Class 10

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.

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.2

Question 1.
Express each number as a product of its prime factors :
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) Prime factors of 140 = 2 × 70
= 2 × 2 × 35
= 2 × 2 × 5 × 7
= 2² × 5 × 7

(ii) Prime factors of 156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 3 × 13
= 2² × 3 × 13

(iii) Prime factors of 3825 = 3 × 1275
= 3 × 3 × 425
= 3 × 3 × 5 × 85
= 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17

(iv) Prime factors of 5005 = 5 × 1001
= 5 × 7 × 143
= 5 × 7 × 11 × 13

(v) Prime factors of 7429 = 17 × 437
= 17 × 19 × 23

The factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70.

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
Prime factors of 26 = 2 × 13
Prime factors of 91 = 7 × 13
∴ LCM of 26 and 91 = 2 × 7 × 13 = 182
and HCF of 26 and 91 = 13
Verification—HCF (26, 91) × LCM (26, 91)
= 13 × 182
= 13 × 2 × 91
= 26 × 91
= Product of given numbers

(ii) 510 and 92
Prime factors of 510 = 2 × 255
= 2 × 3 × 85
= 2 × 3 × 5 × 17 …(i)
and Prime factors of 92 = 2 × 46
= 2 × 2 × 23
= 2² × 23 ……(ii)
LCM (510, 92) = 2² × 3 × 5 × 17 × 23 = 23460
and HCF (510, 92) = 2
Verification—
HCF (510, 92) × LCM (510, 92)
= 2 × 23460
= 2 × 2² × 3 × 5 × 17 × 23
= 2 × 3 × 5 × 17 × 2² × 23
= 510 × 92
= Product of given numnbers

(iii) 336 and 54
Prime factors of 336 = 2 × 168
= 2 × 2 × 84
= 2 × 2 × 2 × 42
= 2 × 2 × 2 × 2 × 21
= 2 × 2 × 2 × 2 × 3 × 7
= 2⁴ × 3 × 7
Prime factors of 54 = 2 × 27
= 2 × 3 × 9
= 2 × 3 × 3 × 3
= 2 × 33
HCF (336, 54) = 2 × 3 = 6
LCM= 2⁴ × 3³ × 7
= 3024
Verification—
HCF (336, 54) × LCM (336, 54)
= 6 × 3024
= 2 × 3 × 2⁴ × 3³ x 7
= 2⁴ × 3 × 7 × 2 × 3³
= 336 × 54
= Product of given numbers

the common factors of 84? The common factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) 12, 15 and 21
Prime factors of 12 = 2 × 2 × 3
Prime factors of 15 = 3 × 5
Prime factors of 21 = 3 × 7
∴ LCM (12, 15 and 21) = 2² × 3 × 5 × 7 = 420
and HCF (12, 15 and 21) = 3

(ii) 17, 23 and 29
Prime factors of 17 = 1 × 17
Prime factors of 23 = 1 × 23
Prime factors of 29 = 1 × 29
∴ LCM(17, 23 and 29) = 17 × 23 × 29 = 11339
and HCF (17, 23 and 29) = 1

(iii) 8, 9 and 25
Prime factors of 8 = 2 × 2 × 2 = (2)³
Prime factors of 9 = 3 × 3 = (3)²
Prime factors of 25 = 5 × 5 = (5)²
∴ LCM (8, 9 and 25) = (2)³ × (3)² × (5)²
= 8 × 9 × 25 = 1800
and HCF (8, 9 and 25) = 1

The prime factorization of 84 = 22•3•7.

Question 4.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
According to the question, the numbers are 306 and 657.
∴ a = 306
b = 657
and HCF = 9 [Given]
We know that
L.C.M = \(\frac{a \times b}{\mathrm{H} \cdot \mathrm{C} \cdot \mathrm{F}}\)
= \(\frac{306 \times 657}{9}\)
= 34 × 657
= 22338
Hence L.C.M. (306, 657) = 22338

What would be the LCM of 12 and 18 24 A72 B144 C36 D120. … multiple among all the three and then whichever number is the least is the lowest common multiple.

Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
Suppose that for any natural number n, n ∈ N, 6ⁿ ends with the digit 0. Hence 6ⁿ will be divisible by 5.
But prime factors of 6 = 2 × 3
∴ The prime factors of (6)ⁿ will be (6)ⁿ = (2 × 3)ⁿ
i.e., it is clear that there is no place of 5 in the prime factors of 6ⁿ.
By Fundamental theorem of Arithematic we know that every composite number can be factorised as a product of prime numbers and this factorisation is unique, i.e., our hypothesis assured in the beginning is wrong. Hence there is no natural number n for which 6ⁿ ends with the digit 0.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1+5 are composite numbers.
Solution:
According to the question
7 × 11 × 13 + 13 = 13 (7 × 11 + 1)
Since 13 is a factor of this number obtained, therefore it is a composite number. Again according to the question
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 (7 × 6 × 4 × 3 × 2 1 1 + 1)
This obtained number is also a composite number because it has also a factor 5. Hence both the given numbers are composite numbers.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Time taken by Sonia to drive one round of the field =18 minutes.
Time taken by Ravi to drive one round of the same field = 12 minutes
In order to find after how much time will they meet again at the starting point, we shall have to find out the DCM-of lS and 12. Therefore
prime factors of 18 = 2 × 9
= 2 × 3 × 3
= 2 × 3²
and prime factors of 12 = 2 × 6
= 2 × 2 × 3
= 2² × 3
Taking the product of the greatest power of each prime factor of 18 and 12
LCM (18, 12) = 2² × 3²
= 4 × 9 = 36
i.e., Sonia and Ravi will meet again of the starting point after 36 minutes.

RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Exercise 4.1.

Inequality solver is an application help you to solve linear inequality and tracing linear equations and root point.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 4
Chapter Name	Linear Equation and Inequalities in Two Variables
Exercise	4.1
Number of Questions Solved	4
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1

Question 1
By comparing \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } ,\frac { { b }_{ 1 } }{ { b }_{ 2 } } \) and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } \) find,
whether the following (RBSESolutions.com) pair of linear equations is consistent or inconsistent.
(i) 2r – 3y = 8; 4c – 6y = 9
(ii) 3x – y = 2; 6x – 2y = 4
(iii) 2x – 2y = 2; 4x – 4y = 5
(iv) \(\frac { 4 }{ 3 } \) + 2y = 8; 2x + 3y = 12
Solution:
(i) Given linear pair of equations
23 – 3y = 8 or 2x – 3y – 8 = 0
and 4x – 6y = 9 or 4x – 6y – 9 = 0
Comparing above equations by a₁ x + b₁y + c1and a₂ x + b₂ y + c₂ = 0,
a₁ = 2, b₁ = – 3, c₁ = – 8
and a₂ = 4, b₂ = – 6, c₂ = – 9.
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } ,\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -3 }{ -6 } =\frac { 1 }{ 2 } ,\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -8 }{ -9 } =\frac { 8 }{ 9 } \)
∴ \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c2 }_{ } } \)
∴ Given linear pair has no solution.
So, given linear pair is inconsistent.

(ii) Given pair of (RBSESolutions.com) linear equations
3x – y= 2
or 3x – y – 2 = 0 …(i)
and 6x – 2y = 4
or 6x – 2y – 4 = 0
or 3x – y – 2 = 0…(ii)
Comparing equations (i) and (ii) by a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 3, b₁ = -1, and c₁ = -2
and a₂ = 3, b₂ = -1 and c₂= -2

Linear pair is coincident, so (RBSESolutions.com) linear pair has infinite solutions.
Thus given pair is consistant.

(iii) Given linear pair
2x – 2y = 2
or 2x – 2y – 2 = 0
or x – y – 1 = 0 …(i)
and 4x – 4y – 5 = 0 …(ii)
Comparing equations (i) and (ii) by a₁x + b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 1, b₁ = -1, and c₁ = -1
and a₂ = 4, b₂ = -4 and c₂= -5

Given linear pair has no solution.
Thus given Linear pair is inconsistent.

(iv) Given (RBSESolutions.com) linear pair
\(\frac { 4 }{ 3 }\)x + 2y = 8
\(\frac { 4 }{ 3 }\)x + 2y – 8 = 0 …(i)
and 2x + 3y = 12
2x + 3y – 12 = 0 …(ii)
Comparing equation (i) and (ii) by pair a₁x + b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0

Given linear pair has infinite solutions, so linear pair is consistent.

Question 2
Solve the following pair of (RBSESolutions.com) linear equations graphically and write nature of solution.
(i) x + y = 3; 3x – 2y = 4
(ii) 2x – y = 4; x + y = -1
(iii) x + y = 5; 2x + 2y = 10
(iv) 3x + y = 2; 2x – 3y = 5
Solution:
(i) Given linear pair
x + y = 3
x + y -3 = 0 ….(i)
3x – 2y = 4
3x – 2y – 4 = 0 ………(ii)
Comparing equation (i) and (ii) by pair a₁x + b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 1, b₁ = 1, and c₁ = -3
and a₂ = 3, b₂ = -2 and c₂= -4

Linear pair has (RBSESolutions.com) unique solution.
Thus, linear pair is consistent.

Graphical Method:
By equation (i),
x + y = 3
x = 3 – y
Putting y = 0, x = 3 – 0 =3
Putting y = 1, x = 3 – 1 = 2
Putting y = 2, x = 3 – 2 = 1
Table 1 for equation (i),

Table 2 for equation (ii),

Plot the points of Table (1) and (2) on graph (RBSESolutions.com) paper and by joining these points, two straight lines are obtained.

From above graph, it is clear that two straight lines cut at point P(2, 1). Thus x = 2 and y = 1 is required solution.

(ii) Given linear pair
2x – y = 4
2x – y – 4 = 0 …..(i)
x + y = -1
x + y + 1 = 0 ….(ii)
Comparing (RBSESolutions.com) equation (i) and (ii) by linear pair
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

Linear pair is consistent which will have unique solutions.

Graphical Method:
By equation (i),
2x – y = 4
Putting x = 0, 2 x 0 – y = 4
y = -4
Putting x = 1, 2 x – y = 4
-y = 4 – 2
Putting x = 2, 2 x 2 – y = 4
4 – y =4
y = 0

By equation (ii),
x + y = -1
Putting x = 0. 0 + y = -1
y = -1
Putting x = 1, +1 + y = -1
y = -1 – 1
y = -2
Putting x = 2,
2 + y = -1
y = -3

By plotting the (RBSESolutions.com) points of Table 1 and 2 we get two straight lines.

From above graph, It is clear that both the straight lines cut each other at point P( 1, – 2).
Thus, x = 1,y = – 2 are required solution

(iii) Given linear pair:
x + y = 5 or x + y – 5 = 0 …(i)
2x + 2y = 10 or 2x + 2y – 10 = 0 ……..(ii)
Comparing above (RBSESolutions.com) pair by general linear pair
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

∴ Lines represented linear pair will be coincident and linear pair will have infinite solutions.
Thus, given linear pair is consistent.

Graphical Method:
By equation (i),
x + y = 5
⇒ x = 5 – y
Putting y = 0, x = 5 – 0 = 5
Putting y = 3, x = 5 – 3 = 2
Putting y = 5, x =5 – 5 = 0

By joining (RBSESolutions.com) the points A(5, 0), B(2, 3) and C(0, 5) on graph paper.
We get a straight line which indicates the equation x + y = 5.
By equation (ii),
2x + 2y = 10
⇒ 2(x + y) = 10
⇒ x + y =5
⇒ x = 5 – y
Putting y = 0, x = 5 – 0 = 5
Putting y = 2, x = 5 – 2 = 3
Putting y = 5, x = 5 – 5 = 0

By joining the (RBSESolutions.com) points A(5, 0), 8(3,2) and C(0, 5) on graph paper we get a straight line which indicates the equation 2x + 2y = 10.

From graph, lis clear that given pair of linear equations are coincident. Thus they have infinitely many solutions.

(iv) Given pair of (RBSESolutions.com) linear equations
3x + y = 2 or 3x + y – 2 = 0
2x – 3y = 5 or 2x – 3y – 5 = 0 …(ii)
Comparing equation (i) and (ii) by pair
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

Thus given equation will have unique solutions.
∴ Given pair is consistent.

Graphical Method:
By equation (i)
3x + y = 2
y = 2 – 3x
Putting x = 0, y = -3 x 0
y = 2
Putting x = -1, y = 2 – 3 x (-1)
y = 2 + 3
y = 5

From Table (1) arid (2), plot the points on (RBSESolutions.com) graph paper and by joining them, we get two straight lines.

From above graph, It is clear that two straight lines intersect each other at point P( 1, – 1)
Thus, x = 1 and y = – 1 is required solution.

Question 3
Solve the following pair of linear equations, (RBSESolutions.com) graphically and find the coordinates of that points where lines represented by these cuts y-axis.
(i) 2x – 5y + 4 = 0; 2x + y – 8 = 0
(ii) 3x + 2 = 12 ; 5x – 2y = 4
Solution:
(i) Given pair of linear equations
2x – 5y + 4 = 0 …(i)
and 2x + y – 8 = 0 …(ii)
From equation (i),

By equation (ii),
2x + y – 8 = 0
or y – 2x + 8 = 0
Putting x = 4, y = – 2 x 4 + 8
= -8 + 8
= 0
Putting x = 3, y = – 2 x 3 + 8
= -6 + 8
= 2
Putting x = 2, y = – 2 x 2 + 8
= -4 + 8
= 4

Plot the points (RBSESolutions.com) from Table (1) and (2) on graph.
By joining these points two straight lines are obtained.

From above graph ¡t is clear that two straight lines intersect each other at point P(3, 2).
∴ Its required solutions are x = 3 and y = 2
and two straight lines cuts the y-axis at (0, 0.8) and (0, 8).

(ii) Given pair of (RBSESolutions.com) linear equations
3x + 2y = 12 …(i)
and 5x – 2y = 4 …(ii)
From equation (i),
3x + 2y = 12


Plot the points from (RBSESolutions.com) Table (1) and (2) on graph paper. By joining these points two straight lines are obtained.

From above graph, it is clear that two straight lines intersect each other at point P(2, 3).
∴ x = 2 and y = 3 are required solutions and two straight lines cuts y-axis at (0, 6) and (0, – 2).

Question 4
Solve the following pair of (RBSESolutions.com) linear equations grapycally and find the coordinates of the triangle so formed with the y-axis and the lines.
4x – 5y = 20,
3x + 5y = 15
Solution:
Given, pair of linear equation
4x – 5y = 20 ………..(i)
and 3x + 5y = 15 ………(ii)
From equation (i),
4x – 5y = 20
5y = 4x – 20


Plot the points obtained from (RBSESolutions.com) Table (1) and (2) on graph paper. By joining these points two straight lines are obtained.

From the above graph it is (RBSESolutions.com) clear that two lines intersect each other at point P(5, 0)
∴ x = 5 and y = 0 are required solution.
(0, 3), (0, – 4) and (5, 0) are co-ordinates of vertices of ΔABP formed by two straight lines at y – axis.

We hope the given RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Exercise 4.1, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	Vedic Mathematics
Exercise	Additional Questions
Number of Questions Solved	20
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions

Question 1.
Add: 112 kg 065 gm + 360 kg 085 gm + 289 kg 872 gm + 156 kg (RBSESolutions.com) 345 gm
Solution

Hint:

1. Write 65 gm as 065 gm and 85 gm as 085 gm
2. Start adding from top of unit coloumn.
3. 5 + 5 = 10 So ekadhik mark on 8. preccecding digit of 51 remainder = 10 – 10 = 0
4. Write remainder 0 + 2 + 5 = 7 at place of answer.
5. Keep it up.

Question 2.
Add: 7534 + 2459 + 1932 + 6547
Solution

Hint:

1. 34 + 59 = 33+ 1 + 59 = 33 + 60 = 93
2. Remaining 93 + 32 = 93 + 7 + 25 – 100 + 25 = 125
   So ekadhiken make on 9.
3. Write remainng 25 + 47 = 22 + 3 + 47 = 22 + 50 = 72 at the place of answer.
4. Remaining (RBSESolutions.com) addition process is an above.

Question 3.
Subtract by Vedic method:
28 km 375 km 46 cm from 37 km 467 km 35 cm.
Solution

Hint:

1. Arrange the column numbers in m-cm.
2. cm. column; 6 cannot be subtracted from 5, thus add 4 the param Mitra digit of 6 to 5.
3. Write sum = 9 at the answer place and mark ekadhk sign on 4 the pre-dissociator digit.
4. 4 = 5, cannot be subtracted from 3, thus add 5 the param Mitra digit to 3, the pre-dissociator digit.
5. Write sum = 8 below and mark the sign of ekadhik on the pre-dissociator.
6. Write 7 – 5 = 1 below.
7. 7 cannot be (RBSESolutions.com) subtracted from 6, So adding 6 + 3 = 9 write below and mark the sign of ekadhik on the pre-disjunction 3.
8. Write 4 – 3 = 0 below.
9. All the next operations will be done in the same way.
   ∴ Answer = 9 km 91 m 89 cm.

Question 4.
Which of the sutra is the best to solve 842 × 858?
Solution

1. Sutra ekadhiken poorvene is not effective (RBSESolutions.com) because product of R.H.S. i.e., 42 × 58 cannotbe got easily.
2. The sutra Nikhilam Adhara cannot be effective bacause if we consider base 1000 then respectively difference will be -158 and -142. The sutra Nikhilam updhara is not effective here as upadhar (sub-base) = 800.
   We get the difference 42 and 58.
3. The sutra Ekanyeune Poorvene is not effective here.
4. The sutra Urdhava-triyagbhayam is effective and the best here.
5. The new option: To get the product of 842 × 858 first we use sutra ekadhikene poorvene and then sutra urdhva-tiryagbhayam.
   

Question 5.
13579 ÷ 975 (dhwahjank method)
Solution

Hint:

1. 13 ÷ 9, first digit (RBSESolutions.com) of quotient = 1, remainder = 4
2. New dividend = 45, corrected dividend = 45 – 1 × 7 = 38
3. 38 ÷ 9, second digit of quotient = 4, remainder = 2
4. New dividend = 27,
   Corrected dividend = 27 – (4 × 7 + 1 × 5) = 27 – 33 = – 6
   Since we get the corrected dividend as negative so the second digit of quotient must be 3 instead of 4. So (iii) and (iv) steps are rejelable
5. Again 38 ÷ 9, Quotuiet second digit = 3, Remainder = 11.
6. New dividend = 1179 So corrected dividend and Last remainder
   = 1179 – (3 × 7 + 1 × 5) × 10 – 3 × 5
   = 1179 – 260 – 15
   = 904

Question 6.
Find a (RBSESolutions.com) square of 41.
Solution

Hint:

1. Make three columns for answer.
2. In first column, square of ten’s digit = 16
3. In third column, square of unit digit = 1
4. In middle column, product of both digit = 1 × 4 = 4
5. In the middle part write the above product once again.
6. The sum is the square of the given number. In the middle and third column only one digit must be written.

So, we have already seen that for completing the square calculator, the most basic requirement is a quadratic polynomial.

Question 7.
Find the (RBSESolutions.com) square of 17 by upsutra Yavaunam.
Solution
17² = 17 + 7/7²
Base = 10, deviation = +7 = 24/₄9 = 289

Question 8.
Find the square of 354 by Dwandwa yoga.
Solution
Five digit groups can be formed of the number 354 – 3, 35, 354, 54 and 4.
Written the Dwandwa yog of these five digit groups, we get

Question 9.
Find a (RBSESolutions.com) square of 12 by Ishta Sankhya Method.
Solution
12² = (12 + 2)(12 – 2) + 2²
= 14 × 10 + 4
= 144

Question 10.
Find the cube of 15.
Solution
15³ = 15 + 2 × 5/3 × 5²/5³
= 25 / ₇5 / ₁₂5
= 3375
Hint:

1. Base = 10, deviation = +5
2. Only single digit in middle and third part.

Question 11.
Find the (RBSESolutions.com) cube of 24.
Solution

Hint:

1. Base = 10, Sub-base = 10 × 2
2. Sub-base digit = 2, deviation = +4

Question 12.
Find the cube of 43 using formula.
Solution

Hint:

1. Base = 10,deviation = 3 × 3 – 10 = -1
2. Value of 2 in first part 20 of second part

Question 13.
Find the (RBSESolutions.com) square root of perfect square number 10329796.

Solution
Hint:

1. 4 pairs of digits in number, so 4 digits in the square root.
2. first square root digit = 3
3. Remainder = 10 – 3² = 1, Take next pair 32, besides 1, so new divident = 132.
4. Divisor = double of 3 = 6
5. 13 of the number 132, can be divided by 6, 2 times so write quotient 2 next to 3 in quotient column.
6. Write 2 next to divisor 6. So, the corrected divisor = 62
7. 132 – 62 × 2 = 132 – 124 = 8 = Remainder
8. Take 97 next to remainder 8, So new dividend = 897 and new divisor = 32 × 2 = 64
9. 89 ÷ 64 = 1 quotient, So write 1 next to quotient 32.
10. Write next to divisor 64, So the corrected divisor = 641.
11. 897 – 641 × 1 = 897 – 641 = 256, take 96 to ths remainder 256.
12. So, the new dividend = 25696 and new divisor = 321 × 2 = 642.
13. 2569 ÷ 642 = 4 quotient, So write 4 next to the quotient 321.
14. Write 4 next to divisor 642, So the corrected divisor = 6424
15. 25696 – 6424 × 4 = 0 remainder.
    Hence, remainder = 0
    Square root = quotient = 3214

Question 14.
Find the (RBSESolutions.com) square root of 41254929 by Dwandwa Yoga method.
Solution

Hint:

1. 4 digit in the square root.
2. 41 – 6² = 5, write before 2 just below of it.
3. New dividend = 52, corrected dividend = 52
4. 52 ÷ 12, quotient digit = 4. Write 4 between 2 and 5 and just below of them.
5. New dividend = 45, corrected dividend = 45 – 4² = 29.
6. 29 ÷ 12,quotient = 2, remainder digit = 5
7. Write 5 between 5 and 4 just below them.
8. New dividend = 54, corrected dividend = 54 – 4 × 2 × 2 = 38
9. 38 ÷ 12, quotient digit = 3 and remainder = 2.
   Write 2 between 4 and 9 just below. Then we have obtained 4 digits in the square root so we are to find the final remainder. After finding complete square root the decimal point and the zeros can be placed. To find the final remainder, as many zeros are placed after the decimal point as the new dividend are to be corrected.
10. Remainder = 29 – 4 × 3 × 2 – 2² = 1. Write 1 between 9 and 2, just below of them.
11. New dividend = 12, remainder = 12 – 2 × 3 × 2 = 0. Write 0 before and just below 9.
12. New dividend = 9, final remainder = 0
    Hence square root = 6423

Question 15.
Find the (RBSESolutions.com) cube root of 355045312441 perfect cube number, by division method.
Solution

Cube root = 7081
Hint:

1. No. of steps = cube root digit no. × 3 – 2
2. Don’t calculate the last three steps to obtain the cube root of a perfect cube number. As soon as we see the unit digit of a number we get the unit digit of its cube root.
3. Using the Dwandwa Yog Method. We can obtain the cube root of a number having so many digits in it.
4. If the cube root has four digits, find the second and third digit by division method. Unit and last digits can be calculated orally.

Question 16.
Simplify 2(x + 1) = 7(x + 1).
Solution
(x + 1) is (RBSESolutions.com) common factor in each term, so x + 1 = 0 ⇒ x = – 1

Question 17.
Simplify (x + 1)(x + 9) = (x + 3) (x + 3)
Solution
Constant terms in both sides are same, so, x = 0

Question 18.
Simplify: \(\frac { m }{ 2x+1 }\) + \(\frac { m }{ 3x+4 }\)
Solution
Both the numerators of fractions due same = m.
By formula 2x + 1 + 3x + 4 = 0
or 5x + 5 = 0
or x = – 1

Question 19.
Simplify : \(\frac { 3x+4 }{ 6x+7 }\) = \(\frac { x+1 }{ 2x+3 }\)
Solution
Sum of numerators (RBSESolutions.com) of both sides = 3x + 4 + x + 1 = 4x + 5 …(i)
Sum of denominators of both sides = 6x + 7 + 2x + 3 = 8x + 10 …(ii)
Ratio of (i) by (ii) = 1 : 2
According to question any sum equation to zero.
From 4x + 5 = 0
or 8x + 10 = 0
⇒ x = \(\frac { -5 }{ 4 }\)

Question 20.
Simplify the equation:

Solution
Sum of the denominator of (RBSESolutions.com) both sides are equal = 2x – 17
According to formula
2x – 17 = 0
⇒ x = \(\frac { 17 }{ 2 }\) = 8\(\frac { 1 }{ 2 }\)

We hope the given RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions, drop a comment below and we will get back to you at the earliest.