My Math Forum Junhao coin collection

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 September 18th, 2018, 12:03 AM #1 Member   Joined: Sep 2018 From: Japan Posts: 35 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, 01:38 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157 $\ell \text{ is the number of local coins started with}$ $f \text{ is the number of foreign coins started with}$ $n = \ell + f$ after giving away the coins we have $\ell \to \dfrac \ell 2$ $f \to f-6$ $\dfrac \ell 2 = (f-6) + 71 = f + 65$ $\ell = 2f + 130$ $\dfrac{\ell}{f} = 7$ $\dfrac{2f+130}{f}=7$ $130=5f$ $f = 26$ $\ell = 7f = 182$ $n = 26 + 182 = 208$ Thanks from xoritos
 September 18th, 2018, 01:39 AM #3 Senior Member   Joined: Apr 2014 From: UK Posts: 892 Thanks: 328 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 Thanks from xoritos
 September 18th, 2018, 02:01 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,988 Thanks: 1855 If Junhao originally had n coins, ((7/2)/8)n = 71 + (1/8)n - 6, so n = 65/((7/2 - 1)/8) = 208. Thanks from xoritos

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