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by Rahul@gurome » Sun Sep 05, 2010 10:44 pm
Please post only one question in one thread.

Q1. If k is not equal to 0, 1, -1, is 1/k > 0?
(1) 1/(k - 1) > 0
(2) 1/(k + 1) > 0

(1) 1/k{1 - (1/k)} > 0
For 1/(k - 1) > 0, 1 - (1/k) > 0, which implies 1 > 1/k or 1/k < 1
So, 1/k will be negative, which answers the main question as "no".
So, (1) is SUFFICIENT.

(2) 1/k{1 + (1/k)} > 0
For 1/(k - 1) > 0, 1 + (1/k) > 0 or 1/k > -1. This means 1/k can take any value greater than -1, so it can be any value viz., -1/2, -1/4, 1, 2 and so on.
So, (2) is NOT SUFFICIENT.

The correct answer is [spoiler](A)[/spoiler].
Rahul Lakhani
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by Rahul@gurome » Sun Sep 05, 2010 10:47 pm
Q2. If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be the form of x^2- (by)^2, where b is an integer

a. ½
b. 1/3
c. ¼
d. 1/5
e. 1/6

Solution:
(x + y)(x + 5y) = x^2 + 6xy + 5y^2, not in the given form.
(x + y)(x - y) = x^2 - y^2, this is one of the possibilities.
(x + y)(5x - y) = 5x^2 + 4xy - y^2, not in the given form.
(x + 5y)(x - y) = x^2 + 4xy - 5y^2, not in the given form.
(x + 5y)(5x - y) = 5x^2 + 24xy - 5y^2, not in the given form.
(x - y)(5x - y) = 5x^2 - 6xy + y^2, not in the given form.
Therefore, required probability = 1/6

The correct answer is [spoiler](E)[/spoiler].
Rahul Lakhani
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by Rahul@gurome » Sun Sep 05, 2010 10:57 pm
Q4. Rasheed bought two kinds of candy bars, chocolate and tofeee that came in pacakges of 2 bars each. He handed out 2/3 of the chocolate bars and 3/5 of the tofeee bars. How many pacakges of chocolate bars did Rasheed buy?

(1) Rasheed bought 1 fewer package of chocolate bars than toffee bars.
(2) Rasheed handed out the same number of each kind of candy bar.

Solution:

Let Rasheed bought 'C" chocolate bars and 'T' toffees. He handed out (2/3)(2C)+ (3/5)(2T).

(1) C = T - 1; we have one equation 2 variables.
So, (1) is NOT SUFFICIENT.

(2) (2/3)(2C) = (3/5)(2T) implies 4C/3 = 6T/5; again we have one equation 2 variables.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), we have 2 equations and 2 variables, which can be solved to get the value of C.
So, combining the statements, we can answer the question.

The correct answer is [spoiler](C)[/spoiler].
Rahul Lakhani
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by Rahul@gurome » Sun Sep 05, 2010 11:16 pm
Q3. In the figure shown, point O is the center of the semicircle and B,C,D lie on the semicircle.If the length of the line segment AB is equal to OC.What is angle BAO ?
1) Angle COD = 60º
2) Angle BCO = 40º

Solution:

(1) Let angle BAO = x and angle OCA = y = angle OBC (OB = OC)
Apply exterior angle theorem in triangle OAC, angle COD = angles (OAC + OCA)
So, 60= x + y
y = 2x, so 60 = x + 2x or 3x = 60 or x = 20º
So, angle BAO = 20º
So, (1) is SUFFICIENT.

(2) angle BCO = angle OBC = 40º (OB = OC)
Let angle BAO = x = angle AOB
Then by exterior angle Theorem, 40º = 2x or x = 20º
So, (2) is SUFFICIENT.

The correct answer is [spoiler](D)[/spoiler].
Rahul Lakhani
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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