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Updated on: 28 Jan 2012, 04:22
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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. Originally posted by RadhaKrishnan on 28 Jan 2012, 04:05. Last edited by Bunuel on 28 Jan 2012, 04:22, edited 1 time in total. Added the OA ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 49206 Re: SI [#permalink] ### Show Tags 28 Jan 2012, 04:20 27 23 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.

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24 Feb 2012, 19:58
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?
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25 Feb 2012, 05:09
[b]Lovely .[/b]
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19 Jun 2012, 08:01
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. _________________ 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) Re: A total of$60,000 was invested for one year. Part of this  [#permalink]

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19 Jun 2012, 08:24
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?
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Joined: 02 Sep 2009
Posts: 49206
Re: A total of $60,000 was invested for one year. Part of this [#permalink] ### Show Tags 19 Jun 2012, 08:27 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. _________________ Intern Joined: 11 Sep 2012 Posts: 6 Location: Norway Concentration: Healthcare, International Business Schools: Insead '14 (A) GMAT Date: 11-19-2013 WE: Brand Management (Pharmaceuticals and Biotech) Re: A total of 6000 was invested [#permalink] ### Show Tags 20 Oct 2012, 07:36 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: 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.
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Posts: 49206
Re: A total of 6000 was invested  [#permalink]

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23 Oct 2012, 07:45
asveaass wrote:
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...?

The amount invested at x% is $$a$$ and the amount invested at y% is $$60,000-a$$. We need to find the value of x. Can you please tell what confuses you here?
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13 Jan 2014, 06:54
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: 260 Re: SI [#permalink] ### Show Tags 07 Jun 2014, 13:34 1 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.
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08 Jun 2014, 04:47
1
russ9 wrote:
Bunuel 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. Answer: C. 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. $$4x = 3y$$ $$\frac{y-6.8}{6.8-x}=\frac{3}{2}$$. For more check: http://www.veritasprep.com/blog/2011/03 ... -averages/ and http://www.veritasprep.com/blog/2011/04 ... ge-brutes/ _________________ Intern Joined: 24 Oct 2014 Posts: 5 Re: A total of$60,000 was invested for one year. Part of this  [#permalink]

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12 Feb 2015, 07:59
hi,

I perfectly understand the mathematical reasoning behind the solution why stmt 1 is not valid BUT:

we have a ratio given for x and y. My question is why I cannot say that A is valid because under the condition that X=3/4Y there is only one division of the 60,000 that with the given ratio between x and y would yield 4,080 dollars??
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17 Jan 2016, 10:43
Bunuel 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. Answer: C. Hello Bunuel, Could you please explain if by asking to find X, we are asked to find the amount of money that was invested at x% or actual rate, denoted in this problem as X? I went ahead ad solved the last equation 4,080=36,000*(x/100)+(60,000-36,000)*(4/3*x/100) and x=8,877......does it mean rate was 88.77%? It just cannot be right. _________________ "You have to learn the rules of the game. And then you have to play better than anyone else". Albert Einstein Intern Joined: 01 Apr 2016 Posts: 11 Re: A total of$60,000 was invested for one year. Part of this  [#permalink]

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01 Jun 2016, 09:50
Let first call T the part of amount invested with interest of x perct

4080 = (1+ x/100)*T + (1+ y/100)*(60000-T)
So we have 3 unknown ; T, x and y
and 1 equation

(1) x = 3y/4

So we have
3 unknow
2 Independent equations
INSUFF

(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.
It means [(1+ x/100)*T ]/[ (1+ y/100)*(60000-T) ]= 3/2
So we have
3 unknow
2 Independent equations
INSUFF

(1+2)
3eq and 3 unknows
SUFF !
Re: A total of $60,000 was invested for one year. Part of this &nbs [#permalink] 01 Jun 2016, 09:50 Go to page 1 2 Next [ 27 posts ] Display posts from previous: Sort by # A total of$60,000 was invested for one year. Part of this

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