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RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management

September 10, 2024 by Fazal Leave a Comment

RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management are part of RBSE Solutions for Class 10 Science. Here we have given Rajasthan Board RBSE Class 10 Science Solutions Chapter 13 Waste and its Management.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 10
Subject Science
Chapter Chapter 13
Chapter Name Waste and its Management
Number of Questions Solved 36
Category RBSE Solutions

Rajasthan Board RBSE Class 10 Science Solutions Chapter 13 Waste and its Management

Textbook Questions Solved

I. Multiple Choice Questions

Question 1:
Which method is ideal for the management of bio-medical waste?
(a) Landfill
(b) Incineration
(c) Recycling
(d) Disposal in water bodies
Answer:
(b) Incineration

Question 2:
Recycling is ideal for which type of waste?
(a) Metallic waste
(b) Medical waste
(c) Household waste
(d) Agricultural waste
Answer:
(a) Metallic waste

Question 3:
Which of the following is a greenhouse gas?
(a) Hydrogen
(b) Carbon monoxide
(c) Carbon dioxide
(d) Sulphur dioxide
Answer:
(c) Carbon dioxide

Question 4:
What is the average amount of waste per person in big cities of India?
(a) 0.1 kg
(b) 0.1 – 0.2 kg
(c) 0.2 – 0.4 kg
(d) 0.4 – 0.6 kg
Answer:
(d) 0.4 – 0.6 kg

Question 5:
Which of the following can be turned into organic manure?
(a) Household waste
(b) Agricultural waste
(c) Both ‘a’ and ‘b’
(d) None of these
Answer:
(c) Both ‘a’ and ‘b’

RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management

Waste and its Management Very Short Answer Type Questions

Question 6:
How is biogas made?
Answer:
Biogas is made bv decomposition of farm waste and some domestic waste.

Question 7:
What is waste?
Answer:
Useless product at the end of any process is called waste.

Question 8:
Write the names of greenhouse gases.
Answer:
Carbon dioxide and methane

Question 9:
What is vermicompost?
Answer:
Compost which is prepared with help from earthworms is called vermicompost.

Question 10:
Which diseases are spread by stagnant water in drainage?
Answer:
Malaria, dengue

Waste and its Management Short Answer Type Questions

Question 11:
Explain waste management.
Answer:
The practice of transportation and dumping of waste, recycling of useful material from waste, etc. are integral part of waste management. Different types of waste need different kinds of management practices. Waste management is highly important for a comfortable and disease-free life in cities and villages.

Question 12:
What do you understand by solid waste?
Answer:
Waste which is in solid form is called solid waste. Management of solid waste requires a different approach. One needs to physically transport solid waste to the dumping site. A huge amount of manpower and capital is required for management of solid waste.

RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management

Question 13:
Differentiate between biodegradable and non-biodegradable waste.
Answer:

Biodegradable Waste Non-biodegradable Waste
Can be decomposed by biological agents. Cannot be decomposed by biological agents.
They are generally environment- friendly. They generally harm the environment.
Examples: fruit peels, stale food, paper, etc. Examples: cans, bottles, plastic, etc.

Question 14:
What do you understand by landfill?
Answer:
Waste is dumped at landfill sites. When amount of waste attains a significant figure, it is covered with soil. After that, waste is allowed to decompose. Pipeline system is installed in landfill to channelize and utilize the methane gas which is produced during decomposition. This gas is utilized for power generation at many places. A landfill site is usually developed outside the city limits.

Question 15:
What is the meaning of recycling?
Answer:
Recyling is the process of treatment of discarded material to make it available fo resue. Certain items from waste are segregated for recycling. Items like plastic, aluminium cans, metals, etc. are sent to recycling unit so that new items can be made from them. This not only helps in reducing the amount of waste but also in conservation of resources.

Question 16:
What is the use of incineration?
Answer:
In this method, waste is burnt under controlled conditions so that production of polluting gases can be minimized. This method is ideal for places where land is in short supply. This method is also ideal for disposing off the medical waste.

Waste and its Management Long Answer Type Questions

Question 17:
Explain different types of waste.
Answer:
Waste can be solid, liquid or gas but on the basis of ability to undergo decomposition, waste is classified into two types.

  • Biodegradable Waste: Waste which can be decomposed by biological factors is called biodegradable waste, e.g. fruit peels, stale food, paper, etc. Biodegradable is generally environment-friendly. Biogas is produced on decomposition of biodegradable waste. This biogas can be utilized for many purposes like as domestic fuel, for power generation, etc. Non-
  • biodegradable Waste: Waste which cannot be decomposed by biological factors is called non-biodegradable waste, e.g. aluminium can, polythene bag, etc. Non-biodegradable waste is generally harmful for environment. However, many non-biodegradable waste can be recycled to make useful products.

Question 18:
Write an essay on waste management.
Answer:
The practice of transportation and dumping of waste, recycling of useful material from waste, etc. are integral part of waste management. Different types of waste need different kinds of management practices. While gaseous waste goes up in air on its own, we need to make proper drainage system for handling liquid waste. For handling solid waste, we need a proper system of transportation and people. Even a small village produces a large amount of waste per day. Daily production of garbage could be in thousands of tons in mega cities.

Methods of Waste Management:

  1. Landfill: Waste is dumped at landfill site. When amount of waste attains a significant figure, it is covered with soil. After that, waste is allowed to decompose. Pipeline system is installed in landfill to channelize and utilize the methane gas which is produced during decomposition. This gas is utilized for power generation at many places. A landfill site is usually developed outside the city limits.
  2. Incineration: In this method, waste is burnt under controlled conditions so that production of polluting gases can be minimized. This method is ideal for places where land is in short supply. This method is also ideal for disposing off the medical waste.
  3. Recycling: Certain items from waste are segregated for recycling. Items like plastic, aluminium cans, metals, etc. are sent to recycling unit so that new items can be made from them. This not only helps in reducing the amount of waste but also in conservation of resources.
  4. Chemical Methods: Many wastes can be managed by using chemical methods.

RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management

Question 19:
Write an essay on sources of waste.
Answer:
Following are different sources of waste:

  • Household Source: Paper, cardboard, fruit peels, vegetable peels, soapy water, rags, etc.
  • Municipality: Municipality looks after management of solid and liquid waste from households, offices and factories. Municipal waste also includes human excreta.
  • Industry and Mining: Many harmful wastes are produced by industries, e.g. harmful chemicals, grease, etc. Mining activity also produces wastes like dust.
  • Agriculture: Hay, dry leaves, husk, cowdung, etc. are examples of farm waste.
  • Medical Waste: Hospitals produce many wastes like syringe, cotton, bandage, catheters, strips, etc. Some of the hospital waste may contain harmful germs.

Question 20:
Make a list of various wastes from your surroundings and classify them.
Answer:

Biodegradable Waste Non-biodegradable Waste
Fruit peels, vegetable peels, dry leaves, stale food, paper, empty cartons, old clothes, jute bags, etc. Empty cans, plastic bottles, bubble wrap, discarded toys, glass items, etc.

Question 21:
What will you do for waste management in your village or neighbourhood?
Answer:
I will take following steps for waste management in my neighbourhood:

  • I will urge people to use garbage bins for throwing garbage. I will organize awareness campaign for this.
  • I will request the municipal authorities for regular maintenance of drainage system.
  • I will take a team of volunteers for ‘Swachhata Abhiyan’ on a weekly basis.
  • I will educate people on benefits of segregating waste so that some items can be recycles.

Waste and its Management Additional Questions Solved

I. Multiple Choice Questions

Question 1:
Which of the following poses the risk of more number of diseases than any other?
(a) Domestic waste
(b) Hospital waste
(c) Farm waste
(d) Industrial waste
Answer:
(b) Hospital waste

Question 2:
Which of the following is biodegradable waste?
(a) Polythene bag
(b) Aluminium can
(c) Fruit peel
(d) Plastic furniture
Answer:
(c) Fruit peel

Question 3:
Which gas is produced on decomposition of farm waste?
(a) Methane
(b) Carbon dioxide
(c) Hydrogen
(d) Nitrogen oxide
Answer:
(a) Methane

RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management

Question 4:
Which epidemic can spread due to stagnant drainage?
(a) Jaundice
(b) Hepatitis
(c) AIDS
(d) Malaria
Answer:
(d) Malaria

Question 5:
Drinking contaminated water can cause which disease?
(a) Jaundice
(b) Cholera
(c) Diarrohea
(d) All of these
Answer:
(d) All of these

Waste and its Management Very Short Answer Type Questions

Question 1:
List some wastes which come from household.
Answer:
Paper, cardboard, fruit peels, vegetable peels, soapy water, rags, etc.

Question 2:
List some wastes from farming.
Answer:
Hay, dry leaves, husk, cowdung, etc.

Question 3:
List some pastes from hospital.
Answer:
Syringe, cotton, bandage, catheters, strips, etc.

Question 4:
What is waste management?
Answer:
The practice of transportation and dumping of waste, recycling of useful material from waste, etc. are integral part of waste management.

Question 5:
Which civic body looks after waste management in towns?
Answer:
Municipality

Waste and its Management Short Answer Type Questions

Question 1:
What is the effect of sewage on water bodies?
Answer:
Sewage from households pollutes water bodies like rivers and ponds. It is harmful for aquatic animals and plants. It also pollutes groundwater.

Question 2:
What is the need for waste management?
Answer:
Waste can keep on accumulating if not managed in time. This will make ugly site around us. Moreover, heaps of garbage can trigger many epidemics. Hence, there is a need for waste management.

Question 3:
Waste management is not efficient in small towns. Why?
Answer:
Municipalities in small towns suffer from shortage of funds. So, the contractors do not get timely payment. Even sanitation workers do not get regular salary. Due to these reasons, waste management is not efficient in small towns.

RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management

Question 4:
What were the suggestions of Shivraman Committee?
Answer:
Following are the suggestions of Shivaraman Committee:

  • Installation of large garbage bins.
  • Proper arrangement of management of human excreta.
  • Proper system for picking up garbage in towns.
  • Using incinerators for disposing off the waste.

Waste and its Management Long Answer Type Questions

Question 1:
What are the harms of waste?
Answer:
Following are the harms of waste:

  • Accumulation of waste around us is not good from the point of aesthetics.
  • Decomposition of waste produces methane and other harmful gases. Methane causes global warming.
  • A heap of waste is the breeding ground for harmful germs, houseflies and mosquitoes. These agents can cause many epidemics.
  • Hospital waste can cause serious infection to a person. HIV and Hepatitis B can spread from coming in contact with infected needle.
  • Plastic waste turns the soil infertile. It chokes the drainage. Stray animals often swallow plastic and die due to choking. Prolonged exposure to plastic increases thalatus in blood which hampers the development of foetus.
  • Sewage from households pollutes water bodies like rivers and ponds. It is harmful for aquatic animals and plants. It also pollutes groundwater.

We hope the given RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Science Solutions Chapter 13 Waste and its Management, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 10 Social Science Chapter 19 Road Safety Education

September 10, 2024 by Fazal Leave a Comment

RBSE Solutions for Class 10 Social Science Chapter 19 Road Safety Education are part of RBSE Solutions for Class 10 Social Science. Here we have given Rajasthan Board RBSE Class 10 Social Science Solutions Chapter 19 Road Safety Education.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 10
Subject Social Science
Chapter Chapter 19
Chapter Name Road Safety Education
Number of Questions Solved 4
Category RBSE Solutions

Rajasthan Board RBSE Class 10 Social Science Solutions Chapter 19 Road Safety Education

Road Safety Education Additional Questions Solved

Question 1.
State the negative effects of the victim of road accidents.
Answer:
The negative effects of victim of road accidents are:

  1. Loss/damage to vehicle or property,
  2. Financial burden on the family.
  3. Administrative or legal action.
  4. Causing major deformity/handicap or death.

Question 2.
Why does attainment of an appropriate age essential for submitting an application for acquiring motor driving license?
Answer:
The age of an applicant of driving license must be minimum of 18 years on the date of submitting application. The applicant must be fully aware of the traffic rules and regulations. It is believed that an adolescent becomes mature and responsible at the age of 18.

Question 3.
Under which act we punished if we do not abide by road safety rule while driving a vehicle?
Answer:
If we do not abide by the road safety rule while driving a vehicle then there is a provision of punishment under Motor Vehicle Act 1988.

Question 4.
State the main causes of road accidents.
Answer:
The main causes of road accidents are:

  1. Driving in drunken state.
  2. Very high speed of the vehicle.
  3. Avoiding the traffic rules.
  4. Not using the helmet and the seat belt.
  5. Carelessness of the driver of the vehicle

We hope the given RBSE Solutions for Class 10 Social Science Chapter 19 Road Safety Education will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Social Science Solutions Chapter 19 Road Safety Education, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

July 2, 2022 by Fazal Leave a Comment

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

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

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

Multiple Choice Questions

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

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

RBSE Solutions

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

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

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

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

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

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

RBSE Solutions

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

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

Very Short Answer Type Questions

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

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

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

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

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

RBSE Solutions

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

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

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

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

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

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

RBSE Solutions

Short Answer Type Questions

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

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

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

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

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

RBSE Solutions

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

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

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

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

Long Answer Type Questions

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

RBSE Solutions

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

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

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

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

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

RBSE Solutions

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

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

RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

April 16, 2022 by Prasanna Leave a Comment

RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 4

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 6 Lines and Angles Ex 6.3

Question 1.
In the figure, sides QP and RQ of APQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110% find ∠PRQ.
RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 1
Answer:
We have :
∠QPR + ∠SPR = 180° (Linear pair)
∠QPR + 135° = 180°
or ∠QPR = 180° – 135° = 45°
Now, ∠TQP = ∠QPR + ∠PRQ [By exterior angle theorem (8)]
⇒ 110° = 45° + ∠PRQ
⇒ ∠PRQ a 110° – 45° = 65°
Hence, ∠PRQ = 65°.

Question 2.
In the figure, ∠X = 62% ∠XYZ = 54°. If YO and ZO are ‘ the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.
RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 2
Answer:
Consider ∆XYZ.
∠YXZ + ∠XYZ + ∠XZY = 180° (Angle-sum property)
⇒ 62° + 54° + ∠XZY = 180° (∵ ∠YXZ = 62% ∠XYZ = 54°)
⇒ ∠XZY = 180° – 62° – 54° = 64°
Since YO and ZO are bisectors of ∠XYZ and ∠XZY, therefore
∠OYZ = \(\frac{1}{2}\) × ∠XYZ = \(\frac{1}{2}\) × 54° = 27°
and, ∠OZY = \(\frac{1}{2}\) × ∠XZY = \(\frac{1}{2}\) × 64° = 32°

Now, in ∆OYZ, we have :
∠YOZ + ∠OYZ + ∠OZY = 180° (Angle -sum property)
∴ ∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° – 27° – 32° = 121°
Hence, ∠OZY = 32° and ∠YOZ = 121°.

RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

Question 3.
In the figure, if AB ∥ DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 3
Answer:
Since AB ∥ DE and transversal AB intersects them at A and E respectively, therefore
∠DEA = ∠BAE (Alternate angles)
⇒ ∠DEC = 35° (∵ ∠DEA = ∠DEC and ∠BAE = 35°)

In ∆DEC, we have : .
∠DCE + ∠DEC + ∠CDE = 180° (Angle-sum property)
⇒ ∠DCE + 35° + 53° = 180°
∠DCE = 180° – 35° – 53° = 92°
Hence, ∠DCE = 92°.

Question 4.
In the figure, if lines PQ and RS intersect at point r, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 4
Answer:
In ∆PRT, we have
∠PRT + ∠RTP + ∠TPR = 180°
(Angle-sum property)
⇒ 40° + ∠RTP + 95° = 180°
⇒ ∠RTP = 180° – 40° – 95° = 45°
Now, ∠STQ = ∠RTP (Vertically opp. angles)
∠STQ = 45° [∵∠RTP – 45° (proved)]

In ∆TQS, we have :
∠SQT + ∠STQ + ∠TSQ = 180° (Angle-sum property)
∠SQT + 45° + 75° = 180° [∵∠STQ = 45° (proved)]
⇒ ∠SQT = 180° – 45° – 75° = 60°
Hence, ∠SQT = 60°.

Question 5.
In the figure, if PQ ⊥ PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 5
Answer:
Using exterior angle property in ASRQ, we have :
∠QRT = ∠RQS + ∠QSR
⇒ 65° = 28° + ∠QSR (∵ ∠QRT = 65°, ∠RQS = 28°)
⇒ ∠QSR = 65° – 28° = 37°
Now, PQ ∥ SR and the transversal PS intersects them at P and S respectively.
∠PSR + ∠SPQ = 180° (Sum of the interior angles on the same side of the transversal is 180°.)
⇒ (∠PSQ + ∠QSR) + 90° = 180°
⇒ y + 37° + 90° = 180°
⇒ y = 180° – 90° – 37° = 53°

In the right triangle SPQ, we have :
∠PQS + ∠PSQ = 90°
⇒ x + 53° = 90°
⇒ x = 90° – 53° = 37°
Hence, x = 37° and y = 53°.

RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

Question 6.
In the figure, the side QR of APQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ~ ∠QPR.
RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 6
Answer:
In ∆PQR, we have :
ext. ∠PRS = ∠P + ∠Q
⇒ \(\frac{1}{2}\)ext. ∠PRS = \(\frac{1}{2}\)∠P + \(\frac{1}{2}\)∠Q
∠TRS = \(\frac{1}{2}\)∠P + ∠TQR ……………..(1)
(∵ QT and RT are bisectors of ∠Q and ∠PRS, respectively.
∴ ∠Q = 2∠TQR and ext. ∠PRS = 2 ∠TRS)

In ∆QRT, we have :
ext. ∠TRS = ∠TQR + ∠T ……………… (2)
From (1) and (2), we get:
\(\frac{1}{2}\)∠P + ∠TQR = ∠TQR + ∠T
⇒ \(\frac{1}{2}\)∠P = ∠T
⇒ ∠QTR = \(\frac{1}{2}\)∠QPR.
Hence Proved.

RBSE Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.2

April 16, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.2 1

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 3 Coordinate Geometry Exercise 3.2

Question 1.
Write the answer of each of the following questions :
(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
Answer:
These are named as the X-axis and Y-axis respectively.

(ii) What is the name of each part of the plane formed by these two lines?
Answer:
These are known as quadrant.

(iii) Write the name of the point where these two lines intersect.
Answer:
These lines intersect at the origin.

RBSE Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.2
Question 2.
See figure, and write the following :
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (- 3, – 5).
(iv) The point identified by the coordinates (2, – 4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii) The coordinates of the point L.
(viii)The coordinates of the point M.
RBSE Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.2 1
Answer:
From the figure :
(i) The coordinates of B are (- 5, 2).
(ii) The coordinates of C are (5, – 5).
(iii) The point identified by the coordinates (- 3, – 5) is E.
(iv) The point identified by the coordinates (2, – 4) is G.
(v) The abscissa of the point D is 6.
(vi) The ordinate of the point if is – 3.
(vii) The coordinates of the point L are (0, 5);
(viii) The coordinates of the point M are (- 3, 0).

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