GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 20 Jul 2018, 19:21

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

If a + 1 = 20/a and b is the average of a set of c consecutive integer

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Senior SC Moderator
User avatar
V
Joined: 14 Nov 2016
Posts: 1322
Location: Malaysia
GMAT ToolKit User Premium Member CAT Tests
If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

Show Tags

New post 22 Mar 2017, 01:13
5
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

59% (02:03) correct 41% (02:12) wrong based on 152 sessions

HideShow timer Statistics

If \(a + 1 = \frac{20}{a}\) and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. \((a^2)(b^2)(c^2)\) is even.

II. \(a + b + c\) is odd.

III. \(ab(c^2 + c)\) is even.


A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Rules for posting in verbal forum | Please DO NOT post short answer in your post!

Advanced Search : https://gmatclub.com/forum/advanced-search/

Expert Post
1 KUDOS received
GMAT Club Legend
GMAT Club Legend
User avatar
P
Joined: 16 Oct 2010
Posts: 8132
Location: Pune, India
Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

Show Tags

New post 22 Mar 2017, 02:05
1
ziyuen wrote:
If \(a + 1 = \frac{20}{a}\) and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. \((a^2)(b^2)(c^2)\) is even.
II. a + b + c is odd.
III. \(ab(c^2 + c)\) is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III



\(a + 1 = \frac{20}{a}\)

\(a^2 + a - 20 = 0\)

a = -5 or 4

a could be even or odd.

c is odd.

b is the average of c consecutive integers. Say 3 consecutive integers (1, 2, 3, then b = 2). Say 5 consecutive integers (1, 2, 3, 4, 5, then b = 3)
b could be even or odd.

I. \((a^2)(b^2)(c^2)\) is even.
a, b and c all could be odd in which case this will be odd. So not necessarily true.

II. a + b + c is odd.
If a is 4 and b and c are odd, this will be even. So not necessarily true.

III. \(ab(c^2 + c)\) is even.
Since c is odd, \(c^2\) is also odd.
Odd + Odd = Even

Since \((c^2 + c)\) is even, and any integer multiplied by an even number gives an even result, \(ab(c^2 + c)\) is certainly even.

Answer (C)
_________________

Karishma
Private Tutor for GMAT
Contact: bansal.karishma@gmail.com

Senior Manager
Senior Manager
avatar
B
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

Show Tags

New post 22 Mar 2017, 02:17
2
ziyuen wrote:
If \(a + 1 = \frac{20}{a}\) and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. \((a^2)(b^2)(c^2)\) is even.
II. a + b + c is odd.
III. \(ab(c^2 + c)\) is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


Hi

\(a + 1 =\frac{20}{a}\)

\(a^2 + a - 20 = 0\)

\((a + 5)(a - 4) = 0\)

\(a= -5\) or \(a=4\)

\(b\) is the meadian of set S, which consists of \(c\) consecutive integers, where \(c\) is odd: S = {1,2,3} --> \(b=2\) or S={1,2,3,4,5} ----> \(b = 3\)

We have: \(c = odd\), \(a = even/odd\) and \(b = even/odd\).

I. \((a^2)(b^2)(c^2)\) is even. We can have \(odd^2*odd^2*odd^2 = odd\) No.

II. \(a + b + c\) is odd. We can have \(even + odd + odd = even\). No.

III. \(ab(c^2 + c)\) is even.

\(a*b*c*(c + 1) = even\). \(c\) is odd hence \(c + 1\) will always be even. Yes.

Answer C.
Senior SC Moderator
User avatar
V
Joined: 14 Nov 2016
Posts: 1322
Location: Malaysia
GMAT ToolKit User Premium Member CAT Tests
Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

Show Tags

New post 23 Mar 2017, 18:53
ziyuen wrote:
If \(a + 1 = \frac{20}{a}\) and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. \((a^2)(b^2)(c^2)\) is even.
II. a + b + c is odd.
III. \(ab(c^2 + c)\) is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


Source : Manhattan Advanced Quant Question No. 2

OFFICIAL SOLUTION



The three statements concern even/odd status, so we need to consider the odd/even status of a, b, and c. Start by solving the quadratic equation for a:

\(a + 1 = \frac{20}{a}\)
\(a^2 + a = 20\)
\(a^2 + a – 20 = 0\)
\(( a + 5)( a – 4) = 0\)
a = –5 or 4

So a can be either odd or even.

We were told that c is odd.

The average of an odd number of consecutive integers is simply equal to the middle term. That middle term must be an integer, but could be odd or even. Thus, b can be either odd or even.

We might have been able to glance ahead and see that \(c^2 + c\) is always even for any integer c. There are at least two rationales: first, since \(c^2\) will have the same odd/even property as c, we are adding either an even and an even or an odd and an odd. The result will always be even. Alternatively, we could factor \(c^2 + c\) into \(c( c + 1)\), which will always give us the even product of an odd and an even.

This problem involves a number of different topics: even/odds, averages, and quadratics. None of these topics is tested at an extremely difficult level, but we have to switch among them, making this hybrid problem more difficult than otherwise.

The correct answer is C.
Attachments

Untitled.jpg
Untitled.jpg [ 44.84 KiB | Viewed 1055 times ]


_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Rules for posting in verbal forum | Please DO NOT post short answer in your post!

Advanced Search : https://gmatclub.com/forum/advanced-search/

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 7312
Premium Member
Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

Show Tags

New post 12 Apr 2018, 06:25
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer &nbs [#permalink] 12 Apr 2018, 06:25
Display posts from previous: Sort by

If a + 1 = 20/a and b is the average of a set of c consecutive integer

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.