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33%
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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-xy/100=4.5
(1) xy=50
(2) y-x-xy/100=4.5
14 Apr 2012, 12:10
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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?
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.
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.
General Discussion
14 Apr 2012, 07:11
Kudos
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The total no. of incidents in 2003 = 1000=T
The incidents in 2004 are x% fewer, so incidents = (1-x/100) *T
The incidents in 2005 are y% more , so incidents = (1-x/100)(1+y/100)*T
= (1-x/100+y/100-xy/100)*T
=(1+(y-x-xy)/100)*T
i) A insufficient , as we require value of x,y
ii) B sufficient , as we have the value of the whole expression.
The incidents in 2004 are x% fewer, so incidents = (1-x/100) *T
The incidents in 2005 are y% more , so incidents = (1-x/100)(1+y/100)*T
= (1-x/100+y/100-xy/100)*T
=(1+(y-x-xy)/100)*T
i) A insufficient , as we require value of x,y
ii) B sufficient , as we have the value of the whole expression.
