Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 1)
1.
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
Answer: Option
Explanation:
| A's 1 day's work = | 1 | ; |
| 15 |
| B's 1 day's work = | 1 | ; |
| 20 |
| (A + B)'s 1 day's work = |  |
1 | + | 1 |  |
= | 7 | . |
| 15 | 20 | 60 |
| (A + B)'s 4 day's work = | |
7 | x 4 | |
= | 7 | . |
| 60 | 15 |
| Therefore, Remaining work = | |
1 - | 7 | |
= | 8 | . |
| 15 | 15 |
Discussion:
344 comments Page 11 of 35.
Arpit kashyap said:
1 decade ago
A's 1 day's work = 1/15,
B's 1 day's work = 1/20,
Since they work together for 4 days so that their left work in fraction = [1 - (1/15 + 1/20)]*4.
= (8/60)*4.
= 8/15.
B's 1 day's work = 1/20,
Since they work together for 4 days so that their left work in fraction = [1 - (1/15 + 1/20)]*4.
= (8/60)*4.
= 8/15.
Ashwini said:
10 years ago
Its very simple 1/1-7/15.
LCM between 1 and 15 is 15.
So = (1/1)*(15/15) - (7/15)*(1/1) {Make both the denominator as 15 i.e LCM}.
= 15/15 - 7/15.
= (15-7)/15 = 8/15.
Hope understood!
LCM between 1 and 15 is 15.
So = (1/1)*(15/15) - (7/15)*(1/1) {Make both the denominator as 15 i.e LCM}.
= 15/15 - 7/15.
= (15-7)/15 = 8/15.
Hope understood!
ANURAG SAXENA said:
1 decade ago
LCM - 15, 20.
LCM = 60 UNIT.
A-- 15 DAY -60/15 = 4 U/D
B -- 20 DAY -60/20 = 3 U/D
A+B = 7 U/D.
4 DAYS = 7x4 = 28 U.
REAMING UNIT = 60-28 = 32 UNIT.
REAMING WORK = 32/60 = 8/15 DAYS.
LCM = 60 UNIT.
A-- 15 DAY -60/15 = 4 U/D
B -- 20 DAY -60/20 = 3 U/D
A+B = 7 U/D.
4 DAYS = 7x4 = 28 U.
REAMING UNIT = 60-28 = 32 UNIT.
REAMING WORK = 32/60 = 8/15 DAYS.
Sundar said:
1 decade ago
@Poorni:
The total work = 1;
Work done by A and B in 4 days = 7/15.
Therefore, the pending work = 1 - 7/15 = 0.533333333... (or) 8/15.
Hope this help you. Have a nice day!
The total work = 1;
Work done by A and B in 4 days = 7/15.
Therefore, the pending work = 1 - 7/15 = 0.533333333... (or) 8/15.
Hope this help you. Have a nice day!
(1)
Analyser said:
1 decade ago
Since work done is 7/15....which is 7 parts out of 15.....
So obviously total work wud be 15/15......which is 1....
Hence we subtract 7/15...from 15/15 or 1..... i.e 1-7/15.
So obviously total work wud be 15/15......which is 1....
Hence we subtract 7/15...from 15/15 or 1..... i.e 1-7/15.
Pankaj Kushwah said:
1 month ago
The answer is 8/15.
Lcm of 15 and 20.
Then it will give you 60.
Now 60/15 = 4, 60/20 = 3.
Together for one day 4 + 3 = 7.
For 4 days = 4 * 7 = 28.
60 - 28 = 32.
32/60 = 8/15.
Lcm of 15 and 20.
Then it will give you 60.
Now 60/15 = 4, 60/20 = 3.
Together for one day 4 + 3 = 7.
For 4 days = 4 * 7 = 28.
60 - 28 = 32.
32/60 = 8/15.
(50)
Debasis Das said:
10 years ago
A's 1 day work = 1/15.
B's 1 day work = 1/20.
Hence, A+B 1 day work = (1/15+1/20) = 7/60.
So, A+B 4 day work = (7/60*4) = 7/15.
Then, remaining work = 1-7/15 = 8/15.
B's 1 day work = 1/20.
Hence, A+B 1 day work = (1/15+1/20) = 7/60.
So, A+B 4 day work = (7/60*4) = 7/15.
Then, remaining work = 1-7/15 = 8/15.
Anjali thakur said:
3 months ago
The answer is 8/15.
Lcm of 15 and 20.
Then it will give you 60.
Now 60/15 = 4, 60/20 = 3.
Together for one day 4 + 3=7.
For 4 days = 4 * 7=28.
60-28 = 32.
32/60 = 8/15.
Lcm of 15 and 20.
Then it will give you 60.
Now 60/15 = 4, 60/20 = 3.
Together for one day 4 + 3=7.
For 4 days = 4 * 7=28.
60-28 = 32.
32/60 = 8/15.
(49)
Anand said:
1 decade ago
Balance work = 8/15.
A = 1/15, B = 1/20.
Total no of days required to complete remaining work.
A = (8/15) / (1/15) = 8 days,
B = (8/15) / (1/20) = 10 2/3 days.
A = 1/15, B = 1/20.
Total no of days required to complete remaining work.
A = (8/15) / (1/15) = 8 days,
B = (8/15) / (1/20) = 10 2/3 days.
Srikanth said:
1 decade ago
In all these problems total work taken as "1" so we have to remove the resulting answer i.e., 7/15 from total work"1" so remaining work is (1-7/15) i.e., 8/15.
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