CBSE Class 10

RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3

Other Exercises

• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3
• RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS
• RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

Question 1.
Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following :

Solution:

Question 2.
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm.

Solution:

Question 3.
Obtain all zeros of the polynomial f(x) = 2x⁴ + x³ – 14x² – 19x – 6, if two of its zeros are -2 and -1.
Solution:

Question 4.
Obtain all zeros of f(x) = x³ + 13x² + 32x + 20, if one of its zeros is -2.
Solution:
f(x) = x³ + 13x² + 32x + 20
One zero = -2 or x = -2
x + 2 is a factor of f (x)
Now dividing f(x) by x + 2, we get

x + 1 = 0 => x = -1
and x + 10 = 0
=> x = -10
-1 and -10
Hence zeros are -10, -1, -2

Question 5.
Obtain all zeros of the polynomial f(x) = x⁴ – 3x³ – x² + 9x – 6, if two of its zeros are – √3 and √3
Solution:

Question 6.
Find all zeros of the polynomial f(x) = 2x⁴ – 2x³ – 7x² + 3x + 6, if its two zeros are \(\surd \frac { 3 }{ 2 }\) and – \(\surd \frac { 3 }{ 2 }\)
Solution:

Question 7.
Find all the zeros of the polynomial x⁴ + x³ – 34x² -4x+ 120, if two of its zeros are 2 and -2. [CBSE 2008]
Solution:

Either x + 6 = 0, then x = -6
or x – 5 = 0, then x = 5
Hence other two zeros are -6, 5
and all zeros are 2, -2, -6, 5

Question 8.
Find all zeros of the polynomial 2x⁴ + 7x³ – 19x² – 14x + 30, if two of its zeros are √2 and -√2
Solution:

Question 9.
Find all the zeros of the polynomial 2x³ + x² – 6x – 3, if two of its zeros are – √3 and √3. (CBSE 2009)
Solution:
Let f(x) = 2x³ + x² – 6x – 3
and two zeros of f(x) are – √3 and √3

Question 10.
Find all the zeros of the polynomial x³ + 3x² – 2x – 6, if two of its zeros are – √2 and √2 (CBSE2009)
Solution:

Question 11.
What must be added to the polynomial f(x) = x⁴ + 2x³ – 2x² + x – 1 so that the resulting polynomial is exactly divisible by x² + 2x – 3 ?
Solution:

Question 12.
What must be subtracted from the polynomial f(x) = x⁴ + 2x³ – 13x² – 12x + 21 so that the resulting polynomial is exactly divisible by x³ – 4x + 3 ?
Solution:

The resulting polynomial is exactly divisible by x² – 4x + 3
Remainder = 0
=> 2x -3 – k = 0
=> k = 2x – 3
(2x – 3) must be subtracted

Question 13.
Given that √2 is a zero of the cubic polynomial 16x³ + √2x² – 10x – 4√2 , find its other two zeroes. [NCERT Exemplar]
Solution:

Question 14.
Given that x – √5 is a factor of the cubic polynomial x³ – 3 √5 x² + 13x – 3 √5 , find all the zeroes of the polynomial. [NCERT Exemplar|
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3
• RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS
• RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

Question 1.
Verify that numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case :
(i) f(x) = 2x³ – x² – 5x + 2 ; \(\frac { 1 }{ 2 }\) , 1, -2
(ii) g(x) = x³ – 4x² + 5x – 2 ; 2, 1, 1
Solution:

Question 2.
Find a cubic polynomial with the sum, sum of product of its zeros taken two at a time and product of its zeros as 3, -1 and -3 respectively.
Solution:

Question 3.
If the zeros of the polynomial f(x) = 2x³ – 15x² + 37x – 30 are in AP. Find them.
Solution:

Question 4.
Find the condition that the zeros of the polynomial f(x) = x³ + 3px² + 3qx + r may be in AP.
Solution:

Question 5.
If the zeros of the polynomial f(x) = ax³ + 3bx² + 3cx + d are in A.P., prove that 2b³ – 3abc + a²d = 0.
Solution:

Question 6.
If the zeros of the polynomial f(x) = x³ – 12x² + 39x + k are in AP, find the value of k.
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3
• RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS
• RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

Question 1.
Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their co-efficients :

Solution:
(i) f(x) = x² – 2x – 8

Question 2.
For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution:
(i) Given that, sum of zeroes (S) = – \(\frac { 8 }{ 3 }\)
and product of zeroes (P) = \(\frac { 4 }{ 3 }\)
Required quadratic expression,

Question 3.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 5x + 4, find the value of \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } -2\alpha \beta\).
Solution:

Question 4.
If α and β are the zeros of the quadratic polynomial p(y) = 5y² – 7y + 1, find the value of \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }\)
Solution:

Question 5.
If α and β are the zeros of the quadratic polynomial f(x) = x² – x – 4, find the value of \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } -\alpha \beta\)
Solution:

Question 6.
If α and β are the zeros of the quadratic polynomial f(x) = x² + x – 2, find the value of \(\frac { 1 }{ \alpha } -\frac { 1 }{ \beta }\)
Solution:

Question 7.
If one zero of the quadratic polynomial f(x) = 4x² – 8kx – 9 is negative of the other, find the value of k.
Solution:

Question 8.
If the sum of the zeros of the quadratic polynomial f(t) = kt² + 2t + 3k is equal to their product, find the value of k.
Solution:

Question 9.
If α and β are the zeros of the quadratic polynomial p(x) = 4x² – 5x – 1, find the value of α²β + αβ².
Solution:

Question 10.
If α and β are the zeros of the quadratic polynomial f(t) = t² – 4t + 3, find the value of α⁴β³ + α³β⁴.
Solution:

Question 11.
If α and β are the zeros of the quadratic polynomial f (x) = 6x⁴ + x – 2, find the value of \(\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha }\)
Solution:

Question 12.
If α and β are the zeros of the quadratic polynomial p(s) = 3s² – 6s + 4, find the value of \(\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta\)
Solution:

Question 13.
If the squared difference of the zeros of the quadratic polynomial f(x) = x² + px + 45 is equal to 144, find the value of p
Solution:

Question 14.
If α and β are the zeros of the quadratic polynomial f(x) = x² – px + q, prove that:

Solution:

Question 15.
If α and β are the zeros of the quadratic polynomial f(x) = x² – p(x + 1) – c, show that (α + 1) (β + 1) = 1 – c.
Solution:

Question 16.
If α and β are the zeros of the quadratic polynomial such that α + β = 24 and α – β = 8, find a quadratic polynomial having α and β as its zeros.
Solution:

Question 17.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 1, find a quadratic polynomial whose zeros are \(\frac { 2\alpha }{ \beta }\) and \(\frac { 2\beta }{ \alpha }\)
Solution:

Question 18.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 3x – 2, find a quadratic polynomial whose zeros are \(\frac { 1 }{ 2\alpha +\beta }\) and \(\frac { 1 }{ 2\beta +\alpha }\)
Solution:

Question 19.
If α and β are the zeroes of the polynomial f(x) = x² + px + q, form a polynomial whose zeros are (α + β)² and (α – β)².
Solution:

Question 20.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 2x + 3, find a polynomial whose roots are :
(i) α + 2, β + 2
(ii) \(\frac { \alpha -1 }{ \alpha +1 } ,\frac { \beta -1 }{ \beta +1 }\)
Solution:

Question 21.
If α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c, then evaluate :


Solution:

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1 are helpful to complete your math homework.

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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

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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.

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