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28 Jan 2012, 04:05
3
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Difficulty:

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53% (02:44) correct 47% (01:44) wrong based on 1481 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 = 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, 04:22, edited 1 time in total. Added the OA Math Expert Joined: 02 Sep 2009 Posts: 37590 Followers: 7398 Kudos [?]: 99558 [17] , given: 11023 Re: SI [#permalink] ### Show Tags 28 Jan 2012, 04:20 17 This post received KUDOS Expert's post 20 This post was BOOKMARKED 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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A total of 6000 was invested [#permalink]

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05 Mar 2012, 13:57
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 . Math Expert Joined: 02 Sep 2009 Posts: 37590 Followers: 7398 Kudos [?]: 99558 [0], given: 11023 Re: A total of 6000 was invested [#permalink] ### Show Tags 05 Mar 2012, 14:06 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.
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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

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) Followers: 0 Kudos [?]: 11 [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, 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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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) Followers: 0 Kudos [?]: 75 [0], given: 1 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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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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19 Jul 2013, 06:31
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Hi guys.

I want to explain my more-abstract solution for this question. A kind of solution one require for answering a DS question in an efficient way.

First lets model what the question put in our table. Consider the part of investment that have x percent interest A and the other part (60k-A). Now what we have is this: Ax+(60k-A)y = 4080

Now lets begin from statement 1: It says that x=(3/4)y. If we put this equation into the original model then we got nothing. Why? because what we got is 1 equation with 2 unknown variables. so cross off statement 1.

Now statement 2: we got A/(60k-A)=3/2. Like the first statement if we plug this equation into original one we still have 1 equation with 2 unknown variables. Now statement 2 is out too.

1+2: from statement 2 we can draw that the 60k investment has 5 part (because the ration of two parts was 3:2) then we can calculate that each part is 12k (60k/5). Now if we plug statement 1 into the original model we got 1 equation with 1 unknown parameter. Problem solved. The answer is C.

My main point here was to stress that for solving DS question try to avoid doing the problem with math concepts only. If you see that you reach to the point that the equation has an unique answer then pick the right answer choice and go ahead!
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28 Jul 2013, 10:11
fozzzy wrote:
What would be the difficulty level of this question? 650 ish or more than that?

My idea, it is a 600-650 problem.
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] ### Show Tags 13 Jan 2014, 06:54 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.

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.

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07 Jun 2014, 13:34
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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. Re: SI [#permalink] 07 Jun 2014, 13:34 Go to page 1 2 Next [ 30 posts ] Similar topics Replies Last post Similar Topics: If Rebecca made$60,000 last year after taxes and put all of this mone 2 23 Feb 2017, 03:51
4 Leo invested a certain amount of money for one year. He invested $1,00 6 15 Jul 2016, 13:54 1 A trustee invested a total of$100,000, more than half of wh 2 07 Sep 2013, 04:08
3 If $5,000 invested for one year at p percent simple annual 8 09 Jul 2012, 04:46 9 A total of$60,000 was invested for one year. But of this 12 22 Mar 2009, 23:18
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