1/5 of chicken John had is equal to 3/4 of Mark.

John sold 150 chickens while Mark bought 160.

The remider is in a ratio of 5:3, for John & Mark respectively.

What was the total number of chickens both had at first.

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- Apr 29th 2009, 07:44 AMpriRatio help
1/5 of chicken John had is equal to 3/4 of Mark.

John sold 150 chickens while Mark bought 160.

The remider is in a ratio of 5:3, for John & Mark respectively.

What was the total number of chickens both had at first. - Apr 29th 2009, 08:00 AMUnenlightened
Let J be the amount John had at the start, and let M be the amount Mark had at the start.

So J/5 = 3M/4

Then if J' is the amount that John had after the buying and selling, and M' is the amount Mark had after buying and selling, then the 5:3 ratio tells us that

3J'/8 = 5M'/8

We also know that J' = J - 150

and M' = M + 160.

So if we replace J' and M' with those

we get:

$\displaystyle \frac{3(J - 150)}{8} = \frac{5(M + 160)}{8}$

So we have two simultaneous equations with two unknowns:

$\displaystyle \frac{3(J - 150)}{8} = \frac{5(M + 160)}{8}$

$\displaystyle \frac{J}{5} = \frac{3M}{4}$

And solve... - Apr 29th 2009, 08:04 AMTwighi
hi!

Let $\displaystyle C_{j} \mbox{ and } C_{m} $ be notations for chickens for John and for Mark.

We have that:

$\displaystyle \frac{1}{5}C_{j}=\frac{3}{4}C_{m} \Rightarrow 4C_{j}-15C_{m}=0 $

We have also:

$\displaystyle \frac{C_{j}-150}{C_{m}+160}=\frac{5}{3} \Rightarrow 3C_{j}-5C_{m}=1250 $

So, you have two equations and two unknowns. Can you proceed from here?