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Manhattan Gmat - Number properties - Integers

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by FridaKräm » Thu Dec 15, 2011 5:11 am
"In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the last four integers?"

I solved this question by calculating the arithmetic mean for the two sets and then multipy them with the number of items in each set (4). Which means:
1->4 4 integers x arithmetic number: 2,5 = 10
5->8 4 integers x arithmetic number: 6,5 = 26

The difference is 26-10 = 16

Since the answer key suggests four different solutions and non is the one I used I wonder if "my" method can be applied on other questions like this one?

Thanks!
Frida
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by GmatMathPro » Thu Dec 15, 2011 5:27 am
FridaKräm wrote:"In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the last four integers?"
I think you may not have typed this quite right.
I solved this question by calculating the arithmetic mean for the two sets and then multipy them with the number of items in each set (4). Which means:
1->4 4 integers x arithmetic number: 2,5 = 10
5->8 4 integers x arithmetic number: 6,5 = 26

The difference is 26-10 = 16

Since the answer key suggests four different solutions and non is the one I used I wonder if "my" method can be applied on other questions like this one?

Thanks!
Frida
Your method is perfectly fine. Notice that it is very similar to the 4th solution they have listed on p. 58. The only difference is they added the numbers directly to find the total, whereas you found the total by multiplying the arithmetic mean by the number of elements. This will always be a valid way to find the total of any group of numbers, as the total is always equal to the average times the number of elements, by definition.
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by neelgandham » Thu Dec 15, 2011 5:32 am
Frida,

'Your' method is flawless. Am I sure ? yes I am .

Conventional Method
Step 1: Let the 8 consecutive integers be n, n+1, n+2, n+3, n+4, n+5, n+6, n+7.
Step 2: Sum of first 4 integers = n + n+1 + n+3 + n+4 = 4n + 8
Step 3: Sum of last 4 integers = n+5 + n+6 + n+7 + n+8 = 4n + 26
Step 4: Difference = 4n + 26 - (4n + 8) = 16

'Your' method
Step 1: Let the 8 consecutive integers be n, n+1, n+2, n+3, n+4, n+5, n+6, n+7.
Step 2: Average of first 4 integers = (n + n+1 + n+3 + n+4)/4 = (4n + 8)/4
Step 3: Average of last 4 integers = (n+5 + n+6 + n+7 + n+8)/4 = (4n + 26)/4
Step 4: Average of first 4 integers * 4 = (4n + 8) (Step 2 of the conventional method)
Step 5: Average of last 4 integers * 4 = (4n + 26) (Step 3 of the conventional method)
Step 6: Difference = 4n + 26 - (4n + 8) = 16 (Step 4 of the conventional method)

So, it is nothing but a conventional method using averages(Involving a couple of steps extra!). Finally it is up-to you to decide, because you would want to solve it in a method you are comfortable solving the problem in.

My way ;)
-1,0,1,2,3,4,5,6 be the integers
sum of first four integers = 2
sum of last four integers = 18
difference = 16 Boom!
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by LalaB » Thu Dec 15, 2011 8:05 am
r r+1 r+2 ....r+7
ignore R since it is constant. see ,that ur last 4 numbers are
4+5+6+7
ur fist 4 numbers are 0+1+2+3

so, (4+5+6+7) - (0+1+2+3)=16
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