Aptitude - Time and Work - Discussion

Discussion Forum : Time and Work - General Questions (Q.No. 6)
6.
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
4 days
5 days
6 days
7 days
Answer: Option
Explanation:

Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.

Then, 6x + 8y = 1 and 26x + 48y = 1 .
10 2

Solving these two equations, we get : x = 1 and y = 1 .
100 200

(15 men + 20 boy)'s 1 day's work = 15 + 20 = 1 .
100 200 4

15 men and 20 boys can do the work in 4 days.

Discussion:
203 comments Page 21 of 21.

ESNALA SIDDA said:   1 year ago
2(3m+4b) = 1/10,
3m+4b = 1/20,
5(3m+5b) = ?
= 5×1/20,
= 1/4,
= 4days.
(40)

Magic said:   1 year ago
First convert it to either boys or men's.
(6m+8b)×10=(26m+48B)×2.
8m = 16B,
m=2B.
One man equals 2 boys.

So, use this in any one of the equation, let's take:
= (6m+8B)×10.
= 6(2B)+8B ×10.
= 200B units.

So, use this trick in and find out the work done boys in the given undefined equation:
= 15m + 20B,
= 15(2B) + 20B,
= 50B.

Calculate total work/ work.
= 200B/50B.
= 4 days.
(43)

A K BURNWAL said:   1 month ago
6M + 8B= 10
26M + 48B= 2
15 M + 20 B= ?
(6M + 8B)10 = (26M+48B)2
M = 2B.
6 * 2B + 8B = 20B.
15 * 2B + 20B = 50 B.
20B * 10 = 50B * ?= 4 DAYS.
(5)


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