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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]

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New post 05 Aug 2006, 07:09
2
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A
B
C
D
E

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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) \(x = \frac{3y}{4}\)
(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.
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New post 05 Aug 2006, 09:28
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
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Re: investment  [#permalink]

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New post 05 Aug 2006, 10:31
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

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New post 05 Aug 2006, 10:35
[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% ?
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New post 05 Aug 2006, 14:40
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...
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Re: investment  [#permalink]

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New post 05 Aug 2006, 15:30
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%
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Re: investment  [#permalink]

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New post 05 Aug 2006, 15:33
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
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Re: investment  [#permalink]

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New post 05 Aug 2006, 15:48
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.

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Re: investment  [#permalink]

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New post 05 Aug 2006, 15:52
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

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Re: interest and equations  [#permalink]

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New post 19 Jan 2008, 09:01
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%
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Re: DS: Interests ratios  [#permalink]

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New post 23 Aug 2009, 19:54
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.
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Re: annual interest  [#permalink]

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New post 29 Dec 2009, 07:31
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
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Re: SI Problem  [#permalink]

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New post 26 Aug 2010, 10:26
3
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.

Answer: C.
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Re: investment  [#permalink]

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Re: SI Problem  [#permalink]

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Re: SI Problem  [#permalink]

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New post 13 May 2011, 18:50
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
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Re: SI Problem  [#permalink]

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New post 13 May 2011, 19:37
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
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Re: SI Problem  [#permalink]

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New post 13 May 2011, 19:38
\(px/100 + (60-p)y/100 = 4080\)

1. Not sufficient


we dont know anything about amount invested at x%

2. Not sufficient.

not enough to find p or x or y.

Together ,

its sufficient

as we can find p,y and x

Answer is C.
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Re: A total of $60,000 was invested for one year. Part of this  [#permalink]

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New post 24 Jul 2014, 15:59
Let A be the part that gets interest at the rate of x% B = 60000 -A that gets interest at y%. There are two unknowns in question stem - the relation between A and B and the relation between x and y
Statement 1 : gives relation between x and y, but we do not know relation between A and B --> insufficient
Statement 2 gives A:B = 3:2, --> A = 36000 and B = 24000 but nothing about x:y, --> Insufficient

Combining 1 and 2 --> Can be solved as equation can be reduced to one variable , hence sufficient
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Re: A total of $60,000 was invested for one year. Part of this  [#permalink]

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Re: A total of $60,000 was invested for one year. Part of this &nbs [#permalink] 05 Aug 2018, 15:05
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