GRE questions

[r2kins]

Q 1: COLUMN A: perimeter of a right triangle with hypotenuse 15 and area 54
COLUMN B: area of a cirle with the radius of [6/sq. rt ( pi)]


A: A Is greater
B: B is greater
C: both equal
D: cannot be found out from given info.

Q 2: the sum of three consecutive odd integers is M, when the least of them is “ t” .What is the sum of three consecutive odd integers, in terms of M when the greatest no: is “ t”.

A: M + 6
B: M + 12
C: M – 12
D : M + 6
E: none of the above

[bigduke]

1. C
2. C

Hell yea! prove me wrong on this one

[omsai]

1.c
2.c

[rashmi]

1) C
2) C

If u need some explanations, let me know.

[Zarah]

would u plz explain it, thank u

[rashmi]

1)assume the height and base of triangle are h and b respectively.
It is given that b^2 + h^2 =15^2 = 225(Pythagorous theorum)
Also given that area= 1/2 *b*h=54 .
Thus b*h=108.

We need to know the perimeter of the triangle.
Now (b+h)^2 = b^2 +h^2 + 2*b*h
=225 + 216
= 441.
Thus b+h =21.
Thus perimeter=b+h+15=36


For the circle,
area = (pi) * r^2
r is given as 6/ sqrt(pi).
Thus area is 36.
So both are equal.

2) Given t + (t+2) + (t+4) =M.

We have to find t +(t-2) + (t-4).Thus the answer.

[lenah]

I still don't see the solution for the second problem. Can anyone explain it to me again, please.
Thanks

[tans]

first take 5,7,9 and add , so the result is 21. so M is 21 and t is 5. Now take 1,3,5 and add. So the result is 9.

9= 21-12=M-12

[lenah]

i finally got it. thank you for your help.