Question

a man sold 12 candies in 10$ had loss of b% then again sold 12 candies at 12$ had profit of b% find the value of b.

Answer 1

Let the CP be 'x' then:
(x-10)*100/x = b% ----------->1
(12- x) * 100 /x = b% ------->2
(x-10)*100 / x = (12-x)*100 / x
x-10 = 12-x
x = 11
So by Submitting value of x into one of above question (1 or 2) we get b
which is 9.09 or 9 Approx

Answer 2

A man sold 12 candies in 10$ had loss of b% then again sold 12 candies at 12$ had profit of b%, the value of b is 9.09.

Given,

candles = 12

Selling prize 1 = $10 with b% loss

Selling prize 2 = $12 with b% profit

To find,

b = ?

Solution,

Let the cost prize be 'x'.

% Loss = $[\frac{loss}{x}]100$

$b = [\frac{x-10}{x} ]100$

% Profit = $[\frac{profit}{x}]100$

$b = [\frac{12-x}{x} ]100$

$[\frac{12-x}{x} ]100 = [\frac{x-10}{x} ]100\\\\12 - x = x - 10\\\\12 + 10 = 2x\\\\22 = 2x\\\\11 = x$

Since, we have value of x, lets substitute it in equation.

$b = [\frac{11-10}{11} ]100\\\\b = [\frac{1}{11}]100\\\\b = 9.09%$

Therefore, the value of b is 9.09%.