
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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October 21st, 2018, 10:54 PM  #1 
Member Joined: Sep 2018 From: Japan Posts: 35 Thanks: 2  Number of coins
In Raju's coin box, there are only 20cent and 50cent coins. There are (1/4) more 20cent coins than 50cent coins. Raju has a total of $54 in his coin box. How many 20cent coins are there in his coin box? My work: I'm not sure how to create an algebraic expression for this. 
October 21st, 2018, 11:29 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281 
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 
October 22nd, 2018, 08:10 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,145 Thanks: 1003 
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 
October 22nd, 2018, 10:14 AM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
$t = \text { number of twenty cent coins.}$ $f = \text { number of fifty cent coins.}$ 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. $\text {One quarter more twenty cent coins than fifty cent} \implies\\ \left ( 1 + \dfrac{1}{4} \right) * f = t.$ $\text { the total value is } \$54 \implies 0.5f + 0.2t = 54.$ Last edited by JeffM1; October 22nd, 2018 at 10:19 AM.  

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