Here you get the CBSE Class 10 Mathematics chapter 1, Real Numbers: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solution of Exercise 1.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Real Numbers. This exercise comprises of only the Very short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Real Numbers:
Very short Answer Type Questions
Question1. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer.
No, every positive integer can’t be of the form 4q + 2
By Euclid’s Lemma: Dividend (b) = divisor (a) × quotient (q) + remainder (r)
Or b = aq + r, 0 ≤ r < a
According to question,
a = 4, b = 4q + r for 0 ≤ r < 4, i.e., r = 0, 1, 2, 3
Therefore, this must be in the form 4q, 4q + 1, 4q + 2 or 4q + 3
Question2. ‘The product of two consecutive positive integers is divisible by 2, Is this statement true or false? Give reasons.
If (n) is an positive integer, then the consecutive integer will be (n + 1)
If (n) is odd, then (n + 1) will be even ⇒Its product, n × (n + 1) is divisible by 2.
If (n) is even, then (n + 1) will be odd ⇒Its product, n × (n + 1) is also divisible by 2.
Hence, the product of two consecutive positive integers is divisible by 2.
Question3. 'The product of three consecutive positive integers is divisible by 6. Is this statement true or false? Justify your answer.
Let three consecutive integers be n, (n + 1) and (n + 2).
If n = 1 then, n + 1 = 2 and n + 2 = 3, so the product of three consecutive will be divisible by both 2 and 3. Also, a number divisible by 2 and 3 will surely be divisible by 6.
Similarly, we can check for any positive integral value of n and we will find one number of these three must be divisible by 2 and another by 3.
Thus, the product of three consecutive positive integers is always divisible by 6.
Question4. Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.
Using Euclid’s lemma, b = aq + r, 0 ≤ r < a, where b is any positive integer,
If a = 3 then, b = 3q + r, 0 ≤ r ≤ 2
So, any positive integer is of the form 3q, 3q + 1 or 3q + 2
Now, (3q)2 = 9q2 = 3m [Where, m = 9k2]
and (3q +1)2 = 9q2 + 6q + 1
= 3 (3q2 + 2q) + 1 = 3m + 1 [ Where, m = 3q2 + 2q]
Also, (3q + 2)2 = 9q2 + 12q + 4 [Using (a + b)2 = a2 + 2ab + b2]
⟹ (3q + 2)2 = 9q2 + 12q + 3 + 1
= 3(3q2 + 4q + 1) + 1 = 3m + 1[Where, m = 3q2 + 4q + 1]
Therefore, the square of any positive integer can be of the form 3mand 3m+1 but not of the form 3m+2.
Question5. A positive integer is of the form 3q+1, q being a natural number. Can you write its square in any form other then 3m + 1, i.e. 3m or 3m+2 for some integer m? Justify your answer.
Because, (3q +1)2 = 9q2 + 6 k + 1 [Using (a + b)2 = a2 + 2ab + b2]
⟹ (3q +1)2 = 3 (3q2 + 2q) + 1 = 3m + 1 [ Where, m = 3q2 + 2q]
(3q +1)2 can’t be expressed in terms of 3m or 3m+2 for some integer m.
Question6. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75.
What is HCF (525, 3000)? Justify your answer.
As the numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75 and the highest number among these is 75 so, the HCF (525, 3000) = 75
By prime factorization method,
525 = 3 × 5 × 5 × 7
3000 = 2 × 2 × 2 × 3 × 5 × 5
Thus, HCF (525, 3000) = 3 × 5 × 5 = 75.
Question7. Explain why 3 × 5 × 7 + 7 is a composite number.
A composite number is a whole number that can be divided evenly by numbers other than 1 or itself.
Now, 3 × 5 × 7 + 7 = 112
Also, 112 = 2 × 2 × 2 × 2 × 7 = 24 × 7
So, it is the product of prime factors 2 and 7 other than 1 and itself so it is a composite number.
Question8. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
No, because HCF is always a factor of LCM but in this case 18 is not a factor of 380.
Question9. Without actually performing the long division find if 987/10500 will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.
Question10. A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form p/q? Give reasons.
As 327.7081 is terminating decimal number hence, it can be written in p/q form where q is of the form 2m × 5n.
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