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

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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 27 Oct 2016, 09:08
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 amounts be A and B.
Stem:
A + B = 60,000
Ax + By = 4,080

We have 4 variables, so we need 4 different equations.

Statement 1, x = 3y/4, gives us the relation between x and y but not x or y themselves

Statement 2, A=3B/2 , in combination with A + B = 60,000, gives us A and B but not x or y

With statements 1 and 2, we have the 4 equations needed.

Answer 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 27 Oct 2016, 10:35
VeritasPrepKarishma - I was trying to use weighted average with statement 1, can you pls help.

We are given that x/y = 3/4 and we can calculate the overall rate by using the 60,000 and the interest earned 4,080. using this won't we be able to get the ratio in which 60,000 would be invested at X and Y%?

Won't we enough data with equation 1 itself to solve it for X%?

TIA
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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 18 Nov 2016, 21:29
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.



Answer: option 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 26 Apr 2017, 02:57
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.



In most DS questions realising that the statements together or individually would lead to the answer is enough and not solving them to the end would save a lot of precious time as average time taken by PS questions are much more.

Here, you can start off with statement B and you can figure out the both the parts in which amount was divided. From there on, suing statement 1 you can either substitute the value of x for or vice versa. That will get you the answer without going through the complex method of solving the problem. If you're still not too sure, you can go ahead and solve the entire problem to be extremely sure.

All the best.
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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 30 Apr 2017, 06:03
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.

My 2 cents.
For this question, I think for non-native speaker as I am, it is important to realize that statement 2 is talking about the amount of x and y.
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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 20 Jun 2017, 22:19
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gauravk wrote:
VeritasPrepKarishma - I was trying to use weighted average with statement 1, can you pls help.

We are given that x/y = 3/4 and we can calculate the overall rate by using the 60,000 and the interest earned 4,080. using this won't we be able to get the ratio in which 60,000 would be invested at X and Y%?

Won't we enough data with equation 1 itself to solve it for X%?

TIA


Using weighted averages,

4080/60,000 = 6.8%

Using stmnt 1:
w1/w2 = (4x/3 - 6.8)/(6.8 - x)
Now here is the problem: We don't know w1/w2 using stmnt 1 alone. So we cannot calculate the value of x.

Using stmnt 2, we get the value of w1/w2 which is 3/2. But here we don't have the ratio of x and y.

So you need both statements to get

3/2 = (4x/3 - 6.8)/(6.8 - x)

And now you can get a unique value of x.

Answer (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 29 Aug 2017, 20:00
The whole trick is you set "a" amount and "60000-a" amount for X% and Y%.
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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 07 Nov 2018, 09:34
St 1 and 2 alone are insuff as both of them result into a linear equation with two variables.

Since these are linear equations in two variables together we can solve for x.

Hence correct ans is C.
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A total of $60,000 was invested for one year. Part of this  [#permalink]

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New post 08 Nov 2018, 23:02
OA:C

Amount invested with \(x\) % interest rate = $ \(z\)

Amount invested with \(y\) % interest rate = $ \(60000-z\)

\(\frac{x}{100}*z*1+\frac{y}{100}*(60000-z)*1 =4080\)....(1)

We have to find the value of \(x\)

(1) \(x = \frac{{3y}}{4} \quad\) ; Putting \(\quad y=\frac{4x}{3}\) in (1) , we get

\(\frac{x}{100}*z*1+\frac{{4x}}{{3*100}}*(60000-z)*1 =4080\)

We still do not know the value of \(z\), So we cannot find out the value of \(x\).

Statement (1) alone is insufficient.

(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 \(2\).

\(\frac{z}{60000-z}=\frac{3}{2}\)

\(z=36000\) $

\(60000-z = 24000\) $

\(\frac{x}{100}*36000*1+\frac{y}{100}*24000*1 =4080\)

We still do not know the value of \(y\) or relation of \(y\) with \(x\), So we cannot find out the value of \(x\).

Statement (2) alone is insufficient.

Combining (1) and (2), we get

\(\frac{x}{100}*36000*1+\frac{{4x}}{{3*100}}*24000*1 =4080\)

Using this equation, We can get the value of \(x\).
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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 09 Nov 2018, 13:30
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 are given that $60,000 was invested for 1 year. We are also given that part of the investment earned x percent simple annual interest and the rest earned y percent simple annual interest. We are also given that the total interest earned was $4,080. Let’s start by defining a variable.

b = the amount that earned x percent simple interest

Using variable b, we can also say:

60,000 – b = the amount that earned y percent simple annual interest

Since we know that the total interest earned was $4,080, we can create the following equation:

b(x/100) + (60,000 – b)(y/100) = 4,080

Note that in the equation above, we express "x percent" as x/100 and "y percent" as y/100 in the same way that we would express, say, 24 percent as 24/100.

Statement One Alone:

x = 3y/4

Although we have an equation with x and y, we still need a third equation to be able to determine the value of x because our equation from the given information has three variables. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

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.

From our given information we know that b is the amount that earned interest at the rate of x percent per year and that 60,000 – b is the amount that earned interest at the rate of y percent per year. Thus, we can create the following equation:

b/(60,000 – b) = 3/2

Without a third equation, statement two alone is not sufficient to determine the value of x.

Statements One and Two Together:

From the given information and statements one and two we have the following 3 equations:

1) b(x/100) + (60,000 – b)(y/100) = 4,080

2) x = 3y/4

3) b/(60,000 – b) = 3/2

Since we have 3 independent equations with variables x, y, and b, we are able to determine the value of x.

Answer: 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 25 Apr 2019, 08:36
VeritasKarishma

I mistakenly read (2) as saying "the amount of interest earned at x" rather than "the amount that earned interest." If it was said in the former way (2) would be sufficient on its own, right? You could find the value of each out of 4080, so you could also determine x% and y% from just (2). Would the GMAT do something tricky like this?

4080/5 = 816
So 2448 at x and 1632 at y.
2448/4080 = .6 and 1632/4080 = .4
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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 25 Apr 2019, 23:13
energetics wrote:
VeritasKarishma

I mistakenly read (2) as saying "the amount of interest earned at x" rather than "the amount that earned interest." If it was said in the former way (2) would be sufficient on its own, right? You could find the value of each out of 4080, so you could also determine x% and y% from just (2). Would the GMAT do something tricky like this?

4080/5 = 816
So 2448 at x and 1632 at y.
2448/4080 = .6 and 1632/4080 = .4


You would still not know the amount invested at x% and the amount invested at y%.

If interest earned is 2448 at x%, what is the amount on which this interest is earned?
You could earn 2448 interest at 10% on 24480.
or you could earn 2448 interest at 5% on 48960.
etc.
x would be different in each case.
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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 26 Apr 2019, 10:18
Thanks, now that I reread your posts and wrote it all out on paper it's very clear.

I wasn't writing things down previously, so in my head I was confusing the ratio of $Amt earned from Amt1 & Amt2 with the ratio of %interest.

Regardless of how the might've question put it, both the total amount and the subsequent interest is divided in a ratio of 3:2 (36k to 24k or 2448 to 1632), this is the weight 1 & weight 2. But this says nothing about the ratio of the x & y%. As you succinctly put it we can accrue the same interest from Amt1 & Amt2 in different amounts based on x & y%

As always thanks for all your amazing blog posts... fundamentally understanding mixtures/weighted average/statistics concepts instead of just applying formulas has been my Achilles heel... it's much worse than combinatorics and probability.
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Re: A total of $60,000 was invested for one year. Part of this   [#permalink] 26 Apr 2019, 10:18

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