Tuesday, August 25, 2009

Question of the day - 25th August, 2009

Question 1:

What is the probability for a family with three children to have a boy and two girls, assuming the probability of having a boy or a girl is equal?

a)1/8
b)¼
c)½
d)3/8
e)5/8

Question 2:

X,Y > 0 ,If X^3 = Y, is Y a fraction?

(1) X^2 is a fraction
(2) X >Y

OAs will be posted soon

Thanks,
Quant-Master
Posted by GMAT at 4:45 AM

5 comments:

  1. AnonymousAugust 25, 2009 at 7:41 AM

    I think it will be 1/8

    ReplyDelete
  2. AnonymousAugust 25, 2009 at 10:19 AM

    1. - 3/8
    2. D

    ReplyDelete
  3. AnonymousAugust 25, 2009 at 1:26 PM

    what's the answer?

    ReplyDelete
  4. GMATAugust 26, 2009 at 5:13 AM

    Solution to Question 1:
    Probability of having a boy out of 3 children is P(b) * P(g) * P(g) =1/2 * 1/2 * 1/2 = 1/8

    Now this can happen in (3!)/2 = 3 ways
    bgg is 1/8
    gbg is 1/8
    ggb is 1/8

    Hence it is 3*(1/8) = 3/8

    Solution to Question 2:

    1. x^2 is a fraction. This means x is a fraction too.
    for example lets say x^2 = a/b where a,b are int and b>a
    x=sqrt(a)/sqrt(b) . since a is less than b, x has to be a fracton.

    now x^3=x^2*x multiplying 2 fractions will always lead to a fraction. Therefore Y is a fraction. Sufficient

    2. x > y
    its given that x^3 = y.
    Any positive raised to a positive power will yield more value if the integer is >1

    since x^3 equals y and also x > y, it means that x is <1
    also, y is less than x and x is less than 1 therefore Y is a fraction. Sufficient.

    Hence D

    Thanks,
    Quant-Master

    ReplyDelete
  5. toucheAugust 26, 2009 at 11:50 PM

    boy girl...
    the probability for a family with three children to have a boy and two girls" does not mean diff in birth order, it means in any order of birth.
    so-- bbb,bbg(inclues bgb,gbb),bgg(includes gbg, gbb),ggg
    hence 1/4

    ReplyDelete

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