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John invest $2,000 at a constant interest rate of annually compound 5%

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John invest $2,000 at a constant interest rate of annually compound 5% [#permalink]

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New post 16 Mar 2016, 17:26
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

63% (01:52) correct 37% (02:16) wrong based on 59 sessions

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John invest $2,000 at a constant interest rate of annually compound 5%. If the amount from the investment after n years is p, p=(2,000)1.05^n, is n>4?

1) p>2,100
2) p<3,000


* A solution will be posted in two days.
[Reveal] Spoiler: OA

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Re: John invest $2,000 at a constant interest rate of annually compound 5% [#permalink]

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New post 17 Mar 2016, 06:36
Interest for 1st year = 100. Interest for subsequent years will also be close to 100

St1: p > 2100 --> interest > 100 --> If interest < 450 (approx.) then n < 4 but if interest > 450 then n > 4. Not sufficient

St2: Clearly not sufficient for the above mentioned reason

Combining St1 and St2, we still have 2 possibilities. Hence insufficient.

Answer: E

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Math Revolution GMAT Instructor
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Re: John invest $2,000 at a constant interest rate of annually compound 5% [#permalink]

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New post 19 Mar 2016, 22:02
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

John invest $2,000 at a constant interest rate of annually compound 5%. If the amount from the investment after n years is p, p=(2,000)1.05^n, is n>4?

1) p>2,100
2) p<3,000


In the original condition, there is 1 variable, which should match with the number of equations. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
Howver, 1) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
2) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
1) & 2) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
The final answer is E
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Re: John invest $2,000 at a constant interest rate of annually compound 5% [#permalink]

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New post 21 Nov 2017, 12:06
MathRevolution wrote:
John invest $2,000 at a constant interest rate of annually compound 5%. If the amount from the investment after n years is p, p=(2,000)1.05^n, is n>4?

1) p>2,100
2) p<3,000


* A solution will be posted in two days.


S1) p>2,100
=> n > 1
if n = 2
p = 2,200+ approx.

if n = 5
p = 2550+ approx.

Insufficient.

S2) p < 3,000
Same case as statement 1.
Insufficient.

Quote:
combining both statements

We have no new information
=> Insufficient

E is the answer.
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Re: John invest $2,000 at a constant interest rate of annually compound 5%   [#permalink] 21 Nov 2017, 12:06
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John invest $2,000 at a constant interest rate of annually compound 5%

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