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A total of $60,000 was invested for one year. Part of this [#permalink] ### Show Tags 28 Jan 2012, 03:05 2 This post received KUDOS 56 This post was BOOKMARKED 00:00 Difficulty: 85% (hard) Question Stats: 53% (02:45) correct 47% (01:44) wrong based on 1433 sessions ### HideShow timer Statistics 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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Jan 2012, 03:22, edited 1 time in total.
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28 Jan 2012, 03:20
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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 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. Answer: C. _________________ Manager Joined: 22 Sep 2011 Posts: 214 GMAT 1: 720 Q49 V40 GMAT 2: Q V Followers: 1 Kudos [?]: 33 [0], given: 3 Re: A total of$60,000 was invested for one year. Part of this [#permalink]

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28 Jan 2012, 04:21
Thanks for the detailed explanation Bunuel.
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25 Feb 2012, 00:43
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shankar245 wrote:
Quote:
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.

Hi bunuel,

Your expert thoughts on a clarification I have got ,

when you have a equation in some problems there is only one solution and we can arrive on a unique value.
such that no other values of x or y can satisy that equation.

In the stmt above i spent 30 secs thinking if a unique solution would be available!
how do we coem to a conclusion tat there is not unique solution and we can say its not sufficeint like stmt 2?

In statement (1) for any value of $$x$$ there will exist some $$a$$ to satisfy this equation (and vise versa), notice that $$a$$ and $$x$$ are not integers so we have no restriction on them whatsoever (mathematically we have an equation of a hyperbola and it has infinitely many solutions for x and a). The same for statement (2).

Most of the time when there might be only one solution for two unknowns you'll have linear equation and a restriction that these unknowns can be integers only (Diophantine equation).

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html

Hope it helps.
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] ### Show Tags 25 Feb 2012, 04:09 [b]Lovely .[/b] Senior Manager Joined: 18 Sep 2009 Posts: 360 Followers: 3 Kudos [?]: 448 [0], given: 2 A total of 6000 was invested [#permalink] ### Show Tags 05 Mar 2012, 12:57 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.

according to stmnt (2) x=3/2, y=2/5.Let A= amount invested in x%,B=60000-A is amount invested in y%.

A(x%)+b(y%)=4080
A(3/5)+(60000-A)(2/5)=4080.
if i solve this equation i am getting negative value. Is the above method correct .
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Re: A total of 6000 was invested [#permalink]

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05 Mar 2012, 13:06
Merging similar topics.

TomB 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. according to stmnt (2) x=3/2, y=2/5.Let A= amount invested in x%,B=60000-A is amount invested in y%. A(x%)+b(y%)=4080 A(3/5)+(60000-A)(2/5)=4080. if i solve this equation i am getting negative value. Is the above method correct . The red part is not correct. We are given that the ratio of two amounts is 3 to 2 not the ratio of interest rates. For complete solutions refer to the posts above. Hope it helps. _________________ Intern Status: Fighting Gravity.. Joined: 25 May 2012 Posts: 29 Location: India GMAT 1: 660 Q47 V35 GMAT 2: 750 Q50 V41 GPA: 3.25 WE: Programming (Computer Software) Followers: 0 Kudos [?]: 10 [0], given: 6 Re: A total of$60,000 was invested for one year. Part of this [#permalink]

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19 Jun 2012, 06:54
One question... Why are we not considering the time for which each investment was made? isnt the formula for Interest earned = P x R X T/100 ??

Apologise if its a dumb question
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] ### Show Tags 19 Jun 2012, 07:01 ankushgrover wrote: One question... Why are we not considering the time for which each investment was made? isnt the formula for Interest earned = P x R X T/100 ?? Apologise if its a dumb question Welcome to GMAT Club. Below is an answer for your question. We are told that "A total of$60,000 was invested for one year", so we can omit multiplying by 1.

For more on this kind of problems please check: math-number-theory-percents-91708.html

Hope it helps.
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19 Jun 2012, 07:27
ankushgrover wrote:
But when we are breaking it into 2 parts 'a' and '60,000 - a' , wouldn't it matter if i invested 'a' for 2 months or 10 months? The interest amount of 4080 can come for a specific duration only, right?

It follows from the stem that both $$a$$ and $$60,000-a$$ were invested for one year.

Hope it's clear.
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Re: A total of 6000 was invested [#permalink]

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20 Oct 2012, 06:36
Could someone please explain in detail why A (amount) is taken into the equation? In my mind, when we have x and y %'s this is what we need to calculate the partial amounts as well...?

Bunuel wrote:
Merging similar topics.

TomB wrote:

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19 Jul 2013, 00:24
From 100 hardest questions.
Bumping for review and further discussion.
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27 Jul 2013, 08:56
What would be the difficulty level of this question? 650 ish or more than that?
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13 Jan 2014, 05:54
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. We're given $$((a)*6000*\frac{x}{100}) + ((1 - a) * 6000 * \frac{y}{100}) = 4080$$ Where a and (1 - a) indicate the fraction of 60,000 that's invested at x and y percent, respectively. What we really need is NOT the values of a, x, and y, we need a RELATIONSHIP between a and (1 - a) and relationship between x and y 1) Only gives us relationship between x and y, we know nothing about a and (1 - a), insufficient 2) Only gives us relationship of a and (1 - a), we know nothing about the relationship between x and y 1 + 2 we're given both relationships, this is sufficient. Answer is C Senior Manager Joined: 15 Aug 2013 Posts: 328 Followers: 0 Kudos [?]: 56 [0], given: 23 Re: SI [#permalink] ### Show Tags 07 Jun 2014, 12:34 1 This post was BOOKMARKED Bunuel wrote: RadhaKrishnan 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 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.

Hi Bunuel,

This makes complete sense in retrospec but I tried to solve using the weighted avg formula:

W1/W2 = A2-Avg/Avg-A1

I realize that the Avg is 6.8% but I was thrown off by the values of W1/W2 and A2, A1. Is w1 and w2 supposed to be the amount that accrues interest rate 1 and 2, or in this case, x and y?

Thanks.
Re: SI   [#permalink] 07 Jun 2014, 12:34

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