Vanilla Cake:

Hi kohjl,

Q18 of paper 2 (last question) is something like this and I think that your son has got the right answer ( 68 ).Here it goes (not in exact words):

Q18
Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim.Jim ate 12 sweets and Ken ate 18 chocolates.The ratio of Jim's sweets to chocolates became 1 : 7 and the ratio of Ken's sweets to chocolates Ken became 1:4.How many sweets did Ken buy?

Suggested solutions
Draw a model to show \"after eating\" scenario, \t\t
\t\t\t
Ratio of Jim's sweets to chocolate (1:7)
Sweets []
Chocos [][][][][][][]

Ratio of Ken's sweets to chocolate (1:4)
Sweets [] +12 (+12 refers to the sweets eaten by Jim)\t\t\t
Chocos [][][][] (+12x4) - x 4 times
\t
Both amount of Chocolates are the same,so
(7-4) units = (12x4) + 18 (18 chocolates were eaten by Ken)
3 units = 66
1 unit = 22\t\t
\t\t\t
Amount of sweets bought by Ken->(unit +12)x2=(22+12) x 2 =68

Or use algebra which is easier.

Assume that sweets that Ken bought was S and chocolates that bought by Jim was C.

Before both of them ate, they had:
Ken -> 0.5C + 0.5S
Jim -> 0.5C + 0.5S
as each of them gave 1/2 to each other.

Jim ate 12 sweets and the ratio of Jim's sweets to chocolates became 1:7.
(0.5S-12)/0.5C = 1/7
3.5S-84=0.5C

Ken ate 18 chocolates and the ratio of Ken's sweets to chocolates became 1:4.
0.5S/(0.5C-18 ) = 1/4
(0.5C-18 ) / 0.5S = 4
0.5C-18 = 2S

3.5S-84-18=2S
3.5S-2S = 84+18
1.5S = 102
S = 102/1.5 = 68

Amount of sweets bought by Ken-> 68

If you were to take this answer and work backwards, will it work? mine ans is 44 since both Jim and Ken did not eat the stuff that they bought, their ratio at the start and at the end would be the same isnt it.. correct me if i am wrong..