Question 27538: Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91.
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! ): Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91.
Let the four consecutive integers be a, (a+1), (a+2) and (a+3)
By data The sum of the two greater integers exceeds three times the first by 91.
That means the sum of (a+2) and (a+3) is greater than 3 times a by 91
That is [(a+2) + (a+3)] > 3a by 91
That is [(a+2) + (a+3)] - 3a = 91
(2a + 5) - 3a = 91
(2a-3a) = 91-5
-a =86 which means
a = -86
The numbers are a, (a+1), (a+2) and (a+3) that is (-86), (-86+1), (-86+2) asnd (-86+3)
that is -86,-85,-84 and -83
Answer: The four consecutive integers are -86, -85,-84 and -83
Verification:
The sum of the two greater integers should exceed three times the first by 91.
The sum of the two greater integers = [(-84) + (-83)] = -167
And 3 times the first is 3X(-86) = -258
And of course (-167) exceeds (-258) by 91 as [(-167) - (-258) ] = (-167+258) = (258-167) = 91
|
|
|
