An article was sold for Rs.75. If the percentage profit was numerically equal to the cost price, the cost price of the article is
(a) Rs. 25
(b) Rs. 50
(c) Rs. 75
(d) Rs. 100

Ans: (b) Rs. 50

Explanation:

Let, CP = x

So, P = x%

Now, P% = \displaystyle\frac{\text{SP - CP}}{\text{CP}}\times 100

⇒ x = \displaystyle\frac{75-x}{x}\times 100

Using back calculation,

x = 25, RHS = \displaystyle\frac{50}{25}\times 100 = 200

x = 50, RHS = \displaystyle\frac{25}{50}\times 100 = 50

x = 50 fits the equation, so this is the correct answer

Note: We could solve this problem using middle term factorization but it may be time consuming for many ones

 On simplification, x = \displaystyle\frac{75-x}{x}\times 100 reduces to

x² + 100x – 7500 = 0

⇒ x² +150x – 50x – 7500 = 0

⇒ (x + 150) (x-50) = 0

100 can be written as 150 – 50 [here 150 and 50 are two factors of 7500]

So, we get x = 50, x = -150 (this is impractical since CP can’t be negative)

Hence, CP = 50

We could have used Sridharacharya Formula also but this could take more time as compared to middle term factorization also

x² + 100x – 7500 = 0

a = 1, b = 100, c = – 7500

∴ x = \displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}

\quad= \displaystyle\frac{-100 \pm \sqrt{100^2-4\cdot 1\cdot(-100)}}{2\cdot 1}

\quad= \displaystyle\frac{-100 \pm \sqrt{10000+30000}}{2}

\quad= \displaystyle\frac{-100 \pm \sqrt{40000}}{2}

\quad= \displaystyle\frac{-100 \pm 200}{2}

\quad= \displaystyle-50 \pm 100

\quad= \displaystyle50, -150

So, x = 50