"> Word Problem 13 | Job Exam Rare Mathematics (JEMrare)

Word Problem 13

Problem 13: A man buys a certain number of oranges at 24 a rupee and an equal number at 30 a rupee. He mixes these at 27 a rupee. What does he gain or loss percent?

a. 1-19/81%                 b. 1-5/7%                   c. 5-1/2%                  d. 5-1/3%

GRE arithmetic questions and answers

Solution:

Suppose the man buys ‘y’ number each rated type of oranges

then for 24 a rupee oranges we can write;

24 Oranges = 1 rupee

One Orange =\frac{1}{24}

y ,Orange =\frac{y}{24}

Similarly for 30 a rupee oranges we write;

y ,Orange =\frac{y}{30}

So, cost prices for y oranges of 24 a rupee type and y oranges of 30 a rupee type are y/24 and y/30 rupees respectively.

Now total cost price for both type of oranges is;

CP=\frac{y}{24}+\frac{y}{30}=\frac{3y}{40}

Now the man  mixes  both types oranges which become 2y in number. He sells these at 27 a rupee rate.

for sale price of 2y oranges at this rate we can write;

27 ornages = 1 rupee

 \Rightarrow 1 orange = \frac{1}{27} rupees

 \Rightarrow 2yOrange = \frac{2y}{27} rupees

so mixed oranges sale price is

 SP = \frac{2y}{27} rupees

Here we can see now that CP is greater than SP i-e total cost price for individual type is greater than sale price for mixed oranges. It means he is earning no profit but a loss !

formula for loss is

Loss=Cost Price-Sale Price

or

L=CP-SP

\Rightarrow L=\frac{3y}{40}-\frac{2y}{27}

\Rightarrow L=\frac{y}{1080}

Now formula for loss %age is

Loss%age=\frac{loss}{cost}\times100

Putting values of loss L and cost CP from above

Loss%age= \frac{\frac{y}{1080}}{\frac{3y}{40}}\times 100

\Rightarrow Loss%age= {\frac{y}{1080}\times \frac{40}{3y}\times 100

After simplification we get;

 Loss%age= 1\frac{19}{81}%

                                                                      So, the option ‘a’ is correct answer

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