Mr. Kwon has just enough money to buy only 20 mangoes or only 30 oranges. If he wants to buy equal numbers of mangoes and oranges together, how many of each type can he buy with the money?

Let x be the amount of money he has.

Let m be price of one mango.

Let o be price of one orange.

Suppose he can just buy n of each fruit.

Then:

n * m + n * o = x

You know 20 mangoes cost x

so

m = x / 20

likewise

o = x / 30

So you have:

n * m + n * o = x

n * x / 20 + n * x / 30 = x Divide both sides by x

n / 20 + n / 30 = 1

3 * n / 3 * 20 + 2 * n / 2 * 30 = 1

3 n / 60 + 2 n / 60 = 1

5 n / 60 = 1

5 n / 5 * 12 = 1

n / 12 = 1 Multiply both sides by 12

n = 12

He can buy 12 mangos and 12 oranges.

To solve this problem, we need to find a common factor between 20 and 30 that will allow us to find the number of mangoes and oranges that Mr. Kwon can buy with his money.

One way to find the common factor is by listing the multiples of each number until we find one in common.

For 20: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200...
For 30: 30, 60, 90, 120, 150, 180, 210...

From these lists, we can see that 60 is the smallest common multiple of 20 and 30.

This means that Mr. Kwon can buy 60 fruits in total, with an equal number of mangoes and oranges.

To find the number of mangoes and oranges Mr. Kwon can buy, we divide the common multiple (60) by the number of fruits he can buy with his money individually (20 mangoes or 30 oranges):

Number of mangoes = 60 ÷ 20 = 3 mangoes
Number of oranges = 60 ÷ 30 = 2 oranges

Therefore, Mr. Kwon can buy 3 mangoes and 2 oranges with his money, allowing him to purchase an equal number of each fruit.