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A total of $60,000 was invested for one year. Part of this [#permalink]
 05 Aug 2006, 08:09
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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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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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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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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
Heman
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[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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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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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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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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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.
Heman
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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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freetheking wrote: 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% OA is C ---your explanation is exceptional and clear
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Re: interest and equations [#permalink]
19 Jan 2008, 10:01
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]
23 Aug 2009, 20:54
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]
29 Dec 2009, 08:31
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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A total of $60,000 was invested for one year [#permalink]
09 Dec 2010, 17:37
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. Could some explain the ans here. I am not sure if the ans is correct
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Re: A total of $60,000 was invested for one year [#permalink]
09 Dec 2010, 17:42
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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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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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