Question

The man solds two pens at Rs.99 each. If he sold one of the pen at a loss of 10% and the other one at a gain i.e., at a profit of 10% then find his loss or gain (in percent).
A. $10\% $ loss
B. $1\% $ loss
C. $9\% $ gain
D. $2\% $ gain

Answer 1

Hint: We have to use the equation for finding the cost price of two pens respectively,
${\text{Cost price(in \% ) = }}\dfrac{{100}}{{[100 \pm (profit\,or\,loss\,percent)}} \times Selling\,price$
We solve this equation using the substitution method. We further relate the investment of the pens using the formula $Total\,\cos t\,price = C.{P_1} + C.{P_2}$ and $Total\,selling\,price = S.{P_1} + S.{P_2}$ & then comparing both the equations with the help of $Total\,loss = (Total\,\cos t\,price - Total\,selling\,price)$ . We also know the formula to find the total profit or loss (in percentage $\% $ ) that is $Loss(in\,\% ) = \dfrac{{Total\,loss}}{{Total\,\cos t\,price}} \times 100$ . After simplification of all the required values we find the required solution.

Complete step by step solution:
Let’s Assume that,
Cost price of pen1 $ = C.{P_1}$
Cost price of pen2 $ = C.{P_2}$
Selling price of pen1 $ = S.{P_1}$
Selling price of pen2 $ = S.{P_2}$
 We have given the Selling Price (S.P.) of two pens are as follows,
i.e., ${(S.P)_1} = {(S.P)_2} = Rs.\,99$ ……..(i)
For the pen 1,
Man receives loss of $10\% $ ,
Thus, $Loss = 10\% $
Hence, we know that,
$ \Rightarrow (C.P) = \dfrac{{100}}{{(100 - 10)}} \times S.P$
Substituting the values in above equation, we get
$\therefore C.{P_1} = \dfrac{{100}}{{90}} \times 99$
$ \Rightarrow C.{P_1} = Rs.110$ ………(ii)
Cost price of pen 1 is Rs. 110 respectively.
Similarly,
For the pen 2,
Man receives gain of $10\% $ ,
Thus, $\Pr ofit = 10\% $
Hence, we know that,
$ \Rightarrow C.{P_2} = \dfrac{{100}}{{(100 + 10)}} \times S.{P_2}$
$\therefore C.{P_2} = \dfrac{{100}}{{110}} \times 99$
$\therefore C.{P_2} = Rs.\,90$
So, Cost price of pen 2 is Rs. 90 respectively.
Now, we can write:
$Total\,\cos t\,price = C.{P_1} + C.{P_2}$
From eq(ii) and (iii)-
$\therefore Total\,\cos t\,price = 110 + 90 = Rs.\,200$ ………..(iv)
Similarly,
$\therefore Total\,selling\,price = S.{P_1} + S.{P_2}$
Now, from eq(i)
$\therefore Total\,selling\,price = 99 + 99 = Rs.198$ ……..(v)
Hence, from (iv) and (v)
As a result,
$Total\,loss = (Total\,\cos t\,price) - (Total\,selling\,price)$
As $(\because Total\,\cos t\,price > Total\,sel\operatorname{li} ng\,price)$
$\therefore Total\,loss = 200 - 198 = Rs.\,2$ ……..(vi)
Hence, from eq(iv) and eq(vi):-
$
  Loss(in\,\% ) = \dfrac{{Total\,Loss}}{{Total\,\cos t\,price}} \times 100 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{{200}} \times 100 = Rs.\,1 \\
 $
     Loss (in %) = 1
Thus, there is a total 1% loss in the entire investment.
Therefore, Correct option is B $1\% loss$ respectively.

So, the correct answer is “Option B”.

Note: We have definite simultaneous equations through the required solution path. If we follow the appropriate sequence you can surely solve the problem quickly. We know some of the formulas which yield the profit, loss lessons in previous standards:
$C.P = \dfrac{{100}}{{(100 - 10)}} \times S.P$ ( $ + , - $ signs varies as per the condition/s i.e., in case of profit, positive $( + )$ sign is imparted and negative $( - )$ sign in case of loss)
$Total\,loss = (Total\,\operatorname{Cos} t\,\Pr ice) - (Total\,Selling\,price)$
$Loss(in\,\% ) = \dfrac{{Total\,loss}}{{Total\,\cos t\,price}} \times 100$
We use them according to the given problem.