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Sunday, July 23 2017 @ 09:47 PM CDT
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 Maths Question 2
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By: shirley wee (offline)  Sunday, April 07 2013 @ 07:09 AM CDT (Read 627 times)  
shirley wee

Hi everybody! I'm stuck with this maths question. May you help me? Please show your working clearly.

Qn 1. There were dark chocolates and milk chocolates in a box. If 10 dark chocolates were removed from the box, the total number of chocolates left would be 6 times the number of dark chocolates left. If 30 milk chocolates were removed from the box, the total number of chocolates left would be 4 times the number of dark chocolates left. How many more milk chocolates than dark chocolates were there in the box?

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By: echeewh (offline)  Sunday, April 07 2013 @ 07:12 PM CDT  
echeewh

hi Shirley,

Following pls find my worked solution:

*** - for alignment purpose


<If 10 D were removed>
there would be (1/6)D and (5/6)M.

D |-----|
M |-----|-----|-----|-----|-----|

<before>

D |-----<-10->| ************** -- 1p + 10
M |-----|-----|-----|-----|-----| *** -- 5p

<after - if 30 M removed>
there would be (1/4)D and (3/4)M

**<--- 1u ---->
D |-----<-10->| *************** -- 1p + 10
M |-----|-----|-----|-|<-- 30 --> ** -- 5p - 30
**<------ 3u ------>


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

1p + 10 = 1u
5p - 30 = 3u

Rearranging these ...

1u - 1p --> 10 -- (1)
5p - 3u --> 30 -- (2)

(1)x3:
3u - 3p --> 30 -- (3)

(2)+(3):
5p - 3p --> 30 + 30
2p --> 60
1p --> 30

M - D (at first) = 5p - 1p - 10 = 4p - 10

4p - 10 --> (4 × 30) - 10 = 120 - 10 = 110

===========

Trust this helps.

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

Cheers,
Edward

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