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 By: richardli (offline)  Sunday, July 08 2012 @ 01:20 AM CDT (Read 818 times)
richardli

At the start of a game, Alice had 75% as many cards as Bob.
Bob had 30% as many cards as Calvin. During the first round,
Alice lost1/3 of her cards to Bob. During the second round, Bob
lost 30% of his card to Calvin. At the end of the game, Calvin had
462 more cards than Alice.

a) how many cards did Bob have at first?
b)what was the percentage increase in Calvin's number of cards
at the end of the game?

Newbie

Registered: 12/31/06
Posts: 10

 By: echeewh (offline)  Sunday, July 08 2012 @ 09:44 PM CDT
echeewh

Hey Richard,

Following pls find my worked solution:

A : B : C

3 : 4 -- (1)
3 : 10 -- (2)

<Apply Repeated / Common Identity concept> means B must have same number of units in these 2 ratios.

(1)x3: (2)x4:

9 : 12 : 40

<Round 1>
A lost (1/3) of her cards to B.

A: 9 - 3 = 6u
B: 12 + 3 = 15u
C: 40u

<Round 2>
B lost 30% (3/10) of his cards to C.

A: 6u
B: 15u - [(3/10) x 15]u = 15u - (9/2)u = (21/2)u
C: 40u + (9/2)u = (89/2)u

Given that C - A = 462, we have ...

(89/2)u - 6u --> 462
89u - 12u --> 462 x 2
77u --> 924
1u --> 12

(a) B (at first) = 12u --> 12 x 12 = 144
(b) % inc in C's cards at end of game

(89/2)u - 40u = (9/2)u
(9/2)u --> (9/2) x 12 = 54
% inc = [54 / (40 x 12)] x 100 = (54/480) x 100
= 11.25%

==================================

Trust this helps.

Do let me know again if there's any clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 623

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