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Question 1161320: Find the sum of three numbers such that the 3rd number is twice the 2nd, and the 2nd exceeds the 1st number which is 80 by 70. Thank you!
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let the 3 numbers be represented by a,b,c.
the third number is equal to twice the second number, therefore:
c = 2b
the second number exceeds the first number, which is 80, by 70.
a = 80
b = 80 + 70 = 150
since c = 2b, then c = 300
a + b + c = 80 + 150 + 300 = 530 is what i get.
530 should be your solution if i did it correctly.
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
Find the sum of three numbers such that the 3rd number is twice the 2nd, and the 2nd exceeds the 1st number which is 80 by 70. Thank you!
1st number: 80
As 2nd exceeds 1st by 70, 2nd is: 80 + 70 = 150
3rd is TWICE the 2nd, so 3rd is: 2(150) = 300
Can you now find the SUM of the 3?
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