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RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Additional Questions.

Board	RBSE
Te×tbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Polynomials
E×ercise	Additional Questions
Number of Questions Solved	57
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Additional Questions

Multiple Choice Questions

Prime factors of 144 are 2x2x2x2, 3×3.

Questions 1.
If 2 is factor of (RBSESolutions.com) polynomial f(x) = x⁴ – x³ – 4x² + kx + 10 then find k.
(A) 2
(B) -2
(C) -1
(D) 1
Solution
f(2) = 2⁴ – 2³ – 4 × 2² + k × 2 + 10
⇒ 0 = 16 – 8 – 16 + 2k + 10
⇒ 0 = 2k + 2
⇒ 2k = -2
⇒ k = -1
Hence option (C) is correct.

Question 2.
Find that polynomial whose zeros are -5 and 4
(A) x² – x – 20
(B) x² + x – 20
(C) x² – x + 20
(D) x² + x + 20
Solution
Let α = – 5 and β = 4
Then α + β = -5 + 4 = -1 and αβ = -5 x 4 = -20
Quadratic polynomial x² – (α + β) x + αβ = 0
⇒ x² – (-1) x + (-20) = 0
⇒ x² + x – 20 = 0
Hence, option (B) is correct.

Question 3.
Expression (x – 3) will be a (RBSESolutions.com) factor of polynomial f(x) = x³ + x² – 17x + 15 if:
(A) f(3) = 0
(B) f(-3) = 0
(C) f(2) = 0
(D) f(-2) = 0
Solution
f(x) = x³ + x² – 17x + 15
f(3) = (3)³ + (3)² – 17(3) + 15 = 27 + 9 – 51 + 15 = 0
f(3) = 0
Hence, option (A) is correct.

Question 4.
If (x – 5) is a factor of (RBSESolutions.com) polynomial x³ – 3x² + kx – 10 then value of k will be:
(A) -8
(B) -7
(C) 5
(D) 8
Solution
Let p(x) = x³ – 3x² + kx – 10
if (x – 5), is a factor of p(x)
then p(x) = 0
(5)³ – 3(5)² + k(5) – 10 = 0
⇒ 125 – 75 + 5k – 10 = 0
⇒ 40 + 5k = 0
⇒ 5k = -40
⇒ k = -8
Thus k = -8
Hence, option (A) is correct.

Question 5.
Zero of 3y³ + 8y² – 1 is
(A) 1
(B) (-1)
(C) 0
(D) None of these
Solution
Let p(y) = 3y³ + 8y² – 1
p(1) = 3 × (1)³ + 8(1)² – 1 = 3 + 8 – 1 = 10 ≠ 0
p(-1) = 3(-1)³ + 8(-1)² – 1 = -3 + 8 – 1 = 4 ≠ 0
and p(0) = 3 × (0)³ + 8(0)² – 1 = 0 + 0 – 3 = -3 ≠ 0
Thus by putting y = 1, -1 the expression (RBSESolutions.com) obtained is not equal to zero.
Hence, the option (D) is correct.

Question 6.
If (x – 1) is a factor of polynomial 2x² + kx + √2, then k will be
(A) 2 + √2
(B) 2 – √2
(C) – (2 + √2)
(D) – (2 – √2)
Solution
Let p{x) = 2x² + kx + √2
If (x – 1), is a factor of p(x)
then p(1) = 0
2(1)² + k(1) + √2 = 0
⇒ 2 × 1 + k + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k + ( 2 + √2) = 0
⇒ k = -(2 + √2)
Hence, option (C) is correct.

Question 7.
One zero of p(x) = 2x + 1 will be
(A) \(\frac { 1 }{ 2 }\)
(B) 3
(C) \(\frac { -1 }{ 2 }\)
(D) 1
Solution
p(x) = 2x + 1
For zeros p(x) = 0
0 = 2x + 1
⇒ x = \(\frac { -1 }{ 2 }\)
Hence, option (C) is correct.

Short Answer Type Questions

Question 1.
If α, β are zeros of any quadratic (RBSESolutions.com) equation, then write the quadratic equation.
Solution
k[x² – (α + β)x + αβ]

Question 2.
If f(x) are two expressions, then write the relationship between their L.C.M. and H.C.F.
Solution
L.C.M. × H.C.F. = f(x) × g(x)

Question 3.
If α and β are zeros of any quadratic polynomial ax² + bx + c, then write the value of α + β and αβ.
Solution
α + β = \(\frac { -b }{ a }\)
and αβ = \(\frac { c }{ a }\)

Question 4.
What is the zero of (RBSESolutions.com) polynomial?
Solution
The value of x for which polynomial f(x) = 0, is called zero of the polynomial.

Question 5.
If α and β are zeros of polynomial f(x) = x² – 5x + k such that α – β = 1, then find k.
Solution
If α, β are zeros of polynomial x² – 5x + k.

Question 6.
Find the condition for which zeros of (RBSESolutions.com) polynomial, p(x) = ax² + bx + c are inverse of each other. (CBSE 2012)
Solution
Let α is first zero of polynomial of p(x) = ax² + bx + c
According to question, second zero of polynomial

Hence, required condition is c = a

Question 7.
If a – b, a + b are zeros of polynomial, x³ – 3x² + x + 1, then find a and b.
Solution
Given polynomial = x³ – 3x² + x + 1

This degree and leading coefficient calculator finds the degree, leading term, and coefficient associated with it absolutely for free.

Question 8.
Find the zeros of quadratic polynomial x² + x – 2 and (RBSESolutions.com) verify the relationship between zeros and coeffcient (M.S.B. Raj. 2016)
Solution
Given quadratic polynomial
f(x) = x² + x – 2 = x² + 2x – x – 2 = x (x + 2) – 1 (x + 2) = (x – 1) (x + 2)
To find zero, f(x) = 0
(x- 1) (x + 2) = 0
x – 1 = 0 or x + 2 = 0
x = 1 or x = -2
Thus 1 and -2 are two zeros of given polynomial
Relation between zeros and coefficient
Sum of zeros = 1 + (- 2) = – 1
and product of zeros = 1 × (-2) = -2
Comparing given polynomial with ax² + bx + c
a = 1, b = 1 and c = -2
Sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -1 }{ 1 }\) = -1
and product of zeros = \(\frac { c }{ a }\) = \(\frac { -2 }{ 1 }\) = -2
Hence, relationship between zeros and coeffi-cient is verified.

Synthetic Division Calculator that can divide polynomials and demonstrate the process in a table format.

Question 9.
Divide x³ – 3x² + 3x – 5 by x – 1 – x² and test (RBSESolutions.com) division algorithm. (M.S.B. Raj. 2013)
Solution

Question 10.
Find all the zeros of (RBSESolutions.com) polynomial x³ – 6x² + 11x – 6 whereas 1 and 2 are its two zeros. (CBSE 2012)
Solution
we know that if a is one zero of polynomial f(x) then (x – α) will be a factor of f(x).
According to equation, polynomial f(x) = x³ – 6x² + 11x – 6 has two zeros 1 and 2.
Thus (x – 1) (x – 2) = (x² – 3x + 2), will be a factor of f(x).
Now dividing polynomial f(x) by x² – 3x + 2

Using division algorithm for polynomial
Dividend = Divisor × Quotient + Remainder
x³ – 6x² + 11x – 6 = (x² – 3x + 2)(x – 3) + 0 = (x – 1)(x – 2)(x – 3)
For zeros of polynomial
f(x) = 0
⇒ (x – 1) (x – 2) (x – 3) = 0
⇒ x – 1 = 0, x – 2 = 0, x – 3 = 0
⇒ x = 1, x = 2, x = 3
Hence, all the zeros of polynomial f(x) are 1, 2, 3.

Question 11.
If α, β are zeros of polynomial p(x) = 2x² + 5x + k which (RBSESolutions.com) satisfies the relation α² + β² + αβ = \(\frac { 21 }{ 4 }\), then find value of k. (CBSE 2012)
Solution

Quadratic Equation

Multiple Choice Equations

Welcome to our step-by-step math roots calculator.

Question 1.
Roots of the equation ax² + bx + c = 0, a ≠ 0 will not be real if
(A) b² < 4ac
(B) b² > 4ac
(C) b² = 4ac
(D) b = 4ac
Solution
Option (A) is correct.

Question 2.
If \(\frac { 1 }{ 2 }\) is one root of (RBSESolutions.com) quadratic equation x² + kx – \(\frac { 5 }{ 4 }\) = 0, then value of k will be [NCERT Exemplar Problem]
(A) 2
(B) -2
(C) \(\frac { 1 }{ 4 }\)
(D) \(\frac { 1 }{ 2 }\)
Solution
Option (A) is correct.

Question 3.
Roots of quadratic equation 2x² – x – 6 = 0 are [CBSE 2012]
(A) -2, \(\frac { 3 }{ 2 }\)
(B) 2, \(\frac { -3 }{ 2 }\)
(C) -2, \(\frac { -3 }{ 2 }\)
(D) 2, \(\frac { 3 }{ 2 }\)
Solution
Given quadratic equation is :
2x² – x – 6 = 0
⇒ 2x² – (4 – 3) x – 6 = 0
⇒ 2x² – 4x + 3x – 6 = 0
⇒ (2x² – 4x) + (3x – 6) = 0
⇒ 2x(x – 2) + 3(x – 2) = 0
⇒ (x – 2)(2x + 3) = 0
⇒ x-2 = 0 or 2x + 3 = 0
⇒ x = 2 or x = \(\frac { -3 }{ 2 }\)
Hence, option (B) is correct.

Question 4.
For which value of k, quadratic (RBSESolutions.com) equation 2x² – kx + k = 0 has equal roots [NCERT Exemplar Problem]
(A) only 0
(B) only 4
(C) only 8
(D) 0, 8
Solution
option (D) is correct.

Question 5.
Root of equation x² – 9 = 0 are
(A) √3
(B) -√3
(C) 9
(D) ±3
Solution
x² – 9 = 0
⇒ x² = 9
⇒ x = ±√9
⇒ x = ± 3
Hence, option (D) is correct.

Question 6.
Quadratic (RBSESolutions.com) equation 2x² – √5 x + 1 = 0 has [NCERT Exemplar Problem]
(A) Two distinct real roots.
(B) Two equal real roots
(C) No real root
(D) More than two real roots
Solution
Option (C) is correct.

Question 7.
Product of root of equation 2x² + x – 6 = 0 will be
(A) -3
(B) -7
(C) 2
(D) 0
Solution
Equation 2x² + x – 6 = 0
Here a = 2, b = 1, c = -6
Product of root = \(\frac { c }{ a }\) = \(\frac { -6 }{ 2 }\) = -3
Hence, option (A) is correct.

Question 8.
Which of the following (RBSESolutions.com) equation has sum of roots 3 ? [NCERT Exemplar Problem]
(A) 2x² – 3x + 6 = 0
(B) -x² + 3x – 3 = 0
(C) -√2 x² – \(\frac { 3 }{ \surd 2 }\) x + 1 = 0
(D) 3x² – 3x + 3 = 0
Solution
Sum of roots of option (A) = \(\frac { -b }{ a }\) = \(\frac { 3 }{ 2 }\)
Sum of roots of option (B) = \(\frac { -b }{ a }\) = 3
Hence, option (B) is correct.

Question 9.
Solution of equation x² – 4x = 0 are
(A) 4, 4
(B) 2, 2
(C) 0, 4
(D) 0, 2
Solution
x² – 4x = 0
⇒ x (x – 4) =0
⇒ x = 0 or x – 4 = 0
⇒ x = 0 or x = 4
Thus x = 0, 4
Hence, option (C) is correct.

Solution 10.
Quadratic (RBSESolutions.com) equation px² + qx + r = 0, p ≠ 0 has equal roots if
(A) p² < 4pr
(B) p² > 4qr
(C) q² = 4pr
(D) p² = 4qr
Solution
Option (C) is correct.

Solution 11.
Root of quadratic equation (x² + 1)² – x = 0 are [NCERT Exemplar Problem]
(A) Four real roots
(B) Two real roots
(C) One real root
(D) No real root
Solution
Option (D) is correct.

Question 12.
If equation x² + 3ax + k = 0 has x = -a as (RBSESolutions.com) solution, then k will be
(A) 2a²
(B) 0
(C) 2
(D) -2a
Solution
Option (A) is correct.

Very Short/Short Answer Type Questions

Question 1.
For quadratic equation ax² + bx + c = 0, a ≠ 0, at which nature of roots depends ?
Solution
Discriminant (D) = b² – 4ac

Question 2.
Describe nature of (RBSESolutions.com) roots of quadratic equation ax² + bx + c = 0, a ≠ 0
Solution
(i) If (b² – 4ac) > 0, then roots will be real and distinct.
(ii) If (b² – 4ac) = 0, then roots will be equal and real.
(iii) If (b² – 4ac) < 0, then roots will be imaginary.

Question 3.
If the sum of two natural numbers is 8 and the product is 15, then find numbers. (CBSE 2012)
Solution
Let the first natural number be x.
Sum of two natural numbers is 8 then other natural numbers will be 8 – x.
According to question.
Product of both natural numbers = 15
⇒ x (8 – x) = 15
⇒ 8x – x² = 15
⇒ x² – 8x + 15 = 0
⇒ x² – (5 + 3)x + 15 = 0
⇒ x² – 5x – 3x + 15 = 0
⇒ (x² – 5x) – (3x – 15) = 0
⇒ x (x – 5) – 3 (x – 5) = 0
⇒ (x – 5) (x – 3) = 0
⇒ x – 5 = 0 or x – 3 = 0
⇒ x = 5 or x = 3
Thus, if First natural no. = 5
then Second natural no. = 8
if First natural no. = 3
or Second natural no. = 8

Question 4.
Find the roots of (RBSESolutions.com) quadratic equation √2 x² + 7x + 5√2 = 0. (CBSE 2013)
Solution
Given quadratic equation is

Question 5.
Solve for x

Solution
Given the quadratic equation is

Question 6.
Find k for which quadratic (RBSESolutions.com) equation (k – 12) x² + 2(k – 12) x + 2 = 0 has equal and real roots.
Solution
Given
(k – 12) x² + 2(k – 12)x + 2 = 0
Comparing it with quadratic equation ax² + bx + c = 0
a = (k – 12),b = 2(k – 12), c = 2
Discriminant (D) = b² – 4ac
= 4(k – 12)² – 4 × (k – 12) × 2
= 4(k – 12) [k – 12 – 2]
= 4(k – 12)(k – 14)
Given equation will have real and equal roots, if discriminant = 0.
D = 0
⇒ 4(k – 12)(k – 14) =0
⇒ k – 12 = 0 or k – 14 = 0
⇒ k = 12 or k = 14

Question 7.
A field is in the shape of a right-angled (RBSESolutions.com) triangle. Its hypotenuse is 1 m larger than twice the smallest side. If its third side is 7 m larger than the smallest side, then find sides of the field.
Solution
Let length of smallest side = x m
hypotenuse = (2x + 1) m and third side = (x + 7) m
By pythagorus theorem,
(hypotenuse)² = sum of square of other both
⇒ (2x + 1)² = x² + (x + 7)²
⇒ 4x² + 4x + 1 = 2x² + 14x + 49
⇒ 2x² – 10x – 48 = 0
⇒ x² – 5x – 24 = 0
⇒ x² – 8x + 3x – 24 = 0
⇒ x(x – 8) + 3(x – 8) = 0
⇒ (x – 8) (x + 3) =0
⇒ x = 8, – 3
⇒ x = 8 [∵ x = -3 not possible]
hypotenuse = 2 × 8 + 1 = 17 m
and third side = 8 + 7 = 15 m.
Hence, the length of sides of the field is 8m, 17 m, and 15 m.

Question 8.
Solve for x:
x² – 4ax – b² + 4a² = 0. [CBSE 2012]
Solution
Given quadratic equation x² – 4ax – b² + 4a² = 0
Comparing it by quadratic (RBSESolutions.com) standard equation Ax² + Bx + C = 0
A = 1, B = -4a and C = -(b² – 4a²)
By Shridhar Acharya formula

Question 9.
Solve for x: √3 x² – 2√2 x – 2√3 = 0. [CBSE 2015]
Solution

Question 10.
If 12 is added to a natural number, (RBSESolutions.com) then it becomes 160 times of its reciprocal. Find that number. [NCERT Exemplar Problem]
Solution
Let x be a natural number. According to the question, when 12 is added to a natural number.
It becomes 160 times its reciprocal
x + 12 = 160 × \(\frac { 1 }{ x }\)
⇒ x² + 12x = 160
⇒ x² + 12x – 160 = 0
⇒ x² + (20 – 8) x – 160 = 0
⇒ x² + 20x – 8x – 160 = 0
⇒ x (x + 20) – 8 (x + 20) = 0
⇒ (x + 20) (x – 8) = 0
if x + 20 = 0, then x = -20
or x – 8 = 0, then x = 8
x is a natural number, so it cannot be negative
x = 8
Hence, number is 8.

Question 11.
If a two digit number is 4 times the sum (RBSESolutions.com) of its digit and three times the product of it digit, then find the number.
Solution
Let digit of ten and unit place are x and y respectively.
number = 10x + y
According to first condition
number = 4 × sum of digit
⇒ 10x + y = 4 × (x + y)
⇒ 10x + y = 4x + 4y
⇒ 10x – 4x = 4y – y
⇒ 6x = 3y
⇒ 2x = y
⇒ y = 2x …(i)
According to second condition
number = 3 × Product of digit
10x + y = 3 × x × y
⇒ 10x + y = 3xy …(ii)
From (RBSESolutions.com) equation (i) putting y = 2x in equation
10x + 2x = 3x × 2x
⇒ 12x = 6x²
⇒ 6x² – 12x = 0
⇒ 6x (x – 2) = 0
when 6x = 0, then x = 0
when x – 2 = 0, then x = 2
Hence, given number is two digit number,
so x ≠ 0 and hence x = 2
⇒ y = 2 × 2 = 4 [∵ y = 2x]
Hence, required number = 10x + y = 10 × 2 + 4 = 24

Long Answer Type Questions

Question 1.
A pillar has to be fitted at the point of the (RBSESolutions.com) boundary of a circular park of diameter 13 m. Such that two gates A and B situated at both ends of diameter having a difference in distance of 7 m from this pillar. Is this possible? If yes, find the distance of the pillar from both gates. (M.S.B. Raj 2013)
Solution
Let pillar is fitted at point C and distance from gate B to point C = x m.

According to the question, the difference (RBSESolutions.com) in the distance from gate A and B to the pillar is 7 m.
AC = (x + 7) m
AB is diameter
∠ACB = 90° (∵ angle in semi circle is right angle)
Now in right angled triangle ACB
AB² = AC² + BC² (By pythagorus Theorem)
⇒ 132 = (x + 7)² + x²
⇒ 169 = x² + 49 + 14x + x²
⇒ 169 = 2x² + 14x + 49
⇒ 2x² + 14x + 49 – 169 = 0
⇒ 2x² + 14x – 120 = 0
⇒ x² + 7x – 60 = 0
Now b² – 4ac = (7)² – 4 × 1 × (-60) = 49 + 240 = 289 > 0
Thus, given quadratic equation has two (RBSESolutions.com) real roots and so pillar can be fitted at boundary of park.
Now x² + 7x – 60 = 0
⇒ x² + (12 – 5)x – 60 = 0
⇒ x² + 12x – 5x – 60 = 0
⇒ (x² + 12x) – (5x + 60) = 0
⇒ x (x + 12) – 5(x + 12) = 0
⇒ (x + 12) (x – 5) = 0
⇒ x + 12 = 0 and x – 5 = 0
⇒ x = – 12 and x = 5
x is distance between pillar and gate B so Ignore x = -12
x = 5 m and x + 7 = 5 + 7 = 12 m.
Hence distance from pillar to gate A = 12 m.
and from pillar to gate B = 5 m.

Question 2.
The difference of square of two (RBSESolutions.com) numbers is 180. Square of the smaller number is 8 times the larger number. Find two numbers.
Solution
Let larger number = x
Smaller number = y
According to first condition of question
x² – y² = 180 …(i)
According to second condition of question
y² = 8x …(ii)
Putting value of y² from equation (ii) in equation (i)
x² – 8x = 180
⇒ x² – 8x – 180 = 0
Comparing it by ax² + bx + c = 0
a = 1, b = -8, c = -180
Then by (RBSESolutions.com) quadratic formula

Thus x = 18 and -10
When x = 18, then from equation (ii)
y² = 8 × 18 = 144
⇒ y = ±√144
⇒ y = ± 12
When x = -10, then (RBSESolutions.com) from equation (ii)
y² = 8 × (-10)
⇒ y² = -80 (not possible)
∴ y = ± 12
y = + 12 and – 12
Hence, required numberes will be 18 and 12 or 18, -12

Question 3.
The sum of areas of two squares is 468 sq.m. If the difference between their perimeter is 24 m, then find sides of both squares.
Solution
Let the side of a square is x m.
Perimeter of that square = 4x m
Difference in peimeter is 24 m
Perimeter of second square = 4x + 24 m
Then, side of (RBSESolutions.com) second square = \(\frac { 4x+24 }{ 4 }\) = \(\frac { 4(x+6) }{ 4 }\) = (x + 6) m
Area of first square = x² sq m
Area of second square = (x + 6)² sq m = x² + 12x + 36 sq m
Sum of areas of both squares = 468 sq m
x² + (x² + 12x + 36) = 468
⇒ 2x² + 12x + 36 – 468 = 0
⇒ 2x²+ 12x – 432 =0
⇒ 2(x² + 6x – 216) = 0
⇒ x² + 6x – 216 = 0
⇒ x² + 18x – 12x – 216 = 0
⇒ x(x + 18) – 12(x + 18) = 0
⇒ (x + 18)(x – 12) = 0
when x + 18 = 0, then x = -18 (not possible)
or x – 12 = 0, then x = 12
∴ x = 12
Side of smaller square = 12 m
and sideof larger square = x + 6 = 12 + 6 = 18 m
Thus sides of both (RBSESolutions.com) squares are 12 m. and 18 m respatively

H.C.F. and L.C.M.

Multiple Choice Questions

Question 1.
H.C.F. of 4x²y and x³y² will be
(A) x²y
(B) x²y²
(C) 4x³y²
(D) 4x²y²
Solution
Option (C) is correct.

Question 2.
H.C.F. of x² – 4 and x² + 4x + 4 will be
(A) (x – 2)
(B) (x – 4)
(C) (x + 2)
(D) (x + 4)
Solution
Option (C) is (RBSESolutions.com) correct.

Question 3.
H.C.F. of 36a⁵b² and 90a³b⁴ will be
(A) 36a³b²
(B) 18a³b²
(C) 90a³b⁴
(D) 180a⁵b⁴
Solution
36a⁵b² = 3 × 3 × 2 × 2 × a⁵ × b²
90a³b⁴ = 3 × 3 × 2 × 5 × a³ × b⁴ = 3 × 3 × 2 × a³ × b²
H.C.F. = 18a³b²
Hence, Option (B) is correct.

Question 4.
L.C.M. of x² – 1 and x² – x – 2 will be
(A) (x² – 1)(x – 2)
(B) (x² – 1)(x + 2)
(C) (x – 1)² (x + 2)
(D) (x + 1)² (x – 2)
Solution
x² – 1 = (x – 1)(x + 1) …(i)
x² – x – 2 = x² – 2x + x – 2 = x (x – 2) + 1 (x – 2) = (x + 1)(x – 2) …(ii)
from eqn (i) and (ii)
(x + 1)(x – 1 )(x – 2)
L.C.M. = (x² – 1)(x – 2)
Hence, Option (A) is correct.

Question 5.
If 5p²q and 15pq²r² are two (RBSESolutions.com) expression 5pq, then L.C.M. will be
(A) 75p²q³r²
(B) 5p²q²r²
(C) 15p²q²r²
(D) 15p³q²r²
Solution

Hence, option (C) is correct.

Question 6.
H.C.F. of expression x² – 1 and x + 1 will be
(A) x – 1
(B) x + 1
(C) (x² – 1)(x + 1)
(D) (x – 1)(x + 1)
Solution
Option (B) is correct.

Question 7.
If (2 + p) and 100 – 25p² are (RBSESolutions.com) two expression, then their L.C.M. will be
(A) 100 – 25p²
(B) 2 + p
(C) 98 – 25p²
(D) (100 – 25p²) (2 + p)
Solution
First expression = (2 + p)
and other expression = 100 – 25p²
= (10 – 5p) (10 + 5p)
= 5(2 – p) × 5(2 + p)
= 25(2 – p) (2 + p)
Thus L.C.M. of two expression = 100 – p²
Hence, Option (A) is correct.

Very Short/Short Answer Type Questions

Question 1.
If one expression is u(x) and (RBSESolutions.com) other is v(x) their H.C.F. is r(x), then find L.C.M.
Solution
L.C.M. = \(\frac { u(x)\times v(x) }{ r(x) }\)

Question 2.
Find the H.C.F. of the following :
(i) x² – 4 and x² + 4x + 4
(ii) 4x⁴ – 16x³ + 12x² and 6x³ + 6x² – 72x
Solution
(i) x² – 4 = (x + 2)(x – 2)
x² + 4x + 4 = (x + 2)²
H.C.F. of coefficient = 1
H.C.F. of other factors = (x + 2)¹ = x + 2
H.C.F. = x + 2
(ii) 4x⁴ – 16x³ + 12x² = 4x²(x² – 4x + 3) = 4x²(x – 1)(x – 3)
6x³ + 6x² – 72x = 6x(x² + x – 12) = 6x(x + 4)(x – 3) = 2x(x – 3) [because H.C.F. of coefficients is 2]
= 2x² – 6x

Question 3.
Find the L.C.M. of the (RBSESolutions.com) following
(i) x² – 1 and x⁴ – 1
(ii) (x + 1)² (x + 5)³ and x² + 10x + 25
(iii) 6(x² – 3x + 2) and 18(x² – 4x + 3)
Solution

Question 4.
L.C.M. of two quadratic (RBSESolutions.com) equations is (x² – y²) (x² + xy + y²) and H.C.F is (x – y). Find expressions
Solution
L.C.M. = (x² – y²) (x² + xy + y²) = (x – y)(x + y)(x² + xy + y²)
and H.C.F. = (x – y)
In L.C.M. and H.C.F. common factor is (x – y)
First expression = (x – y) (x + y) = x² – y²
and second egression = (x – y) (x² + xy + y²)
= x³ – x²y + x²y – xy² + xy² – y³
= (x³ – y³)
Hence, two expression are (x² – y²) and (x³ – y³)

Question 5.
Least common multiples of two (RBSESolutions.com) polynomials is x³ – 3x² + 3x – 2 and the highest common factor is x – 2. If one polynomial is x² – x + 1, find the other.
Solution

Question 6.
Find L.C.M and H.C.F. of the following (RBSESolutions.com) expression
x² + 6x + 9, x² – x – 12, x³ + 4x² + 4x + 3
Solution
Expression x² + 6x + 9 = (x)² + 2 × x × 3 + (3)²

x³ + 4x² + 4x + 3 = (x + 3)(x² + x + 1) …(iii)
common (RBSESolutions.com) factor of eqn. (i), (ii) and (iii) = (x + 3)
Thus, required H.C.F. = (x + 3)
in eqns. (i), (ii) (iii)
= (x + 3)² (x – 4) (x² + x + 1)
= (x² + 6x + 9) (x³ – x² + x – 4x² – 4x – 4)
Hence, required L.C.M = (x² + 6x + 9) (x³ – 5x² – 3x – 4)

We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimal Numbers
Exercise	In Text Exercise
Number of Questions	23
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise

Download and install Partial Fraction Calculator on Windows or MAC and enjoy Partial Fraction Calculator on big screen computers for free.

(Page 16)
Question
(i) You know that area of (RBSESolutions.com) a rectangle = length x breadth. But if length and breadth is given in fraction, then how will you find the area of rectangle?
(ii) Are you agree with that to find the area of a rectangle, we must know how to multiply two or more than fractional numbers?
Solution:
(i) If length and breadth are given in fractional numbers then the numerators are multiplied with numerators and denominators with denominators to find a fraction that is the area of a rectangle:
For example:
length = 4/5 unit and breadth = \(\frac { 3 }{ 7 }\)unit
Then area of rectangle = length x breadth
= \(\frac { 4 }{ 5 }\) x \(\frac { 3 }{ 7 }\) = \(\frac { 4\times 3 }{ 30 }\) = \(\frac { 5 }{ 7 }\)
= \(\frac { 12 }{ 35 }\) sq unit.
(ii) Yes. we are agree with that we must know, how to multiply two or more fractional numbers.

Using a fraction and whole number calculator in situations where you struggle to solve your arithmetic assignments can be useful.

(Page 19)
Question
Complete (RBSESolutions.com) the table:

After completing the table, do you agree (RBSESolutions.com) that value of product of two proper fraction is less than the value of each fraction?
Solution:


yes, we are agree with that the product of two proper fraction is always less than the each fraction.

(Page 20)
Question
Complete (RBSESolutions.com) the table:

Solution:

On completing the table we can say that the product of two improper fractions is greater than each fraction.

(Page 24)
Question
Interchange the numerator (RBSESolutions.com) and denominotor with each of the function \(\frac { 1 }{ 5 }\) and \(\frac { 2 }{ 3 }\).

(ii) Fill in the blanks :

Solution:
(i) After changing the numerator and denominator of \(\frac { 1 }{ 5 }\) required fraction = \(\frac { 5 }{ 1 }\) and after changing the numerator and denominator of \(\frac { 2 }{ 3 }\) required fraction = \(\frac { 3 }{ 2 }\)

(Page 25)
Question
How can you (RBSESolutions.com) read those number
(1) 24.2 = Twenty four decimal two
(ii) 2.04 = Two decimal zero four
(iii) 325.52 = …………..
(iv) 56.32 = ……………
Solution:
(iii) 325.52 = Three hundred twenty five decimal five two.
(iv) 56.32 = Fifty six decimal three two.

How do we write 1/8 as a decimal? To convert 1/8 to a decimal, divide the denominator into the numerator.

(Page 26)
Question
See the following (RBSESolutions.com) table and fill up the blanks:

(ii) We can also write these numbers in their extended form
421.258 = 4 x 100 + 2 x 10 + 1 x 1 + 2 x \(\frac { 1 }{ 10 }\) + 5 x \(\frac { 1 }{ 100 }\) + 8 x \(\frac { 1 }{ 1000 }\)
Similarly write the (RBSESolutions.com) remaining numbers from above table.
120
(iii) Fill in the blanks : 120 m = \(\frac { 120 }{ 1000 }\)km = …… km.
Solution:

(ii) Extended form (RBSESolutions.com) of remaining numbers

(Page 30)
Put decimal in –
Question
1.52 x 1000 = ….
Solution:
1.52 x 1000 = 1520.00

(Page 34)
Question
(i) There are 8 black and 7 white (RBSESolutions.com) strips in a zebra crossing. So tell what part of total strips is the number of white strips?

(ii) On a day 100 people crossed the road by zebra crossing out of which 20 are men, 30 women, 10 children and 40 students. Show all these data’s In decimal.
Solution:
(i) Total strips 8 + 7 = 15
Part of white strips (RBSESolutions.com) related with total strips.

∴ White strips are \(\frac { 7 }{ 5 }\) part of the total strips.

(ii) Total number of people = 100

Do and Learn

(Page 14)

Let us write 5/8 as a decimal using the division method.

Question 1
Find five equivalent (RBSESolutions.com) fractions of \(\frac { 4 }{ 7 }\)
Solution:

∴ Equivalent fractions of \(\frac { 4 }{ 7 }\) are = \(\frac { 8 }{ 14 }\), \(\frac { 12 }{ 21 }\), \(\frac { 16 }{ 28 }\), \(\frac { 20 }{ 35 }\), and \(\frac { 24 }{ 42 }\)

Question 2
Compare and (RBSESolutions.com) write (<,= ,>)

Solution:

(Page 18)
Question
Solve

Solution:

(Page 18)
Question
Solve
(i) 5 x \(\frac { 1 }{ 2 }\) = ?
(ii) 1\(\frac { 4 }{ 9 }\) x 6 = ?
Solution:

(Page 18)
Question
Solve

Solution:

(Page 19)
Question
Find


Solution:

(Page 20)
Question
Find the product of one (RBSESolutions.com) proper and one improper fraction and prepare the table showing theresoult.

Product is greater than proper and smaller then improper fraction.
Solution:

Product is greater than proper and smaller than improper fraction.

Product is greater than proper and smaller than improper fraction.

Product is greater than (RBSESolutions.com) proper and smaller than improper fraction.

Product is greater than proper and smaller than improper fraction.

(Page 24)
Question
Solve

Solution:

(Page 24)
Question
Fill in the blanks

Solution:

(Page 25)
Question
Solve

Solution:

(Page 26)
Question
Which (RBSESolutions.com) number is smaller?
(i) 35.37 and 35.07
(ii) 262.327 and 262.372
Solution:
(j) 35.07
(ii) 262.327

(Page 29)
Question
Find the (RBSESolutions.com) value of
(1) 2.3 x 3.5
(ii) 37 x 5
(iii) 2.4 x 7.35
Solution:

(Page 32)
Question
Divide the given decimal (RBSESolutions.com) numbers by 10,100 and 1000?
(i) 132.4
(ii) 1.03
(iii) 40.033
(iv) 4.321
Solution:

(Page 32)
Question
(i) 6 ÷ 1.2
(ii) 9 ÷ 4.5
(iii) 48 ÷ 0.8
Solution:

(Page 33)
Question
Solve
(i) 7.75 ÷ 0.25
(ii) 5.6 ÷ 1.4
(iii) 42.8 ÷ 0.02
Solution:

We hope the RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise, drop a comment below and we will get back to you at the earliest.