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$10,000 is deposited in a certain account that pays r [#permalink]
14 Jun 2008, 09:10

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

75% (01:49) correct
25% (00:48) wrong based on 181 sessions

$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t. What amount will the deposit grow to in 3 years?

(1) D(1) = 11,000 (2) r=10

Below is a flawed version of the above correct question: 10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000 (2) r=10

Can anybody explain the logic please? Thank you!

This is D indeed.

Rephrased, knowing that t=3, the equation is: D(t)=10,000(1+(r/100))3 - If you know either D(t) or r, you can solve this equation for all variables. _________________

statement 1: D(t) = 11000, but t could be any thing, it could be 10 years or 2 years. insuff

statement 2: suff, because it gives the interest rate and you have the years in the stem, you can find out the total after 3 years.

D(t) means a function. Now, constant values (i.e., 11000) can't be equated with functions because as far as I can think, no one can come up with an equation with a variable that always produces the same value for all values of the variable. Hence, in this question, we must assume that D(t) gives 11000 for some value of t. With that thinking, D should be the correct answer. _________________

statement 1: D(t) = 11000, but t could be any thing, it could be 10 years or 2 years. insuff

statement 2: suff, because it gives the interest rate and you have the years in the stem, you can find out the total after 3 years.

D(t) means a function. Now, constant values (i.e., 11000) can't be equated with functions because as far as I can think, no one can come up with an equation with a variable that always produces the same value for all values of the variable. Hence, in this question, we must assume that D(t) gives 11000 for some value of t. With that thinking, D should be the correct answer.

Guys \(t\) is already give, No? \(3\) years. So \(t\) should be \(3\), No? Hence D _________________

"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

Re: $10,000 is deposited in a certain account that pays r [#permalink]
23 Feb 2012, 22:59

Expert's post

quantum wrote:

10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000 (2) r=10

Can anybody explain the logic please? Thank you!

Two things: Formula should be \(D(t)=10,000(1+\frac{r}{100})^t\) and statement (1) should read \(D(1)=11,000\) (since two statements in DS never contradict and give true information then r=10 must give 11,000 for t=1 as it does). So the question should read:

$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t What amount will the deposit grow to in 3 years?

Question: \(D(3)=10,000(1+\frac{r}{100})^3=?\). Basically the only thing we need is the value of \(r\).

(1) D(1) = 11,000 --> \(D(1)=10,000(1+\frac{r}{100})^1=11,000\) --> we can solve for r. Sufficient.

(2) r=10 --> directly gives the value of r. Sufficient.

Re: $10,000 is deposited in a certain account that pays r [#permalink]
23 Feb 2012, 23:10

quantum wrote:

$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t. What amount will the deposit grow to in 3 years?

(1) D(1) = 11,000 (2) r=10

Below is a flawed version of the above correct question: 10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000 (2) r=10

Bunuel

Hi

The answer is indeeded +1 D

Statement Break Down : You need to know R. t is already know at 3 years

Statement 1 : D(1) : 11,000

Thus you can easily find out 'R' by equating D(1) & the eqn of D(t)

Thus sufficient

Statement 2 : R know hence sufficient

Thus we can find R which is the _________________

Giving +1 kudos is a better way of saying 'Thank You'.

Re: $10,000 is deposited in a certain account that pays r [#permalink]
23 Feb 2012, 23:30

Bunuel wrote:

quantum wrote:

10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000 (2) r=10

Can anybody explain the logic please? Thank you!

Two things: Formula should be \(D(t)=10,000(1+\frac{r}{100})^t\) and statement (1) should read \(D(1)=11,000\) (since two statements in DS never contradict and give true information then r=10 must give 11,000 for t=1 as it does). So the question should read:

$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t What amount will the deposit grow to in 3 years?

Question: \(D(3)=10,000(1+\frac{r}{100})^3=?\). Basically the only thing we need is the value of \(r\).

(1) D(1) = 11,000 --> \(D(1)=10,000(1+\frac{r}{100})^1=11,000\) --> we can solve for r. Sufficient.

(2) r=10 --> directly gives the value of r. Sufficient.

Answer: D.

Hope it's clear.

Agreed. But taking it on the face value, Statement 1 can give an impression that, for any value of t, D(t) gives 11000 ,which I think is not correct. Personally I don;t accept this statement, because in DS category, I do following activities: 1. Is the statement is supported for all situations by the premises/facts given in the question stem. 2. Once the answer for the above is Yes, then I will think, does the statement support all situations of the question.

In this exercise, I get a 'No' to the first part from the question. Hence, the answer should be B.

But, as the statement 1, does not hold good on its face value, I just tried mapping it to the question stem for specific scenario and then got the answer D.

Re: $10,000 is deposited in a certain account that pays r [#permalink]
24 Feb 2012, 00:36

Expert's post

Chembeti wrote:

Bunuel wrote:

quantum wrote:

10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000 (2) r=10

Can anybody explain the logic please? Thank you!

Two things: Formula should be \(D(t)=10,000(1+\frac{r}{100})^t\) and statement (1) should read \(D(1)=11,000\) (since two statements in DS never contradict and give true information then r=10 must give 11,000 for t=1 as it does). So the question should read:

$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t What amount will the deposit grow to in 3 years?

Question: \(D(3)=10,000(1+\frac{r}{100})^3=?\). Basically the only thing we need is the value of \(r\).

(1) D(1) = 11,000 --> \(D(1)=10,000(1+\frac{r}{100})^1=11,000\) --> we can solve for r. Sufficient.

(2) r=10 --> directly gives the value of r. Sufficient.

Answer: D.

Hope it's clear.

Agreed. But taking it on the face value, Statement 1 can give an impression that, for any value of t, D(t) gives 11000 ,which I think is not correct. Personally I don;t accept this statement, because in DS category, I do following activities: 1. Is the statement is supported for all situations by the premises/facts given in the question stem. 2. Once the answer for the above is Yes, then I will think, does the statement support all situations of the question.

In this exercise, I get a 'No' to the first part from the question. Hence, the answer should be B.

But, as the statement 1, does not hold good on its face value, I just tried mapping it to the question stem for specific scenario and then got the answer D.

Isn't this correct approach?

If we take (1) as it was written: D(t) = 11,000 then it would mean that D(t) has the same value no matter the time period and annual interest, which makes no sense at all. Though technically it still would be sufficient as it would mean that for t=3 the answer is also 11,000.

Having said that I recommend not to spend time on flawed version of a question as you won't see such one on the real test. _________________

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