Math, asked by nishantsikarwa83, 1 year ago

A contractor receives a certain sum that he uses for paying wages. His capital, together with the weekly subsidy, would just enable him to pay 42 men for 52 weeks. If he had 60 men at the same wages, his capital together with the weekly subsidy would just suffice for 13 weeks. How many men can be maintained for 26 weeks?

Answers

Answered by Rajeev79
5

Let the contractor's capital be c and his weekly sum of be x

Let payment needed for one man one week = m

His own capital together with the weekly sum enables him to pay 45 men for 52 weeks

c+52x = 45m*52 = 2340m ---(1)

If he had 60 men and the same wages, his capital and weekly sum would suffice for 13 weeks

c+13x = 60m*13 = 780m ---(2)

(1)-(2) => 39x = 1560m

x = 40m ---(A)

Substituting this value in (2),

c + 13*40m = 780m

c = 260m ---(B)

Let n men can be maintained for 26 weeks.

c + 26x = nm * 26

Substituting the values of c and x in the above equation from (A) and (B)

260m +26*40m =nm * 26

1300m = 26nm

1300 = 26n

n = 1300/26 = 50


Rajeev79: Please mark it brainleist
Answered by munawwar15
0

Answer:

50 men

Step-by-step explanation:

If the contractor's capital is c, the weekly sum he receives is s, and if each man receives Rs. X per week. then c+52s = 45*52x,

Similarly, c+13s = 60*13x. Subtracting and solving, we get,

39s = 1560 .

:: s=40x. and c = 260x

If n men are to be maintained for 26 weeks, 260x +

26(40x) = n(26x) . Hnece, n = 50

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