Class X Math Excercise 1.2 Solve
Q1. Express each number as a product of its prime factors:
(i) 140
Solution:
140 = 2 × 70
= 2 × 2 × 35
= 2 × 2 × 5 × 7140=22×5×7
অসমীয়া:
140 ৰ মৌলিক গুণনীয়ক = 22×5×7
(ii) 156
Solution:
156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 3 × 13156=22×3×13
অসমীয়া:
156 = 22×3×13
(iii) 3825
Solution:
3825 = 5 × 765
= 5 × 5 × 153
= 5 × 5 × 3 × 51
= 5 × 5 × 3 × 3 × 173825=32×52×17
অসমীয়া:
3825 = 32×52×17
(iv) 5005
Solution:
5005 = 5 × 1001
= 5 × 7 × 11 × 135005=5×7×11×13
অসমীয়া:
5005 = 5×7×11×13
(v) 7429
Solution:
7429 = 17 × 437
= 17 × 19 × 237429=17×19×23
অসমীয়া:
7429 = 17×19×23
🔹 Q2. Find the LCM and HCF of the following pairs using prime factorisation:
(i) 26 and 91
Prime factors:
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
অসমীয়া:
HCF = 13, LCM = 182
(ii) 510 and 92
Prime factors:
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 22×3×5×17×23=23460
অসমীয়া:
HCF = 2, LCM = 23460
🔹 Q3. Find the LCM and HCF of 306 and 657 by prime factorisation.
Prime factors:
306 = 2 × 3 × 3 × 17
657 = 3 × 3 × 73
HCF = 32=9
LCM = 2×32×17×73=22338
অসমীয়া:
HCF = 9
LCM = 22338
🔹 Q4. Check whether 6ⁿ can end with digit 0 for any natural number n.
Solution:
6ⁿ = (2 × 3)ⁿ
To end with 0, number must contain factor 5.
But 6ⁿ has no factor 5.
👉 Therefore, 6ⁿ cannot end with 0.
অসমীয়া:
0 ত শেষ হ’বলৈ 5 গুণনীয়ক লাগিব।
6ⁿ ত 5 নাই।
👉 অসম্ভৱ।
🔹 Q5. Explain why 7 × 11 × 13 + 13 is a composite number.
Solution:
7 × 11 × 13 + 13
= 13(7 × 11 + 1)
= 13 × 78
Since it has factors other than 1 and itself, it is composite.
অসমীয়া:
13 এটা সাধাৰণ গুণনীয়ক।
অতএব সংখ্যা সংযোজিত (Composite)।
📌 Exam Tips (HSLC)
✔ Prime factorisation steps must be shown
✔ Write HCF & LCM clearly
✔ Proof questions = full marks if steps are neat
✔ Use index form (powers) correctly
