official quant thread for cat08

[naga25french]

Originally Posted by implex

New problem !!
Find the number of ordered pairs(x,y) of integers such that 55x^2+2xy+y^2=2007

a) 2 b) 4 c) 6 d) 8 e) 12

is the answer b) 4 ?
solution:

55x^2+2xy+y^2=2007..

54x^2 + (x + y )^2 = 2007

when x= 3 (x+y)^2 = 1521 =======> 39^2


the ordered pairs can be formed by writing 39^2 in 4 different ways...


ordered pairs are (3,36) ,(-3,-36), (-3 , 42) and (3,-39)

[selebratinglife]

Wrong post

[sandyrocks]

Originally Posted by implex

New Problem!!
The diagonals AC and CE of a regular hexagon ABCDEF are divided by the internal points M and N such that AM/AC=CN/CE=r. Determine r if B, M and N are collinear.
a) 1 b) 1/3 c) 2/3 d) 1/2 e) none of these

hi this is my first post..by midpoint theorem the answer must be 1/2

[nikijpr]

Originally Posted by implex

New problem !!
Find the number of ordered pairs(x,y) of integers such that 55x^2+2xy+y^2=2007

a) 2 b) 4 c) 6 d) 8 e) 12

The ans shud be option a) 2...
55x^2+2xy+y^2 can be written as 54x^2 + (x + y)^2 = 2007 = 54*(3)^2 + (3 + 36)^2
This is satisfied by (x,y)=(3,36) and (-3,-36) only...

[selebratinglife]

Originally Posted by naga25french

is the answer a) 2?

ordered pairs are (3,36) ,(-3,-36)

Total 4 pairs i believe..
Additional 2 are (3,-39)(-3,42)

[naga25french]

Originally Posted by selebratinglife

Total 4 pairs i believe..
Additional 2 are (3,-39)(-3,42)

(3,-39) cannot be an ordered pair...

[sandyrocks]

Originally Posted by selebratinglife

Total 4 pairs i believe..
Additional 2 are (3,-39)(-3,42)

hey i think it must be (3,-42) and not (3,-39)

[selebratinglife]

Originally Posted by implex

New Problem!!
The diagonals AC and CE of a regular hexagon ABCDEF are divided by the internal points M and N such that AM/AC=CN/CE=r. Determine r if B, M and N are collinear.
a) 1 b) 1/3 c) 2/3 d) 1/2 e) none of these

Ans is 2/3....Assuming 'a' to be the side of hexagon, We have BE=2a and BE bisects AC for M and N to be collinear the ratio AM/MC should be greater than 0.5, else N and E would coincide. So checked for 2/3 and it clicked..Else would be none of these. This method can be applied during exam to save time. It can be solved by proper geometry though but is a bit time consuming process.

[selebratinglife]

Originally Posted by sandyrocks

hey i think it must be (3,-42) and not (3,-39)

Aarggh!Righto!You can gauge the amount of torpor I'm in at this time..But the answer nevertheless is 4..Doesn't matter whichever the ordered pair it is

[nikijpr]

Originally Posted by selebratinglife

Ans is 2/3....Assuming 'a' to be the side of hexagon, We have BE=2a and BE bisects AC for M and N to be collinear the ratio AM/MC should be greater than 0.5, else N and E would coincide. So checked for 2/3 and it clicked..Else would be none of these. This method can be applied during exam to save time. It can be solved by proper geometry though but is a bit time consuming process.

Why cant the ans be 1....