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A driver paid n dollars for auto insurance for the year

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A driver paid n dollars for auto insurance for the year

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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?
$$A.\ 12$$
$$B.\ 33\ \frac{1}{3}$$
$$C.\ 36$$
$$D.\ 44$$
$$E.\ 50$$
The OA is D.

Source: Manhattan Prep

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swerve 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?
$$A.\ 12$$
$$B.\ 33\ \frac{1}{3}$$
$$C.\ 36$$
$$D.\ 44$$
$$E.\ 50$$
The OA is D.

Source: Manhattan Prep
> 1997: Premium = n
> 1998: Premium = n(1 + p%)
> 1999: Premium = n(1 + p%) - n(1 + p%)/6 = 5n/6 * (1 + p%)
> 2000: Premium = 5n/6 * (1 + p%) - 5n/36 * (1 + p%) = 25n/36 * (1 + p%)

=> 25n/36 * (1 + p%) = n

25/36 * (1 + p%) = 1

1 + p% = 36/25

p% = 36/25 - 1

p% = 11/25

p = 11/25 * 100 = 44

The correct answer: D

Hope this helps!

-Jay
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swerve 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?
$$A.\ 12$$
$$B.\ 33\ \frac{1}{3}$$
$$C.\ 36$$
$$D.\ 44$$
$$E.\ 50$$
First recognize that INCREASING a value by p percent is the same as multiplying by (100 + p)/100
For example, increasing a value by 7% is the same as multiplying by (100 + 7)/100 [i.e. 1.07]
Likewise, increasing a value by 25% is the same as multiplying by (100 + 25)/100 [i.e. 1.25]

Also, recognize that DECREASING a value by 1/6 is the same as multiplying by 5/6

Okay, let's begin:

Premium for 1997: n
Premium for 1998: [(100 + p)/100]n
Premium for 1999: (5/6)[(100 + p)/100]n
Premium for 2000: (5/6)(5/6)[(100 + p)/100]n

We're told that the premium for 2000 is n.
So, we can write: (5/6)(5/6)[(100 + p)/100]n = n
Divide both sides by n to get: (5/6)(5/6)[(100 + p)/100] = 1
Simplify: (25/36)[(100 + p)/100] = 1
Simplify: (25)(100 + p)/(36)(100) = 1
Simplify: (100 + p)/(36)(4) = 1
So we know that 100 + p = (36)(4)
Simplify: 100 + p = 144
So, p = 44

Answer: D

Cheers,
Brent

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swerve 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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Fabio Skilnik :: www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 30/Sep with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 70% discount!

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swerve 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?
$$A.\ 12$$
$$B.\ 33\ \frac{1}{3}$$
$$C.\ 36$$
$$D.\ 44$$
$$E.\ 50$$
We can create the equation:

n x (100 + p)/100 x 5/6 x 5/6 = n

(100 + p)/100 x 25/36 = 1

(100 + p)/100 = 36/25

25(100 + p) = 100 x 36

100 + p = 4 x 36

100 + p = 144

p = 44

Answer: D

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Scott Woodbury-Stewart Founder and CEO

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Hi All,

We're told that a driver paid N dollars for auto insurance for the year 1997. This annual premium was raised by P percent for the year 1998, then for EACH of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure, leaving the driver’s insurance premium for the year 2000 at N dollars. We're asked for the value of P. This question can be solved with a mix of TESTing VALUES and TESTing THE ANSWERS.

When TESTing THE ANSWERS, it's typically best to start with either Answer B or Answer D. Since Answer D is the 'nicer' of the two options, we'll start there.

Answer D: 44
IF... P=44 and N=100, then....
Insurance in 1997 = $100
Insurance in 1998 = $100 + (44% of $100) = $100 +$44 = $144
Insurance in 1999 = $144 - (1/6)($144) = $144 - $24 = $120
Insurance in 2000 = $120 - (1/6)($120) = $120 - $20 = $100
The value of N is the SAME in 1997 and 2000, so this MUST be the answer.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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