CBSE Class 10 Mathematics Real Numbers Assignment Set 12

Very Short Answer Type Questions (1 Mark Each)

Question 1. Find the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively.
Answer:
The required number divides \( 70 - 5 = 65 \) and \( 125 - 8 = 117 \) exactly, so it is their H.C.F.
\( 65 = 5 \times 13 \) and \( 117 = 9 \times 13 \), so H.C.F. = 13.
In simple words: Subtract the remainders, then take the H.C.F.

Teacher's Note:
1. The divisor must also be larger than the remainders (13 > 8).
2. Same method works for any remainders.

Question 2. If two positive integers a and b are written as \( a = x^3y^2 \) and \( b = xy^3 \), where x, y are prime numbers, then find H.C.F.(a, b).
Answer:
H.C.F. takes the smallest power of each common prime: \( x^1 \) and \( y^2 \).
H.C.F.(a, b) = \( xy^2 \).
In simple words: Take the lower power of each prime.

Teacher's Note:
1. H.C.F. uses the smallest exponent.
2. L.C.M. uses the greatest exponent.

Question 3. If two positive integers p and q can be expressed as \( p = ab^2 \) and \( q = a^3b \), a, b being prime numbers, then find L.C.M.(p, q).
Answer:
L.C.M. takes the greatest power of each prime: \( a^3 \) and \( b^2 \).
L.C.M.(p, q) = \( a^3b^2 \).
In simple words: Take the higher power of each prime.

Teacher's Note:
1. Check: H.C.F. × L.C.M. = \( ab \times a^3b^2 = a^4b^3 = pq \).
2. Powers are compared prime by prime.

Question 4. Find the least number that is divisible by all the numbers from 1 to 10 (both inclusive).
Answer:
Required number = L.C.M.(1, 2, ..., 10). Highest powers of primes up to 10: \( 2^3, 3^2, 5, 7 \).
L.C.M. = \( 8 \times 9 \times 5 \times 7 = \) 2520.
In simple words: Take the highest power of every prime up to 10.

Teacher's Note:
1. Only primes 2, 3, 5, 7 occur.
2. Use 2³ because 8 is in the list, and 3² because 9 is.

Question 5. The decimal expansion of the rational number \( \dfrac{14587}{1250} \) will terminate after how many decimal places?
Answer:
\( 1250 = 2 \times 5^4 \). Multiply numerator and denominator by \( 2^3 \): \( \dfrac{14587 \times 8}{2^4 \times 5^4} = \dfrac{116696}{10^4} = 11.6696 \).
The expansion terminates after 4 decimal places (the larger of the powers of 2 and 5).
In simple words: The number of places is max(m, n) in 2^m 5^n.

Teacher's Note:
1. Here max(1, 4) = 4.
2. 14587 has no factor 2 or 5 in common with 1250, so the fraction is already in lowest terms.

Question 6. Can two numbers have 18 as their H.C.F. and 380 as their L.C.M.? Give reasons.
Answer:
The H.C.F. of two numbers must divide their L.C.M. Here \( 380 \div 18 \) is not an integer (18 does not divide 380, since 380 is not a multiple of 9).
Hence No, two numbers cannot have H.C.F. 18 and L.C.M. 380.
In simple words: H.C.F. must always divide L.C.M.

Teacher's Note:
1. Every H.C.F. is a factor of both numbers, hence of their L.C.M.
2. 380 = 2² × 5 × 19 has no factor 3.

Short Answer Type Questions (2 Marks Each)

Question 7. 'The product of two consecutive positive integers is divisible by 2.' Is this statement true or false? Give reasons.
Answer:
Let the integers be n and n + 1. One of any two consecutive integers is even, so the product \( n(n+1) \) contains the factor 2.
Hence the statement is true.
In simple words: One of two consecutive numbers is always even.

Teacher's Note:
1. If n is even, n is a multiple of 2; if n is odd, n + 1 is.
2. The product of two consecutive integers is always even.

Question 8. Prove that \( \sqrt{3} + \sqrt{7} \) is irrational.
Answer:
Suppose \( \sqrt{3} + \sqrt{7} = r \) is rational. Squaring, \( 10 + 2\sqrt{21} = r^2 \), so \( \sqrt{21} = \dfrac{r^2 - 10}{2} \) is rational.
But 21 is not a perfect square, so \( \sqrt{21} \) is irrational, a contradiction. Hence \( \sqrt{3} + \sqrt{7} \) is irrational.
In simple words: Squaring reduces it to a single square root.

Teacher's Note:
1. The square root of a non-perfect-square integer is irrational.
2. Always state the contradiction.

Question 9. Explain why 3 × 5 × 7 + 7 is a composite number.
Answer:
\( 3 \times 5 \times 7 + 7 = 7(15 + 1) = 7 \times 16 = 112 \).
112 has factors other than 1 and itself (7 and 16), so it is composite.
In simple words: Take 7 common and see the product.

Teacher's Note:
1. A number with more than two factors is composite.
2. 112 = 2⁴ × 7.

Question 10. Without actually performing the long division, find if \( \dfrac{987}{10500} \) will have terminating or non-terminating (repeating) decimal expansion. Give reasons.
Answer:
\( 987 = 3 \times 7 \times 47 \) and \( 10500 = 2^2 \times 3 \times 5^3 \times 7 \). So \( \dfrac{987}{10500} = \dfrac{47}{500} = \dfrac{47}{2^2 \times 5^3} \).
The denominator is of the form \( 2^m \times 5^n \), so the decimal expansion is terminating (0.094).
In simple words: Reduce the fraction first, then look at the denominator.

Teacher's Note:
1. Always reduce to lowest terms before applying the test.
2. 47/500 = 0.094.

Question 11. A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form p/q?
Answer:
327.7081 terminates, so it is rational and \( 327.7081 = \dfrac{3277081}{10^4} \).
In lowest terms, q is of the form \( 2^m \times 5^n \): the prime factors of q are only 2 and/or 5.
In simple words: A terminating decimal has only 2 and 5 in the denominator.

Teacher's Note:
1. Four decimal places means q divides 10⁴.
2. Here q = 10⁴ = 2⁴ × 5⁴ since 3277081 is odd and not a multiple of 5.

Long Answer Type Questions (3 Marks Each)

Question 12. Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.
Answer:
Let a be a positive integer, \( a = 4q + r \) with \( r = 0, 1, 2, 3 \).
r = 0: \( a^3 = 64q^3 = 4(16q^3) = 4m \).
r = 1: \( a^3 = 64q^3 + 48q^2 + 12q + 1 = 4(16q^3 + 12q^2 + 3q) + 1 = 4m + 1 \).
r = 2: \( a^3 = 64q^3 + 96q^2 + 48q + 8 = 4(16q^3 + 24q^2 + 12q + 2) = 4m \).
r = 3: \( a^3 = 64q^3 + 144q^2 + 108q + 27 = 4(16q^3 + 36q^2 + 27q + 6) + 3 = 4m + 3 \).
So every cube is of the form 4m, 4m + 1 or 4m + 3.
In simple words: Test the four remainders 0, 1, 2, 3.

Teacher's Note:
1. Expand (4q + r)³ and take 4 common.
2. The form 4m + 2 never occurs for a cube.

Question 13. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.
Answer:
Any positive integer is \( 5m, 5m+1, 5m+2, 5m+3 \) or \( 5m+4 \).
\( (5m)^2 = 5(5m^2) \); \( (5m+1)^2 = 5(5m^2+2m) + 1 \); \( (5m+2)^2 = 5(5m^2+4m) + 4 \); \( (5m+3)^2 = 5(5m^2+6m+1) + 4 \); \( (5m+4)^2 = 5(5m^2+8m+3) + 1 \).
So a square is of the form 5q, 5q + 1 or 5q + 4, never 5q + 2 or 5q + 3.
In simple words: List the squares for all five remainders.

Teacher's Note:
1. Squares leave remainder 0, 1 or 4 on division by 5.
2. Use this to rule out numbers as perfect squares.

Question 14. If n is an odd integer, then show that \( n^2 - 1 \) is divisible by 8.
Answer:
An odd integer is \( n = 4q + 1 \) or \( 4q + 3 \).
If \( n = 4q+1 \): \( n^2 - 1 = 16q^2 + 8q = 8q(2q+1) \), divisible by 8.
If \( n = 4q+3 \): \( n^2 - 1 = 16q^2 + 24q + 8 = 8(2q^2 + 3q + 1) \), divisible by 8.
Hence \( n^2 - 1 \) is divisible by 8 for every odd n.
In simple words: Write odd numbers as 4q + 1 and 4q + 3.

Teacher's Note:
1. Alternative: n² − 1 = (n − 1)(n + 1), a product of two consecutive even numbers.
2. One of them is a multiple of 4, the other of 2.

Question 15. Using Euclid's division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.
Answer:
The number divides \( 1251 - 1 = 1250 \), \( 9377 - 2 = 9375 \) and \( 15628 - 3 = 15625 \) exactly. Find H.C.F.(1250, 9375, 15625).
\( 9375 = 1250 \times 7 + 625 \), \( 1250 = 625 \times 2 + 0 \), so H.C.F.(1250, 9375) = 625.
\( 15625 = 625 \times 25 + 0 \), so 625 divides 15625.
The required number is 625.
In simple words: Subtract the remainders, then find the H.C.F. of three numbers.

Teacher's Note:
1. H.C.F.(a, b, c) = H.C.F.(H.C.F.(a, b), c).
2. Check that 625 is larger than every remainder.

Question 16. Prove that \( \sqrt{7} \) is irrational.
Answer:
Suppose \( \sqrt{7} = \dfrac{a}{b} \), where a and b are co-prime integers and \( b \ne 0 \). Then \( a^2 = 7b^2 \), so 7 divides \( a^2 \) and hence 7 divides a. Let a = 7c.
Then \( 49c^2 = 7b^2 \), so \( b^2 = 7c^2 \), and 7 divides b. So 7 divides both a and b, contradicting that they are co-prime. Hence \( \sqrt{7} \) is irrational.
In simple words: Assume it is rational and reach a contradiction.

Teacher's Note:
1. Uses: if prime p divides a², then p divides a.
2. The same proof works for the square root of any prime.

Question 17. Show that \( 12^n \) cannot end with the digit 0 or 5 for any natural number n.
Answer:
\( 12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n \). The only primes in its factorisation are 2 and 3.
A number ending in 0 or 5 is divisible by 5, so 5 would have to occur in the factorisation. By the uniqueness of prime factorisation it does not, so \( 12^n \) never ends in 0 or 5.
In simple words: Numbers ending in 0 or 5 are multiples of 5.

Teacher's Note:
1. Use the Fundamental Theorem of Arithmetic.
2. Same idea shows 6ⁿ or 4ⁿ never ends in 0.

Question 18. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
Answer:
Minimum distance = L.C.M.(40, 42, 45). \( 40 = 2^3 \times 5 \), \( 42 = 2 \times 3 \times 7 \), \( 45 = 3^2 \times 5 \).
L.C.M. = \( 2^3 \times 3^2 \times 5 \times 7 = 2520 \) cm = 25.2 m.
In simple words: Complete steps for all means a common multiple, the least one is the L.C.M.

Teacher's Note:
1. Convert at the end if needed: 2520 cm = 25 m 20 cm.
2. Each person takes 63, 60 and 56 steps respectively.

Very Long Answer Type Questions (4 Marks Each)

Question 19. Show that the cube of a positive integer of the form \( 6q + r \), q is an integer and r = 0, 1, 2, 3, 4, 5, is also of the form \( 6m + r \).
Answer:
\( (6q + r)^3 = 216q^3 + 108q^2 r + 18qr^2 + r^3 = 6(36q^3 + 18q^2 r + 3qr^2) + r^3 \).
Now \( r^3 - r = r(r-1)(r+1) \) is a product of three consecutive integers, hence divisible by 6 (shown by checking r = 0,...,5: 0, 0, 6, 24, 60, 120). So \( r^3 = 6k + r \).
Therefore \( (6q+r)^3 = 6(36q^3 + 18q^2 r + 3qr^2 + k) + r = 6m + r \).
In simple words: Expand and show r³ − r is a multiple of 6.

Teacher's Note:
1. r³ − r = (r − 1) r (r + 1), three consecutive integers.
2. Direct check: 0³=0, 1³=1, 2³=8=6+2, 3³=27=24+3, 4³=64=60+4, 5³=125=120+5.

Question 20. Prove that one of any three consecutive positive integers must be divisible by 3.
Answer:
Let the integers be n, n + 1, n + 2 and write n = 3q + r, r = 0, 1, 2.
r = 0: n = 3q is divisible by 3. r = 1: \( n + 2 = 3q + 3 = 3(q+1) \) is divisible by 3. r = 2: \( n + 1 = 3q + 3 = 3(q+1) \) is divisible by 3.
In every case one of the three numbers is divisible by 3.
In simple words: Consider remainders 0, 1, 2 of n.

Teacher's Note:
1. This is why n(n + 1)(n + 2) is always a multiple of 3 (indeed of 6).
2. Used in many divisibility proofs.

Question 21. Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.
Answer:
Write n = 5q + r with r = 0, 1, 2, 3, 4.
r = 0: n is divisible by 5 (n + 4, n + 8, n + 12, n + 16 leave remainders 4, 3, 2, 1).
r = 1: \( n + 4 = 5q + 5 \) is divisible by 5 and no other one is (others leave 1, 4, 3, 2).
r = 2: \( n + 8 = 5q + 10 \). r = 3: \( n + 12 = 5q + 15 \). r = 4: \( n + 16 = 5q + 20 \). In each case exactly one of the five numbers is a multiple of 5.
In simple words: The five numbers have five different remainders mod 5.

Teacher's Note:
1. n, n + 4, ..., n + 16 leave remainders r, r + 4, r + 3, r + 2, r + 1 (mod 5).
2. These are all different, so exactly one is 0.