Ben had 5 times as many marbles as Caleb at first. After Ben gave away 87 marbles and Caleb won 13 marbles, they had the same number of marbles left. How many marbles did each of them have in the end?

Guest Aug 23, 2021

edited by
Guest
Aug 23, 2021

edited by Guest Aug 23, 2021

edited by Guest Aug 23, 2021

#1**0 **

Let 'b' be Ben's amount at first, and 'c' be Caleb's amount at first.

We can form 2 equations: b = 5c and b - 87 = c + 13, so 5c - 87 = c + 13, so 4c = 100, so c = 25, b = 125.

**BUT THE QUESTION ASKS FOR HOW MANY MARBLES THEY HAVE AT THE END.**

Since they both have the same amount left, 125 - 87 = 25 + 13 = **38.**

Guest Aug 23, 2021

#2**0 **

Let us assume that Caleb had x marbles. As Ben had five times as many marbles as Caleb, therefore, Ben had 5x marbles with him.

It is given that Ben gave away 87 marbles, therefore, the number of marbles remaining

with Ben was:

M_{B }= 5x - 87

It is also given that Caleb won 13 new marbles. Therefore, the number of marbles remaining with

Caleb was:

M_{C} = x + 13

It is given that in the end, both Caleb and Ben had the same number of marbles. Therefore, equate the expressions obtained for the number of marbles left with Ben and Caleb and

solve for x.

M_{B} = M_{C}

5x - 87 = x + 13

5x - x = 13 + 87

4x = 100

x = 100/4

x = 25

Therefore, the number of marbles left

with Ben and Caleb were:

M_{B }= 5x - 87

= 5 (25) - 87

= 38

Answer - Each of them had 38 marbles in the end.

KourageKowardlyDog Aug 24, 2021