# Profit and Loss.

A person has 100 kg sugar. some of it he sold at 7% profit and remaining sold at at 17% profit.total profit he got was 10%. Then how much of kg he sold at 17% profit?

• Some effort to solve this problem ought to be shown. What don't you understand?
– zoli
Commented May 11, 2018 at 12:40
• How would you approach this question? Hint: Always when there is some unknown thing (here it's the word "some of it"), it's a good idea to mark it with an $x$. Commented May 11, 2018 at 12:40
• this Questionn was in Hindi language so i translated to English .. X was not give there .. that we have to assume Commented May 11, 2018 at 18:41

Suppose $x$ kg. was sold at $7\%$ profit. It follows that $100-x$ kg. was sold at $17\%$ profit. Can you find the total profit percentage expressed in $x$?

• Is it necessary to put a dot after physical units? Or it's just your convention?
– AHB
Commented May 11, 2018 at 12:58
• @AHB that’s what I was taught. Commented May 11, 2018 at 12:58

Assume that the per kilogram price is $A$. That is, the merchant paid $100A$ Rupees for the sugar.

If the merchant sells the sugar for $B$ Rupees and if the profit profit is $10\%$ then $$B=100A\times1.10$$

How much is $B$? If the merchant made $17\%$ profit on $x$ kilograms of sugar and $7\%$ profit on $100-x$ kilograms of sugar then

$$B=xA\times 1.17+(100-x)A\times1.07.$$That is, we have the following equation $$100A\times 1.10=xA\times 1.17+(100-x)\times 1.07.$$

$A$ cancels out and we have $$110=x1.17+107-x1.07$$ or $$3=x0.1.$$

So, $$x=\frac3{0.1}=30 [kg].$$

• superb .. its really understable.. thank you so much :) Commented May 11, 2018 at 18:51

$x$ kg. we get $7\%$ profit, $100-x$ we get $17\%$ profit and for the entire $100$ kg. we get $10\%$ profit.

hence for $x$ kg. we get $x+\frac{7x}{100}$ and for $(100-x)$ kg. we get $100-x+\frac{17\cdot(100-x)}{100}$ and for $100$ kg. we get $100+\frac{10}{100}$

Combine in equation $$x+\frac{7x}{100}+100-x+\frac{17\cdot(100-x)}{100}=110$$

Solve for $x$ and the solution is $30$.

• yeah thats helped me... thank you Commented May 11, 2018 at 18:55

$x + y = 100$.........(1)

$1.07x + 1.17y = 110$........(2)

Multiply equation (1) by $1.07$

Subtract the new equation (1) from equation (2) to eliminate the x variable and solve for y for the quantity sold at $17\%$ profit.