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 CHIJ 2011 SA2
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By: manhan (offline)  Monday, July 09 2012 @ 04:39 AM CDT (Read 1091 times)  
manhan

Question:

A shopkeeper had some markers in red and blue. If 50 red markers and 25 blue markers were sold each week, there would be 500 red markers left when all the blue markers were sold. If 25 red markers and 50 blue markers were sold each week, there would be 800 red markers left when all the blue markers were sold. How many markers did the shopkeeper have at first?

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Can we use the ""excess - shortage"" method to solve the above? If yes, please advise how.

Thank you.

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By: echeewh (offline)  Monday, July 09 2012 @ 07:07 AM CDT  
echeewh

Hey manhan,

Following pls find the worked solution.

<1st IF>

**<-----2p----><---------500-------->
R |-------|-------|------------------------|
B |-------|
**<-1p->

<2nd IF>

<Process>
R halved; B doubled;
Total R and B markers remain unchanged;

Given that 1p of B remains unchanged (from <1st IF> before and after, and the number of B markers doubled, divide the 1p into 2 equal smaller units (u). Likewise, the 2p of R (from 1st IF) is also divided into 4 equal units (u).


*****<-------------800------------->
**<----4u-----><-----500--------->
R |---|---|---|---|----------------------|
B |---|---|
**<-2u->

Using <Gap and Difference> method, and compare the R model in the 2 IF's, we have ...

4u - 1u --> 800 - 500
3u --> 300
1u --> 100

Total (R + B ) markers at first = 6u + 500 --> (6 x 100) + 500
= 1100

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

Trust this helps.

Do let me know if this is different from the Answerkey or if theres further clarification.

Cheers,
Edward

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By: manhan (offline)  Tuesday, July 10 2012 @ 03:58 AM CDT  
manhan

That was very helpful indeed. Thank you Smile

Any idea if we can use the "Excess and Shortage" method for this question?

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