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 By: geniuskids (offline)  Tuesday, April 30 2013 @ 11:47 PM CDT (Read 1339 times)
geniuskids

Q18 - There are some red and green beans in a container. If Ahmad removes 20 red beans, the ratio of the number of red beans to the number of green been becomes 6:5. If Ahmad removes 10 green beans, the ratio of the number of red beans to the number of green beans becomes 4:3. How many beans are there in the container?

Thank you

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Registered: 11/12/11
Posts: 169

 By: echeewh (offline)  Thursday, May 02 2013 @ 07:30 PM CDT
echeewh

Hey there,

Following pls find my worked solution:

<1st IF>
R : G
6 : 5

<Process1>
R: -20

<Before> {Work Backwards}
R: 6p + 20
G: 5p

<Process2>
G: -10

<2nd IF>
R : G
4 : 3

R: 6p + 20
G: 5p - 10

Combining both sets and using Simultaneous method, we have ...

6p + 20 = 4u
5p - 10 = 3u

Rearranging the above,

4u - 6p = 20 -- (1)
5p - 3u = 10 -- (2)

Eliminating (u), multiply (1) 3x; (2) 4x;

12u - 18p = 60 -- (3)
20p - 12u = 40 -- (4)

(3)+(4):
20p - 18p --> 60 + 40
2p --> 100
1p --> 50

Total (R + G) (before) = 11p + 20
11p + 20 --> (11 × 50) + 20 = 550 + 20 = 570

===========

Trust this helps.

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

Cheers,
Edward

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Registered: 04/21/11
Posts: 627

 By: geniuskids (offline)  Saturday, May 04 2013 @ 10:33 AM CDT
geniuskids

Thanks! Got it!

Active Member

Registered: 11/12/11
Posts: 169

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