Find Tuition/Enrichment Centres

AskQ logo

STUCK ON HOMEWORK?

ASK FOR HELP FROM OUR KIASUPARENTS.COM COMMUNITY!

Question

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! 

Answer

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.

 

 

0 Replies 1 Like

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?

0 Replies 1 Like
Find Tuition/Enrichment Centres