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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 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 20-cent and 50-cent coins. There are (1/4) more 20-cent coins than 50-cent coins. Raju has a total of $54 in his coin box. How many 20-cent coins are there in his coin box? My work: I'm not sure how to create an algebraic expression for this. October 21st, 2018, 11:29 PM #2 Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281 let$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 = 54x = 90$so 90 20 cent coins and 72 50 cent coins Thanks from xoritos 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 Thanks from xoritos October 22nd, 2018, 10:14 AM #4 Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 Quote:  Originally Posted by xoritos 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.\$

Last edited by JeffM1; October 22nd, 2018 at 10:19 AM. Tags coins, number Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Student2018 Elementary Math 9 April 13th, 2018 08:17 AM EvanJ Probability and Statistics 0 November 18th, 2016 06:18 PM shunya Probability and Statistics 3 February 15th, 2016 11:14 AM rogerv New Users 14 February 22nd, 2013 05:58 AM Soha Algebra 6 December 14th, 2006 12:18 PM

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