4 litres of wine are drawn from a cask containing, 40 litres of wine. It is replaced by water. The process is repeated 3 times.
a) What is the final quantity of wine left in the cask ?
b) What is the ratio of wine to water finally ?
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Solution:-
The thought process behind the solution goes something like this:-
To make things simple, Lets just talk about the wine ;)
1) 40 litre wine, 4 litre taken out ---> 36 litre remaining + 4 litre water added
Note we took 10% out. Wine to water ratio = 36:4 = 9:1
2) 40 litre mixture (36 litre wine 4 litre water), 4 litre taken out----> Going by the ratio of wine to water, Amount of wine taken out = 3.6 litre.
Note, here also wine taken out = 10% of existing value.
We can say that with every iteration of the process the amount of wine left in the mixture is 90% of the previous value...
Therefore, the amount of wine left in the mixture after 3 iterations:-
40 * (90/100) (90/100) (90/100)
Final quantity of wine left = 40 * (0.9)^3 = x
Final Ratio of wine to water = x: (40-x)
This question, if attempted through the common methods of arithmetic, will be a source of supreme headache. Digesting the above approach, makes our life much simpler :)
Also, here is a standard solution for a standard problem:-
Hope the article helped,
Till next time,
Take care and bye,
Ashish
for a good small presentation on Alligations and Mixtures.
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