Jill and rick had a total of 90 stickers at first. Jill lost 1/4 of her stickers to rick. In the 2nd game, rick lost 2/5 of his stickers to jill. As a result, the two children have the same number of stickers now. How many stickers did they have at first ?
it will be good if anyone can help to solve using fraction method or ratio method. Models definitely welcome. Thanks!
sushi88
At first,
Jill [——-u——–]
Rick [——————————-(90-u)———————————–]
After Jill gave 1/4u stickers to Rick, Rick’s stickers became (90-u+ (1/4)u) = 90-(3/4)u
Final:
Jill [–(3/4)u-][(2/5)(90-(3/4)u)]
Rick [——-(3/5)(90-(3/4)u)——-]
(3/4)u + (2/5)[ 90-(3/4)u ] = (3/5)[ (90-(3/4)u ]
(3/4)u + 36 – (6/20) u = 54 – (9/20)u <—Multiply both sides by 20,
15u + 720 – 6u = 1080 – 9u
15u – 6u – + 9u = 1080 – 720
18u = 360
u = 20 , 90-u = 90-20 = 70
Hence at first, Jill has 20 stickers and Rick has 70 stickers.
ChiefKiasu
Sorry… I can only do simultaneous equations:
Let J be number of stickers Jill had at first.
Let R be number of stickers Rick had at first.
J + R = 90 … (1)
(3/4)J + (2/5)(R + (1/4)J)) = 45
((3/4)+(1/10))J + (2/5)R = 45
((15+2)/20)J + (2/5)R = 45
(17/20)J + (2/5)R = 45
(5/2)(17/20)J + R = (5/2)45
(85/40)J + R = 112.5 … (2)
(2) – (1):
(85/40)J – J = 22.5
(45/40)J = 22.5
J = (40/45)22.5 = 20 #
So, R = 70 #
Can someone else please do models?
