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wages problem

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wages problem

by hemant_rajput » Wed Jan 30, 2013 11:27 am
Q14 . A contractor receives a certain sum every week for paying wages. His own capital together with the weekly sum enables him to pay 45 men for 52 weeks. If he had 60 men and the same wages his capital and weekly sum would suffice for 13 weeks, how many men can be maintained for 26 weeks?
a. 60
b. 52
c. 50
d. 65

Guys I'm not even sure that this problem is correct or complete. Just let me know if you can solve it with the given information.
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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by vkb001 » Wed Jan 30, 2013 5:30 pm
W = amount received per week
C = capital
S = wage per worker

52W + C = 45S ---- Equation (1)
13W + C = 60S ---- Equation (2)

If X men can be maintained for 26 weeks, then the new equation would be

26W + C = X ---- Equation (3)

Equation (1) - Equation (2):

39W = -15S
==> 13W = -5S ---- Equation (4)

Equation (3) - Equation (2):

13W = X - 60S ----- Equation (5)

Plug Equation (4) in Equation (5):

-5S = X - 60S ==> X = 55S

So, 55 men can be maintained for 26 weeks (with same wages). But, I don't see it in answers (or I have misinterpreted the question).
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by sana.noor » Thu Jan 31, 2013 6:59 am
Answer is 50
C + 52W = 45*52-----equation 1
C + 13W = 60*13-----equation 2
Subtracting equation 2 from equation 1 we will get
C + 52W = 2340
C + 13W = 780
39W =1569
W = 40
putting the value of W in equation 1 we will get C = 260
Now the question ask that how many men can be maintained for 26 weeks
C + 26W = 26*X
260 + 1040 = 26X
X = 50

C is the right choice
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