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by Aman verma » Fri Sep 16, 2011 7:58 am
Q: Two friends undertake to do a job. The second friend started working 2 hours after the first. Five hours after the second friend has begun working, there is still 9/20 of the work to be done. When the assignment is completed, it turns out that first friend has done 60% of the work, while second friend has done the rest of the work. How many hours would it take each one to do the whole job individually ?

a) 10 hours and 12 hours

b) 15 hours and 10 hours

c) 20 hours and 25 hours

d) 18 hours and 20 hours

e) 16 hours and 22 hours

ps: This problem can be very easily solved by plugging in the options but can this be solved algeraically ?
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Source: — Problem Solving |

by cans » Fri Sep 16, 2011 8:21 am
A ->a, B->b. Combined A*B = c = (ab)/(a+b)
A did 3/5 and B 2/5
in 2 hours A did: 2/a
in 5 hours A did: 5/a and B: 5/b
in next x hours, x/a and x/b
Thus (7+x)/a = 3/5 and (5+x)/b = 2/5
also x/a + x/b = 9/20
3 equations, 3 variables....
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by GMATGuruNY » Sat Sep 17, 2011 3:40 am
Aman verma wrote:Q: Two friends undertake to do a job. The second friend started working 2 hours after the first. Five hours after the second friend has begun working, there is still 9/20 of the work to be done. When the assignment is completed, it turns out that first friend has done 60% of the work, while second friend has done the rest of the work. How many hours would it take each one to do the whole job individually ?

a) 10 hours and 12 hours

b) 15 hours and 10 hours

c) 20 hours and 25 hours

d) 18 hours and 20 hours

e) 16 hours and 22 hours

ps: This problem can be very easily solved by plugging in the options but can this be solved algeraically ?
I would plug in the answers, which represent the time for each friend to complete the job alone.

Answer choice C: 20 hours and 25 hours.
Let work = 100 units.
Rate for the first friend = w/t = 100/20 = 5 units per hour.
Rate for the second friend = w/t = 100/25 = 4 units per hour.
Work completed by the first friend in 2 hours = r*t = 5*2 = 10 units.
Combined rate for the two friends = 5+4 = 9 units per hour.
Rate completed by the two friends in the next 5 hours = r*t = 9*5 = 45 units.
Remaining work = 100-10-45 = 45 units.
Remaining work/Total work = 45/100 = 9/20.
Success!
No need to do more work.
Each answer choice offers a different ratio of times.
If we plug in any other time ratio, there will not be 9/20 of the work left to be completed after the initial 7 hours.

The correct answer is C.

For the skeptical:
Time for the two friends to complete the remaining work = w/r = 45/9 = 5 hours.
Total time worked by the first friend = 12 hours.
Work produced by the first friend in 12 hours = r*t = 12*5 = 60 units.
(Work produced by the first friend)/Total work = 60/100 = 60%.
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by Aman verma » Sat Sep 17, 2011 8:25 am
Many Thanks ! OA[spoiler]c)[/spoiler]
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by gmatboost » Sat Sep 17, 2011 9:41 pm
I tried to solve this algebraically, and after an absurd amount of time I finally did. I don't recommend trying.

A = rate of first, B = rate of second, T = time that B worked
I used the equations
7A + 5B = 11/20 (they did 11/20 of work after first 7 hours)
BT = 2/5 (B did 40% of work)
2A + T(A+B) = 1 (together, they did all the work in the allotted time)

After tons of ugly substitution I got to
1400A^2 - 730A + 33 = 0
Which can be factored to (20A - 1)(70A - 33) = 0
So A = 1/20 or A = 33/70
The second is not a valid solution because after two hours almost all the work would have been done already. So A = 1/20 means A does 1/20 per hour, so A takes 20 hours total.

So, umm, don't do this.
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