JEE Mathematics Number Systems MCQs Set 01

Answer: (d) 86

Explanation:
1. Multiply each digit by the power of 5 for its place: \( 3\times5^2 + 2\times5^1 + 1\times5^0 \).
2. The sum is 86.
3. So \( (321)_{5} \) = 86 in the denary system.

Teacher's Note:
1. Place values in base 5 are powers of 5, starting with 1 at the units place.
2. Do not read the digits as a denary number.

Answer: (d) 528

Explanation:
1. Multiply each digit by the power of 5 for its place: \( 4\times5^3 + 1\times5^2 + 3\times5^0 \).
2. The sum is 528.
3. So \( (4103)_{5} \) = 528 in the denary system.

Teacher's Note:
1. Place values in base 5 are powers of 5, starting with 1 at the units place.
2. Do not read the digits as a denary number.

Answer: (d) 11

Explanation:
1. Multiply each digit by the power of 2 for its place: \( 1\times2^3 + 1\times2^1 + 1\times2^0 \).
2. The sum is 11.
3. So \( (1011)_{2} \) = 11 in the denary system.

Teacher's Note:
1. Place values in base 2 are powers of 2, starting with 1 at the units place.
2. Do not read the digits as a denary number.

Answer: (c) 42

Explanation:
1. Multiply each digit by the power of 16 for its place: \( 2\times16^1 + 10\times16^0 \).
2. The sum is 42.
3. So \( (2A)_{16} \) = 42 in the denary system.

Teacher's Note:
1. Place values in base 16 are powers of 16, starting with 1 at the units place.
2. Do not read the digits as a denary number.

Answer: (a) 585

Explanation:
1. Multiply each digit by the power of 8 for its place: \( 1\times8^3 + 1\times8^2 + 1\times8^1 + 1\times8^0 \).
2. The sum is 585.
3. So \( (1111)_{8} \) = 585 in the denary system.

Teacher's Note:
1. Place values in base 8 are powers of 8, starting with 1 at the units place.
2. Do not read the digits as a denary number.

Answer: (a) 17

Explanation:
1. Multiply each digit by the power of 3 for its place: \( 1\times3^2 + 2\times3^1 + 2\times3^0 \).
2. The sum is 17.
3. So \( (122)_{3} \) = 17 in the denary system.

Teacher's Note:
1. Place values in base 3 are powers of 3, starting with 1 at the units place.
2. Do not read the digits as a denary number.

Answer: (b) \( (11101)_{2} \)

Explanation:
1. Divide repeatedly by 2 and note the remainders from the bottom up. The remainders in order are 1, 0, 1, 1, 1.
2. Reading them in reverse gives 11101.
3. So 29 = \( (11101)_{2} \).

Teacher's Note:
1. Remainders are written from the last division to the first.
2. Check by converting back to base 10.

Answer: (c) \( (10112)_{3} \)

Explanation:
1. First convert to denary: \( (340)_{5} \) = 95.
2. Now divide 95 by 3 repeatedly and read the remainders upward.
3. The result is \( (10112)_{3} \), so the answer is 10112 in base 3.

Teacher's Note:
1. To change between two bases other than 10, pass through base 10.
2. Check by converting the answer back.

Answer: (a) \( (23)_{5} \)

Explanation:
1. First convert to denary: \( (1101)_{2} \) = 13.
2. Now divide 13 by 5 repeatedly and read the remainders upward.
3. The result is \( (23)_{5} \), so the answer is 23 in base 5.

Teacher's Note:
1. To change between two bases other than 10, pass through base 10.
2. Check by converting the answer back.

Answer: (a) \( (101111)_{2} \)

Explanation:
1. First convert to denary: \( (57)_{8} \) = 47.
2. Now divide 47 by 2 repeatedly and read the remainders upward.
3. The result is \( (101111)_{2} \), so the answer is 101111 in base 2.

Teacher's Note:
1. To change between two bases other than 10, pass through base 10.
2. Check by converting the answer back.

Answer: (b) \( (10000)_{5} \)

Explanation:
1. The digit 5 does not exist in base 5.
2. Adding 1 to the last digit gives 5, which is written as 0 with a carry of 1, and the carry travels through all four places.
3. So the successor is \( (10000)_5 \), which equals \( 5^4=625 \).

Teacher's Note:
1. The largest 4-digit number in base b is one less than b to the power 4.
2. Adding 1 gives 1 followed by four zeros.

Answer: (b) \( (1321)_{5} \)

Explanation:
1. In denary, \( (323)_{5} \) = 88 and \( (443)_{5} \) = 123.
2. The sum is 211.
3. In base 5 this is 1321, so the answer is \( (1321)_{5} \).

Teacher's Note:
1. Whenever a column sum reaches 5 or more, write the remainder and carry 1.
2. The denary check is a reliable way to verify.

Answer: (b) \( (10001)_{2} \)

Explanation:
1. In denary, \( (1011)_{2} \) = 11 and \( (110)_{2} \) = 6.
2. The sum is 17.
3. In base 2 this is 10001, so the answer is \( (10001)_{2} \).

Teacher's Note:
1. Whenever a column sum reaches 2 or more, write the remainder and carry 1.
2. The denary check is a reliable way to verify.

Answer: (b) \( (10101111)_{2} \)

Explanation:
1. \( (111)_2=7 \) and \( (221)_3=2\times9+2\times3+1=25 \).
2. The product is \( 7\times25=175 \).
3. In base 2 this is 10101111.

Teacher's Note:
1. Convert each factor to denary, multiply, then convert to the required base.
2. The bases of the factors may differ.

Answer: (d) 16

Explanation:
1. Hexadecimal uses sixteen distinct symbols, namely 0 to 9 and A to F.
2. The number of distinct symbols equals the base.
3. So the base is 16.

Teacher's Note:
1. The base of a system equals the number of symbols it uses.
2. The letters A to F stand for 10 to 15.

Answer: (a) 1425

Explanation:
1. Base 5 allows only the digits 0 to 4.
2. The numeral 1425 contains the digit 5, which is not allowed.
3. So 1425 is not a valid base-5 numeral.

Teacher's Note:
1. Every digit must be smaller than the base.
2. Scan each numeral for its largest digit.

Answer: (a) 3

Explanation:
1. The ternary system has base 3 and uses the symbols 0, 1 and 2.
2. So it uses 3 symbols.

Teacher's Note:
1. Ternary means base 3 and quinary means base 5.
2. The symbols are 0 up to one less than the base.