Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 May 2013, 02:13

# A total of $60,000 was invested for one year. Part of this  Question banks Downloads My Bookmarks Reviews Author Message TAGS: Manager Joined: 08 Oct 2005 Posts: 102 Followers: 1 Kudos [?]: 0 [0], given: 0 A total of$60,000 was invested for one year. Part of this [#permalink]  05 Aug 2006, 08:09
00:00

Question Stats:

33% (02:36) correct 66% (01:15) wrong based on 0 sessions
A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the$60,000 for that year was $4,080, what is the value of x ? (1) x = \frac{3y}{4} (2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2. [Reveal] Spoiler: OA GMAT Club team member Joined: 02 Sep 2009 Posts: 11611 Followers: 1802 Kudos [?]: 9600 [1] , given: 829 Re: SI Problem [#permalink] 26 Aug 2010, 11:26 1 This post received KUDOS udaymathapati wrote: A total of$60,000 was invested for one year. But of this amount earned simple annual
interest at the rate of x percent per year, and the rest earned simple annual interest at the
rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?
(1) x = (3/4) y
(2) The ratio of the amount that earned interest at the rate of x percent per year to
the amount that earned interest at the rate of y percent per year was 3 to 2.

Let the amount invested at x% be a, then the amount invested at y% would be 60,000-a.

Given: a\frac{x}{100}+(60,000-a)\frac{y}{100}=4,080 (we have 3 unknowns x, y, and a). Question: x=?

(1) x=\frac{3}{4}y --> y=\frac{4x}{3} --> a\frac{x}{100}+(60,000-a)\frac{4x}{3*100}=4,080 --> still 2 unknowns - x and a. Not sufficient.

(2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2 --> \frac{a}{60,000-a}=\frac{3}{2} --> a=36,000 --> 36,000*\frac{x}{100}+(60,000-36,000)\frac{y}{100}=4,080 --> still 2 unknowns - x and y. Not sufficient.

(1)+(2) From (1) a\frac{x}{100}+(60,000-a)\frac{4x}{3*100}=4,080 and from (2) a=36,000 --> only 1 unknown - x, hence we can solve for it. Sufficient.

_________________
Senior Manager
Joined: 14 Jul 2005
Posts: 410
Followers: 1

Kudos [?]: 3 [0], given: 0

D

From 1 => x/100 * 60,000 + 4x/3 * 100 * 60,000 = 4080

From 2 => 3a + 2a = 4080 => a = 816
Hence x/100 * 60,000 = 3 * 816

Hence you can solve for x using either equation
Manager
Joined: 20 Mar 2006
Posts: 201
Followers: 1

Kudos [?]: 0 [0], given: 0

Re: investment [#permalink]  05 Aug 2006, 11:31
ong wrote:
A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the$60,000 for that year was $4,080, what is the value of x? (1)x = 3y/4 (2)The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2. (1) gives relationship between x and y. we still dont know the number of months to get a numerical value for x Insuff BCE (2) gives the ratio of Interest from x/Interest from y. Cannot determine x from this. CE (1) & (2) No new information about the number of months E Heman Manager Joined: 20 Mar 2006 Posts: 201 Followers: 1 Kudos [?]: 0 [0], given: 0 [#permalink] 05 Aug 2006, 11:35 [quote="gmatornot"]D From 1 => x/100 * 60,000 + 4x/3 * 100 * 60,000 = 4080 (1) I thought SI formula was prinicpal * rate * number of months/100. How are you getting x/100 * 60000. (2) From the question stem i understand that part of the 60000 was invested @ x% and the rest @ y% ? Senior Manager Joined: 14 Jul 2005 Posts: 410 Followers: 1 Kudos [?]: 3 [0], given: 0 [#permalink] 05 Aug 2006, 15:40 heman wrote: gmatornot wrote: D From 1 => x/100 * 60,000 + 4x/3 * 100 * 60,000 = 4080 (1) I thought SI formula was prinicpal * rate * number of months/100. How are you getting x/100 * 60000. (2) From the question stem i understand that part of the 60000 was invested @ x% and the rest @ y% ? You are RIGHT. I completely misread the question. I believe the answer is E as you have identified... Senior Manager Joined: 22 May 2006 Posts: 384 Location: Rancho Palos Verdes Followers: 1 Kudos [?]: 7 [0], given: 0 Re: investment [#permalink] 05 Aug 2006, 16:30 ong wrote: A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?
(1)x = 3y/4
(2)The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.

Let part of amount = A, then the other part = 60000-A
A*(x/100) + (60000-A)*(y/100) = 4080

S1. x=3y/4
A*(3y/400) + (60000-A)*(y/100) = 4080
We need value A to solve this. => insufficient.

S2.
A*(x/100) = (3/5)*4080 = 2448
(60000-A)*(y/100) = (2/5)*4080= 1632
We need value A to solve this. => insufficient.

Combine S1 and S2.
A*(3y/400) = 2448 => A=2448*400/3y=326400/y
(60000-A)*(y/100) = 1632
We have two variables with two eq. thus sufficient.

Hence, C..

(600-3264/y)*(y) = 1632
600y - 3264 = 1632
600y = 1632+3264
600y = 4896
y = 8.16 %
x=3y/4 = 8.16*3/4 = 6.12%
_________________

The only thing that matters is what you believe.

Senior Manager
Joined: 22 May 2006
Posts: 384
Location: Rancho Palos Verdes
Followers: 1

Kudos [?]: 7 [0], given: 0

Re: investment [#permalink]  05 Aug 2006, 16:33
heman wrote:
(1) & (2) No new information about the number of months

E

Heman

If the total interest earned by the $60,000 for that year was$4,080, what is the value of x
_________________

The only thing that matters is what you believe.

Manager
Joined: 20 Mar 2006
Posts: 201
Followers: 1

Kudos [?]: 0 [0], given: 0

Re: investment [#permalink]  05 Aug 2006, 16:48
freetheking wrote:
heman wrote:
(1) & (2) No new information about the number of months

E

Heman

If the total interest earned by the $60,000 for that year was$4,080, what is the value of x

I still dont get it. Wouldnt it be possible for amount A @ x% for n months + amount (60000 - A) @ y % for (12-n) months = 4080.

Then won't we have different value for x and y depending on the time period of investment within a year?

Or does the ? mean that each part invested @ x% and y% was for a year?

I had a similar solution to you except I had a variable for time period and so 2 eqns and 3 variables.

Heman
Senior Manager
Joined: 22 May 2006
Posts: 384
Location: Rancho Palos Verdes
Followers: 1

Kudos [?]: 7 [0], given: 0

Re: investment [#permalink]  05 Aug 2006, 16:52
ong wrote:
A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the$60,000 for that year was $4,080, what is the value of x? x and y are simple annual interest.. dealing with one year investment.. not monthly.. heman wrote: Or does the ? mean that each part invested @ x% and y% was for a year? Yes! Heman _________________ The only thing that matters is what you believe. Manager Joined: 08 Oct 2005 Posts: 102 Followers: 1 Kudos [?]: 0 [0], given: 0 Re: investment [#permalink] 05 Aug 2006, 19:09 freetheking wrote: ong wrote: A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?
(1)x = 3y/4
(2)The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.

Let part of amount = A, then the other part = 60000-A
A*(x/100) + (60000-A)*(y/100) = 4080

S1. x=3y/4
A*(3y/400) + (60000-A)*(y/100) = 4080
We need value A to solve this. => insufficient.

S2.
A*(x/100) = (3/5)*4080 = 2448
(60000-A)*(y/100) = (2/5)*4080= 1632
We need value A to solve this. => insufficient.

Combine S1 and S2.
A*(3y/400) = 2448 => A=2448*400/3y=326400/y
(60000-A)*(y/100) = 1632
We have two variables with two eq. thus sufficient.

Hence, C..

(600-3264/y)*(y) = 1632
600y - 3264 = 1632
600y = 1632+3264
600y = 4896
y = 8.16 %
x=3y/4 = 8.16*3/4 = 6.12%

OA is C ---your explanation is exceptional and clear
Manager
Joined: 01 Jan 2008
Posts: 231
Schools: Booth, Stern, Haas
Followers: 1

Kudos [?]: 37 [0], given: 2

Re: interest and equations [#permalink]  19 Jan 2008, 10:01
marcodonzelli wrote:
A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the$60,000 for that year was $4,080, what is the value of x? (1) x = 3/4 y (2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2. I think it is C First statement is not suff because we don't know which amount was exactly invested, second is not suff since we don't know the rates but combined we can solve it II. 3/5*60000=36000 I. 36000*3/4y+24000*y=4,080 17/4y=4080/12000 => y=8% => x=8*3/4 =6% Manager Joined: 10 Aug 2009 Posts: 139 Followers: 3 Kudos [?]: 46 [0], given: 10 Re: DS: Interests ratios [#permalink] 23 Aug 2009, 20:54 1)n1+ n2 = 60000 n1- amount invested at x%, n2 - amount invested at y%. 2) n1*x%+n2*y% = 4080 To solve this for x, we need two things: 1) the ratio of n1 to n2 (given in the second statement) 2) the ration of x to y (given in the first statement) So, both statements together are sufficient. Senior Manager Joined: 30 Aug 2009 Posts: 296 Location: India Concentration: General Management Followers: 2 Kudos [?]: 65 [0], given: 5 Re: annual interest [#permalink] 29 Dec 2009, 08:31 kirankp wrote: A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?

(1) x = 3y/4

(2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.

will go with C

let "a" be the amount which gets Simple interest at rate of x%/yr. Then SI = ax/100
then "60000 -a " is the amount which gets Simple interest at rate of y%/yr. Then SI = (60000-a)y/100

so we have (ax)/100 + (60000-a)y/100 = 4080

from stmnt 1- we have x = 3y/4 . even after substituting value of x in above equation we will not be able to get the value of x. Insuff

from stmnt 2 - we have (ax/100)/ (60000-a)y/100 =3/2. Again not suff

taking together x = 3y/4 and (ax/100)/ (60000-a)y/100 =3/2 we will get the value of x. hence suff
Senior Manager
Joined: 28 Aug 2010
Posts: 280
Followers: 3

Kudos [?]: 20 [0], given: 11

A total of $60,000 was invested for one year [#permalink] 09 Dec 2010, 17:37 A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?
(1) x = 3y/4
(2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.

Could some explain the ans here. I am not sure if the ans is correct
_________________

Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

GMAT Club team member
Joined: 02 Sep 2009
Posts: 11611
Followers: 1802

Kudos [?]: 9600 [0], given: 829

Re: A total of $60,000 was invested for one year [#permalink] 09 Dec 2010, 17:42 Merging similar topics. _________________ Director Status: GMAT Learner Joined: 14 Jul 2010 Posts: 672 Followers: 21 Kudos [?]: 108 [0], given: 31 Re: investment [#permalink] 13 May 2011, 13:40 Go at by bunnuel a good solution. _________________ I am student of everyone-baten Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html Director Status: GMAT Learner Joined: 14 Jul 2010 Posts: 672 Followers: 21 Kudos [?]: 108 [0], given: 31 Re: SI Problem [#permalink] 13 May 2011, 13:47 good solution by bunnuel. _________________ I am student of everyone-baten Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html SVP Joined: 16 Nov 2010 Posts: 1721 Location: United States (IN) Concentration: Strategy, Technology Followers: 26 Kudos [?]: 228 [0], given: 34 Re: SI Problem [#permalink] 13 May 2011, 19:50 a - x% interest (60000 - a) - y% interest ax/100 + (60000 - a)y/100 = 4080 (1) x = 3y/4 Not Sufficient (2) a/(60000 - a) = 3/2 No information about x and y, insufficient. (1) + (2) Sufficient Answer - C _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) Find out what's new at GMAT Club - latest features and updates VP Status: There is always something new !! Affiliations: PMI,QAI Global,eXampleCG Joined: 08 May 2009 Posts: 1400 Followers: 8 Kudos [?]: 84 [0], given: 10 Re: SI Problem [#permalink] 13 May 2011, 20:37 a tells nothing about P b tells nothing about relative information between x and y. a+b p can be calculated first. together with x and y information, x can be found out. C _________________ Visit -- http://www.sustainable-sphere.com/ Promote Green Business,Sustainable Living and Green Earth !! Re: SI Problem [#permalink] 13 May 2011, 20:37 Similar topics Replies Last post Similar Topics: A total of$60,000 was invested for one year. Part of this 1 01 Dec 2005, 00:59
A total of $60,000 was invested for one year. But of this 2 06 Feb 2006, 22:28 A total of$60,000 was invested for one year. Part of this 2 07 Oct 2008, 07:38
1 A total of $60,000 was invested for one year. Part of this 20 05 Aug 2006, 08:09 3 A total of$60,000 was invested for one year. Part of this 13 28 Jan 2012, 04:05
Display posts from previous: Sort by