d Word Problem 10 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word Problem 10

Problem 10: A man spends equal amount on buying two types of oranges (kinos) at the rate of 5 oranges for a rupee and 10 oranges for 3 rupees. The average cost of the oranges will be;

a) 24 paisas                    b) 26 paisas                        c) 25 paisas                    d) 27 paisas

(math test for jobs)


The man spends certain amount on one type of oranges (Kinos)

1Rs = 5 Kinos

and he also spend same amount on other type of kinos

@!  3Rs = 10 Kinos \Rightarrow 1 Rs = \frac{10}{3}Kinos

Taking sum of both on both side basis

\Rightarrow 1 Rs + 1Rs = 5 Kinos + \frac{10}{3}Kinos

\Rightarrow 2 Rs = \frac{25}{3}Kinos

Or we can write

\Rightarrow \frac{25}{3}Kinos=2 Rs

\Rightarrow 2 Rs = \frac{25}{3}Kinos

Leaving ‘Kinos’ on one side and shifting 25/3 to other side we get;

\Rightarrow 1Kinos=2 \times \frac{3}{25} Rs

\Rightarrow 1Kinos=0.24 RS

Or \Rightarrow 1Kinos=24 Paisas

So average price of Kinos is 24 paisas per kino




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