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October 21st, 2018, 11:54 PM   #1
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Number of coins

In Raju's coin box, there are only 20-cent and 50-cent coins. There are (1/4) more 20-cent coins than 50-cent coins. Raju has a total of $54 in his coin box. How many 20-cent coins are there in his coin box?

My work: I'm not sure how to create an algebraic expression for this.
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October 22nd, 2018, 12:29 AM   #2
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let $x$ be the number of 20 cent coins

If I understand the wording

$\dfrac 4 5 x$ is the number of 50 cent coins

$\dfrac{2}{10}x + \dfrac{5}{10}\dfrac 4 5 x = 54$

$\dfrac 3 5 x = 54$

$x = 90$

so 90 20 cent coins and 72 50 cent coins
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October 22nd, 2018, 09:10 AM   #3
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v = .50 coins ; so 1.25v = .20 coins

50v + 20(1.25)v = 5400
50v + 25 v = 5400
75v = 5400
v = 72 .... as per Sir Romsek
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October 22nd, 2018, 11:14 AM   #4
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Quote:
Originally Posted by xoritos View Post
In Raju's coin box, there are only 20-cent and 50-cent coins. There are (1/4) more 20-cent coins than 50-cent coins. Raju has a total of $54 in his coin box. How many 20-cent coins are there in his coin box?

My work: I'm not sure how to create an algebraic expression for this.
I greatly prefer to set up a variable for each unknown.

$t = \text { number of twenty cent coins.}$

$f = \text { number of fifty cent coins.}$

With two unknowns, we need two equations.

$\text {FIRST, } \left (1 + \dfrac{1}{4} \right ) * f = t \implies t = \dfrac{5f}{4}.$

$\text {SECOND, } 0.5f + 0.2t = 54 \implies 5f + 2t = 540 \implies $ $t = \dfrac{540 - 5f}{2}.$

$\therefore \dfrac{5f}{4} = \dfrac{540 - 5f}{2} \implies 10f = 2160 - 20f \implies\\

30f = 2160 \implies f = 72 \implies t = \dfrac{5 * 72}{4} = 90.$

Notice that the answers are the same as from romsek and denis. Admittedly, my approach requires more algebraic manipulation. But it requires less work to set the problem up because you have one equation for each relevant fact.

$\text {One quarter more twenty cent coins than fifty cent} \implies\\

\left ( 1 + \dfrac{1}{4} \right) * f = t.$

$\text { the total value is } \$54 \implies 0.5f + 0.2t = 54.$
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Last edited by JeffM1; October 22nd, 2018 at 11:19 AM.
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