If the number of cars in city X this year is 520,800...

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If the number of cars in city X this year is 520,800 which is 24% more than there were in city X last year, how many cars were there in city X last year?

A. 395,808
B. 420,000
C. 447,888
D. 496,800
E. 645,795

The OA is B.

How can I solve this PS question? Should I try number by number until I get the correct answer?

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by Jay@ManhattanReview » Mon Oct 23, 2017 6:01 am
Vincen wrote:If the number of cars in city X this year is 520,800 which is 24% more than there were in city X last year, how many cars were there in city X last year?

A. 395,808
B. 420,000
C. 447,888
D. 496,800
E. 645,795

The OA is B.

How can I solve this PS question? Should I try number by number until I get the correct answer?
What is the source of the question? The GMAT does not expect you to do calculation having ugly numbers.

Anyway, here is a way.

Say the number of cars in city X last year = 100, thus, the number of cars in city X this year = 100 + 24 = 124

Thus, the actual number of cars in city X last year = 520800*(100/124) = 520800*(25/31).

Division of 520800 by the prime number 31 is cumbersome, so let's think of another tactic.

520800*(100/124) > 520800*(100/125)

520800*(100/125) = 520800*(4/5) = 104160*4 = 416640

Thus, the correct answer is greater than 416640. But there are at least two options, B and C, to recon.

There is a way to eliminate one of the two options.

520800*(100/124) < 520800*(100/120)

520800*(100/120) = 520800*(5/6) = 86800*5 = 434000.

Thus, the correct answer must be less than 434000.

The only option that satisfies is option B = 420000.

The correct answer: B

Hope this helps!

-Jay

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by [email protected] » Mon Oct 23, 2017 12:55 pm
Hi Vincen,

We're told that the number of cars in city X this year is 520,800 and that total is 24% more than there were in city X last year. We're asked for the number of cars that were there in city X last year. While the numbers in this prompt are not 'pretty', the answer choices are 'spaced out' enough that we can do a bit of estimation and some 'brute force' math to get to the answer.

To start, 24% of 100,000 = 24,000 (so a 24% increase would be 124,000 total cars). Thus, we can quickly map out increases of 100,000 cars...
200,000 + 48,000 = 248,000
300,000 + 72,000 = 372, 000
400,000 + 96,000 = 496,000
500,000 + 120,000 = 620,000

Since 520,800 is far closer to 496,000 than it is to 620,000, the total number of cars last year would be a bit more than 400,000. There's only one answer that matches...

Final Answer: B

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by Scott@TargetTestPrep » Mon Dec 04, 2017 10:47 am
Vincen wrote:If the number of cars in city X this year is 520,800 which is 24% more than there were in city X last year, how many cars were there in city X last year?

A. 395,808
B. 420,000
C. 447,888
D. 496,800
E. 645,795
We are given that the number of cars in city X this year is 520,800, which is 24% more than there were in city X last year. If we let the number of cars in city X last year = c, we can create the following equation:

520,800 = 1.24c

c = 420,000

Answer: B

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by EconomistGMATTutor » Wed Dec 06, 2017 2:24 pm
If the number of cars in city X this year is 520,800 which is 24% more than there were in city X last year, how many cars were there in city X last year?

A. 395,808
B. 420,000
C. 447,888
D. 496,800
E. 645,795
Hi Vincen,
Let's take a look at your question.

Let the number of cards in city X last year be 'N', then
the number of cars in city X is 24% more than x. This can be represented as:
$$N+\ \left(24\%\right)N=520,800$$
$$N+\ \left(\frac{24}{100}\right)N=520,800$$
$$N+\ 0.24N=520,800$$
$$1.24N=520,800$$
$$N=\frac{520,800}{1.24}$$
$$N=\frac{520,800\times100}{124}$$
$$N=\frac{130,200\times100}{31}$$
$$N=4,200\times100$$
$$N=420,000$$

Therefore, the number of cars in city X last year was 420,000.
Hence, Option B is correct.

Hope this helps.
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by GMATGuruNY » Wed Dec 06, 2017 5:38 pm
Vincen wrote:If the number of cars in city X this year is 520,800 which is 24% more than there were in city X last year, how many cars were there in city X last year?

A. 395,808
B. 420,000
C. 447,888
D. 496,800
E. 645,795
We can PLUG IN THE ANSWERS, which represent the number of cars last year.
Since 24% ≈ 25% = 1/4, adding 1/4 to the correct answer must yield a value close to 520,800.
The answer choices range from about 400,000 to about 650,000.
Increasing 400,000 by 1/4 -- in other words, increasing 400,000 by 100,000 -- yields 500,000.
Too small.
Thus, the correct answer must be a bit MORE than 400,000.
Eliminate A, D and E.

Answer choice B: 420,000
420,000 + (1/4)(420,000) = 420,000 + 105,000 = 525,000.
Since increasing 420,000 by 1/4 yields a value a bit more than 520,800, the correct answer cannot be greater than 420,000.
Eliminate C.

The correct answer is B.
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by GMATWisdom » Thu Dec 07, 2017 6:23 am
Vincen wrote:If the number of cars in city X this year is 520,800 which is 24% more than there were in city X last year, how many cars were there in city X last year?

A. 395,808
B. 420,000
C. 447,888
D. 496,800
E. 645,795

The OA is B.

How can I solve this PS question? Should I try number by number until I get the correct answer?
for 100000 with 24% increase it becomes 124000
for 400000..............................496000
hence figure should be more than 400000
25000 with 24% increase becomes (1/4)*124000=31000
therefore for 425000 with 24% increase it becomes 496000+31000=527000
hence initial figure should be less than 425000

Hence B is correct answer

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by Jeff@TargetTestPrep » Mon Dec 11, 2017 4:17 pm
Vincen wrote:If the number of cars in city X this year is 520,800 which is 24% more than there were in city X last year, how many cars were there in city X last year?

A. 395,808
B. 420,000
C. 447,888
D. 496,800
E. 645,795
We are given that the number of cars in city X this year is 520,800, which is 24% more than there were in city X last year. If we let the number of cars in city X last year = c, we can create the following equation:

520,800 = 1.24c

c = 420,000

Answer: B

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[email protected]

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