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

 It is currently 19 Oct 2018, 11:40

### 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

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

# The average (arithmetic mean) of four different positive...

Author Message
TAGS:

### Hide Tags

Intern
Joined: 16 Mar 2013
Posts: 1
The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

Updated on: 17 Mar 2013, 01:30
8
00:00

Difficulty:

45% (medium)

Question Stats:

69% (02:18) correct 31% (02:27) wrong based on 311 sessions

### HideShow timer Statistics

The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

Originally posted by sheyla96 on 16 Mar 2013, 23:15.
Last edited by Bunuel on 17 Mar 2013, 01:30, edited 1 time in total.
Moved to PS forum.
VP
Joined: 24 Jul 2011
Posts: 1489
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740
Re: The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

17 Mar 2013, 01:23
1
Let the second integer be x and the fourth be a.

Then [3x + x + (x+2) + a]/4 = 9
=> 5x + 2 + a = 36
=> 5x + a = 34
=> a = 34 - 5x

From the above equation we can see that a is minimum when x is maximum, provided both are positive
The maximum value that x can take in the above equation while still keeping a positive is x=6
This gives us a= 34 - 30 = 4

Therefore the minimum value that the fourth integer can have is 4. Option B.
_________________

GyanOne | Top MBA Rankings and MBA Admissions Blog

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738

Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 384
Re: The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

27 Oct 2016, 12:21
Starting out...

[w+x+y+z]/4 = 9

w=3x --> 3y-6
x = y-2

Plug back into main equation

3y-6+y-2+y+z=36
5y-8+z=36
5y+z=44

We know multiples of 5 give us either a unit digit of 0 or 5 --> In this case all of the values w,x,y,z can take on are positive, thus we are limited to a unit digit of 0, making z = 4 & y = 8

VP
Joined: 07 Dec 2014
Posts: 1104
Re: The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

27 Oct 2016, 15:03
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

a+b+c+d=36
a+b+c=5b+2
greatest possible value of 5b+2<36=32
36-32=4=least possible value of d
B
Current Student
Joined: 12 Aug 2015
Posts: 2638
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

18 Dec 2016, 14:08
Here is my solution to this one ->

Let the integers be-->
w1
w2
w3
w4

Mean = 9

Using -->

$$Mean = \frac{Sum}{#}$$

Hence sum(4)=9*4=36

As per Question-->
w1=3w3-6
w2=w3-2
w3
w4

Hence w4=44-5w3

To minimise w4 => w3=> 8

Hence w4 => 4

Hence => B

_________________

MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!

Getting into HOLLYWOOD with an MBA!

The MOST AFFORDABLE MBA programs!

STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)

AVERAGE GRE Scores At The Top Business Schools!

Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4094
Location: India
GPA: 3.5
Re: The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

19 Dec 2016, 12:14
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

a + b + c + d = 36

(3c - 6) + (c - 2) + c + d = 36

Or, 5c + d = 44

Now, use Maximization/Minimization...

least possible value of the fourth integer ( ie, d ) is the maximum possible value of 5c

Here maximum possible value of 5c = 40 ; so d = 4

Hence, Correct answer will be (B) 4

_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

03 Jan 2017, 17:49
1
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

We are given that the average of four different positive integers is 9. Thus, the sum of the integers is 4 x 9 = 36.

We can create the following equation in which a = the first integer, b = the second integer, c = the third integer, and d = the fourth integer:

a + b + c + d = 36

We are given that the first of these integers is 3 times the second integer. Thus:

a = 3b

We are also given that the second integer is 2 less than the third integer. Thus:

b = c - 2

b + 2 = c

We can substitute 3b for a and (b + 2) for c in the equation a + b + c + d = 36:

3b + b + b + 2 + d = 36

5b + d + 2 = 36

5b + d = 34

5b = 34 - d

b = (34 - d)/5

Since b is an integer, 34 - d must be a multiple of 5, and since the largest multiple of 5 less than 34 is 30, the smallest value for d is 4.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Manager
Joined: 23 Jul 2015
Posts: 158
The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

04 Jan 2017, 09:40
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

SUM ($$a_1$$,...,$$a_5$$) = 36
$$a_1$$ = 3$$a_2$$
$$a_3$$ = $$a_2$$ + 2

SUM ($$a_1$$,...,$$a_5$$) = 3$$a_2$$ + $$a_2$$ + $$a_2$$ + 2 + $$a_4$$ = 36 --> 5$$a_2$$ + $$a_4$$ = 34 --> Min($$a_4$$) = 4
Non-Human User
Joined: 09 Sep 2013
Posts: 8462
Re: The average (arithmetic mean) of four different positive...  [#permalink]

### Show Tags

20 Apr 2018, 12:07
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.
_________________
Re: The average (arithmetic mean) of four different positive... &nbs [#permalink] 20 Apr 2018, 12:07
Display posts from previous: Sort by