BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
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
GMATOnDemand Start 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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

A

Expert replies
by shashank.ism » Thu Feb 11, 2010 7:13 am
If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a.) A+1
b.) A+5
c.) A+25
d.) 2A
e.) 5A
My Websites:
www.mba.webmaggu.com - India's social Network for MBA Aspirants

www.deal.webmaggu.com -India's online discount, coupon, free stuff informer.

www.dictionary.webmaggu.com - A compact free online dictionary with images.

Nothing is Impossible, even Impossible says I'm possible.
Join the discussion
Source: — Problem Solving |

by sars72 » Thu Feb 11, 2010 8:55 am
1+2+3+4+5 = 15 = A
6+7+8+9+10 = 40

--> a + 25 --> answer choice C

solving algebraically -> x+(x+1)+(x+2)+(x+3)+(x+4) = A
--> 5x+10 = A

sum next 5 integers = (x+5)+(x+6)+(x+7)+(x+8)+(x+9) = 5x+35

(5x+35) - (5x+10) = 25

--> Difference is 25 --> a+25 --> answer choice C
Join the discussion

by sparky_paris » Thu Feb 11, 2010 5:29 pm
Does it have to be positive integers? The answer holds good for any integer
Join the discussion

by money9111 » Thu Feb 11, 2010 8:45 pm
yes +25... all you need to know is that the 1st and 6th number have a difference of 5. the 2nd and 7th number also have a difference of 5 and so on... therefore since there are 5 sets of numbers with a difference of 5... 5*5=25
My goal is to make MBA applicants take onus over their process.

My story from Pre-MBA to Cornell MBA - New Post in Pre-MBA blog

Me featured on Poets & Quants

Free Book for MBA Applicants

Join the discussion

by harsh.champ » Thu Feb 18, 2010 10:16 am
shashank.ism wrote:If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a.) A+1
b.) A+5
c.) A+25
d.) 2A
e.) 5A
I solved by this approach:-
Let the 1st term be t.
2nd term = t+1
SO we get that 5t + 10 = A.
The sum of next 5 consecutive integers is (t+5) + (t+6) + .....
=5t +(5+6+7+8+9)
=(5t + 10) + 25
= [spoiler]A + 25
Hence,C should[/spoiler] be the answer.

Hey money9111,
yes +25... all you need to know is that the 1st and 6th number have a difference of 5. the 2nd and 7th number also have a difference of 5 and so on... therefore since there are 5 sets of numbers with a difference of 5... 5*5=25
I didn't get how you took the above bold-faced statement:- 5*5 = 25
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Join the discussion

by money9111 » Thu Feb 18, 2010 1:02 pm
ok so we have 10 numbers correct?

The 1st 5 numbers are: x, x+1, x+2, x+3, x+4, x+5 - let's call this Group A
The 2nd 5 numbers are: x+6, x+7, x+8, x+9, x+10 - let's call this Group B

We know that these are the numbers because they're consecutive...

so looking at that.. we know that the difference between the 1st number in Group A and the 1st number in the Group B is 5. ((x+6) - x)=5.

The same goes for the 2nd number in Group A and the 2nd number in Group B. ((x+7)-(x+1))=5.

that happens 5 times.. so 5*5=25...
My goal is to make MBA applicants take onus over their process.

My story from Pre-MBA to Cornell MBA - New Post in Pre-MBA blog

Me featured on Poets & Quants

Free Book for MBA Applicants

Join the discussion

by harsh.champ » Thu Feb 18, 2010 2:39 pm
money9111 wrote:ok so we have 10 numbers correct?

The 1st 5 numbers are: x, x+1, x+2, x+3, x+4, x+5 - let's call this Group A
The 2nd 5 numbers are: x+6, x+7, x+8, x+9, x+10 - let's call this Group B

We know that these are the numbers because they're consecutive...

so looking at that.. we know that the difference between the 1st number in Group A and the 1st number in the Group B is 5. ((x+6) - x)=5.

The same goes for the 2nd number in Group A and the 2nd number in Group B. ((x+7)-(x+1))=5.

that happens 5 times.. so 5*5=25...
Okay,I get it now.
Thanks for the shortcut approach.It can really come in very handy.
I guess my problem approach was a bit time taking.
Did you make this approach while solving the question or is this one of a common trick ??
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Join the discussion

by money9111 » Thu Feb 18, 2010 2:59 pm
haha no problem! actually i would never have thought of this approach in a million years...i solved it in about double the time that this approach took... my mgmat instructor told us that 700 level test takers think in that manner... so im like whoa... i better start thinking that way...it totally makes sense though
My goal is to make MBA applicants take onus over their process.

My story from Pre-MBA to Cornell MBA - New Post in Pre-MBA blog

Me featured on Poets & Quants

Free Book for MBA Applicants

Join the discussion

by shashank.ism » Fri Feb 19, 2010 1:17 am
money9111 wrote:yes +25... all you need to know is that the 1st and 6th number have a difference of 5. the 2nd and 7th number also have a difference of 5 and so on... therefore since there are 5 sets of numbers with a difference of 5... 5*5=25
yeah money1119 that is indeed a very good solution. I would have never thought of in this way. It really shows how we can solve a problem in seconds with some juggling...and score much higher...
My Websites:
www.mba.webmaggu.com - India's social Network for MBA Aspirants

www.deal.webmaggu.com -India's online discount, coupon, free stuff informer.

www.dictionary.webmaggu.com - A compact free online dictionary with images.

Nothing is Impossible, even Impossible says I'm possible.
Join the discussion