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September 18th, 2018, 12:03 AM   #1
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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
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September 18th, 2018, 01:38 AM   #2
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$\ell \text{ is the number of local coins started with}$

$f \text{ is the number of foreign coins started with}$

$n = \ell + f$

after giving away the coins we have

$\ell \to \dfrac \ell 2$

$f \to f-6$

$\dfrac \ell 2 = (f-6) + 71 = f + 65$

$\ell = 2f + 130$

$\dfrac{\ell}{f} = 7$

$\dfrac{2f+130}{f}=7$

$130=5f$

$f = 26$

$\ell = 7f = 182$

$n = 26 + 182 = 208$
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September 18th, 2018, 01:39 AM   #3
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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
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September 18th, 2018, 02:01 AM   #4
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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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