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Difficult Work/Rate Problem

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Difficult Work/Rate Problem

by mattocks » Sun Mar 28, 2010 11:23 am
Can someone explain the best way to solve this problem?

13 men and 7 boys can finish a job in 7 days, while 6 boys and 13 women can finish the same job in 6 days. In how many days can 1 man, 1 boy and 1 woman working together finish the same job (assume constant rates each for men, women, and boys)?

a) 42
b) 54
c) 21
d) 63
e) 24
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by thephoenix » Sun Mar 28, 2010 8:47 pm
mattocks wrote:Can someone explain the best way to solve this problem?

13 men and 7 boys can finish a job in 7 days, while 6 boys and 13 women can finish the same job in 6 days. In how many days can 1 man, 1 boy and 1 woman working together finish the same job (assume constant rates each for men, women, and boys)?

a) 42
b) 54
c) 21
d) 63
e) 24
IMO A
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by harshavardhanc » Mon Mar 29, 2010 2:50 am
mattocks wrote:Can someone explain the best way to solve this problem?

13 men and 7 boys can finish a job in 7 days, while 6 boys and 13 women can finish the same job in 6 days. In how many days can 1 man, 1 boy and 1 woman working together finish the same job (assume constant rates each for men, women, and boys)?

a) 42
b) 54
c) 21
d) 63
e) 24
13M + 7B = 1/7 (per day work of 13 men and 7 boys)
13W + 6B = 1/6 (per day work of 13 women and 6 boys)
------------------------------
13M+13W+13B = 13/42

per day work of (M+W+B) = 1/42. Hence, number of days required for 1 man, 1 woman, and 1 boy = 42 days.
Regards,
Harsha
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by pops » Mon Mar 29, 2010 6:42 am
mattocks wrote:Can someone explain the best way to solve this problem?

13 men and 7 boys can finish a job in 7 days, while 6 boys and 13 women can finish the same job in 6 days. In how many days can 1 man, 1 boy and 1 woman working together finish the same job (assume constant rates each for men, women, and boys)?

a) 42
b) 54
c) 21
d) 63
e) 24
Hi,
Trick here would be to notice the pattern of numbers : 13, 7, 6 ....
now make an equation as per given condition:
13 * 1/m + 7 * 1/b = 1/7
13 * 1/w + 6 * 1/b = 1/6

[quickly realise that 7/b + 6/b =13/b and we are asked to know 1/m+1/w+1/b]
so, add above 2:
13 * (1/m+1/w+1/b) = 1/7 + 1/6 = 13/42
hence, the answer = 42 :-)
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by eaakbari » Mon Mar 29, 2010 1:47 pm
Using information given form the equations

13/x + 7/y = 1/7
6/y +13/z = 1/6

Here we have 3 variables and 2 equations which is unsolvable, that means theres some trick, so observing the equations, we realize their sum gives us the answer

Answer A
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by akhpad » Tue Mar 30, 2010 1:35 am
7(13M+7B) = 6(6B+13W) = k(M+B+W)


(13M+7B) = k(M+B+W)/7
(6B+13W) = k(M+B+W)/6
sum these two

13M+13B+13W = k(M+B+W)(1/7 + 1/6)

k = 42
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by kstv » Tue Mar 30, 2010 8:31 am
pops wrote: Trick here would be to notice the pattern of numbers : 13, 7, 6 ....
now make an equation as per given condition:
13 * 1/m + 7 * 1/b = 1/7
13 * 1/w + 6 * 1/b = 1/6
[quickly realise that 7/b + 6/b =13/b and we are asked to know 1/m+1/w+1/b]
so, add above 2:
13 * (1/m+1/w+1/b) = 1/7 + 1/6 = 13/42
hence, the answer = 42 :-)
13 M + 7 B = 7 days ---1 and 6 B + 13 W = 6 days ----2
LCM of 7 and 6 = 42
13M+7B in 42 days can do 6 times the work ---1
......6B + 13 W in 42 days can do 7 times the work-------2 Adding 1 and 2
13M+13B+13W in 42 days can do 13 times the work so dividing by 13
M+B+W can in 42 days can do the the work once.
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