"> word Problem 10 | Job Exam Rare Mathematics (JEMrare)

word Problem 10

Problem 10: A man sold his watch at loss of 5%. He had sold it for Rs 56.25 more he would have gained 10%. Find the cost price of the watch.

a. Rs 300                   b. Rs 375                  c.Rs 400               d. Rs 200

Solution:

We suppose cost price as ‘y’

CP=y

Loss was 5% of CP

L= \frac{5}{100}of CP

= \frac{1}{20}\times y

= \frac{y}{20}

But fromula for loss is

L=CP-SP

\frac{y}{20}=y-SP

\Rightarrow SP= y-\frac{y}{20}=\frac{19y}{20}

So his sale price was 19y/20

Now come to the situation as wished in the question

If had he sold  it with more 56.25 Rs then sale price would have been 19y/20+56.25 Rs

i-e

Wished sale price;

SP= \frac{19y}{20}+56.25

and in this case he would have earned profit of 10% of his CP

Wished profit;

P=\frac{10}{100}of CP

P=\frac{y}{10}

But formula for profit is

P=SP-CP

\frac{y}{10}=\frac{19y}{20}+56.25-y

\frac{y}{10}=56.25-\frac{y}{20}

\frac{y}{10}+\frac{y}{20}=56.25

\Rightarrow\frac{3y}{20}=56.25

\Rightarrow y=\frac{20}{3}\times56.25

\Rightarrow y=375

So, the cost price for the watch is Rs 375

 

 

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