
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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September 17th, 2018, 11:03 PM  #1 
Newbie Joined: Sep 2018 From: Japan Posts: 27 Thanks: 2  Junhao coin collection
Junhao has a coin collection. 7/8 of his coins are local coins and the rest are foreign coins. After he gives away 1/2 of his local coins and 6 foreign coins, he has 71 more local coins than foreign coins left. How many coins did he have at first? My work. (I'm not sure how to solve this) Let number of Foreign coins = F Let number of Local coins = L [1] We know 1/8 of his coins are foreign. [2] He gives away 1/2 of his local coins and 6 foreign coins, and he has 71 more local coins left: 7/64L + 71 = 1/8F  6 Note: how I got 7/64L is 7/8 * 1/8 
September 18th, 2018, 12:38 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,122 Thanks: 1102 
$\ell \text{ is the number of local coins started with}$ $f \text{ is the number of foreign coins started with}$ $n = \ell + f$ after giving away the coins we have $\ell \to \dfrac \ell 2$ $f \to f6$ $\dfrac \ell 2 = (f6) + 71 = f + 65$ $\ell = 2f + 130$ $\dfrac{\ell}{f} = 7$ $\dfrac{2f+130}{f}=7$ $130=5f$ $f = 26$ $\ell = 7f = 182$ $n = 26 + 182 = 208$ 
September 18th, 2018, 12:39 AM  #3 
Senior Member Joined: Apr 2014 From: UK Posts: 886 Thanks: 326 
7/8 of his coins are local coins and the rest are foreign coins. So, L = 7F After he gives away 1/2 of his local coins and 6 foreign coins, he has 71 more local coins than foreign coins left. So, L/2 = F  6 + 71 Therefore: 7F/2 = F  6 + 71 7F = 2F +142  12 5F = 130 F = 26 L =182 Total coins he started with was 208 
September 18th, 2018, 01:01 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,713 Thanks: 1806 
If Junhao originally had n coins, ((7/2)/8)n = 71 + (1/8)n  6, so n = 65/((7/2  1)/8) = 208.


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