Online Education NCERT Solutions for Class 12 English Flamingo Chapter 4 The Rattrap

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The Rattrap NCERT Text Book Questions and Answers

The Rattrap Think as you read 

Question 1.
From where did the peddler get the idea of the world being a rattrap?
Answer:
The poor rattrap peddler struggled for his existence. He could not make both ends meet so he had to resort to both begging and petty thievery. Despite this, his condition was deplorable. Thinking of his miserable plight and his job, as he plodded on, he was struck by the idea that the whole world was nothing but a big rattrap. It baited people in the form of riches and joys, shelter and food, heat and clothing, just as the rattrap offered cheese and pork. Similarly, the moment one was tempted by the material joys, the world closed in on him and trapped him.

Question 2.
Why was he amused by this idea?
Answer:
The world had never been kind to him, so it gave him great joy to think ill of it. It became a treasured pastime for him to think of people he knew who had let themselves be caught in the dangerous trap, and of others who were still moving around the bait, ready to be trapped.

Question 3.
Did the peddler expect the kind of hospitality that he received from the crofter?
Answer:
When the peddler reached the little grey cottage by the roadside, he knocked on the door to ask for shelter for the night. He was surprised because, instead of the sour faces which ordinarily met him, this time he was welcomed by the owner, an old man.

Question 4.
Why was the crofter so talkative and friendly with the peddler?
Answer:
The crofter was an old man without wife or child. He was happy to get someone to talk to, in his loneliness. So, he welcomed the peddler. The old man was just as generous with his confidences as with his porridge and tobacco.

Question 5.
Why did he show the thirty kronor to the peddler?
Answer:
The old man was confiding and trusting. He told the peddler that he was no longer able to work as a day labour, and was supported by what he earned by selling his cow’s milk. The old man showed him the thirty kronor that he had received as payment from the creamery, to convince the peddler.

Question 6.
Did the peddler respect the confidence reposed in him by the crofter?
Answer:
No, the peddler did not respect the confidence reposed in him by the crofter because later, he went up to the window, smashed a pane and got hold of the pouch with the thirty kronor. He took the money and thrust it into his own pocket and went away.

Question 7.
What made the peddler think that he had indeed fallen into a rattrap?
Answer:
At first, the peddler was pleased with the money in his pocket; but he later got lost in the forest. All at once, he recalled his thoughts about the world being a rattrap. He felt that his end was near and that he had been fooled by a bait of thirty kronor.

Question 8.
Why did the ironmaster speak kindly to the peddler and invite him home?
Answer:
The owner of Ramsjo iron mill came into the forge on one of his night rounds of inspection and saw the peddler. The ironmaster walked close up to him, looked him over very carefully and mistook him for his friend, Nils Olof, and invited him home.

Question 9.
Why did the peddler decline the invitation?
Answer:
The peddler did not reveal his identity because he thought that the ironmaster might give him a couple of kronor. But, to go along up to the manor house and be received as his friend scared the peddler. He had stolen the thirty kronor from the old man and he felt that going up to the manor house would make him vulnerable. He merely wanted to sleep in the forge and then quietly sneak away.

Question 10.
What made the peddler accept Edla Willmansson’s invitation?
Answer:
When Edla Willmansson invited him, she looked at him compassionately. She also assured him that he would be allowed to leave just as freely as he came. She requested him to stay with them over Christmas Eve. She said this in such a friendly manner that the rattrap peddler felt reassured. Hence, he accepted Edla Willmansson’s invitation.

Question 11.
What doubts did Edla have about the peddler?
Answer:
When Edla Willmansson looked compassionately at the peddler, she noticed that the man was afraid. She at once knew that he had either stolen something or else had escaped from jail.

Question 12.
When did the ironmaster realise his mistake?
Answer:
When the ironmaster saw the stranger the day after, clean and well dressed, he realized his mistake. The valet had bathed him, cut his hair and shaved him. He was wearing clothes that belonged to the ironmaster. The ironmaster frowned as he realized that in the dimly-lit furnace, he had made a mistake. As the stranger stood there in daylight, it was impossible to mistake him for an old acquaintance.

Question 13.
What did the peddler say in his defence when it was clear that he was not the person the ironmaster had thought he was?
Answer:
When the ironmaster realized his mistake, the stranger made no attempt to hide or pretend. He said that it was not his fault as he had never pretended to be anything but a poor trader. He had pleaded and begged to be allowed to stay in the forge. He offered to put on his rags and go away.

Question 14.
Why did Edla still entertain the peddler even after she knew the truth about him?
Answer:
Edla entertained the peddler even after she knew the truth about him. She said that the whole year long the peddler roamed about without anyone welcoming him. He was forever afraid of being arrested and cross-examined. So, she felt compassion for the homeless tramp and wanted him to enjoy a day of peace with them. She also said that it was their mistake and they ought not to chase away a human being whom they had invited home and had promised Christmas cheer.

Question 15.
Why was Edla happy to see the gift left by the peddler?
Answer:
Edla gave a little cry of joy on seeing the gift, a small rattrap with three wrinkled ten kronor notes. Her deed of kindness had helped the poor peddler atone for his sin. He did not want her to be embarrassed in the Christmas season because of a thief. He said that the rattrap was a Christmas present from a rat who would have been caught in the world’s rattrap, had he not received kindness and honour from her. Ella felt satisfied at having helped someone rise above his faults.

Question 16.
Why did the peddler sign himself as Captain von Stahle?
Answer:
The peddler signed himself as Captain von Stahle because Edla had treated him as if he were a captain. He took away from the incident, the transient honour of having being treated like a captain. It inspired him to do good, and let go of his thieving ways.

The Rattrap Understanding the text

Question 1.
How did the peddler interpret the acts of kindness and hospitality shown by the crofter, the ironmaster, and his daughter?
Answer:
The peddler was granted shelter in a little cottage by the roadside. The crofter gave him food and trusted him by investing confidence in him. However, the peddler betrayed his trust and stole his money. Later, as he was lost in the forest, he felt trapped, and judged the kindness and money that came his way as nothing but a bait to trap him. He was also suspicious of the kindness showed to him by the ironmaster and his daughter. The peddler was convinced that any kindness shown to him was just a bait to trap him.

Question 2.
What are the instances in the story that show that the character of the ironmaster is different from that of his daughter in many ways?
Answer:
Despite a close relationship, the character of the ironmaster is shown to be different from that of his daughter. The ironmaster spoke to the peddler for some time but continued to mistake him for his friend Captain von Stahle, whereas, his observant daughter, who had never met his friend, within minutes, observed that either the peddler had stolen something or else he had escaped from jail.

When the ironmaster discovered that the peddler was not his friend, he wanted to hand him over to the sheriff but his daughter was more compassionate. She said that the peddler roamed without being welcomed anywhere. So, she wanted him to enjoy a day of peace with them. She also said that it was their mistake and they ought not to chase away a human being whom they had asked to come to their home and whom they had promised Christmas cheSr.

At church, they learnt that the rattrap peddler had robbed the crofter. The ironmaster wondered what all he had stolen from their cupboard and blamed the daughter for letting him in. However, she was repaid by the peddler for her trust that helped in reforming his character.

Question 3.
The story has many instances of unexpected reactions from the characters to others’ behaviour. Pick out instances of these surprises.
Answer:
The story does have many instances of unexpected reactions of the characters in response to others’ behaviour. The first was the crofter’s reaction to the peddler. Instead of the sour faces which ordinarily met him, the old man, without wife or child, was happy to get someone to talk to in his loneliness. He was very welcoming. But, the peddler, despite the hospitality, stole his money.

The second was the ironmaster, who mistook the peddler to be his friend Nils Olof, and tried to take him to his house. But, he was quick to turn his back when he realized his mistake.The third was the arrival of the ironmaster’s daughter, who realized that there was something amiss about the peddler but took him in. Even when the father and daughter found out the truth, the daughter stood by him. She wanted him to enjoy a day of peace with them.The peddler, in turn, surprised everyone when he returned his stolen booty. He honoured the trust reposed in him by Edla.

Question 4.
What made the peddler finally change his ways?
Answer:
Due to the trust and compassion shown by the ironmaster’s daughter Edla Willmansson, the peddler became a changed man. This was evident from the package he left for her before leaving. The package contained a small rattrap with three wrinkled ten kronor notes that the peddler left for her. In the rattrap, also lay a letter that revealed how he did not want her to be embarrassed in the Christmas season by a thief. He wanted her to return the money to the old man on the roadside. The rattrap was a Christmas present from him, who had been reformed and touched by the treatment he received from her.

Question 5.
How does the metaphor of the rattrap serve to highlight the human predicament?
Answer:
The peddler considered that the whole world was nothing but a big rattrap. Its only purpose was to set baits for people. It offered riches and joys, shelter and food, heat and clothing, exactly as the rattrap offered cheese and pork, and as soon as one let oneself be lured by the bait, it closed in. The peddler also told the ironmaster that the whole world was nothing but a big rattrap. All the good things that were offered were nothing but cheese rind and bits of pork, set out to drag people into trouble.

None escaped; one person falls into the trap one day and the other the next day. The only thing that could turn a person from rat-like ways was human kindness, something that he had received from Edla. Thus, in the end, the peddler left the rattrap as a Christmas present and called himself a rat who had been spared.

The metaphor of the rattrap highlights the theory of crime and punishment. Crimes, such as theft or giving in to temptation, are compared with a bait and the subsequent imprisonment in a trap. However, the human predicament forms the basis of the story where the story of the rattrap serves to bring out a lesson in moral values.

Question 6.
The peddler comes out as a person with a subtle sense of humour. How does this serve in lightening the seriousness of the theme of the story and also endearing him to us?
Answer:
The peddler has a good sense of humour and this diffuses the seriousness of the theme of the story, he also endears himself to the reader. Although he was a vagabond in pursuit of food and shelter, he found time to meditate on his philosophies and draw pleasure from the anticipated miseries of other people. He imagined people he knew to have landed in the trap and of others who were still circling around the bait.

The humour is brought out when at every occasion he lapsed into his theory of the rattrap. When his identity was discovered by the ironmaster, he said that a day would arrive when the ironmaster too might chase a big piece of pork, and get caught in the trap. He also signs off in the letter as “Captain von Stahle”, revealing how he held on to his sense of humour till the end.

The Rattrap Talking about the text

Discuss the following in groups of four. Each group can deal with one topic and present the views of the group to the whole class.

Question 1.
The reader’s sympathy is with the peddler right from the beginning of the story. Is the sympathy justified?
Answer:
The reader’s sympathy is with the peddler right from the beginning of the story for various reasons. It is justified.He is a victim of circumstances. He made rattraps with the material he got by begging in stores or at big farms. But even so, the business was not profitable. He had to resort to both begging and petty thievery for a living. His clothes were in rags, his cheeks were sunken, and hunger gleamed in his eyes. When he knocked on a door to ask for shelter at night, he was often refused and he “met sour faces”. Even when he stole the money from the crofter, he at once realized that his philosophy about the world as a rattrap had turned true for him. He had been caught with the bait of money. He regretted it and knew his own turn had come.
OR
It is not justified.
There were many who had lost their jobs. He stole money from the crofter who was hospitable to him. Secondly, he had the opportunity to tell the ironmaster of his true identity which he did not. He knew all along that if he did something wrong, he would be trapped, yet, he could not resist the bait.

Question 2.
The story also focuses on human loneliness and the need to bond with others. Discuss.
Answer:
The above statement is true for many characters in the chapters:
(a) The crofter, because of loneliness, lets the peddler in.

(b) The ironmaster at once bonded with the peddler without confirming his identity—“But of course it is you, Nils Olof!” he said. “How you do look!” and again “.. .We were just saying that it was too bad we didn’t have any company for Christmas.”

(c) Edla Willmansson sensed that all was not well, yet insisted that the peddler spend Christmas with them. When the ironmaster wanted to turn the peddler out, the daughter said, “I think he ought to stay with us today. I don’t want him to go.”

(d) The peddler, while leaving the ironmaster’s house left a message, “Written with friendship and high regard”.
The peddler was reformed by the kindness shown by Edla Willmansson. He wanted to be nice to her in return and did not want to embarrass her by his theft. He returned the stolen money, asking her to forward it to the old man. Human warmth and bonding enabled this change in him.

Question 3.
Have you known/heard of an episode where a good deed or an act of kindness has changed a person’s view of the world?
Answer:
The Bishop’s Candlesticks is one such episode that takes up on a similar theme.

The Rattrap Working with words

Question 1.
The man selling rattraps is referred to by many terms such as “peddler”, “stranger”, etc. Pick out all such references to him. What does each of these labels indicate about the context or the attitude of the people around him?
Answer:
He is referred to as a vagabond, intruder, tramp, ragamuffin and poor hungry wretch. These labels indicate the context or the attitude of the people around him. The people

  • Had no respect for him.
  • Felt he was a burden.
  • Did not care to know him or his problems.
  • Could pity him, but were not really compassionate.

Question 2.
You came across the words plod, trudge, stagger in the story. These words indicate movement accompanied by weariness. Write down five other words with a similar meaning.
Answer:
drag oneself, footslog, lumber, plod along, slog, stumble, traipse

The Rattrap Extra Questions and Answers

The Rattrap Short Answer Questions

Question 1.
What did the rattrap peddler do for a living?
Answer:
The rattrap peddler was an iron worker who lost his job to machines in an age of industrialization. Hence, he resorted to selling small rattraps of wire, and he obtained the required materials by begging. His business was not a profitable one, so he had no option but to beg and steal. His clothes were in rags, and his cheeks were sunken; he looked starved. He roamed alone like a vagabond.

Question 2.
How did the peddler get the idea of the world being a rattrap?
Answer:
The peddler struggled for his existence as selling rattraps was not a profitable business. He often resorted to begging and stealing. This led him to believe that the world was a big rattrap that sets baits for people. The moment one was tempted to reach out for the bait, the trap closed in on him.

Question 3.
Why was he amused by the idea of the rattrap?
Answer:
The peddler was amused by the idea of the rattrap because the world had been unkind to him. No one helped him; on the contrary, everyone shunned him. Hence, the peddler got joy in thinking ill of others. His treasured pastime consisted of thinking of people caught in the trap.

Question 4.
How was the crofter different from the kind of people the peddler usually met?
Answer:
The crofter was different from the kind of people the peddler usually met. The crofter invited him into his house, and unlike the usual people, he did not shun him. When the peddler went begging, the crofter offered him porridge, supper, and tobacco. He played cards with the peddler. Not only was he hospitable, but reposed a lot of trust in the peddler. He showed peddler where he kept his money.

Question 5.
Why did crofter show the thirty kronor to the peddler?
Answer:
The crofter showed the thirty kronor to the peddler because he was generous with his confidences. Moreover, he had told the peddler that he could not labour, but a cow supported him. When the peddler looked at him in disbelief, just to prove his point, he showed him the thirty kronor.

Question 6.
How did the peddler discover Ramsjo Ironworks?
Answer:
After stealing the crofter’s money, the peddler realized that he was lost. He had been walking around in the same area. Soon it became dark, and his fears and apprehensions increased. He felt he was caught in a rattrap. He sank down on the ground in gloom and dejection. When he put his head on the ground, he heard a hard, thumping sound. He was certain that the sound was produced by hammer strokes. So he mustered strength and moved in the direction of the sound.

Question 7.
What sight did the peddler see at the Ramsjo Ironworks?
Answer:
At the Ramsjo Ironworks, the peddler noticed that the master smith and his helper were sitting near the furnace. They were heating pig iron to put on the anvil. They wore nothing but a long shirt and a pair of wooden shoes. He also saw them get up every few minutes to stir the glowing mass with a long iron bar.

Question 8.
What was the blacksmith’s reaction to the peddler’s arrival at the forge?
Answer:
When the peddler entered the forge, the blacksmith failed to notice him right away as it was noisy and they were busy working. When the peddler stood close to the furnace, the blacksmith showed no reaction. It was a usual sight for the blacksmith to see vagabonds take shelter for the night at the forge. So he just glanced casually at the peddler and nodded a haughty consent, permitting him to spend the night.

Question 9.
How was the blacksmith’s reaction to the peddler different from that of the ironmaster? Why was it different?
Answer:
The blacksmith’s reaction was one of sheer indifference. He glanced casually and nodded a haughty consent. But the ironmaster noticed the tall stranger at once. He walked up to him, looked him over and called out, “Nils Olof!” The reactions were different because it was a usual sight for the blacksmith who often saw vagabonds come to spend the night there. But the ironmaster had mistaken him for his friend and hence invited him home.

Question 10.
Why did the rattrap peddler not reveal his identity to the ironmaster? Why did the peddler decline the invitation?
Answer:
The rattrap peddler realized the ironmaster’s mistake. He expected the ironmaster to give him a couple of kronor. He, however, felt discomforted to visit the ironmaster’s home with the mistaken identity. He also felt guilty having stolen thirty kronor. He felt going to the ironmaster’s house would be like throwing himself into the lion’s den. He just wanted to sleep in the forge that night and then sneak away.

Question 11.
What were the peddler’s feelings on his way to the ironmaster’s manor?
Answer:
On his way to the ironmaster’s manor, the peddler felt confidence in Edla. He followed her to the carriage. However, while travelling in the carriage, he had evil premonitions and regretted taking the crofter’s money. He could see himself in the trap. He felt he would never get out of it.

Question 12.
When the peddler was bathed and dressed, the ironmaster was not pleased. Why?
Answer:
The peddler had reached the ironmaster’s house looking like a vagabond. There, the valet helped the peddler dress. He had bathed him, cut his hair, and shaved him. The peddler was dressed in the ironmaster’s clothes. But when the peddler was bathed and dressed, the ironmaster was shocked. He then realized that this man was not his friend. In the dim light, he had mistaken him for his friend.

Question 13.
After the peddler’s identity was discovered, how did he spend the rest of his stay at the ironmaster’s house?
Answer:
After the peddler’s identity was discovered, he wondered why Edla had supported him. He sat and ate quietly. Through the morning he slept, and at noon they woke him for his share of the Christmas fare, after which he slept again. Feeling safe, he slept soundly like he had never slept before. He awoke in the evening, and after he had had his dinner, he thanked each one in the house and wished them for the night.

Question 14.
What did the ironmaster and Edla learn about the peddler at the church? How did they react?
Answer:
At the church, Edla received news of how the rattrap peddler had robbed the old crofter. The ironmaster was angry with Edla for letting the peddler stay. He was certain that the peddler might have had stolen things from the house while they were at the church. Edla was dejected, as well as embarrassed. She realized her mistake and was speechless.

Question 15.
What did the ironmaster and Edla discover when they went home?
Answer:
When ironmaster and Edla reached home, they expected the peddler to have escaped after robbing them. However, when they reached home, they discovered that the peddler had gone away, but much to their surprise, he had left a gift for Edla. The gift was a small rattrap with three wrinkled ten-kronor notes. Edla’s compassion had helped the poor peddler atone for his sin. The peddler also left a note saying he would have been caught in the world’s rattrap had he not been inspired and moved by Edla’s generosity.

The Rattrap Long Answer Questions

Question 1.
“The old man (crofter) was just as generous with his confidences as with his porridge and tobacco.” Justify.
Answer:
The crofter was as generous with his confidences as with his porridge and tobacco. It was out of a sense of hospitality that he looked after the vagabond as a guest. He had offered him porridge, supper and a big slice from his tobacco roll. After that they had played cards until bedtime. He discussed his life with the peddler and told him about his days of prosperity. He took him into confidence and told him that he earned a living by selling the milk his cow produced. The crofter had received a payment of thirty kronor. He also showed him the three ten-kronor bills and where he kept his money. The crofter was liberal with his hospitality and trust.

Question 2.
How did nature play against the peddler after he stole the crofter’s money?
Answer:
Nature seemed to penalize the peddler for stealing the crofter’s money. The peddler was initially pleased to get the money. He decided to steer away from the public highway for the fear of getting caught. So he went into the woods. The first hours of travel were easy, but later, the forest seemed like a labyrinth. The paths twisted back and forth strangely. The peddler walked on without coming to the end of the wood, He then realized that he had been walking around in the same area. At once, he recalled the rattrap theory he was baited with the thirty kronor.

Question 3.
Was the ironmaster a generous man? Why or why not?
Answer:
The ironmaster was a generous man. He mistook the rattrap peddler for an old acquaintance, Captain von Stahle, and coaxed him into going home with him. When the peddler refused, he sent his daughter to get him. He was also anxious that the peddler was looked after and comfortably dressed at his home. He also promised to try to get him a job. However, he felt cheated by the peddler once he found out his real identity.
OR
The ironmaster was not generous. He was a lonely man whose wife was dead and whose sons were abroad. He lived with his daughter. He had invited the peddler, mistaking him as an old friend, to alleviate his loneliness. Later, when he discovered his true identity, he wanted to get rid of him. In fact after discovering the truth about the peddler at the church, he was so angry and worried that he wanted to hand him over to the sheriff.

Question 4.
What kind of a girl was Edla? Give reasons for your answer.
Answer:
Edla was not pretty, but looked modest and shy. She was very hospitable and went to the forge to welcome her father’s guest. Her father was confident of her powers of persuasion and was sure that she would convince the guest to stay with them. Edla was also concerned and sensitive. She handed her big fur coat to the peddler to make him feel more comfortable and to cover his rags. She was polite and gentle.

She was, however, a keen observer. Unlike her father, she noticed that the peddler was afraid. She knew at once that either he had stolen something or had escaped from jail. She had the maturity to hide her emotions. She also assured the peddler that he would be allowed to leave just as freely as he had come. But, out of concern, she made him stay till Christmas.

Question 5.
What were the attitudes of the ironmaster and Edla toward the peddler before and after they discovered his identity?
Answer:
Before discovering the peddler’s identity, the ironmaster addressed the peddler as Nils Loft. He did not take the peddler’s reaction into account. He sensed that the peddler must have been uncomfortable because of the economic disparity between the two of them. The ironmaster also made sure that he was supplied with good meal and clothes. But once he discovered the case of mistaken identity, he immediately wanted to hand the peddler over to the sheriff. After he heard of the peddler in the church, he was furious and wondered what he had stolen from their cupboard. He blamed his daughter for letting him in.

Edla saw the peddler for the first time when she came to fetch him home. She noticed at once that the man was afraid. She guessed that he was either a thief or a runaway culprit. Despite this, she requested him to stay.

She was spontaneous and friendly, and coaxed the peddler into spending the Christmas with them. He also accepted the fur coat, and wore it over his rags. Edla expressed her surprise about his miserable plight and noticed that there was nothing to show that he was an educated man. When in the church, she realized that he was a thief; she never once expressed her fear. Her trust and compassion helped in reforming the peddler.

Question 6.
Was the case of mistaken identity the peddler’s fault?
Answer:
It was his fault.
The peddler let the ironmaster believe that he was an old acquaintance. He did so in order to get a couple of kronor from the ironmaster. He told the ironmaster that he was not doing well in life. His second chance to clear himself came when Edla came to invite him home. He kept quiet deliberately in the hope of material gain. He realized the risk of going to the manor house. Even when his identity was revealed, he behaved in a self-righteous fashion.
OR
It was not his fault.
The poor peddler stuggled to survive.
He had lost all ways of earning his livelihood and lived a life of hardship. He did not deliberately cheat them but merely kept quiet. He did try to avoid going to the iron master’s place many times, but starved as he was, and in rags, he was tempted to spend a comfortable day.

Question 7.
The story is both entertaining and philosophical. Discuss.
Answer:
The story has entertaining merit. The peddler has a pronounced sense of humour despite odds. He tends to philosophize even in the most difficult times. The reader is also kept in a ‘cliff hanging situation’ throughout the story. Every moment, the reader waits to know what would happen next. The story has entertaining moments like when the peddler’s identity is revealed after the valet had bathed him.
The philosophical message of the story is brought out in the theory of the rattrap— life is one big rattrap which closes in on an individual when he gives in to a temptation or steals or commits a crime. The story also elaborates the philosophy of second chances, stating that everyone should get another chance.

The Rattrap Value Based Question

Question 1.
The warmth of compassion extends its rays around the world, engraving mankind with its characteristic. Selma Lagerlof supports the theory of compassion in “Rattrap” which depicts the powerful and positive impact of such care. Discuss.
Answer:
The story revolves around the theory that life is one big rattrap. This implies that if one takes something wrongfully, the person will end up getting trapped in life as a consequence. The peddler felt that the whole world was a big rattrap that sets baits for people. The peddler earned his living by petty thievery. However, he turns over a new leaf when he receives compassion and trust from the ironmaster’s daughter. The protagonist believed till then in the dismal side of human nature.

His first “true” Christmas at the ironmaster’s . house egged him to change his ways and honour people’s trust in him. The story validates the concept that compassion revolves around humankind and the consideration of others. The peddler makes amends by returning the money he had previously stolen from the old man who had sheltered him. The tale also throws light on the value of second chances, stating that everyone should get another chance in life.

MCQ Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots with Answers

Online Education MCQ Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots with Answers

Check the below Online Education NCERT MCQ Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots with Answers Pdf free download. MCQ Questions for Class 8 Maths with Answers were prepared based on the latest exam pattern. We have provided Cubes and Cube Roots Class 8 Maths MCQs Questions with Answers to help students understand the concept very well. https://ncertmcq.com/mcq-questions-for-class-8-maths-with-answers/

Students can also refer to NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots for better exam preparation and score more marks.

Online Education for Cubes and Cube Roots Class 8 MCQs Questions with Answers

Cube And Cube Roots Class 8 MCQ Question 1.
Which of the following numbers is a perfect cube ?
(a) 125
(b) 36
(c) 75
(d) 100.

Answer

Answer: (a)
Hint:
125 = 5 × 5 × 5 = 5³.


Class 8 Maths Chapter 7 MCQ Question 2.
Which of the following numbers is a cube number ?
(a) 1000
(b) 400
(c) 100
(d) 600.

Answer

Answer: (a)
Hint:
1000 = 10 × 10 × 10 = 10³.


MCQ Questions For Class 8 Maths Chapter 7 Question 3.
Which of the following numbers is not a perfect cube ?
(a) 1331
(b) 512
(c) 343
(d) 100.

Answer

Answer: (d)
Hint:
100 = 2 × 2 × 5 × 5 = 2² × 5².


Class 8 Maths Ch 7 MCQ Question 4.
Which of the following numbers is not a cube number ?
(a) 10000
(b) 3125
(c) 64
(d) 729.

Answer

Answer: (a)
Hint:
10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.
= 24 × 54 = 2³ × 2 × 5³ × 5.


Cubes And Cube Roots Class 8 MCQ Question 5.
The cube of an odd natural number is
(a) even
(b) odd
(c) may be even, may be odd
(d) prime number.

Answer

Answer: (b)
Hint:
3 × 3 × 3 = 27 (odd).


MCQ Of Cube And Cube Roots Class 8 Question 6.
The cube of an even natural number is
(a) even
(b) odd
(c) may be even, may be odd
(d) prime number.

Answer

Answer: (a)
Hint:
6 × 6 × 6 = 216 (even).


Class 8 Math Chapter 7 MCQ Question 7.
The one’s digit of the cube of the number 111 is
(a) 1
(b) 2
(c) 3
(d) 9.

Answer

Answer: (a)
Hint:
1 × 1 × 1 = 1.


Class 8 Cube And Cube Roots MCQ Question 8.
The one’s digit of the cube of the number 242 is
(a) 2
(b) 4
(c) 6
(d) 8.

Answer

Answer: (d)
Hint:
2 × 2 × 2 = 8.


Maths Class 8 Chapter 7 MCQ Question 9.
The one’s digit of the cube of the number 123 is
(a) 3
(b) 6
(c) 9
(d) 7.

Answer

Answer: (d)
Hint:
3 × 3 × 3 = 27.


Ch 7 Maths Class 8 MCQ Question 10.
The one’s digit of the cube of the number 144 is
(a) 1
(b) 2
(c) 3
(d) 4.

Answer

Answer: (d)
Hint:
4 × 4 × 4 = 64.


MCQ Questions For Class 8 Maths Cube And Cube Roots Question 11.
The one’s digit of the cube of the number 50 is
(a) 1
(b) 0
(c) 5
(d) 4.

Answer

Answer: (b)
Hint:
0 × 0 × 0 = 0.


MCQ Class 8 Maths Chapter 7 Question 12.
The one’s digit of the cube of the number 326 is
(a) 2
(b) 3
(c) 6
(d) 4.

Answer

Answer: (c)
Hint:
6 × 6 × 6 = 216.


Class 8 Ch 7 Maths MCQ Question 13.
The one’s digit of the cube of the number 325 is
(a) 2
(b) 5
(c) 3
(d) 6.

Answer

Answer: (b)
Hint:
5 × 5 × 5 = 125.


Ncert Class 8 Maths Chapter 7 MCQ Question 14.
The one’s digit of the cube of the number 347 is
(a) 3
(b) 4
(c) 7
(d) 1.

Answer

Answer: (a)
Hint:
7 × 7 × 7 = 343.


MCQ Questions For Class 8 Maths Chapter 7 With Answers Question 15.
The one’s digit of the cube of the number 68 is
(a) 1
(b) 2
(c) 6
(d) 8.

Answer

Answer: (b)
Hint:
8 × 8 × 8 = 512.


Class 8 Maths Cube And Cube Roots MCQ Question 16.
The one’s digit of the cube of the number 249 is
(a) 2
(b) 4
(c) 9
(d) 1.

Answer

Answer: (c)
Hint:
9 × 9 × 9 = 729.


Question 17.
What is the one’s digit in the cube root of the cube number 1331 ?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a)
Hint:
1 × 1 × 1 = 1.


Question 18.
What is the one’s digit in the cube root of the cube number 1000000 ?
(a) 0
(b) 1
(c) 2
(d) 9.

Answer

Answer: (a)
Hint:
0 × 0 × 0 = 0.


Question 19.
What is the one’s digit in the cube root of the cube number 1728 ?
(a) 1
(b) 2
(c) 3
(d) 9.

Answer

Answer: (b)
Hint:
2 × 2 × 2 = 8.


Question 20.
What is the one’s digit in the cube root of the cube number 2197 ?
(a) 1
(b) 2
(c) 3
(d) 7.

Answer

Answer: (c)
Hint:
3 × 3 × 3 = 27.


Question 21.
What is the one’s digit in the cube root of the cube number 2744 ?
(a) 1
(b) 2
(c) 3
(d) 4.

Answer

Answer: (d)
Hint:
4 × 4 × 4 = 64.


Question 22.
What is the one’s digit in the cube root of the cube number 3375 ?
(a) 2
(b) 3
(c) 5
(d) 4.

Answer

Answer: (c)
Hint:
5 × 5 × 5 = 125.


Question 23.
What is the one’s digit in the cube root of the cube number 4096 ?
(a) 2
(b) 6
(c) 4
(d) 9.

Answer

Answer: (b)
Hint:
6 × 6 × 6 = 216.


Question 24.
What is the one’s digit in the cube root of the cube number 4913 ?
(a) 7
(b) 9
(c) 3
(d) 6.

Answer

Answer: (a)
Hint:
7 × 7 × 7 = 343.


Question 25.
What is the one’s digit in the cube root of the cube number 5832 ?
(a) 2
(b) 4
(c) 6
(d) 8.

Answer

Answer: (d)
Hint:
8 × 8 × 8 = 512.


Question 26.
What is the one’s digit in the cube root of the cube number 6859 ?
(a) 7
(b) 8
(c) 9
(d) 6.

Answer

Answer: (c)
Hint:
9 × 9 × 9 = 729.


Question 27.
What is the one’s digit in the cube root of the cube number 8000 ?
(a) 0
(b) 2
(c) 4
(d) 8.

Answer

Answer: (a)
Hint:
0 × 0 × 0 = 0.


Question 28.
The number of zeroes at the end of the cube of the number 20 is
(a) 1
(b) 2
(c) 3
(d) 6.

Answer

Answer: (c)
Hint:
∵ Number of zeroes at the end of the number 20 = 1
∴ Number of zeroes at the end of its cube = 3 × 1 = 3.


Question 29.
The number of zeroes at the end of the cube root of the cube number 1000 is
(a) 1
(b) 2
(c) 3
(d) 4.

Answer

Answer: (a)
Hint:
∵ Number of zeroes at the end of the cube = 3
∴ Number of zeroes at the end of the cube root = \(\frac{3}{3}\) = 1.


Question 30.
The number of zeroes at the end of the cube of the number 100 is
(a) 1
(b) 2
(c) 4
(d) 6.

Answer

Answer: (d)
Hint:
∵ Number of zeroes at the end of the number 100 = 2
∴ Number of zeroes at the end of its cube = 3 × 2 = 6.


Question 31.
The number of zeroes at the end of the cube root of the cube number 8000000 is
(a) 1
(b) 2
(c) 3
(d) 6.

Answer

Answer: (b)
Hint:
∵ Number of zeroes at the end of the cube = 6
∴ Number of zeroes at the end of the cube root = \(\frac{6}{3}\) = 2.


Question 32.
Find the smallest number by which the number 108 must be multiplied to obtain a perfect cube.
(a) 2
(b) 3
(c) 4
(d) 5.

Answer

Answer: (a)
Hint:
108 = 2 × 2 × 3 × 3 × 3 = 2 × 2 × 3³.


Question 33.
Find the smallest number by which the number 250 must be divided to obtain a perfect cube.
(o) 2
(b) 3
(c) 4
(d) 5.

Answer

Answer: (a)
Hint:
250 = 5 × 5 × 5 × 2 = 5³ × 2.


Question 34.
Find the smallest number by which the number 72 must be multiplied to obtain a perfect cube.
(a) 2
(b) 3
(c) 4
(d) 6.

Answer

Answer: (b)
Hint:
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3 × 3.


Question 35.
Find the smallest number by which the number 375 must be divided to obtain a perfect cube.
(a) 2
(b) 3
(c) 5
(d) 4.

Answer

Answer: (b)
Hint:
375 = 3 × 5 × 5 × 5 = 3 × 5³.


Question 36.
Find the smallest number by which the number 100 must be multiplied to obtain a perfect cube.
(a) 5
(b) 2
(c) 4
(d) 10.

Answer

Answer: (d)
Hint:
100 = 2 × 2 × 5 × 5.


Question 37.
Find the smallest number by which the number 10000 must be divided to obtain a perfect cube.
(a) 2
(b) 5
(c) 10
(d) 100

Answer

Answer: (c)
Hint:
10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
= 2³ × 2 × 5³ × 5.


Question 38.
Find the smallest number by which the number 200 must be multiplied to obtain a perfect cube.
(a) 2
(b) 10
(c) 5
(d) 100.

Answer

Answer: (c)
Hint:
200 = 2 × 2 × 2 × 5 × 5 = 2³ × 5 × 5.


Question 39.
Find the smallest number by which the number 625 must be divided to obtain a perfect cube.
(a) 3
(b) 5
(c) 25
(d) 125.

Answer

Answer: (b)
Hint:
625 = 5 × 5 × 5 × 5 = 5³ × 5.


Question 40.
Find the smallest number by which the number 128 must be multiplied to obtain a perfect cube.
(a) 2
(b)4
(c) 3
(d) 8.

Answer

Answer: (b)
Hint:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³ × 2.


Question 41.
Find the smallest number by which the number 256 must be divided to obtain a perfect cube.
(a) 2
(b) 4
(c) 8
(d) 16.

Answer

Answer: (b)
Hint:
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³ × 2 × 2.


Question 42.
Find the smallest number by which the number 36 must be multiphed to obtain a perfect cube.
(a) 6
(b) 2
(c) 3
(d) 4.

Answer

Answer: (a)
Hint:
36 = 2 × 2 × 3 × 3.


Question 43.
Find the smallest number by which the number 1296 must be divided to obtain a perfect cube.
(a) 6
(b) 2
(c) 4
(d) 3.

Answer

Answer: (a)
Hint:
1296 =2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2³ × 2 × 3³ × 3.


Question 44.
Find the smallest number by which the number 392 must be multiplied to obtain a perfect cube.
(a) 3
(b) 5
(c) 7
(d) 6.

Answer

Answer: (c)
Hint:
392 = 2 × 2 × 2 × 7 × 7 = 2³ × 7 × 7.


Question 45.
Find the smallest number by which the number 2401 must be divided to obtain a perfect cube.
(a) 7
(b) 6
(c) 5
(d) 9.

Answer

Answer: (a)
Hint:
2401 = 7 × 7 × 7 × 7 = 7³ × 7.


Question 46.
Find the smallest number by which the number 121 must be multiplied to obtain a perfect cube.
(a) 7
(b) 9
(c) 11
(d) 5.

Answer

Answer: (c)
Hint:
121 = 11 × 11.


Question 47.
Find the smallest number by which the number 88 must be divided to obtain a perfect cube.
(a) 11
(b) 5
(c) 7
(d) 9.

Answer

Answer: (a)
Hint:
88 = 2 × 2 × 2 × 11 = 2³ × 11.


Question 48.
The volume of a cube is 64 cm³. The edge of the cube is
(a) 4 cm
(b) 8 cm
(c) 16 cm
(d) 6 cm.

Answer

Answer: (a)
Hint:
Edge = \(\sqrt[3]{64}\) = \(\sqrt[3]{2 × 2 × 2 × 2 × 2 × 2}\)
= \(\sqrt[3]{2^3×2^3}\) = 2 × 2 = 4.


Question 49.
Apala makes a cuboid of plasticine of sides 5 cm, 4 cm, 2 cm. How many such cuboids will be needed to form a cube ?
(a) 20
(b) 25
(c) 10
(d) 16.

Answer

Answer: (b)
Hint:
Volume = 5 × 4 × 2 = 5 × 2 × 2 × 2
= 5 × 2³.


Question 50.
Which of the following is false ?
(a) Cube of any odd number is odd
(b) A perfect cube does not end with two zeroes
(c) The cube of a single digit number may be a single digit number
(d) There is no perfect cube which ends with 8.

Answer

Answer: (d)
Hint:
1728 = 12³.


We hope the given NCERT MCQ Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots with Answers Pdf free download will help you. If you have any queries regarding Cubes and Cube Roots CBSE Class 8 Maths MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Online Education for RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6

Online Education for RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6

These Solutions are part of Online Education RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6

Other Exercises

Question 1.
Find the sum of the following arithmetic progressions :
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 1
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 2
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 3
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 4
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 5
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 6
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 7

Question 2.
Find the sum to n term of the A.P. 5, 2, – 1, 4, -7, …,
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 8

Question 3.
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 – 6n.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 9

Question 4.
Find the sum of last ten terms of the A.P.: 8, 10, 12, 14,…, 126. [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 10

Question 5.
Find the sum of the first 15 terms of each of the following sequences having nth term as
(i) an = 3 + 4n
(ii) bn = 5 + 2n
(iii) xn = 6 – n
(iv) yn = 9 – 5n
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 11
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 12
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 13
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 14

Question 6.
Find the sum of first 20 terms of the sequence whose nth term is an = An + B.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 15

Question 7.
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 – 3n. [CBSE 2004]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 16
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 17

Question 8.
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 7 – 3n. [CBSE 2004]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 18

Question 9.
If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, …, is 116. Find the last term.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 19
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 20

Question 10.
(i) How many terms of the sequence 18, 16, 14, … should be taken so that their sum is zero ?
(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?
(iii) How many terms of the A.P. 9, 17, 25,… must be taken so that their sum is 636 ? [NCERT]
(iv) How many terms of the A.P. 63, 60, 57, ……… must be taken so that their sum is 693 ? [CBSE 2005]
(v) How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero? [CBSE 2016]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 21
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 22
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 23
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 24
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 25
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 26

Question 11.
Find the sum of the first
(i) 11 terms of the A.P. : 2, 6, 10, 14,…
(ii) 13 terms of the A.P. : -6, 0, 6, 12,…
(iii) 51 terms of the A.P.: whose second term is 2 and fourth term is 8.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 27
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 28
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 29

Question 12.
Find the sum of
(i) the first 15 multiples of 8
(ii) the first 40 positive integers divisible by
(a) 3, (b) 5, (c) 6
(iii) all 3-digit natural numbers which are divisible by 13. [CBSE 2006C]
(iv) all 3-digit natural numbers, which are multiples of 11. [CBSE 2012]
(v) all 2-digit natural numbers divisible by 4. [CBSE 2017]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 30
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 31
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 32
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 33
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 34
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 35
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 36

Question 13.
Find the sum :
(i) 2 + 4 + 6 + ……….. + 200
(ii) 3 + 11 + 19 + ………. + 803
(iii) (-5) + (-8) + (-11) + ……. + (-230)
(iv) 1 + 3 + 5 + 7 + …….. + 199
(v) 7 + 10\(\frac { 1 }{ 2 }\) + 14 + ……… + 84
(vi) 34 + 32 + 30 + ………. + 10
(vii) 25 + 28 + 31 + ……….. + 100 [CBSE 2006C]
(viii) 18 + 15\(\frac { 1 }{ 2 }\) + 13 + ……… + (-49\(\frac { 1 }{ 2 }\))
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 37
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 38
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 39
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 40
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 41
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 42

Question 14.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?
Solution:
First term of an A.P. (a) = 17
Last term (l) = 350
Common difference (d) = 9
Let n be the number of terms Then an = a + (n – 1) d
=> 350 = 17 + (n – 1) x 9
=> 350 = 17 + 9n – 9
=> 9n = 350 – 17 + 9 = 342
n = 38
Number of terms = 38
Now Sn = \(\frac { n }{ 2 }\) [a + l]
= \(\frac { 38 }{ 2 }\) [17 + 350] = 19 (367) = 6973

Question 15.
The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 43
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 44

Question 16.
The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 45

Question 17.
If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms ?
Solution:
12th term of an A.P. = -13
Sum of first 4 terms = 24
Let a be the first term and d be the common difference
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 46

Question 18.
Find the sum of n terms of the series
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 47
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 48
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 49

Question 19.
In an A.P., if the first term is 22, the common difference is -4 and the sum to n terms is 64, find n.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 50

Question 20.
In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms ?
Solution:
In an A.P.
5th term = 30
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 51

Question 21.
Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Solution:
In an A.P.
No. of terms = 51
Second term a2 = 14
and third term a3 = 18
Let a be the first term and d be the common
difference, then
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 52
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 53

Question 22.
If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
Solution:
Let a be the first term and d be the common difference of an A.P.
Sum of 7 terms = 49
and sum of 17 terms = 289
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 54
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 55

Question 23.
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 56

Question 24.
In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences. [CBSE 2008]
Solution:
In an A.P.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 57

Question 25.
In an A.P., the first term is 22, nth term is -11 and the sum to first n terms is 66. Find n and d, the common difference. [CBSE 2008]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 58
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 59

Question 26.
The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find its common difference. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 60

Question 27.
The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 61

Question 28.
The sum of first q terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P. [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 62
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 63

Question 29.
If the 10th term of an A.P. is 21 and the sum of its first ten terms is 120, find its nth term. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 64
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 65

Question 30.
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 66
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 67

Question 31.
The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 68
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 69

Question 32.
The nth term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 70
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 71

Question 33.
In an A.P., the sum of first ten terms is -150 and the sum of its next ten terms is -550. Find the A.P. [CBSE 2010]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 72
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 73

Question 34.
Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term. [CBSE 2012]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 74

Question 35.
In an A.P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P. [CBSE 2010]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 75

Question 36.
The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ? [NCERT]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 76

Question 37.
Find the number of terms of the A.P. -12, -9, -6,…, 21. If 1 is added to each term of this A.P., then find the sum ofi all terms of the A.P. thus obtained. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 77
= 6 x 9 = 54
If we add 1 to each term, then the new sum of so formed A.P.
= 54 + 1 x 12 = 54 + 12 = 66

Question 38.
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 78

Question 39.
The sum of first n terms of an A.P. is 5n – n2. Find the nth term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 79

Question 40.
The sum of the first n terms of an A.P. is 4n2 + 2n. Find the nth term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 80

Question 41.
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P. [CBSE 2013]
Solution:
Let a be the first term and d be common difference
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 81

Question 42.
The sum of first n terms of an A.P. is 5n2 + 3n. If its mth term is 168, find the value of m. Also, find the 20th term of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 82

Question 43.
The sum of first q terms of an A.P. is 63q – 3q2. If its pth term is -60, find the value of p, Also, find the 11th term of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 83

Question 44.
The sum of first m terms of an A.P. is 4m2 – m. If its nth term is 107, find the value of n. Also, find the 21st term of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 84

Question 45.
If the sum of the first n terms of an A.P. is 4n – n2, what is the first term ? What is the sum of first two terms ? What is the second term ? Similarly, find the third, the tenth and the nth terms. [NCERT]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 85

Question 46.
If the sum of first n terms of an A.P. is \(\frac { 1 }{ 2 }\) (3n2 + 7n), then find its nth term. Hence write its 20th term. [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 86
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 87

Question 47.
In an A.P., the sum of first n terms is \(\frac { { 3n }^{ 2 } }{ 2 } +\frac { 13 }{ 2 } n\). Find its 25th term. [CBSE 2006C]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 88
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 89
a25 = 8 + (25 – 1) x 3 = 8 + 24 x 3 = 8 + 72 = 80
Hence 25th term = 80

Question 48.
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 90

Question 49.
Find the sum of first n odd natural numbers.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 91

Question 50.
Find the sum of all odd numbers between
(i) 0 and 50
(ii) 100 and 200
Solution:
(i) Odd numbers between 0 and 50 are = 1, 3, 5, 7, …, 49 in which
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 92

Question 51.
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 93
Hence proved.

Question 52.
Find the sum of all integers between 84 and 719, which are multiples of 5.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 94

Question 53.
Find the sum of all integers between 50 and 500, which are divisible by 7.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 95

Question 54.
Find the sum of all even integers between 101 and 999.
Solution:
All integers which are even, between 101 and 999 are = 102, 104, 106, 108, … 998
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 96

Question 55.
(i) Find the sum of all integers between 100 and 550, which are divisible by 9.
(ii) all integers between 100 and 550 which are not divisible by 9.
(iii) all integers between 1 and 500 which are multiplies of 2 as well as of 5.
(iv) all integers from 1 to 500 which are multiplies 2 as well as of 5.
(v) all integers from 1 to 500 which are multiplies of 2 or 5.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 97
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 98
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 99
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 100
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 101
= 250 x 251 + 505 x 50 – 25 x 510
= 62750 + 25250 – 12750
= 88000 – 12750
= 75250

Question 56.
Let there be an A.P. with first term ‘a’, common difference d. If an denotes its nth term and S the sum of first n terms, find.
(i) n and S , if a = 5, d = 3 and an = 50.
(ii) n and a, if an = 4, d = 2 and Sn = -14.
(iii) d, if a = 3, n = 8 and Sn = 192.
(iv) a, if an = 28, Sn = 144 and n = 9.
(v) n and d, if a = 8, an = 62 and Sn = 210.
(vi) n and an, if a = 2, d = 8 and Sn = 90.
(vii) k, if Sn = 3n2 + 5n and ak = 164.
Solution:
In an A.P. a is the first term, d, the common difference a is the nth term and Sn is the sum of first n terms,
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 102
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 103
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 104
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 105
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 106
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 107

Question 57.
If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 – S4). [NCERT Exemplar, CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 108

Question 58.
A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5m/minute every succeeding minute. After how many minutes, the policeman will catch the thief? [CBSE 2016]
Solution:
Let total time be 22 minutes.
Total distance covered by thief in 22 minutes = Speed x Time
= 100 x n = 100n metres
Total distance covered by policeman
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 109

Question 59.
The sums of first n terms of three A.P.S are S1, S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2. [CBSE 2016]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 110
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 111

Question 60.
Resham wanted to save at least 76500 for sending her daughter to school next year (after 12 months). She saved ₹450 in the first month and raised her savings by ₹20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?
Solution:
Given : Resham saved ₹450 in the first month and raised her saving by ₹20 every month and saved in next 12 months.
First term (a) = 450
Common difference (d) = 20
and No. of terms (n) = 12
We know sum of n terms is in A.P.
Sn = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
Sn = \(\frac { 12 }{ 2 }\) [2 x 450 + (12 – 1) x 20]
=> Sn = 6[900 + 240]
=> Sn = 6720
Here we can see that Resham saved ₹ 6720 which is more than ₹ 6500.
So, yes Resham shall be able to send her daughter to school.

Question 61.
In a school, students decided to plant 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 double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 112

Question 62.
Ramkali would need ₹ 1800 for admission fee and books etc., for her daughter to start going to school from next year. She saved ₹ 50 in the first month of this year and increased her monthly saving by ₹ 20. After a year, how much money will she save? Will she be able to fulfill her dream of sending her daughter to school? [CBSE 2014]
Solution:
Admission fee and books etc. = ₹ 1800
First month’s savings = ₹ 50
Increase in monthly savings = ₹ 720
Period = 1 year = 12 months
Here a = 50, d = 20 and n = 12
S12 = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
= \(\frac { 12 }{ 2 }\) [2 x 50 + (12 – 1) x 20]
= 6[100 + 11 x 20]
= 6[100 + 220]
= 6 x 320 = ₹ 1920
Savings = ₹ 1920
Yes, she will be able to send her daughter.

Question 63.
A man saved ₹ 16500 in ten years. In each year after the first he saved ₹ 100 more than he did in the preceding year. How much did he save in the first year ?
Solution:
Savings in 10 years = ₹ 16500
S10 = ₹ 16500 and d = 7100
Sn = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
16500= \(\frac { 10 }{ 2 }\) [2 x a + (10 – 1) x 100]
16500 = 5 (2a + 900)
16500 = 10a + 4500
=> 10a = 16500 – 4500 = 12000
a = 1200
Saving for the first year = ₹ 1200

Question 64.
A man saved ₹ 32 during the first year, ₹ 36 in the second year and in this way he increases his savings by ₹ 4 every year. Find in what time his saving will be ₹ 200.
Solution:
Savings for the first year = ₹ 32
For the second year = ₹ 36
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 113

Question 65.
A man arranges to pay off a debt of ₹ 3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one – third of the debt unpaid, find the value of the first installment.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 114
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 115

Question 66.
There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.
Solution:
Number of trees = 25
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 116
Distance between one to other tree = 5 m
Distance between first near and the well = 10 m
Now in order to water the first tree, the gardener has to cover 10m + 10m = 20m
and to water the second tree, the distance to covered is 15 + 15 = 30 m
To water the third tree, the distance to cover is = 20 + 20 = 40 m
The series will be 20, 30, 40, ……….
where a = 20, d = 30 – 20 = 10 and n = 25
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 117

Question 67.
A man is employed to count ₹ 10710. He counts at the rate of ₹ 180 per minute for half an hour. After this he counts at the rate of ₹ 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 118
=> (n – 59) (n – 60) = 0
Either n – 59 = 0, then n – 59 or n – 60 = 0, then n = 60
Total time = 59 + 30 = 89 minutes or = 60 + 30 = 90 minutes

Question 68.
A piece of equipment cost a certain factory ₹ 600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost ?
Solution:
Cost of a piece of equipment = ₹ 600,000
Rate of depreciation for the first year = 15%
for the second year = 13.5%
for the third year = 12.0% and so on
The depreciation is in A.P.
whose first term (a) = 15
and common difference (d) = 13.5 – 15.0 = -1.5
Period (n) = 10
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 119

Question 69.
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 prize.
Solution:
Total sum = ₹ 700
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 120

Question 70.
If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3 (S20 – S10). [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 121

Question 71.
Solve the question: (-4) + (-1) + 2 + 5 + … + x = 437. [NCERT Exemplar]
Solution:
Given equation is,
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 122
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 123

Question 72.
Which term of the A.P. -2, -7, -12, … will be -77 ? Find the sum of this A.P. up to the term -77.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 124

Question 73.
The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n. [NCERT Exemplar]
Solution:
Given that, first term of the first A.P. (a) = 8
and common difference of the first A.P. (d) = 20
Let the number of terms in first A.P. be n
Sum of first n terms of an A.P., Sn
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 125
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 126

Question 74.
The students of a school decided to beautify the school on the annual day by fixing colourful on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the position of the middle most flag Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag? [NCERT Exemplar]
Solution:
Given that, the students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school.
Given that, the number of flags = 27
and distance between each flag = 2 m.
Also, the flags are stored at the position of the middle most flag i. e., 14th flag and Ruchi was given the responsibility of placing the flags.
Ruchi kept her books, where the flags were stored i.e., 14th flag and she could carry only one flag at a time.
Let she placed 13 flags into her left position from middle most flag i.e., 14th flag.
For placing second flag and return his initial position distance travelled = 2 + 2 = 4 m.
Similarly, for placing third flag and return his initial position, distance travelled = 4 + 4 = 8 m.
For placing fourth flag and return his initial position, distance travelled = 6 + 6 = 12 m.
For placing fourteenth flag and return his initial position, distance travelled = 26 + 26 = 52 m.
Proceed same manner into her right position from middle most flag i.e., 14th flag.
Total distance travelled in that case = 52 m.
Also, when Ruchi placed the last flag she return her middle position and collect her books.
This distance also included in placed the last flag.
So, these distances from a series.
4 + 8 + 12 + 16 + … + 52 [for left]
and 4 + 8 + 12 + 16 + … + 52 [for right] .
Total distance covered by Ruchi for placing these flags
= 2 x (4 + 8 + 12 + … +52)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 127
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 128
Hence, the required is 728 m in which she did cover in completing this job and returning back to collect her books.
Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position
= (2 + 2 + 2 + … + 13 times)
= 2 x 13 = 26 m
Hence, the required maximum distance she travelled carrying a flag is 26 m.

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Online Education NCERT Solutions for Class 9 Hindi Sparsh Chapter 15

Online Education NCERT Solutions for Class 9 Hindi Sparsh Chapter 15 नए इलाके में … खुशबू रचते हैं हाथ

These Solutions are part of Online Education NCERT Solutions for Class 9 Hindi. Here we have given NCERT Solutions for Class 9 Hindi Sparsh Chapter 15 नए इलाके में … खुशबू रचते हैं हाथ.

पाठ्य-पुस्तक के प्रश्न-अभ्यास

(1) नए इलाके में

प्रश्न 1.
निम्नलिखित प्रश्नों के उत्तर दीजिए-
(क) नए बसते इलाके में कवि रास्ता क्यों भूल जाता है? [CBSE]
(ख) कविता में कौन-कौन से पुराने निशानों का उल्लेख किया गया है?
अथवा
कवि अपने गंतव्य तक पहुँचने के लिए कौन-कौन-सी निशानियाँ ढूँढता है? [CBSE]
(ग) कवि एक घर पीछे या दो घर आगे क्यों चल देता है? [CBSE]
(घ) “वसंत का गया पतझड़’ और ‘बैसाख का गया भादों को लौटा’ से क्या अभिप्राय है?
(ङ) कवि ने इस कविता में समय की कमी की ओर क्यों इशारा किया है? [CBSE]
(च) इस कविता में कवि ने शहरों की किस विडंबना की ओर संकेत किया है?
अथवा
यह कविता किस ओर इशारा करती है, स्पष्ट कीजिए। (CBSE]
उत्तर:
(क) नए इलाके में कवि इसलिए रास्ता भूल जाता है, क्योंकि-

  • यहाँ रोज़ नए मकान बनते रहते हैं।
  • पुराने मकान ढहाकर नए मकान बनाए जाते हैं।
  • नए मकान बनाने के लिए पुराने पेड़ काटने से निशानी नष्ट हो जाती है।
  • खाली जमीन पर कोई नया मकान बन जाता है।

(ख) कविता में निम्नलिखित पुराने निशानों का उल्लेख हुआ है-

  • पीपल का पेड़
  • ढहा घर या खंडहर
  • जमीन का खाली टुकड़ा
  • बिना रंग वाले लोहे के फाटक वाला इकमंजिला मकान

(ग) कवि एक घर आगे या दो घर पीछे इसलिए चल देता है, क्योंकि नए बस रहे उस इलाके में एक ही दिन में काफ़ी बदलाव आ जाता है। वह अपने घर को पहचान नहीं पाता है कि वह सवेरे किस घर से गया था।

(घ)  ‘वसंत का गया पतझड़’ और ‘बैसाख का गया भादों को लौटा’ से यह अभिप्राय है कि वहाँ एक ही दिन में इतना कुछ नया बन गया है, जितना बनने में पहले नौ-दस महीने या साल भर लगते थे। सुबह का निकला कवि जब शाम को वापस आता है तो एक ही दिन में नौ-दस महीने के बराबर का बदलाव दिखाई देता है।

(ङ) कवि ने कविता में समय की कमी की ओर इसलिए संकेत किया है क्योंकि तेज़ी से आ रहे बदलाव के कारण मनुष्य की व्यस्तता भी बढ़ती जा रही है। इससे उसके पास समय की कमी होती जा रही है।

(च) इस कविता में कवि ने शहरों की उस विडंबना की ओर संकेत किया है, जिसमें शहरों में हो रहे बदलाव, खाली जमीनों में टूटे मकानों की जगह इतने नित नए मकान बनते जा रहे हैं कि सुबह घर से निकले आदमी को शाम के समय अपना मकान खोजना पड़ता है, फिर भी उसे अपना मकान नहीं मिल पाता है।

प्रश्न 2.
व्याख्या कीजिए-
(क) यहाँ स्मृति का भरोसा नहीं
       एक ही दिन में पुरानी पड़ जाती है दुनिया
उत्तर:
नगरों में बसने वाली नई बस्तियाँ इस तरह तेजी से बढ़ती चली जा रही हैं कि आदमी को अपना घर तक ढूँढना कठिन हो गया है। वह कुछ ही दिन बाद अपनी बस्ती में लौटकर आए तो रास्ते तक भूल जाता है। उसकी पुरानी निशानियाँ देखते ही देखते नष्ट हो जाती हैं। इसलिए उसकी पुरानी स्मृतियाँ और निशानियाँ किसी काम नहीं आतीं। दुनिया इतनी तेजी से बदल-बन रही है कि जो निर्माण एक दिन पहले किया जाता है, दूसरे दिन तक पुराना पड़ चुका होता है। उसके बाद नए-नए निर्माण और खड़े हो जाते हैं।

(ख)  समय बहुत कम है तुम्हारे पास।
        आ चला पानी ढहा आ रहा अकास
        शायद पुकार ले कोई पहचाना ऊपर से देखकर
उत्तर:
देखिए व्याख्या क्र. 2..

योग्यता-विस्तार

प्रश्न 1.
पाठ में हिंदी महीनों के कुछ नाम आए हैं। आप सभी हिंदी महीनों के नाम क्रम से लिखिए।
उत्तर:
हिंदी महीनों के नाम-

  1. चैत्र,
  2. बैसाख,
  3. ज्येष्ठ,
  4. आषाढ़,
  5. श्रावण,
  6. भाद्रपक्ष,
  7. आश्विन,
  8. कार्तिक,
  9. मार्गशीर्ष,
  10. पौष,
  11. माघ,
  12. फाल्गुन

(2) खुशबू रचते हैं हाथ

प्रश्न 1.
निम्नलिखित प्रश्नों के उत्तर दीजिए-
(क) “खुशबू रचनेवाले हाथ’ कैसी परिस्थितियों में तथा कहाँ-कहाँ रहते हैं?
(ख) कविता में कितने तरह के हाथों की चर्चा हुई है?
(ग) कवि ने यह क्यों कहा है कि ‘खुशबू रचते हैं हाथ’?
(घ) जहाँ अगरबत्तियाँ बनती हैं, वहाँ का माहौल कैसा होता है?
(ङ) इस कविता को लिखने का मुख्य उद्देश्य क्या है?
उत्तर:
(क) खुशबू रचनेवाले हाथ अत्यंत कठोर परिस्थितियों में गंदी बस्तियों में, गलियों में, कूड़े के ढेर के इर्द-गिर्द तथा नाले के किनारे रहते हैं। वे अस्वच्छ एवं प्रदूषित वातावरण में जीवन बिताते हैं। वे इस दुर्गंधमय वातावरण में रहने को विवश हैं। वे सामाजिक और आर्थिक विषमता के शिकार हैं। दूसरों को खुशबू देने का काम करने । वाले इस प्रकार बदहाली का जीवन बिताते हैं।

(ख) कविता में निम्नलिखित तरह के हाथों की चर्चा हुई है-

  1. उभरी नसोंवाले अर्थात् वृद्ध हाथ।
  2. घिसे नाखूनोंवाले हाथ श्रमिक वर्ग को प्रतीक है।
  3. पीपल के पत्ते जैसे नए-नए हाथ अर्थात् छोटे बच्चों के कोमल हाथ।
  4. जूही की डाल जैसे खुशबूदार हाथ अर्थात् नवयुवतियों के सुंदर हाथ।
  5. गंदे कटे-पिटे हाथ।
  6. जखम से फटे हुए हाथ।

(ग) कवि ने ऐसा इसलिए कहा है क्योंकि इन गरीब मजदूरों के हाथ सुगंधित अगरबत्तियों का निर्माण करते हैं। तथा हमारे जीवन को सुख-सुविधाएँ उपलब्ध कराकर खुशबू से महकाते हैं जिससे ऐसा लगता है कि अत्यंत प्रदूषित वातावरण में रहकर भी इनके हाथ हमारे लिए सुख-सुविधाओं से भरी वस्तुओं का निर्माण करते हैं। जिससे समस्त प्राणियों के जीवन में सुगंध फैल जाती है। ये लोग स्वयं बदहाली का जीवन बिताकर दूसरे लोगों के जीवन में खुशहाली लाते हैं। इन शब्दों द्वारा कवि ने श्रमिकों के श्रम का गुणगान किया है।

(घ) जहाँ अगरबत्तियाँ बनती हैं वहाँ का वातावरण अत्यंत गंदगी भरा होता है। चारों ओर नालियाँ तथा कूड़े-करकट का ढेर जमा होता है। चारों ओर बदबू फैली होती है। ये सुगंधित अगरबत्तियाँ बनाने वाले ऐसे गंदे वातावरण में रहकर भी दूसरों के जीवन में खुशबू बिखेरते हैं पर ऐसे वातावरण में, ऐसी भयावह स्थितियों में रहनी इनकी विवशता है।

(ङ) इस कविता को लिखने का मुख्य उद्देश्य यह है कि हमारे समाज में सुंदरता की रचना करनेवाले गरीब
और उपेक्षित लोगों की ओर हमारा ध्यान आकर्षित करना है ताकि आम लोग इन गरीब मजदूरों के जीवन की वास्तविकता को जान लें और समाज में फैली विषमताओं तथा भेदभावों को मिटाने की कोशिश करें। मजदूरों और कारीगरों की दुर्दशा का चित्रण करना तथा लोगों में उनके उद्धार की चेतना जगाना भी है। कवि अगरबत्तियाँ बनानेवाले कारीगरों का प्रदूषित वातावरण में रहना दिखाकर यह कहना चाहता है कि इनके जीवन स्तर को ऊँचा उठाने के लिए हम सबको मिलकर प्रयास करना चाहिए ताकि इन्हें भी जीवन जीने के लिए। स्वच्छ वातावरण मिल सके।

प्रश्न 2.
व्याख्या कीजिए-
(क)
(i)  पीपल के पत्ते-से नए-नए हाथ
      जूही की डाल-से खुशबूदार हाथ
उत्तर:
अगरबत्ती बनाने वाले हाथों में कुछ के हाथ पीपल के नए-नए पत्तों के समान कोमल हैं। आशय यह है कि कुछ नन्हे-नन्हे बच्चे भी अगरबत्ती बनाने के काम में लगे हुए हैं। कुछ हाथ ऐसे हैं जिनमें से जूही की डालों जैसी खुशबू आती है। आशय यह है कि कुछ सुंदर युवतियाँ भी अगरबत्तियाँ बनाने में लगी हुई हैं।

(ii) दुनिया की सारी गंदगी के बीच
      दुनिया की सारी खुशबू
      रचते रहते हैं हाथे
उत्तर:
यद्यपि अगरबत्ती बनाने वाले कारीगर दुनिया भर को सुगंधित अगरबत्ती प्रदान करते हैं और वातावरण में सुगंध फैलाते हैं किंतु उन्हें स्वयं दुनिया भर की गंदगी के बीच रहना पड़ता है। उनके चारों ओर गंदगी का ही साम्राज्य रहता है। वे शोषित हैं, पीड़ित हैं।

(ख) कवि ने इस कविता में ‘बहुवचन’ का प्रयोग अधिक किया है? इसका क्या कारण है?
उत्तर:
कविता में ‘हाथ’ के लिए बहुवचन का प्रयोग किया गया है। इसके माध्यम से कवि बताना चाहता है कि यहाँ एक कारीगर या एक मजदूर की बात नहीं की जा रही। यह समस्या सब मज़दूरों की है।

(ग) कवि ने हाथों के लिए कौन-कौन से विशेषणों का प्रयोग किया है?
उत्तर:
कवि ने हाथों के लिए निम्नलिखित विशेषणों का प्रयोग किया है-

उभरी नसोंवाले
घिसे नाखूनोंवाले
पीपल के पत्ते-से नए-नए
जूही की डाल-से खुशबूदार
गंदे कटे-पिटे
ज़ख्म से फटे हुए।

योग्यता-विस्तार

प्रश्न 1.
अगरबत्ती बनाना, माचिस बनाना, मोमबत्ती बनाना, लिफ़ाफ़े बनाना, पापड़ बनाना, मसाले कूटना आदि लघु उद्योगों के विषय में जानकारी एकत्रित कीजिए।
उत्तर:
आस पड़ोस में रहने वाले किसी मज़दूर या कर्मचारी से बात करके जानिए और उनकी फैक्ट्री में जाकर देखिए। संभव हो तो घर में बनाने का प्रयास कीजिए।

Hope given NCERT Solutions for Class 9 Hindi Sparsh Chapter 15 are helpful to complete your homework.

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Online Education for RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A

Online Education for RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A

These Solutions are part of Online Education RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A.

Other Exercises

Simplify

Question 1.
Solution:
21 – 12 ÷ 3 x 2
= 21 – 4 x 2
= 21 – 8
= 13. Ans

Question 2.
Solution:
16 + 8 ÷ 4 – 2 x 3
= 16 + 2 – 2 x 3
= 16 + 2 – 6
= 18 – 6
= 12. Ans.

Question 3.
Solution:
13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – (10)
= 13 – 10
= 3 Ans.

Question 4.
Solution:
19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6]
= 19 – 10
= 9. Ans

Question 5.
Solution:
36 – [18 – {14 – (15 – 4 ÷ 2 x 2)}]
= 36 – [18 – {14 – (15 – 2 x 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3]
= 36 – 15
= 21. Ans.

Question 6.
Solution:
\(27-[18-\{ 16-(5-\overline { 4-1 } )\} ] \)
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14]
= 27 – 4
= 23. Ans.

Question 7.
Solution:
\(4\frac { 4 }{ 3 } \div \frac { 3 }{ 5 } of5+\frac { 4 }{ 5 } \times \frac { 3 }{ 10 } -\frac { 1 }{ 5 } \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 7.1

Question 8.
Solution:
\(\left( \frac { 2 }{ 3 } +\frac { 4 }{ 9 } \right) of\frac { 3 }{ 5 } \div 1\frac { 2 }{ 3 } \times 1\frac { 1 }{ 4 } -\frac { 1 }{ 3 } \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 8.1

Question 9.
Solution:
\(7\frac { 1 }{ 3 } \div \frac { 2 }{ 3 } of2\frac { 1 }{ 5 } +1\frac { 3 }{ 8 } \div 2\frac { 3 }{ 4 } -1\frac { 1 }{ 2 } \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 9.1

Question 10.
Solution:
\(5\frac { 1 }{ 7 } -\left\{ 3\frac { 3 }{ 10 } \div \left( 2\frac { 4 }{ 5 } -\frac { 7 }{ 10 } \right) \right\} \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 10.1

Question 11.
Solution:
\(9\frac { 3 }{ 4 } \div \left[ 2\frac { 1 }{ 6 } +\left\{ 4\frac { 1 }{ 3 } -\left( 1\frac { 1 }{ 2 } +1\frac { 3 }{ 4 } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 11.1
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 11.2

Question 12.
Solution:
\(4\frac { 1 }{ 10 } -\left[ 2\frac { 1 }{ 2 } -\left\{ \frac { 5 }{ 6 } -\left( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } -\frac { 4 }{ 15 } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 12.1

Question 13.
Solution:
\(1\frac { 5 }{ 6 } +\left[ 2\frac { 2 }{ 3 } -\left\{ 3\frac { 3 }{ 4 } \left( 3\frac { 4 }{ 5 } \div 9\frac { 1 }{ 2 } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 13.1

Question 14.
Solution:
\(4\frac { 4 }{ 5 } \div \left\{ 2\frac { 1 }{ 5 } -\frac { 1 }{ 2 } \left( 1\frac { 1 }{ 4 } -\overline { \frac { 1 }{ 4 } -\frac { 1 }{ 5 } } \right) \right\} \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 14.1

Question 15.
Solution:
\(7\frac { 1 }{ 2 } -\left[ 2\frac { 1 }{ 4 } \div \left\{ 1\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \left( \frac { 3 }{ 2 } -\overline { \frac { 1 }{ 3 } -\frac { 1 }{ 6 } } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 15.1

Hope given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Students can access the Online Education CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

Online Education CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 for Practice

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

  1. The question paper consists of 14 questions divided into 3 sections A, B, C.
  2. All questions are compulsory.
  3. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
  4. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
  5. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study-based questions.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

SECTION – A
(12 Marks)

Question 1.
Three cubes of a metal whose edges are in the ratio 3 : 4 : 5, are melted and converted into a single cube whose diagonal is 12\(\sqrt{3}\) cm. Find the edges of the three cubes.
OR
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the cylinder is 1628 sq. cm, find the volume of the cylinder. (2)
Answer:
Given: Ratio of edges of three cubes = 3 : 4 : 5
∴ Let the three edges be 3x, 4x and 5x respectively.

Also,
Diagonal of bigger cube = 12\(\sqrt{3}\)cm
∴ Edge of bigger cube = \(\frac{12 \sqrt{3}}{\sqrt{3}}\) = 12cm
[Diagonal of a cube = \(\sqrt{3}\) x Edge of the cube]

Now,
Volume of bigger cube = Volume of three cubes
⇒ (12)3 = (3x)3 + (4x)3 + (5x)3
⇒ 12 x 12 x 12 = 27x3 + 64x3 + 125x3
⇒ 12 x 12 x 12 = 216x3
⇒ x3 = \(\frac{12 \times 12 \times 12}{216}\) = 2 x 2 x 2
⇒ x = 2
∴ Edges of three cubes are 3x, 4x and 5x i.e. 6 cm, 8 cm and 10 cm respectively.
OR
Let r, h be the radius of base and height of solid right circular cylinder.
Then, according to the question r + h = 37 cm
and TSA of cylinder = 1628 cm2
⇒ 2πr(h + r) = 1628
⇒ 2 x \(\frac{22}{7}\) x r x 37 = 1628
⇒ r = \(r=\frac{1628 \times 7}{2 \times 22 \times 37}=7\) = 7
So, h = 37 – r
= 37 – 7
= 30 cm

∴ Volume of cylinder = πr2h
= \(\frac{22}{7}\) x (7)2 x 30
= 4620 cm3

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 2.
Mr. Rawat, HR manager of XYZ company, prepared a report on the number of leaves taken by its employees during the year 2020-21.

Number of leaves Number of employees
0-10 7
10-20 13
20-30 10
30-40 12
40-50 8
50-60 11

Find the frequency of the class, succeeding the modal class of the above distribution. (2)
Answer:
We know, class with maximum frequency is called modal class.
Here, class with maximum frequency i.e. 13 is 10 – 20
∴ Modal class = 10 – 20
So, class succeeding the modal class is 20-30. Hence, the frequency of required class 20 – 30 is 10.

Question 3.
Prove that tangents drawn at the ends of diameter of a circle are parallel to each other.
Answer:
Given: A circle with centre O and diameter AOB at which tangents PQ and MN are drawn.

To prove: PQ || MN
Proof: We know, tangent is perpendicular to radius at the point of contact.
∴ PQ ⊥ OA and MN ⊥ OB
∴ ∠OAP = 90° and ∠OBN = 90°
⇒ ∠OAP = ∠OBN
But ∠OAP and ∠OBN form a pair of alternate angles between the lines PQ and MN, which are intersected by a transversal AB.
∴ PQ || MN
Hence proved.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 4.
A ladder, leaning against a wall, makes an angle of 60° with it. If the length of the ladder is 15 m, find the height of the wall. (2)
Answer:
Let AB be the ladder and BC be the wall.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 5
Hence, height of the wall is 7.5 m.

Question 5.
The average score of boys in an examination of a school is 71 and that of the girls is 73. If the average score of the school in the examination is 71.8, then find the ratio of number of girls to the number of boys, who appeared in the examination. (2)
Answer:
Let the number of boys in the school be x and the number of girls be y.
Now, average score of boys = 71
∴ Total score of boys = 71 × x
= 71x
Also, average score of girls = 73
Total score of girls =73 × y
= 73y

Now,
Average score of school
\(=\frac{\text { Tota score of school }}{\text { Total number of students in school }}\)
⇒ 71.8 = \(\frac{71 x+73 y}{x+y}\)
⇒ 71.8 (x + y) = 71x + 73y
⇒ 71.8x – 71x = 73y – 71.8y
⇒ 0.8x = 1.2y
⇒ \(\frac{y}{x}=\frac{0.8}{1.2}\)
= \(\frac{2}{3}\)

Hence, the ratio of number of girls to the number of boys is 2 : 3.

Caution
In average score of school both boys and girls will be included, as a school constitute both boys and girls.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 6.
The following figure shows the observation of point C from point A. Find the angle of depression of point C from the point A.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 1
A tower of height h m casts a shadow 90 m long and at the same time, a pole 80 m long, casts a shadow 120 m long. Find the value of h. (2)
Answer:
In ΔABC
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 6
Now, angle of depression of point C from point A
= ∠XAC
= ∠ACB [Alternate angles]
30°

Caution
Here, we need to calculate ∠XAC, so calculating ∠ACB is not the final answer.
OR
Let AB be the tower and BC be its shadow. Similarly, let PQ be the pole and QR be its shadow.

Also, let angle of elevation of the Sun be θ.
∴ AB = h, BC = 90 m, PQ = 80 m, QR = 120 m and ∠C = ∠R = 0.
So, in ΔPQR
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 7
Hence, the value of h is 60.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

SECTION – B
(12 Marks)

Question 7.
The sum of first seven terms of an A.P. is 63 and the sum of its next seven terms is 161. Find the tenth term of the A.P. (3)
Answer:
Let a be the first term and d be the common difference of the A.P.
We have, S7 = 63
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 8
Solving equations (i) and (ii), we get
a = 3, d = 2
Now, 10th term, a10 = a + 9d = 3 + 9 x 2
= 21
Hence, tenth term of the A.P. is 21.

Caution
Here, 161 is the sum of next seven terms of the A.P. i.e. sum of 8th to 14th terms of the A.P., so apply the formula wisely.

Question 8.
Draw a line segment XY of length 5.4 cm and mark a point P on it such that PX = \(\frac{3}{5}\)XY. (3)
Answer:
We have,
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 9

Steps of construction:

  • Draw a line segment XY of length 5.4 cm.
  • From point X, draw a ray XM such that ∠YXM is acute.
  • On ray XM, mark 2 + 3 e. 5 points, namely X1, X2, X3, X4 and X5 at equal distances.
  • Join X5Y.
  • From point X3, draw a line X3P, parallel to X5Y, intersecting XY at P.

Thus, point P divides XY in the ratio 3 : 2.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 9.
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point R, 40 m vertically above X, the angle of elevation of the top Q of tower PQ is 45°. Find the height of the tower PQ. [Use \(\sqrt{3}\) = 1.732] (3)
Answer:
We have, RX = 40 m, ∠QRM = 45°, ∠QXP = 60°.
Here, RX = PM = 40 m and RM = XP
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 10
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 11
Hence, height of the tower PQ is 94.64 m.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 10.
Solve for x:
\(\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5 x}, x \neq 0,-1,2\)
OR
In a class test, the sum of Shrishti’s marks in maths and science is 45. If she had 1 more mark in maths and 1 less in science, the product of marks would have been 500. Find her marks in maths and science. (3)
Answer:
We have,
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 12
Let the Shrishti’s marks in maths be x.
Then, her marks in science = 45 – x
Now, according to the question,
(x + 1) [(45 – x) – 1] = 500
⇒ (x + l) (44 – x) = 500
⇒ 44x + 44 – x2 – x = 500
⇒ x2 – 43x + 456 = 0
⇒ x2 – 24x – 19x + 456 = 0
⇒ x(x- 24) – 19(x- 24) = 0
⇒ (x – 19) (x – 24) = 0
x = 19, 24
if x = 19, then (45 – x) = 45 – 19 = 26
If x = 24, then (45 – x) = 45 – 24 = 21

Hence, she either got 19 marks in maths 26 marks in science or 24 marks in maths 21 marks in science.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

SECTION – C
(16 Marks)

Question 11.
Rohan has a solid wax cylinder of diameter 21 cm and height 21 cm. From it, he curved out a conical cavity of maximum volume. From the remaining wax, he formed a solid sphere. Find the diameter of the sphere. (4)
Answer:
For a conical cavity of maximum volume Diameter of cavity = Diameter of cylinder
= 21 cm
and,Height of cavity = Height of cylinder
= 21 cm
∴ Radius of cavity (r) = \(\frac{21}{2}\) cm
Let the radius of the sphere be R.
Since, volume of remaining wax = Volume of cylinder – Volume of cone
And the spherical ball is formed from this remaining wax,
∴ Volume of sphere = Volume of cylinder – Volume of cone
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 13
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 14
∴ Diameter of shere = 2R = 2 x \(\frac{21}{2}\) = 21cm.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 12
In the given figure, quadrilateral ABCD is circumscribing a circle with centre O and AD ⊥ AB. If the radius of incircle is 10 cm, then find the value of x.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 2
PA and PB are the two tangents to a circle with centre O in which OP is equal to diameter of the circle. Prove that ΔAPB is equilateral. (4)
Answer:
We know, tangents drawn from an external point to a circle are equal in length.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 15
∴ AP = AS, BP = BQ,
CQ = CR and DR = DS
Now, in quadrilateral OPAS,
OS = OP = Radii, AP = AS and ∠PAS = 90°
i.e. adjacent sides are equal and an angle is a right angle.
∴ OPAS is a square.
∴ AP = AS = 10 cm
Now, CR = CQ = 27 cm
So, BP = BQ
= BC – CQ
= 38 – 27
= 11cm
So, x = AB = AP + BP
= 10 + 11
= 21
Hence, the value of x is 21.
OR
Given: A circle with centre 0 in which tangent PA and PB are drawn from an external point P.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 16
Also, OP is equal to diameter of the circle.
To prove: ΔAPB is equilateral.
Construction: Join OA and OB.
Proof: Let radius of the circle be r.
∴ OA = OB = r and OP = 2r
Now, ∠OAP = ∠OBP = 90°
[Tangent ⊥ Radius]
Let ∠OPA = 0
Then, in right angled triangle AOP
\(\sin \theta=\frac{\mathrm{OA}}{\mathrm{OP}}=\frac{r}{2 r}=\frac{1}{2}=\sin 30^{\circ}\)
⇒ θ = 30°
Since, tangents are equally inclined to the line segment joining the centre of the circle to their common external point.
∴ ∠OPA = ∠OPB = \(\frac{1}{2}\) ∠APB
or, ∠APB = 2∠OPA = 2 x 30° = 60°
We know, tangents drawn from an external point to a circle are equal in length.
∴ In ΔAPB,
PA = PB
⇒ ∠PBA = ∠PAB = x° (say)
So, \(x^{\circ}=\frac{180^{\circ}-\angle \mathrm{APB}}{2}=\frac{180^{\circ}-60^{\circ}}{2}\)
= 60°
i.e. ∠PBA = ∠PAB = ∠APB = 60°
∴ ΔAPB is an equilateral triangle.

Hence, proved.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 13.
Case Study-1
A loan is an amount which one receives from a bank with the assurance of returning it in the future along with some fixed annual interest.

Deepak took a loan of ₹ 1,18,000 from HDFC bank to invest in his start-up. He has to repay this loan by paying every month, starting with the first installment of ₹ 1000. Since, his start-up was going well, he decided to increase the installment of loan by ₹ 100 every month.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 3
Based on this information answer the following question.
(A) What amount of ban he still have to pay after the 30th installment? (2)
(B) What amount will be paid to the bank as the last installment of ban? (2)
Answer:
(A) Since installments paid by Deepak, each successive month, increases uniformly by a fixed amount, so his series of installments form an A.P. with a = 1000 and d = 100.
∴ Amount paid till 30 installments
= S30
= \(\frac{30}{2}\) [2 x 1000 + (30 – 1) x 100]
∵ [Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
= 15 x 4900
= 73500

So, amountto be pay after 30th installment = Total amount to be paid – A amount paid till 30 installments = ₹ (1,18,000-73,500)
= ₹ 44,500

(B) Let the loan of ₹ 1,18,000 be repaid to the bank in n months.
Then, Sn = 118000
⇒ \(\frac{n}{2}\) [2a + (n – 1)d]
= 118000
⇒ \(\frac{n}{2}\) [2 x 1000 + (n – 1) x 100] = 118000
⇒ \(\frac{n}{2}\) [1900+ 100n] = 118000
⇒ n(950 + 50n) = 118000
⇒ 50n2 + 950n – 2360 = 0
or n2 + 19n – 2360 = 0
⇒ (n + 59) (n – 40) = 0
⇒ n = -59, 40
Since, number of months cannot be negative
∴ n = 40
Amount paid in last installment = a40
= a + 39d
= 1000 + 39 x 100
= 4900
Hence, the amount paid in the last installment of the loan is ₹ 4900.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions

Question 14.
Case Study-2
‘Swachh Bharat Abhiyan’ is a country-wide campaign initiated by our Honourable Prime Minister of India, Mr. Narendra Singh Modi in the year 2014 to eliminate open defecation, to improve solid waste management and to accelerate the efforts to achieve universal sanitisation.

As part of the ‘Swachh Bharat Abhiyan’, some houses of a locality in Agra decided to clean up and beautify a Primary School of their locality by planting a number of plants. They involved the school kids and the local community in doing so.

CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 4
Here is the data indicating the number of plants contributed by different houses:

Number of plants contributed Number of houses
1-3 10
4-6 8
7-9 X
10-12 7
13-15 12
16-18 4

(A) If the mean number of plants contributed be 8.9, then how many houses contributed 7 to 9 plants? (2)
(B) Find the median number of plants contributed. (2)
Answer:
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 17
Hence, 9 houses contributed 7 to 9 plants.

(B)

Class Frequency Cumulative Frequency
0.5 – 3.5 10 10
3.5 – 6.5 8 18
6.5 – 9.5 X = 9 27
9.5 – 12.5 7 34
12.5 -15.5 12 46
15.5 -18.5 4 50
N = 50

Here, \(\frac{N}{2}=25\)
Cumulative frequency just greater than 25 is 27 which belongs to class 6.5 – 9.5.
∴ Median class = 6.5 – 9.5.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 with Solutions 18

Online Education ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1

Online Education ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1

Question 1.
The length and breadth of a rectangular field are in the ratio 9 : 5. If the area of the field is 14580 square metres, find the cost of surrounding the field with a fence at the rate of ₹3·25 per metre.
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 1

Question 2.
A rectangle is 16 m by 9 m. Find a side of the square whose area equals the area of the rectangle. By how much does the perimeter of the rectangle exceed the perimeter of the square?
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 2

Question 3.
Two adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between longer sides is 12 cm, find the distance between shorter sides.
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 3

Question 4.
Rajesh has a square plot with the measurement as shown in the given figure. He wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹50 per m2.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 4
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 5

Question 5.
A flooring tile has a shape of a parallelogram whose base is 18 cm and the corresponding height is 6 cm. How many such tiles are required to cover a floor of area 540 m2? (If required you can split the tiles in whatever way you want to fill up the comers).
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 6

Question 6.
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round?
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 7
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 8
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 9

Question 7.
In the adjoining figure, the area enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 10
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 11

Question 8.
A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent into the form of a circle, find the area of the circle.
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 12

Question 9.
From the given figure, find
(i) the area of ∆ ABC
(ii) length of BC
(iii) the length of altitude from A to BC
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 13
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 14

Question 10.
A rectangular garden 80 m by 40 m is divided into four equal parts by two cross-paths 2.5 m wide. Find
(i) the area of the cross-paths.
(ii) the area of the unshaded portion.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 15
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 16

Question 11.
In the given figure, ABCD is a rectangle. Find the area of the shaded region.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 17
Solution:
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 18

Question 12.
In the adjoining figure, ABCD is a square grassy lawn of area 729 m2. A path of uniform width runs all around it. If the area of the path is 295 m2, find
(i) the length of the boundary of the square field enclosing the lawn and the path.
(ii) the width of the path.
al Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 19
Solution:
al Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 20
al Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 21

ML Aggarwal Class 8 Solutions for ICSE Maths

Online Education for Class 7 Sanskrit Grammar Book Solutions अव्ययाः

We have given detailed Online Education NCERT Solutions for Class 7 Sanskrit Grammar Book अव्ययाः Questions and Answers come in handy for quickly completing your homework.

Online Education for Sanskrit Vyakaran Class 7 Solutions अव्ययाः

अव्यय – जो शब्द विभक्ति, वचन, लिंग एवं काल के अनुसार परिवर्तित नहीं होते हैं, अव्यय कहलाते हैं। इन शब्दों में कोई व्यय (विकार) नहीं होता, इसीलिए उन्हें ‘अव्यय’ कहा जाता है।
यथा- बालकः अधुना पठति।
वयम् अधुना पठामः।
मीता अधुना पठति।

यहाँ ‘अधुना’ में लिंग अथवा पुरुष से कोई परिवर्तन नहीं आया।
सः कुत्र गच्छति?
सः कुत्र अगच्छत्?
ते कुत्र अगच्छन्?
यहाँ ‘अधुना और कुत्र’ में काल अथवा वचन के कारण कोई विकार नहीं आया। अतः ‘अधुना’, ‘कुत्र’ जैसे शब्द अव्यय कहलाते हैं। अब निम्नलिखित अव्ययों को देखें और प्रयोग समझें।
Class 7 Sanskrit Grammar Book Solutions अव्ययाः 1
Class 7 Sanskrit Grammar Book Solutions अव्ययाः 2
Class 7 Sanskrit Grammar Book Solutions अव्ययाः 3

अभ्यासः

प्रश्न 1.
(क) अव्ययपदं चित्वा लिखत। (अव्ययपद चुनकर लिखिए।) Pick out the indeclinables.
1. जलम् नीचैः वहति।
2. अलं कोलाहलेन।
3. शशक: शीघ्रं धावति।
4. वृद्धः शनैः चलति।
5. अहम् अपि तेन सह गमिष्यामि।
उत्तरम्-
नीचैः, अलम्, शीघ्रम्, शनैः, अपि, सह।

(ख) उचितेन अव्यय-पदेन रिक्तस्थानपूर्ति कुरुत- (उचित अव्यय पद द्वारा रिक्त स्थान पूर्ति कीजिए- Fill in the blanks with the appropriate indeclinables.)
यत्र-तत्र, यदा-तदा, एकदा ह्यः, तत्र
1. ___________ अहं विदेशम् अगच्छम्।
2. ___________ बालकाः ___________ कोलाहलः।
3. ___________ अहम् जन्तुशालाम् अगच्छम्।
4. ___________ अहम् विविधान् जन्तून् अपश्यम्।
5. ___________ वृष्टिः भवति ___________ मयूरः नृत्यति।
उत्तरम्-
1. एकदा
2. यत्र-तत्र
3. ह्यः
4. तत्र
5. यदा-तदा

(ग) प्रदत्तविकल्पेभ्यः उचितं पदं चित्वा वाक्यानि पूरयत-
(दिए गए विकल्पों से उचित पद चुनकर वाक्य पूरे कीजिए- Pick out the appropriate word from the options given and complete the sentences.)
1. अधुना वयम् क्रीडामः, त्वम् ___________ आगच्छ। (एव, अपि, मा)
2. ___________ वयम् खो-खो खेलम् खेलिष्यामः। (अद्य, ह्यः, श्व:)
3. ते बालकाः ___________ इतस्ततः भ्रमन्ति। (मृषा, वृथा, कुतः)
4. विद्यालयात् ___________ बसयानम् तिष्ठति। (अधुना, सर्वदा, बहि:)
5. पुत्र! अधुना त्वं ___________ आगच्छसि? (कुत्र, कुतः, कदा)
उत्तरम्-
1. अपि
2. श्वः
3. वृथा
4. बहिः
5. कुतः

Online Education Simple Present Tense Worksheet Exercises for Class 3 CBSE with Answers

Simple Present Tense Class 3
This grammar section explains Online Education English Grammar in a clear and simple way. There are example sentences to show how the language is used. https://ncertmcq.com/simple-present-tense-exercises-for-class-3/

Online Education for Simple Present Tense Worksheet Exercises for Class 3 CBSE with Answers PDF

Simple Present Tense Worksheet
Fundamentals

  • Verbs have forms called tenses that tell us when the action happens.
  • We use simple present tense to talk about the actions or events happening in the present.
  • If the action happens regularly we use the simple present tense.
  • The simple present tense is also used to state facts.
  • Add s or-es with singular nouns.

Simple Present Tense Class 3
Presentation

Read the picture story.
Simple Present Tense For Class 3

Simple Present Tense Worksheet For Class 3
A. Answer the following questions according to the story.

1. What do Mrs. and Mr. Paul do in the garden? ___________________
2. Who goes to river nearby? ___________________
3. Who brings water from the river? ___________________

Simple Present Tense Worksheet With Answers For Class 3
B. Fill in each blank with a verb from the story.

1. Mrs. and Mr. Paul ___________________ their garden.
2. Mr. Paul ___________________ up early everyday.
3. Mrs. Paul ___________________ the plants.

Simple Present Tense Worksheets for Grade 3 with Answers PDF

Simple Present Tense Worksheet Class 3
A. Tick the correct word.

1. Ram (love, loves) chocolates.
2. The girl (dance, dances) well.
3. The teacher (teach, teaches) us grammar.
4. The baby (cry, cries) for milk.
5. They (grow, grows) vegetables.
6. Birds (fly, flies) in the sky.

Tense Worksheet For Class 3
B. Write the correct form of the verbs in the following sentences.

1. She ___________ (drive) very well.
2. Water ___________ (boil) at 100 degrees.
3. I ___________ (work) hard.
4. My sister ___________ (speak) English.
5. They ___________ (open) the store at 8.00 am.
6. The box ___________ (contain) food.

Simple Present Tense Exercise For Class 3
C. Read the passage about Gopi’s morning routine.

I get up at 6 o’clock and brush my teeth. First of all I take a bath. Then, I take a cup of tea and some biscuits. I study my lessons and have a meal. I polish my shoes. I wear my uniform and comb my hair. I carry my lunch and leave for school.

Now tell Gopi’s Morning Routine in your words.
Gopi gets up at 6 o’clock. First of all he _________________________________________________________________________
Tenses Worksheet For Class 3

Online Education HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom

Online Education HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom

These Solutions are part of Online Education HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom

Question 1.
Both helium (He) and beryllium (Be) have two valence electrons. Whereas He represents a noble gas element, Be does not. Assign reason.
Answer:
The element He (Z = 2) has two electrons present in the only shell i.e., K-shell. Since this shell can have a maximum of two electrons only therefore, He is a noble gas element. The element Be (Z = 4) has electronic configuration as : 2, 2. Although the second shell has also two electrons but it is not complete. It can still accommodate six more electrons. Therefore, the element beryllium does not represent a noble gas element.

More Resources

Question 2.
Study the data given below and answer the questions which follow :
HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom image - 1

  1. Write the mass number and atomic number of the particles A, B, C and D.
  2. Which represent a pair of isotopes ?

Answer:

  1. Particle A : Mass number = 7 ; Atomic number = 3
    Particle B : Mass number = 17 ; Atomic number = 9
    Particle C : Mass number =16, Atomic number = 8
    Particle D : Mass number =18, Atomic number = 8
  2. Particles C and D represent a pair of isotopes since they have same atomic number.

Question 3.
Which of the two will be chemically more reactive ; element X with atomic number 17 or element Y with atomic number 16 ?
Answer:
The electronic configuration of the two elements are as follows :
X(Z = 16): K (2), L(8), M(6) ;
Y(Z = 17): K(2), L(8), M(7)
The element X needs two electrons in the M shell to have the noble gas configuration of element, Ar (Z = 18). However, the element Y needs only one electron to achieve this. This means that the element Y has a greater urge or desire to take up one electron from an outside atom. It is therefore, more reactive than the element X which needs two electrons.

Question 4.
The number of protons, neutrons and electrons in particles from A to E are given below :
HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom image - 2

  1. Which one is a cation ?
  2. Which one is an anion ?
  3. Which represent pair of isotopes ?

Answer:

  1. B is a monovalent cation (B+)
  2. E is a monovalent anion (E)
  3. A and D represent pair of isotopes.

Question 5.
An atom of an element has three electrons in the third shell which is the outermost shell. Write

  1. the electronic configuration
  2. the atomic number
  3. number of protons
  4. valency
  5. the name of the element
  6. its nature whether metal or non-metal. (CBSE 2012)

Answer:
The third shell is M shell. If the atom of the element has three electrons in the third shell, this means that K and L shells are already filled.

  1. Electronic configuration : 2, 8, 3.
  2. Atomic number = No. of electrons =13
  3. Number of protons = No. of electrons =13
  4. Valency of the element = 3
  5. The element with Z = 13 is aluminium (Al)
  6. It is a metal.

Hope given HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

Online Education for Villa for Sale Summary in English by Sacha Guitry

We have decided to create the most comprehensive Online Education English Summary that will help students with learning and understanding.

Online Education for Villa for Sale Summary in English by Sacha Guitry

Villa for Sale Summary in English

This one-act play is set in France. The cast consists of five characters. The action takes place during the course of an evening and revolves around the sale of a villa. The owner of the villa, Juliette, is anxious to sell it as she needs the money. She stays alone with her maid. As the play starts, Juliette is expecting a customer, the agency people are sending. As she and her maid discuss the prospective sale of the villa, her maid discloses that she has got a role in a film, like many other local people and suggests Juliette try for a film role, too, as it will help ease her financial problems. Juliette is not very keen to act in films and angrily tells her maid that she is ‘not quite so hard up as that yet! ’

While the two are talking, Jeanne comes in with her husband Gaston. From the moment they enter, Gaston expresses his displeasure about the villa and finds a number of faults with it. In fact he shows no desire in buying a house at all as he feels Jeanne’s parents and nieces and nephews will spend summers with them. On the other hand, Jeanne finds the villa excellent and tries to persuade Gaston to buy the villa for her.

Juliette counts the many advantages the house possesses to please the customers. The house has electricity, gas, water, telephone, and drainage. The bathroom is beautifully fitted and the roof was entirely repaired the previous year. She says she will sell the villa entirely furnished with all the fixtures, just as it is, with the exception of one little picture signed by the famous painter, Corot. While Juliette asks for two hundred and fifty thousand francs, and quotes two hundred thousand francs as her last price, Gaston bargains for the house with Juliette and quotes the ridiculously low price of sixty thousand francs, knowing it will not get accepted. Juliette rejects the offer but Jeanne expresses a desire to see the upper floor before she leaves and Juliette eagerly takes her upstairs.

While Gaston waits for the two women to return, the actual customer Juliette was waiting for, an American lady, Mrs Al Smith, comes in. Her hurry to buy the villa without even looking at it, at whatever the price asked for, is exploited by Gaston. This episode is the most entertaining part of the play. In a moment, the reluctant buyer is transformed into a skillful seller. Gaston shows himself to be ingenious and totally unscrupulous. He counts the plus points of the villa. But all his salesmanship is not needed as Mrs Al Smith has already made up her mind to buy the villa. She is a big star and wants the villa as it is near Paramount Studios, where she is going to shoot some films.

Gaston is able to sell the villa to her for three hundred thousand francs. When Jeanne returns she expresses her opinion about the unsuitability of the house. But now Gaston, who has already sold the house to Mrs A1 Smith, coxes Juliette to talk about the additional facilities the villa offers and buys the villa, including the Corot, for two hundred thousand pounds. Thus, Gaston makes a profit of a hundred thousand francs and a painting by Corot with his resourcefulness.

Villa for Sale Summary Questions and Answers

Question 1.
Copy and complete the following paragraph about the theme of the play using the clues given in the box below. Remember that there are more clues than required.
Answer:
Juliette, the owner of a Villa wants to sell it as she is in need of money. Moreover, she is not in favour of the house. Jeanne and Gaston, a couple, visit her with the aim of buying the Villa. While Jeanne is enthusiastic about buying, Gaston detests the idea as he does not want his in laws in that house. Also, he finds the asking price of 300 thousand francs to be expensive. When Jeanne and Juliette go around the house, another customer walks in and starts talking to Gaston taking him to be Juliette’s husband. Gaston strikes a deal with the customer by which he is able to give 200 thousand francs to the owner and keep one hundred thousand francs for himself.

Question 2.
Answer the following questions briefly.

a. Why does Jeanne want to buy a villa?
Answer:
Jeanne wants to buy a villa for her parents. Buying a villa will ensure that her parents and her sister’s children can live with them.

b. Why is Gaston not interested in buying the villa in the beginning?
Answer:
Gaston is not interested in buying the villa in the beginning because he does not want his parents-in-law and sister- in-law’s children to live with them. He also feels that the asking price of the villa is too high for its size.

c. Mrs. Al.Smith makes many statements about the French. Pick out any two and explain them.
Answer:
Mrs Al Smith has typically American prejudices against the French. She says ‘You French people have a cute way – of doing business! ’ This reflects her distrust for the French way of doing business. She is extremely judgmental about it because the price of the villa has not been written on the signboard and she is astute enough to realize that the buyer may, thus, be exploited.

She also says ‘Frenchmen usually have to consult about ten people before they get a move on.’ This, again is a generalized and prejudiced judgement against all French people wherein Mrs A1 Smith feels that French people cannot conduct business quickly and efficiently because they talk to a lot of people before making a decision.

d. Juliette says “…………. now I have only one thought that is to get the wretched place off my hands. I would sacrifice it at any price”. Does she stick to her words? Why/Why not?
Answer:
No, Juliette does not stick to her words. She says that she will sacrifice the villa at any price, but refuses to go below 200 thousand francs as the selling price.

e. Who is better in business – Juliette or Gaston? Why?
Answer:
Gaston is a much better businessman because not only does he sell a house that is not even his yet, but he makes a profit of 100 thousand francs and a painting by Corot, from the deal.

f. Do you like/dislike Gaston? Give your reasons.
Answer:
Encourage the students to come up with their own answers.

Yes, I like Gaston because of his intelligence and ingenuity. He is smart and quick, and can remain cool under pressure. This is an admirable characteristic in any human being and I like him for it.

No, I don’t like Gaston because while he is sharp, he is unscrupulous and lack of scruples is not something that can be pardoned in any human being.

Question 3.
Read the following extracts and answer the questions that follow by choosing the correct options.

A. But the sign has been hanging on the gate for over a month now and I am beginning to be afraid that the day I bought it was when I was the real fool.

a. Why is Juliette disappointed?
Answer:
She is unable to find a suitable buyer for her villa.

b. Why does she call herself a fool?
Answer:
She had bought the villa for more than it was worth.

B. ‘But your parents would take possession of it, every year from the beginning of spring until the end of September. What’s more they would bring the whole tribe of your sister’s children with them.’

a. What does Gaston mean by ‘take possession’?
Answer:
Her parents would stay with them for a long time.

C. ‘While you were upstairs, I have been thinking a lot about your Papa and Mamma.

a. What is the discrepancy between what Gaston said earlier and what he says now?
Answer:
Earlier he did not want Juliette’s parents to stay with them but now he is showing concern for them.

b. What does the above statement reveal about Gaston’s character?
Answer:
He is an opportunist.

Question 4.
Select words from the box to describe the characters in the play as revealed by the following lines. You may take the words from the box given on the next page.
Answer:

Lines from the play Speaker Quality revealed
a. One hundred thousand francs if necessary and that’s only twice what it cost me. Juliette greedy
b. If you don’t want the house, tell me so at once and we’ll say no more about it. Jean Overbearing
c. No! Iam very fond of your family, but not quite so fond as that. Gaston Haughty and Disapproving
d. Quite so. I have, but you haven’t. Gaston Boastful
e. I have never cared such a damned little about anybody’s opinion Gaston Haughty and Boastful
f. On the principle of people who like children and haven’t any can always go and live near a school. Gaston Witty
g. The garden is not very large, but you see. it is surrounded by other gardens. Juliette Clever and Materialistic
h. I will be philanthropic and let you have it for two hundred thousand. Juliette Cunning
i. I have been thinking a lot about your Papa and Mamma. You see, I am really unselfish. Gaston Clever