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 By: kfrchong (offline)  Wednesday, October 08 2014 @ 01:38 AM CDT (Read 1344 times)
kfrchong

Need help to solve the below question. Thank you.

The ratio of the number of 20-cent coins to 50-cent coins in a jar was 7:2. After Ashwin added 51 20-cent coins and 51 50-cents coins to the jar, the ratio of the number of 20-cents coins to 50-cents coins in a jar became 9:5. What was the total value of the 50-cent coins in the jar at first?

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Registered: 07/10/14
Posts: 2

 By: echeewh (offline)  Wednesday, October 08 2014 @ 08:20 AM CDT
echeewh

Hey <kfrchong>,

Following is my worked solution :

*** - for alignment purpose

Given 51 20-cent coins and 51 50-cent coins were to the jar, we make use of the <Unchanged Difference / Constant Difference > concept. I.e. difference in coins at first is the same as the difference after.

<Before>
20-cent : 50-cent Diff
7 ****** : 2 ******** 5 -- (1)

<After>
20-cent : 50-cent Diff
9 ****** : 5 ******** 4 -- (2)

Common multiple in Diff (5, 4) is 20. Hence, multiply (1)x4; (2)x5. We have ...

<Before>
20-cent : 50-cent Diff
28 ***** : 8 ******* 20 -- (3)

<After>
20-cent : 50-cent Diff
45 ***** : 25 ****** 20 -- (4)

hence,
45u - 28u = 17u
17u --> 51
1u --> 51 ÷ 17 = 3

Number of 50-cent coins (at first):
8u --> 8 × 3 = 24

Value of 50-cent coins (at first):
24 × 50 = \$12.00

======

Trust this helps.

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

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: kfrchong (offline)  Wednesday, October 08 2014 @ 06:47 PM CDT
kfrchong

Thank you, Edward!

Newbie

Registered: 07/10/14
Posts: 2

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