Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 26)
26.
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :
Answer: Option
Explanation:
| (A + B + C)'s 1 day's work = | 1 | ; |
| 6 |
| (A + B)'s 1 day's work = | 1 | ; |
| 8 |
| (B + C)'s 1 day's work = | 1 | . |
| 12 |
 (A + C)'s 1 day's work |
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So, A and C together will do the work in 8 days.
Discussion:
53 comments Page 3 of 6.
Vick said:
1 decade ago
A's = (A+B+C) - (B+C) = 1/6-1/8 = 1/24.
B's = (A+b+C) - (A+B) 1/6-1/12 = 1/12.
A+C's (1/24) + (1/12) = 1/8.
Therefore they can complete in 8 days.
B's = (A+b+C) - (A+B) 1/6-1/12 = 1/12.
A+C's (1/24) + (1/12) = 1/8.
Therefore they can complete in 8 days.
Hussain said:
1 decade ago
Hey guys please say me.
How 1/12+1/24 = 1/8.
I did not understand the logic the lcm of these two numbers is 24.
How 1/12+1/24 = 1/8.
I did not understand the logic the lcm of these two numbers is 24.
Jeeva909 said:
1 decade ago
Hi am little week in maths, so please somebody explain me the answer how this [2x1/6]-[1/8+1/12] came?
Monu said:
1 decade ago
Why can't we simply subtract A+B's 1 day work from B+C's 1 day work to get A+C's 1 day work?
Nav said:
7 years ago
Why can't we simply subtract A+B's 1-day work from B+C's 1-day work to get A+C's 1-day work?
Basu said:
1 decade ago
How its come on 1/6-1/2=1/12
1/6-1/8= 1/24
1/12+1/24= 1/8
1/6-1/8= 1/24
1/12+1/24= 1/8
Anishkumar said:
1 decade ago
Assume A+C as x
so now
2(A+B+C)=1/8+1/12+1/x
2(1/6) =5/24+1/x
finally
x =8
so now
2(A+B+C)=1/8+1/12+1/x
2(1/6) =5/24+1/x
finally
x =8
Pandeswari Durai Pandi said:
5 years ago
@Nanthu.
Thank you for using LCM method it makes me to understand it easily.
Thank you for using LCM method it makes me to understand it easily.
Gregory wachiye said:
9 years ago
This is awesome. I have gained a lot with the help of this explanations.
Suraj said:
9 years ago
In original solution why they take 1/6*2 from where 2 is occur?
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