2008/02/25
2008/02/17
2008/02/13
Problem 10 : Probability (2000) - Combination Topic
( This problem is taken from Indonesian Entrance Test of State University- 2000 )
The number of triangles that can be drawn from 7 points with no three collinear points is ...
a. 30 b. 35 c. 42 d. 70 e. 210
Answer :
It's a combination case. The formula is nCr = n! / [ (n-r)!r!].
Since the triangle has 3 points so from it's problem we have :
7C3 = 7!/ [ (7-3)!.3!]
= 7.6.5.4! / 4! 3.2.1
= 7.6.5 /6
= 7.5
= 35 Answer : B
2008/02/12
2008/02/07
Problem 6 : Quadratic Equation-2002
( This problem is taken from Indonesian Entrance Test of State University - Year 2002)
If the quadratic equation ( p + 1 ) x^2 - 2 ( p + 3 ) x + 3p = 0 has two equal roots, then the constant p = ...
a. -3 and 3/2 b. -3/2 and 3 c. 1 and 3 d. 2 and -3 e. 3 and -9
Answer :
We know that if the roots are equal then the value of determinant ( D ) = b^2 - 4ac = 0
From the quadratic equation that is given we have a = ( p + 1) ; b = -2 ( p + 3 ) ; c = 3p.
So D = b^2 - 4ac = [ -2(p + 3)]^2 - 4.( p + 1).3p = 0
4 (p + 3)^2 - 12p( p + 1) = 0
4 ( p^2 + 6p + 9) -12p^2-12p = 0
4p^2 + 24p + 36 -12p^2 -12p = 0
-8p^2 + 12 p + 36 = 0
Dividing by -4 we have : 2p ^2 - 3p - 9 = 0
( 2p + 3 )( p - 3) = 0
2p + 3 = 0 or p-3 = 0
2p = -3 p = 3
p = -3/2 Answer : B
2008/02/02
Problem 4 : Linear Equation - ( 2002 )
( This problem is taken from Indonesian Entrance Test Of State University - Year 2002 )
If x and y satisfy the equations system 2/x + 1/y = 1 and 1/x - 2/y = 8 then 1/ (x+y) = ....
a. -3/2 b. 5/6 c. 6/5 d. 5 e. 6
Answer :
We take 1/x = p and 1/y = q then it's problem becomes 2p + q = 1 ...(1) and p - 2q =8...(2).
Eliminating q, we obtain : 4p + 2q = 2 .....2x(1)
p - 2q = 8.....1x(2)
------------- +
5p = 10
p = 2
1/x = 2 -----> x = 1/2
p = 2 -----> 2p + q = 1
2.2 + q = 1
4 + q = 1
q = 1- 4 = -3
1/y = -3
y = -1/3
The value of 1/( x+y) is 1/[ 1/2 + (-1/3) ] = 1/ (3/6 - 2/6 ) = 1/ ( 1/ 6) = 6 Answer : E
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