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GMAT Prep - PS - Geometry

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by naveenhv » Mon Jan 04, 2010 9:35 am
Can somebody help me with this?


Image


I selected the answer as root 3. But the answer is 1. I can't figure out how.

My reasoning is that P( - root3,1) is in the second quadrant. So Q (s,t) is in the first quadrant. So y-coordinate will not change but the x-coordinate will change the sign. Is this wrong? Please help.
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by naveenhv » Mon Jan 04, 2010 9:56 am
Another from the same set:


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by naveenhv » Mon Jan 04, 2010 11:15 am
Another from the same set.Can somebody please help me how to approach this one:



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by ace_gre » Mon Jan 04, 2010 4:21 pm
2) If n is a positive integer product of all integers 1 to n inclusive and multiple of 990,what is the least possible value of n?

As far as I know there is no direct formula to calculate product of n integers. So I started with multiplying integers.

1*2*3*4*5*6=720
720*7=5040. Then I realized that 990 *1 is not possible.
Then I factorized 990. 3*3*11*10. I realized that this is 9*10*11, where 11 is prime. So the product has to be atleast until n=11. (See that 10=5*2 and 9=3*3 can be obtained from other digits like 2,5,6,9..But not 11).

So n must atleast be equal to 11.

3) Since speeds are the same 50mph, this becomes a direct proportion problem.
car uses 1 gallon for 30 miles.
How many gallons does the car use for 250(=5*50) miles?
25/3 gallons.

1 full tank=12 gallons
fraction = 25/3 gallons

Fraction = (25/3)/12= 25/36.
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by papgust » Mon Jan 04, 2010 6:31 pm
Check this out for 1st question,
https://www.beatthegmat.com/what-is-the- ... 27181.html
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by naveenhv » Tue Jan 05, 2010 7:19 am
ace_gre wrote:2) If n is a positive integer product of all integers 1 to n inclusive and multiple of 990,what is the least possible value of n?

As far as I know there is no direct formula to calculate product of n integers. So I started with multiplying integers.

1*2*3*4*5*6=720
720*7=5040. Then I realized that 990 *1 is not possible.
Then I factorized 990. 3*3*11*10. I realized that this is 9*10*11, where 11 is prime. So the product has to be atleast until n=11. (See that 10=5*2 and 9=3*3 can be obtained from other digits like 2,5,6,9..But not 11).

So n must atleast be equal to 11.

3) Since speeds are the same 50mph, this becomes a direct proportion problem.
car uses 1 gallon for 30 miles.
How many gallons does the car use for 250(=5*50) miles?
25/3 gallons.

1 full tank=12 gallons
fraction = 25/3 gallons

Fraction = (25/3)/12= 25/36.
Hey ace_gre,

Thanks for the answers. I understood the 3 rd problem for fraction of the fuel tank.

Can you explain the second series problem? I am kind of not understanding it still.

Thanks
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by mmslf75 » Tue Jan 05, 2010 7:27 am
naveenhv wrote:
ace_gre wrote:2) If n is a positive integer product of all integers 1 to n inclusive and multiple of 990,what is the least possible value of n?

As far as I know there is no direct formula to calculate product of n integers. So I started with multiplying integers.

1*2*3*4*5*6=720
720*7=5040. Then I realized that 990 *1 is not possible.
Then I factorized 990. 3*3*11*10. I realized that this is 9*10*11, where 11 is prime. So the product has to be atleast until n=11. (See that 10=5*2 and 9=3*3 can be obtained from other digits like 2,5,6,9..But not 11).

So n must atleast be equal to 11.

3) Since speeds are the same 50mph, this becomes a direct proportion problem.
car uses 1 gallon for 30 miles.
How many gallons does the car use for 250(=5*50) miles?
25/3 gallons.

1 full tank=12 gallons
fraction = 25/3 gallons

Fraction = (25/3)/12= 25/36.
Hey ace_gre,

Thanks for the answers. I understood the 3 rd problem for fraction of the fuel tank.

Can you explain the second series problem? I am kind of not understanding it still.

Thanks
naveen,

Multiple of 990 will have atleast ( one 2, one 5 , two 3s & one 11 )
Now, as u see, if u take 10 as the answer, u imply that

1*2*.... *10 = somevalue == But there is NO "11" as a factor there.

So take next nmber that will give u a 11 in the product.

1* 2 ...*10* 11 ===> there u are

Therfore 11 isthe answer..simple. ;-)
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by naveenhv » Tue Jan 05, 2010 7:36 am
mmslf75 wrote:
naveenhv wrote:
ace_gre wrote:2) If n is a positive integer product of all integers 1 to n inclusive and multiple of 990,what is the least possible value of n?

As far as I know there is no direct formula to calculate product of n integers. So I started with multiplying integers.

1*2*3*4*5*6=720
720*7=5040. Then I realized that 990 *1 is not possible.
Then I factorized 990. 3*3*11*10. I realized that this is 9*10*11, where 11 is prime. So the product has to be atleast until n=11. (See that 10=5*2 and 9=3*3 can be obtained from other digits like 2,5,6,9..But not 11).

So n must atleast be equal to 11.

3) Since speeds are the same 50mph, this becomes a direct proportion problem.
car uses 1 gallon for 30 miles.
How many gallons does the car use for 250(=5*50) miles?
25/3 gallons.

1 full tank=12 gallons
fraction = 25/3 gallons

Fraction = (25/3)/12= 25/36.
Hey ace_gre,

Thanks for the answers. I understood the 3 rd problem for fraction of the fuel tank.

Can you explain the second series problem? I am kind of not understanding it still.

Thanks
naveen,

Multiple of 990 will have atleast ( one 2, one 5 , two 3s & one 11 )
Now, as u see, if u take 10 as the answer, u imply that

1*2*.... *10 = somevalue == But there is NO "11" as a factor there.

So take next nmber that will give u a 11 in the product.

1* 2 ...*10* 11 ===> there u are

Therfore 11 isthe answer..simple. ;-)

Thanks
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