+0

# Fraction Problem

0
281
4

Cayden and Jayden had a total of $784. After Cayden spent 1/6 of his money on food and Jayden spent$124 on toys, they had the same amount left. How much money did Cayden have at first?

Jun 22, 2021

#1
+4

Let's say cayden had x dollar. So Jayden had (784-x) dollar from 1st information.

From 2nd information:

x-x/6=(784-x)-124,

or, x*11/6=660

or, x=360.

So, cayden had 360$at first. Jun 22, 2021 edited by Tgfromthe4 Jun 22, 2021 #2 +4 Let, Cayden had$x and Jayden had             $y. According to the question, x + y = 784 ...............(1) and, x - $${x \over 6}\frac{}{}$$ = y - 124 5x - 6y = -744 ................(2) Solving equation (1) and (2) eq(1) *6 + eq(2) 6x + 6y = 4704 5x - 6y = -744 ............................................... 11x = 3960 x = 360 Therefore, Cayden had$360 at first.

Jun 23, 2021
edited by apsiganocj  Jun 23, 2021
edited by apsiganocj  Jun 23, 2021
#3
+4

Let the amount clayden has be X

Given :

sum of money clayden and jayden have=$784 So x+ amount jayden has =$784

Amount jayden has=$784-x Clayden spent $${1 \over 6}\frac{}{}$$ =$${1 \over 6}\frac{}{}$$* X So amount left with clayden=x-$$x = {x \over 6}{}^{}\frac{}{}$$ Amount left with clayden =$${5x \over 6}\frac{}{}$$ Jayden spends$124

So amount left with jayden = $784-$124-x

So amount left with jayden=$660-x Given :both have same amount left Hence - 660-x=$${5x \over 6}\frac{}{}$$ 660*6-6x=5x 660*6=11x 60*6=x 360=x Amount clayden has =$360

Jun 26, 2021
#4
+4

c+j =784

left money

$${5c \over 6}\frac{}{}$$ = j- 124

$${5c \over 6}\frac{}{}$$=784-c-124

$${11c \over 6}\frac{}{}$$=660

c=360 (ans)

Jun 26, 2021