
By: LeowML (offline) Friday, June 08 2012 @ 07:58 AM CDT (Read 1644 times)



LeowML 
12)Roy, Sam and Ted had a sum of money. Roy had 80% of Sam's money. Sam had 60% of Ted's. After Roy gave $12 to Sam, he had 5/7 of what Sam had. How mouch more did Ted have than Sam in the end?

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Registered: 12/31/06 Posts: 11





By: awyw1201 (offline) Friday, June 08 2012 @ 11:38 AM CDT



awyw1201 
R:S80:100=4:5(1)
S:T60:100=3:5(2)
Looking at the two sets of ratio, since S is found in both, we need to make that an equal quantity. The l
L.C.M. Of 3 and 5 is 15. So,
(1) x 312:15
(2) x515:25
Thus, R:S:T12:15:25(3)
When R gave $12 to S, R has5/7 of what S has. This means their total remains unchanged.
R:S=5:7(4)
Thus R+S=12+15=27(at first)
But now R+S=5+7=12 (now)
L.C.M of 27 and 12 is 108, so
(3) x 448:60:100
(4) x 945:63
Comparing the two ratios, we have 4845=3u$12
1u$4
10060=40u$4x 40=$160

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Registered: 07/08/11 Posts: 87





By: echeewh (offline) Friday, June 08 2012 @ 08:23 PM CDT



echeewh 
Hey there,
Correction
Pls ignore the last line of the above solution provided by awyw1201.
10060=40u$4x 40=$160
This is not answering to what the question is asking for:
How much more did Ted have than Sam in the end?
It should be as follows:
At the end, T  S = 100u  63u = 37u
> 37 x 4 = $148
======================================
Cheers,
Edward

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Registered: 04/21/11 Posts: 627





By: awyw1201 (offline) Saturday, June 09 2012 @ 12:05 AM CDT



awyw1201 
Oops, must b too tired, blurry eyes. Thanks for the correction, Edward. Appreciate that

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Registered: 07/08/11 Posts: 87





By: LeowML (offline) Sunday, June 10 2012 @ 07:04 PM CDT



LeowML 
Thanks to both of you for your kind assistance.

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Registered: 12/31/06 Posts: 11



