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19 Mar 2005, 10:21
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If 15 men or 24 women or 35 boys do a piece of work in 12 days working 8 hrs a day, how many men must be associated with 12 women and 6 boys to do another piece 9/2 times as great in 30 days working 6 hrs a day
(a) 8 men
(b) 10 men
(c) 12 men
(d) 4 men
(e) 15 men
(a) 8 men
(b) 10 men
(c) 12 men
(d) 4 men
(e) 15 men
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19 Mar 2005, 12:07
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anirban16
I am getting a wrong answer - maybe the aproach is wrong, or some calculation mistake. Can someone please point out?
------------------
Rate of work of men : women: boys = 1/15:1/24:1/35
1 man = 8/5 women = 7/3 boys.
1 woman = 5/8 men, 1 boy = 3/7 men.
First work = 15*12*8 man hours = 1440 man hours
Second work = 9/2 * 1440 = 720*9 = 6480 man hours.
Workforce = 12 women + 6 boys = 15/2 + 18/7 men = 141/14 men
Time = 180 hours
If additional men required = x, then
(141/14 + x) 180 = 6480
141/14 + x = 648/18 = 36
x is approx 26.
19 Mar 2005, 12:56
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"A"
If x is number of men for the ques then
women doing the job alone in the second case will take = ((96*24)/12)9/2
= 96*9 hrs
similarly for boys = ((96*35)/6)9/2 = 9*8*35
similarly for men = ((96*15)/X)9/2
9/2 = 180/96*9 + 180/9*8*35 + 180*2X/96*15*9
X = 7.9 ~ 8...
don't know if it is right
If x is number of men for the ques then
women doing the job alone in the second case will take = ((96*24)/12)9/2
= 96*9 hrs
similarly for boys = ((96*35)/6)9/2 = 9*8*35
similarly for men = ((96*15)/X)9/2
9/2 = 180/96*9 + 180/9*8*35 + 180*2X/96*15*9
X = 7.9 ~ 8...
don't know if it is right 
