Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In an increasing set of 7 consecutive integers, the sum of t [#permalink]
05 Nov 2012, 11:02

1

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

95% (02:08) correct
5% (01:00) wrong based on 21 sessions

In an increasing set of 7 consecutive integers, the sum of the last 4 integers is 34. What is the sum of the first 3 integers in the set?

A. 12 B. 15 C. 18 D. 36 E. 120

Now my doubt is:

A question from OG 13 page num 164 ( Q no. 90), this question i solved in the following manner

Sum of (1 to 5) = 15 Sum of last 5 (6 to 10) is 40 Differnce is 25 As per question we know sum of first 5 so we can obtain sum of last 5 by adding 25 to it. The method is also explained as alternative method in the OG

If the same concept we apply in the problem mentioned above Sum of (1,2,3) = 6 Sum of (4,5,6,7) = 22 difference is 16 So if we subtract 16 from 34 ie 18 should be the answer.

One difference her i found is that the difference of 16 is not common. So what is the principle and logic here. Method x,x+1.... is known to me Pls explain my doubt.

Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
05 Nov 2012, 11:11

Expert's post

Archit143 wrote:

In an increasing set of 7 consecutive integers, the sum of the last 4 integers is 34. What is the sum of the first 3 integers in the set?

A. 12 B. 15 C. 18 D. 36 E. 120

Now my doubt is:

A question from OG 13 page num 164 ( Q no. 90), this question i solved in the following manner

Sum of (1 to 5) = 15 Sum of last 5 (6 to 10) is 40 Differnce is 25 As per question we know sum of first 5 so we can obtain sum of last 5 by adding 25 to it. The method is also explained as alternative method in the OG

If the same concept we apply in the problem mentioned above Sum of (1,2,3) = 6 Sum of (4,5,6,7) = 22 difference is 16 So if we subtract 16 from 34 ie 18 should be the answer.

One difference her i found is that the difference of 16 is not common. So what is the principle and logic here. Method x,x+1.... is known to me Pls explain my doubt.

I'm happy to help with this.

The trouble is: this trick (finding the sum of the last set minus the sum of the first set) only works if the two sets are of equal size. Once the sets are of unequal sizes, the difference will change radically --- think if the first three integers are {101, 102, 103} and the last four are {104, 105, 106, 107} --- the the sum of the latter will be over a hundred more than the sum of the former!

I would say --- it's a good start to add 4+5+6+7 = 22. We know that 34 is 12 more than 22, so that means each of the four terms needs to be 12/4 = 3 units bigger --- instead of {4, 5, 6, 7}, those last four will be "up three" from those: {7, 8, 9, 10}. Then the first three are {4, 5, 6}, and their sum if 15.