The sum of three numbers is 70. The third number is 2 times the second. The first number is 10 less than the second. What are the numbers?
Let's call the second number "x".
According to the problem, the third number is 2 times the second:
Third number = 2x
The first number is 10 less than the second:
First number = x - 10
And we know that the sum of all three numbers is 70:
First number + Second number + Third number = 70
Now we can substitute the expressions we found for the first, second, and third numbers into this equation:
(x - 10) + x + (2x) = 70
Simplifying:
4x - 10 = 70
Adding 10 to both sides:
4x = 80
Dividing both sides by 4:
x = 20
So the second number is 20.
Now we can use the expressions we found earlier to find the first and third numbers:
First number = x - 10 = 20 - 10 = 10
Third number = 2x = 2(20) = 40
So the three numbers are 10, 20, and 40.
Let's solve the problem step by step.
Step 1: Let's define the three numbers.
Let the second number be "x".
Therefore, the first number would be (x - 10) since it is 10 less than the second number.
And, the third number would be 2x since it is 2 times the second number.
Step 2: Write the equation based on the given information.
According to the problem, the sum of the three numbers is 70.
So, we can write the equation as:
(x - 10) + x + 2x = 70.
Step 3: Simplify the equation.
Combine like terms on the left side:
4x - 10 = 70.
Step 4: Isolate the variable.
Add 10 to both sides of the equation:
4x = 80.
Step 5: Solve for x.
Divide both sides by 4:
x = 20.
Step 6: Find the other two numbers.
The first number is (x - 10):
First number = 20 - 10 = 10.
The third number is 2x:
Third number = 2 * 20 = 40.
Therefore, the numbers are 10, 20, and 40.