Maths Class 10 RD Sharma Solutions

RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
The exponent of 2 in the prime factorisation of 144, is
(a) 4
(b) 5
(c) 6
(d) 3
Solution:
(a)

Question 2.
The LCM of two numbersls 1200. Which of the following cannot be their HCF ?
(a) 600
(b) 500
(c) 400
(d) 200
Solution:
(b) LCM of two number = 1200
Their HCF of these two numbers will be the factor of 1200
500 cannot be its HCF

Question 3.
If n = 2³ x 3⁴ x 4⁴ x 7, then the number of consecutive zeroes in n, where n is a natural number, is
(a) 2
(b) 3
(c) 4
(d) 7
Solution:
(c) Because it has four factors n = 2³ x 3⁴ x 4⁴ x 7
It has 4 zeroes

Question 4.
The sum of the exponents of the prime factors in the prime factorisation of 196, is
(a) 1
(b) 2
(c) 4
(d) 6
Solution:
(c)

Question 5.
The number of decimal places after which the decimal expansion of the rational number \(\frac { 23 }{ { 2 }^{ 2 }\times 5 }\) will terminate, is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b)

Question 6.

(a) an even number
(b) an odd number
(c) an odd prime number
(d) a prime number
Solution:
(a)

Question 7.
If two positive integers a and b are expressible in the form a = pq² and b = p²q ; p, q being prime numbers, then LCM (a, b) is
(a) pq
(b) p³q³
(c) p³q²
(d) p²q²
Solution:
(c) a and b are two positive integers and a =pq² and b = p³q, where p and q are prime numbers, then LCM=p³q²

Question 8.
In Q. No. 7, HCF (a, b) is
(a) pq
(b) p³q³
(c) p³q²
(d) p²q²
Solution:
(a) a = pq² and b =p³q where a and b are positive integers and p, q are prime numbers, then HCF =pq

Question 9.
If two positive integers tn and n arc expressible in the form m = pq³ and n = p³q², where p, q are prime numbers, then HCF (m, n) =
(a) pq
(b) pq²
(c) p³q³
(d) p²q³
Solution:
(b) m and n are two positive integers and m = pq³ and n = pq², where p and q are prime numbers, then HCF = pq²

Question 10.
If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
(a) 2
(b) 3
(c) 4
(d) 1
Solution:
(c) LCM of a and 18 = 36
and HCF of a and 18 = 2
Product of LCM and HCF = product of numbers
36 x 2 = a x 18
a = \(\frac { 36\times 2 }{ 18 }\) = 4

Question 11.
The HCF of 95 and 152, is
(a) 57
(b) 1
(c) 19
(d) 38
Solution:
(c) HCF of 95 and 152 = 19

Question 12.
If HCF (26, 169) = 13, then LCM (26, 169) =
(a) 26
(b) 52
(f) 338
(d) 13
Solution:
(c) HCF (26, 169) = 13
LCM (26, 169) = \(\frac { 26\times 169 }{ 13 }\) = 338

Question 13.

(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b)

Question 14.
The decimal expansion of the rational \(\frac { 14587 }{ 1250 }\) number will terminate after
(a) one decimal place
(b) two decimal place
(c) three decimal place
(d) four decimal place
Solution:
(d)

Question 15.
If p and q are co-prime numbers, then p² and q² are
(a) co prime
(b) not co prime
(c) even
(d) odd
Solution:
(a) p and q are co-prime, then
p² and q² will also be coprime

Question 16.
Which of the following rational numbers have terminating decimal ?

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i) and (iv)
Solution:
(d) We know that a rational number has terminating decimal if the prime factors of its denominator are in the form 2^m x 5ⁿ
\(\frac { 16 }{ 225 }\) and \(\frac { 7 }{ 250 }\) has terminating decimals

Question 17.
If 3 is the least prime factor of number a and 7 is the least prime factor of number b, then the least prime factor of a + b, is
(a) 2
(b) 3
(c) 5
(d) 10
Solution:
(a) 3 is the least prime factor of a
7 is the least prime factor of b, then
Sum of a a and b will be divisible by 2
2 is the least prime factor of a + b

Question 18.
\(3.\bar { 27 }\) is
(a) an integer
(b) a rational number
(c) a natural number
(d) an irrational number
Solution:
(b) \(3.\bar { 27 }\) is a rational number

Question 19.
The smallest number by which √27 should be multiplied so as to get a rational number is
(a) √27
(b) 3√3
(c) √3
(d) 3
Solution:
(c)

Question 20.
The smallest rational number by which \(\frac { 1 }{ 3 }\) should be multiplied so that its decimal expansion terminates after one place of decimal, is
(a) \(\frac { 3 }{ 10 }\)
(b) \(\frac { 1 }{ 10 }\)
(c) 3
(d) \(\frac { 3 }{ 100 }\)
Solution:
(a) The smallest rational number which should be multiplied by \(\frac { 1 }{ 3 }\) to get a terminating

Question 21.
If n is a natural number, then 9²ⁿ – 4²ⁿ is always divisible by
(a) 5
(b) 13
(c) both 5 and 13
(d) None of these
Solution:
(c)

Question 22.
If n is any natural number, then 6ⁿ – 5ⁿ always ends with
(a) 1
(b) 3
(c) 5
(d) 7
Solution:
(a) n is any natural number and 6ⁿ – 5ⁿ
We know that 6ⁿ ends with 6 and 5ⁿ ends with 5
6ⁿ – 5ⁿ will end with 6 – 5 = 1

Question 23.
The LCM and HCF of two rational numbers are equal, then the numbers must be
(a) prime
(b) co-prime
(c) composite
(d) equal
Solution:
(d) LCM and HCF of two rational numbers are equal Then those must be equal

Question 24.
If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is
(a) 203400
(b) 194400
(c) 198400
(d) 205400
Solution:
(b) Sum of LCM and HCF of two numbers = 1260
LCM = 900 more than HCF
LCM = 900 +HCF
But LCM = HCF = 1260
900 + HCF + HCF = 1260
=> 2HCF = 1260 – 900 = 360
=> HCF = 180
and LCM = 1260 – 180 = 1080
Product = LCM x HCF = 1080 x 180 = 194400
Product of numbers = 194400

Question 25.
The remainder when the square of any prime number greater than 3 is divided by 6, is
(a) 1
(b) 3
(c) 2
(d) 4
Solution:
(a)

Question 26.
For some integer m, every even integer is of the form
(a) m
(b) m + 1
(c) 2m
(d) 2m + 1
Solution:
(c) We know that, even integers are 2, 4, 6, …
So, it can be written in the form of 2m Where, m = Integer = Z
[Since, integer is represented by Z]
or m = …, -1, 0, 1, 2, 3, …
2m = …, -2, 0, 2, 4, 6, …
Alternate Method
Let ‘a’ be a positive integer.
On dividing ‘a’ by 2, let m be the quotient and r be the remainder.
Then, by Euclid’s division algorithm, we have
a – 2m + r, where a ≤ r < 2 i.e., r = 0 and r = 1
=> a = 2 m or a = 2m + 1
When, a = 2m for some integer m, then clearly a is even.

Question 27.
For some integer q, every odd integer is of the form
(a) q
(b) q + 1
(c) 2q
(d) 2q + 1
Solution:
(d) We know that, odd integers are 1, 3, 5,…
So, it can be written in the form of 2q + 1 Where, q = integer = Z
or q = …, -1, 0, 1, 2, 3, …
2q + 1 = …, -3, -1, 1, 3, 5, …
Alternate Method
Let ‘a’ be given positive integer.
On dividing ‘a’ by 2, let q be the quotient and r be the remainder.
Then, by Euclid’s division algorithm, we have
a = 2q + r, where 0 ≤ r < 2
=> a = 2q + r, where r = 0 or r = 1
=> a = 2q or 2q + 1
When a = 2q + 1 for some integer q, then clearly a is odd.

Question 28.
n² – 1 is divisible by 8, if n is
(a) an integer
(b) a natural number
(c) an odd integer
(d) an even integer
Solution:
(c)


At k = -1, a = 4(-1)(-1 + 1) = 0 which is divisible by 8.
At k = 0, a = 4(0)(0 + 1) = 4 which is divisible by 8.
At k = 1, a = 4(1)(1 + 1) = 8 which is divisible by 8.
Hence, we can conclude from above two cases, if n is odd, then n² – 1 is divisible by 8.

Question 29.
The decimal expansion of the rational number \(\frac { 33 }{ { 2 }^{ 2 }\times 5 }\) will terminate after
(a) one decimal place
(b) two decimal places .
(c) three decimal places
(d) more than 3 decimal places
Solution:
(b)

Question 30.
If two positive integers a and b are written as a = x³y² and b = xy³ ; x, y are prime numbers, then HCF (a, b) is
(a) xy
(b) xy²
(c) x³y³
(d) x²y²
Solution:
(b)

[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]

Question 31.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(a) 10
(b) 100
(c) 504
(d) 2520
Solution:
(d) Factors of 1 to 10 numbers
1 = 1
2 = 1 x 2
3 = 1 x 3
4 = 1 x 2 x 2
5 = 1 x 5
6 = 1 x 2 x 3
7 = 1 x 7
8 = 1 x 2 x 2 x 2
9 = 1 x 3 x 3
10 = 1 x 2 x 5
LCM of number 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 1 x 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520

Question 32.
The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is
(a) 13
(b) 65
(c) 875
(d) 1750
Solution:
(a) Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70 – 5), 117 = (125 – 8), which is divisible by the required number.
Now, required number = HCF of 65, 117
[For the largest number]
For this, 117 = 65 x 1 + 52 [Dividend = divisor x quotient + remainder]
=> 65 = 52 x 1 + 13
=> 52 = 13 x 4 + 0
HCF = 13
Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.

Question 33.
If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is
(a) 4
(b) 2
(c) 1
(d) 3
Solution:
(b) By Euclid’s division algorithm,
b = aq + r, 0 ≤ r < a [dividend = divisor x quotient + remainder]
=> 117 = 65 x 1 + 52
=> 65 = 52 x 1 + 13
=> 52 = 13 x 4 + 0
HCF (65, 117)= 13 …(i)
Also, given that HCF (65, 117) = 65m – 117 …..(ii)
From equations (i) and (ii),
65m – 117 = 13
=> 65m = 130
=> m = 2

Question 34.
The decimal expansion of the rational number \(\frac { 14587 }{ 1250 }\) will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Solution:
(d)

Hence, given rational number will terminate after four decimal places.

Question 35.
Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy
(a) 1 < r < b
(b) 0 < r ≤ b
(c) 0 ≤ r < b
(d) 0 < r < b
Solution:
(c) According to Euclid’s Division lemma, for a positive pair of integers there exists unique integers q and r, such that
a = bq + r, where 0 ≤ r < b

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQSare helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
State Euclid’s division lemma.
Solution:
Euclid’s division lemma:
Let a and b be any two positive integers, then there exist unique integers q and r such that
a = bq + r, 0 ≤ r < b
If b\a, then r = 0, otherwise x. satisfies the stronger inequality 0 < r < b.

Question 2.
State Fundamental Theorem of Arithmetic.
Solution:
Fundamental Theorem of Arithmetics :
Every composite number can be expressed (factorised) as a product of primes and this factorization is unique except for the order in which the prime factors occur.

Question 3.
Write 98 as product of its prime factors.
Solution:

Question 4.
Write the exponent of 2 in the prime factorization of 144.
Solution:

Question 5.
Write the sum of the exponents of prime factors in the prime factorization of 98
Solution:
98 = 2 x 7 x 7 = 2¹ x 7²
Sum of exponents = 1 + 2 = 3

Question 6.
If the prime factorization of a natural number n is 2³ x 3² x 5² x 7, write the number of consecutive zeros in n.
Solution:
n = 2³ x 3² x 5² x 7
Number of zeros will be 5² x 2² = 10² two zeros

Question 7.
If the product of two numbers is 1080 and their H.C.F. is 30, find their L.C.M.
Solution:
Product of two numbers = 1080
H.C.F. = 30

Question 8.
Write the condition to be satisfied by q so that a rational number \(\frac { p }{ q }\) has a terminating decimal expansion. [C.B.S.E. 2008]
Solution:
In the rational number \(\frac { p }{ q }\) , the factorization of denominator q must be in form of 2^m x 5ⁿ where m and n are non-negative integers.

Question 9.
Write the condition to be satisfied by q so that a rational number \(\frac { p }{ q }\) has a non-terminating decimal expansion.
Solution:
In the rational number \(\frac { p }{ q }\) , the factorization of denominator q, is not in the form of 2^m x 5ⁿ where m and n are non-negative integers.

Question 10.
Complete the missing entries in the following factor tree.

Solution:

Question 11.
The decimal expression of the rational number \(\frac { 43 }{ { 2 }^{ 4 }\times { 5 }^{ 3 } }\) will terminate after how many places of decimals. [C.B.S.E. 2009]
Solution:
The denominator of \(\frac { 43 }{ { 2 }^{ 4 }\times { 5 }^{ 3 } }\) is 24 x 53 which is in the form of 2^m x 5ⁿ where m and n are positive integers
\(\frac { 43 }{ { 2 }^{ 4 }\times { 5 }^{ 3 } }\) has terminating decimals
The decimal expansion of \(\frac { 43 }{ { 2 }^{ 4 }\times { 5 }^{ 3 } }\) terminates after 4 (the highest power is 4) decimal places

Question 12.
Has the rational number \(\frac { 441 }{ { 2 }^{ 5 }\times { 5 }^{ 7 }\times { 7 }^{ 2 } }\) of a terminating or a non terminating decimal representation ? [CBSE 2010]
Solution:

Question 13.
Write whether \(\frac { 2\surd 45+3\surd 20 }{ 2\surd 5 }\) on simplification gives a rational or an irrational number. [CBSE 2010]
Solution:

Question 14.
What is an algorithm ?
Solution:
Algorithm : An algorithm is a series of well defined slips which gives a procedure for solving a type of problem.

Question 15.
What is a lemma ?
Solution:
A lemma is a proven statement used for proving another statement.

Question 16.
If p and q are two prime numbers, then what is their HCF ?
Solution:
If p and q are two primes, then their HCF will be 1 as they have no common factor except 1.

Question 17.
If p and q are two prime numbers, then what is their LCM ?
Solution:
If p and q are two primes, their LCM will be their product.

Question 18.
What is the total number of factors of a prime number ?
Solution:
Total number of factors of a prime number are 2, first 1 and second the number itself.

Question 19.
What is a composite number ?
Solution:
A composite number is a number which can be factorised into more than two factors.

Question 20.
What is the HCF of the smallest composite number and the smallest prime number ?
Solution:
We know that 2 is the smallest prime number and 4 is the smallest composite number
HCF of 2 and 4 = 2

Question 21.
HCF of two numbers is always a factor of their LCM (True / False).
Solution:
True.

Question 22.
π is an irrational number (True / False).
Solution:
True as value of π is neither terminating nor repeating.

Question 23.
The sum of two prime numbers is always a prime number (True / False).
Solution:
False. Sum of two prime numbers can be a composite number
e.g. 3 and 5 are prime numbers but their sum 3 + 5 = 8 is a composite number.

Question 24.
The product of any three consecutive natural numbers is divisible by 6 (True / False).
Solution:
True.

Question 25.
Every even integer is of the form 2m, where m is an integer (True / False).
Solution:
True, as 2m is divisible by 2.

Question 26.
Every odd integer is of the form 2m – 1, where m is an integer (True / False).
Solution:
True, as 2m is an even number but if we subtract 1 from it, it will be odd number.

Question 27.
The product of two irrational numbers is an irrational number (True / False).
Solution:
False, as it is not always possible that the product of two irrational number be also an irrational number, it may be a rational number
for example
√3 x √3 = 3, √7 x √7 = 7

Question 28.
The sum of two irrational numbers is an irrational number (True / False).
Solution:
False, as it is not always possible that the sum of two irrational is also an irrational number, it may be rational number also.
For example
(2 + √3) + (2 – √3) = 2 + √3 + 2 – √3 = 4

Question 29.
For what value of n, 2ⁿ x 5ⁿ ends in 5.
Solution:
In 2ⁿ x 5ⁿ ,
There is no such value of n, which satisifies the given condition.

Question 30.
If a and b are relatively prime numbers, then what is their HCF ?
Solution:
a and b are two prime numbers
Their HCF =1

Question 31.
If a and b are relatively prime numbers, then what is their LCM ?
Solution:
a and b are two prime numbers
Their LCM = a x b

Question 32.
Two numbers have 12 as their HCF and 350 as their LCM (True / False).
Solution:
HCF of two numbers = 12
and LCM is 350
False, as the HCF of two numbers is a factor of their LCM and 12 is not a factor of 350

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution:



Since, the denominator is of the form 2^m x 5ⁿ, the rational number has a terminating decimal expansion.

Question 2.
Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2^m x 5ⁿ, where m and n are non-negative integers:

Solution:

Question 3.
Write the denominator of the rational number \(\frac { 257 }{ 5000 }\) in the form 2^m x 5ⁿ , where m, n are non-negative integers. Hence, write the decimal expansion, without actual division.
Solution:

Question 4.
What can you say about the prime factorisations of the denominators of the following rationals :
Solution:
(i) 43.123456789
This decimal fraction is terminating Its denominator will be factorised in the form of 2^m x 5ⁿ where m and n are non-negative integers.

This decimal fraction is non-terminating repeating decimals.
The denominator of their fraction will be not in the form of 2^m x 5ⁿ where m and n are non-negative integers.

This decimal fraction is terminating
Its denominator will be factorised in the form of 2^m x 5ⁿ where m and n are non-negative integers
(iv) 0.120120012000120000 ………..
This decimal fraction in non-terminating non recurring
Its denominator will not be factorised in the form of 2^m x 5ⁿ where m and n are non negative integers

Question 5.
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 \(\frac { p }{ q }\) ? Give reasons. [NCERT Exemplar]
Solution:
327.7081 is terminating decimal number. So, it represents a rational number and also its denominator must have the form 2^m x 5ⁿ.

Hence, the prime factors of q is 2 and 5.

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
Show that the following numbers are irrational
(i) \(\frac { 1 }{ \surd 2 }\)
(ii) 7 √5
(iii) 6 + √2
(iv) 3 – √5
Solution:



But it contradics that because √5 is irrational
3 – √5 is irrational

Question 2.
Prove that following numbers are irrationals :
(i) \(\frac { 2 }{ \surd 7 }\)
(ii) \(\frac { 3 }{ 2\surd 5 }\)
(iii) 4 + √2
(iv) 5 √2
Solution:




5 √2 is an irrational number

Question 3.
Show that 2 – √3 is an irrational number. [C.B.S.E. 2008]
Solution:
Let 2 – √3 is not an irrational number

√3 is a rational number
But it contradicts because √3 is an irrational number
2 – √3 is an irrational number
Hence proved.

Question 4.
Show that 3 + √2 is an irrational number.
Solution:
Let 3 + √2 is a rational number

and √2 is irrational
But our suppositon is wrong
3 + √2 is an irrational number

Question 5.
Prove that 4 – 5√2 is an irrational number. [CBSE 2010]
Solution:
Let 4 – 5 √2 is not are irrational number
and let 4 – 5 √2 is a rational number
and 4 – 5 √2 = \(\frac { a }{ b }\) where a and b are positive prime integers

√2 is a rational number
But √2 is an irrational number
Our supposition is wrong
4 – 5 √2 is an irrational number

Question 6.
Show that 5 – 2 √3 is an irrational number.
Solution:
Let 5 – 2 √3 is a rational number
Let 5 – 2 √3 = \(\frac { a }{ b }\) where a and b are positive integers

and √3 is a rational number
Our supposition is wrong
5 – 2 √3 is a rational number

Question 7.
Prove that 2 √3 – 1 is an irrational number. [CBSE 2010]
Solution:
Let 2 √3 – 1 is not an irrational number
and let 2 √3 – 1 a ration number
and then 2 √3 – 1 = \(\frac { a }{ b }\) where a, b positive prime integers

√3 is a rational number
But √3 is an irrational number
Our supposition is wrong
2 √3 – 1 is an irrational number

Question 8.
Prove that 2 – 3 √5 is an irrational number. [CBSE 2010]
Solution:
Let 2 – 3 √5 is not an irrational number and let 2 – 3 √5 is a rational number
Let 2 – 3 √5 = \(\frac { a }{ b }\) where a and b are positive prime integers

\(\Longrightarrow \frac { 2b-a }{ 3b } =\surd 5\)
√5 is a rational
But √5 is an irrational number
Our supposition is wrong
2 – 3 √5 is an irrational

Question 9.
Prove that √5 + √3 is irrational.
Solution:
Let √5 + √3 is a rational number
and let √5 + √3 = \(\frac { a }{ b }\) where a and b are co-primes

√3 is a rational number
But it contradics as √3 is irrational number
√5 + √3 is irrational

Question 10.
Prove that √2 + √3 is an irrational number.
Solution:
Let us suppose that √2 + √3 is rational.
Let √2 + √3 = a, where a is rational.
Therefore, √2 = a – √3
Squaring on both sides, we get

which is a contradiction as the right hand side is a rational number while √3 is irrational.
Hence, √2 + √3 is irrational.

Question 11.
Prove that for any prime positive integer p, √p is an irrational number.
Solution:
Suppose √p is not a rational number
Let √p be a rational number
and let √p = \(\frac { a }{ b }\)
Where a and b are co-prime number


But it contradicts that a and b are co-primes
Hence our supposition is wrong
√p is an irrational

Question 12.
If p, q are prime positive integers, prove that √p + √q is an irrational number
Solution:


Hence proved.

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. x H.C.F. = Product of the integers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
26 = 2 x 13
91 = 7 x 13
H.C.F. = 13
and L.C.M. = 2 x 7 x 13 = 182
Now, L.C.M. x H.C.F. = 182 x 13 = 2366
and 26 x 91 = 2366
L.C.M. x H.C.F. = Product of integers

Question 2.
Find the L.C.M. and H.C.F. of the following integers by applying the prime factorisation method :
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
(iv) 40, 36 and 126
(v) 84, 90 and 120
(vi) 24,15 and 36
Solution:

Question 3.
Given that HCF (306, 657) = 9, Find LCM (306, 657). [NCERT]
Solution:
HCF of 306, 657 = 9

Question 4.
Can two numbers have 16 as their H.C.F. and 380 as their L.C.M. ? Give reason.
Solution:
H.C.F. of two numbers = 16
and their L.C.M. = 380

We know the H.C.F. of two numbers is a factor of their L.C.M. but 16 is not a factor of 380 or 380 is not divisible by 16
It can not be possible.

Question 5.
The H.C.F. of two numbers is 145 and their L.C.M. is 2175. If one number is 725, find the other.
Solution:
First number = 725
Let second number = x
Their H.C.F. = 145
and L.C.M. = 2175

Second number = 435

Question 6.
The H.C.F. of two numbers is 16 and their product is 3072. Find their L.C.M.
Solution:
H.C.F. of two numbers = 16
and product of two numbers = 3072

Question 7.
The L.C.M. and H.C.F. of two numbers are 180 and 6 respectively. If one of the number is 30, find the other number.
Solution:
First number = 30
Let x be the second number
Their L.C.M. = 180 and H.C.F. = 6
We know that
first number x second number = L.C.M. x H.C.F.
30 x x = 180 x 6
⇒ x = \(\frac { 180\times 6 }{ 30 }\) = 36
Second number = 36

Question 8.
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Solution:
L.C.M. of 520 and 468

= 2 x 2 x 9 x 10 x 13 = 4680
The number which is increased = 17
Required number 4680 – 17 = 4663

Question 9.
Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
Solution:
Dividing by 28 and 32, the remainders are 8 and 12 respectively
28 – 8 = 20
32 – 12 = 20
Common difference = 20
Now, L.C.M. of 28 and 32

= 2 x 2 x 7 x 8 = 224
Required smallest number = 224 – 20 = 204

Question 10.
What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case ?
Solution:
L.C.M. of 35, 56, 91

= 5 x 7 x 8 x 13 = 3640
Remainder in each case = 7
The required smallest number = 3640 + 7 = 3647

Question 11.
A rectangular courtyard is 18 m 72 cm long and 13 m 20 cm broad. It is to be paved with square tiles of same size. Find the least possible number of such tiles.
Solution:
Length of rectangle = 18 m 72 cm = 1872 cm
and breadth = 13 m 20 cm = 1320 cm
Side of the greatest size of square tile = H.C.F. of 1872 and 1320

Question 12.
Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Solution:
Greatest number of 6 digits = 999999
Now L.C.M. of 24, 15 and 36

We get quotient = 2777
and remainder = 279
Required number = 999999 – 279 = 999720

Question 13.
Determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Solution:

Required number will be = 110000 – 800 =109200

Question 14.
Find the least number that is divisible by ail the numbers between 1 to 10 (both inclusive).
Solution:
The required least number which is divisible by 1 to 10 will be the L.C.M. of 1 to 10
L.C.M. of 1 to 10

Question 15.
A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again ?
Solution:
Circumference of a circular field = 360 km
Three cyclist start together who can cycle 48, 60 and 72 km per day round the field
L.C.M. of 48, 60, 72

They will meet again after 720 km distance

Question 16.
In a morning walk, three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that they can cover the distance in complete steps ?
Solution:
Measures of steps of three persons = 80 cm, 85 cm and 90 cm
Minimum required distance covered by them = L.C.M. of 80 cm, 85 cm, 90 cm

= 2 x 5 x 8 x 9 x 17
= 12240 cm
= 122.40 m
= 122 m 40 cm

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.