Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Help - Problem Solving Question! sitting GMAT soon

Expert replies
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 :)
Join the discussion
Source: — Problem Solving |

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
Join the discussion

by Karishma123 » Tue Jul 06, 2010 7:24 pm
Thanks Heaps
Join the discussion

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
Join the discussion

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
Join the discussion