Question

A group of men decided to do a work in 10 days, but five of them absented themselves. If the rest of the group finished the work in 12 days, find the original number of men? (a) 20 men (b) 30 men (c) 40 men (d) 50 men (e) None of these

Answer 1

In time and work problems the important point to be remebered is (number of MENDAYS is always constant.)i.e product of men into days always remains constant.

if one variable among men or days changes the other varible changes in accordance to make the value of remains constant.

let us assume X be the number of initial persons. so mendays = 10* X( as told above), now 5 are absent.i.e (X-5) are present.as i told men days should be always constant

10* X= (X-5)*12

10X= 12X-60

2X=60 implies X=30.7 years ago

Answer 2

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume that, original number of men in the group was x .

Than, we can say that,

x men completes the work in 10 days. while (x - 5) men will complete the same work in 12 days.

So,

→ M1 = x men

→ D1 = 10 days.

→ M2 = (x - 5) men .

→ D2 = 12 Days.

We know That, M1 * D1 = M2 * D2 .

Putting values we get,

→ x * 10 = (x - 5) * 12

→ 10x = 12x - 60

→ 12x - 10x = 60

→ 2x = 60

→ x = 30 Men. (b) (Ans.)

Hence, the original number of men were 30 .