JEE Mathematics Number Systems MCQs Set 03

Answer: (b) The distance between them is \(|b-a|\)

Explanation:
1. If \(a>b\), the distance is \(a-b\); if \(a\lt b\), the distance is \(b-a\).
2. In both cases it equals \(|b-a|\) (which is also \(|a-b|\)).
3. The other expressions fail: for \(a=-3,\ b=2\) the distance is 5, but \(|a|-|b|=1\).

Teacher's Note:
1. The distance between two points is always non-negative.
2. Absolute value turns a difference into a distance.

Answer: (a) If \(a\lt b\), it is not always true that \(|a|\lt|b|\); for example \(a=-3,\ b=2\)

Explanation:
1. Take \(a=-3\) and \(b=2\). Then \(a\lt b\).
2. But \(|a|=3\) and \(|b|=2\), so \(|a|\gt|b|\).
3. Hence \(a\lt b\) does not imply \(|a|\lt|b|\); a single counterexample is enough.

Teacher's Note:
1. Absolute value ignores sign, so order is not preserved for negative numbers.
2. One counterexample disproves a general statement.

Answer: (d) Two numbers neither of which is divisible by 5 can have a sum divisible by 5, for example 18 and 7

Explanation:
1. Take 18 and 7. Neither is divisible by 5.
2. Their sum is 25, which is divisible by 5.
3. So the claim that the sum cannot be divisible by 5 is false.

Teacher's Note:
1. Remainders add: 3 + 2 = 5 gives a multiple of 5.
2. Look for numbers with remainders 1 and 4, or 2 and 3.

Answer: (c) neither number need be divisible by 5; for example, 12 + 13 = 25

Explanation:
1. Take 12 and 13. Their sum 25 is divisible by 5.
2. Neither 12 nor 13 is divisible by 5, so the statement 'each must be divisible' is false.
3. The remainders are 2 and 3, not 1, and neither number is a multiple of 5, so the other three statements are all false for this pair.

Teacher's Note:
1. Use remainders: the remainders of the two numbers must add up to 0 or 5.
2. A single counterexample is enough to reject an 'always' claim.

Answer: (b) The sum of two numbers can fail to be divisible by 5 even though one of them is divisible by 5; for example 10 + 7 = 17

Explanation:
1. Take 10 and 7. The sum 17 is not divisible by 5.
2. But 10 is divisible by 5, so 'neither number is divisible by 5' is false.
3. So the first statement is correct. The others are false: 10 is a multiple of 5 but 10 + 7 is not, and 10 and 7 are not both multiples of 5.

Teacher's Note:
1. A multiple of 5 added to a non-multiple is never a multiple of 5, so this example is consistent.
2. Test statements with small numbers.

Answer: (a) at least one of the numbers is not divisible by 5

Explanation:
1. If both numbers were divisible by 5, their sum would be divisible by 5.
2. So at least one of them is not divisible by 5.
3. The other options fail: \(15+8=23\) has exactly one multiple of 5, but \(12+7=19\) has none, so neither 'exactly one' nor 'neither' always holds.

Teacher's Note:
1. This is the contrapositive of: if both are divisible by 5, so is the sum.
2. Contrapositive statements are always logically equivalent.

Answer: (d) \(u\) is a multiple of 4

Explanation:
1. \(u\) is even only if \(x,y\) are both even or both odd.
2. Then \(x+y\) and \(x-y\) are both even, so \(u=(x+y)(x-y)\) is a multiple of \(2\times2=4\).
3. It need not be a multiple of 8 or 6: \(x=2,\ y=0\) gives \(u=4\).

Teacher's Note:
1. Difference of two squares with the same parity is a multiple of 4.
2. If x and y have different parity, u is odd.

Answer: (c) an even multiple of 3, that is a multiple of 6

Explanation:
1. One of the three consecutive numbers \(n-1,\ n,\ n+1\) is divisible by 3; the two primes greater than 5 are not, so \(3\mid n\).
2. Both primes are odd, so \(n\) is even.
3. Hence \(n\) is a multiple of 6. Examples: \(n=12\) (11, 13) and \(n=30\) (29, 31). Neither 12 nor 30 is a multiple of both 5 and 4.

Teacher's Note:
1. Twin primes beyond 3 and 5 are of the form 6k plus or minus 1.
2. Test with n = 12 and n = 30.

Answer: (b) Zero is neither positive nor negative

Explanation:
1. A number is positive if it is greater than 0 and negative if it is less than 0.
2. Zero is neither greater nor less than itself, so it is neither.

Teacher's Note:
1. Zero is the boundary between positive and negative numbers.
2. Zero is a whole number, an integer and a rational number.

Answer: (a) \(a=0\) and \(b=0\)

Explanation:
1. Both \(|a|\ge0\) and \(|b|\ge0\).
2. A sum of two non-negative numbers is zero only if both are zero.
3. So \(a=0\) and \(b=0\).

Teacher's Note:
1. Absolute values are never negative.
2. The same idea gives: a squared plus b squared equals 0 implies a = b = 0.

Answer: (d) \((51)_8\)

Explanation:
1. \(41=8\times5+1\).
2. So the digits are 5 and 1, giving \((51)_8\).
3. Check: \((51)_8=5\times8+1=41\).

Teacher's Note:
1. Divide by 8 and read the quotient then remainder.
2. Octal digits go from 0 to 7.

Answer: (c) \((10212)_3\)

Explanation:
1. \((404)_5=4\cdot25+0+4=104\).
2. \(104=81+18+3+2=1\cdot3^4+0\cdot3^3+2\cdot3^2+1\cdot3+2\), so it is \((10212)_3\).

Teacher's Note:
1. Convert through base 10.
2. Check: 81 + 0 + 18 + 3 + 2 = 104.

Answer: (b) \(A-B\) and \(B-A\) are generally different; for \(A=\{1,2,3,4\},\ B=\{3,4,5,6\}\) they are \(\{1,2\}\) and \(\{5,6\}\)

Explanation:
1. For \(A=\{1,2,3,4\}\) and \(B=\{3,4,5,6\}\): \(A-B=\{1,2\}\) and \(B-A=\{5,6\}\).
2. These are not equal, so set difference is not commutative.
3. If \(A\subset B\) properly, \(A-B=\varnothing\) but \(B-A\ne\varnothing\).

Teacher's Note:
1. Set difference is not commutative.
2. A-B and B-A are always disjoint.

Answer: (a) 0

Explanation:
1. For all real \(x\), \(|x-1|\ge0\), so \(|x-1|+2\ge2\).
2. Hence \(\big||x-1|+2\big|=|x-1|+2\ge2>1\).
3. The equation cannot hold, so there is no real solution.

Teacher's Note:
1. Find the minimum value of the inner expression first.
2. A quantity that is at least 2 can never equal 1.

Answer: (d) both their sum and their difference

Explanation:
1. Let the numbers be \(am\) and \(bm\).
2. Their sum is \((a+b)m\) and their difference is \((a-b)m\), both multiples of \(m\).
3. So \(m\) divides both.

Teacher's Note:
1. The same factor can be taken out of a sum or a difference.
2. This is the basis of the Euclidean algorithm.

Answer: (c) always odd

Explanation:
1. Odd numbers have the form \(2m+1\) and \(2n+1\).
2. Their product is \((2m+1)(2n+1)=4mn+2m+2n+1=2(2mn+m+n)+1\).
3. This is of the form \(2k+1\), so it is odd.

Teacher's Note:
1. Odd times odd is odd; even times anything is even.
2. Write numbers in the forms 2k and 2k plus 1.

Answer: (b) \(1\dfrac14\)

Explanation:
1. \(1\dfrac{9}{16}=\dfrac{25}{16}\).
2. \(\sqrt{25/16}=\dfrac54=1\dfrac14\).

Teacher's Note:
1. Convert the mixed number to an improper fraction before taking the root.
2. The wrong answer 1 3/4 comes from taking the root of 1 and 9/16 separately.

Answer: (a) composite

Explanation:
1. The three consecutive numbers \(2^n-1,\ 2^n,\ 2^n+1\) include exactly one multiple of 3.
2. \(2^n\) is not a multiple of 3, so one of \(2^n\pm1\) is a multiple of 3, and it exceeds 3 when \(n>2\).
3. If the other is prime, the multiple of 3 must be composite. For \(n=4\): 15 is composite and 17 is prime.

Teacher's Note:
1. Among three consecutive integers one is a multiple of 3.
2. n greater than 2 ensures both numbers exceed 3.

Answer: (d) a multiple of 11

Explanation:
1. Let the digits be \(a,\ a+c,\ c\). The number is \(100a+10(a+c)+c=110a+11c\).
2. \(110a+11c=11(10a+c)\), a multiple of 11.
3. It need not be a multiple of 9, 7 or 13: \(132=11\times12\) fails all three.

Teacher's Note:
1. Divisibility by 11: alternate digit sums differ by a multiple of 11.
2. Here a + c minus the middle digit equals 0.

Answer: (c) 0 or \(-1\)

Explanation:
1. A fraction is undefined when its denominator is zero.
2. \(x(x+1)=0\) gives \(x=0\) or \(x=-1\).

Teacher's Note:
1. Set the denominator equal to zero.
2. The domain is all real numbers except 0 and -1.

Answer: (b) 26

Explanation:
1. \((222)_3=2\cdot9+2\cdot3+2\).
2. \(=18+6+2=26=3^3-1\).

Teacher's Note:
1. Three 2's in base 3 equal 3^3 - 1.
2. Use place values 9, 3, 1.

Answer: (a) \((1122)_3\)

Explanation:
1. \(44=3\times14+2\), \(14=3\times4+2\), \(4=3\times1+1\), \(1=3\times0+1\).
2. The remainders read upward give \((1122)_3\). Check: \(27+9+6+2=44\).

Teacher's Note:
1. Repeated division by 3, remainders read from the bottom up.
2. Verify by expansion.

Answer: (d) \((100)_5\)

Explanation:
1. \(25=5^2\), so it is one 25, no 5's and no units.
2. Hence \(25=(100)_5\).

Teacher's Note:
1. Any power of the base b is written as 1 followed by zeros in base b.
2. Quinary digits are 0 to 4.

Answer: (c) 455

Explanation:
1. \((707)_8=7\cdot64+0\cdot8+7\).
2. \(=448+7=455\).

Teacher's Note:
1. Place values in base 8 are 1, 8, 64.
2. Octal numbers contain only the digits 0 to 7.

Answer: (b) \((3222)_5\)

Explanation:
1. \((34)_5=19\) and \((43)_5=23\), so the product is \(437\).
2. \(437=3\cdot125+2\cdot25+2\cdot5+2=(3222)_5\).

Teacher's Note:
1. Convert to base 10, multiply, and convert back.
2. Digits in base 5 never exceed 4.