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 RGS Prelim 2011 - Question 18
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By: thayal (offline)  Wednesday, July 18 2012 @ 07:28 AM CDT (Read 1604 times)  
thayal

HI Maths gurus..

Could you help to answere the question:
Beaker A and beaker B contains some water. If 46.5 of water is drained out from beaker A , teh volume of the water in beaker A will be 60% that of the water in beaker B.if 35.2 ml of water is drained oout from beaker B , teh volume of the water n beaker B will be 85% that of water in beaker A.
what is the total volume of water in beaker A and beaker B?

thanks
thayal

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By: echeewh (offline)  Wednesday, July 18 2012 @ 09:21 PM CDT  
echeewh

Hey thayal,

Following pls find my worked solution (very similar to that provided in the Answerkey)

<RGPS Prelim 2011 P2 Q18>

**** - buffer as spaces for alignment purpose

60% = (3/5); 85% = (17/20)

<Before>

****************46.5
A |----|----|----|<----->|***--> (3p + 46.5)
B |----|----|----|----|----|***--> (5p)

<After>

***<--------20u-------->
A |----|----|----|<----->|***--> (3p + 46.5)
B |----|----|----|----|----|***--> (5p - 35.2)
******************<--|
******************35.2
***<----17u------>

Hence, we have ...

3p + 46.5 --> 20u
5p - 35.2 --> 17u

Re-arranging these 2 equations, we have ...

20u - 3p --> 46.5 -- (1)
5p - 17u --> 35.2 -- (2)

Eliminating (u) (since we need to find out the initial volume of A and B ), we have ...

(1)x17: 340u - 51p --> 790.5 -- (3)
(2)x20: 100p - 340u --> 704 -- (4)

(3)+(4):
100p - 51p --> 704 + 790.5
49p --> 1494.5
1p --> 30.5

Total volume in A and B (at first)
= 8p + 46.5 --> (8 x 30.5) + 46.5
= 290.5 ml

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

Trust this helps.

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

Cheers,
Edward

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