Real numbers questions are given here to help the students of Classes 9 and 10 for their exams. Real numbers is one of the important concepts from the examination point of view. This article will get real numbers questions based on the latest NCERT curriculum. Practising these questions will help you to improve your problem-solving skills.
What are Real Numbers?
Real numbers are the collection of all rational and irrational numbers. A unique real number corresponds to every point on the number line. Also, there is a unique point on the number line corresponding to each real number.
| A number ‘r’ is called a rational number if it can be written in the form p/q, where p and q are integers and q ≠ 0. | A number ‘s’ is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. |
| The decimal expansion of a rational number is either terminating or non-terminating recurring. Moreover, a number whose decimal expansion is terminating or non-terminating recurring is rational. | The decimal expansion of an irrational number is non-terminating non-recurring. Moreover, a number whose decimal expansion is non-terminating non-recurring is irrational. |
Also, check: Real numbers
Real Numbers Questions and Answers
1. Find which of the variables x, y, z and u represent rational numbers and which irrational numbers:
(i) x2 = 5 (ii) y2 = 9 (iii) z2 = .04 (iv) u2 = 17/4
Solution:
(i) x2 = 5
On simplifying, we get;
x = ± √5
Therefore, x is an irrational number.
(ii) y2 = 9
On simplifying, we get;
y = ± 3
Therefore, y is a rational number.
(iii) z2 = .04
On simplifying, we get;
z = ± 0.2
Therefore, z is a rational number.
(iv) u2 = 17/4
On simplifying, we get;
u = ± √17/2
As we know, √17 is irrational.
Therefore, u is an irrational number.
2. Locate √9.3 on the number line.
Solution:
Below are the steps to locate √9.3 on the number line.
Step 1: Draw a line AB = 9.3 cm
Step 2: From the point B, add 1 cm and mark it as C
Step 3: Mark the point of bisection by a compass and say it as ‘O’
Step 4: Taking AO as the radius, draw a semi-circle.
Step 5: From B, draw a perpendicular AB touching the semi-circle and mark as D
Step 6: Draw an arc on the number line by taking compass pointer on B and pencil on D
Step 7: The point which intersects the number line is the square root of 9.3, i.e. BE = √9.3
3. Find the value of a, if 6/(3√2 – 2√3) = 3√2 – a√3.
Solution:
Given,
6/(3√2 – 2√3) = 3√2 – a√3….(i)
Consider LHS,
6/(3√2 – 2√3)
Rationalizing the denominator, we get;
[6/(3√2 – 2√3)] [(3√2 + 2√3)/ (3√2 + 2√3)]= [6(3√2 + 2√3)]/ [(3√2)2 – (2√3)2]
= [6(3√2 + 2√3)]/ (18 – 12)
= 6(3√2 + 2√3)/6
= 3√2 + 2√3….(ii)
From (i) and (ii),
-a = 2
a = -2
4. Simplify the following:
(i) (17)⅕ (3)⅕
(ii) (13 + 23 + 33)½
Solution:
(i) (17)⅕ (3)⅕
= (17 × 3)⅕
= (51)⅕
(ii) (13 + 23 + 33)½
= (1 + 8 + 27)½
= (36)½
= (62)½
= 6
5. If a = 5 + 2√6 and b = 1/a, what will be the value of a2 + b2?
Solution:
Given,
a = 5 + 2√6
b = 1/a
= 1/(5 + 2√6)
Rationalizing the denominator, we get;
b = [1/(5 + 2√6)] [(5 – 2√6)/(5 – 2√6)]
= (5 – 2√6)/ [(5)2 – (2√6)2]
= (5 – 2√6)/ (25 – 24)
= 5 – 2√6
Now,
a2 + b2 = (a + b)2 – 2ab
= (5 + 2√6 + 5 – 2√6)2 – 2(5 + 2√6)(5 – 2√6)
= (10)2 – 2[(5)2 – (2√6)2]
= 100 – 2(1)
= 98
| Euclid’s Division Lemma:
Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b. Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes, and this expression (factorization) is unique, apart from the order in which the prime factors occur. Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. |
6. Using Euclid’s division algorithm, find which of the following pairs of numbers are co-prime:
(i) 231, 396
(ii) 847, 2160
Solution:
Let us find the HCF of each pair of numbers.
(i) 396 > 231
By Euclid’s division lemma,
396 = 231 × 1 + 165
231 = 165 × 1 + 66
165 = 66 × 2 + 33
66 = 33 × 2 + 0
Here, the remainder is 0.
Therefore, HCF(396, 231) = 33.
Hence, numbers are not co-prime.
(ii) 2160 > 847
By Euclid’s division lemma,
2160 = 847 × 2 + 466
847 = 466 × 1 + 381
466 = 381 × 1 + 85
381 = 85 × 4 + 41
85 = 41 × 2 + 3
41 = 3 × 13 + 2
3 = 2 × 1 + 1
2 = 1 × 2 + 0
Here, the remainder is 0.
Therefore, HCF(2160, 847) = 1
Hence, the numbers 2160 and 847 are co-prime.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
From the given,
Time taken by Sonia to drive one round = 18 minutes
Time taken by Ravi to drive one round = 12 minutes
Suppose they both start at the same point and at the same time and go in the same direction, then the time taken by them to meet again at the starting point is the LCM of 18 and 12.
Prime factorization of 18 = 2 × 3 × 3 = 2 × 32
Prime factorization of 12 = 2 × 2 × 3 = 22 × 3
As we know, LCM is the product of the greatest power of each prime factor involved in the numbers.
So, LCM of 18 and 12 = 22 × 32 = 4 × 9 = 36
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.
8. Prove that one of any three consecutive positive integers must be divisible by 3.
Solution:
Let n, n + 1, and n + 2 be the three consecutive positive integers, where n is an integer.
By Euclid’s division lemma, we have a = bq + r; 0 ≤ r < b
Let a = n and b = 3 such that n = 3q + r …(i)
Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Substituting r = 0 in (i), we get;
n = 3q
⇒ n is divisible by 3.
n + 1 = 3q + 1
⇒ n + 1 is not divisible by 3.
n + 2 = 3q + 2
⇒ n + 2 is not divisible by 3.
Substituting r = 1 in (i), we get;
n = 3q + 1
⇒ n is not divisible by 3.
n + 1 = 3q + 1+ 1 = 3q + 2
⇒ n + 1 is not divisible by 3.
n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1)
⇒ n + 2 is divisible by 3.
Substituting r = 2 in (i), we get;
n = 3q + 2
⇒ n is not divisible by 3.
n + 1 = 3q + 2 +1 = 3q + 3 = 3(q + 1)
⇒ n + 1 is divisible by 3.
n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1
⇒ n + 2 is not divisible by 3.
Therefore, for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3.
Hence proved.
9. Show that 5 + √3 is irrational.
Solution:
Let us assume that 5 + √3 is a rational number.
We can find coprime a and b (b ≠ 0) such that 5 + √3 = a/b.
So,
√3 = (a/b) – 5
√3 = (a – 5b)/b
Since a and b are integers, we get (a/b) – 5 is rational, and so √3 is rational.
But this contradicts the fact that √3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 + √3 is rational.
Therefore, we conclude that 5 + √3 is irrational.
10. Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
Solution:
Let a be an odd integer and b = 6.
By Euclid’s division algorithm,
a = 6m + r for some integer m ≥ 0
Here, r = 0, 1, 2, 3, 4, 5 since 0 ≤ r < 6.
So, a = 6m or, 6m + 1 or, 6m + 2 or, 6m + 3 or, 6m + 4 or 6m + 5
Let us consider a = 6m + 1 or, 6m + 3 or 6m + 5 since we know that 6m, 6m + 2 and 6m + 4 are even integers.
For a = 6m + 1,
(6m + 1)2 = 36m2 + 12m + 1
= 6(6m2 + 2m) + 1
= 6q + 1, where q is some integer and q = 6m2 + 2m.
For a = 6m + 3,
(6m + 3)2 = 36m2 + 36m + 9
= 6(6m2 + 6m + 1) + 3
= 6q + 3, where q is some integer and q = 6m2 + 6m + 1
For a = 6m + 5,
(6m + 5)2 = 36m2 + 60m + 25
= 6(6m2 + 5m + 4) + 1 = 6q + 1, where q is some integer and q = 6m2 + 5m + 4.
Therefore, the square of an odd integer is of the form 6q + 1 or 6q + 3, for some integer q.
Hence proved.
Video Lesson on Numbers
Practice Questions on Real Numbers
- If a = 2 + √3, find the value of a – 1/a.
- Show that 0.142857142857… = 1/7.
- Find three rational numbers between –1 and –2.
- Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.
- Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628, leaving remainders 1, 2 and 3, respectively.
Comments