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Mixture and Alligation #7

7. 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine did the cask originally hold?
A. 26 litresB. 24 litres
C. 30 litresD. 32 litres

Answer: Option B

Explanation:

Let initial quantity of wine $=x$ litre

After a total of 4 operations, quantity of wine
$=x\left(1-\dfrac{y}{x}\right)^n\\= x\left(1-\dfrac{8}{x}\right)^4$

Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65

$\Rightarrow \dfrac{x\left(1-\dfrac{8}{x}\right)^4}{x} = \dfrac{16}{81}\\\Rightarrow \left(1-\dfrac{8}{x}\right)^4 = \left(\dfrac{2}{3}\right)^4\\\Rightarrow \left(1-\dfrac{8}{x}\right) = \dfrac{2}{3}\\\Rightarrow \left(\dfrac{x-8}{x}\right) = \dfrac{2}{3}\\\Rightarrow 3x-24=2x\\\Rightarrow x=24$

Comments(5)

profilePriyanka bulla
2016-05-08 10:33:03 
The formula is x[1-(y/x)]^n .So y do v divide it by x again ?
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profileAkhtar
2016-05-08 18:48:15 
x[1-(y/x)]^n = quantity of wine after n operations.

x[1-(y/x)]^n / x = ratio of "quantity of wine after n operations" to "initial quantity of wine"
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profilenalin
2015-08-01 08:59:57 
Wine left/total vol = $\left(1-\dfrac{y}{x}\right)^n$

initial total vol of wine $x$

So
$\dfrac{16}{16+65}=\left(1-\dfrac{8}{x}\right)^4\\
\dfrac{16}{81}=\left(\dfrac{x-8}{x}\right)^4\\
\left(\dfrac{2}{3}\right)^4=\left(\dfrac{x-8}{x}\right)^4\\
\dfrac{2}{3}=\dfrac{x-8}{x}\\
x=24$
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profileAvinash
2013-09-19 21:41:25 
After a total of 4 operations,  quantity of wine : quantity of water = 16 : 65

Assume that after 4 operations, quantity of wine = 16 liter and quantity of water = 65 liter

Then initial quantity was 16 + 65 = 81

i.e., quantity of wine after 4 operations / initial quantity = 16/81

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profileshine
2013-09-19 14:52:26 
why n ques 7 its 16/81
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