I simply solved it by the following method,

m+n = 12q +8 equation 1

m-n = 12q +6 equation 2


since we know that to a find range of equation 1 and equation 2 we need to apply the following rule

1 ---> 12 * 1 + 8 = 20
12 * 2 + 8 = 32
12 * 3 + 8 = 44
12 * 4 + 8 = 56----etc

2 12 * 1 + 6 = 18
12 * 2 + 6 = 30
12 * 3 + 6 = 42
12 * 4 + 6 = 52

now we can use the first values of the above lists 20 * 18 =360/6 leads to 60 remainder 1 and can check with other values as well -- so the answer is A

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wouldn't adding m+n to m-n result in m = 12a + 12b + 14, IShanGRE? (the quote function isnt working on my browser atm)

Sorry for inconvenience, we have updated forum server due to which some error are floating, we will get over these error soon.

Thanks for your patience.

Just choose numbers for m & n.

Choosing m = 7 & n = 1 fullfil all the provided conditions.

Ans: A) 1 when, mn = 7 is divided by 6.

I chose m-n = 18 ( 18%12=6) ; m+n = 20 (20%12=8).

thus the equations :
m - n = 18....(1)
m+ n = 12...(2)
---------------------------
m=19 and n=1

Hence,
mn/6 = 19/6 Gives remainder as 1.

The answer is A.

The remainder when m + n is divided by 12 is 8

Theory: Dividend = Divisor*Quotient + Remainder

m + n -> Dividend
12 -> Divisor
a -> Quotient (Assume)
8 -> Remainder
=> m + n = 12*a + 8 = 12a + 8 ...(1)

The remainder when m - n is divided by 12 is 6

Let Quotient be b

=> m - n = 12b + 6 ...(2)

Adding (1) and (2) we get

m + n + m - n = 12a + 8 + 12b + 6 = 12*(a+b) + 14
=> 2m = 12*(a+b) + 14
=> m = 6*(a+b) + 7 = 6*(a+b+1) + 1

(1) - (2) we get

m + n -( m - n) = 12a + 8 -( 12b + 6) = 12*(a-b) + 2
=> 2n = 12*(a-b) + 2
=> n = 6*(a-b) + 1

=> m*n = (6*(a+b+1) + 1) * (6*(a-b) + 1)
=> on right hand side all terms will be a multiple of 6 except 1*1

=> m*n = multiple of 6 + 1
=> m*n will give 1 remainder when divided by 6

So, Answer will be A
Hope it helps!

Watch the following video to learn the Basics of Remainders