CBSE Class 10 – Page 4 – MCQ Questions

Footprints without Feet Extra Questions and Answers Class 10 English Footprints Without Feet

February 10, 2023February 9, 2023 by Prasanna

Here we are providing Footprints without Feet Extra Questions and Answers Class 10 English Footprints Without Feet, Extra Questions for Class 10 English was designed by subject expert teachers.

Footprints without Feet Extra Questions and Answers Class 10 Footprints Without Feet

Footprints without Feet Extra Questions and Answers Short Answer Type

Question 1.
Why were the two boys in London surprised and fascinated?
Answer:
The two boys in London saw fresh muddy footprints appearing on the steps of a house but the barefooted man was not visible. So, they were surprised and fascinated.

Question 2.
What did Griffin do inside the store?
Answer:
Griffin broke open the boxes and wrappers and dressed himself in warm clothes. He ate and drank to his heart’s content. In the end, he laid down on the pile of quilts to sleep.

Question 3.
What did Griffin do in the shop of a theatrical company?
Answer:
He wore bandages round his forehead and put on dark glasses, a false nose, big bushy side whiskers and a large hat. He attacked the shopkeeper from behind and robbed him of all his money.

Question 4.
What was the “curious episode” that took place in the clergyman’s study?
Answer:
One morning, the clergyman and his wife were awakened by noise. They saw no person. They were shocked to find that their money had been stolen from the desk.

Question 5.
The landlord’s wife was convinced that Griffin was an “eccentric scientist”. What made her think of Griffin in these terms?
Answer:
Griffin remained indoors and did not mix up with others. His habits were strange and his appearance was fantastic. All these things of Griffin made the landlady think that he was an “eccentric scientist.

Question 6.
Griffin was rather a lawless person. Pick out instances which prove that he was lawless.
Answer:
He set fire to a house which didn’t belong to him. He tools away clothes without paying for them. He hit a shopkeeper and stole his money.

Question 7.
What did the scientist do when he became furious? Why were the people in the bar horrified? What happened to the constable?
Answer:
The scientist took off his bandages and spectacles and became headless. The people in the bar were shocked to see a headless man. Griffin hit the constable Jaffers hard and made him senseless.

Question 8.
How do you assess Griffin as a scientist?
Answer:
Griffin was a brilliant scientist. He invented some drugs and made himself as an invisible person. But he misused his scientific discovery and became a lawless person. Griffin was, thus not a true scientist.

Question 9.
Why did the landlord want to eject Griffin? Why did Griffin set the house on fire?
Answer:
The landlord wanted to eject Griffin from his house because his activities were whimsical. Griffin set the house on fire to take his revenge on the landlord.

Question 10.
Who was Mr Jaffers? What was his surprise?
Answer:
Mr JafFers was a police constable. His surprise was that he was ordered to arrest a lawless person who was totally invisible.

Question 11.
Why were they surprised when they opened the room?
Answer:
Both the clergyman and his wife were surprised because the room was empty. They searched everywhere but couldn’t find anyone. Yet the desk had been opened and the housekeeping money was missing.

Question 12.
What did the scientist do when Mrs Hall confronted him?
Answer:
When Mrs Hall confronted the scientist he threatened her and threw off his bandages, whiskers, spectacles, and even the nose in a minute. The people were horrified to see the headless man.

Question 13.
How did Griffin escape?
Answer:
Griffin began throwing off his bandages, whiskers, spectacles and even the nose. Then he became even more invisible by throwing off all the garments one after another. There followed a male and none knew where he was.

Question 14.
Do you think Griffin’s discovery was of any use to humanity?
Answer:
No, Griffin’s discovery, through remarkable, was of no use to anyone except himself. He began robbing money and other things and even assaulting people after he became invisible.

Footprints without Feet Extra Questions and Answers Long Answer Type

Question 1.
‘Brilliant scientist though he was, Griffin was a lawless person. Does this statement approve the apprehension that science in wrong hands is Devil’s Pandora Box?
Answer:
Griffin was a brilliant scientist. He had discovered a rare drug that could make a human body invisible. This made Griffin an arrogant lawless person. He broke the law more them once but never for a good reason. He could dedicate his discovery to his country but he didn’t do that. His lawlessness made the law helpless. All his actions prove that science in devil’s hand is disastrous.

Question 2.
‘Griffin could use his discovery for welfare of the people but misuses it to take revenge.’ Discuss with reference with the story Footprints without Feet.
OR
Do you think Griffin misused a scientific discovery or he took advantages of his discovery?
OR
Griffin misused his invention, instead of using it for the betterment of the society. Do you think moral values are important along with intellectual abilities? Discuss.
Answer:
Griffin was a scientist who had made a discovery due to which’he could be invisible. But he was rather a lawless scientist, who misused a scientific discovery for his petty, selfish gains. He set his landlord’s house on fire, when he asked him to vacate it. He entered the store invisible, stole clothes and food.

At another shop, he attacked the shopkeeper from behind and ran away after robbing his money. , At Iping, he stole the clergyman’s money from his desk and hit Mrs Hall by throwing a chair on her. He hurled blows on the police constable Mr Jaffers and knocked him down. In this way, Griffin misused his scientific discovery without making benefit to the mankind.

Question 3.
The modem world has created great scientific inventions. But very often these inventions are more misused than used for the benefit of man. Griffin had made wonderful discovery; but be used it for unlawful means. Write a paragraph on “Science is being more misused than used”.
Answer:
Science is being more misused than used. The present time is highly dependent on science and technology. In the 21st century, survival without gadgets and electronics is unimaginable. But there are various instances where science is being misused to create panic and terror in the society.

Scientific experiments on various deadly disease are let out to cause mass destruction. Technology is used more for warfare purpose, to kill and to terrorize. The resulting products of science have been valuable to human society and have greatly raised the quality of life. But they have also led to the growth of means to destroy human society.

Question 4.
“Scientific discoveries have made life easier but insecure.” Explain with reference to the story ‘Footprints without Feet’ written by H.G. Wells.
Answer:
There is no doubt that scientific discoveries have made life easier. Take any sphere of human activity, electricity, telecommunications, electronics, computer, etc, these have totally changed man’s life. We can interact and walk on moon. Trip to Mars is next on our agenda. We can fly like birds. Supersonic planes fly at more than the speed of sound. Wonderful drugs can overcome deadly diseases.lSurgery can replace defective limbs.

Despite all these discoveries, human life is becoming insecure. Weapons of mass destruction are a constant danger. The fear of chemical, or biological warfare looms large. Diseases like AIDS and SARS can destroy human race.

Question 5.
“A lawless scientist is a curse to the society.” Discuss.
Answer:
It is true, that a lawless scientist is a curse to the society. A true scientist uses his discoveries for the good of the society and welfare of the people. It helps the nations to progress. A lawless scientist increases the miseries and sufferings of society. In this story, we find how Griffin misuses his discovery.

He causes pain and suffering to others. He causes destruction. Suppose the nuclear weapons fall in the hands of a lawless scientist, he can cause widespread destruction. He may ruin the achievements of other scientists. That’s why the leaders of the world are worried about chemical weapons falling into the hands of lawless heads of governments or terrorists. If this happens, all the scientists of the world will be doomed.

Question 6.
How did the invisible man first become visible?
Answer:
Griffin was completely invisible until he happened to step in some mud, which caused him to leave footprints as he walked. Two boys noticed these footprints and followed him as long as the prints were visible. Griffin, after getting rid of them, went into a big London store to put on some warm clothes. After the store was shut, he got himself dressed in an overcoat and a wide-brimmed hat. He also put on a pair of shoes. This made him visible to the people who came to work at the store the next day.

Question 7.
Why was he wandering the streets?
Answer:
Griffin was a lawless person. His landlord disliked him and tried to get rid of him. In revenge, Griffin set fire to his house. He had to remove his clothes to get away without being noticed. This was why he had become a homeless wanderer—without clothes and money.

Question 8.
Why does Mrs Hall find the scientist eccentric?
Answer:
The arrival of a stranger at an inn in winter was in itself a strange occurrence. In addition, the stranger had an uncommon appearance. Inspite of Mrs Hall’s repeated attempts to be friendly, he would respond in a closed manner. He told her that he had no desire to talk and wanted to live in solitude. He did not wish to be disturbed in his work. For all these strange reasons, she regarded him as an eccentric scientist.

Question 9.
What curious episode occurs in the study?
Answer:
A clergyman and his wife were awakened by some strange noises in the study very early in the morning. Creeping downstairs, they heard the chink of money being taken from the clergyman’s desk. He and his wife looked under the desk and behind the curtains, and even up the chimney. There wasn’t a sign of anybody. Yet the desk had been opened and the housekeeping money was missing.

Question 10.
What other extraordinary things happened at the Inn?
Answer:
The landlord and his wife were surprised to see the door of the scientist’s room open. Usually it is shut and locked and Griffin becomes furious if anyone enters his room. So they did not want to miss this opportunity. They peeped round the door, and finding nobody inside, decided to enter and investigate. The bedclothes were cold, showing that the scientist must have been up for some time now. All of a sudden Mrs Hall heard a sniff close to her ear. A moment later that hat on the bed suddenly leapt up and dashed itself onto her face.

Then the bedroom chair became alive. Springing into the air, it charged straight at her legs. As she and her husband turned away in terror, the extraordinary chair pushed them both out of the room and then appeared to slam and lock the door after them. Mrs Hall almost fell down the stairs in hysterics. She was, by now, convinced that the room was haunted by spirits and the stranger had somehow caused these to enter into the furniture.

Question 11.
“Griffin was rather a lawless person.” Comment.
Answer:
It is rightly said that Griffin was a lawless person. A person who follows law and order of the country can not think of damaging other’s property. But Griffin set fire to the house and became a homeless wanderer. He also robbed a shopkeeper of all the money he could find. He also stole the housekeeping money from the clergyman’s desk. Griffin also attacked Mrs Hall when she entered his room. All these activities and instances prove that he was not a law abiding citizen.

Question 12.
How would you access Griffin as a scientist?
Answer:
After repeated experiments, Griffin had discovered how to make the human body invisible. This was a big achievement. However, he misused his discovery for personal gains and for hurting others. Thus, though he can be termed as a brilliant scientist, he was not a noble one.

Question 13.
Would you like to become invisible? What advantages and disadvantages do you foresee, if you did?
Answer:
Yes, I would like to become invisible. It would be fun. I would be able to see everyone without being seen. I would be able to enjoy all activities. I may go to movie hall without paying for it. I would also help the police in nabbing the criminals. But being invisible would be a disadvantage also. I will be able to see all but nobody would see me. I will not get any recognition for my work.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

February 1, 2022February 1, 2022 by Dattu

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-5-ex-5-3/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 5
Chapter Name	Arithmetic Progressions
Exercise	Ex 5.3
Number of Questions Solved	20
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Ex 5.3 Class 10 Maths Question 1.
Find the sum of the following APs:
(i) 2, 7, 12,…… to 10 terms.
(ii) -37, -33, -29, …… to 12 terms.
(iii) 0.6, 1.7, 2.8, ……, to 100 terms.
(iv) \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), …….., to 11 terms.
Solution:
Ex 5.3 Class 10 Maths
Exercise 5.3 Class 10 Maths

Exercise 5.3 Class 10 Maths Question 2.
Find the sums given below:
(i) 7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84
(ii) 34 + 32 + 30 + … + 10
(iii) -5 + (-8) + (-11) + ….. + (-230)
Solution:

5.3 Class 10 Maths Question 3.
In an AP:
(i) given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) given a₃ = -15, S₁₀ = 125, find d and a₁₀.
(v) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) given a_n = 4, d = 2, S_n = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
Ex5.3 Class 10 Maths
Class 10 Ex 5.3 Maths
Class 10 Maths Chapter 5 Exercise 5.3
Ex 5.3 Maths Class 10

Ex5.3 Class 10 Maths Question 4.
How many terms of AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Exercise 5.3 Maths Class 10

Class 10 Ex 5.3 Maths Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Class 10 Maths Ch 5 Ex 5.3

Class 10 Maths Chapter 5 Exercise 5.3 Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 10

Ex 5.3 Maths Class 10 Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:

5.3 Maths Class 10 Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 12

Exercise 5.3 Maths Class 10 Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Ex 5.3 Class 10 Maths Solutions
Class 10 Maths Ch 5 Ex 5.3

Class 10 Maths Ex 5.3 Question 10.
Show that a₁, a₂, ……. a_n,…… form an AP where an is defined as below:
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
Ncert Solutions For Class 10 Maths Chapter 5 Exercise 5.3

Ncert Solutions For Class 10 Maths Chapter 5 Exercise 5.3 Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:

Class 10th Exercise 5.3 Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:

Exercise 5.3 Class 10 Solutions Question 13.
Find the sum of the first 15 multiples of 8.
Solution:

Exercise 5.3 Class 10th Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Let odd numbers between 0 and 50 be 1, 3, 5, 7,…….., 49.
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 19

Class 10 Maths Chapter 5 Exercise 5.3 Solution Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:
₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
Class 10 Maths Chapter 5 Exercise 5.3 Solution

Class 10 Maths Ch 5 Ex 5.3 Chapter 5 Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let 1st prize be of ₹ a
2nd prize be ₹ (a – 20) and
3rd prize be ₹ (a – 20 – 20) = ₹ (a – 40)
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 21

Class 10 Chapter 5 Exercise 5.3 Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, eg. a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Let the trees be planted 1, 2, 3, 4, 5 , …… 12
Here, a = 1, d = 1, n = 12
Total number of trees planted by each section
S₁₂ = \(\frac { 12 }{ 2 }\) [2a + (n – 1) d] = 6 [2 x 1 + (12 – 1) x 1]
= 6 [2 + 11] = 6 x 13 = 78
Total number of trees planted by 3 sections = 78 x 3 = 234

Class 10 Maths Chapter 5 Exercise 5.3 Solution Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?
(Take π = \(\frac { 22 }{ 7 }\))
[Hint: Length of successive semicircles is l₁, l₂, l₃, l₄, … with centres at A, B, respectively.]
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 22
Solution:

5.3 Exercise Class 10 Question 19.
200 logs are stacked in the following manner 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 24
Solution:

Class 10th Exercise 5.3 Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 x 5 + 2 x (5 + 3)]
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 26

Solution:
Distance between the first potato and the bucket = 5 m
Distance between next 2 potatoes = 3 m each
So, series is 5 m, 8 m, 11 m,
Here, a = 5 m, d = (8 – 5) m = 3 m
Total distance travelled for 10 potatoes = 2 [5 + 8 + 11 + …….. + 10 terms]
= 2[\(\frac { 10 }{ 2 }\){2 x 5 + (10 – 1) 3}]
= 2[5{10 + 27}] = 2(37 x 5) = 37 x 10 = 370 m.

We hope the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

February 1, 2022February 1, 2022 by Dattu

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-4-ex-4-4/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 4
Chapter Name	Quadratic Equations
Exercise	Ex 4.4
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

Ex 4.4 Class 10 Maths Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² -3x + 5 = 0
(ii) 3x² – 4√3x + 4 = 0
(iii) 2x²-6x + 3 = 0
Solution:
(i) 2x² – 3x + 5 = 0
This is of the form ax² + bx + c = 0,
where a = 2, b = -3 and c = 5
Discriminant, D = b²-4ac
= (-3) ²-4 x 2 x 5 = 9 – 40 = -31
Since, D < 0
Hence, no real roots exist.

(ii) 3x² – 4√3x + 4 = 0
Ex 4.4 Class 10 Maths
(iii) 2x²-6x + 3 = 0

4.4 Class 10 Maths Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(1) 2x² + kx + 3 = 0
(2) kx (x – 2) + 6 = 0
Solution:
Exercise 4.4 Class 10 Maths

Exercise 4.4 Class 10 Maths Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Solution:
Let breadth of the rectangular mango grove be x m
Then, the length of rectangular mango grove be 2xm
According to question,
x x 2x = 800
⇒ 2x² = 800
⇒ x² = 400
⇒ x = ±20 [-20 is rejected]
Hence, breadth = 20 m and length = 2 x 20 = 40 m
So, it is possible to design a rectangular mango grove whose length is twice its breadth.

Ex4.4 Class 10 Maths Question 4.
Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the present age of one friend be x years
Then, the present age of other friend be (20 -x) years
4 years ago, one friend’s age was (x – 4) years
4 years ago, other friend’s age was (20 -x – 4) = (16 -x) years
According to question,
(x-4) (16-x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
This is of the form ax² + bx + c = 0, where, a = 1, b = -20 and c = 112
Discriminant, D = b² – 4ac
= (-20)² – 4 x 1 x 112 = 400 – 448 = – 48 < 0
Since, no real roots exist.
So, the given situation is not possible.

Class 10 Maths Chapter 4 Exercise 4.4 Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Solution:
Let the length of rectangular park be x m and breadth be y m.
Given: area = 400 m²

We hope the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

February 1, 2022February 1, 2022 by Dattu

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-3-ex-3-7/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Pair of Linear Equations in Two Variables
Exercise	Ex 3.7
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Ex 3.7 Class 10 Maths Question 1.
The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the ages of Ani and Biju be x years and y years respectively.
If Ani is older than Biju
x – y =3
If Biju is older than Ani
y – x = 3
-x + y =3 [Given]
Ex 3.7 Class 10 Maths
Subtracting equation (i) from equation (ii), we get:
3x – 57
⇒ x = 19
Putting x = 19 in equation (i), we get
19-y = 3
⇒ y = 16
Again subtracting equation (iv) from equation (iii), we get
3x = 63
⇒ x = 21
Putting x = 21 in equation (iii) we get
21 -y= -3
⇒ y = 24
Hence, Ani’s age is either 19 years or 21 years and Biju’s age is either 16 years or 24 years.

Exercise 3.7 Class 10 Maths Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?
Solution:
Let the two friends have ₹ x and ₹ y.
According to the first condition:
One friend has an amount = ₹(x + 100)
Other has an amount = ₹ (y – 100
∴ (x + 100) =2 (y – 100)
⇒ x + 100 = 2y – 200
⇒ x – 2y = -300       …(i)
According to the second condition:
One friend has an amount = ₹(x – 10)
Other friend has an amount =₹ (y + 10)
∴ 6(x – 10) = y + 10
⇒ 6x – 60 = y + 10
⇒ 6x-y = 70                                        …(ii)
Multiplying (ii) equation by 2 and subtracting the result from equation (i), we get:
x – 12x = – 300 – 140
⇒ -11x = -440
⇒ x = 40
Substituting x = 40 in equation (ii), we get
6 x 40 – y = 70
⇒ -y = 70- 24
⇒ y   = 170
Thus, the two friends have ₹ 40 and ₹ 170.

3.7 Class 10 Maths Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the original speed of the train be x km/h
and the time taken to complete the journey be y hours. ‘
Then the distance covered = xy km

Case I: When speed = (x + 10) km/h and time taken = (y – 2) h
Distance = (x + 10) (y – 2) km
⇒ xy = (x + 10) (y – 2)
⇒ 10y – 2x = 20
⇒ 5y – x = 10
⇒ -x + 5y = 10 …(i)

Case II: When speed = (x – 10) km/h and time taken = (y + 3) h
Distance = (x – 10) (y + 3) km
⇒ xy = (x – 10) (y + 3)
⇒ 3x- 10y = 30 …(ii)
Multiplying equation (i) by 3 and adding the result to equation (ii), we get
15y – 10y = 30 f 30
⇒ 5y = 60
⇒ y   = 12
Putting y = 12 in equation (ii), we get
3x- 10 x 12= 30
⇒ 3x   = 150
⇒ x   = 50
∴ x = 50 and y = 12
Thus, original speed of train is 50 km/h and time taken by it is 12 h.
Distance covered by train = Speed x Time
= 50 x 12 = 600 km.

Ex 3.7 Maths Chapter 3 Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of rows be x and the number of students in each row be y.
Then the total number of students = xy
Case I: When there are 3 more students in each row
Then the number of students in a row = (y + 3)
and the number of rows = (x – 1)
Total number of students = (x – 1) (y + 3)
∴ (x – 1) (y + 3) = xy
⇒ 3x -y =3 …(i)
Case II: When 3 students are removed from each row
Then the number of students in each row = (y-3)
and the number of rows = (x + 2)
Total number of students = (x + 2) (y – 3)
∴ (x + 2) (y – 3) = xy
⇒ -3x + 2y = 6 …(ii)
Adding the equations (i) and (ii), we get
-y + 2y = 3 + 6
⇒ y = 9
Putting y = 9 in the equation (ii), we get
-3x + 18 = 6
⇒ x = 4
∴ x = 4 and y = 9
Hence, the total number of students in the class is 9 x 4 = 36.

Ex3.7 Class 10 Chapter 3 Maths Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Let ∠A = x° and ∠B = y°.
Then ∠C = 3∠B = (3y)°.
Now ∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180°
⇒ x + 4y = 180° …(i)
Also, ∠C = 2(∠A + ∠B)
⇒ 3y – 2(x + y)
⇒ 2x – y = 0° …(ii)
Multiplying (ii) by 4 and adding the result to equation (i), we get:
9x = 180°
⇒ x = 20°
Putting x = 20 in equation (i), we get:
20 + 4y = 180°
⇒ 4y = 160°
⇒ y = \(\frac { 160 }{ 40 }\) = 40°
∴ ∠A = 20°, ∠B = 40° and ∠C = 3 x 40° = 120°.

Class 10 Ex 3.7 Chapter 3 Maths Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
5x – y = 5 …(i)
3x-y = 3 …(ii)
For graphical representation:
From equation (i), we get: y = 5x – 5
When x = 0, then y -5
When x = 2, then y = 10 – 5 = 5
When x = 1, then y = 5 – 5 = 10
Thus, we have the following table of solutions:
Exercise 3.7 Class 10 Maths
From equation (ii), we get:
⇒ y = 3x – 3
When x = 0, then y = -3
When x = 2, then y = 6 – 3 = 3
When x = 1, then y = 3 – 3 = 0
Thus, we have the following table of solutions:

Plotting the points of each table of solutions, we obtain the graphs of two lines intersecting each other at a point C(1, 0).
Ex 3.7 Maths Chapter 3
The vertices of ΔABC formed by these lines and the y-axis are A(0, -5), B(0, -3) and C(1, 0).

Exercise 3.7 Class 10 Maths Chapter 3 Question 7.
Solve the following pairs of linear equations:
Ex 3.7 Maths Chapter 3
Solution:
(i) The given equations are
px + qy = p – q …(1)
qx – py = p + q …(2)
Multiplying equation (1) byp and equation (2) by q and then adding the results, we get:
x(p² + q²) = p(p – q) + q(p + q)
Ex3.7 Class 10 Chapter 3 Maths
Class 10 Ex 3.7 Chapter 3 Maths

(ii) The given equations are
ax + by = c …(1)
bx – ay = 1 + c …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
abx + b²y = cb …(3)
abx + a²y = a(1+ c) …(4)
Subtracting (3) from (4), we get:
Exercise 3.7 Class 10 Maths
Exercise 3.7 Class 10 Maths Chapter 3

(iii) The given equations may be written as: bx – ay = 0 …(1)
ax + by = a² + b² …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
b²x + aby = 0 ….(3)
a²x + aby = a(a² + b²) …..(4)
Adding equation (3) and equation (4), we get:
(a² + b²)x = a (a² + b²) a(a² + b²)
Class 10 Maths Ex 3.7

(iv) The given equations may be written as:
(a – b)x + (a + b)y = a² – 2ab – b² …(1)
(a + b)x + (a + b)y = a² + b² …(2)
Subtracting equation (2) from equation (1), we get:
(a – b)x – (a + b)x
= (a² – 2ab – b²) – (a² + b²)
⇒ x(a – b- a-b) = a² – 2ab – b² – a² – b²⇒ -2bx = -2ab – 2b²⇒ 2bx = 2b² + 2ab
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 11
Class 10 Maths Chapter 3 Exercise 3.7

(v) The given equations may be written as:
76x – 189y = -37 …(1)
-189x + 76y = -302 …(2)
Multiplying equation (1) by 76 and equation (2) by 189, we get:
5776x – 14364y = -2812 …(3)
-35721x + 14364y = -57078 …(4)
Adding equations (3) and (4), we get:
5776x – 35721x = -2812 – 57078
⇒ – 29945x = -59890
⇒ x = 2
Putting x = 2 in equation (1), we get:
76 x 2 – 189y = -37
⇒ 152 – 189y = -37
⇒ -189y = -189
⇒ y = 1
Thus, x = 2 and y = 1 is the required solution.

Class 10 Maths Ex 3.7 Pair of Linear Equations in Two Variables Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
Class 10 3.7
Solution:
Exercise 3.7 Class 10th
ncert solutions for class 10 maths chapter 3 exercise 3.7

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

February 1, 2022February 1, 2022 by Dattu

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-3-ex-3-3/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Pair of Linear Equations in Two Variables
Exercise	Ex 3.3
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Ex 3.3 Class 10 Maths Chapter 3 Question 1.
Solve the following pair of linear equations by the substitution method,
(i) x + y = 14, x – y = 4
(ii) s – t = 3, s/3 + t/2 = 6
(iii) 3x – y = 3, 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3

Solution:
From equation (i),
x + y – 14 ⇒ y = 14x
Putting the value ofy in equation (ii), we get
x – (14 – x) = 4 ⇒ x – 14 + x = 4 ⇒ 2x = 4 + 14
2x = 18 ⇒ x = 9
Now, puttingx = 9 in equation (i), we have
9 + y = 14 ⇒ y = 14 – 9 ⇒ y = 5
so, x = 9, y = 5
Exercise 3.3 Class 10 Maths
∴ y can have infinite real values
∴ x can have infinite real values because x = \(\frac { y+3 }{ 3 }\)

Ex3.3 Class 10 Maths NCERT Solutions

Exercise 3.3 Class 10 Maths Question 2.
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
Equations are 2x + 3y = 11
and 2x – 4y = -24
From equation (i)
2x = 11 – 3y
Putting this value in equation (ii), we get
11 – 3y – 4y = -24 ⇒ 11 – 7y = -24 ⇒ – 7y = – 35
y = \(\frac { 35 }{ 7 }\) ⇒ y = 5
Putting y = 5 in equation (i). we have
2x + 3 x 5 = 11 ⇒ 2x + 15 = 11 ⇒ 2x = 11 – 15 ⇒ 2x = -4 ⇒ x = -2
Now. putting the value of x andy in equation
y = mx + 3 ⇒ 5 = -2m + 3 ⇒ 2 = -2m ⇒ m = -1

3.3 Class 10 Maths Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball,
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km₹ How much does a person have to pay for travelling a distance of 25 km₹
(v) A fraction becomes 9/2, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes 5/6, Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let 1st number be x and 2nd number be y.
Let x >y
1st condition :
x – y = 26
2nd condition :
x = 3y
Putting x = 3y in equation (i)
3y – y = 26 ⇒ 2y = 26 ⇒ y = 13
From (ii)
x = 3 x 13 = 39
∴ One number is 13 and the other number is 39.

(ii) Let one angle be x and its supplementary angle = y
Let x > y
1st Condition :
x + y = 180°
2nd Condition :
x – y = 18° ⇒ X = 18° + y
From equation (ii), putting the value ofx in equation (i),
18° + y + y = 180° ⇒ 18° + 2y = 180°
2y = 162° ⇒ y = 81°
From (ii) x = 18° + 81° = 99° ⇒ x = 99°
∴ One angle is 81° and another angle is 99°.

(iii) Let cost of 1 bat = ₹x and cost of 1 ball = ₹y
1st Condition:
7x + 6y = 3800
2nd Condition:
3x + 5y = 1750
From equation (ii), we get
3.3 Maths Class 10 Pair of Linear Equations in Two Variables
putting x = 1750-5y/3 in equation (i), we get
Cost of one bat = ₹ 500 and cost of one ball = ₹ 50.

(iv) Let fixed charges be ₹.v and charge for per km be ₹y.
A.T.Q.
1st Condition :
x + lOy = 105
2nd Condition :
x + 15y = 155
From equation (i), we get
x= 105 – 10y
Putting this value in equation (ii), we have
105 – 10y + 15y = 155 ⇒ 105 + 5y = 155
⇒ 5y = 155 – 105 ⇒ 5y = 50 ⇒ y = 10
Now, puttingy = 10 in equation (i), we have
x + 10(10) = 105 ⇒ x + 100 = 105 ⇒ x = 5
Fixed charges is ₹ 5 and charges per km is ₹ 10.

3rd Condition :
For distance of 25 km
x + 25y = 5 + 25(10) = 5 + 250 = 255
Amount paid for travelling 25 km is ₹ 255.

(v) Let numerator be x and denominator be y.
∴ Fraction is x/y
A.T.Q.
1st condition :

2nd condition :

(vi) Let present age of Jacob be x years and that of his son bey years.
A.T.Q.
1st Condition :
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15 ⇒ x – 3y = 15 – 5 ⇒ x – 3y = 10
2nd Condition:
x – 5 = 7(y – 5) ⇒ x – 5 = 7y – 35 ⇒ x = 7y – 35 + 5
⇒ x = 7y – 30
Putting the value of ‘x’ in equation (i), we get
7y – 30 – 3y = 10
4y – 30 = 10
4y = 40 y = 10 ⇒ y = 10
putting the value of y in equation(ii), we get
x = 7(10) – 30 = 70 – 30 ⇒ x = 40
Hence, the present age of Jacob is 40 years and that of his son is 10 years.

We hope the NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3, drop a comment below and we will get back to you at the earliest.