# Thread: Problem based on ratio

1. ## Problem based on ratio

Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..

The answer is 4:11. But I am not getting how to get it.

2. Originally Posted by siddscool19
Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..

The answer is 4:11. But I am not getting how to get it.

If we assume the first ship has a volume of VL then:

In ship one there will be $\displaystyle V\frac{2}{2+3}$ of coke and so $\displaystyle V(1-\frac{2}{2+3})$ of soda

In ship two there will be $\displaystyle 2V\frac{3}{10}$ of coke and $\displaystyle 2V(1-\frac{3}{10})$ of soda

In ship three there will be $\displaystyle 3V\frac{1}{5}$ of coke and $\displaystyle 3V(1-\frac{1}{5})$

Add the sum of the coke from the three ships and factor out V.
$\displaystyle V(\frac{2}{5} + 2\frac{3}{10} + \frac{1}{5}) = \frac{8V}{5}$

Add the sum of the soda from the three ships and factor out V

$\displaystyle V(1-\frac{2}{5} + 2(1-\frac{3}{10}) + 3(1-\frac{1}{5})) = \frac{22V}{5}$

To find the ratio we divide the cola by the soda:

$\displaystyle \frac{8V}{5} \div \frac{22V}{5} = \frac{8V}{5} \times \frac{5}{22V} = \frac{8}{22} = \frac{4}{11}$

Now we have a ratio of $\displaystyle \frac{4}{11} : 1$ so we multiply by 11 to get an integer ratio of 4:11

(if you're having trouble visualising it try putting V=5)

3. Originally Posted by siddscool19
Three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda.In the first vessel ratio of coke and soda is 2:3 , in the second 3:7, and in third 1:4.If the liquid in all the three vessels were mixed in a bigger container, then the resulting ratio of coke and soda is..
The first vessel is coke:soda = 2:3, or five parts.

The second vessel is coke:soda = 3:7, or ten parts.

The third vessel is coke:soda = 1:4, or five parts.

Overall, their volumes are 1st:2nd:3rd = 1:2:3, but "5:10:5" doesn't match this. The first bit, the "5:10", matches the "1:2", but the second "5" doesn't match the "3". So make it match:

. . . . .third vessel: 3:12, or fifteen parts

Now you have:

The first vessel is coke:soda = 2:3, or five parts.

The second vessel is coke:soda = 3:7, or ten parts.

The third vessel is coke:soda = 3:12, or fifteen parts.

This gives you the right ratio, assuming all the "parts" are of the same volume, so the mixture will then be:

. . . . .fourth vessel: (2+3+3) : (3+7+12)

...or 8:22, which simplifies as 4:11.

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# three vessels having volumes in the ratio of 1:2:3 are full of a mixture of coke and soda

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