Celebrating Algebra

Algebra Level 2

I was in a party with a glass containing alcohol and water in the ratio of 9:1. Then I added 50 ml of water in my glass making the alcohol to water of the mixture 4:1. What was the initial amount of water in the glass in ml?


The answer is 40.

Abhay Tiwari by Abhay Tiwari , 28, India

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2 solutions

Ashish Menon
Jun 8, 2016

Let the initial volume be x x , then:-
Initial amount of water = x 9 + 1 = x 10 \dfrac{x}{9 + 1} = \dfrac{x}{10}
Final amount of water = 1 4 + 1 ( x + 50 ) = x + 50 5 \dfrac{1}{4 + 1}\left(x + 50\right) = \dfrac{x + 50}{5}

So, setting up the equation:-
x 10 + 50 = x + 50 5 x + 500 10 = x + 50 5 x + 500 = 2 x + 100 x = 400 \begin{aligned} \dfrac{x}{10} + 50 & = \dfrac{x + 50}{5}\\ \dfrac{x + 500}{10} & = \dfrac{x + 50}{5}\\ x + 500 & = 2x + 100\\ x & = 400 \end{aligned}

So, original amount of water = x 10 = 400 10 = 40 \dfrac{x}{10} = \dfrac{400}{10} = \color{#3D99F6}{\boxed{40}} .

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Nice and easy. ;)

Abhay Tiwari - 5 years ago

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Thanks! :)

Ashish Menon - 5 years ago
Hung Woei Neoh
Jun 8, 2016

Let the initial volume of water be x x

The initial volume of alcohol = 9 x 9x

After 50 50 ml of water was added, the ratio became 4 : 1 4:1 . This means

9 x x + 50 = 4 1 9 x = 4 x + 200 5 x = 200 x = 40 \dfrac{9x}{x+50} = \dfrac{4}{1}\\ 9x=4x+200\\ 5x=200\\ x=\boxed{40}

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