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A driver paid n dollars for auto insurance for the year 1997

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A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 19 Aug 2013, 02:41
5
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

70% (02:27) correct 30% (02:41) wrong based on 429 sessions

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A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure. If the driver’s insurance premium for the year 2000 was again n dollars, what is the value of p?

A. 12
B. \(33\frac{{1}}{{3}}\)
C. 36
D. 44
E. 50

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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 19 Aug 2013, 02:46
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Stiv wrote:
A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure. If the driver’s insurance premium for the year 2000 was again n dollars, what is the value of p?

A. 12
B. \(33\frac{{1}}{{3}}\)
C. 36
D. 44
E. 50


Premium in 1997 = n;
Premium in 1998 = n(1+p/100);
Premium in 1999 = 5/6*n(1+p/100);
Premium in 2000 = 5/6*5/6*n(1+p/100).

Given that premium in 2000 = 5/6*5/6*n(1+p/100)=n --> 5/6*5/6*(1+p/100)=1 --> 1+p/100=36/25 --> p/100=11/25=44/100 --> p=44.

Answer: D.

Hope it's clear.
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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 19 Aug 2013, 03:45
1
Stiv wrote:
A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure. If the driver’s insurance premium for the year 2000 was again n dollars, what is the value of p?

A. 12
B. \(33\frac{{1}}{{3}}\)
C. 36
D. 44
E. 50


1997, initial = n
1998, n + n p% = n(1+p%)
1999 and 2000, 1/6 decreased in each , 1-1/6 = 5/6 remains after year in consecutive two years.

So finally it became, 5/6 . 5/6 . n(1+p%)

From question,

25/36 n(1+p%) = n
or, 1+p% = 36/25
or, p% = 36/25 - 1 = 11/25
or, p = 100 . 11/25 = 44 (Answer D)
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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 11 Nov 2013, 06:30
Bunuel wrote:
Stiv wrote:
A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure. If the driver’s insurance premium for the year 2000 was again n dollars, what is the value of p?

A. 12
B. \(33\frac{{1}}{{3}}\)
C. 36
D. 44
E. 50


Premium in 1997 = n;
Premium in 1998 = n(1+p/100);
Premium in 1999 = 5/6*n(1+p/100);
Premium in 2000 = 5/6*5/6*n(1+p/100).

Given that premium in 2000 = 5/6*5/6*n(1+p/100)=n --> 5/6*5/6*(1+p/100)=1 --> 1+p/100=36/25 --> p/100=11/25=44/100 --> p=44.

Answer: D.

Hope it's clear.


COuld you have guessed the number mentally without solving ? I mean can you a draw a conclusion by seeing the numbers ?
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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 07 Dec 2013, 23:54
2
1
ygdrasil24 wrote:
COuld you have guessed the number mentally without solving ? I mean can you a draw a conclusion by seeing the numbers ?


In this question I would just out try the answers. For example I started with n=100$ and answer E, because it's the only round number in the answer choices.
So in 1998 we have 150$, then decrease by 1/6 (25$) - > 125$

Decrease 125$ by 1/6 and it will be slightly more than 100$ so I would pick answer D without recalculating, unless of course you have plenty of time
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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 15 Feb 2015, 22:30
1
Hi All,

This question is perfect for TESTing THE ANSWERS. To be efficient, we have to pay careful attention to the details and take the proper notes though.

When TESTing THE ANSWERS, it's usually best to start with either Answer B or Answer D. Looking at the given options, Answer D seems the like the "nicer" number. Looking at the question, the answer choices are a reference to the value of P.

We're given a "timeline" of facts for a series of years:

1) In 1997, a driver paid N dollars for auto insurance. Let's set N = 100

1997 = $100 for auto insurance.

2) In 1998, this premium was increased by P%

Using P = 44, the premium rose 44% of 100 = $44

1998 = $144 for auto insurance.

3) For each of the years 1999 and 2000, the premium DECREASED 1/6 from the previous year:

1999 = $144 - (1/6)(144) = 144 - 24 = $120 for auto insurance

2000 = $120 - (1/6)(120) = 120 - 20 = $100 for auto insurance

4) The driver's premium in 2000 was AGAIN N dollars.....

Notice here that the premium in 2000 is what it was in 1997: N dollars. This is a MATCH for what was described, so P MUST be 44%.

Final Answer:

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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 28 Jun 2017, 13:29
How do we interpret the "decreased by 1/6" phrase? I thought this means to multiply the figure by 1/6. How is that wrong and how is 5/6 right?

Thankyou.
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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 14 Sep 2018, 06:32
Stiv wrote:
A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure. If the driver’s insurance premium for the year 2000 was again n dollars, what is the value of p?

A. 12
B. \(33\frac{{1}}{{3}}\)
C. 36
D. 44
E. 50



\(? = p\)

\(97:\,\,n\,\,\,\,\,\, \to \,\,\,\,\,\,98:\,\,\,\left( {1 + \frac{p}{{100}}} \right)\,\,n\,\,\,\, \to \,\,\,\,99:\,\,\,\frac{5}{6}\,\,\,\left( {1 + \frac{p}{{100}}} \right)\,\,n\,\,\,\,\, \to \,\,00:\,\,\,{\left( {\frac{5}{6}} \right)^{\,2}}\,\left( {1 + \frac{p}{{100}}} \right)\,\,n\)

\({\left( {\frac{5}{6}} \right)^{\,2}}\,\left( {1 + \frac{p}{{100}}} \right)\,\,n = n\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1 + \frac{p}{{100}} = {\left( {\frac{6}{5}} \right)^{\,2}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = p = 100\left( {\frac{{36}}{{25}} - \frac{{1 \cdot \boxed{25}}}{{\boxed{25}}}} \right) = 4 \cdot 11 = 44\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: A driver paid n dollars for auto insurance for the year 1997  [#permalink]

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New post 14 Sep 2018, 06:34
TheMastermind wrote:
How do we interpret the "decreased by 1/6" phrase? I thought this means to multiply the figure by 1/6. How is that wrong and how is 5/6 right?

Thankyou.


Hi TheMastermind!

You would be right to multiply by 1/6 IF it was said there was a decrease TO 1/6 of the previous value.

When there is a decrease BY 1/6 of the previous value, what is left of the previous value is 1-1/6 = 5/6 of that value.

Regards,
Fabio.
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Re: A driver paid n dollars for auto insurance for the year 1997 &nbs [#permalink] 14 Sep 2018, 06:34
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A driver paid n dollars for auto insurance for the year 1997

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