+0  
 
+1
48
1
avatar

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?

 Apr 13, 2021
 #1
avatar+59 
+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



 

 Apr 13, 2021
edited by SirAkko  Apr 13, 2021

18 Online Users