Q1 - IMO - B

Q2 - IMO - A

Q3 - IMO - E

1) IMO B

1: q=3 , we do not know what we have to do with the table, so Insufficient.
2: tells about how to calculate other variables other than q

r = q+q = 2q
s=q+r=3q (as r = 2q)
t=q+s=4q........after this we kinda know that all the variables can be calculated in terms of q , so we can stop and mark B as answer as it does n't matter if z=20q because answer Yes/No is still sufficient. Just for clarity I am calculating...
u=q+r=3q
v=s+u=3q+3q=6q
w =t+v =10q
y=v+x=10q
z=w+y = 20q , Sufficient

2)IMO A

whether rt = even?-->rephrase the question Is one of them(r and s) even ?
1: r+t = odd means one of them must even and other must be odd , sufficient

2: r^t is odd 3^3 can be odd and 3^2 can be odd, so we do no know if one of them is , for sure, even. Insufficient.

3) IMO C

What is the value of a^4 - b^4? --> rephrase the question--> (a^4 - b^4)=(a^2 + b^2)(a^2 - b^2)= (a^2 + b^2)(a - b)(a + b)


1: a^2 - b^2 = 16, we do not know whats the value of (a^2 + b^2) , so Insufficient.
2: a + b = 8 , we do not know whats the value of (a^2 + b^2)(a-b) , so Insufficient.

Combining 1 and 2, we get

a^2 - b^2 = 16 => (a - b)(a + b) = 16

From st 2 substitute the value of (a + b) in the above equation : (a - b) * 8 = 16 => (a - b) = 2

now we have two equation[(a - b) = 2 & a + b = 8 ] and two variable , so we can solve for a and b (5 & 3 respectively ) and consecutively (a^4 - b^4) , So sufficient.

Wow, Q:3 seemed to be a tough one...thanks so much

3 a^4 - b^4 = (a^2-b^2)(a^2+b^2)

1 a^2-b^2 = 16 tells nothing about (a^2+b^2) not sufficient.

2 a+b = 8 not sufficient.

1+2 a+b = 8 means a^2-b^2 = 16 giving a-b = 2
solving a+b = 8 and a-b = 2 we have a=5 and b = 3

thus C

For question 2 the answer might be C
from statement 2, r^t is odd implies r is odd(since even^ anything is even and vice-versa)
In statemebt 1, r +t is odd implies t is even (odd + odd is even and odd + even = odd)
From 1 and 2, we know rt is even.