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

Redeem

Arithmetic Progression

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Sun May 15, 2011 6:10 pm

Arithmetic Progression

by vikram.sumer » Wed Apr 04, 2012 6:58 pm
Ques: A person saves each year Rs 100 more than he saved in the preceding year, and he
saves Rs 200 the first year. How many years would it take for his savings, not including
interest, to amount to Rs 23000 ?

OA: 20

I know my strategy is wrong, but could not figure Why? Please advise

My Thoughts: This question is like a series: 200,300,400........23000. We need to find what is term in this series which corresponds to 23000

Sn = 100(n)+x

X= 100

23,000 = 100(n)+ 100

X= 229
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Wed Apr 04, 2012 8:19 pm
This is an arithmetic series, so the sum can found using the formula:
Sn = (n/2)[2a + (n - 1)d], where a = first term, d = common difference, n = number of terms

Here, let us assume that n = total number of years
First term, a = 200, common difference, d = 100
Then, 23000 = (n/2)[2 * 200 + (n - 1)100]
23000 = n[200 + (n - 1)50]
23000 = 200n + 50n² - 50n
23000 = 50n² + 150n
n² + 3n - 460 = 0
(n - 20)(n + 23) = 0
n = 20 (n = -23 is not possible, as the no. of years cannot be negative)
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Senior | Next Rank: 100 Posts
Posts: 61
Joined: Mon Jan 18, 2010 11:09 pm
Thanked: 8 times

by Pharo » Wed Apr 04, 2012 10:33 pm
vikram.sumer wrote:Ques: A person saves each year Rs 100 more than he saved in the preceding year, and he
saves Rs 200 the first year. How many years would it take for his savings, not including
interest, to amount to Rs 23000 ?

OA: 20

I know my strategy is wrong, but could not figure Why? Please advise

My Thoughts: This question is like a series: 200,300,400........23000. We need to find what is term in this series which corresponds to 23000

Sn = 100(n)+x

X= 100

23,000 = 100(n)+ 100

X= 229
Hey dude!

Your mistake is you are solving for the wrong thing. Yes it is a series but the question is not asking what n (year) he would save 23000. The question is asking how long it will take him to save up to 23000. As you demonstrated above, he will save 23000 on his 229th year (haha good luck living that long :P)

So you have to add 200+300+400+...+((100)nth year + 100) and solve for the n:

Sum ((100)n + 100)) from 1 to x (i.e from year one to x) = 100 * sum(n){1:x} + sum(100){1:x}
= 100*((x *(x+1))/2) + 100x ;factor out the first nasty looking term
= 50x^2 + 50x + 100x
= 50x^2 + 150x ; this is how much he saves in x years. Set this to 23000 to find how many years it will take him to save 23000
; 50x^2 + 150x = 23000; divide everything by 50 to make the numbers smaller (smaller = easier)
x^2 + 3x = 460 ; put everything on one side
x^2 + 3x - 460 = 0; find to factors of 460 such that k*m = -460 AND k-m = 3;
x^2 + 3x + ((-20)*23) = 0; --> and the answer is 20 :)
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Thu Apr 05, 2012 4:21 am
vikram.sumer wrote:Ques: A person saves each year Rs 100 more than he saved in the preceding year, and he
saves Rs 200 the first year. How many years would it take for his savings, not including
interest, to amount to Rs 23000 ?

OA: 20

I know my strategy is wrong, but could not figure Why? Please advise

My Thoughts: This question is like a series: 200,300,400........23000. We need to find what is term in this series which corresponds to 23000

Sn = 100(n)+x

X= 100

23,000 = 100(n)+ 100

X= 229
The values here are evenly spaced integers.
Given evenly spaced values:
Average = (biggest+smallest)/2.
Sum = number*average.

The GMAT would provide answer choices, which we could plug in for the number of years needed to accrue 23,000 in savings.

Let's say that the answer choices were as follows:
5 10 15 20 30

Answer choice C: 15 years
Biggest value = 200 + 14(100) = 1600.
Average = (1600+200)/2 = 900.
Sum = 15*900 = 13,500.
Too small.
Eliminate A, B and C.

Answer choice D: 20 years
Biggest value = 200 + 19(100) = 2100.
Average = (2100+200)/2 = 1150.
Sum = 20*1150 = 23,000.
Success!
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Post new topic Post Reply
• Page 1 of 1