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Fazal

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions

December 27, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions

RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers Additional Questions.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 6
Subject Maths
Chapter Chapter 6
Chapter Name Decimal Numbers
Exercise Additional Questions
Number of Questions 17
Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers Additional Questions

Multiple Choice Questions

Question 1.
Place value of 1 in (RBSESolutions.com) number 412 is :
(i) 1
(ii) 10
(iii) 100
(iv) 1000

Question 2.
Place value of 9 in number 137.29 is :
(i) 9
(ii) 90
(iii) 0.9
(iv) 0.09

Question 3.
Value of 3 tenth will be :
(i) 0.3
(ii) 3
(iii) 0.03
(iv) 30

RBSE Solutions

3/5 as a decimal is 0.6

Question 4.
Value of 47 hundredth (RBSESolutions.com) will be :
(i) 0.047
(ii) 470
(iii) 0.47
(iv) 47

Question 5.
Value of 9 thousandth will be :
(i) 0.9
(ii) 0.09
(iii) 0.009
(iv) 0.0009

Question 6.
Value of 3 tens 7 hundredth will be :
(i) 3.7
(ii) 30.7
(iii) 3.70
(iv) 30.07

Question 7.
Value of 4 mm in cm (RBSESolutions.com) will be :
(i) 0.4
(ii) 4
(iii) 40
(iv) 0.04

Question 8.
Simple form of fraction 0.52 will be :
(i) \(\frac { 12 }{ 25 } \)
(ii) \(\frac { 13 }{ 25 } \)
(iii) \(\frac { 14 }{ 25 } \)
(iv) \(\frac { 15 }{ 21 } \)

Answers
1. (ii)
2. (iv)
3. (i)
4. (iii)
5. (iii)
6. (iv)
7. (i)
8. (ii)

Fill in the blanks

(i) First place from right (RBSESolutions.com) side of decimal point called ………………….
(ii) Number on the left side of decimal point called …………………. number.
(iii) Second place from right side of decimal point called ………………….
(iv) Third place from right side of decimal point called ………………….
(v) Number come after decimal point read by their ………………….

Solution.
(i) tenth
(ii) whole
(iii) hundredth
(iv) thousandth
(v) name.

RBSE Solutions

9/20 as a decimal is 0.45

Very Short Answer type Questions

Question 1.
Write the place value of each (RBSESolutions.com) digit in number 127.914.
Solution.
On writing the place value of each digit of number 127.914 –
Place value of 1 = 1 × 100 = 100
Place value of 2 = 2 × 10 = 20
Place value of 7 = 7 × 1 = 7
Place value of 9 = 9 × \(\frac { 1 }{ 10 } \) = 0.9
Place value of 1 = 1 × \(\frac { 1 }{ 100 } \) = 0.01
Place value of 4 = 4 × \(\frac { 1 }{ 1000 } \) = 0.004

Question 2.
How we read 370.156?
Solution
370.156 – Three hundred seventy point one five six.

Question 3.
Write the following (RBSESolutions.com) numbers in decimal form :
(i) 6 unit 2 tenth 5 thousandth
(ii) 3 hundred 1 tens 2 hundredth
Solution.
On writing given numbers in decimal form –
(i) 6.205
(ii) 310.02

Question 4.
Represent numbers 0.4,1.7 and 2.9 on number line.
Solution.
On representing numbers 0.4, 1.7 and 2.9 on number line.
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions image 1

Short/Long Answers Type Questions

Question 1.
Represent the following (RBSESolutions.com) in decimal form :
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions image 2
Solution.
On representing each in decimal form :
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions image 3

RBSE Solutions

Question 2.
Convert the following in fraction (RBSESolutions.com) and write in simple form :
(i) 2.75
(ii) 0.045
(iii) 0.02
(iv) 0.6
Solution.
On converting each in fraction and write them in simple form –
RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions image 4

Question 3.
Sohan bought 7.423 kg rice and 6.129 kg pulse. How (RBSESolutions.com) much total kg things Sohan bought?
Solution :
Sohan bought rice = 7.423 kg
Sohan bought pulse = 6.129 kg
∴ Total things he bought = (7.423 + 6.129) = 13.552 kg

Question 4.
Sunita has 93.12 l oil. She sold 67.19 l oil from it. How much oil left with her?
Solution.
Sunita has oil = 93.12 l
Oil sold out = 67.19 l
∴ Remaining oil = (93.12 – 67.19) = 25.93 l

RBSE Solutions

We hope the RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 6 Decimal Numbers Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions

July 2, 2022 by Fazal Leave a Comment

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.

RBSE Solutions

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.

RBSE Solutions

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.
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 1

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 2
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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 3

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 4

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 5
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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 6

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.

RBSE Solutions

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.

RBSE Solutions

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

RBSE Solutions

Question 4.
Find the roots of (RBSESolutions.com) quadratic equation √2 x2 + 7x + 5√2 = 0. (CBSE 2013)
Solution
Given quadratic equation is
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 7

Question 5.
Solve for x
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 8
Solution
Given the quadratic equation is
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 9
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 10

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 11

RBSE Solutions

Question 9.
Solve for x: √3 x2 – 2√2 x – 2√3 = 0. [CBSE 2015]
Solution
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 12

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

RBSE Solutions

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.
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 13
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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 14
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

RBSE Solutions

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.

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 15
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

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 16

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 17

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
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions 18
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)

RBSE Solutions

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 9 Maths Chapter 3 Polynomial Additional Questions

July 2, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions.

Board RBSE
Class Class 9
Subject Maths
Chapter Chapter 3
Chapter Name Polynomial
Exercise Additional Questions
Number of Questions Solved 36
Category RBSE Solutions

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

Multiple Choice Questions

Question 1.
A polynomial (RBSESolutions.com) of degree n in x has atmost:
(A) n terms
(B) (n – 1) terms
(C) (n + 1) terms
(D) \(\frac { n }{ 2 }\) terms
Solution
C

Question 2.
The zeroes of the (RBSESolutions.com) polynomial p(x) = x(x² – 1) are:
(A) 0, 1
(B) 0, – 1
(C) 0, – 1, 1
(D) ± 1
Solution
C

RBSE Solutions

Question 3.
If
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 1
is equal to
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 2
Solution
B

Question 4.
√7 is a polynomial (RBSESolutions.com) of degree:
(A) 0
(B) 7
(C) \(\frac { 1 }{ 2 }\)
(D) 2
Solution
A

Question 5.
The coefficient of x in (x – 3)(x – 4) is:
(A) 7
(B) 1
(C) – 7
(D) 12
Solution
C

Question 6.
If p(x) = x² -3√2x + 1, then p(3√2) is equal to:
(A) 3√2
(B) 3√2 – 1
(C) 6√2 – 1
(D) 1
Solution
D

Question 7.
Which of the following is (RBSESolutions.com) a polynomial in one variable:
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 3
Solution
A

Question 8.
The degree of the zero polynomial is:
(A) 0
(B) 1
(C) any real number
(D) not exist
Solution
D

RBSE Solutions

Question 9.
If x91 + 91 is divided by x + 1, then (RBSESolutions.com) the remainder is:
(A) 0
(B) 90
(C) 92
(D) None of these
Solution
B

Question 10.
Zero of the polynomial p(x) = √3x + 3 is
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 4
Solution
A

Very Short Answer Type Questions

Question 1.
Show that \(\frac { 1 }{ 2 }\) is zero (RBSESolutions.com) of the polynomial 2x2 + 7x – 4.
Solution.
Let p(x) = 2x2 + 7x – 4
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 5

Question 2.
Find the (RBSESolutions.com) remainder when
x4 + x3 – 2x2 + x + 1 is divided by x – 1.
Solution.
Let p(x) = x4 + x3 – 2x2 + x + 1
zero of x – 1 is 1
So, P(1) = (1)4 + (1)3 – 2(1)2 + (1) + 1 = 1 + 1 – 2 + 2 = 2
Hence, 2 is the remainder, when x4 + x3 – 2x2 + x + 1 is divided by x – 1.

Question 3.
If x – 1 is a factor of p(x) = 2x2 + kx + √2 then find (RBSESolutions.com) the value of k.
Solution.
x – 1 is a factor of p(x) = 2x2 + kx + √2
p(1) = 0 (by factor theorem)
⇒ 2(1)2 + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
k = -(2 + √2)

Question 4.
Find the (RBSESolutions.com) value of m, if (x + 3) is a factor of 3x2 + mx + 6.
Solution.
Let p(x) = 3x2 + mx + 6
If x + 3 is a factor of p(x), then
p(-3) = 0
p(-3) = 3(- 3)2 + m(-3) + 6 = 0
⇒ 3 x 9 – 3m + 6 = 0
⇒ 27 – 3m + 6 = 0
⇒ 3m = 33
⇒ m = 11

Question 5.
Verify that 1 is not a zero (RBSESolutions.com) of the polynomial 4x4 – 3x3 + 2x2 – 5x + 1.
Solution.
Let p(x) = 4x4 – 3x3 + 2x2 – 5x + 1
Now
P(1) = 4(1)4 – 3(1)3 + 2(1)2 – 5(1) + 1 = 4 – 3 + 2 – 5 + 1 = -1
p(1) ≠ 0
1 is not a zero of p(x).

RBSE Solutions

Remainder Theorem Calculator. Various types of online remainder theorem calculators are available online.

Question 6.
Using remainder theorem, find the (RBSESolutions.com) value of k so that (4x2 + kx – 1) leaves the remainder 2 when divided by (x – 3).
Solution.
Let p(x) = 4x2 + kx – 1
Here we are given that p(3) = 2
i.e. 4(3)2 + k(3) -1 = 2
⇒ 4 x 9 + 3k – 1 = 2
⇒ 36 + 3k = 3
⇒ 3k = -33
⇒ k = -11

Question 7.
If (x – 2) is a factor of the (RBSESolutions.com) polynomial x4 – 2x3 + ax – 1, then find the value of a.
Solution.
(x – 2) is a factor of the polynomial p(x) = x4 – 2x3 + ax – 1, then according to factor theorem p(2) = 0
⇒ (2)4 – 2(2)3 + a(2) – 1 = 0
⇒ 2a – 1 = 0
⇒ a = \(\frac { 1 }{ 2 }\)

Question 8.
Find the value of
(x – y)3 + (y – z)3 + (z – x)3.
Solution.
Let a = x – y, b = y – z, c = z – x
Here a + b + c = x – y + y – z + z – x = 0
a3 + b3 + c3 = 3abc
⇒ (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y)(y – z)(z – x).

Question 9.
If a, b, c are all non-zero and a + b + c = 0 then find the (RBSESolutions.com) value of \(\frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab }\)
Solution.
We know that a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a + b + c then a3 + b3 + c3 = 3abc …(1)
Dividing equation (1) by abc, we get
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 6

Question 10.
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 7
Solution.
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 8

RBSE Solutions

Short Answer Type Questions

Question 1.
Find the remainder (RBSESolutions.com) obtained on dividing p(x) = x3 + 1 by x + 1.
Solution.
By long division method
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 9
Here we find that remainder is zero.
Also p(-1) = (-1)3 + 1 = -1 + 1 = 0,
which is equal to the (RBSESolutions.com) remainder obtained by actual division.

Question 2.
Find the value of
4a2 + b2 + 25c2 + 4ab – 10bc – 20ca when a = 1, b = 2 and c = 3.
Solution.
We have,
4a2 + b2 + 25c2 + 4ab – 10bc – 20ca = (2a)2 + (b)2 + (-5c)2 – 2(2a)(b) + 2(b)(-5c) + 2(-5c)(2a) = (2a + b – 5c)2
It is given that a = 1, b = 2 and c = 3,
so (2a + b – 5c)2 = (2 x 1 + 2 – 5 x 3)2 = (2 + 2 – 15)2 = (4 – 15)2 = (-11)2 = 121

Question 3.
If 3a – 2b = 11 and ab = 12, then find (RBSESolutions.com) the value of 27a3 – 8b3.
Solution.
We know that
(a – b)3 = a3 – b3 – 3ab(a – b)
Using this identity,
(3a – 2b)3 = (3a)3 – (2b)3 -3 x 3a x 2b (3a – 2b)
⇒ (3a – 2b)3 = 27a3 – 8b3 – 18ab(3a – 2b)
Now substituting 3a – 2b = 11 and ab = 12,
we get
(11)2 = 27a3 – 8b3 – 18 x 12 x 11
⇒ 1331 = 27a3 – 8b3 – 2376
⇒ 27a3 – 8b2 = 1331 + 2376
⇒ 27a3 – 8b3 = 3707

Question 4.
Use factor (RBSESolutions.com) theorem, show that (x + √2) is a factor of (2√2x2 + 5x + √2).
Solution.
Let p(x) = (2√2x2 + 5x + √2)
If x + √2 is a factor of p(x), then according to factor theorem p(√2) = 0.
p(-√2) = 2√2(-√2)2 + 5(-√2) + √2 = 2√2 x 2 – 5√2 + √2 = 5√2 – 5√2 = 0
p(-√2) = 0
(x + √2) is a factor p(x) i.e. (2√2 x2 + 5x + √2).

Question 5.
If a + b + c = 9 and ab + bc + ca = 23 then find (RBSESolutions.com) the value of a2 + b2 + c2.
Solution.
We have, a + b + c = 9 Squaring both sides, we get
(a + b + c)2 = (9)2
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 81
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81
⇒ a2 + b2 + c2 + 2 x 23 = 81
⇒ a2 + b2 + c2 = 81 – 46
⇒ a2 + b2 + c2 = 35 Thus, a2 + b2 + c2 = 35.

RBSE Solutions

Question 6.
Simplify by using (RBSESolutions.com) suitable identity
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 10
Solution.
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 11

Question 7.
Factorise
(i) 216a3 – 125
(ii) a3 – b3 – a + b
Solution.
(i) we have,
216a3 – 125 = (6a)3 – (5)3
[a3 – b3 = (a – b)(a2 + ab + b2)]
= (6a – 5) {(6a)2 + 6a x 5 + (5)2}
= (6a – 5)(36a2 + 30a + 25)
(ii) We have,
a3 – b3 – a + b = a3 – b3 – (a – b)
= (a – b)(a2 + ab + b2) – (a – b)
[a3 – b3 = (a – b)(a2 + ab + b2)]
= (a – b)(a2 + ab + b2 – 1)

Question 8.
If p = 4 – q show that p3 + q3 + 12pq = 64.
Solution.
We have,
p = 4 – q ⇒ p + q = 4 …(i)
Cubing both sides, (RBSESolutions.com) we get
(p + q)3 = (4)3
⇒ p3 + 3p2q + 3pq2 + q3 = 64
⇒ p3 + q3 + 3pq(p + q) = 64
⇒ p3 + q3 + 3pq x 4 = 64 [using (i)]
⇒ p3 + q3 + 12pq = 64

Question 9.
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 12
Solution.
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 13

Long Answer Type Questions

Question 1.
If (x – 3) and (x – \(\frac { 1 }{ 2 }\)) are both (RBSESolutions.com) factors of ax2 + 5x + b, show that a = b.
Solution.
Let p(x) = ax2 + 5x + b
x – 3 is a factor of p(x).
p(3) = 0
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 14

RBSE Solutions

Question 2.
For what values of a and b so that (RBSESolutions.com) the polynomial x3 + 10x2 + ax – 6 is exactly divisible by (x – 1) and (x + 2).
Solution.
Let p(x) = x3 + 10x2 + ax + b
p(x) i.e. x3 + 10x2 + ax + 6 is exactly divisible by (x – 1) and (x + 2)
Therefore, p(1) and p(-2) must equal to zero.
p(1) = (1)3 + 10(1)2 + a x 1 + b = 0
⇒ 1 + 10 + a + b = 0
⇒ a + b = – 11 …(i)
Also, p(-2) = 0
(-2)3 + 10(-2)2 + a x (- 2) + b = 0
-8 + 40 – 2a + b = 0
⇒ – 2a + b = – 32 …(ii)
Solving (i) and (ii), we get
a = 7 and b = -18.

Question 3.
If ax3 + bx2 + x – 6 has x + 2 as a factor and leaves (RBSESolutions.com) a remainder 4 when divided by (x – 2), find the values of a and b.
Solution.
Let p(x) = ax3 + bx2 + x – 6
(x + 2) is a factor of p(x)
⇒ p(-2) = 0 [∴ x + 2 = 0 ⇒ x = -2]
⇒ a(-2)3 + b(-2)2 + (-2) – 6 = 0
⇒ -8a + 4b – 2 – 6 = 0
⇒ -8a + 4b = 8
⇒ -2a + b = 2 …(i)
It is given that p(x) leaves the (RBSESolutions.com) remainder 4 when it divided by (x – 2) i.e. p(2) = 4.
⇒ a(2)3 + b(2)2 + (2) – 6 = 4
⇒ 8a + 4b – 4 = 4
⇒ 8a + 46 = 8
⇒ 2a + b = 2 …(ii)
Solving (i) and (ii), we get a = 0 and b = 2.

Question 4.
Evaluate
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 15
Solution.
In the given expression, we see that both numerator and denominator (RBSESolutions.com) is in the form a3 + b3 + c3 = 3abc because a + b + c = 0.
From numerator, we see that a2 – b2 + b2 – c2 + c2 – a2 = 0
⇒ (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3 = 3 (a2 – b2)(b2 – c2)(c2 – a2)
Similarly, from denominator,
a – b + b – c + c – a = 0
(a – b)3 + (b – c)3 + (c – a)3 = 3(a – b)(b – c)(c – a)
Value of the expression
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 16

Question 5.
If the polynomials (3x3 + ax2 + 3x + 5) and (4x3 + x2 – 2x + a) leaves the same (RBSESolutions.com) remainder when divided by (x – 2), find the value of a. Also the find the remainder in each-case.
Solution.
Let the given polynomials are p(x) = 3x3 + ax2 + 3x + 5 and f(x) = 4x3 + x2 – 2x + a
According to question p(2) = f(2)
3(2)3 + a(2)2 + 3(2) + 5 = 4(2)3 + (2)2 – 2(2) + a
⇒ 3 x 8 + 4a + 6 + 5 = 32 + 4 – 4 + a
⇒ 24 + 4a + 11 = 32 + a
⇒ 4a – a = 32 – 35
⇒ 3a = -3
⇒ a = -1
As the remainder is same, so p(2) or f(2) would (RBSESolutions.com) be same when a = – 1
p(2) = 3(2)3 + (-1)(2)2 + 3 x 2 + 5 [∴ a = -1]
= 24 – 4 + 6 + 5
= 35 – 4 = 31
Thus, a = – 1 and p(2) or f(2) = 31

Question 6.
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a).
Solution.
We have,
L.H.S.
= (a + b + c)3 – a3 -b3 – c3
= {(a + b + c)3 – (a)3} – (b3 + c3)
= (a + b + c – a){(a + b + c)2 + a(a + b + c) + a2} – (b + c){b2 – bc + c2)
[∵ x3 – y3 = (x – y)(x2 + xy + y2) and x3 + y3 = (x + y)(x2 – xy + y2)]
= (b + c){a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + ab + ac + a2} – (b + c)(b2 – bc + c2)
= (b + c)[3a2 + b2 + c2 + 3ab + 2bc + 3ca – b2 + bc – c2]
= (b + c)[3a2 + 3ab + 3bc + 3ca]
= 3(b + c)[a2 + ab + bc + ca]
= 3(b + c)[a(a + b) + c(a + b)]
= 3(b + c)(a + b)(a + c)
= 3(a + b)(b + c)(c + a)
= R.H.S.

RBSE Solutions

Question 7.
(i) For what (RBSESolutions.com) value of m is x3 – 2mn2 + 16 divisible by (x + 2)?
(ii) Show that (2x – 3) is a factor of x + 2x3 – 9x + 12.
Solution.
(i) Let p(x) = x3 – 2mn2 + 16
p(x) will be divisible by (x + 2) if
p(-2)= 0
p(-2) = (-2)3 – 2m(-2)2 + 16 = -8 – 8m + 16 = – 8m + 8
p(-2) = 0
⇒ -8m + 8 = 0
⇒ 8m = 8
⇒ m = 1
(ii) We know that (2x – 3) will be a factor of x + 2x3 – 9x + 12 if p(x) on dividing by 2x – 3, leaves (RBSESolutions.com) a remainder zero
RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions 17
So, the remainder obtained (RBSESolutions.com) on dividing p(x) by 2x – 3 is zero.
Hence, (2x – 3) is a factor of x + 2x3 – 9x + 12

We hope the given RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 6 Maths Chapter 5 Fractions Additional Questions

July 1, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 6 Maths Chapter 5 Fractions Additional Questions

RBSE Solutions for Class 6 Maths Chapter 5 Fractions Additional Questions is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 5 Fractions Additional Questions.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 6
Subject Maths
Chapter Chapter 5
Chapter Name Fractions
Exercise Additional Questions
Number of Questions 17
Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 5 Fractions Additional Questions

Multiple Choice Questions

Question 1.
Parts of (RBSESolutions.com) fractions :
(i) 1
(ii) 2
(iii) 3
(iv) 4.

Question 2.
Improper fraction in the following :
(i) \(\frac { 4 }{ 7 } \)
(ii) \(\frac { 2 }{ 3 } \)
(iii) \(\frac { 6 }{ 5 } \)
(iv) \(\frac { 7 }{ 8 } \)

Question 3.
Proper fraction in the following :
(i) \(\frac { 1 }{ 4 } \)
(ii) \(\frac { 8 }{ 3 } \)
(iii) \(\frac { 5 }{ 2 } \)
(iv) \(\frac { 4 }{ 3 } \)

RBSE Solutions

The first step would be to write the mix number 223 as an improper fraction.

Question 4.
Unit fraction in the (RBSESolutions.com) following :
(i) \(\frac { 4 }{ 3 } \)
(ii) \(\frac { 3 }{ 4 } \)
(iii) \(\frac { 5 }{ 5 } \)
(iv) \(\frac { 1 }{ 3 } \)

Question 5.
Bigger fraction in \(\frac { 5 }{ 7 } \) and \(\frac { 6 }{ 7 } \):
(i) \(\frac { 5 }{ 7 } \)
(ii) \(\frac { 6 }{ 7 } \)
(iii) \(\frac { 1 }{ 7 } \)
(iv) \(\frac { 7 }{ 6 } \)

Question 6.
Smaller fraction in \(\frac { 7 }{ 9 } \) and \(\frac { 8 }{ 9 } \) :
(i) \(\frac { 7 }{ 9 } \)
(ii) \(\frac { 8 }{ 9 } \)
(iii) \(\frac { 1 }{ 9 } \)
(iv) \(\frac { 9 }{ 8 } \)

Question 7.
Bigger fraction (RBSESolutions.com) in \(\frac { 1 }{ 4 } \) and \(\frac { 1 }{ 7 } \) –
(i) \(\frac { 1 }{ 3 } \)
(ii) \(\frac { 1 }{ 2 } \)
(iii) \(\frac { 1 }{ 7 } \)
(iv) \(\frac { 1 }{ 4 } \)

Question 8.
Simplified form of \(\frac { 25 }{ 35 } \) is
(i) \(\frac { 25 }{ 35 } \)
(ii) \(\frac { 2 }{ 3 } \)
(iii) \(\frac { 5 }{ 7 } \)
(iv) \(\frac { 4 }{ 5 } \)

Answers.
1. (ii)
2. (iii)
3. (i)
4. (iv)
5. (ii)
6. (i)
7. (iv)
8. (iii)

Fill in the blanks

(i) In a fraction, the total (RBSESolutions.com) number of divisions of a unit is called ……………….
(ii) In a fraction, some parts taken from divisions of a unit are called ……………….
(iii) In a mixed fraction one part is ………………. and other part is a ……………….
(iv) Equivalent fraction of \(\frac { 4 }{ 9 } \) will be ……………….
(v) Fraction with like denominators are called ………………. fraction.

Answers.
(i) denominator
(ii) numerator
(iii) whole number, fraction
(iv) 8/18
(v) like

RBSE Solutions

Very Short Answer Type Questions

Question 1.
\(\frac { 1 }{ 7 } \),\(\frac { 1 }{ 4 } \),\(\frac { 1 }{ 9 } \),\(\frac { 1 }{ 5 } \) arrange in (RBSESolutions.com) ascending order
Solution.
Ascending order : \(\frac { 1 }{ 9 } \) < \(\frac { 1 }{ 7 } \) < \(\frac { 1 }{ 5 } \) < \(\frac { 1 }{ 4 } \)

Question 2
What will be the sum of \(\frac { 8 }{ 7 } \) + \(\frac { 9 }{ 2 } \)
Solution.
\(\frac { 8 }{ 7 } \) + \(\frac { 9 }{ 2 } \) = \(\frac { 2\times 8+7\times 9 }{ 14 } \)
= \(\frac { 16+63 }{ 14 } \) = \(\frac { 79 }{ 14 } \)

Question 3
Convert mixed (RBSESolutions.com) fraction 7\(\frac { 1 }{ 2 } \) in improper fraction
Solution.
RBSE Solutions for Class 6 Maths Chapter 5 Fractions Additional Questions image 1
Question 4.
Find add of \(\frac { 1 }{ 2 } \) + \(\frac { 2 }{ 2 } \) + \(\frac { 3 }{ 2 } \)
Solution.
∴ \(\frac { 1 }{ 2 } \) + \(\frac { 2 }{ 2 } \) + \(\frac { 3 }{ 2 } \) = \(\frac { 1+2+3 }{ 2 } \)
= \(\frac { 6 }{ 2 } \) = 3

Short/Long Answers Type Questions

Question 1.
Find five equivalent (RBSESolutions.com) fractions of \(\frac { 2 }{ 5 } \).
Solution.
First equivalent fraction of \(\frac { 2 }{ 5 } \) = \(\frac { 2 }{ 5 } \) × \(\frac { 2 }{ 2 } \) = \(\frac { 4 }{ 10 } \)
Second equivalent fraction of \(\frac { 2 }{ 5 } \) = \(\frac { 2 }{ 5 } \) x \(\frac { 3 }{ 3 } \) = \(\frac { 6 }{ 15 } \)
Third equivalent fraction of \(\frac { 2 }{ 5 } \) = \(\frac { 2 }{ 5 } \) x \(\frac { 4 }{ 4 } \) = \(\frac { 8 }{ 20 } \)
Fourth equivalent fraction of \(\frac { 2 }{ 5 } \) = \(\frac { 2 }{ 5 } \) x \(\frac { 5 }{ 5 } \) = \(\frac { 10 }{ 25 } \)
Fifth equivalent fraction of \(\frac { 2 }{ 5 } \) = \(\frac { 2 }{ 5 } \) x \(\frac { 6 }{ 6 } \) = \(\frac { 12 }{ 30 } \)
Thus, five equivalent fractions of \(\frac { 2 }{ 5 } \) are \(\frac { 4 }{ 10 } \),\(\frac { 6 }{ 15 } \),\(\frac { 8 }{ 20 } \),\(\frac { 10 }{ 25 } \) and \(\frac { 12 }{ 30 } \).

Question 2.
Reena ate \(\frac { 3 }{ 5 } \) part of a fruit and Kamal ate \(\frac { 2 }{ 5 } \) part of that fruit. How much fruit they ate together?
Solution.
Reena ate part of fruit = \(\frac { 3 }{ 5 } \)
Kamal ate part of fruit = \(\frac { 2 }{ 5 } \)
∴ They both ate total fruit = \(\frac { 3 }{ 5 } \) + \(\frac { 2 }{ 5 } \) = \(\frac { 5 }{ 5 } \) = 1

Question 3.
Sohan has \(\frac { 5 }{ 8 } \) bananas. He (RBSESolutions.com) gave \(\frac { 1 }{ 2 } \) bananas to Monika. How many bananas Sohan has left with?
Solution.
Sohan has bananas = \(\frac { 5 }{ 8 } \)
And he gave bananas to monika bananas = \(\frac { 1 }{ 8 } \)
∴ Remaining bananas with Sohan = \(\frac { 5 }{ 8 } \) – \(\frac { 1 }{ 8 } \)
= \(\frac { 5-1 }{ 8 } \) = \(\frac { 4 }{ 8 } \) = \(\frac { 1 }{ 2 } \)

Question 4.
Represent \(\frac { 3 }{ 8 } \) on number line.
Solution.
∵ \(\frac { 3 }{ 8 } \), is bigger than 0 and less than 1. So, number \(\frac { 3 }{ 8 } \) comes between 0 and 1.
RBSE Solutions for Class 6 Maths Chapter 5 Fractions Additional Questions image 2

RBSE Solutions

We hope the RBSE Solutions for Class 6 Maths Chapter 5 Fractions Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 5 Fractions 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

July 1, 2022 by Fazal Leave a Comment

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.

RBSE Solutions

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:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 19
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:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 19a
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 19b
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:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 20
Solution:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 20a
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 :
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 24
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 = \(\frac { 3 }{ 2 }\)
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 24a

(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.

RBSE Solutions

(Page 26)
Question
See the following (RBSESolutions.com) table and fill up the blanks:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 26
(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:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 26a
(ii) Extended form (RBSESolutions.com) of remaining numbers
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 26b

(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?
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 34
(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.
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 34a
∴ White strips are \(\frac { 7 }{ 5 }\) part of the total strips.

(ii) Total number of people = 100
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise 34b

RBSE Solutions

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:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K1
∴ 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 (<,= ,>)
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K2
Solution:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K2a
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K2b

(Page 18)
Question
Solve
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K3
Solution:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K3a

(Page 18)
Question
Solve
(i) 5 x \(\frac { 1 }{ 2 }\) = ?
(ii) 1\(\frac { 4 }{ 9 }\) x 6 = ?
Solution:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K4

RBSE Solutions

(Page 18)
Question
Solve
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K5
Solution:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K5a

(Page 19)
Question
Find
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K6
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K6a
Solution:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K6b

(Page 20)
Question
Find the product of one (RBSESolutions.com) proper and one improper fraction and prepare the table showing theresoult.
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K7
Product is greater than proper and smaller then improper fraction.
Solution:
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K7a
Product is greater than proper and smaller than improper fraction.
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K7b
Product is greater than proper and smaller than improper fraction.
RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers In Text Exercise K7c
Product is greater than (RBSESolutions.com) proper and smaller than improper fraction.

Product is greater than proper and smaller than improper fraction.

RBSE Solutions

(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

RBSE Solutions

(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:

RBSE Solutions

(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.

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