1.1

Q No. 1:

Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225
Answer No. 1 (i):
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
✅ HCF = 45

(ii) 196 and 38220
Answer No. 1 (ii):
38220 = 196 × 195 + 0
✅ HCF = 196

(iii) 867 and 225
Answer No. 1 (iii):
867 = 225 × 3 + 192
225 = 192 × 1 + 33
192 = 33 × 5 + 27
33 = 27 × 1 + 6
27 = 6 × 4 + 3
6 = 3 × 2 + 0
✅ HCF = 3

(iv) 272 and 1032
Answer No. 1 (iv):
1032 = 272 × 3 + 216
272 = 216 × 1 + 56
216 = 56 × 3 + 48
56 = 48 × 1 + 8
48 = 8 × 6 + 0
✅ HCF = 8

(v) 405 and 2520
Answer No. 1 (v):
2520 = 405 × 6 + 90
405 = 90 × 4 + 45
90 = 45 × 2 + 0
✅ HCF = 45

(vi) 155 and 1385
Answer No. 1 (vi):
1385 = 155 × 8 + 145
155 = 145 × 1 + 10
145 = 10 × 14 + 5
10 = 5 × 2 + 0
✅ HCF = 5

(vii) 384 and 1296
Answer No. 1 (vii):
1296 = 384 × 3 + 144
384 = 144 × 2 + 96
144 = 96 × 1 + 48
96 = 48 × 2 + 0
✅ HCF = 48

(viii) 1848 and 3058
Answer No. 1 (viii):
3058 = 1848 × 1 + 1210
1848 = 1210 × 1 + 638
1210 = 638 × 1 + 572
638 = 572 × 1 + 66
572 = 66 × 8 + 44
66 = 44 × 1 + 22
44 = 22 × 2 + 0
✅ HCF = 22

Q No. 2:

Show that any positive odd integer is of the form 6q + 3 or 6q + 5, where q is some integer.
Answer No. 2:
Let any integer n be of the form n = 6q + r, where 0 ≤ r < 6
Then possible values of r are: 0, 1, 2, 3, 4, 5

So, n can be:

• 6q → even
• 6q + 1 → odd
• 6q + 2 → even
• 6q + 3 → odd
• 6q + 4 → even
• 6q + 5 → odd

Therefore, the odd numbers are of the form:
6q + 1, 6q + 3, or 6q + 5

Among them, 6q + 3 and 6q + 5 always give odd values greater than 1.
✅ Hence, any positive odd integer can be of the form 6q + 3 or 6q + 5

Q No. 3:

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 the same number of columns. What is the maximum number of columns in which it can march?
Answer No. 3:
We need to find the HCF of 616 and 32

616 = 32 × 19 + 8
32 = 8 × 4 + 0
✅ HCF = 8

➡️ Maximum number of columns = 8

Q No. 4:

Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1
Answer No. 4:
Let any positive integer be n
By Euclid’s lemma: n = 3q + r, where r = 0, 1, or 2

Now squaring:

• If n = 3q, then n² = 9q² = 3m
• If n = 3q + 1, then n² = (3q + 1)² = 9q² + 6q + 1 = 3m + 1
• If n = 3q + 2, then n² = (3q + 2)² = 9q² + 12q + 4 = 3m + 1

✅ So, the square of any number is either 3m or 3m + 1

Q No. 5:

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 No. 5:
Let n = 3q + r, where r = 0, 1, or 2

Now cube it:

• If n = 3q, then n³ = 27q³ = 9m
• If n = 3q + 1, then
  n³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9m + 1
• If n = 3q + 2, then
  n³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9m + 8

✅ Hence, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8

Q No. 6:

Himadri has 625 Indian postal stamps and 325 International stamps. She wants to display them in identical groups with no stamps left out. What is the greatest number of groups she can display the stamps?
Answer No. 6:
We need to find HCF of 625 and 325

625 = 325 × 1 + 300
325 = 300 × 1 + 25
300 = 25 × 12 + 0
✅ HCF = 25

➡️ Greatest number of groups = 25

Q No. 7:

Two ropes are of length 64 cm and 80 cm. Both are to be cut into pieces of equal length. What should be the maximum length of the pieces?
Answer No. 7:
Find HCF of 64 and 80

80 = 64 × 1 + 16
64 = 16 × 4 + 0
✅ HCF = 16

➡️ Maximum length of the pieces = 16 cm

1.2

Q No. 1:

Express each number as a product of its prime factors:

(i) 140
Answer No. 1 (i):
140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

(ii) 156
Answer No. 1 (ii):
156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

(iii) 3825
Answer No. 1 (iii):
3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

(iv) 5005
Answer No. 1 (iv):
5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429
Answer No. 1 (v):
7429 = 17 × 19 × 23 = 17 × 19 × 23

Q No. 2:

Find the LCM and HCF of the following pairs and verify LCM × HCF = Product of the two numbers:

(i) 26 and 91
Answer No. 2 (i):
26 = 2 × 13
91 = 7 × 13
➡️ HCF = 13
➡️ LCM = 2 × 7 × 13 = 182
✅ 26 × 91 = 2366
✅ HCF × LCM = 13 × 182 = 2366 ✔️

(ii) 510 and 92
Answer No. 2 (ii):
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
➡️ HCF = 2
➡️ LCM = 2² × 3 × 5 × 17 × 23 = 23460
✅ 510 × 92 = 46920
✅ HCF × LCM = 2 × 23460 = 46920 ✔️

(iii) 336 and 54
Answer No. 2 (iii):
336 = 2⁴ × 3 × 7
54 = 2 × 3³
➡️ HCF = 2 × 3 = 6
➡️ LCM = 2⁴ × 3³ × 7 = 3024
✅ 336 × 54 = 18144
✅ HCF × LCM = 6 × 3024 = 18144 ✔️

Q No. 3:

Find the LCM and HCF of the following integers using prime factorisation:

(i) 12, 15, and 21
Answer No. 3 (i):
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
➡️ HCF = 3
➡️ LCM = 2² × 3 × 5 × 7 = 420

(ii) 17, 23, and 29
Answer No. 3 (ii):
All are prime numbers.
➡️ HCF = 1
➡️ LCM = 17 × 23 × 29 = 11339

(iii) 8, 9, and 25
Answer No. 3 (iii):
8 = 2³
9 = 3²
25 = 5²
➡️ HCF = 1
➡️ LCM = 2³ × 3² × 5² = 1800

Q No. 4:

Given that HCF (306, 657) = 9, find LCM(306, 657)
Answer No. 4:
Using the identity: HCF × LCM = Product of the numbers
➡️ LCM = (306 × 657) ÷ 9
= 201042 ÷ 9 = 22338

Q No. 5:

Check whether 6n can end with digit 0 for any natural number n.
Answer No. 5:
For a number to end in 0, it must be divisible by 10, i.e., by 2 and 5.
Since 6 = 2 × 3, any multiple of 6 is not divisible by 5.
✅ So, 6n cannot end with 0 for any natural number n.

Q No. 6:

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer No. 6:

First Expression:
7 × 11 × 13 + 13 = (7 × 11 × 13) + 13 = 1001 + 13 = 1014
1014 is divisible by 13 ⇒ 13 × 78 = 1014
✅ Hence, 1014 is composite

Second Expression:
7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
5040 + 5 = 5045
5045 is divisible by 5 ⇒ 5 × 1009 = 5045
✅ Hence, 5045 is composite

Q No. 7:

Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round. When will they meet again at the starting point?
Answer No. 7:
We need to find the LCM of 18 and 12
18 = 2 × 3²
12 = 2² × 3
➡️ LCM = 2² × 3² = 36
✅ They will meet again after 36 minutes

Q No. 8 (i):

The soldiers in the regiment can be stood in some rows consisting of 15, 20 and 25 soldiers. Find the least number of soldiers in the regiment.
Answer No. 8 (i):
Find LCM of 15, 20, and 25
15 = 3 × 5
20 = 2² × 5
25 = 5²
➡️ LCM = 2² × 3 × 5² = 300
✅ Least number of soldiers = 300

Q No. 8 (ii):

A bell rings every 18 seconds, another every 60 seconds. After how many seconds will they ring together again?
Answer No. 8 (ii):
Find LCM of 18 and 60
18 = 2 × 3²
60 = 2² × 3 × 5
➡️ LCM = 2² × 3² × 5 = 180
✅ Bells will ring together after 180 seconds

Q No. 8 (iii):

A radio station plays ‘Assam Sangeet’ once every 2 days, another every 3 days. How many times in 30 days will both play on the same day?
Answer No. 8 (iii):
Find LCM of 2 and 3 = 6
In 30 days, the song will play together: 30 ÷ 6 = 5 times

1.3

Q No. 1:

Prove that √5 is an irrational number.

Answer No. 1:
We will prove this by contradiction.

Step 1: Assume √5 is rational.
That means √5 = a/b, where a and b are integers with no common factors (in lowest terms), and b ≠ 0.

Step 2: Square both sides:
√5 = a/b
⇒ 5 = a² / b²
⇒ a² = 5b²

Step 3: This implies a² is divisible by 5.
So, a must also be divisible by 5.
Let a = 5k for some integer k.

Step 4: Substitute a = 5k into the equation:
a² = 25k²
So, 25k² = 5b²
⇒ b² = 5k²
⇒ b is also divisible by 5.

Step 5: Contradiction
Both a and b are divisible by 5, which contradicts the assumption that a/b is in lowest terms.

✅ Hence, √5 is irrational.

Q No. 2:

Prove that 3 + 2√5 is irrational.

Answer No. 2:
We will again use proof by contradiction.

Step 1: Assume 3 + 2√5 is rational.
Let 3 + 2√5 = r, where r is rational.

Step 2: Rearranging:
2√5 = r – 3
⇒ √5 = (r – 3)/2

Since r – 3 is rational, and 2 is rational, their division is rational.
So, √5 would be rational.

Step 3: Contradiction
But we already know √5 is irrational (proved in Q1).

✅ So, our assumption is wrong.
Hence, 3 + 2√5 is irrational.

Q No. 3:

Prove that the following are irrational numbers:

(i) 1/√2
Answer No. 3 (i):
Assume 1/√2 is rational.
Then, 1/√2 = p/q (p, q are integers, q ≠ 0)
⇒ √2 = q/p, which is rational.

But we know √2 is irrational.
So, contradiction.

✅ Therefore, 1/√2 is irrational.

(ii) 7√5
Answer No. 3 (ii):
7 is rational, √5 is irrational.
Multiplying a non-zero rational number with an irrational number gives an irrational result.

✅ So, 7√5 is irrational.

(iii) 6 + √2
Answer No. 3 (iii):
Assume 6 + √2 is rational.
Then, √2 = (6 + √2) – 6 = rational – rational = rational.

But √2 is irrational, so contradiction.

✅ Hence, 6 + √2 is irrational.

1.4

Q No. 1:

Without actually performing the long division, determine whether each of the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion. Also give a reason.

(i) 13/3125
(ii) 17/8
(iii) 64/455
(iv) 15/1600
(v) 29/343
(vi) 121/2⁵
(vii) 77/210
(viii) 6/15
(ix) 35/50
(x) 27/210

Answer No. 1:

(i) 13/3125
3125 = 5⁵ → only prime factor is 5
✅ Terminating (denominator has only 5)

(ii) 17/8
8 = 2³ → only prime factor is 2
✅ Terminating

(iii) 64/455
455 = 5 × 7 × 13 → contains 7 and 13
❌ Non-terminating repeating

(iv) 15/1600
1600 = 2⁶ × 5² → only prime factors are 2 and 5
✅ Terminating

(v) 29/343
343 = 7³ → contains 7
❌ Non-terminating repeating

(vi) 121/2⁵
2⁵ = 32 → only prime factor is 2
✅ Terminating

(vii) 77/210
210 = 2 × 3 × 5 × 7 → contains 3 and 7
❌ Non-terminating repeating

(viii) 6/15
6/15 = 2/5 → denominator 5
✅ Terminating

(ix) 35/50
35/50 = 7/10 → 10 = 2 × 5
✅ Terminating

(x) 27/210
210 = 2 × 3 × 5 × 7 → contains 3 and 7
❌ Non-terminating repeating

Q No. 2:

Write down the decimal expressions of those rational numbers in question 1 above which have terminating decimal expansions.

Answer No. 2:

• (i) 13 ÷ 3125 = 0.00416
• (ii) 17 ÷ 8 = 2.125
• (iv) 15 ÷ 1600 = 0.009375
• (vi) 121 ÷ 32 = 3.78125
• (viii) 6 ÷ 15 = 2 ÷ 5 = 0.4
• (ix) 35 ÷ 50 = 7 ÷ 10 = 0.7

Q No. 3:

The following real numbers have decimal expansions as given below. 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 No. 3:

1. 0.125
   → Terminating decimal → Rational
   → Can be written as 1/8 (q = 8 = 2³)
   ✅ Rational; q has only 2 as a prime factor
2. 0.333…
   → Repeating decimal → Rational
   → Can be written as 1/3
   ❌ Prime factor of denominator is 3 (≠ 2 or 5)
3. 0.142857142857…
   → Repeating decimal → Rational
   → Can be written as 1/7
   ❌ Denominator has 7
4. π = 3.14159… (non-terminating, non-repeating)
   ❌ Irrational
5. √2 = 1.4142135… (non-terminating, non-repeating)
   ❌ Irrational