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 By: Tping (offline)  Wednesday, August 28 2013 @ 09:47 AM CDT (Read 685 times)
Tping

Q1)
Frederic had a candy bag which contained 35 chocolate bars and 50 sweets.
Ryan had another candy bag which contained 45 chocolate bars and 10 sweets.
After Frederic gave Ryan some chocolate bars and sweets,
40% of Frederic's candy bag contained chocolate bars and
30% of Ryan's candy bag contained sweets.
How many chocolate bars and sweets did Frederic give to Ryan altogether?

Junior

Registered: 01/26/12
Posts: 25

 By: echeewh (offline)  Wednesday, August 28 2013 @ 02:53 PM CDT
echeewh

Hello <Tping>,

My worked solution is as follows:

This is a typical example of an <Internal Transfer> process {F gave R some chocolate bars and sweets}. Thus we can apply the <Unchanged Total> concept, i.e. the total number of chocolate bars (C) <before> and <after> remains the same. Likewise, the total number of sweets (S).

<before>

FC = 35, FS = 50
RC = 45, RS = 10

Total number of C: 35 + 45 = 80
Total number of S: 50 + 10 = 60

<process>
F gave R some chocolate bars and sweets

<after>

FC : FS
40 : 60
2 : 3

RC : RS
70 : 30
7 : 3

Applying <Unchanged Total> concept, we have ...

<C>
2p + 7u = 80 -- (1)

<S>
3p + 3u = 60 -- (2)

Eliminating (p), we have ...
(1)x3:
6p + 21u = 240 -- (3)

(2)x2:
6p + 6u = 120 -- (4)

(3)-(4):
21u - 6u = 240 - 120
15u --> 120
1u --> 120 ÷ 15 = 8

Hence,
<After>
RC = 7u --> 7 × 8 = 56
RS = 3u --> 3 × 8 = 24

Given that at first,
RC = 45, RS = 10

F gave R:
C: 56 - 45 = 11
S: 24 - 10 = 14

Total given: 11 + 14 = 25

Thus, Frederic gave 25 chocolate bars and sweets altogether to Ryan.

=========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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