NIMCET 2027 Number System Previous Year Questions (PYQs) – Page 2 of 7
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Showing 61 questions
How many numbers from 1 to 100 are such each of which is divisible by 8 and whose at least one digit is 8?
1
2
3
4
Solution
Multiples of 8 up to 100:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Now check which contain digit 8:
8
48
80
88
Count = 4
The decimal equivalent of the hexadecimal operation A10 + B21 is
1
2
3
4
Solution
The result of multiplication of the numbers (10101)2 and (11101)2 in hexadecimal form is
1
2
3
4
Solution
If all digits 6 in the numbers from 1 to 100 are replaced by 9,
by how much does the algebraic sum change?
1
2
3
4
Solution
Each digit 6 → 9, increase = 3.
Count how many times digit 6 appears from 1 to 100.
Two-digit numbers (10–99):
Tens place = 6 → numbers 60–69 → 10 times
Units place = 6 → appears once in each decade (16, 26, ..., 96) → 9 times
Total digit 6 occurrences = $10 + 9 = 19$
Additionally, number 6 itself contributes 1 more occurrence.
So total 6s = $19 + 1 = 20$
Increase in sum = $20 \times 3 = 60$
But this does not match options → Check again carefully.
Correct counting:
Tens place occurrences:
60–69 → 10
Units place occurrences:
6, 16, 26, 36, 46, 56, 66, 76, 86, 96 → 10
Total = 20 occurrences.
Increase = $20 \times 3 = 60$
But total sum change involves positional value:
A “6” in tens place increases number by 30
A “6” in ones place increases by 3
Count tens-place 6s = 10 → total = $10 \times 30 = 300$
Count ones-place 6s = 10 → total = $10 \times 3 = 30$
Total change = $300 + 30 = 330$
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