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Here's a question in my math book. A man had enough money to buy 20 mangoes or 30 oranges. If he wants to buy an equal amount of each, how many would he buy? I don't really care about the answer, just how to get it. Thanks!

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  • $\begingroup$ I do not think that you have enough information to solve it. Check again, perhaps you are given something more. (or I am wrong) $\endgroup$
    – Jimmy R.
    Mar 31, 2014 at 20:37

2 Answers 2

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Hint: He has $x$ dollars. A mango costs $\frac x{20}$, while an orange costs $\frac x{30}$. So a (mango + orange) costs how much? How many (mango + orange)s can he buy?

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  • $\begingroup$ Sorry if this sounds stupid. When you add up x/20 + x/30 you get 5x/60. 5x/60=x? I still don't get it. $\endgroup$
    – Isaacium
    Mar 31, 2014 at 22:26
  • $\begingroup$ Divide the amount of money available by the cost: $x \div \frac{5x}{60}$. $\endgroup$ Apr 1, 2014 at 0:14
  • $\begingroup$ Correct, you get $\frac {5x}{60}=\frac x{12}$. With $x$ dollars you can buy $12$ of those. $\endgroup$ Apr 1, 2014 at 0:36
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Love the (mango + orange) solution. A mechanical solution:

Let $x$ be the amount of money he has. Let $m$ be price of one mango. Let $r$ be price of one orange. Suppose he can just buy $n$ of each fruit. Then $$nm +nr = x\,.$$ We know 20 mangoes cost $x$ so $m = x/20$. Likewise $r = x/30$. So we have $$\frac{nx}{20} + \frac{nx}{30} = x\,.$$ The $x$'s cancel, so $$\frac{n}{20} + \frac{n}{30} = 1\,.$$ Solve for $n$. The form of the result points to the (mango + orange) solution.

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