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

Redeem

Number Series

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Fri Jan 08, 2010 3:58 am

Number Series

by ahinye » Sat Feb 13, 2010 2:20 pm
If S is a series of x consecutive odd integers and Z is the larget integer in S, what is the smallest integer in S.

1) Z-2x
2)Z-x+1
3)Z-2(x-1)
4)x-6+Z
5)Z- (x/2)


How can I solve this algebraically?
Top
Source: — Problem Solving |
User avatar
Senior | Next Rank: 100 Posts
Posts: 90
Joined: Fri Jan 22, 2010 1:10 pm
Location: New Jersey
Thanked: 13 times
Followed by:4 members
GMAT Score:640

by Mom4MBA » Sat Feb 13, 2010 2:49 pm
If S is a series of x consecutive odd integers and Z is the larget integer in S, what is the smallest integer in S.
Let the first number be 2n+1,
the series will be like
2n+1
2n+3
2n+5
:
:
Z

numbers 1,3,5,.....are in A.P. to calculate last number use [last number = first number + common difference (number of terms -1)]
L = 1 + 2(x-1) = 2x-1

so we get Z = 2n + 2x-1
Z+1 = (2n+1) +2x-1..............adding 1 on both sides
2n+1 = Z+1-2x+1 = Z-2x+2 = Z-2(x-1)

Answer C
Stay focused
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Feb 14, 2010 11:40 am
ahinye wrote:If S is a series of x consecutive odd integers and Z is the larget integer in S, what is the smallest integer in S.
1) Z-2x
2)Z-x+1
3)Z-2(x-1)
4)x-6+Z
5)Z- (x/2)
How can I solve this algebraically?
Here's another approach:

t1 = a (this is the fist term)
t2 = a + 2 (odd numbers increase by 2)
t3 = a + 2 + 2 (the third term is 2 greater than then second term)
t4 = a + 2 + 2 + 2
t5 = a + 2 + 2 + 2 + 2
t6 = a + 2 + 2 + 2 + 2 + 2 (I've listed many terms so we can see the pattern: the nth term is equal to a plus n-1 2's
So, tx = a + 2(x-1)
The question tells us that tx = z, so we can write a + 2(x-1) = z
Our goal is to find the value of the smallest/first term (a), so we solve a + 2(x-1) = z for a to get:
a = z - 2(x-1)
The answer is C
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Fri Jan 08, 2010 3:58 am

by ahinye » Sun Feb 14, 2010 12:24 pm
Thank you for this explanation. Looks easier when it has been layed out logically. Was going round in circles and had to post something. thanks for all the responses thus far. I prefer to solve with algebra than plugging numbers where possible
Top
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Mon Feb 15, 2010 9:24 am
Agreed .But plugging in numbers are pretty easy.

take ,5,7,9.....

HereZ=9,x=3

We know that our smallest number is 5.

Plug in the values of Z & X.

1.Z-2x
--->9- 2*3=3( Not 5, which is required)

2.Z-x+1
---> 9-3+1=7(Not the answer we want)

3.Z-2(x-1)
--->9-2(3-1)=5(Bingo! we got answer smallest integer)
Top
User avatar
Legendary Member
Posts: 1022
Joined: Mon Jul 20, 2009 11:49 pm
Location: Gandhinagar
Thanked: 41 times
Followed by:2 members

by shashank.ism » Mon Feb 15, 2010 10:10 pm
ahinye wrote:If S is a series of x consecutive odd integers and Z is the largest integer in S, what is the smallest integer in S.

1) Z-2x
2)Z-x+1
3)Z-2(x-1)
4)x-6+Z
5)Z- (x/2)


How can I solve this algebraically?
It is a very simple question of AP
where a= S, Tn= Z and n c.d. = 2 (for odd numbers)
thus Z= S+(x-1)2 --> [spoiler]S= Z-2(x-1) Ans3[/spoiler]
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.
Top
User avatar
Legendary Member
Posts: 1022
Joined: Mon Jul 20, 2009 11:49 pm
Location: Gandhinagar
Thanked: 41 times
Followed by:2 members

by shashank.ism » Mon Feb 15, 2010 10:11 pm
gmatmachoman wrote:Agreed .But plugging in numbers are pretty easy.

take ,5,7,9.....

HereZ=9,x=3

We know that our smallest number is 5.

Plug in the values of Z & X.

1.Z-2x
--->9- 2*3=3( Not 5, which is required)

2.Z-x+1
---> 9-3+1=7(Not the answer we want)

3.Z-2(x-1)
--->9-2(3-1)=5(Bingo! we got answer smallest integer)
I don't think plugin is at all required here. Just go with simple concept of AP and u will get ans in seconds.. Though plug in really helps a lot many a times..
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.
Top
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 am
Location: India
Thanked: 64 times
Followed by:6 members
GMAT Score:760

by harsh.champ » Mon Feb 15, 2010 11:36 pm
ahinye wrote:Thank you for this explanation. Looks easier when it has been layed out logically. Was going round in circles and had to post something. thanks for all the responses thus far. I prefer to solve with algebra than plugging numbers where possible
Well,I would say it should be decided upon seeing the question which technique to use:-In some cases that I have personally found,solving by the formal proper approach takes too much time and even the calculation process is very cumbersome.

But yeah,ofcourse,in case of even 10% doubt I would say adopt the formal approach.It is better not to change the strategy on the exam day.

It is better to start practising these techniques on a day-to-day basis so that the shortcut approaches becomes second nature to us.

In this question,going by the soln. that brent has posted,i guess it takes equal time by using both the techniques.
So any approach will do fine.
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
Top
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Wed Nov 11, 2009 11:23 am
Thanked: 8 times
Followed by:1 members

by amittilak » Tue Feb 16, 2010 12:21 pm
I think for many questions its a personal choice wether to use algebra or plug in numbers. For this problem, I would use algebra:
We have consecutive odd numbers "X", with largest number "Z"
Now, for evenly spaced sets,
The number of terms is calculated as:
X = (Z - L)/n + 1
n = common difference/spacing ( for Odd/even number n = 2)
L = SMALLEST ODD NUMBER OF THE SET,
Z = LARGEST ODD NUMBER OF THE SET
X = (Z - L)/2 + 1
Thus,
2X = Z - L + 2
Rearranging,
L = Z - 2X - 2
L = Z - 2(X - 1)
Top
Post new topic Post Reply
• Page 1 of 1