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A total of $60,000 was invested for one year. Part of this [#permalink]
05 Aug 2006, 07:09

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

54% (02:44) correct
46% (01:23) wrong based on 94 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 = \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.

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

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% ?

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...

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.

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

Re: interest and equations [#permalink]
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%

Re: DS: Interests ratios [#permalink]
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.

Re: annual interest [#permalink]
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

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.

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

gmatclubot

Re: A total of $60,000 was invested for one year. Part of this
[#permalink]
24 Jul 2014, 15:59