Question

A trader sells a total 315 TV sets. He sells black and white TV sets at a loss of \[6\%\] and color TV sets at a profit of $15\%$ thus he gains \[9\%\] on the whole. What is the number of black and white sets which he has sold? (Solve this question assuming that cost price of coloured T.V. is equal to cost price of black and white T.V. )

Answer 1

Hint: To find the number of black and white sets which have been sold, let us find the selling price of colour TVs, black and white TVs and the total selling price. We will be using equations $SP=\left( \dfrac{100+Gain\%}{100} \right)CP$ and $SP=\left( \dfrac{100-Loss\%}{100} \right)CP$ . Let us assume that the Cost price (CP) of colour TV and black and white TV sets $=C$ . Also let us assume $X$ to be the number of colour TVs sold. Therefore, the number of black and white TVs sold is $=315-X$ . Substituting these in \[\text{selling price of X colour TVs+Selling price of (315-X) black and white TVs}=\text{selling price of 315 TVs}\text{.}\] we will get the value of $X$ from which we can find the number of colour and black and white TVs sold.

Complete step by step answer:
We need to find the number of black and white sets which have been sold.
Given that the Cost price (CP) of colour TV and black and white TV sets are similar that is $=C$
Now let us find the selling price of colour TV.
We know that selling price, $SP=\left( \dfrac{100+Gain\%}{100} \right)CP$
It is given that color TV sets are sold for a profit of $15\%$ .
Hence, $SP=\left( \dfrac{100+15}{100} \right)\times C$
This can be written as
$SP=\left( \dfrac{115}{100} \right)\times C$
Now let us find the selling price of black and white TV.
It is given that black and white TV sets are sold for a loss of \[6\%\] .
Hence, we will use the formula
 $SP=\left( \dfrac{100-Loss\%}{100} \right)CP$
Let us now substitute the values. We will get
$SP=\left( \dfrac{100-6}{100} \right)\times C$
This can be written as
$SP=\left( \dfrac{94}{100} \right)\times C$
It is given that the total TV sets sold total \[=315\]
Let $X$ be the number of colour TVs sold.
Then, a number of black and white TVs sold $=315-X$ .
The total profit earned $=9\%$ .
Now, let us find the total selling price. That is
$SP=\left( \dfrac{100+9}{100} \right)C$
Let us simplify this. We will get
$SP=\left( \dfrac{109}{100} \right)\times C$
Let us write the total selling price as\[\text{selling price of X colour TVs+Selling price of (315-X) black and white TVs}=\text{selling price of 315 TVs}\text{.}\] Let us now substitute the values. We will get
$\Rightarrow X\left( \dfrac{115}{100} \right)\times C+(315-X)\times \left( \dfrac{94}{100} \right)\times C=315\times \left( \dfrac{109}{100} \right)\times C$
Now, let us take $C$ outside from LHS and cancel it from RHS. We will get
$\Rightarrow X\left( \dfrac{115}{100} \right)+(315-X)\times \left( \dfrac{94}{100} \right)=315\times \left( \dfrac{109}{100} \right)$
Now let us solve this expression.
$\Rightarrow 1.15X+315\times 0.94-0.94X=315\times 1.09$
Let us collect the constant terms to one side. We will get
$\Rightarrow 1.15X-0.94X=315\times 1.09-315\times 0.94$
Let us simplify this. We will get
$\Rightarrow 1.15X-0.94X=343.35-296.1$
Let us subtract the values. We will get
$\Rightarrow 0.21X=47.25$
From, this $X$ can be found as
$X=\dfrac{47.25}{0.21}=225$
Number of black and white TV sets $=315-225=90$

Hence, number of colour TVs sold is $225$ and the number black and white TVs sold is $90$.

Note: Do not substitute gain or loss by converting it into its fractional form in equations$SP=\left( \dfrac{100+Gain\%}{100} \right)CP$ and $SP=\left( \dfrac{100-Loss\%}{100} \right)CP$ , that is, if the gain is $15\%$ do not do this: $SP=\left( \dfrac{100+\dfrac{15}{100}}{100} \right)CP$ . We can also solve this question by assuming \[X\] to be the number of black and white TVs sold. This is shown below:
Let $X$ be the number of black and white TVs sold.
Then, number of colour TVs sold $=315-X$
Let us write the total selling price as\[\text{selling price of X colour TVs+Selling price of (315-X) black and white TVs}=\text{selling price of 315 TVs}\text{.}\]We can now substitute the values.
\[\Rightarrow X\times \left( \dfrac{94}{100} \right)\times C+(315-X)\times \left( \dfrac{115}{100} \right)\times C=315\times \left( \dfrac{109}{100} \right)\times C\]
Now, let us take $C$ outside from LHS and cancel it from RHS. We will get
\[\Rightarrow X\left( \dfrac{94}{100} \right)+(315-X)\times \left( \dfrac{115}{100} \right)=315\times \left( \dfrac{109}{100} \right)\]
Now let us solve this expression.
$\Rightarrow 0.94X-1.15X+315\times 1.15=315\times 1.09$
Let us collect the constant terms to one side. We will get
$\Rightarrow 0.94X-1.15X=315\times 1.09-315\times 1.15$
Let us simplify this. We will get
$\Rightarrow 0.94X-1.15X=343.35-362.25$
Let us subtract the values. We will get
$\Rightarrow -0.21X=-18.9$
From, this $X$ can be found as
$X=\dfrac{-18.9}{-0.21}=90$
Hence, the number of black and white TV sets sold $=90$
Number of black and white TV sets $=315-90=225$