completing the work at 5:30 PM), hence they all can start
to play at 5:30 PM. That  the only time they all will meet

the answer is approximately 230 gallons and if u want exact then it comes 228.25

explanation is given below:
3 person together fill tank in 25 min hence, calculate the time of person C that in how much time he empty the tank so,

(1/A)+(1/B)-(1/C)=1/25

i.e, (1/30)+(1/35)-(1/C)=1/25 .......... we get,

1/C=23/1050,

while, its given the C can empty 5 gallon per min. therefore

capacity of tank = 5*1050/23

and capacity of tank will get if we solve above expression,
we get 228.25