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### CBSE Class 10 Maths Chapter 1 Real Numbers Objective Questions

CBSE Class 10 Maths Chapter 1 Real Numbers Objective Questions

Want to assess how thorough you are with the concepts of “Real Numbers”? Are you worried about how you would fare if asked MCQs from this chapter? With the CBSE Class 10 exam pattern undergoing a major overhaul for the coming academic year, you can now expect more objective type questions for the exams. So, we have compiled here some questions from the chapter and arranged them topic wise for your easy convenience. Find the CBSE Class 10 Maths Chapter 1 Real Numbers Objective Questions given in this article.

Our focus is on providing solutions that can simplify the complex fundamentals of the subject. We believe in giving access to quality solutions that can help you score high marks in the exams. Below is the link to download the CBSE Class 10 Maths Chapter 1 Real Numbers Objective Questions.

The chapter discusses real numbers and their applications. It is an important topic of CBSE Class 10 Maths. While the topic is taught in Class 9 as well, in Class 10 it goes more in depth into the concept. Here, we have given the list of sub-topics that are covered in chapter 1 of CBSE Class 10 Maths Objective Questions . They are as follows:

1.1 Introduction to Real Numbers (We have 5 MCQs From This Topic)

1.2 Revisiting Irrational Numbers (We have 5 MCQs Covering This Topic)

1.3 Revisiting Rational Numbers and Their Decimal Expansions (5 MCQs Listed)

1.4 The Fundamental Theorem of Arithmetic (5MCQs Given From This Topic)

On dividing 1056 by 23, we got 21 as remainder.

⇒ If we add 23 – 21 = 2 to the dividend 1056, we will get a number completely divisible by 23.

Solution: Let the smaller number be x.

∴ Larger number is (270 + 1365) = 1635

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r where 0≤r

Basically, it can be observed that remainder can never be more than the divisor, and is a non-negative integer (could be zero).

Solution: There exist integers m and n such that a = 7m + p and b = 6n + q such that

Therefore, the maximum value of p + q will be 11.

Solution: If p=d×q+r, (p>q) where p, q, d, r are integers and for a given (p, d), there exist a unique (q, r), then HCF (p, d) = HCF (d, r). Because this relation holds true, the Euclid’s Division Algorithm exists in a step by step manner. So, to find the HCF (1008, 20), we use Euclid’s division lemma at every step.

Step 1: 1008=20×50+8 ⇒ HCF(1008, 20) = HCF(20,8) ⇒ a could be 8

Step 2: 20=8×2+4 ⇒ HCF(20, 8) = HCF(8,4) ⇒ b could be 4

Since 1008=20×q+a where q and a are positive integers satisfy Euclid’s Division Lemma, we must have 0≤a<20. So a is surely 8 and b is 4.

Solutions: We know that the square of a natural number never ends in 2, 3, 7 and 8.

∴ 42437 can’t be the square of a natural number.

Solution: For a number to be divisible by 8, the last three digits must be divisible by 8.

Solution: If the number is divisible by another number, it will be divisible by its factors too.

For the number to be divisible by both 2 and 5, the last digit should be 0

So we can rewrite the number as 653×0

Now For the number to be divisible by 8; last 3 digits should be divisible by 8

So, x can be either 2 or 6 since 320 and 360 are divisible by 8

If x=2; then the number becomes 65320 which is not divisible by 80

If x=6; then the number becomes 65360 which is divisible by 80

Answer: (A) 3.1415926535… (Non-repeating and non-terminating)

Solution: Non-terminating and non-repeating decimals are irrational.

If the denominator of a rational number is in the form of  where n and m are non-negative integers, then the rational number is terminating.

∴ 0.2 has terminating decimal and is a rational number.

3.1415926535… is non-terminating and non-repeating.

∴ 1/ 0.2 has terminating decimal and is a rational number.

(0.2) 2 has terminating decimal and is a rational number.

Solution : ‘s’ is called irrational if it cannot be written in the form of pq, where p and q are integers and q ≠ 0

To get the denominator in powers of 10, multiply both numerator and denominator by 2 4

We know a rational number is expressed in simplest form p/q and if q can be expressed as

×, then it is a terminating decimal

Clearly, 2 and 5 are not the factors of 23

8 can be expressed in terms of its primes as 2×2×2

Since q can be expressed in the form of., we can say 23/8 is a terminating decimal.

Answer: (A)both have terminating decimal expansion

Solution: If the denominator of a rational number is in the form of., where m and n are non-negative integers, then the rational number has terminating decimal expansion.

Answer: (D)Rational Number, Prime factors of q will have either 2 or 5 or both

Solution: It is rational because decimal expansion is terminating. Therefore, it can be expressed in p/q form where factors of q are of the form. and n and m are non-negative integers.

Solution : Non-terminating repeating (recurring)

Solution: Consider first two numbers 1848 and 3058, where 3058 > 1848.

Let us find the HCF of the numbers 1331 and 22.

∴ HCF of the three given numbers 1848, 3058 and 1331 is 11.

Solution: Maximum number of columns = HCF of 616 and 32

Solution: Finding the biggest number that will divide both 324 and 144 is same as finding the HCF of both.

Prime factorizing the two numbers we get,

Now, taking the common factors between them gives us the HCF.

Solution: Suppose the number be x. Since it divides 542 and 128 leave a remainder 2, using the Euclid’s division algorithm for 542 and x, 542 = ax + 2, where a is an integer. So, ax = 542 – 2 = 540, i.e. x is a factor of 540. Similarly x is a factor of (128 – 2).

The HCF of 542 – 2 = 540, 128 – 2 = 126:

3, 6, 9 are factors of 18. 12 is the only number which isn’t a factor and hence doesn’t satisfy the above property.

Solution: So p is 5 and the number is 900.

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