# Logic help.

• December 9th 2012, 07:18 PM
hisajesh
Logic help.
2 men and 7 boys can do a piece of work in 14 days,
3 men and 8 boys can do it in 11 days.
8 men and 6 boys can do this 3 times of work in how many days?

Logic: I found either the first or the second sentence is redundant, but when I ignored one and tried solving, it gives me wrong answers, Kindly help how to approach?
• December 9th 2012, 09:39 PM
Prove It
Re: Logic help.
Let x represent the amount of work that can be done by a man, and let y represent the amount of work that can be done by a boy. Then \displaystyle \begin{align*} 2x + 7y = 14 \end{align*} and \displaystyle \begin{align*} 3x + 8y = 11 \end{align*}. Solve these two equations simultaneously for x and y, then use this to answer the remainder of your question.
• December 9th 2012, 10:52 PM
Soroban
Re: Logic help.
Hello, hisajesh!

I assume that all the men work at the same constant rate
and that all the boys work at the same constant rate.

Quote:

2 men and 7 boys can do a piece of work in 14 days,
3 men and 8 boys can do it in 11 days.
8 men and 6 boys can do this 3 times of work in how many days?

Let $m$ = the number of days it takes one man to do the job..
In one day, a man can do $\tfrac{1}{m}$ of the job.
In one day, 2 men can do $\tfrac{2}{m}$ of the job.

Let $b$ = the number of days it takes one boy to do the job.
In one day, a boy can do $\tfrac{1}{b}$ of the job.
In one day, 7 boys can do $\tfrac{7}{b}$ of the job.

So, in one day, 2 men and 7 boys can do $\tfrac{2}{m} + \tfrac{7}{b}$ of the job.
. . But this equals $\tfrac{1}{14}$ of the job.
Hence: . $\tfrac{2}{m} + \tfrac{7}{b} \:=\:\tfrac{1}{14}$

In one day, 3 men can do $\tfrac{3}{m}$ of the job.
In one day, 8 boys can do $\tfrac{8}{b}$ of the job.

So in one day, 3 men and 8 boys can do $\tfrac{3}{m} + \tfrac{8}{b}$ of the job.
. . But this equals $\tfrac{1}{11}$ of the job.
Hence: . $\tfrac{3}{m} + \tfrac{8}{b} \:=\:\tfrac{1}{11}$

We have a system of equations: . $\begin{Bmatrix}\frac{2}{m} + \frac{7}{b} \:=\:\frac{1}{14} & [1] \\ \\[-4mm] \frac{3}{m} + \frac{8}{b} \:=\:\frac{1}{11} & [2] \end{Bmatrix}$

$\begin{array}{ccccccc}\text{Multiply [1] by 3:} & \frac{6}{m} + \frac{21}{b} &=& \frac{3}{14} \\ \\[-4mm] \text{Multiply [2] by -}2\!: & \text{-}\frac{6}{m} - \frac{16}{b} &=& \text{-}\frac{2}{11} \end{array}$

Add: . $\tfrac{21}{b} - \tfrac{16}{b} \:=\:\tfrac{3}{14} - \tfrac{2}{11} \quad\Rightarrow\quad \tfrac{5}{b} \:=\:\tfrac{5}{154} \quad\Rightarrow\quad b \:=\:154$

Substitute into [1]: . $\tfrac{2}{m} + \tfrac{7}{154} \:=\:\tfrac{1}{14} \quad\Rightarrow\quad \tfrac{2}{m} \:=\:\tfrac{2}{77} \quad\Rightarrow\quad m \:=\:77$

Now we know the following:

In one day, a man can do $\tfrac{1}{77}$ of the jpb.
In one day, 8 men can do $\tfrac{8}{77}$ of the job.
In $x$ days, 8 men can do $\tfrac{8x}{77}$ of the job.

In one day, a boy can do $\tfrac{1}{154}$ of the job.
In one day, 6 boys can do $\tfrac{6}{154}$ of the job.
In $x$ days, 6 boys can do $\tfrac{6x}{154}$ of the job.

In $x$ days, 8 men and 6 boys can do: $\tfrac{8x}{77} + \tfrac{6x}{154}$ of the job.
But this equals 3 jobs.

Hence: . $\frac{8x}{77} + \frac{6x}{154} \:=\:3$

Multiply by 77: . $8x + 3x \:=\:231 \quad\Rightarrow\quad 11x \:=\:231 \quad\Rightarrow\quad x \:=\:21$