Question:
A money box contains only 10-cent
and 20-cent coins. There are 33
coins with a total value of $4.60.
Please help give the answer.
x+y=33
10x+20y = 460
As for the answer, I guess that would depend on the question, eh?
A money box contains only 10-cent
and 20-cent coins. There are 33
coins with a total value of $4.60.
How many coins of each?
No. of 10-cent coins =
No. of 20-cent coins =
My bad, I didn't copy the whole question.
Well then solve it Kayla.
like
10 x + 10 y = 330
10 x + 20 y = 460
--------------------------- subtract
-10 y = - 130
y = 13
etc
A money box contains only 10-cent
and 20-cent coins. There are 33
coins with a total value of $4.60.
How many coins of each?
To solve this problem, we can set up a system of equations based on the information given. Let's denote the number of 10-cent coins as 'x' and the number of 20-cent coins as 'y'.
From the problem, we know that there are a total of 33 coins. Therefore, we have:
x + y = 33 -- Equation 1
We are also told that the total value of the coins is $4.60. Since each 10-cent coin is worth $0.10 and each 20-cent coin is worth $0.20, we can write the equation for the total value as:
0.10x + 0.20y = 4.60 -- Equation 2
Now we can solve this system of equations.
First, let's multiply Equation 1 by 0.10 to make it easier to eliminate x:
0.10x + 0.10y = 3.30 -- Equation 3
Now we can subtract Equation 3 from Equation 2 to eliminate x:
(0.10x + 0.20y) - (0.10x + 0.10y) = 4.60 - 3.30
0.20y - 0.10y = 1.30
0.10y = 1.30
Now divide both sides by 0.10:
y = 1.30 / 0.10
y = 13
Now substitute the value of y back into Equation 1 to find the value of x:
x + 13 = 33
x = 33 - 13
x = 20
Therefore, there are 20 10-cent coins and 13 20-cent coins in the money box.