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# Can you solve this?

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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
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#1
+18368
+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
+75302
+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

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