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A certain auto manufacturer sold 3% fewer vehicles in...

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by AAPL » Mon Mar 19, 2018 6:36 am
A certain auto manufacturer sold 3% fewer vehicles in 2007 than in 2006. If the manufacturer sold 2.1 million vehicles in 2006, how many vehicles, to the nearest 10,000, did the manufacturer sell in 2007?

A. 63,000
B. 2,000,000
C. 2,030,000
D. 2,040,000
E. 2,300,000

The OA is D.

Let's assume in 2006 he has sold X. Therefore in 2007, he should have sold 97/100×X.

97/100×2100=2037
-->2,037,000 sold in 2007. When rounded to the nearest 10,000 :

2,040,000. Option D.

Is there another strategic approach to this PS question? Can any experts help, please?
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by [email protected] » Mon Mar 19, 2018 2:34 pm
Hi AAPL,

We're told that a certain auto manufacturer sold 3% fewer vehicles in 2007 than in 2006 and that the manufacturer sold 2.1 million vehicles in 2006. We're asked for the number of vehicles, to the nearest 10,000, that the manufacturer sold in 2007. Since we're given all of the necessary 'numbers' to work with, this question really just comes down to doing the proper calculations.

You might find it helpful to do the math in small 'steps'....
1% of 1,000,000 = 10,000, and....
3% of 1,000,000 = 30,000 and
3% of 2,100,000 = (2.1)(30,000) = 63,000

Thus, the number of cars sold in 2007 would be approximately 2,100,000 - 60,000 = 2,040,000

Final Answer: D

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by Jeff@TargetTestPrep » Tue Mar 20, 2018 4:09 pm
AAPL wrote:A certain auto manufacturer sold 3% fewer vehicles in 2007 than in 2006. If the manufacturer sold 2.1 million vehicles in 2006, how many vehicles, to the nearest 10,000, did the manufacturer sell in 2007?

A. 63,000
B. 2,000,000
C. 2,030,000
D. 2,040,000
E. 2,300,000
The number of vehicles sold in 2006 is 2.1 x 0.97 = 2.037 million ≈ 2,040,000.

Answer: D

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by swerve » Wed Mar 21, 2018 9:37 am
Hi AAPL,

Let's assume in 2006 the manufacturer sold X.

In 2007 sold X-3/100(X).

If sales in 2006, X = 2.1 million find X-3/100(X)

(2.1*10^6) - 3 / 100(2.1*10^6)

2100000 - 63000 = 2,037,000 --> 2,040,000 approximately.

Regards!
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by deloitte247 » Sat Mar 24, 2018 12:08 pm
3% of 2.1 million = $$\frac{3}{100}\cdot2,100,100=63,000$$
$$Number\ of\ vechicle\ sold\ in\ 2007\ is\ 63,000\ \left(3\%\right)\ fewer\ than\ 2.1million\ vechicles\ sold\ in\ 2006$$
$$Therefore,\ number\ of\ vechicle\ sold\ in\ 2007=\ 2100000-63000=2,037,000$$
$$when\ rounded\ to\ the\ nearest\ 10,000=2040000\ vehicles\ sold\ in\ 2007$$
$$Option\ D\ is\ right$$
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