Fazal

RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 15
Chapter Name	Comparison of Quantities
Exercise	In Text Exercise
Number of Questions	16
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise

Do and Learn

(Page 159)
Question 1
Find the ratio of actual length and breadth (RBSESolutions.com) of mathematics book of class VII.
Solution:
Length of book = 27 cm.
Breadth = 15 cm

Calculate How Many Weeks Between Dates

Question 2
Find the ratio of length and breadth of national flag by taking help from your teacher.
Solution:
Different size of national flag (RBSESolutions.com) length and breadth ratio as follows :

Question 3
Find the ratio of length and breadth (RBSESolutions.com) of your classroom by measuring it
Solution:
Let the length and breadth of room is 30 feet and 20 feet.
Ratio = 30 : 20 = \(\frac { 30 }{ 20 }\) = \(\frac { 3 }{ 2 }\) = 3 : 2

Question 4
Measure your height and measure the length by expanding both the hand completely. Now find out the ratio between these two quantities.
Solution:
Length of both hands = 60 cm
Self height = 120 cm
Ratio = 120 : 60
\(\frac { 120 }{ 60 }\) = \(\frac { 2 }{ 1 }\) = 2 : 1

(Page 162)

Percentage Difference calculator is a free online tool to find the percent difference between two numbers.

Question 1
25 students of class tell their (RBSESolutions.com) interests about games.
Kabbadi – 4 students
Cricket – 11 students
Chess – 6 students
Tennis – 3 students
Other games – 1 students
Express the number of students in percentage according to the interest in each game.
Solution:
Total students = 25

Percentage Difference calculator – online basic math function tool to calculate how much percent difference between two quantities or observations.

Question 2
Following results are obtained on taking (RBSESolutions.com) advice of 250 students of preference of menu of mid day meal is selected school of Jalore Panchayat.

Express the percentage of preference of each type of menu from the above results.
Solution:
Total number (RBSESolutions.com) of students = 250

(Page 163)
Question 1
Convert the following (RBSESolutions.com) fractions into per-centage.
(i) \(\frac { 5 }{ 8 }\)
(ii) \(\frac { 5 }{ 3 }\)
Solution:

Question 2
Convert the decimal(RBSESolutions.com) fractions into percentage.
(i) 0.5
(ii) 0.08
(iii) 0.225
(iv) 6.5
Solution:

Question 3
Convert the percentage (RBSESolutions.com) into simple fraction and decimal fraction
(i) 36%
(ii) 12\(\frac { 1 }{ 2 }\)%
(iii) 3.6%
Solution:

(Page 166)

convert CGPA into percentage? Read this blog to understand an easy process to calculate percentage from CGPA.

Question 1
The population of a village is increased (RBSESolutions.com) to 15000 from 12000 in last 10 years. Then what is the percentage growth of population ?
Solution:
Population of village = 12000
Increased population of village in last 10 year = 15000
Increase in population = (15000 – 12000) = 3000

Question 2
Express rate of growth or loss in the (RBSESolutions.com) form of percentage in the following :
(1) The cost of electricity per unit has increased to ₹6.00 from ₹3.50
(2) The cost of 100 envelopes has decreased to ₹80 from ₹100.
Solution:
(1) Original rate per unit = ₹3.50
Increased rate per unit = ₹6
increase = 6 – 3.50 = ₹2.50

(2)100 envelopes cost = ₹100
New price = ₹80
Decrease in price = 100 – 80 = ₹20
Decrease % = \(\frac { 20 }{ 100 }\) x 100% = 20%

(Page 168)
Question 1
Mahaveer bought 5 bags of sugar for ₹16000. He (RBSESolutions.com) spent ₹200 for taxi fare, ₹120 for labour charges and ₹200 for transporttation charges. Find the selling price of per kg of sugar so as to earn a profit of ₹3 on each kg.
Solution:
Cost of 5 bags of sugar = ₹16000
Other expenses = ₹200 + ₹120 + ₹200
= ₹520
∴ Total C.P. of 5 bags of sugar
= ₹16000 + ₹520 = ₹16520
∵ Profit on 1 kg = ₹3
Profit on 5 bags or 500 kg sugar
= 500 x 3 = ₹1500
So, Profit = S.P – C.P
or S.P = C.P + Profit
= 16520+ 1500 = ₹18020
So, S.P. of 5 bags = ₹18020
S.P. of 1 bag = \(\frac { 1820 }{ 5 }\) = ₹3604

Question 2
Manoj bought a second hand car for ₹1,50,000. He spent ₹60,000 on Its (RBSESolutions.com) engine repair and ₹15000 on replacing the tyre tubes. Manoj sold this car to Jitendra for ₹2,10,000 Calculate loss or gain in this business.
Solution:
Cost of car = ₹1,50,000
Expense on engine = ₹60,000
Expense on tyre & tube = ₹15,000
Total cost price = ₹2,25,000
Selling price = ₹2,10,000
Manoj has a loss
So, Loss = C.P – S.P
= 2,25,000 – 2,10,000
Loss = ₹15,000
Manoj has a loss of ₹ 15,000

Percentage Difference calculator – online basic math function tool to calculate how much percent difference between two quantities or observations.

(Page 170)
Ashok has borrowed ₹50000/- for the construction (RBSESolutions.com) of his house. This borrowed amount is known as Principal. He repays ₹55000 to that institute after one year.
Ashok paid ₹5,000/- extra on ₹50,000. This extra amount is known as interest.
The amount of interest depends on following things:
1. Borrowed money (Principal)
2. Time (Period for which the amount is borrowed).
3. Rate (The extra amount paid on per hundred) which is determined on the basis of per mounth/per annum.
You can find the amount you have to pay at the end of the year by adding sum borrowed and the interest
i.e., Amount = Principal + Interest

Question 1
How much interest would Ashok have pay after 2 years if he (RBSESolutions.com) was not able to return the money to the instutute after 1 year ?
Solution:
Because after one year he pays ₹5000/-
as interest so in 2 years he has to pay ₹10000/- as interest.

Question 2
How much money in (RBSESolutions.com) total has to be paid including interest?
Solution:
Amount = Principal + Interest
= 50,000 + 10,000
= ₹60,000/-

We hope the RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities In Text Exercise, drop a comment below and we will get back to you at the earliest.

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