Math, asked by ANMOL9035, 1 year ago

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.

Answers

Answered by ayushpratapsingh33
23
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
Answered by swethassynergy
0

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%.

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