"> Word Problem 14 | Job Exam Rare Mathematics (JEMrare)

Word Problem 14

Problem 14: A dealer bought two horses for Rs 480. He sold one at a loss of 15% and the other at a gain of 19% and then he found that each horse was sold for the same price. Find the cost price of each.

a. Rs 350 and 150                 b. Rs 280 and 200                    c. Rs 250 and 230            d. Rs 300 and 180

Solution:

Cost Price of two horses =CP=Rs 480

Loss = L= 15%

Profit=P= 19%

Selling price of one horse=Selling price of other

Cost Price of each horse=?

We start by supposing cost prices of two horses to be ‘y’ and ‘z’ Rs

Now first horse is sold at 15% loss, so,

Selling PricHorse1=\frac{85y}{100}

Second horse is sold at 19% profit, so,

Selling PricHorse2=\frac{119z}{100}

Since given that both horses sold at same price

\Rightarrow \frac{119z}{100}=\frac{85y}{100}

\Rightarrow 119z=85y

\Rightarrow y=\frac{119z}{85}-------(1)

But we are given that total cost is Rs 480

so,

\Rightarrow y+z=480---(2)

Now putting value of y from eq 1 in eq 2

\Rightarrow 480=\frac{119z}{85}+z

\Rightarrow 480=\frac{119z+85z}{85}

\Rightarrow 480 \times 85=119z+85z

\Rightarrow z=\frac{480\times 85}{204}=200

S, the cost price of second horse is found to be Rs 200. Put this value of z in eq 2

\Rightarrow 480=y+200

\Rightarrow y=480-200=280

So the cost price of 1st horse is Rs 280

So, the option ‘b’ is correct answer

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