circular gears

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LevelOne
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Topic: circular gears
PostFri Jul 03, 2009 2:06 pm
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1) Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute, and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than gear P?

A. 6
B. 8
C. 10
D. 12
E. 15

2) The value of 10^8 - 10^2/10^7 - 10^3 is closest to which of the following?

A. 1
B. 10
C. 10^2
D. 10^3
E. 10^4

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mehravikas
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PostFri Jul 03, 2009 9:21 pm
Answer to Question 1 -

P = 10 rev per minute -> in one second it makes 10/60 = 1/6 rev per sec
Q = 40 rev per min -> in 1 sec = 40/60 = 2/3 rev per sec

1/6 * t + 6 = 2/3 * t

solve for t and you'll get 12. Answer D
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LevelOne
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PostFri Jul 03, 2009 11:37 pm
OA is D. how about exponents in (2)?
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rahulg83
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PostSat Jul 04, 2009 6:19 am
Quote:
2) The value of 10^8 - 10^2/10^7 - 10^3 is closest to which of the following?

A. 1
B. 10
C. 10^2
D. 10^3
E. 10^4
i suppose question asks for value of (10^8 - 10^2)/(10^7 - 10^3), it can be written as 10^2{(10^6)-1}/10^3{(10^4)-1}, now 1 can be neglected in comparison with 10^6 or 10^4

So our expression will now reduce to (10^2*10^6)/(10^3*10^4)=10 ans..
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LevelOne
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PostSat Jul 04, 2009 6:31 am
rahulg83, thanks again.
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mehravikas
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PostSun Jul 05, 2009 2:45 pm
Rahul,

can you explain how can you neglect 1 please?

Thanks,
Vikas

rahulg83 wrote:
Quote:
2) The value of 10^8 - 10^2/10^7 - 10^3 is closest to which of the following?

A. 1
B. 10
C. 10^2
D. 10^3
E. 10^4
i suppose question asks for value of (10^8 - 10^2)/(10^7 - 10^3), it can be written as 10^2{(10^6)-1}/10^3{(10^4)-1}, now 1 can be neglected in comparison with 10^6 or 10^4

So our expression will now reduce to (10^2*10^6)/(10^3*10^4)=10 ans..
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kath_the_great4
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PostSun Jul 05, 2009 8:15 pm
1 is insignificant when you are dealing with numbers in the thousands...we are not looking for an exact response here.
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