Question

a) square of the avg of two positive nos b) avg of the square of the two positive nos...................which one of a) and b) is greater or c)relation cant be determined. if i take 1 and 1 then a) and b) are equal and for other nos its b) which is greater . So i think it answer is C) but in book its given as B)..so wat is the ans?

Answer 1

Use 1 and 3. Square of the average is 4. Average of the square of them is 5. Therefore, B is greater. You can test this with any positive numbers where one of them is greater than 1 and the other greater than 0.

Answer 2

but wat if i take 1 and 1? it is givin a) = b)

Answer 3

Excellent point, and I honestly can't give you an answer, what is your book called, I can attempt to find an explanation.

Answer 4

actually its not a book but its a question in gre guide exam... http://www.greguide.com/free-gre-full-length-tests/gre-full-test-12.html

Answer 5

wat do u suggest then bro? coz ans dey ve stated as b).

Answer 6

lets call one of the numbers x and other y.
then a) gives \((\frac{x+y}{2})^2=\frac{x^2+y^2+2xy}{4}\)
and b) gives \(\frac{x^2+y^2}{2}\)
subtracting a) from b) gives: \(\frac{x^2+y^2}{2}-\frac{x^2+y^2+2xy}{4}=\frac{x^2+y^2-2xy}{4}=\frac{(x-y)^2}{4}\)
Now you can see that this quantity is always greater then or equal to zero if x and y are positive numbers

Answer 7

if x=y, then this quantity is zero

Answer 8

therefore, if x=y then (a) and (b) are the same value
otherwise (b) is greater than (a)

Answer 9

so if the question stated that the two number are "distinct" then b) is the answer
otherwise answer should be c)

Answer 10

yup that what i was thinking. ty

Answer 11

yw :)