CAT
Exam
Numerical Ability
Time and Work
Two pipes a and b can fill a tank in 15min and 20 min respectively. both the pipes are opended together but after 4 min.pipe a is turned off. what is the total time required to fill the tank
Read Solution (Total 3)
-
- a's 1 min work = 1/15
b's 1 min work = 1/20
a+b's 1 min work = 1/15 + 1/20 = 35/ 300 = 7/60
a+b's 4 min work = 4 * 7/60 = 7/15
Remaining tank yet to be filled = 1-7/15 = 8/15
B ll fill
1 min ---- 1/20
x --------8/15
so 8/15*20 = 32/3 min
So totally 4 + 32/3 = 44/3 = 14 2/3 mins ===> 14mins 40 sec
Ans : 14 min 40 secs - 10 years agoHelpfull: Yes(6) No(1)
- a's 1 min work will be 4
b's 1 min work will be 3
Total work will be 60
Now a+b work for 4 min so total work done = 7*4=28
so work left=60-28==>32
Now time taken by b to complete remaining work=32/3
which is 10 min 40 sec and previouslly they worked for 4 min
So total time will be [14min 40sec] - 10 years agoHelpfull: Yes(1) No(0)
- As both work for 4mins,work done by both for 4mins is 4(1/15 + 1/20)=28/60
Let X be the remaining time
only b worked for Xmin = (1/20)X
so, 28/60 + X/20 = 1
X/20=32/60
X=44/3 min = 14 min.40 secs.
- 10 years agoHelpfull: Yes(0) No(0)
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