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 By: psiewpeng (offline)  Monday, August 19 2013 @ 09:48 PM CDT (Read 1067 times)
psiewpeng

Hi.

Three children sold some fun fair tickets. Alex sold 5/8 of the tickets. Bei Shan sold 36 tickets fewer than Alex. Chai Lian sold 20 tickets. How many tickets did they sell altogether? ( answer- 64 tickets)

Newbie

Registered: 02/23/10
Posts: 2

 By: echeewh (offline)  Monday, August 19 2013 @ 11:26 PM CDT
echeewh

Hello <psiewpeng>,

Following is my worked solution:

A: (5/8)total
B: (5/8)total - 36
C: 20

A + B + C = (8/8)total
(5/8)total + (5/8)total - 36 + 20 = (8/8)total
5u + 5u - 36 + 20 = 8u
10u - 8u = 36 - 20
2u --> 16
1u --> 8

Total = 8u --> 8 × 8 = 64

==========

Trust this helps.

Do let me know again if there's further clarifications.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 623

 By: psiewpeng (offline)  Thursday, August 22 2013 @ 02:09 AM CDT
psiewpeng

Thanks Edward. However if I need to represent it using model diagram, can you help? Thanks

Newbie

Registered: 02/23/10
Posts: 2

 By: echeewh (offline)  Thursday, August 22 2013 @ 04:27 AM CDT
echeewh

Hey

As requested, here is the <Model> alternative:

*** - for alignment purpose

|----|----|----|----|----|----|----|----|
<--------- A --------><--><-20-->
******************** B *** C

From model,
B + C sold the remaining 3u of tickets

Given C sold 20 and B sold 36 tickets less than A,
C = 20
B = 5u - 36

B + C = 3u
5u - 36 + 20 = 3u
5u - 3u = 36 - 20
2u --> 16
1u --> 8

Total tickets sold = 8u
8u --> 8 × 8 = 64

==========

Trust this helps.

Do let me know again if there's further clarifications.

Cheers,
Edward.

Active Member

Registered: 04/21/11
Posts: 623

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