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 By: JessicaTan (offline)  Saturday, April 28 2012 @ 06:18 AM CDT (Read 736 times)
JessicaTan

Anuar had 40 more twenty-cents coins than one-dollar coins. He exchanged all the twenty-cent coins for one-dollar coins. He then had 20 one-dollar coins altogether. What is the value of all the twenty-cents coins he had at first? (answer is \$10) kindly show me the working clearly. Thank you very much.

Regular Member

Registered: 10/17/09
Posts: 82

 By: echeewh (offline)  Sunday, April 29 2012 @ 09:32 PM CDT
echeewh

Hey Jessica,

Kindly refer to the worked solution below:

Total value of all the coins at first = \$20
At first, value of 40 more 20cent coins = \$8
Same unit (qty) of both 20cent and \$1 coins = \$20 - \$8 = \$12
using Number x Value method, we have ...
( 1u x 20 ) + ( 1u x 100 ) --> 1200 ( converting money to cents , and 1u -> same number/qty of both types of coins )
20u + 100u --> 1200
120u --> 1200
1u --> 1200 / 120 = 10
So the value of the \$1 coins = 1u x \$1 = 10 x \$1 = \$10
Since total value of both types of coins is \$20, the value of all the 20cent coins = \$20 - \$10 = \$10

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

Trust the above helps.

Any clarification , pls do not hesitate to ask.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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