Thursday, December 17, 2009
17th Dec Gre Quants Question
3. Given x((75+y) + (75-y)) = 900.
Col A: xy
Col B: 100
4. Which of the following cannot be expressed as the sum of three consecutive integers?
A. 0
B. 1
C. 2
D. 3
E. 5
5. Given a < 0 < b < c
Col A: ac/b
Col B: ac
6. If 2^(2x+1) − 2^2x = 2^1000, then what is the value of x?
7. Col A: (x^x)^x
Col B: x^(x^x)
8. How many numbers among nine consecutive positive numbers are divisible by 9?
9. Given a set of numbers: {1/2, 1/8, 2, 8}
Col A: Median of the set
Col B: Mean of the set
10. There are 10 set of numbers. Each set contains numbers whose unit’s digit represent the set number. For example, if the set number is 1, the numbers in it are 21, 31, 51, and so on.. If the set number is 5 the numbers are 55, 75 and so on. So, if we take the cube of the numbers in set 7, then it represents which of the following set?
A. 3
B. 4
C. 5
D. 6
E. 7
11. Given that two points (0, 2) and (2, 0) lie on the circle.
Col A: Radius of the circle
Col B: 2
12. If x^2 + y^2 = 2xy, then
Col A: x
Col B: y
13. Given a set of three numbers {x, x^2, x^3}; -1 < x < 0. What is the ascending order of the set?
14. Given 7 < xy < 13, where x and y are positive integers. Find the total number of different possible values for XY?
Thursday, December 17, 2009
17th Dec Gre Quants Question
3. Given x((75+y) + (75-y)) = 900.
Col A: xy
Col B: 100
4. Which of the following cannot be expressed as the sum of three consecutive integers?
A. 0
B. 1
C. 2
D. 3
E. 5
5. Given a < 0 < b < c
Col A: ac/b
Col B: ac
6. If 2^(2x+1) − 2^2x = 2^1000, then what is the value of x?
7. Col A: (x^x)^x
Col B: x^(x^x)
8. How many numbers among nine consecutive positive numbers are divisible by 9?
9. Given a set of numbers: {1/2, 1/8, 2, 8}
Col A: Median of the set
Col B: Mean of the set
10. There are 10 set of numbers. Each set contains numbers whose unit’s digit represent the set number. For example, if the set number is 1, the numbers in it are 21, 31, 51, and so on.. If the set number is 5 the numbers are 55, 75 and so on. So, if we take the cube of the numbers in set 7, then it represents which of the following set?
A. 3
B. 4
C. 5
D. 6
E. 7
11. Given that two points (0, 2) and (2, 0) lie on the circle.
Col A: Radius of the circle
Col B: 2
12. If x^2 + y^2 = 2xy, then
Col A: x
Col B: y
13. Given a set of three numbers {x, x^2, x^3}; -1 < x < 0. What is the ascending order of the set?
14. Given 7 < xy < 13, where x and y are positive integers. Find the total number of different possible values for XY?
17 comments:
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Anonymous said...
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1. A
2. 450 [9*1*10*5]
3. D
4. [1 2 5]---> (x+1) + x + (x-1) = 3x where x=INT
5. D
6. 500
7. D
8. 1
9. B
10. A
11. C
12. C
13. x x^3 x^2
14. 5 [8---12]
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December 20, 2009 at 8:08 PM
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akash said...
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Can someone explain how 11th is C .. ?
the way i thought is..
assume center as (x,y) .
so dt of (x,y) from (2,0) = dt of (x,y) from (0,2)
(x-2)^2 + (y)^2 = x^2 + (y-2)^2,
we get x=y..
so checking with (1,1) (2,2) (3,3) as centers, radius will be root(2),2,root(10) .
plz correct if i m wrong ..!
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December 20, 2009 at 8:10 PM
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suhani said...
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2. As the number is odd, in unit place there can be any of [1 3 5 7 9], in tenth place can be [0 1 2 3 4 5 6 7 8 9] in hundredth place can be [6] and in thousandth place can be [1 2 3 4 5 6 7 8 9]. So total number of possibilities 9*1*10*5 = 450
6. 2^(2x+1) - 2^2x = 2^2x.2^1 - 2^2x = 2^2x(2-1) = 2^2x = 2^1000
so, x = 500
12. x^2 -2xy + y^2 = 0 => (x-y)^2=0=>(x-y) = 0=>x=y
14. As x & y are integer, we can think xy as an integer Z. now, 7 < Z < 13. There are only 5 integer in this range [8 9 10 11 12 ].
Hope it helps. I also have exam on Dec 22. Pray 4 me.
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December 20, 2009 at 8:11 PM
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Anonymous said...
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I can't figure out number six. I think I'm suppose to factor out a 2^2x, but I', not sure.
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December 21, 2009 at 7:39 AM
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Anonymous said...
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finally figured out how to factor it...2^2x(2-1)
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December 21, 2009 at 7:51 AM
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Paddy said...
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Ans of 13th is (x^3,x,x^2)
Put x= -0.5
x^2= +0.25
x^3= -0.125 -
January 21, 2010 at 4:27 AM
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Anonymous said...
- This comment has been removed by a blog administrator.
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March 15, 2010 at 3:39 AM
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JotaPéCe said...
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I do not agree with 14th, because zero is not considered positive nor negative. For this, the only value x can take is 1, and y 1-2. The correct answer should be, then, 2, not 5 (11 and 12)
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April 27, 2010 at 4:03 PM
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nisha said...
- This comment has been removed by a blog administrator.
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May 27, 2010 at 10:38 PM
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Anonymous said...
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13. isnt the answer x^3, x, x^2? 'coz x is obviously a negative value and the highest power of a negative value will be the least no. therefore x^3 first, then x^2 will give a positive no. which is supposed to be the greatest in the list and then the tacit...am i right?
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August 5, 2010 at 10:21 AM
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GRE Test said...
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Thanks for your kind! I think it's really helpful for GRE test takers. More useful resource : masteryourgre.com . I think it's good.
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November 14, 2010 at 7:54 PM
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ozlemderici said...
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Is there a chance that you post 2010 questions and answers?
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January 4, 2011 at 2:56 AM
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Will said...
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January 5, 2011 at 11:25 PM
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Will said...
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Problem 13 is x, x^3, x^2 because the problem says list them in ascending order. Because x is between -1 and 0 it is a negative number, x^2 is a positive and x^3 is a negative.
=> x^2 is on the right.
Now to assign which is smaller of x and x^3: Think as if it were positive number between 0 and 1, since any number that cube between 0 and 1 will be a smaller number than the number by itself
(i.e. 0.1 > (0.1)^3 : 0.1 > .001
or 0.9 > (0.9)^3 : 0.9 > 0.729)
the larger number will actually be the smallest number because it is negative therefore for -0.1 we have the ascending order as
-0.1 < -.001 < .01 -
January 5, 2011 at 11:42 PM
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Will said...
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12 is C
Yes, x^2 -2xy + y^2 = 0 => (x-y)^2
but how do we know its not (y-x)^2.
Also let x = y than x^2 + x^2 = 2xx equal 2x^2 = 2x^2 -
January 5, 2011 at 11:49 PM
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Will said...
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Akash your right about prob 11, the answer is D. Try plotting it on a piece of graph paper for clarity.
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January 6, 2011 at 12:17 AM
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Anonymous said...
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what about question 3rd, the ans is very ambiguous.
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February 5, 2011 at 12:41 AM
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17 comments:
1. A
2. 450 [9*1*10*5]
3. D
4. [1 2 5]---> (x+1) + x + (x-1) = 3x where x=INT
5. D
6. 500
7. D
8. 1
9. B
10. A
11. C
12. C
13. x x^3 x^2
14. 5 [8---12]
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Can someone explain how 11th is C .. ?
the way i thought is..
assume center as (x,y) .
so dt of (x,y) from (2,0) = dt of (x,y) from (0,2)
(x-2)^2 + (y)^2 = x^2 + (y-2)^2,
we get x=y..
so checking with (1,1) (2,2) (3,3) as centers, radius will be root(2),2,root(10) .
plz correct if i m wrong ..!
Your Ad Here
2. As the number is odd, in unit place there can be any of [1 3 5 7 9], in tenth place can be [0 1 2 3 4 5 6 7 8 9] in hundredth place can be [6] and in thousandth place can be [1 2 3 4 5 6 7 8 9]. So total number of possibilities 9*1*10*5 = 450
6. 2^(2x+1) - 2^2x = 2^2x.2^1 - 2^2x = 2^2x(2-1) = 2^2x = 2^1000
so, x = 500
12. x^2 -2xy + y^2 = 0 => (x-y)^2=0=>(x-y) = 0=>x=y
14. As x & y are integer, we can think xy as an integer Z. now, 7 < Z < 13. There are only 5 integer in this range [8 9 10 11 12 ].
Hope it helps. I also have exam on Dec 22. Pray 4 me.
Your Ad Here
I can't figure out number six. I think I'm suppose to factor out a 2^2x, but I', not sure.
finally figured out how to factor it...2^2x(2-1)
Ans of 13th is (x^3,x,x^2)
Put x= -0.5
x^2= +0.25
x^3= -0.125
I do not agree with 14th, because zero is not considered positive nor negative. For this, the only value x can take is 1, and y 1-2. The correct answer should be, then, 2, not 5 (11 and 12)
13. isnt the answer x^3, x, x^2? 'coz x is obviously a negative value and the highest power of a negative value will be the least no. therefore x^3 first, then x^2 will give a positive no. which is supposed to be the greatest in the list and then the tacit...am i right?
Thanks for your kind! I think it's really helpful for GRE test takers. More useful resource : masteryourgre.com . I think it's good.
Is there a chance that you post 2010 questions and answers?
Problem 13 is x, x^3, x^2 because the problem says list them in ascending order. Because x is between -1 and 0 it is a negative number, x^2 is a positive and x^3 is a negative.
=> x^2 is on the right.
Now to assign which is smaller of x and x^3: Think as if it were positive number between 0 and 1, since any number that cube between 0 and 1 will be a smaller number than the number by itself
(i.e. 0.1 > (0.1)^3 : 0.1 > .001
or 0.9 > (0.9)^3 : 0.9 > 0.729)
the larger number will actually be the smallest number because it is negative therefore for -0.1 we have the ascending order as
-0.1 < -.001 < .01
12 is C
Yes, x^2 -2xy + y^2 = 0 => (x-y)^2
but how do we know its not (y-x)^2.
Also let x = y than x^2 + x^2 = 2xx equal 2x^2 = 2x^2
Akash your right about prob 11, the answer is D. Try plotting it on a piece of graph paper for clarity.
what about question 3rd, the ans is very ambiguous.
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