UGCNET Questions

To find the smallest number to be added, we first determine the remainder when 625 is divided by 7. Since 7 × 89 = 623, the remainder is 625 − 623 = 2. A number is divisible by 7 when the remainder becomes zero, so we must add a number that brings the remainder 2 up to 7. Therefore, we add 5 (because 2 + 5 = 7), making 625 + 5 = 630, which is divisible by 7.

Option A:
Option A correctly uses the idea that the required adjustment is 7 minus the remainder. With a remainder of 2, adding 5 eliminates the remainder and yields a number (630) that lies exactly on the 7-times table.

Option B:
Option B, 2, would produce 627, and 627 ÷ 7 still leaves a remainder because 7 × 89 = 623 and 7 × 90 = 630. Thus, 627 is not divisible by 7.

Option C:
Option C, 4, gives 629, which also falls between multiples of 7 and leaves a nonzero remainder, so it does not satisfy the condition.

Option D:
Option D, 6, would lead to 631, which again does not align with a multiple of 7, because the nearest multiples are 630 and 637.

The days of the week repeat every 7 days. We find the effective shift by computing 50 mod 7. Since 50 = 7 × 7 + 1, the remainder is 1, meaning the day moves forward by 1 from Wednesday. One day after Wednesday is Thursday, so 50 days from now it will be Thursday.

Option A:
Option A, Thursday, is exactly one day after Wednesday and corresponds to a remainder of 1 in the modulo-7 calculation, so it matches the correct reasoning.

Option B:
Option B, Saturday, is three days after Wednesday and would be correct only if the remainder were 3, not 1.

Option C:
Option C, Sunday, is four days after Wednesday and would require a remainder of 4, which does not arise from 50 ÷ 7.

Option D:
Option D, Monday, is five days after Wednesday, again inconsistent with the actual remainder of 1.