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Class 10

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 10 Maths Chapter 1 Real Numbers Ex 1.2

June 1, 2022 by Safia Leave a Comment

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

RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

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

RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

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

RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

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.

RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

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 17 Measures of Central Tendency Additional Questions

May 28, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions

RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 10
Subject Maths
Chapter Chapter 17
Chapter Name Measures of Central Tendency
Exercise Additional Questions
Number of Questions Solved 37
Category RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions

Multiple Choice Questions
Question 1.
Which of the following is not the (RBSESolutions.com) measure of central tendency:
(A) Mean
(B) Median
(C) Mode
(D) Standard devitation
Solution :
(D) is correct.

Question 2.
In calculating mean of grouped data we assumed that frequencies: [NCERT Exemplar Problem]
(A) equally distribution In all classes
(B) centralized on sign of groups
(C) centralized on upper limits of classes
(D) centralized on lower limits of classes.
Solution :
(B) is correct.

RBSE Solutions

Question 3.
If xi are mid-point of class-intervals of grouped data fi are their (RBSESolutions.com) corresponding frequencies and \(\overline { x }\) is mean, then Σ(fixi – \(\overline { x }\)) equals: [NCERT Exemplar Problem]
(A) 0
(B) -1
(C) 1
(D) 2
Solution :
(A) is correct.

Question 4.
Cumulative frequency table is helpful in: [NCERT Exemplar Problem]
(A) Mean
(B) Mode
(C)Median
(D) None of these
Solution :
(C) is correct.

Question 5.
In given formula \(\overline { x }\) = a+h\(\left( \frac { { \Sigma f }_{ i }{ u }_{ i } }{ { \Sigma f }_{ i } } \right)\), value of ui will be : [NCERT Exemplar Problem]
(A) h(xi – a)
(B) \(\frac { { x }_{ i }-a }{ h }\)
(C) \(\frac { { a-x }_{ i } }{ h }\)
(D) \(\frac { { x }_{ i }+a }{ h }\)
Solution :
Thus (B) is correct.

Question 6.
For distribution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Sum of lower limits of median class and (RBSESolutions.com) modal class is: [NCERT Exemplar Problem]
(A) 15
(B) 25
(C) 30
(D) 35
Solution :
Cumulative frequency just above 33 is 37 and its corresponding class interval is 10 – 15.
So modal class = 10 – 15.
Thus, sum of lower limit of median class and modal class
= 10 + 15 = 25, so (B) is correct.

Age difference calculator that can be used to find the number of years, months, and days between two persons.

Question 7.
For distribution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Difference between upper limit of (RBSESolutions.com) median class and lower limit of modal class is.
(A) 0
(B) 19
(C) 20
(D) 38
Solution :
In given data, maximum frequency is 20 and its corresponding class-interval is 125 – 145.
Thus modal class will be 125 – 145.
For median,
cumulative frequency just above 33 is 37 and its corresponding class-interval is 10 – 15.
So median class = 10 – 15
cumulative frequency table
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Here N = 66 ⇒ \(\frac { N }{ 2 }\) = \(\frac { 66 }{ 2 }\) = 33
Difference between upper limit of median (RBSESolutions.com) class and lower limit of modal class
= 145 – 125 = 20.
Cumulative frequency just above 33.5 is 42
whose corresponding class interval is 125 – 145.
So median class = 125 – 145 =20
Thus, (C) is correct.

RBSE Solutions

Question 8.
To find mean from grouped data on the formula \(\overline { x }\) = a + \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) di is deviation of (RBSESolutions.com) following from a. [NCERT Exemplar Problem]
(A) Lower limits of classes
(B) Upper limits of classes
(C) Mid-points of classes
(D) Frequencies of group sign
Solution :
(C) is correct

Question 9.
The average of n observation is \(\overline { X }\). If we add 1 in first term, 2 in second term similarly 3, 4, 5, ………., n, then new mean will be:
(A) \(\overline { X }\) + n
(B) \(\overline { X }\) + \(\frac { n }{ 2 }\)
(C) \(\overline { X }\) + \(\frac { n+1 }{ 2 }\)
(D) None of these
Solution :
(C) is correct.

Question 10.
Number of students in a school (RBSESolutions.com) according to their age are as follows:
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Their mode is:
(A) 41
(B) 12
(C) 3
(D) 17.
Solution :
From above table it is clear that (RBSESolutions.com) frequency 41 is maximum and its corresponding age group of 12. So its mode will be 12.
Thus (B) is correct

Question 11.
The average of statistical data is called:
(A) Arithmetic Mean
(B) Median
(C) Mode
(D) Frequency
Solution :
(A) is correct.

Question 12.
If 9 is mean of 5, 7, 9, x then find x.
(A) 11
(B) 15
(C) 18
(D) 16.
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
⇒ 36 = 21 + x
⇒ x = 36 – 21 = 15
Thus, (B) is correct.

RBSE Solutions

Short/Long Answer Type Questions

Question 1.
Find median of (RBSESolutions.com) following data
25, 34, 31, 23, 22, 26, 35, 28, 20, 32
Solution :
Arranging given data in ascending order
20, 22, 23, 25, 26, 28, 31, 32, 34, 35
Here total terms (n) = 10 (even number)
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, median = 27

Question 2.
The mean of 20 observation is 20. If 2 is added in each of first ten observation, then find mean of 20 new observations.
Solution :
Sum of 20 observations 20 × 20 = 400
2 is added in 10 observation so total increase = 10 × 2 = 20
Sum of total 20 observations = 400 + 20 = 420
Thus new arithmetic mean = \(\frac { 420 }{ 20 }\) = 21

Question 3.
If \(\overline { x }\) is mean of x1, x2 ….. xn, then find (RBSESolutions.com) mean of mx1, mx2 ….. mxn.
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, required mean = m\(\overline { x }\)

Question 4.
If \(\overline { x }\) is mean of x1, x2 ….. xn, then prove that if a is added to (RBSESolutions.com) each term then mean will be (\(\overline { x }\)+a)
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions

Question 5.
If median of 17, 26, 60, 45, 33 is 33 and if 17 is replaced by 27 then (RBSESolutions.com) what will be new median?
Solution :
If 17 replaced by 27 then numbers will be 27, 26, 60, 45, 33 writing in ascending order 26, 27, 33, 45, 60.
Here n =5 (odd number)
Median = \({ \left( \frac { n+1 }{ 2 } \right) }^{ th }\) term
= \({ \left( \frac { 5+1 }{ 2 } \right) }^{ th }\) term = \(\left( \frac { { 6 }^{ th } }{ 2 } \right) \) term
= 3rd term = 33
∴ Median = 33

RBSE Solutions

Question 6.
If arithmetic mean of following data (RBSESolutions.com) is 15 then find p.
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 1
Solution :
Table for calculating A.M
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, A.M. (\(\overline { x }\)) = \(\frac { \Sigma fx }{ \Sigma f } \)
15 = \(\frac { 695+10P }{ 27+P }\)
⇒ (27 + p)15 = 695 + 10P
⇒ 405 + 15P = 695 + 10P
⇒ 15P – 10P = 695 – 405
⇒ 5P = 290
⇒ P = \(\frac { 290 }{ 5 }\) = 58

Question 7.
Find mean of the following (RBSESolutions.com) distribution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 2
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } }\) = \(\frac { 412.5 }{ 75 }\) = 5.5

Question 8.
Consider the following distribution of daily wages of 40 workers of a factory (RBSESolutions.com) and by using suitable method, find mean daily wages of the workers of this factory :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 3
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Arithmetic mean = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } }\) = \(\frac { 1980 }{ 40 }\) = 49.50
Thus, mean of daily wages = ₹ 49.50

Question 9.
Calculate the mean of the (RBSESolutions.com) following data : [NCERT Exemplar Problem]
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
Here class-interval is discontinuous but if we convert it into continuous form, mid-values will not change so we will not change class-intervals.
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } }\) = 12.93

Question 10.
Weight of 30 students of a class at the time of medical (RBSESolutions.com) test is recorded as per table given below
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Find mean weight of students. [Board of Secondary Education Raj 2012]
Solution :
Let assumed mean (A) = 57.5
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) = 57.5 + \(\frac { -10 }{ 30 }\) = 57.167 = 57.17 kg.
Thus, mean weight of students = 57.17 kg.

RBSE Solutions

Question 11.
To find concentration (part per million) of (SO2) Sulphur dioxide in air (RBSESolutions.com) data ¡s collected from 30 colonies of a city which are given below:
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Find mean of concentration of SO2 in Air.
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Arithmetic mean (\(\overline { x }\)) = \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) = \(\frac { 2.96 }{ 30 }\) = 0.0986 or 0.099 (approx)
Thus, mean of concentration of SO2 in Air = 0.099 part per million.

Question 12.
Following distribution respresents weight of 30 students in a class. Find (RBSESolutions.com) median weight of students.
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Here, N = 30 and \(\frac { N }{ 2 }\) = \(\frac { 30 }{ 2 }\) = 15 C.f. just above 15 is 19 and corresponding class-mterval is 55 – 60.
∴ Median class = 55 – 60
∴ l = 55, \(\frac { N }{ 2 }\) = 15, C = 13, f = 6 and h = 5
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, median weight of students = 56.67 kg(approx)

Question 13.
Following data represents life period observations (RBSESolutions.com) of electrical instruments.
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 4
Find mode of life period of instruments.
Solution :
Here maximum frequency is 61 and its corresponding class-interval is 60 – 80. So modal class = 60 – 80.
l = 60, f1 = 61, f0 = 52, f2 = 38 and h = 20
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, mode of life period is 65.625 hrs

Question 14.
Find mode of the (RBSESolutions.com) following data : (CBSE 2013)
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
Here maximum frequency is 12 whose class is 60 – 80. So modal class 60 – 80
∴ l = 60, h = 20, f1 = 12, f0 = 10, f2 = 6
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, required mode = 65

Question 15.
Find mode of the following (RBSESolutions.com) data. (CBSE 2014)
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
Here maximum frequency = 155
whose class is 95 – 100
∴ l = 95, f1 = 155, f0 = 85, f2 = 110, h = 5
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, required mode = 98.04

RBSE Solutions

Question 16.
Find unknown frequency in the (RBSESolutions.com) following data of N = 100 and median = 32. (CBSE 2013)
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
Let f1 and f2 are unknown frequence, then
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, 75 + f1 + f2 = 100
f1 + f2 = 100 – 75 = 25
Given: Median = 32
So, median class is 30 – 40.
Thus, l = 30, h = 10, f = 30, C = 35 + f1
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
15 – f1 = 6
f1 = 15 – 6 = 9
From eqn.(i), f2 = 16
Thus, f1 = 9, f2 = 16

Question 17.
If mean of following (RBSESolutions.com) data is 21.5, then find K. (CBSE 2012)
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 5
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
∵ Arithmetic mean = \(\frac { { \Sigma f }_{ i }{ x }_{ i } }{ { \Sigma f }_{ i } }\)
21.5 = \(\frac { 255+35K }{ 15+K }\)
⇒ 322.5 + 21.5 = 255 + 35K
⇒ 322.5 – 255 = 35K – 21.5K
⇒ 13.5K = 67.5
K = \(\frac { 67.5 }{ 13.5 }\) = 5
∴ K = 5

Question 18.
If median of given data is 28.5 then find the (RBSESolutions.com) values of x and y. [Board of Secondary Education Raj. 2014]
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
cumulative frequency table
But sum of frequency Σfi = N = 80 which is equal (RBSESolutions.com) to last term of c.f. group
∴ 45 + x + y = 80 or x + y = 80 – 45
⇒ x + y = 35 …(i)
Now \(\frac { N }{ 2 }\) = \(\frac { 80 }{ 2 }\) = 40
and median of distribution = 28.5 which lies in class-interval 20 – 30.
Median class = 20 – 30
l = 20, f = 20, c = 5 + x and h = 10
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
57 = 75 – x or x = 18
Putting value of x in eqn. (i), 18 + y = 35
Thus, y = 17.

Question 19.
If mean of given distribution is 50, then find the (RBSESolutions.com) values of x and y. [Board of Secondary Raj. 2014]
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
But sum of frequencies is Σfi = N = 120 which is equal (RBSESolutions.com) to last term of cumulative frequency group.
68 + x + y = 120
x + y = 120 – 68
x + y = 52
Now, \(\frac { N }{ 2 }\) = \(\frac { 120 }{ 2 }\) = 60
And median of distribution is equal to 50, which lies in class-interval 40 – 60.
Median class = 40 – 60
l = 40, f = 32, c = 17 + x and h = 20
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
⇒ 50 = 320 + 215 – 5x
⇒ 400 – 535 = – 5x
⇒ 5x = 135
⇒ x = 27 ……(i)
From equation (i) and (ii)
27 + y = 52
y = 25

RBSE Solutions

Question 20.
Median of following data is 525. If sum of frequencies is 100, then (RBSESolutions.com) find values of x and y. (Board of Secondary Raj. 2012)
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Given: Σf = N = 100
Thus, 76 + x + y = 100
⇒ x + y = 100 – 76 = 24 …..(i)
Median is 525 which (RBSESolutions.com) lies in 500 – 600.
∴ l = 500; f = 20, c = 36 + x, and h = 100
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 6
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
⇒ 25 = (14 – x) × 5
⇒ 14 – x = \(\frac { 25 }{ 5 }\) = 5
⇒ x = 14 – 5
∴ x = 9
From equation (i), 9 + y = 24
⇒ y = 24 – 9
∴ y = 15
Thus, x = 9 and y = 15.

Question 21.
If median of the following distribution is 27, then (RBSESolutions.com) find the values of x and y. [CBSE 2012]
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Given: N = 68
⇒ 47 + x + y = 68
⇒ x + y = 68 – 47
⇒ x + y = 21 ……..(i)
Now \(\frac { N }{ 2 }\) = \(\frac { 68 }{ 2 }\) = 34
Median of distribution is 27 which (RBSESolutions.com) lies is class-interval 20 – 30.
∴ Median class 20 – 30
l = 20, c = 5 + x, f = 20 and h = 10
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
⇒ 69 – x = 2 × 27
⇒ x = 69 – 54 = 15
Putting value of x in eqn. (i),
15 + y = 21
⇒ y = 21 – 15 = 6
Thus, x = 15, y = 6

Question 22.
Following table shows the age of (RBSESolutions.com) patients admit in Hospital :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 7
Find mode and mean. Compare two central measures and explain. them. (NCERT)
Solution :
For Mode : Here maximum frequency is 23 and its corresponding class is 35 – 45.
So, modal class 35 – 45
∴ l = 35, f0 = 21, f1 = 23, f2 = 14, h = 10
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, mode = 36.8 yrs.
For Mean : Let assumed (RBSESolutions.com) mean A = 40
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
∵ Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) = 40 – \(\frac { 370 }{ 80 }\) = 40 – 4.625 = 35.375
Thus, mode of data = 36.8 years. and mean = 35.375 years.
Maximum age of patients admit in hospital is 36.8 years. where as average age of patients admit is hospital is 35.57 yrs.

Question 23.
Following table represents literacy rate of 35 cities (RBSESolutions.com) in percentage.
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions 8
Find mean literacy rate.
Solution :
Let assumed mean (A) 70
Class size (h) = 55 – 45 = 10
By step deviation method
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Mean (\(\overline { x }\)) = A + \(\left( \frac { { \Sigma f }_{ i }{ u }_{ i } }{ { \Sigma f }_{ i } } \right)\) × h
⇒ \(\overline { x }\) = 70 + \(\frac { -2 }{ 35 }\) × 10 = 70 + \(\frac { -20 }{ 35 }\)
= 70 + (-0.57) = 70 – 0.57 = 69.43
Thus, mean literacy rate = 69.43%

RBSE Solutions

Question 24.
The distribution of marks obtained by 30 students in (RBSESolutions.com) Maths examination as follows :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Find mean from assumed mean method for given data also find mode.
Solution :
Let assumed mean (A) = 47.5
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
∵ Arithmetic mean (\(\overline { x }\)) = A + \(\frac { { \Sigma f }_{ i }{ d }_{ i } }{ { \Sigma f }_{ i } } \) = 47.5 + \(\frac { 435 }{ 30 }\) = 47.5 + 14.5 = 62
From given table, if is clear that maximum (RBSESolutions.com) frequency is 7 and its corresponding class is 40 – 55.
Thus, modal class will be 40 – 55.
∵ l = 40, f0 = 3, f1 = 7, f2 = 6, and h = 15
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Thus, mean = 62 and mode = 52

Question 25.
Following frequency distribution represents electricity cost of 68 users in a colony. Find mean, median and mode from these data:
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Solution :
For median :
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Here, N = 68 \(\frac { N }{ 2 }\) = \(\frac { 68 }{ 2 }\) = 34
Cumulative frequency (RBSESolutions.com) just above 34 is 42 and its corresponding class is 125 – 145.
∴ Median class = 125 – 145, here l = 125, N = 68, f = 20, C = 22, and h = 20
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
For mean : Let assumed mean (A) = 135, h = 20
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Mean = A + \(\frac { { \Sigma f }_{ i }{ u }_{ i } }{ { \Sigma f }_{ i } }\) × h = 135 + \(\frac { 7 }{ 68 }\) × 20
= 135 + 0.1029 × 20 = 135 + 2.05 = 137.05
For mode : In given data, maximum (RBSESolutions.com) frequency is 20 and its corresponding class is 125 – 145.
Modal class = 125 – 145
Here l = 125, f1 = 20, f0 = 13, f2 = 14 and h = 20
RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions
Hence, median = 137, mean = 137.05 and mode = 135.77

We hope the given RBSE Solutions for Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 Measures of Central Tendency Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1

May 28, 2022 by Fazal Leave a Comment

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.

RBSE Solutions

(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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 1
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 2
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 3
Given linear pair has infinite solutions, so linear pair is consistent.

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 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),
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 5
Table 2 for equation (ii),
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 6
Plot the points of Table (1) and (2) on graph (RBSESolutions.com) paper and by joining these points, two straight lines are obtained.
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 7
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.

RBSE Solutions

(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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 8
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 9
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 10
By plotting the (RBSESolutions.com) points of Table 1 and 2 we get two straight lines.
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 11
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 12
∴ 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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 13
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 14
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.
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 15
From graph, lis clear that given pair of linear equations are coincident. Thus they have infinitely many solutions.

RBSE 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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 16
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 17
From Table (1) arid (2), plot the points on (RBSESolutions.com) graph paper and by joining them, we get two straight lines.
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 18
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.

RBSE Solutions

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),
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 19
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 20
Plot the points (RBSESolutions.com) from Table (1) and (2) on graph.
By joining these points two straight lines are obtained.
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 21
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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 22
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 23
Plot the points from (RBSESolutions.com) Table (1) and (2) on graph paper. By joining these points two straight lines are obtained.
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 24
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).

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 25
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 26
Plot the points obtained from (RBSESolutions.com) Table (1) and (2) on graph paper. By joining these points two straight lines are obtained.
RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 27
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.

RBSE Solutions

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

May 28, 2022 by Fazal Leave a Comment

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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q1
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.

RBSE Solutions

Question 2.
Add: 7534 + 2459 + 1932 + 6547
Solution
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q2
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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q3
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.
    RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q4

Question 5.
13579 ÷ 975 (dhwahjank method)
Solution
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q5
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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q6
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.

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q8

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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q11
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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q12
Hint:

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

RBSE Solutions

Question 13.
Find the (RBSESolutions.com) square root of perfect square number 10329796.
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q13
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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q14
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
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q15
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

RBSE Solutions

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:
RBSE Solutions for Class 10 Maths Chapter 1 Vedic Mathematics Additional Questions Q20
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.

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