ISI 2012 ME I Doubts

[Ayushya Kaul]

Hey, any help regarding these:

23. Given x  y  z, and x + y + z = 9, the maximum value of x + 3y +
5z is
(A) 27,
(B) 42,
(C) 21,
(D) 18.


25. Suppose there is a multiple choice test which has 20 questions. Each
question has two possible responses - true or false. Moreover, only
one of them is correct. Suppose a student answers each of them randomly.Which one of the following statements is correct?
(A) The probability of getting 15 correct answers is less than the probability of getting 5 correct answers,
(B) The probability of getting 15 correct answers is more than theprobability of getting 5 correct answers,
(C) The probability of getting 15 correct answers is equal to the probability of getting 5 correct answers,
(D) The answer depends on such things as the order of the questions.

26. From a group of 6 men and 5 women, how many different committees
consisting of three men and two women can be formed when it is known
that 2 of the men do not want to be on the committee together?
(A) 160,
(B) 80,
(C) 120,
(D) 200.

20. If a^2 + b^2 + c^2 = 1, then ab + bc + ca is,
(A) 0:75,
(B) Belongs to the interval [1; 0:5],
(C) Belongs to the interval [0:5; 1],
(D) None of the above.

[Sinistral]

23) possible only when x=y=z. so (A) 27

25) P(getting 15 correct ans) = 20C15 (1/2)^20
      P(getting 5 correct ans) =  20C5 (1/2)^20
      clearly both the above probs are equal . so (C)

26) first lets keep aside those 2 men who dont want to come together.
     now choosing  3Men out of remaining 4Men = 4C3
     choosing 2 women out of 5 women = 5C2
     so, 4C3 * 5C2            --------------- (1 )

     now lets select one of those two disapproving men and choose remaining 2 men of the committee from remaining 4 tolerant men = 4C2
      above can be done in 2 ways as we can select either of 2 disapproving men.
      so 2* 4C2
        and ofcourse choosing 2 women out of 5 women = 5C2
          hence required prob (in this case) =2* 4C2*5C2 ------------ (2)
adding (1) and (2) will give (A) 160

20)
(a+b+c)^2 >= 0
==> a^2 + b^2 + c^2  +2(ab + bc + ca) >=0
                1               +2(ab + bc + ca) >=0
                                    ab + bc + ca  >= -1/2


hence (D) None of the above.

[Ayushya Kaul]

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[Ankit Agrawal]

Q 23.) Given
x>=y>=z---(1)
x+y+z=9---(2)
Max: x+3y+5z----(3)

Notice that the coefficients of y and z are 3 and 5. This means that in order to maximise the value of the equation (3), you must allocate as many numbers as possible to, first, z and then to y. Assume that condition (1) is not there. Then, You would have allotted  9 to z, 0 to y and 0 to x. But you have this condition, where you have to make sure that z is at most equal to y and x. Hence you would try to allocate the maximum number possible to allocate to z,i.e., 3.

Considering condition (1) and (2), you will allocate 3 each to y and x as well.

The value of (3) that you get is: 3+9+15=27.

Hence the answer is A.

[Sinistral]

Using method of lagrange multiplier gives D=0 ==> no conclusion.
may be some theorem of LP can be used to prove it formally.

this is how I do. i choose the middle point & go around this middle point slowly to check the behaviour of the function. also I check it at extreme points.
so the moment one starts to go away from (3,3,3) f starts declining.
hence I put x=3,y=3,z=3. (basically I did it by inspection)

had this ques been asked in subjective portion I would have been in a fix.

[Ayushya Kaul]

Ankit and Sinistral. Thank you so much :D