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Question: A man bought two goats for Rs. 3000. He sold one at loss of 25% and the other at profit of 50%. If e...

A man bought two goats for Rs. 3000. He sold one at loss of 25% and the other at profit of 50%. If each goat was sold for the same price, the cost price of the goat which was sold at a loss was:
(a) Rs. 1648
(b) Rs. 1256
(c) Rs. 2000
(d) Rs. 2048

Explanation

Solution

Hint:Here, we are given total cost price of two goats i.e. Rs. 3000 along with that, he made loss by selling 1 goat and made profit by selling another goat at the same selling price. So, we need to make equation on both loss and profit percentage by using the formula
SP=(100loss%)100×CPSP=\dfrac{\left( 100-loss\% \right)}{100}\times CP and SP=(100+profit%)100×CPSP=\dfrac{\left( 100+profit\% \right)}{100}\times CP

Complete step-by-step answer:
We have a total cost price of Rs. 3000. Let us assume the cost price of goat 1 as x1{{x}_{1}} and cost price of goat 2 as x2{{x}_{2}} . So, we can say that
x1+x2=3000{{x}_{1}}+{{x}_{2}}=3000 …………………………(i)
Now, it is said that the selling price of both goats are the same. So, SP1=SP2S{{P}_{1}}=S{{P}_{2}} where SP1S{{P}_{1}} and SP2S{{P}_{2}} are selling price of goat 1 and 2 respectively.
Using the formula for selling price, we will make an equation for both goats.
SP1S{{P}_{1}} at loss of 25% will be SP1=(100loss%)100×CPS{{P}_{1}}=\dfrac{\left( 100-loss\% \right)}{100}\times CP
SP1=(10025)100×x1S{{P}_{1}}=\dfrac{\left( 100-25 \right)}{100}\times {{x}_{1}} Where CP =x1={{x}_{1}} , Loss%=25Loss\% = 25
On solving, we get
SP1=(75)100×x1S{{P}_{1}}=\dfrac{\left( 75 \right)}{100}\times {{x}_{1}}
SP1=0.75x1=34x1S{{P}_{1}}=0.75{{x}_{1}}=\dfrac{3}{4}{{x}_{1}} ……………………………(ii)
Similarly,
SP2S{{P}_{2}} at profit of 50% will be SP2=(100+profit%)100×CPS{{P}_{2}}=\dfrac{\left( 100+profit\% \right)}{100}\times CP
SP2=(100+50)100×x2S{{P}_{2}}=\dfrac{\left( 100+50 \right)}{100}\times {{x}_{2}} Where CP =x2={{x}_{2}} , profit%=50profit\% =50
SP2=(150)100×x2S{{P}_{2}}=\dfrac{\left( 150 \right)}{100}\times {{x}_{2}}
SP2=1.5x2=32x2S{{P}_{2}}=1.5{{x}_{2}}=\dfrac{3}{2}{{x}_{2}} ……………………………..(iii)
Now given that the selling price is the same. Therefore, comparing equation (ii) and (iii) we get
34x1=32x2\Rightarrow \dfrac{3}{4}{{x}_{1}}=\dfrac{3}{2}{{x}_{2}}
On equating, we get
x1=2x2\Rightarrow {{x}_{1}}=2{{x}_{2}} …………………………(iv)
Thus, we get the CP of goat 1 is twice the CP of goat 2.
Now doing substitution of equation (iv) in equation (i), we get
2x2+x2=30002{{x}_{2}}+{{x}_{2}}=3000
3x2=30003{{x}_{2}}=3000
x2=1000{{x}_{2}}=1000
Putting the above value in equation (iv) to get the value of x1{{x}_{1}} .
x1=2×1000{{x}_{1}}=2\times 1000
x1=2000{{x}_{1}}=2000
Thus, we are asked to find the cost price of goat 1 which is x1{{x}_{1}} and is equal to Rs. 2000.
Hence, option (c) is correct.

Note: Students should know the formula of profit and loss percentage and be careful while doing calculations. There are also chances of making mistakes by taking 25% as profit and 50% as loss. Also, the formula to find profit is equal to selling price minus cost price i.e SPCPSP-CP and to find loss, the formula is cost price minus selling price i.e CPSPCP-SP..