If two numbers are each divided by the same divisor, then the remainders are 6 and 7, respectively. If the sum of the two numbers be divided by the same divisor, then the remainder is 5. The divisor is:

Correct option is B

Given :
Remainder 1 = 6
Remainder 2=7
Remainder 3=5

Concept used:

Dividend = Divisor × Quotient + Remainder

​Solution:​
​
Let the divisor be ‘d’ and the quotient be q1 for the dividend x, q2 for the dividend y and q3 for the dividend (x + y)

x = q1 × d + 6
y = q2 × d + 7

And (x + y) = q3 × d + 5

Since q1, q2, q3 are whole numbers and in all cases, the remainder are 6, 7 and 5 then it is obvious that the divisor d > 7

Now,

(x – 6), (y – 7) and (x + y – 5) all will be completely divisible by d ----(1)

{(x – 6) + (y – 7)} is also completely divisible by d

Again

(x + y – 13) is also completely divisible by d ----(1)

Now,

from (1) and (2) we can write {(x + y – 5) – (x + y – 13)} = 8, which must be divisible by ‘d’ or ‘d’ be a factor of 8

Since d > 8, t

The value of ‘d’ must be 8

∴ The required divisor is 8 

Alternative Method:

Given :
Remainder 1 = 6
Remainder 2=7
Remainder 3=5
Formula Used :
Divisor = Remainder 1 + Remainder 2 - Remainder 3
Solution:
Divisor = Remainder 1 + Remainder 2 - Remainder 3 = 6+7- 5 =8