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Kaplan Math Workbook seventh edition

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Kaplan Math Workbook seventh edition

by wayneyau1214 » Wed Aug 29, 2012 7:56 pm
Hey guys, I've been stuck on this question for a while and couldn't figure out how to do it even after I looked at the answers. It's from the Kaplan Math workbook.

Problem:
The sum of three consecutive integers is 312. What is the sum of the next three consecutive integers?
1. 315
2. 321
3. 330
4. 415
5. 424

Solution:
1. 321

Explanation:
Suppose we call the three original integers x, x+1, and x+2. Their sum is 312, so x+(x+1)+(x+2)=312 or 3x+3=312. The next three integers are x+3, x+4, and x+5. What is the value of (x+3)+(x+4)+(x+5)? It's 3x+12.3x+12 is 9 greater than 3x+3.3x+3=312, so 3x+12=312+9, or 321.

What I don't understand is how they came up with 3x+12.3x+12 and 3x+3.3x+3=312. Where did 12.3 and 3.3 come from and why is there an extra "x" in the equation all of a sudden?
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by vk_vinayak » Wed Aug 29, 2012 10:17 pm
wayneyau1214 wrote:Hey guys, I've been stuck on this question for a while and couldn't figure out how to do it even after I looked at the answers. It's from the Kaplan Math workbook.

Problem:
The sum of three consecutive integers is 312. What is the sum of the next three consecutive integers?
1. 315
2. 321
3. 330
4. 415
5. 424

Solution:
1. 321

Explanation:
Suppose we call the three original integers x, x+1, and x+2. Their sum is 312, so x+(x+1)+(x+2)=312 or 3x+3=312. The next three integers are x+3, x+4, and x+5. What is the value of (x+3)+(x+4)+(x+5)? It's 3x+12.3x+12 is 9 greater than 3x+3.3x+3=312, so 3x+12=312+9, or 321.

What I don't understand is how they came up with 3x+12.3x+12 and 3x+3.3x+3=312. Where did 12.3 and 3.3 come from and why is there an extra "x" in the equation all of a sudden?
Suppose there are six CONSECUTIVE integers starting with x. They will be
x, x+1, x+2, x+3, x+4, x+5

From the problem, we are given that x + (x+1) + (x+2) = 312. i.e. 3x + 3 =312 --------(A)
Now, one way to solve this problem is find out x, and then find out the sum of x+3, x+4, x+5

OR, we can use a short-cut:
We need to find the sum x+3, x+4, x+5. Add them up (x+3)+(x+4)+(x+5) = 3x+12

From A, we can see that 3x+3=312, and we are asked to find 3x+12.
3x+12 can be written as (3x+3)+9 = 312 + 9 = 321
- VK

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by \'manpreet singh » Thu Aug 30, 2012 2:18 am
let x,x+1,x+2,x+3,x+4,x+5 be the six consequitive numbers
given that:
3x+3=312 ----------1

Required 3x+12=(3x+3)+9 --------------2

Put the value of 1 in eq 2
3x+12=312 +9=321

Ans=321
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by Brent@GMATPrepNow » Sun Sep 23, 2012 6:46 am
wayneyau1214 wrote:
The sum of three consecutive integers is 312. What is the sum of the next three consecutive integers?
1. 315
2. 321
3. 330
4. 415
5. 424
Here's a solution that requires no algebra:

Consider these 6 consecutive integers: 13, 14, 15, 16, 17, 18
Notice that the first red number (16), is 3 greater than the first blue number (13).
Similarly, the second red number (17), is 3 greater than the second blue number (14).
And the third red number (18), is 3 greater than the third blue number (15).

The same applies to these 6 consecutive integers: 45, 46, 47, 48, 49, 50, in fact it applies to any 6 consecutive integers.

So, if we're told that the sum of three consecutive integers is 312, then each of the next 3 numbers will be 3 greater than their earlier counterparts.

So, to find the sum of the next 3 numbers all we need to do is add three 3's to the earlier sum.

312 + 3 + 3 + 3 = 321

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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