Problem Solving

Hi friends,

I was not able to solve this problem. Can someone please solve it..

A store owner has a stock of 100 shirts. He sold some shirts for a 10 day period at a profit of 20%. Thereafter, he put the shirts on a sale, and sold the shirts at a lower profit of 10%. If he made an overall profit of 13%, how many shirts did he sell in the sale?

Thank you so much

The shirts will be sold in the ratio

(20-13)/(13-10) = 7/3

So number of shirts sold at 20% profit = 30
at 10% profit = 70

We can create an equation based on total sale price.

x = base price
n = num shirts sold at 20% profit

1.2*x*n + 1.1*x*(100-n) = 1.13*100*x
.1n = 3
n = 30

So, num shirts sold in the sale = 70

Thanks a lot....

Also:
is this like those mixture problems. Can it be solved using a table format? if so, how can we solve it using a table?

Yup i solved it using the mixtures process.

Not a table but an "X" On the top left write 20 on top right write 10
At the middle of the X write 13
On the bottom write the difference i.e. 20-13 = 7
and 13-10 = 7 (Dont worry about the difference OR consider Mod subtraction

You'll get 3 and 7. carry as it is to the top so that you now get ratio of 20:10 is 3:7

Hope this helps. Am not at home else would have attached an image. Will hopefully add that later in the evening.

Thank you Neo2000, I understand your method.

Pretty simple method to solve

Assume the cost of each shirt as $100

100 shirts (x sold at a profit of 20% and y sold at a profit of only 10%)

For x Shirts the Selling price is $120 (100 + 20%)

For y Shirts the Selling Price is $110 (100 + 10%)

x + y = 100

Therefore y = 100 - x

Now profit % = 13

Formula for Profit % = (Selling Price - Cost Price / Cost Price) * 100

13 = (120x + 110(100-x) - 10,000 (100* $100))/10,000

1300 = 120 x + 11000 - 110x - 10,000

1300 = 10x + 1000

x = 30

Therefore Y = 100 - 30 = 70