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 Help - Red Swastika P6 Maths (2011 Prelim)
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By: linjimt (offline)  Tuesday, July 24 2012 @ 11:02 AM CDT (Read 1919 times)  
linjimt

Please help on workings for these 3 problems. Thanks.

Q15:
Eugene, Freddy and George has some playing cards. Freddy had 200 more playing cards than Eugene. George had 3/4 the number of cards Freddy had. After George lost 1/6 of his cards to Eugene, he has 320 fewer cards than Eugene. What was the ratio of the number of cards Eugene had to the number of cards Freddy has to the number of cards George had in the end?
Ans:97:104:65

Q16:
Laura and Ahmad have some stamps. If Laura gives 80 stamps to Ahmad, he will have five times the number of stamps as she has. If Ahmad gives 120 stamps to Laura, she will have 1/3 as many stamps as Ahmad. How many stamps does Laura have at first?
Ans:480 stamps.

Q17:
A shopkeeper had some blue and red pens in his bookshop. There were twice as many red pens as blue pens. The shopkeeper sold the pens in bundles of 2 red pens and 3 blue pens. After selling all the blue pens, he still had 120 red pens left. How many pens did he have in his bookshop at first?
Ans:270 pens.

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By: echeewh (offline)  Tuesday, July 24 2012 @ 10:05 PM CDT  
echeewh

Hey lin,

Pls refer to separate post / thread for worked solution to Q15.

Cheers,
Edward

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By: linjimt (offline)  Tuesday, July 24 2012 @ 11:03 PM CDT  
linjimt

Thanks Edward.

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By: echeewh (offline)  Wednesday, July 25 2012 @ 08:14 PM CDT  
echeewh

Hi Lin,

Following pls find the worked solutions for Q16, Q17. Both questions make use of similar method to solve ( Simultaneous method - aka Parts and Units ).

Q16.

*** - buffer space for alignment purpose
Models are not drawn to scale.


<After>

L |--------|
A |--------|--------|--------|--------|--------|

<Before>

**********<80>
L |--------|------| --> (1p + 80)
A |--------|--------|--------|--------|--|<-80> --> (5p - 80)

<Process>
A: -120 L: +120

<After1>

***<---------1u-------->
L |--------|------|<120> --> (1p + 200)
A |--------|--------|-------|--|<-120> --> (5p - 200)
***<----------3u------------>

From the <After1> model, we have ...

1u --> 1p + 200
3u --> 5p - 200

Rearranging these equations, we have ...

1u - 1p --> 200 --> (1)
5p - 3u --> 200 --> (2)

As question is interested in knowing L's stamps at first <Before>, we shall eliminate (u) as shown:

(1)x3:
3u - 3p --> 600 --> (3)

(2)+(3):
5p - 3p --> 200 + 600
2p --> 800
1p --> 400


L's stamps at first = 1p + 80 --> 400 + 80 = 480

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

Q17.

<Before>
R : B
2 : 1

<Sold>
2 : 3

<After>
120 R left; 0 B left

2p - 2u --> 120 -- (1)
1p - 3u --> 0 -- (2)

Eliminating (u), we have ...

(1)x3:
6p - 6u --> 360 -- (3)
(2)x2:
2p - 6u --> 0 -- (4)

(3)-(4):
6p - 2p --> 360
4p --> 360
1p --> 90

Total pens at first = 2p + 1p = 3p --> 3 x 90 = 270

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

Trust the above helps.

Do let me know again should there be any clarification.

Cheers,
Edward

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By: linjimt (offline)  Friday, July 27 2012 @ 10:04 AM CDT  
linjimt

Thank you very much.
It's clearer now.

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