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Context: Ahamd has some 10 cent, 20 cent and 50 cent coins. There are 3 times as many 50 cent coins as 2o cent coins. The number of 50 cent coins is 3/5 of the number of 10 cent coins. The total amount of the 50 cent and 10 cent coins Ahmad had is $84. What is the total number of 50. cent and and 10 cent coins? Answer question

Question

Context: Ahamd has some 10 cent, 20 cent and 50 cent coins. There are 3 times as many 50 cent coins as 2o cent coins. The number of 50 cent coins is 3/5 of the number of 10 cent coins. The total amount of the 50 cent and 10 cent coins Ahmad had is $84. What is the total number of 50. cent and and 10 cent coins? Answer question

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Solution

Let's denote the number of 10 cent coins as x, the number of 20 cent coins as y, and the number of 50 cent coins as z.

From the problem, we have the following equations:

  1. z = 3y (There are 3 times as many 50 cent coins as 20 cent coins)

  2. z = 3/5 * x (The number of 50 cent coins is 3/5 of the number of 10 cent coins)

  3. 0.10x + 0.50z = 84 (The total amount of the 50 cent and 10 cent coins Ahmad had is $84)

We can substitute equation 1) into equation 2) to get: 3y = 3/5 * x, which simplifies to y = x/5.

Now we can substitute y = x/5 into equation 1) to get: z = 3 * (x/5), which simplifies to z = 3x/5.

Finally, we substitute z = 3x/5 into equation 3) to get: 0.10x + 0.50 * (3x/5) = 84, which simplifies to 0.10x + 0.30x = 84, and further simplifies to 0.40x = 84.

Solving for x, we get x = 84 / 0.40 = 210.

Substituting x = 210 into z = 3x/5, we get z = 3 * 210 / 5 = 126.

So, the total number of 50 cent and 10 cent coins Ahmad had is 210 + 126 = 336.

This problem has been solved

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