CBSE Class 10 Mathematics Real Numbers Assignment Set 11

Long Answer Type Questions

Question 1. If a, b, c, d are positive rationals such that \( a + \sqrt{b} = c + \sqrt{d} \), then prove that either a = c and b = d or b and d are squares of rationals.
Answer:
Case 1: a = c. Then \( \sqrt{b} = \sqrt{d} \), so b = d.
Case 2: a ≠ c. Then a − c = k is a non-zero rational number, and \( a = c + k \). Substituting, \( c + k + \sqrt{b} = c + \sqrt{d} \), so \( k + \sqrt{b} = \sqrt{d} \). ... (i)
Squaring both sides: \( k^2 + 2k\sqrt{b} + b = d \), so \( \sqrt{b} = \dfrac{d - k^2 - b}{2k} \).
Since d, k and b are rational, the right-hand side is rational. So \( \sqrt{b} \) is rational, which means b is the square of a rational number.
From (i), \( \sqrt{d} = k + \sqrt{b} \) is a sum of rationals, so \( \sqrt{d} \) is rational and d is the square of a rational number.
Hence either a = c and b = d, or b and d are squares of rational numbers.
In simple words: Either the rational parts and the root parts match, or both roots are themselves rational.

Teacher's Note:
1. Split the proof into the cases a = c and a ≠ c.
2. In the second case isolate the root and square it.

Question 2. Prove that \( \sqrt{7} + \sqrt{11} \) is an irrational number.
Answer:
Assume, to the contrary, that \( \sqrt{7} + \sqrt{11} = \frac{a}{b} \), where a and b are integers and b ≠ 0.
Squaring both sides: \( 7 + 11 + 2\sqrt{77} = \frac{a^2}{b^2} \), so \( \sqrt{77} = \dfrac{1}{2}\left(\dfrac{a^2}{b^2} - 18\right) = \dfrac{a^2 - 18b^2}{2b^2} \).
Since a and b are integers, \( \frac{a^2 - 18b^2}{2b^2} \) is rational, so \( \sqrt{77} \) would be rational.
But \( \sqrt{77} \) is irrational (77 = 7 × 11 is not a perfect square), a contradiction.
Hence \( \sqrt{7} + \sqrt{11} \) is irrational.
In simple words: If the sum were rational, its square would force √77 to be rational.

Teacher's Note:
1. Square the assumed rational sum.
2. Isolate the single remaining root \( \sqrt{77} \).

Question 3. Show that (n² − 1) is divisible by 8, if n is an odd integer.
Answer:
Any odd positive integer is of the form 4q + 1 or 4q + 3 (Euclid's lemma with b = 4).
Case I: n = 4q + 1. \( n^2 - 1 = (4q+1)^2 - 1 = 16q^2 + 8q = 8q(2q + 1) = 8r \), where r = q(2q + 1) is an integer.
Case II: n = 4q + 3. \( n^2 - 1 = (4q+3)^2 - 1 = 16q^2 + 24q + 8 = 8(2q^2 + 3q + 1) = 8(2q + 1)(q + 1) = 8r \), where r = (2q + 1)(q + 1) is an integer.
In both cases n² − 1 is divisible by 8.
In simple words: Split odd numbers as 4q + 1 and 4q + 3, then take out 8.

Teacher's Note:
1. Odd numbers leave remainder 1 or 3 on division by 4.
2. A quick check: 3² − 1 = 8, 5² − 1 = 24.

Question 4. Show that only one of the numbers n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.
Answer:
By Euclid's lemma, n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4. Consider each case.
n = 5q: n is divisible by 5. The other numbers leave remainders 4, 3 (since 5q + 8 = 5(q + 1) + 3), 2 (5q + 12 = 5(q + 2) + 2) and 1 (5q + 16 = 5(q + 3) + 1). So only n is divisible by 5.
n = 5q + 1: n + 4 = 5q + 5 = 5(q + 1) is divisible by 5; n, n + 8, n + 12, n + 16 leave remainders 1, 4, 3, 2.
n = 5q + 2: n + 8 = 5q + 10 = 5(q + 2) is divisible by 5; the others leave remainders 2, 1, 4, 3.
n = 5q + 3: n + 12 = 5q + 15 = 5(q + 3) is divisible by 5; the others leave remainders 3, 2, 1, 4.
n = 5q + 4: n + 16 = 5q + 20 = 5(q + 4) is divisible by 5; the others leave remainders 4, 3, 2, 1.
In every case exactly one of the five numbers is divisible by 5.
In simple words: The five numbers leave five different remainders, so exactly one leaves 0.

Teacher's Note:
1. The numbers n, n + 4, n + 8, n + 12, n + 16 leave all five different remainders on division by 5.
2. Check with n = 1: 1, 5, 9, 13, 17 and only 5 is divisible by 5.

Question 5. For every positive integer n, prove that n² − n is divisible by 2.
Answer:
Case I: n is even, n = 2q. Then n² − n = 4q² − 2q = 2q(2q − 1) = 2r, where r = q(2q − 1) is an integer.
Case II: n is odd, n = 2q + 1. Then n² − n = (2q + 1)² − (2q + 1) = 4q² + 2q = 2q(2q + 1) = 2r, where r = q(2q + 1) is an integer.
So n² − n is divisible by 2 for every positive integer n.
In simple words: n² − n = n(n − 1) is a product of two consecutive integers, so it is even.

Teacher's Note:
1. Treat even n and odd n separately.
2. The product of two consecutive integers is always even.

Integer Type Questions

Question 6. How many positive integers n are there for which \( \sqrt{n-1} + \sqrt{n+1} \) is rational?
Answer:
Suppose \( \sqrt{n-1} + \sqrt{n+1} = r \) is rational (r > 0). Then \( \sqrt{n+1} - \sqrt{n-1} = \dfrac{(n+1) - (n-1)}{r} = \dfrac{2}{r} \) is also rational.
Adding and subtracting, both \( \sqrt{n+1} \) and \( \sqrt{n-1} \) are rational, so n − 1 and n + 1 would both be perfect squares of integers. Two perfect squares never differ by 2 (the squares 0, 1, 4, 9, … differ by 1, 3, 5, …).
For n = 1 the expression is \( 0 + \sqrt{2} \), which is irrational. So no positive integer works.
Answer: 0
In simple words: No positive integer makes the sum rational.

Teacher's Note:
1. Use the difference of the two roots, which is 2 divided by the sum.
2. Check n = 1 separately.

Question 7. What will be the last digit of the expression \( 4 + 9^2 + 4^3 + 9^4 + 4^5 + 9^6 + \cdots + 4^{99} + 9^{100} \)?
Answer:
Odd powers of 4 end in 4 and even powers of 9 end in 1, so each pair (4^odd + 9^even) ends in 4 + 1 = 5.
There are 50 such pairs, from (4 + 9²) to (4⁹⁹ + 9¹⁰⁰).
The last digit of the sum is that of 50 × 5 = 250, which is 0.
In simple words: Each pair ends in 5, and 50 fives end in 0.

Teacher's Note:
1. Find the repeating last digits of 4ⁿ and 9ⁿ.
2. Add the last digits of the 50 pairs.

Question 8. Find the HCF of 96 and 294, using Euclid's division algorithm.
Answer:
294 = 96 × 3 + 6
96 = 6 × 16 + 0
The last divisor is 6, so HCF (96, 294) = 6.
In simple words: Divide repeatedly until the remainder is 0.

Teacher's Note:
1. Start by dividing the larger number by the smaller.
2. The last non-zero remainder, or the last divisor, is the HCF.

Question 9. A number on being divided by 342 leaves a remainder 47. What would be the remainder if the same number is divided by 19?
Answer:
N = 342q + 47 = 342q + 38 + 9 = 19(18q + 2) + 9.
Since 342 = 19 × 18 and 47 = 19 × 2 + 9, the remainder on division by 19 is 9.
In simple words: 342 is a multiple of 19, so only the remainder 47 matters.

Teacher's Note:
1. Write 47 in terms of 19 as 2 × 19 + 9.
2. Check with N = 47: 47 ÷ 19 leaves 9.

Question 10. Two runners P and Q stop after 3 hrs and 5 hrs respectively while running. After ab hours (where ab is a two-digit number) both will pause together for the first time, if they started running at the same time. Find the value of a + b.
Answer:
They pause together after the LCM of 3 and 5 hours: LCM (3, 5) = 15 hours.
So the two-digit number ab = 15, giving a = 1 and b = 5.
a + b = 6.
In simple words: They meet again after the LCM, 15 hours.

Teacher's Note:
1. Here ab means the digits a and b, not a × b.
2. The LCM of two co-prime numbers is their product.

Question 11. If the L.C.M and H.C.F of two numbers are 168 and 28 respectively, find the number of possible such pairs.
Answer:
Let the numbers be 28x and 28y, where x and y are co-prime. Then LCM × HCF = product, so 168 × 28 = 28x × 28y, giving xy = 6.
Co-prime pairs with product 6: (1, 6) and (2, 3). The pairs of numbers are (28, 168) and (56, 84).
2 pairs
In simple words: Write the numbers as HCF times co-prime numbers and use the product rule.

Teacher's Note:
1. x and y must be co-prime, so (1, 6) and (2, 3) are the only choices.
2. Both pairs give LCM 168.

Question 12. Find the total number of factors of 3025.
Answer:
\( 3025 = 5^2 \times 11^2 \).
Number of factors = (2 + 1)(2 + 1) = 9.
In simple words: Add 1 to each power and multiply.

Teacher's Note:
1. Factorise 3025 = 55².
2. A perfect square always has an odd number of factors.

Question 13. Find the HCF of the first 1000 even natural numbers.
Answer:
The first 1000 even natural numbers are 2, 4, 6, …, 2000. Every one of them is a multiple of 2, and 2 itself is in the list.
HCF = 2
In simple words: 2 divides all of them and is one of them.

Teacher's Note:
1. The HCF cannot exceed the smallest number, 2.
2. 2 divides every even number.

Question 14. Find the number of possible values of z, if a is a natural number, and \( 8^a - 7^a \) ends with the digit z.
Answer:
The last digit of 8^a repeats as 8, 4, 2, 6 and that of 7^a as 7, 9, 3, 1 with period 4.
For a = 1, 2, 3, 4: 8 − 7 = 1, 64 − 49 = 15, 512 − 343 = 169 and 4096 − 2401 = 1695, so the last digits are 1, 5, 9, 5. Because both sets of last digits repeat with period 4, this pattern repeats.
Possible values of z: 1, 5 and 9, so there are 3 possible values.
In simple words: Last digits repeat every 4 powers, giving only 1, 5, 9.

Teacher's Note:
1. Check a = 1 to 4 and see the pattern repeat.
2. The set of possible digits is {1, 5, 9}.

Question 15. What is the minimum possible sum of the two numbers, if the L.C.M and the H.C.F of the two numbers are 924 and 7 respectively?
Answer:
Let the numbers be 7a and 7b, where a and b are co-prime. LCM × HCF = product, so 924 × 7 = 49ab and ab = 132.
Co-prime pairs with product 132: (1, 132), (3, 44), (4, 33) and (11, 12). The sums a + b are 133, 47, 37 and 23.
Minimum sum of the numbers = 7 × 23 = 7 × 11 + 7 × 12 = 77 + 84 = 161.
In simple words: Use the co-prime pair with the smallest sum.

Teacher's Note:
1. Factorise 132 = 2² × 3 × 11 to list co-prime pairs.
2. Numbers closest to each other give the smallest sum.