Question

A man sold two pipes at Rs 12 each. On one he gained 20% and on other he lost 20%. What is his gain / loss in the whole transaction?
(a) Gain of Rs. 3
(b) Loss of Rs. 3
(c) Loss of Rs. 1
(d) Gain of Rs. 1

Answer 1

Hint: Take the cost price of both pipes as $C{{P}_{1}}$ and $C{{P}_{2}}$. Find $C{{P}_{1}}$ using formula for gain % as gain of 20% is given for pipe 1. Now, find $C{{P}_{2}}$ using the formula of Loss% as loss of 20% is given on ${{P}_{2}}$ . Now compare total selling price and cost price to find out gain/loss.

Complete step-by-step answer:
It is said that a man sold two pipes. Now both the pipes were sold at Rs 12 each. Hence, the selling price of each pipe is Rs. 12. Now, he gained 20% with the first pipe which we can mark as ${{P}_{1}}$. Then he lost 20% with the second pipe, which we can mark as ${{P}_{2}}$.
Let us first find the cot price of pipe 1, ${{P}_{1}}$ . We know the formula for finding gain,
Gain $=$ selling price $-$ cost price $=SP-C{{P}_{1}}$ …………………..(i)
For ${{P}_{1}}$ , gain is 20% and $SP=Rs\text{ 12}$
Thus, gain percentage $\text{=}\left( \dfrac{gain}{Cost\text{ price}} \right)\times 100$
$\text{=}\left( \dfrac{SP-C{{P}_{1}}}{C{{P}_{1}}} \right)\times 100$
$\Rightarrow \text{20=}\left( \dfrac{12-C{{P}_{1}}}{C{{P}_{1}}} \right)\times 100$
Now let us apply cross multiplication property and simplify it.
$C{{P}_{1}}\text{=}\left( \dfrac{12-C{{P}_{1}}}{20} \right)\times 100$
$C{{P}_{1}}\text{=}\left( 12-C{{P}_{1}} \right)\times 5$
$C{{P}_{1}}+5C{{P}_{1}}=60$ $\Rightarrow 6C{{P}_{1}}=60$
$\therefore C{{P}_{1}}=Rs.10$
Thus, we got the cost price of the first pipe ${{P}_{1}}$ as Rs. 10.
Now it is said that pipe 2, ${{P}_{2}}$ encounter loss of 20%. The cost price of pipe 2 is $C{{P}_{2}}$. We know the formula of loss as
Loss $=$ Cost price $-$ selling price $=C{{P}_{2}}-SP$ …………………………(ii)
$\therefore Loss\%=\dfrac{loss}{\operatorname{C}ost\text{ price}}\times 100$
i.e. $Loss\%=\left( \dfrac{C{{P}_{2}}-SP}{C{{P}_{2}}} \right)\times 100$
$20=\left( \dfrac{C{{P}_{2}}-12}{C{{P}_{2}}} \right)\times 100$
Let us apply cross multiplication property and simplify it.
$C{{P}_{2}}=\left( \dfrac{C{{P}_{2}}-12}{20} \right)\times 100=\left( C{{P}_{2}}-12 \right)5$
$=5C{{P}_{2}}-60$
$5C{{P}_{2}}-C{{P}_{2}}=60$
$4C{{P}_{2}}=60$ $\Rightarrow C{{P}_{2}}=\dfrac{60}{4}=Rs.15$
Hence, we got the cost price of pipe 2, ${{P}_{2}}$ as Rs. 15
Now, let us see the total cost price $=C{{P}_{1}}+C{{P}_{2}}=Rs.10+Rs.15=Rs.25$
Total selling price $=$ SP of 2 pipes $=12+12=Rs.24$
Now from this SP is less than CP. Thus, we have encountered a loss. Now, let us find the loss percentage.
$Loss \%=\left( \dfrac{CP-SP}{CP} \right)\times 100=\left( \dfrac{25-24}{25} \right)\times 100$
$=1\times 4=4\%$
Form (ii),
Loss $=$ Cost price $-$ selling price $=25-24=Rs.1$
Hence, in the whole transaction he lost Rs. 1. Hence, there was a loss of Rs. 1.
$\therefore $ option (C) is the correct answer.

Note: We can also find the required loss % as $\dfrac{{{y}^{2}}}{100}$ where $y=20\%$
Required loss% $=\dfrac{{{\left( 20 \right)}^{2}}}{100}=4\%$
SP of 2 pipes $=Rs.24$
$\left( 100-4 \right)\%=24$
i.e. $4\%=\dfrac{24}{96}\times 4=\dfrac{96}{96}=1$
i.e. loss of Rs. 1.