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# S95-03

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Math Expert
Joined: 02 Sep 2009
Posts: 50009

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16 Sep 2014, 01:49
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Difficulty:

55% (hard)

Question Stats:

59% (01:17) correct 41% (01:52) wrong based on 71 sessions

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The workers at a large construction company reported $$x$$ percent fewer safety incidents in 2004 than in 2003, and $$y$$ percent more incidents in 2005 than in 2004. If the workers reported a total of 1,000 incidents in 2003, how many incidents did the workers report in 2005?

(1) $$xy = 50$$

(2) $$y - x - \frac{xy}{100} = 4.5$$

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16 Sep 2014, 01:49
Official Solution:

We need to determine the number of incidents reported in 2005. The problem tells us that 1,000 incidents were reported in 2003. In 2004, $$x$$ percent fewer incidents were reported, so $$(100 - x) \times 1,000$$ incidents were reported. Since a percentage is equivalent to a fraction with a denominator of 100, we can express the number of incidents reported in 2004 as $$\frac{100 - x}{100} \times 1,000$$. In 2005, $$y$$ percent more incidents were reported than in 2004. This can be expressed as $$\frac{100 + y}{100}(\frac{100 - x}{100} \times 1,000)$$. It is this target expression that we must find a value for.

Multiply the numerators and denominators: $$\frac{100 + y}{100}(\frac{100 - x}{100} \times 1,000) = \frac{1,000(100 + y )(100 - x)}{10,000}$$.

Cancel 1,000 from the numerator and denominator and FOIL what remains: $$\frac{(100 + y)(100 - x)}{10} = \frac{10,000 - 100x + 100y - xy}{10}$$.

Evaluating this expression will require either solving for $$x$$ and $$y$$ or finding a way to substitute a number for all variable terms.

Statement 1 says that $$xy = 50$$. This does not allow us to solve for either $$x$$ or $$y$$. We can substitute it into our target expression to get $$\frac{10,000 - 100x + 100y - 50}{10}$$, but doing so does not eliminate the need to solve for $$x$$ and $$y$$. Statement 1 is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 says that $$y - x - \frac{xy}{100} = 4.5$$. Note that this equation has an $$x$$ term, a $$y$$ term, and a term containing $$xy$$. Thus, substitution will likely be possible.

Divide the original expression by $$10$$ to eliminate the fraction: $$\frac{10,000 - 100x + 100y - xy}{10} = 1,000 - 10x + 10y - \frac{xy}{10}$$.

Now factor out an additional $$10$$ and reorder the terms in the expression: $$10(100 + y - x - \frac{xy}{100})$$.

The left side of the equation in statement 2 is contained in this expression. Thus, we know that we can find a value for the expression. Statement 2 alone provides sufficient information to answer the question.

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Joined: 07 Dec 2009
Posts: 94
GMAT Date: 12-03-2014

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26 Nov 2014, 11:13
1
hi Bunuel,

How would one attempt the question within 2 mins. even though I got the answer It took me 4 complete minutes!!

Thanks
Intern
Joined: 05 Aug 2016
Posts: 38

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14 Sep 2016, 10:42
2003 :1000
2004 : {(100-x )/100}*1000=(100-x)*10
2005 : {(100+y )/100}*(100-x)*10

In 2005, unknown contains terms y-x-xy/100
Therefore, option B is sufficient
Manager
Joined: 01 Sep 2016
Posts: 68

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14 Sep 2016, 12:19
tough one.not sure if I can complete this during the test

Bunuel wrote:
Official Solution:

We need to determine the number of incidents reported in 2005. The problem tells us that 1,000 incidents were reported in 2003. In 2004, $$x$$ percent fewer incidents were reported, so $$(100 - x) \times 1,000$$ incidents were reported. Since a percentage is equivalent to a fraction with a denominator of 100, we can express the number of incidents reported in 2004 as $$\frac{100 - x}{100} \times 1,000$$. In 2005, $$y$$ percent more incidents were reported than in 2004. This can be expressed as $$\frac{100 + y}{100}(\frac{100 - x}{100} \times 1,000)$$. It is this target expression that we must find a value for.

Multiply the numerators and denominators: $$\frac{100 + y}{100}(\frac{100 - x}{100} \times 1,000) = \frac{1,000(100 + y )(100 - x)}{10,000}$$.

Cancel 1,000 from the numerator and denominator and FOIL what remains: $$\frac{(100 + y)(100 - x)}{10} = \frac{10,000 - 100x + 100y - xy}{10}$$.

Evaluating this expression will require either solving for $$x$$ and $$y$$ or finding a way to substitute a number for all variable terms.

Statement 1 says that $$xy = 50$$. This does not allow us to solve for either $$x$$ or $$y$$. We can substitute it into our target expression to get $$\frac{10,000 - 100x + 100y - 50}{10}$$, but doing so does not eliminate the need to solve for $$x$$ and $$y$$. Statement 1 is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 says that $$y - x - \frac{xy}{100} = 4.5$$. Note that this equation has an $$x$$ term, a $$y$$ term, and a term containing $$xy$$. Thus, substitution will likely be possible.

Divide the original expression by $$10$$ to eliminate the fraction: $$\frac{10,000 - 100x + 100y - xy}{10} = 1,000 - 10x + 10y - \frac{xy}{10}$$.

Now factor out an additional $$10$$ and reorder the terms in the expression: $$10(100 + y - x - \frac{xy}{100})$$.

The left side of the equation in statement 2 is contained in this expression. Thus, we know that we can find a value for the expression. Statement 2 alone provides sufficient information to answer the question.

Current Student
Joined: 11 Oct 2016
Posts: 20
Location: Russian Federation
GMAT 1: 740 Q47 V44
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07 Nov 2016, 01:51
bhatiavai wrote:
hi Bunuel,

How would one attempt the question within 2 mins. even though I got the answer It took me 4 complete minutes!!

Thanks

It's an old post but since i've been solving this now, maybe it will help someone...

You can solve it within 2 minutes by getting the expression 1000 (100-x)(100+y) which is 1000 (10000+100y-100x-xy) and noticing that statement 2 has the same variables in the same order with the same signs: y - x - xy/100. I think you don't need to go any further - you can already see that it has a solution.
Manager
Joined: 14 Jun 2016
Posts: 67
Location: India
GMAT 1: 610 Q49 V21
WE: Engineering (Manufacturing)

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01 Aug 2017, 09:09
fandango wrote:
bhatiavai wrote:
hi Bunuel,

How would one attempt the question within 2 mins. even though I got the answer It took me 4 complete minutes!!

Thanks

It's an old post but since i've been solving this now, maybe it will help someone...

You can solve it within 2 minutes by getting the expression 1000 (100-x)(100+y) which is 1000 (10000+100y-100x-xy) and noticing that statement 2 has the same variables in the same order with the same signs: y - x - xy/100. I think you don't need to go any further - you can already see that it has a solution.

We can use the formula for successive increase of x%, followed by decrease of y% is given by effective increase = x-y-xy/100.
Hence we can estimate the ultimate value without much of calculation. Final value= Intial Value* effective increase in %
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Intern
Joined: 24 Jul 2017
Posts: 5

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06 Dec 2017, 23:03
Hi Bunuel,

I solved this question in under 1:30 minutes, by:
declaring (i) INSUFFICIENT - as only the value of xy is given, and by no means we can identify the separate values of x and y to find out the final value post the successive change.
(ii) SUFFICIENT - just by looking at it, since the direct formula of Successive change is given with its value.

My question is, was my decision to declare (i) insufficient too hasty ? And was my reasoning to it, correct ?

Intern
Joined: 26 Apr 2018
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04 May 2018, 08:40
In my opinion you basically have to know that the formula for successive percentage change looks like Statement 2 because for me it would be hard to do the Math behind it in 2 min.
Math Expert
Joined: 02 Sep 2009
Posts: 50009

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04 May 2018, 09:53
Bunuel wrote:
The workers at a large construction company reported $$x$$ percent fewer safety incidents in 2004 than in 2003, and $$y$$ percent more incidents in 2005 than in 2004. If the workers reported a total of 1,000 incidents in 2003, how many incidents did the workers report in 2005?

(1) $$xy = 50$$

(2) $$y - x - \frac{xy}{100} = 4.5$$

The workers at a large construction company reported x percent fewer safety incidents in 2004 than in 2003, and y percent more incidents in 2005 than in 2004. If the workers reported a total of 1,000 incidents in 2003, how many incidents did the workers report in 2005?

The total # of incident is 2003 was 1,000;
The total # of incident in 2004 was x percent fewer than in 2003, so in 2004 there were $$1,000*(1-\frac{x}{100})$$ incidents;
The total # of incident in 2005 was y percent more than in 2004, so in 2005 there were $$1,000*(1-\frac{x}{100})*(1+\frac{y}{100})=1,000(1+\frac{y}{100}-\frac{x}{100}-\frac{xy}{100*100)}=1,000(1+\frac{1}{100}(y-x-\frac{xy}{100}))$$ incidents.

So, as you can see we need to find the value of $$y-x-\frac{xy}{100}$$.

(1) xy=50. Not sufficient to find the required value.

(2) y-x-xy/100=4.5 --> we are directly given the value of $$y-x-\frac{xy}{100}$$. Sufficient.

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# S95-03

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