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 By: LeowML (offline)  Friday, June 08 2012 @ 07:58 AM CDT (Read 1633 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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Posts: 11

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

R:S--80:100=4:5----(1)

S:T--60:100=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 3----12:15

(2) x5----15:25

Thus, R:S:T----12:15:25---(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 4----48:60:100

(4) x 9---45:63

Comparing the two ratios, we have 48-45=3u----\$12
1u---\$4

100-60=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.

100-60=40u----\$4x 40=\$160

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

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