Short Questions & Answers
CBSE Board
Class 10
English Medium
Mathematics
Important Short Questions with Answers for Class 10 Real Numbers (2 & 3 Marks)
Frequently asked 2-mark and 3-mark questions in CBSE Board examinations with complete step-by-step marking scheme answers.
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Frequently asked 2-mark and 3-mark questions in CBSE Board examinations with complete step-by-step marking scheme answers.
Questions & Step-by-Step Solutions (2)
Short Answer Question (SAQ • 2-3 Marks)
Q1.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Solution / Detailed Answer:
Let a be any positive odd integer and b = 4.
By Euclid's division algorithm, a = 4q + r, for some integer q >= 0, and r = 0, 1, 2, 3.
Therefore, a can be 4q, 4q + 1, 4q + 2, or 4q + 3.
Since a is odd, a cannot be 4q or 4q + 2 (both are divisible by 2).
Therefore, any positive odd integer is of the form 4q + 1 or 4q + 3.
By Euclid's division algorithm, a = 4q + r, for some integer q >= 0, and r = 0, 1, 2, 3.
Therefore, a can be 4q, 4q + 1, 4q + 2, or 4q + 3.
Since a is odd, a cannot be 4q or 4q + 2 (both are divisible by 2).
Therefore, any positive odd integer is of the form 4q + 1 or 4q + 3.
Short Answer Question (SAQ • 2-3 Marks)
Q2.
Given that HCF(306, 657) = 9, find LCM(306, 657).
Solution / Detailed Answer:
We know that for any two positive integers a and b:
LCM(a, b) * HCF(a, b) = a * b
LCM(306, 657) * 9 = 306 * 657
LCM(306, 657) = (306 * 657) / 9 = 34 * 657 = 22,338.
Hence, LCM(306, 657) = 22,338.
LCM(a, b) * HCF(a, b) = a * b
LCM(306, 657) * 9 = 306 * 657
LCM(306, 657) = (306 * 657) / 9 = 34 * 657 = 22,338.
Hence, LCM(306, 657) = 22,338.
Curriculum Details
- Board / Portal: CBSE Board
- Class: Class 10
- Subject: Mathematics
- Chapter: Real Numbers
- Format: Short Questions & Answers
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