Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 23)
23.
A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
Answer: Option
Explanation:
| 2(A + B + C)'s 1 day's work = |  |
1 | + | 1 | + | 1 |  |
= | 15 | = | 1 | . |
| 30 | 24 | 20 | 120 | 8 |
| Therefore, (A + B + C)'s 1 day's work = | 1 | = | 1 | . |
| 2 x 8 | 16 |
| Work done by A, B, C in 10 days = | 10 | = | 5 | . |
| 16 | 8 |
| Remaining work = | |
1 - | 5 | |
= | 3 | . |
| 8 | 8 |
| A's 1 day's work = | |
1 | - | 1 | |
= | 1 | . |
| 16 | 24 | 48 |
| Now, | 1 | work is done by A in 1 day. |
| 48 |
| So, | 3 | work will be done by A in | |
48 x | 3 | |
= 18 days. |
| 8 | 8 |
Discussion:
61 comments Page 1 of 7.
Kishore said:
1 decade ago
Why shld we consider 2(a+b+c) at first ?
Dinesh said:
1 decade ago
See dude...
ab, bc, ac
So add three (a+b) + (b+c) + (a+c) = so we need (a+b+c).
Add all.
2a + 2b + 2c
2 (a + b + c) = now we add na.
ab, bc, ac
So add three (a+b) + (b+c) + (a+c) = so we need (a+b+c).
Add all.
2a + 2b + 2c
2 (a + b + c) = now we add na.
Shanthi said:
1 decade ago
Yes the same question what kishore asked! why we should consider
2 (a+b+c) ?
2 (a+b+c) ?
Ramkumar said:
1 decade ago
Ramkumar:.
Good question but easy to understand.
Good question but easy to understand.
Lalitha said:
1 decade ago
Wow can we take 1/24 (1/16-1/24) ?
Gurjit said:
1 decade ago
why subtract 1/16-1/24 on the above equation a=1/30
Suji said:
1 decade ago
a is not equal to 30...dude
a+b+c=1/16
b+c=1/24
solving above eqn
a+b+c-(b+c)=1/16-1/24..........
a+b+c=1/16
b+c=1/24
solving above eqn
a+b+c-(b+c)=1/16-1/24..........
Naveen kumar said:
1 decade ago
Hi friends, let me explain one simple method to solve this problem.
It is lengthy process, but you will understand easily.
(a+b)'s 1 day work = 1/30.
(b+c)'s 1 day work = 1/24
(c+a)'s 1 day work = 1/20.
solving that equations we will get
a =1/48 , b = 7/240 , c = 1/80
now,
(a+b+c)'s one day work = 1/48 + 7/240 + 1/80.
= 15/240.
= 1/16.
i.e,
(a+b+c)'s 10 days work = 10/16
=5/8
remaining work = (1- 5/8)
= 3/8
3/8 portion of work is completed by a alone....
now,
a completes 1/48 portion of work in 1 day....
a completes 3/8 portion of work in _ days.....?
1/48------> 1
3/8-------> x
by cross multyplying,we will get
x /48 = 3/8
x = 18..
So, A alone can complete the work in 18 days.
It is lengthy process, but you will understand easily.
(a+b)'s 1 day work = 1/30.
(b+c)'s 1 day work = 1/24
(c+a)'s 1 day work = 1/20.
solving that equations we will get
a =1/48 , b = 7/240 , c = 1/80
now,
(a+b+c)'s one day work = 1/48 + 7/240 + 1/80.
= 15/240.
= 1/16.
i.e,
(a+b+c)'s 10 days work = 10/16
=5/8
remaining work = (1- 5/8)
= 3/8
3/8 portion of work is completed by a alone....
now,
a completes 1/48 portion of work in 1 day....
a completes 3/8 portion of work in _ days.....?
1/48------> 1
3/8-------> x
by cross multyplying,we will get
x /48 = 3/8
x = 18..
So, A alone can complete the work in 18 days.
Meena said:
1 decade ago
We want short cut please anybody solve this shortcut method. Why taken 2 (a+b+C).
Logu said:
1 decade ago
2(A + B + C)'s 1 day's work = 1/8.
Then (A + B + C)'s 1 day's work = 2/8.
How come it is 1/16 ?
Then (A + B + C)'s 1 day's work = 2/8.
How come it is 1/16 ?
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