Latest Elitmus Aptitude Question SOLUTION: 4 pipes can fill a reservoir in 20,30,40 and 60 hrs respectively. The first was opened at 6 am , second at 7 am, third at 8 am and fourth at 9 am. When will the reservoir be filled

let reservoir will fill in x hours
(x)/20+(x-1)/30+(x-2)/40+(x-2)/60=1
x*1/8=17/15
x = (8*17)/15 = 136/15 = (9+1/15)hr = 9 hr+4 Min
so resevior will fill 6am + 9 hr+4 Min = 3:04 pm

6 am- Empty tank
7 am = 1/20 full
8 am= 1/20+(1/20+1/30)
=1/15 full

9 am= 1/15+(1/20+1/30+1/40)
=7/40 full

10 am=7/40+(1/20+1/30+1/40+1/60)
7/40+(1/8) since 4 pipes can fill 1/8 part of the tank in 1 hr
=3/10 full

so d tank to be filled is 7/10

7/10*8/1= 28/5= 5 hrs 36 mins


so the tank ll be filled at 3.36pm

let tank will fill in n hours
n/20+(n-1)/30+(n-2)/40+(n-2)/60=1
n*1/8=34/30
n=9.066 hours
the tank will filled 9+6=15 means 3pm ans...

6am-7am--->one in on.amount filled,so=1/20.
7am-8am--->two in work.amount filled=2*20+1/30.
8am-9am--->three in work.amount filled=3*20+2/30+1/40.
9am-10am-->four in work.amount filled=4*20+3/30+2/40+1/60.
adding them up gives 22/60,that in four hrs.
in one hr,(1/20)+(1/30)+(1/40)+(1/60) is filled and it equals to 5/8.

so,
22/60=amount filled in 4hrs,5/8 filled in 1 hr.

22/60+x(5/8)=1,
x(5/8)=1-22/60=38/60,
x=(38/60)/(5/8)=1.0133

so totally 4+1.0133 hr=5.0133 hrs.

1st can fill 1/20 in 1hr
two can fill 1/12 in 1hr
three can fill 13/120 in 1 hr
all four can fill 15/120 in 1 hr

after 9am reservoir left 1-1/20-1/12-13/120
= 81/120
and now 81/120 part of reservoir can fill by all pipes in 5.4hr

so tank can fill fully in 8.4 hr

@Sunil Kumar Banjara: the equation would be something like this:

(x)/20+(x-1)/30+(x-2)/40+(x-3)/60=1... you wrote it as (x)/20+(x-1)/30+(x-2)/40+(x-2)/60=1

@roshan jha is correct it should be (x-3/60) rather than (x-2/60) @sunil kumar banjara....

Imagine 4 pipes as
A=20
B=30
C=40
D=60
total work/LiterS of water (LCM OF ALL)=120

Then A can do 6 work in 1 hour i.e 120/20=6
similarly B=4 ,ie 120/30
C=3, ie 120/40
D=2 ,ie 120/60

image time line as
6am 7am 8am 9am 10am 11am 12 am 1pm 2pm 3pm
A's work 6 , 6 6 6 6 6 6 6 6
B's work 4 4 4 4 4 4 4 4
C 3 3 3 3 3 3 3
D 2 2 2 2 2 2
-------------------------------------------------------------------------------------------------------------------------------------------------
total 6 10 13 15 15 15 15 15 15...seriesGo on
Total=6+10+13+15+15+15+15+15+15=119 liters at 3pm
but LCM earlier=120 liters
1 hour = 15 liter
1 minutes=60/15=4
so the answer will be 3:04pm

let the time be t hours and the first tap was opened
in t hours first pipe fills up t*1/20 part....
4t+3(t-1)+2(t-2)+(t-3)=60
t=7hrs
t/20+(t-1)/30+(t-2)/40+(t-3)/60=89/120=1:34pm

Frm 6 to 9 a hs wrkd 3hrs hence a wrkd 3/20 similarly b=2/30 and c=1/40 so at 9 the wrk remaining =1-3/20-2/30-1/40=87/120
Aftr 9 all will wrk together a+b+c+d=1/20+1/30+1/40+1/60=15/120
So time taken aftr 9am=(87/120)×(120/15)=3hrs48min therefore at time (9+3)hr 48 min

Ans: 3:39 pm

let reservoir will fill in x hrs
(x)/20 + (x-1)/30 + (x-2)/40 + (x-3)/60=1
solving we get, x= 136/15 = 9+1/15 hrs = 9 hours and 1/15*60 min = 9 hr+4min
start at 6am + 9 hr and 4 mins = 3:04pm

1st pipe 1 hr work=1/20 ,2nd pipe 1hr work=1/30,3rd pipe 1 hr work=1/40,4th pipe 1 hr work=1/60
(1st +2nd+3rd+4th) 1 hr work=1/20 + 1/30 + 1/40+1/60=>1/8
now, work done by every pipe before 9 am =3/20+2/30+1/40+0
=>29/120
work left=1-29/120=>91/120
so, (1st + 2nd +3rd+4th) pipe complete wok=>1 in 8 hr
so ,(1st + 2nd +3rd+4th) pipe complete wok=>91/120 in (91*8)/120 hr=>6hr 4 min
so required time=3:04 pm