Sum of 3 consecutive integers is 312

Its B

n+(n+1)+(n+2)= 312. Solve for N & find the next 3 Integers & add them up.




Hello,
Please help me understand this:

---
The sum of three consecutive integers is 312, whats the sum of the next three consecutive integers?

315

321

330

415

424

Source: Kaplan GMAT math workbook 6th edition - Page 32[/quote][/spoiler]

[/spoiler]

ceo1000 wrote:Hello,
Please help me understand this:

---
The sum of three consecutive integers is 312, whats the sum of the next three consecutive integers?

315

321

330

415

424

Source: Kaplan GMAT math workbook 6th edition - Page 32

When numbers are evenly spaced, median = average.
Thus, the median of the 3 integers = 312/3 = 104.
Thus, the 3 integers are 103, 104, and 105.
Sum of the next 3 consecutive integers = 106+107+108 = 321.

The correct answer is B.

thanks to all for your help!

ceo1000 wrote:Hello,
Please help me understand this:

---
The sum of three consecutive integers is 312, whats the sum of the next three consecutive integers?

315

321

330

415

424

Source: Kaplan GMAT math workbook 6th edition - Page 32

The sum of three consecutive numbers is 312, there fore the sum of next three consecutive numbers must be

312 + 3 x 3 = 321

mk101 wrote:
The sum of three consecutive numbers is 312, there fore the sum of next three consecutive numbers must be

312 + 3 x 3 = 321

mk101 has it dead right--this is definitely the fastest solution to the problem. I think it could use a little clarification where it's coming from, however: if the first number from the first set is N, followed by N + 1 and N +2, then the first number of the next set will be N + 3--exactly three higher. Similarly, the second number of the second set will be exactly 3 greater than the corresponding number in the original, as will the third number. Since each of the three numbers one the new list will be 3 greater than it's counterpart on the old list, the new total must be 3x3=9 greater than the old one.

Here's the detailed Math based on everyone's input:

Long Version:

n+(n+1)+(n+2)=312
3n + 3 = 312
3n = 312 - 3
3n = 309
n= 309/3
n= 103
-----------------------
103+(103+1)+(103+2)=312
103+104+105=312
312=312
Three consecutive integers = 103, 104 and 105
------------------------
The next three consecutive integers = 106, 107 and 108
106 + 107 + 108 = 321

Short Version:

312 + 3 X 3 = 321

3 X 3 = "# of Consecutive Integers" X "Sum of Increment Per Integer (n+(n+1)+(n+2) = 3n+3)"