NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 1.1
Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Answer
(i) 225 = 135 × 1 + 90
⇒ 135 = 90 × 1 + 45
⇒ 90 = 45 × 2 + 0
⇒ HCF = 45
(ii) 38220 = 196 × 195 + 0 → HCF = 196
(iii) 867 = 255 × 3 + 102
⇒ 255 = 102 × 2 + 51
⇒ 102 = 51 × 2 + 0
⇒ HCF = 51
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Answer
By Euclid's division lemma, any positive integer a = 6q + r, where 0 ≤ r < 6. So r = 0, 1, 2, 3, 4, or 5.
a = 6q (even), 6q+1 (odd), 6q+2 (even), 6q+3 (odd), 6q+4 (even), 6q+5 (odd).
Since even values of r give even numbers, any positive odd integer must be of the form 6q + 1, 6q + 3, or 6q + 5.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer
Maximum number of columns = HCF(616, 32)
616 = 32 × 19 + 8
⇒ 32 = 8 × 4 + 0
⇒ HCF = 8
They can march in a maximum of 8 columns.
Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Answer
Let a = 3q, 3q+1, or 3q+2 (by Euclid's division lemma with divisor 3).
a² = (3q)² = 9q² = 3(3q²) = 3m where m = 3q²
a² = (3q+1)² = 9q²+6q+1 = 3(3q²+2q)+1 = 3m+1 where m = 3q²+2q
a² = (3q+2)² = 9q²+12q+4 = 3(3q²+4q+1)+1 = 3m+1 where m = 3q²+4q+1
Hence, the square of any positive integer is of the form 3m or 3m+1.
Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Answer
Let a = 3q, 3q+1, or 3q+2.
a³ = (3q)³ = 27q³ = 9(3q³) = 9m
a³ = (3q+1)³ = 27q³+27q²+9q+1 = 9(3q³+3q²+q)+1 = 9m+1
a³ = (3q+2)³ = 27q³+54q²+36q+8 = 9(3q³+6q²+4q)+8 = 9m+8
Hence proved.
Exercise 1.2
Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Answer
(i) 140 = 2² × 5 × 7
(ii) 156 = 2² × 3 × 13
(iii) 3825 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers: (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Answer
(i) 26 = 2 × 13, 91 = 7 × 13. HCF = 13, LCM = 2 × 7 × 13 = 182. Check: 13 × 182 = 2366 = 26 × 91 ✓
(ii) 510 = 2 × 3 × 5 × 17, 92 = 2² × 23. HCF = 2, LCM = 2² × 3 × 5 × 17 × 23 = 23460. Check: 2 × 23460 = 46920 = 510 × 92 ✓
(iii) 336 = 2⁴ × 3 × 7, 54 = 2 × 3³. HCF = 2 × 3 = 6, LCM = 2⁴ × 3³ × 7 = 3024. Check: 6 × 3024 = 18144 = 336 × 54 ✓
Find the LCM and HCF of the following integers by applying the prime factorisation method: (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Answer
(i) 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. HCF = 3, LCM = 2² × 3 × 5 × 7 = 420
(ii) 17, 23, 29 are all prime numbers. HCF = 1, LCM = 17 × 23 × 29 = 11339
(iii) 8 = 2³, 9 = 3², 25 = 5². HCF = 1, LCM = 2³ × 3² × 5² = 1800
Given that HCF(306, 657) = 9, find LCM(306, 657).
Answer
LCM = (306 × 657) / HCF = (306 × 657) / 9 = 201042 / 9 = 22338
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Answer
For a number to end with 0, it must have both 2 and 5 as prime factors. 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. The prime factorisation of 6ⁿ does not contain 5 as a factor for any value of n. Therefore, 6ⁿ can never end with the digit 0.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer
7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78 = 13 × 78. Since it has factors other than 1 and itself, it is composite.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009. Since it has 5 as a factor (besides 1 and itself), it is composite.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Answer
They will meet at the starting point after LCM(18, 12) minutes.
18 = 2 × 3², 12 = 2² × 3. LCM = 2² × 3² = 36 minutes
Exercise 1.3
Prove that √5 is irrational.
Answer
Assume √5 is rational. Then √5 = a/b where a, b are coprime integers (HCF = 1).
Squaring: 5 = a²/b²
⇒ a² = 5b²
⇒ a² is divisible by 5
⇒ a is divisible by 5 (if p divides a², then p divides a, for prime p).
Let a = 5c. Then (5c)² = 5b²
⇒ 25c² = 5b²
⇒ b² = 5c²
⇒ b² is divisible by 5
⇒ b is divisible by 5.
But if both a and b are divisible by 5, they have a common factor 5, contradicting that HCF(a,b) = 1.
This contradiction proves that √5 is irrational.
Prove that 3 + 2√5 is irrational.
Answer
Assume 3 + 2√5 is rational = a/b (where a, b are integers, b ≠ 0).
Then 2√5 = a/b – 3 = (a – 3b)/b
⇒ √5 = (a – 3b)/2b
Since a and b are integers, (a – 3b)/2b is rational. But √5 is irrational (proved above).
A rational number cannot equal an irrational number — contradiction.
Hence, 3 + 2√5 is irrational.
Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
Answer
(i) Assume 1/√2 = a/b (rational). Then √2 = b/a, which is rational — contradicts that √2 is irrational. Hence 1/√2 is irrational.
(ii) Assume 7√5 = a/b (rational). Then √5 = a/7b, which is rational — contradicts that √5 is irrational. Hence 7√5 is irrational.
(iii) Assume 6 + √2 = a/b (rational). Then √2 = a/b – 6 = (a – 6b)/b, which is rational — contradicts that √2 is irrational. Hence 6 + √2 is irrational.
Exercise 1.4
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
Answer
A rational number p/q has a terminating decimal if q (in simplest form) = 2ⁿ × 5ᵐ.
(i) 13/3125 = 13/5⁵ → Terminating
(ii) 17/8 = 17/2³ → Terminating
(iii) 64/455 = 64/(5 × 7 × 13)
⇒ Non-terminating repeating (has 7, 13)
(iv) 15/1600 = 3/320 = 3/(2⁶ × 5)
⇒ Terminating
(v) 29/343 = 29/7³
⇒ Non-terminating repeating
(vi) 23/(2³ × 5²) → Terminating
(vii) 129/(2² × 5⁷ × 7⁵)
⇒ Non-terminating repeating (has 7⁵)
(viii) 6/15 = 2/5 → Terminating
(ix) 35/50 = 7/10 = 7/(2 × 5)
⇒ Terminating
(x) 77/210 = 11/30 = 11/(2 × 3 × 5)
⇒ Non-terminating repeating (has 3)
Write down the decimal expansions of those rational numbers in Q1 which have terminating decimal expansions.
Answer
(i) 13/3125 = 13/5⁵ = 13 × 2⁵/10⁵ = 416/100000 = 0.00416
(ii) 17/8 = 17/2³ = 17 × 5³/10³ = 2125/1000 = 2.125
(iv) 15/1600 = 15/(2⁶ × 5²) = 15 × 5⁴/(2⁶ × 5⁶) = 9375/10⁶ = 0.009375
(vi) 23/(2³ × 5²) = 23 × 5/(2³ × 5³) = 115/1000 = 0.115
(viii) 6/15 = 2/5 = 0.4
(ix) 35/50 = 7/10 = 0.7
The following real numbers have decimal expansions as given. In each case, decide whether they are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q?
Answer
(i) 43.123456789 — Rational (terminating). q = 10⁹ = 2⁹ × 5⁹. Prime factors of q are only 2 and 5.
(ii) 0.120120012000120000... — Irrational. The pattern has increasing zeros, so it is non-terminating and non-repeating.
(iii) 43.123456789123456789... — Rational (non-terminating repeating with period 9). q will have prime factors other than just 2 and 5.
Important Extra Questions
State the Fundamental Theorem of Arithmetic.
Answer
Every composite number can be expressed as a product of prime numbers, and this factorisation is unique, apart from the order in which the prime factors occur. For example, 36 = 2² × 3² — this is the only way to express 36 as a product of primes (ignoring order).
State Euclid's Division Lemma.
Answer
For any two positive integers a and b, there exist unique integers q (quotient) and r (remainder) such that a = bq + r, where 0 ≤ r < b. This is the basis of Euclid's division algorithm for finding HCF.
Find the HCF and LCM of 6, 72 and 120 using prime factorisation.
Answer
6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5
HCF = 2¹ × 3¹ = 6 (take lowest powers of common factors)
LCM = 2³ × 3² × 5 = 360 (take highest powers of all factors)
Show that √3 is irrational.
Answer
Assume √3 = a/b (coprime). Then a² = 3b², so 3 divides a. Let a = 3c. Then 9c² = 3b², so b² = 3c², meaning 3 divides b. But then both a and b are divisible by 3, contradicting that they are coprime. Hence √3 is irrational.