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Help - Problem Solving Question! sitting GMAT soon

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by Karishma123 » Tue Jul 06, 2010 5:30 pm
Hi.

Please help me solve this question

If X is the sum of even integers from 120 to 180 and Y is the number of even integers from 120 to 180 inclusive, what is the value of X+Y?

A) 4000
B) 4500
C) 4650
D) 4681
E) 9300

Thanks Heaps :)
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by jeremy8 » Tue Jul 06, 2010 6:16 pm
Here's how I do it.

120 to 180 included is 61 integers (180-120+1; +1 because of the "included" part), half of which are even. Since 120 and 180 are already even in this set (the extremes), we know we have 31 even integers out of 61.

Then you find the average of 120 and 180, which is 150. So you have X=31*150=4650.
Then Y is the number of even integers, and that's 31, as we found out already, so X+Y is 4650+31=4681
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by Karishma123 » Tue Jul 06, 2010 7:24 pm
Thanks Heaps
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by Rich@VeritasPrep » Tue Jul 06, 2010 7:28 pm
Just to illustrate an alternative approach...

You could recognize that adding the number of even integers from 120 to 180 to the sum of the even integers from 120 to 180 amounts to the same thing as adding 1 to each of the numbers in the list and adding the result.

Here's what I mean:

(120+122+124+...+178+180)+(# of numbers in the list)

= (120+1)+(122+1)+(124+1)+...+(178+1)+(180+1)

Notice that all the 1's add up to the number of numbers in the list.

= 121+123+125+...+179+181

The next part is exactly as jeremy8 outlined: Since there are 61 numbers from 121 to 181, there are 31 odd numbers in this range, and we can pair them up (or average, as jeremy8 did), and you'd get 15 pairs and a leftover number:

= (121+181) + (123+179) + (125+177) + ... + (149+153) + 151

= 302 * 15 + 151

= 4681
Rich Zwelling
GMAT Instructor, Veritas Prep
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by raul_200435 » Tue Jul 06, 2010 9:17 pm
There is another way to it also,
Remember, the formulae for sum of consecutive even numbers is n(n+1)
eg: sum of 2,4,6,8,10 (which are five consecutive even number is n=((2+10)/2))-1)=5, where 2 is the first number in the series & 10 is the last number therefore n(n+1)=5*(5+1)=30)

Apply the formulae here. The sum of consecutive even number till 180 is n(n+1)=90*91=8190, where ((n=2+180)/2)-1=90
The sum of consecutive even number till 120 is n(n+1)=60*61=3660
Take the difference of the sum 8190-3660+120=4650. (Remember while deducting the two sums we r also reducing 120 frm the sum which has to be added back.
Here we got the value of x=4650.
Now y=(180/2)-(120/2)+1=31 (Here we have to add 1 back because we have excluded 120 while calculating the difference. Remember x/y=z implies there are z multiples of of y between 0 & x).

Therefore we get x+y=4650+31=4681
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