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

Question

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

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

Bunuel · Accepted Answer

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.

Asifpirlo · Answer

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)

ygdrasil24 · Answer

COuld you have guessed the number mentally without solving ? I mean can you a draw a conclusion by seeing the numbers ?

xhimi · Answer

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

EMPOWERgmatRichC · Answer

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:  D

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

profileusername · Answer

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.

fskilnik · Answer

? = p

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

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

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

Regards,
Fabio.

fskilnik · Answer

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.

rushimehta · Answer

Just another quick approach with minimal calculations ...

Concept --> When a number decreases by 1/n, it must increase by 1/n-1 to go back to the same number... E.g. - 100 decreased by 1/5 is 80, but increasing 80 by 1/4 we get 100 again.

Here, 
n After a p% increase, gives us n+(0.p)n... the questions says that the number is now decreased by 1/6 successively, to give us n again.

That means, (by using backward calculation logic) to get n+(0.p)n, we must increase "n" by 1/5 successively...

As in many percentage questions, taking the base number (n) as 100, to make calculations easier... and then increasing 100 by 1/5 successively...

1st increase => 100 + 1/5(100) = 120 
2nd increase => 120 + 1/5(120) = 144

We get 144, which is equal to n+(0.p)n, we see that 100 needs to be increased by 44% to get to 144... thus p% is 44...

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

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