GMATPrep: video cameras

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ugoyal: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Mon Apr 19, 2010 4:10 amFollowed by: 1 member

by ugoyal » Sun Oct 31, 2010 8:08 am

Q. A manufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more video cameras in 1995 than in 1994. If the manufacturer produced 1,000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995 ?

(1) xy = 20

(2) x + y + xy/100 = 9.2

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Source: Beat The GMAT — Data Sufficiency | Sun Oct 31, 2010 8:08 am

clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Sun Oct 31, 2010 9:30 am

usually such problems result in B answer. but here i got C perhaps i am wrong
we need to know what is
1000(1+x/100)*(1+y/100)=?, simplify
1000+10(x+y)+xy/10=?
1 st is insuff as x,y,may be any numbers, but i can`t understand how to solve with B only
both are obviously suff xy=20, from here we can extract value for x+y, x+y+20/100=9,2. and find the value for 1995
waiting for oa

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sun Oct 31, 2010 10:36 pm

Solution:
If 1000 cameras were manufactured in 1993,
then 1000(1 + x/100) cameras were manufactured in 1994 and so
1000(1+x/100)(1+y/100) cameras were manufactured in 1995.
We need to know the value of 1000(1+x/100)(1+y/100).
First consider (1) alone.
The information xy = 20 is not sufficient to know the answer.
So (1) alone is not sufficient.
Next consider (2) alone.
It says x + y + xy/100 = 9.2.
Now 1000(1+x/100)(1+y/100) = 1000*(100+x)/100*(100 + y)/100 = 1/10(10000 +100x + 100y + xy) = 10(100 + x+ y + xy/100) = 10(100 + 9.2) = 1092.
So (2) alone is sufficient to answer the question.

The correct answer is (B).

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Nov 01, 2010 3:54 am

ugoyal wrote:Q. A manufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more video cameras in 1995 than in 1994. If the manufacturer produced 1,000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995 ?

(1) xy = 20

(2) x + y + xy/100 = 9.2

Since the number of cameras produced in 1993 is given, all we need to know is the percent increase from 1993 to 1995.
The percent increase -- up x% in 1994 and then up another y% in 1995 -- is going to be the same no matter how many cameras were produced in 1993. To make the math easier, let's plug in that in 1993 the number of cameras produced was 100.

If 100 cameras were produced in 1993, the increase in the number of cameras produced in 1994 would be (x/100)*100 = x.
So the total number produced in 1994 would be 100+x.

In 1995, the increase in the number of cameras produced would be (y/100)(100+x) = y + xy/100.

Thus, the total increase from 1993 to 1995 would be (increase in 1994) + (increase in 1995) = x + y + xy/100.
Since we started with 100 cameras, this expression also represents the percent increase from 1993 to 1995.
So to have sufficient information, we need to know the value of x + y + xy/100.

Statement 1:
If x=1 and y=20, x + y + xy/100 = 1 + 20 + (1*20)/100 = 21.2.
If x=4 and y=5, x + y + xy/100 = 4 + 5 + (4*5)/100 = 9.2.
Since the expression can take on different values, insufficient.

Statement 2:
Tells us that x + y + xy/100 = 9.2.
Sufficient.

The correct answer is B.

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HPengineer: Master | Next Rank: 500 Posts; Posts: 165; Joined: Wed Mar 24, 2010 8:02 amFollowed by: 1 member; Thanked: 2 times

by HPengineer » Mon Dec 13, 2010 8:06 pm

Rahul,

i got to this point and become stuck by the algebra

1/10(10000 +100x + 100y + xy) = 10(100 + x+ y + xy/100) = 10(100 + 9.2) = 1092.

can you break it down a bit more?

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