One number is 9 times a first number. A third number is 100 more than the first number. If the sum of three numbers is 980, find the numbers

Question

I don't know the answer 

Answer 1

Monet,

a+b+c=980  We want to have just one variable to solve, so let's do that.

The way the question is written, ?=9a, where ? can be ambiguously either b or c. It seems to me that it would more likely be b. Nonetheless, we can solve either way.

 

Write down your knowns. We are sure that c=100+a

• if the "one number" is the second, then the knowns are 
• a+b+c=980
• b=9a
• c=100+a

So, substitute b & c into the first equation to get a+9a+(100+a)=980. Combine like terms and solve for a. Then, plug in the value of a into b=9a. Then plug in now known both b&c values into a+b+c=980

 

• if the "one number" is the the third #, then the equations are
• a+b+c=980
• 9a=100+a
• c=100+a

So, Solve the second equation for a. Then, plug a into the 3rd equation to find c. You'll then know both a&c to plug the values into the 1st eqn.

 

Again, I think they want you to solve via the 1st option. 

 

Best,

Marina

Answer 2

Monet,

 

Since everything is based on the first number, let's call the first number x. Then we have three numbers, x, 9x and x+100. If the sum of the three numbers is 980, we can set up the equation:

x+9x+x+100=980

11x=880

x=80

 

So one number is 80, another is 80*9=720, and the third is 100+80=180. Checking our work, 80+180+720=980.

 

Hope that helps!