Tuesday, July 7, 2009

7th july Gre Quants Questions

2. What is the least common factor of 123 × 255
A. 3
B. 7
C. 17
& so on….

3. If x, y are positive integers and (x^2 + y^2)/2 = xy, then
Col A: x
Col B: y

4. If |J + 2|< 5 and |〖(k + 1)〗^2|>16, then
Col A: JK
Col B: 15
(Similar to this)

5. If 2^(2n+1)- 2^(2n )= 2^1000, then what is the value of n?

6. Col A: Standard Deviation of u, v and w
Col B: Standard Deviation of u+3, v+3 and w+3

7. If x > 0, y > x and x^y= 64, then
Col A: x
Col B: 2

8. Given figures of a regular pentagon and an equilateral triangle. If perimeter of pentagon is equal to twice the perimeter of equilateral triangle, then
Col A: Side length of the pentagon
Col B: Side length of equilateral triangle

4 comments:

Anonymous said...

2-a
3-d
4-d
7-d

Anonymous said...

2. 1 if in option else a in given options
3.c
4.d
5.500
6.c
7.d
8.a

Anonymous said...

I think answer to 7 is "c" since x > 0 and y > x therefore they can't be negative and only 2^6 = 64 satisfies the condition of y > x.

Anonymous said...

Hi...
For 7th question,
Please think in this way...
(root of 2)^12 = 64
x= Root pf 2 = 1.141 >1.
Y= 12 >x
Got it

Tuesday, July 7, 2009

7th july Gre Quants Questions

2. What is the least common factor of 123 × 255
A. 3
B. 7
C. 17
& so on….

3. If x, y are positive integers and (x^2 + y^2)/2 = xy, then
Col A: x
Col B: y

4. If |J + 2|< 5 and |〖(k + 1)〗^2|>16, then
Col A: JK
Col B: 15
(Similar to this)

5. If 2^(2n+1)- 2^(2n )= 2^1000, then what is the value of n?

6. Col A: Standard Deviation of u, v and w
Col B: Standard Deviation of u+3, v+3 and w+3

7. If x > 0, y > x and x^y= 64, then
Col A: x
Col B: 2

8. Given figures of a regular pentagon and an equilateral triangle. If perimeter of pentagon is equal to twice the perimeter of equilateral triangle, then
Col A: Side length of the pentagon
Col B: Side length of equilateral triangle

4 comments:

Anonymous said...

2-a
3-d
4-d
7-d

Anonymous said...

2. 1 if in option else a in given options
3.c
4.d
5.500
6.c
7.d
8.a

Anonymous said...

I think answer to 7 is "c" since x > 0 and y > x therefore they can't be negative and only 2^6 = 64 satisfies the condition of y > x.

Anonymous said...

Hi...
For 7th question,
Please think in this way...
(root of 2)^12 = 64
x= Root pf 2 = 1.141 >1.
Y= 12 >x
Got it