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 By: shirley wee (offline)  Wednesday, May 09 2012 @ 07:09 AM CDT (Read 783 times)
shirley wee

C7. If Carl gives away 4 sweets, the ratio of the number of sweets Carl has to the number of sweets Pete has is 2:5. If Pete gives away 4 sweets, the ratio will becomes 8:13. How many sweets does each of them have at first?

C11. Trianne and Hito had some stickers in the ratio 3:5. After Trianne bought 60 more stickers and Hito bought 20 more stickers, the ratio became 5:7. Find the number of stickers Trianne had at first.

Chatty

Registered: 02/28/12
Posts: 68

 By: echeewh (offline)  Wednesday, May 09 2012 @ 09:45 PM CDT
echeewh

Hey Shirley,

Following pls find the worked solutions:

C7.

Apply Simultaneous method here...(aka Compare & Replace / Elimination method )

<Before>
C: 2p + 4
P: 5p

<Process>
P gave away 4.

<After>
C: 2p + 4
P: 5p - 4

Ratio C : P = 8 : 13

Hence we have ...

2p + 4 --> 8u --- (1)
5p - 4 --> 13u --- (2)

Given that question is interested in number of sweets at first, we shall eliminate (u).

(1) x13: 26p + 52 --> 104u --- (3)
(2) x8: 40p - 32 --> 104u --- (4)

(4) - (3): 40p - 26p - 32 - 52 --> 0
14p --> 84
1p --> 6

At first, C = 2p + 4 --> (2 x 6) + 4 = 16
P = 5p --> 5 x 6 = 30

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

C11.

Apply Cross-Multiply method here ...

<Before>

T : H
3 : 5

<Process>

+60 +20

<After>

5 : 7

(3u + 60) / 5 = (5u + 20) / 7
7 x (3u + 60) = 5 x (5u + 20)
21u + 420 = 25u + 100
25u - 21u --> 420 - 100
4u --> 320
1u --> 80

Number of stickers T had at first = 3u --> 3 x 80 = 240

*Notes: Simultaneous method can also be employed here. Also, the standard <Before> and <After> model concept by drawing can also be used; however, it can be very 'challenging' to some students.

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

Trust this helps. Anything else, pls do clarify.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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