Ken and Ben shared some sweets. Ken took 2/7 of the sweets and Ben took the rest. If Ken took 18 fewer sweets than Ben, how many sweets did they have altogether at first?

Guest Apr 13, 2021

#1**+2 **

From the problem, we have this information:

Kenn took 2/7 of the treats

and Ben took 5/7 of the treats

*If Kenn took 18 fewer sweets than Ben, how many sweets did they have altogether at first?*

First, we figure out how much more Benn took from the pile than Kenn.

We take Ben's total and subtract it from Kenn's total. Putting that in number form it looks like this:

**5/7 - 2/7**

Thankfully, it has the same denominator so we can subtract 5 from 2 which is 3.

Now we know that Ben has 18 (or 3/7) more treats than Kenn so we divide 18 and 3 to figure out 1/7 of the fraction which is 6

**(because 6 multiplied by 3 is 18)**

So now we know 1/7 is 6 we just have to multiply 6 by 7 to get our total. Multiplying it gets you 42

**So in total, they had 42**

Hope this helps! :D

SirAkko Apr 13, 2021