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LevelOne
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Posted: Fri Jul 03, 2009 1:06 pm Post subject: circular gears |
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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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Posted: Fri Jul 03, 2009 8:21 pm Post subject: |
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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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Posted: Fri Jul 03, 2009 10:37 pm Post subject: |
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| OA is D. how about exponents in (2)? |
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rahulg83
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Posted: Sat Jul 04, 2009 5:19 am Post subject: |
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| 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
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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.. _________________ You know what you want...keep moving towards it |
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LevelOne
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Posted: Sat Jul 04, 2009 5:31 am Post subject: |
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| rahulg83, thanks again. |
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mehravikas
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Posted: Sun Jul 05, 2009 1:45 pm Post subject: |
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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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Posted: Sun Jul 05, 2009 7:15 pm Post subject: |
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| 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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