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S95-03

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S95-03  [#permalink]

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New post 16 Sep 2014, 01:49
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A
B
C
D
E

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  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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Re S95-03  [#permalink]

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New post 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.


Answer: B
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Re: S95-03  [#permalink]

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New post 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
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Re: S95-03  [#permalink]

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New post 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
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Re: S95-03  [#permalink]

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New post 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.


Answer: B
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Re: S95-03  [#permalink]

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New post 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.
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Re: S95-03  [#permalink]

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New post 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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Re: S95-03  [#permalink]

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New post 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 ?


Please help.
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Re: S95-03  [#permalink]

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New post 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.
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Re: S95-03  [#permalink]

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New post 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.

Answer: B.
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New to the Math Forum?
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: S95-03 &nbs [#permalink] 04 May 2018, 09:53
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