Q1.
Consider a/b = b/c. This means that a = (b^2)*c. Now, since a is a perfect square and b^2 is obviously a perfect square, then c, which is a/(b^2) must also be a perfect square.

Q2.
S being an odd or an even number depends on the nature of n. As you may know, n(n + 1) is always even (the product of two consecutive numbers is always even), but when you divide it by 2, you may have either an odd or an even number. Examples:
n = 4 makes S = 10 (even)
n = 5 makes S = 15 (odd)
S being a prime number is the easiest to eliminate. The two examples above are more than sufficient for that.
S must not be a prime number is the correct choice here and this happens because n(n + 1) is even. When you divide it by 2, you have two cases:
a. n is even - this makes n/2 an integer and, since n+1 is also an integer, you have that S is the product of two integers
b. n + 1 is even - this makes (n + 1)/2 an integer and n is also an integer, and you follow the same reasoning as above.
E can be again easily eliminated with the two examples above.

Q3.
Say x = your number. Then you have that x = 12q + 7 = 6*2*q + 6 + 1 = 6(2q + 1) + 1. This is why the remainder will be indeed 1 when you divide x by b.

Q4. x has the following format: MN, with M, N digits from 1 to 9 (N can also be 0). The tens position of MN + 10 will be M + 1, since you added 10. An example would be:
MN = 34
MN + 10 = 44 - the tens position is (initial position + 1)
This means that the new sum of digits will be M + N + 1 = (M + N) + 1 = 10.
But we must not forget of the special case when M is 9. In this case, MN + 10 will have the "10N" format. Example:
MN = 94
MN + 10 = 104.
This makes the sum of digits N + 1. Now, since M is considered 9 in this special case, this means that from M + N = 9 you get that N = 0. This makes the sum of digits for the special case 1.
So the answer will be 10 or 1.

I don't understand Q4 really:$

But damnnn you're good hahaha, got some tips for me to be as good as you

Q4.

I think the examples given for 4 may be a little off or I am missing something which is quite possible

MN or x whatever-> the sum of the digits of this 2 digit number must be 9

Eg:

54
36
27

Add 10 to these and the number u get will have digit sum equalt to 10


Take 90 for example

90+10 = 100 sum of the digits of 100 is 1

Hence 1 or 10

I chose those examples without taking into consideration the fact that the sum of digits was 9, but merely to illustrate what happens when you add 10 to a two-digit number. Maybe that's what was a bit confusing.

Got u ;all's well!

Good luck, DanaJ.

Regards,
Cramya

For Q4. Only 2 conditions are possible...

Condition 1 when no. comes with '0', say 90 then 90+10 = 100 => digit sum =1

Condition 2 when no. comes without '0', say 45 then 45+10 = 55 => digit sum =10

Answer E.

Ah thanks thanks :D

Question 4. is little bit tricky

Good post.