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AtifS
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The net value of a certain stock increased at a constant 05 Mar 2010, 23:58
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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92% (00:43) correct 8% (00:37) wrong based on 123 sessions

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The net value of a certain stock increased at a constant rate during the ten-year period between 1990 and 2000. What was the value of the stock in the year 1998?

(1)In 1991, the value of the stock was 130 U.S dollars.
(2)In 1992, the value of the stock was 149.5 U.S dollars.
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C
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Bunuel
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The net value of a certain stock increased at a constant 06 Mar 2010, 05:00
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AtifS
The net value of a certain stock increased at a constant rate during the ten-year period between 1990 and 2000. What was the value of the stock in the year 1998?

(1)In 1991, the value of the stock was 130 U.S dollars.
(2)In 1992, the value of the stock was 149.5 U.S dollars.

OA
Show Spoiler
C
Please! solve & explain.
Source--> Posted by Andrea on FB group "The Daily Gmat"
Show more

To solve this question no calculation is needed. one should just know the concept of geometric progression. A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. Thus, the general form of a geometric sequence is:

\(a\), \(ar\), \(ar^2\), \(ar^3\), ... where \(a\) is the first term of the sequence and \(r\) is a common ratio.

The \(n_{th}\) term of a geometric sequence with initial value \(a\) and common ratio \(r\) is given by:

\(a_n=ar^{n-1}\).

Basically if you know any two terms (\(a_n\) and \(a_m\)) OR if you know any term and common ratio (\(a_n\) and \(r\)) you will be able to calculate any missing value of given sequence. There is one more option though: if you know the sum of the sequence and either any term or common ratio you also will be able to calculate any missing value, but this is different issue.

Now, what the above has to do with the original question? We are told that "value of a certain stock increased at a constant rate during the ten-year period", so we have the sequence where the first term is the value of stock in 1990 and the common ratio is constant rate. We are asked to calculate the value of stock in 1998 or the 9th term of the sequence, \(a_9\). To do this we need either any two terms or any term and common ratio (constant rate).

Statement (1) tells us the value of the stock in 1991, \(a_2\). But only one term is not sufficient to calculate \(a_9\).
Statement (2) tells us the value of the stock in 1992, \(a_3\). But only one term is not sufficient to calculate \(a_9\).

Together we have the value of two terms, hence we can determine any missing value, in our case \(a_9\). Sufficient.

To demonstrate how it can be done:
\(a_9=ar^{9-1}=ar^8=a_2r^7\)
\(a_2=130\)
\(a_3=149.5=ar^2=a_2r\) --> \(149.5=130r\) -> \(r=\frac{149.5}{130}={1.15}\) --> \(a_9=a_2r^7=130*1.15^7\)

Answer: C.

Hope it helps.
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The net value of a certain stock increased at a constant 06 Mar 2010, 06:46
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Thanks bunuel, it was a nice explanation. In last 2 or 3 weeks I've become one of you fans. You've been really inspirational especially in Quant I've learned a lot from you and hope to learn more.
Once again Thanks!
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The net value of a certain stock increased at a constant 11 Jun 2014, 01:33
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Bunuel
AtifS
The net value of a certain stock increased at a constant rate during the ten-year period between 1990 and 2000. What was the value of the stock in the year 1998?

(1)In 1991, the value of the stock was 130 U.S dollars.
(2)In 1992, the value of the stock was 149.5 U.S dollars.

OA
Show Spoiler
C
Please! solve & explain.
Source--> Posted by Andrea on FB group "The Daily Gmat"

To solve this question no calculation is needed. one should just know the concept of geometric progression. A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. Thus, the general form of a geometric sequence is:

\(a\), \(ar\), \(ar^2\), \(ar^3\), ... where \(a\) is the first term of the sequence and \(r\) is a common ratio.

The \(n_{th}\) term of a geometric sequence with initial value \(a\) and common ratio \(r\) is given by:

\(a_n=ar^{n-1}\).

Basically if you know any two terms (\(a_n\) and \(a_m\)) OR if you know any term and common ratio (\(a_n\) and \(r\)) you will be able to calculate any missing value of given sequence. There is one more option though: if you know the sum of the sequence and either any term or common ratio you also will be able to calculate any missing value, but this is different issue.

Now, what the above has to do with the original question? We are told that "value of a certain stock increased at a constant rate during the ten-year period", so we have the sequence where the first term is the value of stock in 1990 and the common ratio is constant rate. We are asked to calculate the value of stock in 1998 or the 9th term of the sequence, \(a_9\). To do this we need either any two terms or any term and common ratio (constant rate).

Statement (1) tells us the value of the stock in 1991, \(a_2\). But only one term is not sufficient to calculate \(a_9\).
Statement (2) tells us the value of the stock in 1992, \(a_3\). But only one term is not sufficient to calculate \(a_9\).

Together we have the value of two terms, hence we can determine any missing value, in our case \(a_9\). Sufficient.

To demonstrate how it can be done:
\(a_9=ar^{9-1}=ar^8=a_2r^7\)
\(a_2=130\)
\(a_3=149.5=ar^2=a_2r\) --> \(149.5=130r\) -> \(r=\frac{149.5}{130}={1.15}\) --> \(a_9=a_2r^7=130*1.15^7\)

Answer: C.

Hope it helps.
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Hi,
Why have u used GP?
Cant it be AP also?
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The net value of a certain stock increased at a constant 11 Jun 2014, 01:42
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abid1986
Bunuel
AtifS
The net value of a certain stock increased at a constant rate during the ten-year period between 1990 and 2000. What was the value of the stock in the year 1998?

(1)In 1991, the value of the stock was 130 U.S dollars.
(2)In 1992, the value of the stock was 149.5 U.S dollars.

OA
Show Spoiler
C
Please! solve & explain.
Source--> Posted by Andrea on FB group "The Daily Gmat"

To solve this question no calculation is needed. one should just know the concept of geometric progression. A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. Thus, the general form of a geometric sequence is:

\(a\), \(ar\), \(ar^2\), \(ar^3\), ... where \(a\) is the first term of the sequence and \(r\) is a common ratio.

The \(n_{th}\) term of a geometric sequence with initial value \(a\) and common ratio \(r\) is given by:

\(a_n=ar^{n-1}\).

Basically if you know any two terms (\(a_n\) and \(a_m\)) OR if you know any term and common ratio (\(a_n\) and \(r\)) you will be able to calculate any missing value of given sequence. There is one more option though: if you know the sum of the sequence and either any term or common ratio you also will be able to calculate any missing value, but this is different issue.

Now, what the above has to do with the original question? We are told that "value of a certain stock increased at a constant rate during the ten-year period", so we have the sequence where the first term is the value of stock in 1990 and the common ratio is constant rate. We are asked to calculate the value of stock in 1998 or the 9th term of the sequence, \(a_9\). To do this we need either any two terms or any term and common ratio (constant rate).

Statement (1) tells us the value of the stock in 1991, \(a_2\). But only one term is not sufficient to calculate \(a_9\).
Statement (2) tells us the value of the stock in 1992, \(a_3\). But only one term is not sufficient to calculate \(a_9\).

Together we have the value of two terms, hence we can determine any missing value, in our case \(a_9\). Sufficient.

To demonstrate how it can be done:
\(a_9=ar^{9-1}=ar^8=a_2r^7\)
\(a_2=130\)
\(a_3=149.5=ar^2=a_2r\) --> \(149.5=130r\) -> \(r=\frac{149.5}{130}={1.15}\) --> \(a_9=a_2r^7=130*1.15^7\)

Answer: C.

Hope it helps.

Hi,
Why have u used GP?
Cant it be AP also?
Show more

Note that the question says " value of a certain stock increased at a constant rate"
Say, something like stock increases at a constant rate of 10%. So every year, the stock value will get multiplied by 110/100. This gives us a GP. Had the value of the stock increased by a fixed amount every year, then it would have been AP.
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