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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

There are 25 necklaces such that the first necklace contains [#permalink]

Show Tags

07 Nov 2009, 10:37

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

53% (03:40) correct
47% (03:42) wrong based on 85 sessions

HideShow timer Statictics

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

The answer is B. I would be happy to skip this problem on the GMAT, if it comes. It has taken me ages to solve this

Any way Please comment whether this solution is faster

We have a sequence of 5, 7, 10, 14, 19…. = (4+1), (4+3), (4+6)……. The sum of the sequence is = (25*4)+1+3+6+….25(25+1)/2 (Generic term is x(x+1)/2)

1+3+6+….25(25+1)/2 = Sum of first 25 triangular number (The formula of sum of n triangular number is ½[Sum [n^2] +Sum[n]], when n=25; ½[Sum [n^2] +Sum[n]] =2925)

Total = 25*5 + 24*2 + 23*3 + ....3*23+ 2*24 + 25 = 26*5 + definae Integral of x(26-x)dx from 2 to 24 = 2961 ( it took me 5 min to do by hand, i checked with Mathcad for the integral part)

Im trying to think up a faster way to solve this but nothing seems to be striking Im sure there must be one else this problem is way to time consuming to be on the gmat... _________________

N = 5 + (5+2) + (5+2+3) + (5+2+3+4) + + (5+2+3+4+ 25) By rearranging the terms, we find N = (25*5) + (24*2) + (23*3) + (22*4) + (1*25) N = 125 + Sum ((25-k)*(k+1)) from k = 1 to 24 N = 125 + Sum (24k-(k^2)+25) from k = 1 to 24 By developing and applying the formulas Sum (k^2) = 1/3 N*(N+0,5)*(N+1) (for k = 1 to N) We find N = 3025

N = 5 + (5+2) + (5+2+3) + (5+2+3+4) + + (5+2+3+4+ 25) By rearranging the terms, we find N = (25*5) + (24*2) + (23*3) + (22*4) + (1*25) N = 125 + Sum ((25-k)*(k+1)) from k = 1 to 24 N = 125 + Sum (24k-(k^2)+25) from k = 1 to 24 By developing and applying the formulas Sum (k^2) = 1/3 N*(N+0,5)*(N+1) (for k = 1 to N) We find N = 3025

which formula is applied in the above higlighted step....please elaborate it more....

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get: \(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

in the above solution i understood each n every step... but i have only one concern .....from where it wud enter in our minds during the exam that v got to subtract '4' from each term to get the sequence.... is there any way to observe this thing....i know it wud come only by practice but still i wanted to know that how it wud strike our minds to subtract 4 from each term.....phewwww!!!!!!!!

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

Bunuel, how did you think about reducing 4 in each number to find perfect squares? What's the link between perfect squares and that sequence?

Re: There are 25 necklaces such that the first necklace contains [#permalink]

Show Tags

03 Jul 2014, 17:21

Bunuel wrote:

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

How did you know after 7 to go on to 10 and so forth? I know you probably used the i more beads info but can you show the plugging in the #'s to get 10 if this is what you did?

Re: There are 25 necklaces such that the first necklace contains [#permalink]

Show Tags

03 Jul 2014, 17:28

Actually Bunuel,

I am also confused on how exactly you arrived at 100 + 1 + (3+6) + (10+15) + (21+28)+... I know you said subtract 4 from each term and then 4 x 25 = 100 but I don't see where all these terms are coming from.

Also I don't get the step after which is 100 + 1^2 + 3^2 etc... where is the 1^2 , 3^2 coming from?

Re: There are 25 necklaces such that the first necklace contains [#permalink]

Show Tags

05 Jul 2014, 06:59

Expert's post

sagnik242 wrote:

Bunuel wrote:

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

How did you know after 7 to go on to 10 and so forth? I know you probably used the i more beads info but can you show the plugging in the #'s to get 10 if this is what you did?

Re: There are 25 necklaces such that the first necklace contains [#permalink]

Show Tags

05 Jul 2014, 07:03

Expert's post

sagnik242 wrote:

Actually Bunuel,

I am also confused on how exactly you arrived at 100 + 1 + (3+6) + (10+15) + (21+28)+... I know you said subtract 4 from each term and then 4 x 25 = 100 but I don't see where all these terms are coming from.

Also I don't get the step after which is 100 + 1^2 + 3^2 etc... where is the 1^2 , 3^2 coming from?

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...