Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 16)
16.
X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
Answer: Option
Explanation:
| Work done by X in 4 days = |  |
1 | x 4 |  |
= | 1 | . |
| 20 | 5 |
| Remaining work = | |
1 - | 1 | |
= | 4 | . |
| 5 | 5 |
| (X + Y)'s 1 day's work = | |
1 | + | 1 | |
= | 8 | = | 2 | . |
| 20 | 12 | 60 | 15 |
| Now, | 2 | work is done by X and Y in 1 day. |
| 15 |
| So, | 4 | work will be done by X and Y in | |
15 | x | 4 | |
= 6 days. |
| 5 | 2 | 5 |
Hence, total time taken = (6 + 4) days = 10 days.
Discussion:
68 comments Page 2 of 7.
Shubham said:
5 years ago
@All.
Simply, the solution is
X takes 20days.
Y takes 12 days.
20 - 60 - 3
12 - 60- 5
So consider X makes 3 chairs and y makes 5 in one day. X worked for 4 days so (3*4) = 12,
So remained chairs 60 - 12 = 48.
Now x+y =3+5=8 so this will complete in (48/8) = 6,
So total 6 + 4 = 10 days.
Simply, the solution is
X takes 20days.
Y takes 12 days.
20 - 60 - 3
12 - 60- 5
So consider X makes 3 chairs and y makes 5 in one day. X worked for 4 days so (3*4) = 12,
So remained chairs 60 - 12 = 48.
Now x+y =3+5=8 so this will complete in (48/8) = 6,
So total 6 + 4 = 10 days.
(5)
Larisa said:
2 years ago
EFFICIENCY:
X = 20 = 3 > 60 (total work)
Y = 12 = 5 >60 (total work).
X started the work & work for 4 days with 3 ( per day work done).
4×3 = 12 ( work done by X in 4 days).
60 - 12 = 48 work remains.
48 ÷ 8 ( 5+3 * & y per day efficiency),
= 6 days + 4 days = 10 days.
X = 20 = 3 > 60 (total work)
Y = 12 = 5 >60 (total work).
X started the work & work for 4 days with 3 ( per day work done).
4×3 = 12 ( work done by X in 4 days).
60 - 12 = 48 work remains.
48 ÷ 8 ( 5+3 * & y per day efficiency),
= 6 days + 4 days = 10 days.
(8)
Sonu jangra said:
10 years ago
x can do a work in 20 days.
y can do a work in 12 days.
To find total work, take LCM of 20 and 12 i.e. 60.
So x+y total 1 day's work = (1/20+1/1/12) = 2/15.
But x work alone = 1/4 day's. Remain work = (1-1/4) = 3/4.
Now, work will be done by x and y = 2/15*3/4 = 1/10.
So 10 day's.
y can do a work in 12 days.
To find total work, take LCM of 20 and 12 i.e. 60.
So x+y total 1 day's work = (1/20+1/1/12) = 2/15.
But x work alone = 1/4 day's. Remain work = (1-1/4) = 3/4.
Now, work will be done by x and y = 2/15*3/4 = 1/10.
So 10 day's.
Santosh said:
1 decade ago
Well a simple logic
1day work x =1/20
y=1/12
x first start and works upto 4 days=4*1/20
then both x and y works upto some day to completework=a(1/20+1/12)
add (4*1/20)+a(1/20+1/12)=1
we will get the days where both x and y worked=a=6
So total days to complete work =4+6=10.
1day work x =1/20
y=1/12
x first start and works upto 4 days=4*1/20
then both x and y works upto some day to completework=a(1/20+1/12)
add (4*1/20)+a(1/20+1/12)=1
we will get the days where both x and y worked=a=6
So total days to complete work =4+6=10.
Jay said:
2 years ago
Let 'p' be the one we should find which is ,
The number of days of total work which is basically X.
Y came after 4 days which is (total days-4) or (p-4).
X and Y can do a piece of work in 20 days and 12 days.
(p/20) + (p-4)/12 = 1
(12*p)+(20*p)-80 = 240.
32 *p =320.
p=10.
The number of days of total work which is basically X.
Y came after 4 days which is (total days-4) or (p-4).
X and Y can do a piece of work in 20 days and 12 days.
(p/20) + (p-4)/12 = 1
(12*p)+(20*p)-80 = 240.
32 *p =320.
p=10.
(5)
Ashish said:
1 decade ago
Work done by A in 1 day=1/20.
Work done by B in 1 day=1/12.
(A+B) 'S 1 DAY WORK=1/20+1/12=8/60.
So A can work in 4 days =4*1/20=1/5.
So remaining work=1-1/5=4/5.
So 4/5 of work done by (A+B) =4/5*60/8.
=6 DAYS.
TOTAL DAY TAKEN WILL BE=6+4=10 DAYS.
Work done by B in 1 day=1/12.
(A+B) 'S 1 DAY WORK=1/20+1/12=8/60.
So A can work in 4 days =4*1/20=1/5.
So remaining work=1-1/5=4/5.
So 4/5 of work done by (A+B) =4/5*60/8.
=6 DAYS.
TOTAL DAY TAKEN WILL BE=6+4=10 DAYS.
Bikram Mondal said:
5 years ago
X = 20
Y = 12
L.C.M = 60 (Total work)
X efficiency 60÷20 = 3
Y efficiency 60÷12 = 5
Work done by X in 4days = 3 x 4 = 12
X&Y total work efficiency = 3 + 5 = 8
Remaining work completed = 60 - 12/8 = 6
Total time taken = (6+4)days = 10 days.
Y = 12
L.C.M = 60 (Total work)
X efficiency 60÷20 = 3
Y efficiency 60÷12 = 5
Work done by X in 4days = 3 x 4 = 12
X&Y total work efficiency = 3 + 5 = 8
Remaining work completed = 60 - 12/8 = 6
Total time taken = (6+4)days = 10 days.
(10)
Nagu said:
2 decades ago
2/15 is work is done by X and Y in 1 day.
And 4/5 is Remaining work
suppose X and Y together take A days to complete the remaining wotk then
A*2/15 = 4/5
now A = 4/5 * 15/2
so A = 6 days
i think you understood the logic behind that
And 4/5 is Remaining work
suppose X and Y together take A days to complete the remaining wotk then
A*2/15 = 4/5
now A = 4/5 * 15/2
so A = 6 days
i think you understood the logic behind that
Ajith said:
1 decade ago
I this way also it can be thought about:
Assume total days for completion = N.
X 's 1 day's effort = 1/20.
Y's 1 day's effort = 1/12.
(1/20)*N = (1/12)*(N-4)----[i.e, X's N days work = Y's (N-4) days work].
By solving, N = 10 days.
Assume total days for completion = N.
X 's 1 day's effort = 1/20.
Y's 1 day's effort = 1/12.
(1/20)*N = (1/12)*(N-4)----[i.e, X's N days work = Y's (N-4) days work].
By solving, N = 10 days.
Sahib said:
1 decade ago
Take it this way
1/5 work x does in 4 days
Remaining 4/5 is to be found
X+y does 1/20+1/12 in 1 day
Therefore x+y=2/15 w/d
Therefore 4/5 w = 2/15(w/d)/4/5(w)
We get 1/6(1/d)
Therefore 6 days
Therefore total wrk in 10 days
1/5 work x does in 4 days
Remaining 4/5 is to be found
X+y does 1/20+1/12 in 1 day
Therefore x+y=2/15 w/d
Therefore 4/5 w = 2/15(w/d)/4/5(w)
We get 1/6(1/d)
Therefore 6 days
Therefore total wrk in 10 days
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