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A total of $60,000 was invested for one year. Part of this

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A total of $60,000 was invested for one year. Part of this [#permalink] New post 28 Jan 2012, 03:05
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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Jan 2012, 03:22, edited 1 time in total.
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Re: SI [#permalink] New post 28 Jan 2012, 03:20
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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.

Answer: C.
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 28 Jan 2012, 04:21
Thanks for the detailed explanation Bunuel.
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 24 Feb 2012, 18:58
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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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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 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] New post 25 Feb 2012, 04:09
[b]Lovely :) .[/b]
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A total of 6000 was invested [#permalink] New post 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] New post 05 Mar 2012, 13:06
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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 $60,000 was invested for one year. Part of this [#permalink] New post 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 :oops:
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 19 Jun 2012, 07:01
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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 :oops:


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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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 19 Jun 2012, 07:24
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] New post 19 Jun 2012, 07:27
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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] New post 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:
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] New post 23 Oct 2012, 06:45
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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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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 19 Jul 2013, 00:24
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 19 Jul 2013, 05:31
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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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 27 Jul 2013, 08:56
What would be the difficulty level of this question? 650 ish or more than that?
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Re: A total of $60,000 was invested for one year. Part of this [#permalink] New post 28 Jul 2013, 09:11
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] New post 13 Jan 2014, 05:54
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.

Answer is C
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Re: SI [#permalink] New post 07 Jun 2014, 12:34
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.

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, 12:34
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