Real Numbers Class 10 Maths CBSE Important Questions with Solutions

Important Questions for Class 10 Maths Chapter 1 Real Numbers with solutions includes all the important topics with detailed explanation that aims to help students to score more marks in Board Exams 2020. Students who are preparing for their Class 10 exams must go through Important Questions for Class 10 Math Chapter 1 Real Numbers.

Important Questions for Class 10 Maths Chapter 1 Real Numbers

Expert teachers at CBSETuts.com collected and solved 2 Marks and 4 mark important questions for Class 10 Maths Chapter 1 Real Numbers. All the solutions given in this page are solved based on CBSE marking scheme and NCERT guidelines.

2016

Short Answer Type Questions I [2 Marks]

Question 1.
Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker in exact number of times.
Solution:
Maximum capacity of a container, which can measure the petrol in exact number of times.

Question 2.
Find the value of: (-1) + (-1)²ⁿ +(-l) ²ⁿ⁺¹ + (-l)⁴ⁿ⁺¹ , where n is any positive odd integer.
Solution:

Question 3.
Find whether decimal expansion of 13/64 is a terminating or non-terminating decimal. If it terminates, find the number of decimal places its decimal expansion has.
Solution:

Short Answer Type Question II [3 Marks]

Question 4.
Explain whether the number 3 x 5 x 13 x 46 + 23 isa prime number or a composite number.
Solution:

Long Answer Type Question [4 Marks]

Question 5.
Prove that the product of any three consecutive positive integers is divisible by 6. Solution: Let three consecutive numbers are n, n + 1, n + 2
Solution:
1st Case: If n is even
This means n + 2 is also even.
Hence n and n + 2 are divisible by 2
Also, product of n and (n + 2) is divisible by 2.
.’. n(n + 2) is divisible by 2.
This conclude n(n + 2) (n + 1) is divisible by 2 …(i)
As, n, n + 1, n + 2 are three consecutive numbers. n(n + 1) (n + 2) is a multiple of 3.
This shows n(n + 1) (n + 2) is divisible by 3. …(ii)
By equating (i) and (ii) we can say
n(n + 1) (n + 2) is divisible by 2 and 3 both.
Hence, n(n + 1) (n + 2) is divisible by 6.
2nd Case: When n is odd.
This show (n + 1) is even
Hence (n + 1) is divisible by 2. …(iii)
This conclude n(n + 1) (n + 2) is an even number and divisible by 2.
Also product of three consecutive number is a multiple of 3.
n(n + 1)(n + 2) is divisible by 3. …(iv)
Equating (iii) and (iv) we can say
n(n + 1) (n + 2) is divisible by both 2 and 3 Hence, n(n + 1)(n + 2) is divisible by 6.

2015

Short Answer Type Questions I [2 Marks]

Question 6.
Apply Euclid’s division algorithm to find HCF of numbers 4052 and 420.
Solution:

Question 7.

Show that (√3+√5)² is an irrational number.
Solution:

Short Answer Type Question [3 Marks]

For example, LCM of 15 and 20 is 60, and LCM of 5 and 7 is 35.

Question 8.
Three bells toll at intervals of 12 minutes, 15 minutes and 18 minutes respectively. If they start tolling together, after what time will they next toll together?
Solution:
LCM of 12, 15, 18 = 2²x 3² x 5
=4x9x5 = 180
So, next time the bells will ring together after 180 minutes.

2014

Short Answer Type Questions I [2 Marks]

Question 9.
If HCF of 144 and 180 is expressed in the form 13m – 3, find the value of m.
Solution:
On applying Euclid’s division algorithm,
180 = 144 x 1 + 36
144 = 36 x 4 + 0
At the last stage, the divisor is 36.
∴ HCF of 144 and 180 is 36.
∵ 36 = 13 x 3 – 3
So, m = 3

Question 10.
Show that 9ⁿ can not end with digit 0 for any natural number n.
Solution:
Since prime factorisation of 9ⁿ is given by 9ⁿ= (3 x 3)ⁿ = 3271.
Prime factorisation of 9″ contains only prime number 3.
9 may end with the digit 0 for some natural number V if 5 must be in its prime factorisation, which is not present.
So, there is no natural number N for which 9ⁿ ends with the digit zero.

Question 11.
Determine the values otp and q so that the prime factorisation of2520 is expressible as 23 X y X q x 7.
Solution:
Prime factorisation of 2520 is given by
2520 = 23 x 32 x 5 x 7
Given that 2520 = 23 x 3p x q x 7
On comparing both factorisation we get p = 2 and q = 5.

Question 12.
Show that 2√2 is an irrational number.
Solution:

Question 13.
Show that any positive odd integer is of the form 4m + 1 or 4m + 3, where m is some integer.
Solution:

Short Answer Type Questions II [3 Marks]

Question 14.
By using, Euclid’s algorithm, find the largest number which divides 650 and 1170.
Solution:
Given numbers are 650 and 1170.
On applying Euclid’s division algorithm,
we get 1170 = 650 x 1 + 520
650 = 520 x 1 + 130
520 = 130 x 4 + 0
∵ At the last stage, the divisor is 130.
∴ The HCF of 650 and 1170 is 130.

Question 15.
Show that reciprocal of 3+2√2 is an irrational number
Solution:

Long Answer Type Question [4 Marks]

Question 16.
Find HCF of 378,180 and 420 by prime factorisation method. Is HCF x LCM of three numbers equal to the product of the three numbers?
Solution:
378 = 2 x 33 x 7
180 = 22 x 32 x 5
420 = 22 x 3 x 5 x 7
∴ HCF (378, 180, 420) = 2 x 3 = 6.
No. HCF (p, q, r) x LCM (p, q, r) ≠ p x q x r. where p, q, r are positive integers.

2013

Short Answer Type Questions I [2 Marks]

Question 17.
Find the HCF of 255 and 867 by Euclid’s division algorithm
Solution:
Given numbers are 255 and 867.
On applying Euclid’s division algorithm, we have
867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
∵ At the last stage, the divisor is 51
∴ The HCF of 255 and 867 is 51.

Question 18.
Find the HCF (865, 255) using Euclid’s division lemma.
Solution:
Given numbers are 255 and 865.
On applying Euclid’s division algorithm, we have
865 = 255 x 3 + 100
255 = 100 x 2 + 55
100 = 55 x 1 + 45
55 = 45 x 1 + 10
45 = 10 x 4 + 5
10 = 5 x 2 + 0
∵At the last stage, the divisor is 5
∴ The HCF of 255 and 865 is 5.

Short Answer Type Questions II [3 Marks]

Question 19.
Find HCF of 65 and 117 and find a pair of integral values of m and n such that HCF = 65m + 117n.
Solution:
Given numbers are 65 and 117.
On applying Euclid’s division algorithm, we get
117 = 65 x 1 + 52
65 = 52 x 1 + 13
52 = 13 x 4 + 0
∵At the last stage, the divisor is 13.
∴ The HCF of 65 and 117 is 13.
The required pair of integral values of m and n is
(2,-1) which satisfies the given relation HCF = 65m + 117n.

Question 20.
By using Euclid’s algorithm, find the largest number which divides 650 and 1170
Solution:

Since prime factorisation of 9ⁿ is given by 9ⁿ= (3 x 3)ⁿ = 3271.
Prime factorisation of 9″ contains only prime number 3.
9 may end with the digit 0 for some natural number V if 5 must be in its prime factorisation, which is not present.
So, there is no natural number N for which 9ⁿ ends with the digit zero.

 

2012

Short Answer Type Question I [2 Marks]

Question 21.
If  find the values of m and n where m and n are non-negative integers.Hence write its decimal expansion without actual division.
Solution:

Short Answer Type Questions II [3 Marks]

Question 22.
Express the number 0.3178 in the form of rational number a/b.
Solution:

Question 23.
Using Euclid’s division algorithm, find whether the pair of numbers 847,2160 are coprimes or not.
Solution:

Question 24.
The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then find the other number.
Solution:

2011

Short Answer Type Questions I [2 Marks]

Question 25.
Prove that 15 + 17√3 is an irrational number.
Solution:

Question 26.
Find the LCM and HCF of 120 and 144 by using Fundamental Theorem of Arithmetic.
Solution:
120 = 23 x 3 x 5
144 = 24 x 32
∴ HCF = 23 x 3 = 24
LCM = 24 x 5 x 32 = 720

Short Answer Type Questions II [3 Marks]

Question 27.
An army contingent of 1000 members is to march behind an army band of 56 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
1000 =2x2x2x5x5x5
56 = 2x2x2x7
HCF of 1000 and 56 = 8
Maximum number of columns = 8.

Question 28.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3 where q is a positive integer.
Solution:

Question 29.
Prove that 2√3/5 is irrational
Solution:

2010

Very Short Answer Type Questions [1 Mark]

Question 30.
Has the rational number  a terminating or a non-terminating decimal repressentation
Solution:

Question 31.
Write whether  on simplification gives a rational or an irrational number.
Solution:

Question 32.
The HCF of 45 and 105 is 15. Write their LCM.
Solution:

Short Answer Type Questions II [3 Marks]

Question 33.
Prove that 2-3√5 is an irrational number.
Solution:

Question 34.
Prove that 2√3 – 1 is an irrational number.
Solution:

Let 2√3 – 1 is a  irrational number.

Question 35.
Prove that √2 is irrational.
Solution:

Question 36.
Prove that 7 – 2√3 is an irrational number.
Solution:

Question 37.
Show that 5 + 3√2 is an irrational number.
Solution:

Let us assume 5 + 3√2 is an irrational number.
There exists coprime integers a and b (b≠0)

2009

Very Short Answer Type Questions [1 Mark]

Question 38.
The decimal expansion of the rational number  will terminate after how many places of decimals.
Solution:

Question 39.
Find the [HCF X LCM] for the numbers 100 and 190.
Solution:
HCF x LCM = one number x another number
= 100 x 190 = 19000

Question 40.
Find the [HCF and LCM] for the numbers 105 and 120. [All India]
Solution:
105 = 5 x 7 x 3
120 = 2x2x2x3x5
HCF = 3 X 5 = 15
LCM = 5x7x3x2x2x2 = 840

Question 41.
Write whether the rational number  will have a terminating decimal expansion or a non terminating repeating decimal expansion.
Solution:

Question 42.
The HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, write the other number.
Solution:

Short Answer Type Questions II [3 Marks]

Question 43.
Show that 5 – 2√3 is an irrational number.
Solution:

Question 44.
Show that 3 + 5√2 is an irrational number.
Solution:

Question 45.
Show that the square of any positive odd integer is of the form 8m + 1, for some integer m.
Solution:

Question 46.
Prove that 7 + 3√2 is not a rational number.
Solution: