d Word Problem 3 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Word Problem 3

Problem 3: A garrison have provision for certain number of days, after 10 days 1/5th of the men desert, and it is found that the provision will now last as long as before. How long was that ?

a) 25 days         b) 50 days                  c) 15 days                  d) 5 days



Solution:

We suppose that in the beginning , provision was sufficient for ‘x’ days and there were ‘y’ men in the garrison. So after 10 days 1/5th of ‘y’ left away – then to see the situation , we arrange data as;

Now it has been said that provision will again last for same number of days as in the beginning, so the data before 11th day and on 11th day can be arranged as;

Now this is problem of inverse proportion- i-e-if men are less in number, more days the food will go on, so we have put arrow signs in opposite directions;

Now, solving for x and y

\frac{y}{\frac{4y}{5}}=\frac{x}{x-10}

\Rightarrow y(x-10)=x(\frac{4y}{5})

\Rightarrow xy-10x=\frac{4xy}{5}

\Rightarrow 5(xy-10x)=4xy

\Rightarrow 5xy-50x=4xy

\Rightarrow 5xy-4xy=50x

\Rightarrow xy=50x

\Rightarrow y=50

So, the provision was sufficient for 50 days

 

 

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