+0

# Can you solve this?

+1
34
2

A market had only cookies and donuts.The donuts were the 3/5 of the market and were 5 more than cookies.How many cookies were there?

Guest Apr 19, 2017
Sort:

#1
+18015
+1

The donuts were the 3/5 of the market and were 5 more than cookies.

Let d = donuts

$$\begin{array}{|lrcll|} \hline (1) & c &=& d-5 \\\\ (2) & \frac{3}{5}\cdot ( c+d) &=& d \quad & | \quad c=d-5 \\ & \frac{3}{5}\cdot ( d-5+d) &=& d \\ & \frac{3}{5}\cdot ( 2d-5 ) &=& d \\ & 3\cdot ( 2d-5 ) &=& 5d \\ & 6d-15 &=& 5d \\ & \mathbf{d} & \mathbf{=} & \mathbf{15} \\\\ & c &=& d-5 \\ & c &=& 15-5 \\ & \mathbf{c} & \mathbf{=} & \mathbf{10} \\ \hline \end{array}$$

heureka Apr 19, 2017
#2
+72830
+1

Since the donuts were 3/5 of the total....let  the number be represented by (3/5)T where T is the total of donuts and cookies

Then the number of cookies must have been (2/5)T

And we know that adding 5 to the number of cookies = the number of donuts

So we have

(3/5)T   =  (2/5)T + 5       subtract  (2/5)T from both sides

1/5T  =  5                      multiply through by 5

T  =  25

So.....the number of cookies  =  (2/5)(T) = (2/5)(25)  = 10

CPhill Apr 19, 2017

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.   See details