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The moneybox have 600 coins with 50 cent and 20 cent pieces .
How many cents were in the box

User KBog
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1 Answer

1 vote

Answer:

Let's use algebra to solve the problem.

Let x be the number of 50 cent pieces in the moneybox.

Then the number of 20 cent pieces in the moneybox is 600 - x (since there are 600 coins in total).

The value of x 50 cent pieces is 50x cents.

The value of (600 - x) 20 cent pieces is 20(600 - x) = 12000 - 20x cents.

The total value of the coins is the sum of these two values:

50x + (12000 - 20x) = 3000 + 30x

So there are 3000 + 30x cents in the moneybox.

To find the total number of cents, we need to evaluate this expression for x:

3000 + 30x

We don't know the value of x, but we do know that there are 600 coins in the moneybox, so we can set up another equation:

x + (600 - x) = 600

Simplifying this equation, we get:

x + 600 - x = 600

Simplifying further, we get:

600 = 600

This equation is always true, which means we can solve for x in terms of 600:

x = 600 - (600 - x)

x = 600 - (600 - x)

x = 600 - (600 - x)

x = 600 - 600 + x

x = x

So x can be any value between 0 and 600.

To find the total number of cents in the moneybox, we need to evaluate the expression 3000 + 30x for any value of x between 0 and 600.

The minimum value of this expression occurs when x = 0:

3000 + 30(0) = 3000

The maximum value of this expression occurs when x = 600:

3000 + 30(600) = 21000

So there are between 3000 and 21000 cents in the moneybox, depending on how many 50 cent pieces there are.

Hope This Helps!

User Sejanus
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