a man has a total of 90. he has six more 5 peso coins than 1 peso coins. how many coins of each type does the man have?
Well, 90/2=45 and it says he has 6 more 5 peso coin, so 45+6=? and 45-6=?. Finish solving the problem.
To solve this problem, we can use a system of equations. Let's assign variables to the unknowns:
Let x be the number of 1 peso coins.
Let y be the number of 5 peso coins.
According to the problem, we know that the man has a total of 90 coins. This can be written as an equation:
x + y = 90 ....(Equation 1)
We are also given that the man has six more 5 peso coins than 1 peso coins. This can be written as another equation:
y = x + 6 ....(Equation 2)
Now, we can solve this system of equations to find the values of x and y.
Substitute Equation 2 into Equation 1:
x + (x + 6) = 90
Simplify the equation:
2x + 6 = 90
Subtract 6 from both sides:
2x = 84
Divide both sides by 2:
x = 42
Now that we have the value of x, we can substitute it back into Equation 2 to solve for y:
y = x + 6
y = 42 + 6
y = 48
Therefore, the man has 42 coins of 1 peso and 48 coins of 5 peso.