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Chief Seattle’s Speech Summary by Chief Seattle’s

Chief Seattle who was the native American leader of the Suquamish and Duwamish tribes of Washington, delivered his speech in January of 1854. His speech was given to his people when American colonisers wanted to buy native land of his tribe and in return offered them amnesty, and the right to live there. As an old Chief who had seen natives killed, he reluctantly accepted the offer since he believed, turning it down will only result in the total annihilation of his tribe.

His speech was both consolatory in nature, helping his people to understand what was going on, their weak position in the political climate and helping them to understand the transition they were now forced to make. He even mentions how Canada’s borders are now being controlled by King George but their ‘Big Chief and ‘Father’ is now a White man who sits in Washington. The speech delivered by Seattle is considered a legendary speech by Native Americans as it sums up their plight in front of American Colonisers.

The speech delivered by Seattle was published in the Seattle Sunday Star, October 29, 1887, by Dr. Henry A. Smith. Smith is said to have taken notes as Chief Seattle spoke in the Suquamish dialect, and then transcribed the text in English from his notes. The last two sentences of the text here given have been considered for many years to have been part of the original, but are now known to have been added by an early 20th century historian and ethnographic writer, A.C. Ballard. The most important fact to note is that there is no verbatim transcript in existence.

All known texts are second hand. Smith makes it very clear that his version is not exact a copy, but rather the best he could put together from notes taken at the time. There is an undecided historical argument on which native dialect the Chief would have used, Duwamish on Suquamish. Either way all agree the speech was translated into Chinook Jargon on the spot, since Chief Seattle never learned to speak English.

Chief Seattle’s Speech Summary Introduction

Chief Seattle was an important figure in the early American History. He was the Chief of the Suquamish and Duwamish tribes, fighting for the rights of his native people in the face of American Colonisers. He was the powerful leader of Red Indians in the Puget sound area. He wanted to put his people at ease and corporate with the White settlers at the same time. White people get the land they wanted from Seattle and Seattle brings comfort to his own tribe by talking to them about the real nature of living to them.

The speech which was delivered by Seattle in January, 1854, was a formal response to Governor Stevens’ proposal’ to acquire the land of Red Indians. Smith published the speech from what he claimed were his notes taken at the time.

According to H.A. Smith, Chief Seattle is in front of a crowd gathered by Governor Isaac Stevens to discuss the sale of native land to the white men. While delivering his speech, Seattle thanks the white men for showing sympathetic attitude and generosity by allowing him and his people a place to live peacefully and comfortably.

Chief Seattle begins his speech with a positive change in the native Red Indian’s fives. He says that nature has been kind and sympathetic for his people for many centuries. But the situation will not remain same. He feels that the things which seem to be fine at present, but the scenario will change in the future as the Great Chief (George Washington, the first President of the United States of America) wishes to put the land of the native Americans with words of feigned ‘good will and friendship.’ Seattle says that the Great Chief is in little need of their help and friendship, as his people (the whites) are strong and powerful if compared to the natives, but it is the greatness of the Washington Chief as he has extended the hand of friendship.

Then the Chief Seattle recollects the time when his people were large in number, but now, their number has been reduced. He compares the Whites to the grass that covers the vast prairies, large in number. His people are few and they resemble the scattering trees of a storm swept plain. Seattle says that the proposition seems to be just, kind and generous as the Red man no longer has rights. In Seattle’s opinion, the offer appears to be wise since the native Americans are less in number and don’t require a vast territory.

Once again, Seattle compares the past situation of the tribal people when they were in their glorious phase to the waves of a wind-ruffed sea. He means to express how lively and energetic his people were then. But, instead of mourning over their untimely decay, he wants to look forward. Then, Seattle points out that youth is impulsive and the youngmen often indulge in revengeful acts considering them to be gainful. During war or other revengeful acts, they even lose their own lives, but their families have to bear the loss. The members of their families have to suffer from utter sufferings and sorrows. So, he suggests that his natives and the white settlers should make a peaceful atmosphere and they should never think of hostility.

Chief Seattle mentions George Washington as their ‘good father’ and says that their ‘good father’ has promised his natives if they surrender or sell their land to the White settlers, he will protect them from foreign enemies like Haidas and Tsimshians. His vigorous, energetic and brave warriors will provide them strength. Seattle also hopes that the White Chiefs brave men will provide the natives strengths. His ship would fill their harbours so that Hidas and Tsimshians will cease to frighten the natives. Then, in true sense, he would be their father and they would be his children.

But then a doubt lingers in the mind of Seattle and he feels that God of the White people is not their God. The God of the Whites loves and protects his people and hates the tribal people (the red children). He has forsaken his Red children. Seattle, then, refers to his own God, the Great Spirit who also seems to be forsaking them. His natives seem to be orphans who can look nowhere for help. Thus he remarks that they are two distinct races having separate origins and separate destinies. A great difference is found between the whites and his people.

Then, Seattle praises the sacredness which is related not only, with the ashes of their ancestors but also with this land which is their resting place. The natives love to stay in the land where their ancestors’ memories are alive whereas the whites wander far from their ancestors’ graves. He, then, says that the white man’s religion was inscribed upon tablets of stone by the iron fingers of their God so that they could not forget whereas the Red man could never comprehend or remember it.

But the religion of his people is the traditions of their ancestors. It constitutes the dreams of their old men and is written in the hearts of their people. The ancestors of the Europeans (after their death) cease to love them and the land of their nativity, but on the other hand, the ancestors (or the dead) of his race never forget their beautiful world that gave them their being. They still love its valleys, rivers, magnificent mountains and lakes. Even after their death, they long to show their affection to the gloomy hearts and often return to visit, guide, console and comfort them.

After this, Chief Seattle says that there is great difference between these two races (native Americans and the Whites). They can not develop any attachment with each other. They can not live together. Hence he compares them like day and night. The Red men flee away at the approach of the whites as the morning mist flees before the morning sun. But the white man’s poposition seems fair and he thinks that his people will except it. He hopes that then they will dwell apart in place.

Then, Seattle says that it is a matter of little importance where his people the remnant of their days, but it is certain that the number of his people will reduce. No single star of hope can be seen hovering above his horizon. The winds moan and grain fate follows them. Their situation is similar to a wounded doe that is being hunted down. In a few more year, the race of his people will disappear. He also warns that some day the white settlers too will face the same destiny. It is the order of nature that everything will see decline sooner or later. As the Red tribal people of America have been reduced to their meager existence, the white people will see the same fate whatever distant it may be. Though God has favoured the white people more than the tribal people, they can not be exempted from the common destiny of decay and doom.

Chief Seattle says that they will accept the proposal of acquisition of land of the Governor of Washington but puts forward a condition-that they will not be prevented from visting the ashes (tombs) of their ancestors as the land is sacred to them. Every hill, every valley, every plain and even rivers which are known as lifeless, are sacred to his people.

Seattle says that the native Americans would be transported to a reality beyond what is felt by the senses. The shores, the pathless woods, the field would never be empty of their spirits. This land will make them eternal. They will be a part of land forever. Their death would not be death but only a gateway to the eternal world. They will only change their world and hence will become immortal. He ends his speech with the assertion that there is no death, only a change of world.

Chief Seattle’s Speech Summary Stanzawise Word-Meanings

Yonder = at a distance, but within a view. Compassion = a sensation of scftrow excited by the distress or misfortunes; pity. External = everlasting; endless. Untold = not related; not revealed. Overcast = filled or abounding with clouds. The Great chief at Washington = George Washington. White Chief = Governor Isaac Stevens. Greeting = expression of kindness or joy. Vast prairies = large open area of grassland specially found in North America. Resemble = to be like or similar to. Scattering = a small number of something. Presume = to suppose or assume.

Red man = native American. Extensive = expanded; large. Ruffled = disturbed. Dwell on = linger over. Mourn over = grieve over. Untimely decay = destroyed before the usual time. Reproach = to attribute blame to. Paleface = a derogatory term for a white person (said to have been used by North American Indians). Impulsive = without forethought; actuated by impulse or by transient feelings. Disfigure = to mar the figure of. Relentless = stern; insensible to the distress of others. Restrain = to hold back from. Push = to drive or impel by force or pressure. Hostilities = animosities. Bristling = vigorous and energetic. Harbors = seaports. Haidas = members of a seafaring group of North American Indian who lived on the Pacific coast of British Columbia and Southwestern Alaska.

Tsimshians = members of a penutian people who lived on rivers in British Columbia. For-saken = left entirely; abandoned. Waxstronger = spread from one to another area. Ebbing = receding; going out; falling. Receding = moving back; retreating; withdrawing. Teeming multitudes = a large number of people. Firmament = the region of the air; the sky or heavens. Distinct = different. Ashes = the remains of the human body when burnt, or when returned to dust by natural decay. (Perhaps here it is used for tombs). Hallowed = sacred; worthy of religious veneration. Iron finger = finger of God. Comprehend = to take into the mind; to understand.

Sachemes = chiefs of a tribe of the American Indian. Portals = doors or gates; ways of entrance or exit. Verdant = covered with growing plants or grass; green; fresh; flourishing. Magnificent = grand in appearance. Sequestered = secluded. Vales = valleys. Yearn = dream of, a strong longing. Fond = loving; affectionate. Flees = makes a quick exit. Mist = fog. Remnant = remaining; yet left. Hovers = hangs fluttering in the air. Horizon = the apparent junction of the earth and sky. Moan = to make a low prolonged sound of grief or pain; to groan softly and continuously. Trail = track, path. Stolidly = without being upset. Doom = destruction. Untimely fate = premature death or doom. Decay = destruction; death. Ponder = to think; to muse. Privilege = a peculiar benefit, advantage, or favour. Molestation = disturbance; annoyance. Estimation = an approximate calculation; opinion. Grove = a small growth of trees. Swelter = to be overcome and faint with heat. Eventide = the time of evening; evening. Myth = a traditional story accepted as history. Solitude = state of being alone; a lovely life; loveliness. Throng = a multitude of persons or of living beings; a crowd.

Chief Seattle’s Speech Summary About the Writer

Chief Seattle or Sealth was a leader of the Suquamish and Duwamish Native American tribes. He was a prominent figure among his people. He pursued a path of accommodation to white settlers, forming a personal relationship with David Swinson ‘Doc’ Maynard. Maynard was an advocate of Native American rights whose friendship with Chief Seattle was important in the formation of city of Seattle. When the first plots for the village were filed on May 23, 1853, due to Maynard’s prompting, it was for the ‘Town of Seattle.’

. Beyond leadership skills and the gift of oratory, Chief Seattle had the desire for the two vastly different cultures to coexist in peace. He both observed and played a part in the birth of a small village named after him, that has since grown into a large metropolis known for its innovation, openness, diversity and love for creation.

Chief Seattle was born around 1786 on or near Blake Island, Washington near present day Seattle. His father, Schweabe, was a leader of the Suquamish tribe of Agate Pass, between Brainbridge Island and the mainland of Washington state’s Kitsap Peninsula across Puget Sound from the present city of Seattle. Seattle’s mother was Sholitza, the daughter of a Duwamish chief, from near the lower Green River area. As the line of descent traditionally ran through the mother, Seattle was considered Duwamish. Both the Suquamish and Duwamish are Coast Salish people. Seattle’s given name at birth was Sealth.

Seattle grew up speaking two different dialects of Lushootseed and was blessed with skill-sets from the two different tribes. Once he was made the chief of the Duwamish tribe. It is believed that he sighted the ships from the Vancouver expedition, as they explored the region around the Salish Sea, which is now known as Puget Sound. From a very young age, he earned the standing of an authoritative personality and was known for his leadership qualities.

By 1833, when the Hudson’s Bay Company founded Fort Nisqually near the head of Puget Sound, Seattle had a solid reputation as an intelligent and formidable leader with a compelling voice. He was also known as an orator and when he addressed an audience, his voice is said to have carried from his camp to the Stevens Hotel at First and Marrion, a distance of three-quarters of a mile. He was tall and broad for a Puget Sound native at nearly six feet; Hudson’s Bay Company traders gave him to nickname LeGros (The Big One).

In 1847, Seattle helped lead the Suquamish in an attack upon the Chemakum stronghold of Tsetsibus, near Port Townsend, that effectively wiped out this rival group. The death of one of his sons during the raid affected him deeply, for not long after that he was baptized into the Roman Catholic Church and give the baptismal name Noah. He is believed to have received his baptism by the Oblates of Mary Immaculate at their Joseph of Newmarket Mission, founded near the new settlement of Olympia in 1848. This conversion was a turning point for Seattle and the Duwamish, as it marked the end of his fighting days and his emergence as leader known as ‘friend to the whites’.

White settlers began arriving in the Puget Sound area in 1846, and in the area that later became the city of Seattle, in 1851. Seattle welcomed the settlers and sought out friendships with those with whom he could do business. His initial contact was with San Francisco merchant, Charles Fay, with whom he organized a fishery on Elliott Bay in the summer of 1851. When Fay returned to San Francisco, Chief Seattle moved South to Olympia. Here he took up with David S. ‘Doc’ Maynard.

Seattle helped protect the small band of settlers. Because of his friendship and assistance, it was Maynard who advocated for naming the settlement ‘Seattle’ after Chief Seattle. When the first plots for the village were filed on May 23, 1853, it was for the ‘Town of Seattle’.

Seattle served as native spokesman during the treaty council held at Point Elliott (later Mukilteo), from December 27, 1854, to January 9, 1855. While he voiced misgivings about ceding title to some 2.5 million acres of land, he understood the futility of opposing a force so much larger than his own people. In signing the treaty and retaining a reservation for the suquamish but not for the Duwamish, he lost the support of the latter. This unhappiness soon led to the Yakima Indian War of 1855-1857. Seattle kept his people out of Battle ofi Seattle (1856). Afterwards he unsuccessfully sought clemency for the war leader, Leschi. On the reservation, he attempted to curtail the influence of whiskev sellers and he interceded between the whites and the natives. Off the reservation, he participated in meetings to resolve native disputes.

Seattle maintained his friendship with Maynard and cultivated new relationships with other settlers. He was unwilling to lead his tribe to the reservation established, since mixing Duwamish and Snohomish was likely to lead to bloodshed. Maynard persuaded the government of the necessity of allowing Seattle to remove to his father’s longhouse on Agate Passage, ‘Old Man House’. Seattle frequented the town named after him. He died on June 7, 1866, on the Suquamish reservation at Port Madison, Washington.

Treasure Trove Poems and Short Stories Workbook Answers

Morning Star Total History and Civics Class 9 ICSE Solutions Book PDF Download 2022

Total History and Civics Class 9 ICSE Morning Star Answers Pdf

History

• Chapter 1 The Harappan Civilisation
• Chapter 2 The Vedic Period
• Chapter 3 Jainism and Buddhism
• Chapter 4 The Mauryan Empire
• Chapter 5 The Sangam Age
• Chapter 6 The Age of the Guptas
• Chapter 7 The Medieval India: The Cholas
• Chapter 8 Medieval India: The Delhi Sultanate
• Chapter 9 Medieval India: The Mughal Empire
• Chapter 10 Medieval India: Composite Culture
• Chapter 11 The Modem Age in Europe: Renaissance
• Chapter 12 The Modem Age in Europe: The Reformation
• Chapter 13 The Modem Age in Europe: Industrial Revolution

 Civics

• Chapter 14 Our Constitution
• Chapter 15 Salient Features of the Constitution—I
• Chapter 16 Salient Features of the Constitution—II
• Chapter 17 Elections
• Chapter 18 Local Self-Government: Rural
• Chapter 19 Local Self-Government: Urban

More Resources for Class 9 ICSE Solutions

• Total History and Civics Class 9 ICSE Morning Star Solutions (Morning Star)
• ICSE Solutions for Class 9 History and Civics (Avichal, Candid & Goyal Brothers Prakashan)
• ICSE Solutions for Class 9 Geography (Morning Star, Candid, Pitambar and others)
• ICSE Solutions for Class 9 Hindi
• Selina ICSE Solutions for Class 9 Maths (Selina Publishers)
• Selina ICSE Solutions for Class 9 Physics (Selina Publishers)
• Selina ICSE Solutions for Class 9 Chemistry (Selina Publishers)
• Selina ICSE Solutions for Class 9 Biology (Selina Publishers)
• New Simplified Chemistry Class 9 ICSE Solutions (Dr. Viraf J.Dalal, Allied Publishers)
• A New Approach to ICSE Physics Part 1 Class 9 Solutions (Goyal Brothers Prakashan)

Treasure Trove A Collection of ICSE Poems Workbook Answers Chapter 10 Notes -Nine Gold Medals – ICSE Class 10, 9 English

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About the Poem

Sports is not only about winning medals. They are also about learning the values of cooperation, sharing, competing and complementing. In this poem ‘Nine Gold Medals’, the poet, David Roth has presented the idea of empathy and how human values are as important as the spirit of competition. The poem presents the situation of a race, where the contestants leave aside their desire to win the medal to help a smaller and weaker contestant.

They all go hand-in-hand to the finishing line. Empathy is the capacity to understand another person’s experience from his point of view. Simply stated, empathy is the ability to ‘put oneself in another’s shoes’. That is exactly what the eight contestants had done. One look at the fallen contestant had forced them to think ‘what would I have felt if I had fallen?’ and they knew exactly what they had to do. By awarding gold medals to all nine contestants, the authorities honoured their display of empathy, helpful nature and human values.

About the Poet

Kerrville New Folk award winner (1986), landslide top vote-getter at the Falcon Ridge Folk Festival’s ‘Most Wanted Showcase’ (1996), and NAIRD INDIE nominee – Digging Through My Closet, singer/songwriter album of the year (1994), David Roth has often been cited for his entertaining stage presence, accomplished musicianship, and powerful singing and subject matter.

David Lee Roth (born October 10, 1954) is an American rock vocalist, songwriter, actor, author, and former radio personality. In 2007, he was inducted into the Rock and Roll Hall of Fame. Roth is best known as the original (1974-1985) and current (2006- present) lead singer of the Southern California-based hard rock band Van Halen. He is also known as a successful solo artist, releasing numerous RIAA-certified Gold and Platinum records. After more than two decades apart, Roth re-joined Van Halen in 2006 for a North American tour that became the highest grossing in the band’s history and one of the highest grossing of that year. In 2012, Roth and Van Halen released the critically successful comeback album, A Different Kind of Truth.

In addition to performing at music festivals, clubs and venues across the U.S. and Canada, David leads singing, songwriting, and performance workshops and is a presenter and emcee at a wide variety of conferences and retreats. He has been the artist-in ­residence for several years at New York’s Omega Institute, one of the country’s leading adult education centers, and has recorded six albums of his work.

Central Idea

Sports is not only about winning medals. They are also about learning the values of cooperation, sharing, competing and complementing. In this poem ‘Nine Gold Medals’, the poet, David Roth has presented the idea of empathy and how human values are as important as the spirit of competition. The poem presents the situation of a race, where the contestants leave aside their desire to win the medal to help a smaller and weaker contestant. They all go hand-in-hand to the finishing line.

Word Meanings

1. Spectators – (here)persons watching especially an event or sports without taking part
2. Block – the two starting blocks on the ground that runners push their feet against at the beginning of a race
3. Resolved – determined
4. Poised – ready
5. Pistol – (here) a starting pistol used to signal the start of a race
6. Stumbled – (here) hit his foot against something when he began to run and almost fell
7. Staggered – lost balance
8. Asphalt – black tarred road
9. Anguish – pain and disappointment
10. Dashed – destroyed

Paraphrase

The hundred-yard race is about to begin. The athletes take position at the starting blocks. They begin to run immediately after the starting pistol is fired. However, one of them is unable to run and falls on the track. The action has begun and already one episode has taken place. Notice how eight contestants are strong and run forward, while the ninth, who is the smallest, falls down. He cries out with the pain of disappointment. He has trained hard but does not get the opportunity to show his talent. All his dreams of winning the medal are broken and destroyed.

When the remaining eight contestants saw him fall, they, instead of continuing the race, came to the help of their fellow contestant. All the athletes had dreamt of winning the medal. However, they readily for got their dream and came forward to help the boy to his feet. Then all the nine contestants walked hand-in-hand to the finish line. The audience was so moved by the exemplary behaviour of the contestants that it stood up and clapped. There were now nine winners, instead of one, and each was given a gold medal. All the contestants displayed empathy turning the Special Olympics into a really ‘special’ one.

      Summary

Olympics are held once every four years. Athletes from all over the world train hard to participate in this event. Winning a medal in the Olympics is the ultimate goal of every athlete of the world. However, the setting or the scene of this poem is that of ‘Special Olympics’. In these Olympics, differently-abled persons, who have some problem/s in a particular part of the body, participate in various sports events. The contestants put in a lot of preparation and practice. Everyone hopes to win a medal. The spectators are as excited as the contestants. They cheer and encourage the contestants.

Of all the events in Olympics, the hundred-meter race is the most prestigious. The athlete, who wins it, is remembered as the fastest man in the world. So, for Special Olympics mentioned in the poem this is the final event, hence the most prestigious. The hundred-yard race is about to begin. The athletes take position at the starting blocks. They begin to run immediately after the starting pistol is fired. However, one of them is unable to run and falls on the track. The action has begun and already one episode has taken place. Notice how eight contestants are strong and run forward, while the ninth, who is the smallest, falls down. He cries out with the pain of disappointment. He has trained hard but does not get the opportunity to show his talent. All his dreams of winning the medal are broken and destroyed.

When the remaining eight contestants saw him fall, they, instead of continuing the race, came to the help of their fellow contestant. All the athletes had dreamt of winning the medal. However, they readily for got their dream and came forward to help the boy to his feet. Then all the nine contestants walked hand-in-hand to the finish line. The audience was so moved by the exemplary behaviour of the contestants that it stood up and clapped. There were now nine winners, instead of one, and each was given a gold medal. All the contestants displayed empathy turning the Special Olympics into a really ‘special’ one. Empathy is the capacity to understand another person’s experience from his point of view. Simply stated, empathy is the ability to ‘put oneself in another’s shoes’. That is exactly what the eight contestants had done. One look at the fallen contestant had forced them to think ‘what would I have felt if I had fallen?’ and they knew exactly what they had to do. By awarding gold medals to all nine contestants, the authorities honoured their display of empathy, helpful nature and human values.

Critical Appreciation

Alliteration

1. In a line in stanza 4, the consonant sound /s/ has been repeated in order to bring about a musical effect.
   But the smallest among them, he stumbled and staggered.
   This repetition of the same sound is called alliteration.
   Here the sound /s/ has been repeated. Find another such line from stanza 5.
2. No specific rhyme scheme has been followed in the poem. Yet the poem has a rhythm of its own. Read it aloud to feel the rhythm. Here are a few examples.
   • And a banner above that said ‘Special Olympics’
     Could not have been more on the mark.
   • And a standing ovation and nine beaming faces
     Said more than these words ever will.

(stanza 8)

3. Poetry says a lot in a few words. Here too, the poet has used the technique of not expressing directly and encouraging the readers to infer meanings on their own.

For More Resources

 

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 15 Linear Inequations (Including Number Lines). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Linear Inequations Exercise 15A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If the replacement set is the set of natural numbers, solve.
(i) x – 5 < 0
(ii) x + 1 < 7
(iii) 3x – 4 > 6
(iv) 4x + 1 > 17
Solution:
(i) x – 5 < 0
x – 5 + 5 <0 + 5 ………(Adding 5)
=> x < 5
Required answer = {1, 2, 3, 4}
(ii) x + 1 ≤ 7 => x + 1 – 1 ≤ 7 – 1 (Subtracting 1)
=> x ≤ 6
Required answer = {1, 2, 3, 4, 5, 6}
(iii) 3x – 4 > 6
3x – 4 + 4 > 6 + 4 (Adding 4)
=> 3x > 10
\(\frac { 3x }{ 3 }\) > \(\frac { 10 }{ 3 }\) …(Dividing by 3)
=> x > \(\frac { 10 }{ 3 }\)
=> x > \(3\frac { 1 }{ 3 }\)
Required answer = { 4, 5, 6, …}
(iv) 4x + 1 ≥ 17
=> 4x + 1 – 1 ≥ 17 – 1 (Subtracting)
=> 4x ≥ 16
=> \(\frac { 4x }{ 4 }\) ≥ \(\frac { 16 }{ 4 }\) (Dividing by 4)
=> x ≥ 4
Required answer = {4, 5, 6, …}

Question 2.
If the replacement set = {-6, -3, 0, 3, 6, 9}; find the truth set of the following:
(i) 2x – 1 > 9
(ii) 3x + 7 < 1
Solution:
(i) 2x – 1 > 9
⇒ 2x – 1 + 1 > 9 + 1 (Adding 1)
⇒ 2x > 10
⇒ x > 5 (Dividing by 2)
⇒ x > 5
Required answer = {6, 9}
(ii) 3x + 7 ≤ 1
⇒ 3x + 7 – 7 ≤ 1 – 7 (Subtracting 7)
⇒ 3x ≤ – 6
⇒ x ≤ – 2
Required Answer = {-6, -3}

Question 3.
Solve 7 > 3x – 8; x ∈ N
Solution:
7 > 3x – 8
=> 7 – 3x > 3x – 3x – 8 (Subtracting 3x)
=> 7 – 7 – 3x > 3x – 3x – 8 – 7 (Subtracting 7)
=> -3x > -15
=> x < 5 (Dividing by -3)
Required Answer = {1, 2, 3, 4}
Note : Division by negative number reverses the inequality.

Question 4.
-17 < 9y – 8 ; y ∈ Z
Solution:
-17 < 9y – 8
=> -17 + 8 < 9y – 8 + 8 (Adding 8)
=> -9 < 9y
=> -1 < y (Dividing by 9)
Required number = {0, 1, 2, 3, 4, …}

Question 5.
Solve 9x – 7 ≤ 28 + 4x; x ∈ W
Solution:
9x – 1 ≤ 28 + 4x
=> 9x – 4x – 7 ≤ 28 + 4x – 4x (Subtracting 4x)
=> 5x – 7 ≤ 28
=> 5x – 7 + 7 ≤ 28 + 7 (Adding 7)
=> 5x ≤ 35
=> x ≤ 7 (Dividing by 5)
Required answer = {0, 1, 2, 3, 4, 5, 6, 7}

Question 6.
Solve : \(\frac { 2 }{ 3 }\)x + 8 < 12 ; x ∈ W
Solution:

Question 7.
Solve -5 (x + 4) > 30 ; x ∈ Z
Solution:

Question 8.
Solve the inquation 8 – 2x > x – 5 ; x ∈ N.
Solution:

x = 1, 2, 3, 4 (x ∈ N)
Solution set = {1, 2, 3, 4}

Question 9.
Solve the inequality 18 – 3 (2x – 5) > 12; x ∈ W.
Solution:

Question 10.
Solve : \(\frac { 2x+1 }{ 3 }\) + 15 < 17; x ∈ W.
Solution:

Question 11.
Solve : -3 + x < 2, x ∈ N
Solution:

Question 12.
Solve : 4x – 5 > 10 – x, x ∈ {0, 1, 2, 3, 4, 5, 6, 7}
Solution:

Solution set = {4, 5, 6, 7}
Question 13.
Solve : 15 – 2(2x – 1) < 15, x ∈ Z.
Solution:

Question 14.
Solve : \(\frac { 2x+3 }{ 5 }\) > \(\frac { 4x-1 }{ 2 }\) , x ∈ W.
Solution:

Linear Inequations Exercise 15B – Selina Concise Mathematics Class 8 ICSE Solutions

Solve and graph the solution set on a number line :
Question 1.
x – 5 < -2 ; x ∈ N
Solution:

Question 2.
3x – 1 > 5 ; x ∈ W
Solution:

Question 3.
-3x + 12 < -15 ; x ∈ R.
Solution:

Question 4.
7 > 3x – 8 ; x ∈ W
Solution:

Question 5.
8x – 8 < – 24 ; x ∈ Z
Solution:

Question 6.
8x – 9 > 35 – 3x ; x ∈ N
Solution:

Question 7.
5x + 4 > 8x – 11 ; x ∈ Z
Solution:

Question 8.
\(\frac { 2x }{ 5 }\) + 1 < -3 ; x ∈ R
Solution:

Question 9.
\(\frac { x }{ 2 }\) > -1 + \(\frac { 3x }{ 4 }\) ; x ∈ N
Solution:

Question 10.
\(\frac { 2 }{ 3 }\) x + 5 ≤ \(\frac { 1 }{ 2 }\) x + 6 ; x ∈ W
Solution:

Question 11.
Solve the inequation 5(x – 2) > 4 (x + 3) – 24 and represent its solution on a number line.
Given the replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.
Solution:

Question 12.
Solve \(\frac { 2 }{ 3 }\) (x – 1) + 4 < 10 and represent its solution on a number line.
Given replacement set is {-8, -6, -4, 3, 6, 8, 12}.
Solution:

Question 13.
For each inequation, given below, represent the solution on a number line :
(i) \(\frac { 5 }{ 2 }\) – 2x ≥ \(\frac { 1 }{ 2 }\) ; x ∈ W
(ii) 3(2x – 1) ≥ 2(2x + 3), x ∈ Z
(iii) 2(4 – 3x) ≤ 4(x – 5), x ∈ W
(iv) 4(3x + 1) > 2(4x – 1), x is a negative integer
(v) \(\frac { 4 – x }{ 2 }\) < 3, x ∈ R
(vi) -2(x + 8) ≤ 8, x ∈ R
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 19 Representing 3-D in 2-D. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Representing 3-D in 2-D Exercise 19 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If a polyhedron has 8 faces and 8 vertices, find the number of edges in it.
Solution:
Faces = 8
Vertices = 8
using Eulers formula,
F + V – E = 2
8 + 8 – E = 2
-E = 2 – 16
E= 14

Question 2.
If a polyhedron has 10 vertices and 7 faces, find the number of edges in it.
Solution:
Vertices = 10
Faces = 7
Using Eulers formula,
F + V – E = 2
7 + 10 – E = 2
-E = -15
E = 15

Question 3.
State, the number of faces, number of vertices and number of edges of:
(i) a pentagonal pyramid
(ii) a hexagonal prism
Solution:
(i) A pentagonal pyramid
Number of faces = 6
Number of vertices = 6
Number of edges = 10

(ii) A hexagonal prism
Number of faces = 8
Number of vertices = 12
Number of edges = 18

Question 4.
Verily Euler’s formula for the following three dimensional figures:

Solution:
(i) Number of vertices = 6
Number of faces = 8
Number of edges = 12
Using Euler formula,
F + V – E = 2
8 + 6 – 12 = 2
2 = 2 Hence proved.

(ii) Number of vertices = 9
Number of faces = 8
Number of edges = 15
Using, Euler’s formula,
F + V – E = 2
9 + 8 – 15 = 2
2 = 2 Hence proved.

(iii) Number of vertices = 9
Number of faces = 5
Number of edges = 12
Using, Euler’s formula,
F + V – E = 2
9 + 5 – 12 = 2
2 = 2 Hence proved.

Question 5.
Can a polyhedron have 8 faces, 26 edges and 16 vertices?
Solution:
Number of faces = 8
Number of vertices = 16
Number of edges = 26
Using Euler’s formula
F + V – E
⇒ 8 + 16 – 26 ≠ -2
⇒ -2 ≠ 2
No, a polyhedron cannot have 8 faces, 26 edges and 16 vertices.

Question 6.
Can a polyhedron have:
(i) 3 triangles only ?
(ii) 4 triangles only ?
(iii) a square and four triangles ?
Solution:
(i) No.
(ii) Yes.
(iii) Yes.

Question 7.
Using Euler’s formula, find the values of x, y, z.

Solution:

Question 8.
What is the least number of planes that can enclose a solid? What is the name of the solid.
Solution:
The least number of planes that can enclose a solid is 4.
The name of the solid is Tetrahedron.

Question 9.
Is a square prism same as a cube?
Solution:
Yes, a square prism is same as a cube.

Question 10.
A cubical box is 6 cm x 4 cm x 2 cm. Draw two different nets of it.
Solution:

Question 11.
Dice are cubes where the sum of the numbers on the opposite faces is 7. Find the missing numbers a, b and c.

Solution:

Question 12.
Name the polyhedron that can be made by folding each of the following nets:

Solution:
(i) Triangular prism. It has 3 rectangles and 2 triangles.
(ii) Triangular prism. It has 3 rectangles and 2 triangles.
(iii) Hexagonal pyramid as it has a hexagonal base and 6 triangles.

Question 13.
Draw nets for the following polyhedrons:

Solution:
Net of hexagonal prism:

Net of pentagonal pyramid:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 20 Area of Trapezium and a Polygon. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Area of Trapezium and a Polygon Exercise 20A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the area of a triangle, whose sides are :
(i) 10 cm, 24 cm and 26 cm
(ii) 18 mm, 24 mm and 30 mm
(iii) 21 m, 28 m and 35 m
Solution:
(i) Sides of ∆ are
a = 10 cm
b = 24 cm
c = 26 cm

(ii) Sides of ∆ are
a = 18 mm
b = 24 mm
c = 30 mm

(iii) Sides of ∆ are
a = 21 m
b = 28 m
c = 35 m

Question 2.
Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm ; find :
(i) area of the triangle
(ii) height of the triangle corresponding to 8 cm side.
Solution:

Question 3.
The sides of a triangle are 16 cm, 12 cm and 20 cm. Find :
(i) area of the triangle ;
(ii) height of the triangle, corresponding to the largest side ;
(iii) height of the triangle, corresponding to the smallest side.
Solution:
Sides of ∆ are
a = 20 cm
b = 12 cm
c = 16 cm

Question 4.
Two sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8 m side is 6 m; find :
(i) area of the triangle ;
(ii) height of the triangle corresponding to 6.4 m side.
Solution:

= 9/2 = 4.5 m
Hence (i) 14.4 m² (ii) 4.5 m

Question 5.
The base and the height of a triangle are in the ratio 4 : 5. If the area of the triangle is 40 m²; find its base and height.
Solution:
Let base of ∆ = 4x m
and height of ∆ = 5x m
area of ∆ =40 m²

Question 6.
The base and the height of a triangle are in the ratio 5 : 3. If the area of the triangle is 67.5 m²; find its base and height.
Solution:
Let base = 5x m
height = 3x m

base = 5x = 5 x 3 = 15 m
height = 3x = 3 x 3 = 9 m

Question 7.
The area of an equilateral triangle is 144√3 cm²; find its perimeter.
Solution:
Let each side of an equilateral triangle = x cm

Question 8.
The area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.
Solution:
Let each side of the equilateral traingle = x

Question 9.
A field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40m, DC = 50 m and angle A = 90°. Find the area of the field.
Solution:
Since ∠A = 90°
By Pythagorus Theorem,
In ∆ABD,

Question 10.
The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96 cm. Find its area.
Solution:
Let the sides of the triangle ABC be 4x, 5x and 3x
Let AB = 4x, AC = 5x and BC = 3x
Perimeter = 4x + 5x + 3x = 96
=> 12x = 96

Question 11.
One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.
Solution:
In isosceles ∆ABC
AB = AC = 13 cm But perimeter = 50 cm

Question 12.
The altitude and the base of a triangular field are in the ratio 6 : 5. If its cost is ₹ 49,57,200 at the rate of ₹ 36,720 per hectare and 1 hectare = 10,000 sq. m, find (in metre) dimensions of the field,
Solution:
Total cost = ₹ 49,57,200
Rate = ₹ 36,720 per hectare
Total area of the triangular field

Question 13.
Find the area of the right-angled triangle with hypotenuse 40 cm and one of the other two sides 24 cm.
Solution:
In right angled triangle ABC Hypotenuse AC = 40 cm
One side AB = 24 cm

Question 14.
Use the information given in the adjoining figure to find :
(i) the length of AC.
(ii) the area of a ∆ABC
(iii) the length of BD, correct to one decimal place.

Solution:

Area of Trapezium and a Polygon Exercise 20B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the length and perimeter of a rectangle, whose area = 120 cm² and breadth = 8 cm
Solution:
area of rectangle = 120 cm²
breadth, b = 8 cm
Area = l x b
l x 8 = 120
l = 120/8 = 15 cm
Perimeter = 2 (l+b) = 2(15+8) = 2 x 23 = 46 cm
Length = 15 cm; Perimeter = 46 cm

Question 2.
The perimeter of a rectangle is 46 m and its length is 15 m. Find its :
(i) breadth
(ii) area
(iii) diagonal.
Solution:
(i) Perimeter of rectangle = 46 m
length, l = 15 m
2 (l+b) = 46
2(15 + b) = 46
15+b = 46/2 = 23
b = 23 – 15
b = 8 m

Question 3.
The diagonal of a rectangle is 34 cm. If its breadth is 16 cm; find its :
(i) length
(ii) area
Solution:

AC² = AB²+BC² (By Pythagoras theorem)
(34)² = l² + (16)²
1156 = l² + 256
l² = 1156 – 256
l² = 900
l = √900 = 30 cm
area = l x b = 30 x 16 = 480 cm²
(i) 30 cm (ii) 480 cm²

Question 4.
The area of a small rectangular plot is 84 m². If the difference between its length and the breadth is 5 m; find its perimeter.
Solution:
Area of a rectangular plot = 84 m²
Let breadth = x m
Then length = (x + 5) m
Area = l x b
x(x + 5) = 84
x² + 5x – 84 = 0
=> x²+ 12x – 7x – 84 = 0
=> x(x + 12) – 7(x + 12) = 0
=> (x + 12) (x – 7) = 0
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 7 = 0, then x = 7
Length = x + 5 = 7 + 5 = 12m
and breadth = x = 7 m
Perimeter = 2(l + b) = 2(12+ 7) = 2 x 19 m = 38 m

Question 5.
The perimeter of a square is 36 cm; find its area
Solution:
Perimeter of Square = 36 cm

Question 6.
Find the perimeter of a square; whose area is : 1.69 m²
Solution:
Area of square= 1.69 m²
Side = √area = √1.69 = 1.3 m
Perimeter = 4 x side = 4 x 1.3 = 5.2 m

Question 7.
The diagonal of a square is 12 cm long; find its area and length of one side.
Solution:
Let side of square = a cm
diagonal = 12 cm
By Pythagoras Theorem, a² + a² = (12)²
2a² = 144
a² = 72
Area of square = a² = 72 cm²
a² = 72
a = √72 = 8.49 cm

Question 8.
The diagonal of a square is 15 m; find the length of its one side and perimeter.
Solution:
Diagonal of square = 15 m
Let side of square = a
a² + a² = (15)² = 225
a² = 225/2 = 112.50
a = √112.50 = 10.6 m
Perimeter = 4 x a = 10.6 x 4 = 42.4 m

Question 9.
The area of a square is 169 cm². Find its:
(i) one side
(ii) perimeter
Solution:
Let each side of the square be x cm.
Its area = x² = 169 (given)
x = √169
x = 13 cm
(i) Thus, side of the square = 13 cm
(ii) Again perimeter = 4 (side) = 4 x 13 = 52 cm

Question 10.
The length of a rectangle is 16 cm and its perimeter is equal to the perimeter of a square with side 12.5 cm. Find the area of the rectangle.
Solution:
Length of the rectangle = 16 cm
Let its breadth be x cm
Perimeter = 2 (16 + x) = 32 + 2x
Also perimeter = 4(12.5) = 50 cm.
According to statement,
32 + 2x = 50
=> 2x = 50 – 32 = 18
=> x = 9
Breadth of the rectangle = 9 cm.
Area of the rectangle (l x b)= 16 x 9 = 144 cm²

Question 11.
The perimeter of a square is numerically equal to its area. Find its area.
Solution:
Let each side of the square be x cm.
Its perimeter = 4x,
Area =x²
By the given condition 4x = x²
=> x² – 4x = 0
=> x (x – 4) = 0
=> x = 4 [x ≠ 0]
Area = x² = (4)² = 4 x 4 = 16 sq.units.

Question 12.
Each side of a rectangle is doubled. Find the ratio between :
(i) perimeters of the original rectangle and the resulting rectangle.
(ii) areas of the original rectangle and the resulting rectangle.
Solution:
Let length of the rectangle = x
and breadth of the rectangle = y
(i) Perimeter P = 2(x + y)
Again, new length = 2x
New breadth = 2y

Question 13.
In each of the following cases ABCD is a square and PQRS is a rectangle. Find, in each case, the area of the shaded portion.
(All measurements are in metre).

Solution:
(i) Area of the shaded portion
= Area of the rectangle PQRS – Area of square ABCD
= 3.2 x 1.8 – (1.4)² (∵ PQ = 3.2 and PS = 1.8) Side of square AB = 1.4
= 5.76 – 1.96 = 3.80 = 3.8 m²
(ii) Area of the shaded portion = Area of square ABCD – Area of rectangle PQRS
= 6 x 6 – (3.6) (4.8) = 36 – 17.28 = 18.72 m²

Question 14.
A path of uniform width, 3 m, runs around the outside of a square field of side 21 m. Find the area of the path.
Solution:
According to the given information the figure will be as shown alongside.
Clearly, length of the square field excluding path = 21 m.

Area of the square side excluding the path = 21 x 21 = 441 m²
Again, length of the square field including the path = 21 + 3 + 3 = 27 m
Area of the square field including the path = 27 x 27 = 729 m²
Area of the path = 729 – 441 = 288 m²

Question 15.
A path of uniform width, 2.5 m, runs around the inside of a rectangular field 30 m by 27 m. Find the area of the path.
Solution:
According to the given statement the figure will be as shown alongside.

Clearly, the length of the rectangular field including the path = 30 m.
Breadth = 27 m.
Its Area = 30 x 27 = 810 m²
Width of the path = 2.5 m
Length of the rectangular field including the path = 30 – 2.5 – 2.5 = 25 m.
Breadth = 27 – 2.5 – 2.5 = 22m
Area of the rectangular field including the path = 25 x 22 = 550 m²
Hence, area of the path = 810 – 550 = 260 m².

Question 16.
The length of a hall is 18 m and its width is 13.5 m. Find the least number of square tiles, each of side 25 cm, required to cover the floor of the hall,
(i) without leaving any margin.
(ii) leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of Rs. 6 per tile
Solution:
(i) Length of hall (l) = 18 m and breadth (b) = 13.5 m

Question 17.
A rectangular field is 30 m in length and 22m in width. Two mutually perpendicular roads, each 2.5 m wide, are drawn inside the field so that one road is parallel to the length of the field and the other road is parallel to its width. Calculate the area of the crossroads.
Solution:
Length of rectangular field (l) = 30 m and breadth (b) = 22m
width of parallel roads perpendicular to each other inside the field =2.5m

Area of cross roads = width of roads (Length + breadth) – area of middle square
= 2.5 (30 + 22) – (2.5)²
= 2.5 x 52 – 6.25 m²
= (130 – 6.25) m = 123.75 m²

Question 18.
The length and the breadth of a rectangular field are in the ratio 5 : 4 and its area is 3380 m². Find the cost of fencing it at the rate of ₹75 per m.
Solution:
Ratio in length and breadth = 5 : 4
Area of rectangular field = 3380 m²
Let length = 5x and breadth = 4x
5x x 4x = 3380
=> 20x²= 3380
x² = 3380/20 = 169 = (13)2
x = 13
Length = 13 x 5 = 65 m
Breadth =13 x 4 = 52 m
Perimeter = (l + b) = 2 x (65 + 52) m = 2 x 117 = 234 m
Rate of fencing = ₹ 75 per m
Total cost = 234 x 75 = ₹ 17550

Question 19.
The length and the breadth of a conference hall are in the ratio 7 : 4 and its perimeter is 110 m. Find:
(i) area of the floor of the hall.
(ii) number of tiles, each a rectangle of size 25 cm x 20 cm, required for flooring of the hall.
(iii) the cost of the tiles at the rate of ₹ 1,400 per hundred tiles.
Solution:
Ratio in length and breadth = 7 : 4
Perimeter = 110 m

Area of Trapezium and a Polygon Exercise 20C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape.
If the width DC, of the swimming pool is 6.4cm, depth (AD) at the shallow end is 80 cm and depth (BC) at deepest end is 2.4m, find Its area of the cross-section.

Solution:
Area of the cross-section = Area of trapezium ABCD

= 1024 cm² or = 10.24 sq.m.

Question 2.
The parallel sides of a trapezium are in the ratio 3 : 4. If the distance between the parallel sides is 9 dm and its area is 126 dm² ; find the lengths of its parallel sides.
Solution:

Let parallel sides of trapezium be
a = 3x
b = 4x
Distance between parallel sides, h = 9 dm
area of trapezium = 126 dm²

Question 3.
The two parallel sides and the distance between them are in the ratio 3 : 4 : 2. If the area of the trapezium is 175 cm², find its height.
Solution:
Let the two parallel sides and the distance between them be 3x, 4x, 2x cm respectively
Area = \(\frac { 1 }{ 2 }\) (sum of parallel sides) x (distance between parallel sides)
= \(\frac { 1 }{ 2 }\) (3x + 4x) x 2x = 175 (given)
=> 7x x x = 175
=> 7x² = 175
=> x² = 25
=> x = 5
Height i.e. distance between parallel sides = 2x = 2 x 5 = 10 cm

Question 4.
A parallelogram has sides of 15 cm and 12 cm; if the distance between the 15 cm sides is 6 cm; find the distance between 12 cm sides.
Solution:

BQ = \(\frac { 15 }{ 2 }\) = 7.5 cm

Question 5.
A parallelogram has sides of 20 cm and 30 cm. If the distance between its shorter sides is 15 cm; find the distance between the longer sides.
Solution:

Question 6.
The adjacent sides of a parallelogram are 21 cm and 28 cm. If its one diagonal is 35 cm; find the area of the parallelogram.
Solution:

Question 7.
The diagonals of a rhombus are 18 cm and 24 cm. Find:
(i) its area ;
(ii) length of its sides.
(iii) its perimeter;
Solution:
Diagonal of rhombus are 18 cm and 24 cm.
area of rhombus = \(\frac { 1 }{ 2 }\) x Product of diagonals
= \(\frac { 1 }{ 2 }\) x 18 x 24
= 216 cm²
(ii) Diagonals of rhombus bisect each other at right angles.

Question 8.
The perimeter of a rhombus is 40 cm. If one diagonal is 16 cm; find :
(i) its another diagonal
(ii) area
Solution:
(i) Perimeter of rhombus = 40 cm
side = \(\frac { 1 }{ 4 }\) x 40 = 10 cm
One diagonal = 16 cm
Diagonals of rhombus bisect each other at right angles.

Question 9.
Each side of a rhombus is 18 cm. If the distance between two parallel sides is 12 cm, find its area.
Solution:
Each side of the rhombus = 18 cm
base of the rhombus = 18 cm
Distance between two parallel sides = 12 cm
Height = 12 cm
Area of the rhombus = base x height = 18 x 12 = 216 cm²

Question 10.
The length of the diagonals of a rhombus is in the ratio 4 : 3. If its area is 384 cm², find its side.
Solution:
Let the lengths of the diagonals of rhombus are 4x, 3x.

Question 11.
A thin metal iron-sheet is rhombus in shape, with each side 10 m. If one of its diagonals is 16 m, find the cost of painting its both sides at the rate of ₹ 6 per m².
Also, find the distance between the opposite sides of this rhombus.
Solution:
Side of rhombus shaped iron sheet = 10 m and one diagonals (AC) = 16 m
Join BD diagonal which bisects AC at O
The diagonals of a rhombus bisect each other at right angle

Question 12.
The area of a trapezium is 279 sq.cm and the distance between its two parallel sides is 18 cm. If one of its parallel sides is longer than the other side by 5 cm, find the lengths of its parallel sides.
Solution:
Area of trapezium = 279 sq.cm
Distance between two parallel lines (h) = 18 cm

Question 13.
The area of a rhombus is equal to the area of a triangle. If base of ∆ is 24 cm, its corresponding altitude is 16 cm and one of the diagonals of the rhombus is 19.2 cm. Find its other diagonal.
Solution:
Area of a rhombus = Area of a triangle Base of triangle = 24 cm
and altitude = 16 cm

Question 14.
Find the area of the trapezium ABCD in which AB//DC, AB = 18 cm, ∠B = ∠C = 90°, CD = 12 cm and AD = 10 cm.
Solution:
In trapezium ABCD,

Area of Trapezium and a Polygon Exercise 20D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the radius and area of a circle, whose circumference is :
(i) 132 cm
(ii) 22 m
Solution:
(i) Circumference of circle = 132 cm
2πr = 132
2 x 22/7 x r = 132

Question 2.
Find the radius and circumference of a circle, whose area is :
(i) 154 cm²
(ii) 6.16 m²
Solution:

Question 3.
The circumference of a circular table is 88 m. Find its area.
Solution:
Circumference of circle = 88 m
2πr = 88 m

Question 4.
The area of a circle is 1386 sq.cm ; find its circumference.
Solution:
Area of circle = 1386 cm²
πr² = 1386

Question 5.
Find the area of a flat circular ring formed by two concentric circles (circles with same centre) whose radii are 9 cm and 5 cm.
Solution:

Question 6.
Find the area of the shaded portion in each of the following diagrams :

Solution:

Question 7.
The radii of the inner and outer circumferences of a circular running track are 63 m and 70 m respectively. Find :
(i) the area of the track ;
(it) the difference between the lengths of the two circumferences of the track.
Solution:

= 2x 22/7 x 63 = 396 m
Difference between lengths of two circumferences = 440 – 396 = 44 m
Hence (i) 2926 m² (ii) 44 m

Question 8.
A circular field cf radius 105 m has a circular path of uniform width of 5 m along and inside its boundary. Find the area of the path.
Solution:

Question 9.
There is a path of uniform width 7 m round and outside a circular garden of diameter 210 m. Find the area of the path.
Solution:

Question 10.
A wire, when bent in the form of a square encloses an area of 484 cm². Find :
(i) one side of the square ;
(ii) length of the wire ;
(iii) the largest area enclosed; if the same wire is bent to form a circle.
Solution:
(i) Area of Square = 484 cm²
Side of Square = √Area = √484 = 22 cm
(ii) Perimeter, i.e. length of wire = 4 x 22 = 88 cm
(iii) Circumference of circle = 88 cm
2πr = 88

Question 11.
A wire, when bent in the form of a square; encloses an area of 196 cm². If the same wire is bent to form a circle; find the area of the circle.
Solution:
Area of Square = 196 cm²
Side of Square = √Area = √196 = 14 cm
Perimeter of Square = 4 x 14 cm
i.e. length of wire = 56 cm
Circumference of circle = 56 cm
2πr = 56

= 2744/11
249.45 cm²

Question 12.
The radius of a circular wheel is 42 cm. Find the distance travelled by it in :
(i) 1 revolution ;
(ii) 50 revolutions ;
(iii) 200 revolutions ;
Solution:
(i) Radius of wheel, r = 42 cm
Circumference i.e. distance travelled in 1 revolution = 2πr = 2 x 22/7 x 42 = 264 cm
(ii) Distance travelled in 50 revolutions = 264 x 50 = 13200 cm = 132 m
(iii) Distance travelled in 200 revolutions = 264 x 200 = 52800 cm = 528 m
Hence (i) 264 cm (ii) 132 m (iii) 528 m

Question 13.
The diameter of a wheel is 0.70 m. Find the distance covered by it in 500 revolutions. If the wheel takes 5 minutes to make 500 revolutions; find its speed in :
(i) m/s
(ii) km/hr.
Solution:
Diameter = 0.70 m
Radius, r = 0.35 m
Distance covered in 1 revolution, i.e. circumference = 2πr = 2 x 22/7 x 0.35 = 2.20 m
Distance covered in 500 revolutions = 2.20 x 500 = 1100 m
Time taken = 5 minutes = 5 x 60 = 300 sec.

Question 14.
A bicycle wheel, diameter 56 cm, is making 45 revolutions in every 10 seconds. At what speed in kilometre per hour is the bicycle travelling ?
Solution:

Question 15.
A roller has a diameter of 1.4 m. Find :
(i) its circumference ;
(ii) the number of revolutions it makes while travelling 61.6 m.
Solution:
Diameter = 1.4 m

Question 16.
Find the area of the circle, length of whose circumference is equal to the sum of the lengths of the circumferences with radii 15 cm and 13 cm.
Solution:
In a circle
Circumference = Sum of circumferences of two circle of radii 15 cm and 13 cm
Now circumference of first smaller circle = 2πr

Question 17.
A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed.
Solution:
Length of wire = 108 cm
Let r be the radius of the semicircle
πr+ 2r = 108

Question 18.
In the following figure, a rectangle ABCD enclosed three circles. If BC = 14 cm, find the area of the shaded portion (Take π = 22/7)

Solution:
In rectangle ABCD, BC = 14 cm

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

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Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square of :
(i) 59
(ii) 63
(iii) 15
Solution:
(i) Square of 59= 59 x 59 = 3481
(ii) Square of 6.3 = 6.3 x 6.3 = 39.69
(iii) Square of 15 = 15 x 15 = 225

Question 2.
By splitting into prime factors, find the square root of :
(i) 11025
(if) 396900
(iii) 194481
Solution:

Question 3.
(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be mutliplied so that the product is a perfect square?
Solution:

On grouping the prime factors of 2592 as shown; on factor i.e. 2 is left which cannot be paired with equal factor.

The given number should be multiplied by 2 to make the given number a perfect square.

On grouping the prime factors of 12748 as shown; one factor i.e. 3187 is left which cannot be paired with equal factor.

The given number should be multiplied by 3187.

Question 4.
Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.
Solution:

Question 5.
Find the square root of :
(i) 0.1764
(ii) \(96\frac { 1 }{ 25 }\)
(iii) 0.0169
Solution:

Question 6.
Evaluate


Solution:

Question 7.

Solution:

Question 8.
A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all ₹ 1,296
Solution:
Let the number of days he had spent = x
Number of rupees spent in each day = x
Total money spent = x x x = x² = 1,296 (given)

Question 9.
Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.
Solution:
Total number of students = 745
Students left after standing in arrangement = 16
No. of students who were to be arranged = 745 – 16 = 729
The number of rows = no. of students in each row
No. of rows = √729

Question 10.
13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.
Solution:

Question 11.
Find the smallest perfect square divisible by 3, 4, 5 and 6.
Solution:
L.C.M. of 3, 4, 5, 6 = 2 x 2 x 3 x 5 = 60

in which 3 and 5 are not in pairs L.C.M. = 2 x 3 x 2 x 5 = 60
We should multiple it by 3 x 5 i.e. by 15
Required perfect square = 60 x 15 = 900

Question 12.
If √784 = 28, find the value of:
(i) √7.84 + √78400
(ii) √0.0784 + √0.000784
Solution:

Squares and Square Roots Exercise 3B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Solution:

Question 2.
Find the square root of:
(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Solution:

Question 3.
Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Solution:

Required square root = 0.78 upto two places of decimals.

Question 4.
Find the square root of each of the following correct to two decimal places:
(i) \(3\frac { 4 }{ 5 }\)
(ii) \(6\frac { 7 }{ 8 }\)
Solution:

Question 5.
For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Solution:

Question 6.
For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Solution:
(i) 511
Taking square root of 511, we find that 27 has been left We see that 511 is greater than (22)²

On adding the required number to 511, we get (23)² i.e., 529
So, the required number = 529 – 511 = 18
(ii) 7172
Taking square root of 7172, we find that 116 has been left
We see that 7172 is greater than (84)²

Taking square root of 55078, we find that 322 has been left
We see that 55078 is greater than (234)²
On adding the required number to 55078, we get (235)² i.e., 55225
Required number = 55225 – 55078 = 147

Question 7.
Find the square root of 7 correct to two decimal places; then use it to find the value of \(\sqrt { \frac { 4+\sqrt { 7 } }{ 4-\sqrt { 7 } } }\) correct to three significant digits.
Solution:
√7 = 2.645 = 2.65

Question 8.
Find the value of √5 correct to 2 decimal places; then use it to find the square root of \(\sqrt { \frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } }\) correct to 2 significant digits.
Solution:

Question 9.
Find the square root of:
(i) \(\frac { 1764 }{ 2809 }\)
(ii) \(\frac { 507 }{ 4107 }\)
(iii) \(\sqrt { 108\times 2028 }\)
(iv) 0.01 + √0.0064
Solution:

Question 10.
Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.
Solution:
Square root of 7.832

√7.832 = 2.80 upto two decimal places
= 2.8 upto two significant places

Question 11.
Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.
Solution:
Clearly, if 49 is subtracted from 1205, the number will be a perfect square.

Question 12.
Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Solution:

Question 13.
Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Solution:
Clearly; if 12 is subtracted from 2037, the remainder will be a perfect square.

Question 14.
Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Solution:

Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699
Solution:
We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9
So, the following numbers can be squares : 3051, 5684, and 50699 i.e., (i), (iii), and (v)

Question 2.
Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64
Solution:
The square of the following numbers will have 1 at their units place as (1)² = 1, (9)² = 81
81 and 139 i.e., (i) and (iii)

Question 3.
Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Solution:
The square of the following numbers will not have 1 at their units place : as only (1)² = 1, (9)² = 81 have 1 at then units place
322, 572, 2652 i.e., (i), (ii) and (v)

Question 4.
Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98
Solution:
The squares of the following numbers, Will not have 6 at their units place as only (4)² = 16, (6)² = 36 has but its units place 35, 23 and 98 i.e., (i), (ii), and (v)

Question 5.
Which of the following numbers will have 6 at their unit’s place :
(i) 262
(ii) 492
(iii) 342
(iv) 432
(v) 2442
Solution:
The following numbers have 6 at their units place as (4)² = 16, (6)² = 36 has 6 at their units place 262, 342, 2442 i.e., (i), (iii) and (v)

Question 6.
If a number ends with 3 zeroes, how many zeroes will its square have ?
Solution:
We know that if a number ends with n zeros, then its square will have 2n zeroes at their ends
A number ends with 3 zeroes, then its square will have 3 x 2 = 6 zeroes

Question 7.
If the square of a number ends with 10 zeroes, how many zeroes will the number have ?
Solution:
We know that if a number ends with n zeros Then its square will have 2n zeroes Conversely, if square of a number have 2n zeros at their ends then the number will have n zeroes
The square of a number ends 10 zeroes, then the number will have \(\frac { 10 }{ 2 }\) = 5 zeroes

Question 8.
Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Solution:
No, it is not possible for the square of a number, to have 5 zeroes which is odd because the number of zeros of the square must be 2n zeroes i.e., even number of zeroes.

Question 9.
Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Solution:
A number having 2,3,7 or 8 at the unit place is never a perfect square.

Question 10.
State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(iii) 76
(iv) 75
Solution:
(i) 23 – odd
(ii) 54 – even
(iii) 76 – odd
(iv) 75 – even

Question 11.
Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Solution:
No, number has an even number of zeroes.

Question 12.
Evaluate:
(i) 37² – 36²
(ii) 85² – 84²
(iii) 101² – 100²
Solution:

Question 13.
Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53
Solution:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23
= Sum of first 12 odd natural numbers = 122 = 144
(ii) 1+3 + 5 + 7 + 9 + ……….. + 39 + 41
= Sum of first 21 odd natural numbers = 212 = 441
(iii) 1 + 3 + 5 + 7 + 9 + ……………. + 51 + 53
= Sum of first 27 odd natural number = 272 = 729

Question 14.
Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Solution:
The three sets of Pythagorean triplets such that each set has numbers less than 30 are 3, 4 and 5; 6, 8 and 10; 5, 12 and 13
Proof:

Selina Concise Mathematics class 7 ICSE Solutions – Profit, Loss and Discount

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

1. The Cost Price (C.P.) of an article is the price at which the article is bought.
2. The Selling Price (S.P.) of an article is the price at which the article is sold.
3. If Selling Price of an article is more than its cost price ; it is sold at a profit (gain)
   Profit = Selling Price-Cost Price
   i.e., Profit (gain) = S.P. – C.P. and S.P. = C.P. + Gain
4. If Selling Price of an article is less than its cost price ; it is sold at a loss.
   Loss = Cost Price – Selling Price
   i.e., Loss = C.P. – S.P. and S.P. = C.P. — Loss
5. Profit percent and loss percent are always calculated on cost price (C.P.) only.
   i.e., (i) Profit % = \(\frac { Profit }{ C.P. }\) x 100% and
   (ii) Loss % = \(\frac {Loss }{ C.P. }\) x 100%
6. Selling Price = Marked price – Discount
   i.e., S.P. = M.P.—(piscount
   Note : (i) Discount is calculated on marked price (M.P.)
   (ii) Marked price is also written as List price.

EXERCISE 9 (A)

Question 1.
Find the gain or loss percent, if
(i) C.P. = Rs. 200 and S.P.: = Rs. 224
(ii) C.P. = Rs. 450 and S.P. = Rs. 400
(iii) C.P. = Rs. 550 and gain = Rs . 22
(iv) CP. = Rs. 216 and loss = Rs. 72
(v) S.P. = Rs. 500 and loss : = Rs. 100
(vi) S.P. = Rs. 12 and profit = Rs. 4
(vii) C.P. = Rs. 5 and gain = 60 P

Solution:

Question 2.
Find the selling price, if:
(i) C.P. = Rs. 500 and gain = 25%
(ii) C.P. = Rs. 60 and loss = 12 1/2%
(iii) C.P. = Rs. 150 and loss = 20%
(iv) C.P. = Rs. 80 and gain = 2.5%

Solution:

Question 3.
Rohit bought a tape-recorder for Rs. 1,500 and sold it for Rs. 1,800. Calculate his profit or loss percent.

Solution:

Question 4.
An article bought for Rs. 350 is sold at a profit of 20%. Find its selling price.

Solution:

Question 5.
An old machine is bought for Rs. 1,400 and is sold at a loss of 15%. Find its selling price.

Solution:

Question 6.
Oranges are bought at 5 for Rs. 10 and sold at 6 for Rs. 15. Find profit or loss as percent.

Solution:

Question 7.
A certain number of articles are bought at 3 for Rs. 150 and all of them are sold at 4 for Rs. 180. Find the loss or gain as percent.

Solution:

Question 8.
A vendor bought 120 sweets at 20 p each. In his house, 18 were consumed and he sold the remaining at 30 p each. Find his profit or loss as percent.

Solution:

Question 9.
The cost price of an article is Rs. 1,200 and selling price is \(\frac { 5 }{ 4 }\) times of its cost price. Find:
(i) selling price of the article
(ii) profit or loss as percent.

Solution:

Question 10.
The selling price of an article is Rs. 1,200 and cost price is \(\frac { 5 }{ 4 }\) times of its selling price,
find :
(i) cost price of the article ;
(ii) profit or loss as percent.

Solution:

EXERCISE 9 (B)

Question 1.
Find the cost price, if:
(i) S.P. = Rs. 21 and gain = 5%
(ii) S.P. = Rs. 22 and loss = 12%
(iii) S.P. = Rs. 340 and gain = Rs. 20
(iv) S.P. = Rs. 200 and loss = Rs. 50
(v) S.P. = Re. 1 and loss = 5 p.

Solution:

Question 2.
By selling an article for Rs. 810, a loss of 10 percent is suffered. Find its cost price.

Solution:

Question 3.
By selling a scooter for Rs. 9,200, a man gains 15%. Find the cost price of the scooter.

Solution:

Question 4.
On selling an article for Rs. 2,640, a profit of 10 percent is made. Find
(i) cost price of the article
(ii) new selling price of it, in order to gain 15%

Solution:

Question 5.
A T.V. set is sold for Rs. 6800 at a loss of 15%. Find
(i)cost price of the T.V. set.
(ii)new selling price of it, in order to gain 12%

Solution:

Question 6.
A fruit seller bought mangoes at Rs. 90 per dozen and sold them at a loss of 8 percent. How much will a customer pay for.
(i) one mango
(ii) 40 mangoes

Solution:

Question 7.
By selling two transistors for Rs. 00 each, a shopkeeper gains 20 percent on one transistor and loses 20 percent on the other.
Find :
(i) C.P. of each transistor
(ii) total C.P. and total S.P. of both the transistors
(iii) profit or loss percent on the whole.

Solution:

Question 8.
Mangoes are bought at 20 for Rs. 60. If
they are sold at 33\(\frac { 1 }{ 3 }\) percent profit.
Find:
(i) selling price of each mango.
(ii) S.P. of 8 mangoes.

Solution:

Question 9.
Find the cost price of an article, which is sold for Rs. 4050 at a loss of 10%. Also, find the new selling price of the article which must give a profit of 8%.

Solution:

Question 10.
By selling an article for ₹825, a man loses \(\frac { 1 }{ 3 }\) equal to j of its selling price.
Find :
(i) the cost price of the article,
(ii) the profit percent or the loss percent made, if the same article is sold for ₹1265.

Solution:

Question 11.
Find the loss or gain as percent, if the C.P. of 10 articles, all of the same kind, is equal to S.P. of 8 articles.

Solution:

Question 12.
Find the loss or gain as percent, if the C.P. of 8 articles, all of the same kind, is equal to S.P. of 10 articles.

Solution:

Question 13.
The cost price of an article is 96% of its selling price. Find the loss or the gain as percent on the whole.

Solution:

Question 14.
The selling price of an article is 96% of its cost price. Find the loss or the gain as percent on the whole.

Solution:

Question 15.
Hundred oranges are bought for ₹350 and all of them are sold at the rate of ₹48 per dozen. Find the profit percent or loss percent made.

Solution:

Question 16.
Oranges are bought at 100 for ?80 and all of them are sold at ₹80 for ₹100. Find the loss or gain as percent in this transaction.

Solution:

Question 17.
An article is bought for ₹5,700 and ₹1,300 is spent on its repairing, transportion, etc. For how much should this article be sold in order to gain 20% on the whole.

Solution:

EXERCISE 9 (C)

Question 1.
A machine is marked at ₹5000 and is sold at a discount of 10%. Find the selling price of the machine.

Solution:

Question 2.
shopkeeper marked a dinner set for ₹1000. He sold it at ₹900, what percent discount did he give ?

Solution:

Question 3.
A pair of shoes marked at ₹320, are sold at a discount of 15 percent.
Find :
(i) discount
(ii) selling price of the shoes.

Solution:

Question 4.
The list price of an article is ₹450 and it is sold for ₹360.
Find :
(i) discount
(ii) discount percent

Solution:

Question 5.
A shopkeeper buys an article for₹300. He increases its price by 20% and then gives 10% discount on the new price. Find:
(i) the new price (marked price) of the article.
(ii) the discount given by the shopkeeper.
(iii) the selling price.
(iv) profit percent made by the shopkeeper.

Solution:

Question 6.
A car is marked at Rs. 50,000. The dealer gives 5% discount on first Rs. 20,000 and 2% discount on the remaining Rs. 30,000.
Find :
(i) the total discount.
(ii) the price charged by the dealer.

Solution:

Question 7.
A dealer buys a T.V. set for Rs. 2500. He marks it at Rs. 3,200 and then gives a discount of 10% on it.
Find :
(i) the selling price of the T.V. set
(ii) the profit percent made by the dealer.

Solution:

Question 8.
A sells his goods at 15% discount. Find the price of an article which is sold for Rs. 680.

Solution:

Question 9.
A shopkeeper allows 20% discount on the marked price of his articles. Find the marked price of an article for which he charges Rs. 560.

Solution:

Question 10.
An article is bought for Rs. 1,200 and Rs. 100 is spent on its transportation, etc.
Find :
(i) the total C.P. of the article.
(ii) the selling price of it in order to gain 20% on the whole.

Solution:

Question 11.
40 pens are bought at 4 for Rs. 50 and all of them are sold at 5 for Rs. 80
Find :
(i) C.P. of one pen.
(ii) S/P. of one pen.
(iii) Profit made by selling one pen.
(iv) Profit percent made by selling one pen.
(v) C.P. of 40 pens
(vi) S.P. of 40 pens.
(vii) Profit made by selling 40 pens.
(viii) Profit percent made by selling 40 pens. Are the results of parts (iv) and (viii) same? What conclusion do you draw from the above result ?

Solution:

Question 12.
The C.P. of 5 identical articles is equal to S.P. of 4 articles. Calculate the profit percent or loss percent made if all the articles bought are sold.

Solution:

Question 13.
The C.P. of 8 pens is same as S.P. of 10 pens. Calculate the profit or loss percent made, if all the pens bought are considered to be sold

Solution:

Question 14.
A certain number of articles are bought at Rs. 450 per dozen and all of them are sold at a profit of 20%. Find the S.P. of:
(i) one article
(ii)seven articles.

Solution:

Question 15.
An article is marked 60% above the cost price and sold at 20% discount. Find the profit percent made.

Solution:

Selina Concise Mathematics class 7 ICSE Solutions – Fundamental Concepts (Including Fundamental Operations)

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

1. Constants and Variables : The numbers which has fixed value is called constant and same at English alphabet which can be assigned any value according to the requirement is called variables.
2. Term : A term is a number, (constant), a variable or a combination of numbers and variables.
3. Algebraic Expression : An algebraic expression is a collection of one or more terms, which are separated from each other by addition (+) or subtraction (-) signs.
4. Types of algebraic expressions :
   (i) Monomial : It has only one term
   (ii) Binomial : It has two terms
   (iii) Trinomial : It has three terms
   (iv) Multinomial : It has more than three terms
   (v) Polynomial : It has two or more than two terms.
   Note : An expression of the type \(\frac { 2 }{ 5 }\) does not form a monomial unless JC is not equal to zero.
5. Product: When two or more quantities are multiplied together, the result is called their product.
6. Factors : Each of the quantities (numbers or variables) multiplied together to form a term is called a factor of the given term.
7. Co-efficient: In a monomial, any factor or group of factors of a term is called the co-efficient of the remaining part of the monomial.
8. Degree of a monomial: The degree of a monomial is the exponent of its variable or the sum of the exponents of its variables.
9. Degree of a polynomial: The degree of a polynomial is the degree of its highest degree term.
10. Like and unlike terms : Terms having the same literal co-efficients or alphabetic letters are called like terms ; whereas the terms with different literal co-efficients are called unlike terms.
11. Addition and subtraction : Addition and subtraction of only like terms is possible by adding or subtracting the numerical co-efficients.
12. Multiplication and division :
    (A) Multiplication :
    (i) Multiplications of monomials.
    (a) Multiply the numerical co-efficient together
    (ii) Multiply the literal co-efficients separately together.
    (iii) Combine the like terms.
    (B) Division :
    (i) Dividing a polynomial by a monomial Divide each term of the polynomial by monomial and simplify each fractions.
    (ii) While dividing one polynomial by another polynomial ; arrange the terms of both the dividend and the divisior both in descending or in ascending order of their powers and then divide.

SOME IMPORTANT POINTS

TYPES OF BRACKETS:
The name of different types of brackets and the order in which they are removed is shown below:
(a) ____ ; Bar (Vinculum) bracket
(b) ( ); Circular bracket .
(c) { } ; Curly bracket and then
(d) [ ]; square bracket

EXERCISE 11 (A)

Question 1.
Separate constant terms and variable terms from tile following :

Solution:

Constant is only 8 others are variables

Question 2.
Constant is only 8 others are variables
(i) 2x ÷ 15
(ii) ax+ 9
(iii) 3x² × 5x
(iv) 5 + 2a-3b
(v) 2y – \(\frac { 7 }{ 3 }\) z÷x
(vi) 3p x q ÷ z
(vii) 12z ÷ 5x + 4
(viii) 12 – 5z – 4
(ix) a³ – 3ab² x c

Answer:

Question 3.
Write the coefficient of:
(i) xy in – 3axy
(ii) z² in p²yz²
(iii) mn in -mn
(iv) 15 in – 15p²

Solution:
(i) Co-efficient of xy in – 3 axy = – 3a
(ii) Co-efficient of z² in p²yz² = p²y
(iii) Co-efficient of mn in – mn = – 1
(iv) Co-efficient of 15 in – 15p² is -p²

Question 4.
For each of the following monomials, write its degree :
(i) 7y
(ii) – x²y
(iii) xy²z
(iv) – 9y²z³
(v) 3 m³n⁴
(vi) – 2p²q³r⁴

Solution:
(i) Degree of 7y = 1
(ii) Degree of – x²y = 2+1=3
(iii) Degree of xy²z = 1 + 2 + 1 = 4
(iv) Degree of – 9y²z³ = 2 + 3 = 5
(v) Degree of 3m³n⁴ = 3 + 4 = 7
(vi) Degree of – 2p²q³r⁴ = 2 + 3 + 4 = 9

Question 5.
Write the degree of each of the following polynomials :
(i) 3y³-x²y² + 4x
(ii) p³q² – 6p²q⁵ + p⁴q⁴
(iii) – 8mn⁶+ 5m³n
(iv) 7 – 3x²y + y²
(v) 3x – 15
(vi) 2y²z + 9yz³

Solution:
(i) The degree of 3y³ – x²y²+ 4x is 4 as x²
y² is the term which has highest degree.
(ii) The degree of p³q² – 6p²q⁵-p⁴q⁴ is 8 as p⁴ q⁴ is the term which has highest degree.
(iii) The degree of- 8mn⁶ + 5m³n is 7 as – 8mx⁶ is the term which has the highest degree.
(iv) The degree of 7 – 3x² y + y² is 3 as – 3x²y is the term which has the highest degree.
(v) The degree of 3x – 15 is 1 as 3x is the term which is highest degree.
(vi) The degree of 2y² z + 9y z³ is 4 as 9yz³ has the highest degree.

Question 6.
Group the like term together :
(i) 9x², xy, – 3x², x² and – 2xy
(ii) ab, – a²b, – 3ab, 5a²b and – 8a²b
(iii) 7p, 8pq, – 5pq – 2p and 3p

Solution:
(i) 9x², – 3x² and x² are like terms
xy and – 2xy are like terms
(ii) ab, – 3ab, are like terms,
– a²b, 5a²b, – 8a²b are like terms
(iii) 7p, – 2p and 3p are like terms,
8pq, – 5pq are like terms.

Question 7.
Write numerical co-efficient of each of the followings :
(i) y
(ii) -y
(iii) 2x²y
(iv) – 8xy³
(v) 3py²
(vi) – 9a²b³

Solution:
(i) Co-efficient of y = 1
(ii) Co-efficient of-y = – 1
(iii) Co-efficient of 2x2y is = 2
(iv) Co-efficient of – 8xy3 is = – 8
(v) Co-efficient of Ipy2 is = 3
(vi) Co-efficient of – 9a2b3 is = – 9

Question 8.
In -5x³y²z⁴; write the coefficient of:
(i) z²
(ii) y²
(iii) yz²
(iv) x³y
(v) -xy²
(vi) -5xy²z
Also, write the degree of the given algebraic expression.

Solution:
-5x³y²z⁴
(i) Co-efficient of z2 is -5x³y²z²
(ii) Co-efficient of y2 is -5x³z⁴
(iii) Co-efficient of yz² is -5x³yz²
(iv) Co-efficient of x³y is -5yz⁴
(v) Co-efficient of -xy² is 5x²z⁴
(vi) Co-efficient of -5xy²z is x²z³
Degree of the given expression is 3 + 2 + 4 = 9

EXERCISE 11 (B)

Question 1.
Fill in the blanks :
(i) 8x + 5x = ………
(ii) 8x – 5x =……..
(iii) 6xy² + 9xy² =……..
(iv) 6xy² – 9xy² = ………
(v) The sum of 8a, 6a and 5b = ……..
(vi) The addition of 5, 7xy, 6 and 3xy = …………
(vii) 4a + 3b – 7a + 4b = ……….
(viii) – 15x + 13x + 8 = ………
(ix) 6x²y + 13xy² – 4x²y + 2xy² = ……..
(x) 16x² – 9x² = and 25xy² – 17xy²=………

Solution :

Question 2.
Add :
(i)- 9x, 3x and 4x
(ii) 23y², 8y² and – 12y²
(iii) 18pq – 15pq and 3pq

Solution:

Question 3.
Simplify :
(i) 3m + 12m – 5m
(ii) 7n² – 9n² + 3n²
(iii) 25zy—8zy—6zy
(iv) -5ax² + 7ax² – 12ax²
(v) – 16am + 4mx + 4am – 15mx + 5am

Solution:

Question 4.
Add : 
(i) a + i and 2a + 3b
(ii) 2x + y and 3x – 4y
(iii)- 3a + 2b and 3a + b
(iv) 4 + x, 5 – 2x and 6x

Solution:

Question 5.
Find the sum of:
(i) 3x + 8y + 7z, 6y + 4z- 2x and 3y – 4x + 6z
(ii) 3a + 5b + 2c, 2a + 3b-c and a + b + c.
(iii) 4x²+ 8xy – 2y² and 8xy – 5y² + x²
(iv) 9x² – 6x + 7, 5 – 4x and 6 – 3x²
(v) 5x² – 2xy + 3y² and – 2x² + 5xy + 9y²
and 3x² -xy- 4y²
(vi) a² + b² + 2ab, 2b² + c² + 2bc
and 4c²-a² + 2ac
(vii) 9ax – 6bx + 8, 4ax + 8bx – 7
and – 6ax – 46x – 3
(viii) abc + 2 ba + 3 ac, 4ca – 4ab + 2 bca
and 2ab – 3abc – 6ac
(ix) 4a² + 5b² – 6ab, 3ab, 6a² – 2b²
and 4b² – 5 ab
(x) x² + x – 2, 2x – 3x² + 5 and 2x² – 5x + 7
(xi) 4x³ + 2x² – x + 1, 2x³ – 5x²– 3x + 6, x² + 8 and 5x³ – 7x

Solution:

Question 6.
Find the sum of:
(i) x and 3y
(ii) -2a and +5
(iii) – 4x² and +7x
(iv) +4a and -7b
(v) x³+3x²y and 2y²
(vi) 11 and -by

Solution:

Question 7.
The sides of a triangle are 2x + 3y, x + 5y and 7x – 2y, find its perimeter.

Solution:

Question 8.
The two adjacent sides of a rectangle are 6a + 96 and 8a – 46. Find its, perimeter.

Solution

Question 9.
Subtract the second expression from the first:

Solution:

Question 10.
Subtract:

Solution:

Question 11.
Subtract – 5a² – 3a + 1 from the sum of 4a² + 3 – 8a and 9a – 7.

Solution:

Question 12.
By how much does 8x³ – 6x² + 9x – 10 exceed 4x³ + 2x² + 7x -3 ?

Solution:

Question 13.
What must be added to 2a³ + 5a – a² – 6 to get a² – a – a³ + 1 ?

Solution:

Question 14.
What must be subtracted from a² + b² + lab to get – 4ab + 2b² ?

Solution:

Question 15.
Find the excess of 4m² + 4n² + 4p² over m²+ 3n² – 5p²

Solution:

Question 16.
By how much is 3x³ – 2x²y + xy² -y³ less than 4x³ – 3x²y – 7xy² +2y³

Solution:

Question 17.
Subtract the sum of 3a² – 2a + 5 and a² – 5a – 7 from the sum of 5a² -9a + 3 and 2a – a² – 1

Solution:

Question 18.
The perimeter of a rectangle is 28x³+ 16x² + 8x + 4. One of its sides is 8x² + 4x. Find the other side

Solution:

Question 19.
The perimeter of a triangle is 14a² + 20a + 13. Two of its sides are 3a² + 5a + 1 and a² + 10a – 6. Find its third side.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.
Simplify:

Solution:

EXERCISE 11 (C)

Question 1.
Multiply:

Solution:

Question 2.
Copy and complete the following multi-plications :

Solution:

Question 3.
Evaluate :

Solution:

Question 4.
Evaluate:

Solution:

Question 5.
Evaluate :

Solution:

Question 6.
Multiply:

Solution:

Question 7.
Multiply:

Solution:

EXERCISE 11 (D)

Question 1.
Divide:

Solution:

Question 2.
Divide :

Solution:

Question 3.
The area of a rectangle is 6x²– 4xy – 10y² square unit and its length is 2x + 2y unit. Find its breadth

Solution:

Question 4.
The area of a rectangular field is 25x² + 20xy + 3y² square unit. If its length is 5x + 3y unit, find its breadth, Hence find its perimeter.

Solution:

Question 5.
Divide:

Solution:

EXERCISE 11 (E)

Simplify
Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

EXERCISE 11 (F)

Enclose the given terms in brackets as required :

Question 1.
 x – y – z = x-{…….)

Solution:
x – y – z = x – (y + z)

Question 2.
x² – xy² – 2xy – y² = x² – (…….. )

Solution:
x² – xy² – 2xy – y²
= x² – (xy² + 2xy + y²)

Question 3.
4a – 9 + 2b – 6 = 4a – (…….. )

Solution:
4a – 9 + 2b – 6
= 4a – (9 – 2b + 6)

Question 4.
x² -y² + z² + 3x – 2y = x² – (…….. )

Solution:
x² – y² + z² + 3x – 2y
= x² – (y² – z² – 3x + 2y)

Question 5.
– 2a² + 4ab – 6a²b² + 8ab² = – 2a (……… )

Solution:
 – 2a² + 4ab – 6a²b² + 8ab²
= – 2a (a – 2b + 3ab² – 4b²)

Simplify :

Question 6.
2x – (x + 2y- z)

Solution:
2x-(x + 2y-z) = 2x – x – 2y + z
= x – 2y + z

Question 7.
p + q – (p – q) + (2p – 3q)

Solution:
p + q – (p – q) + (2p- 3q)
= p + q – p + q + 2p – 3q = 2p – q

Question 8.
9x – (-4x + 5)

Solution:
9x – (-4x + 5) = 9x + 4x – 5
= 13x- 5

Question 9.
6a – (- 5a – 8b) + (3a + b)

Solution:
6a – (- 5a – 8b) + (3a + b)
= 6a + 5a + 8b + 3a + b
= 6a + 5a + 3a + 8b + b
= 14a + 9b

 Question 10.
(p – 2q) – (3q – r)

Solution:
(p-2q) – (3q – r) =p – 2q – 3q + r =p – 5q + r

Question 11.
9a (2b – 3a + 7c)

Solution:
9a (2b – 3a + 7c)
= 18ab – 27a² + 63ca

Question 12.
-5m (-2m + 3n – 7p)

Solution:
-5m (-2m + 3n- 7p)
= – 5m x (-2m) + (-5m) (3n) – (-5m) (7p)
= 10m² – 15mn + 35 mp.

Question 13.
-2x (x + y) + x²

Solution:
– 2x (x + y) + x²
= -2x x x + (-2x)y + x²
= – 2x² – 2xy + x²
= – 2x² + x² – 2xy = – x² – 2xy

Question 14.

Solution:

Question 15.
8 (2a + 3b – c) – 10 (a + 2b + 3c)

Solution:
8 (2a + 3b -c)- 10 (a + 2b + 3c)
= 16a + 24b – 8c – 10a – 20b- 30c
= 16a – 10a + 24b – 20b – 8c – 30c
= 6a + 4b – 38c

Question 16.

Solution:

Question 17.
5 x (2x + 3y) – 2x (x – 9y)

Solution:
5x (2x + 3y) – 2x (x – 9y)
= 10x² + 15xy – 2x² + 18xy
= 10x² – 2x²+ 15xy+ 18xy
= 8x² + 33 xy

Question 18.
a + (b + c – d)

Solution:
a + (b + c – d) = a + (b + c – d)
= a + b + c – d

Question 19.
5 – 8x – 6 – x

Solution:
5 – 8x – 6 – x
= 5 – 6 –  8x – x
= -1 -7x

Question 20.
2a + (6- \(\overline { a-b }\) )

Solution:
2a + (6 – \(\overline { a-b }\) )
= 2a + (b – a + b)
= 2a + b – a + b
= a + 2b

Question 21.
3x + [4x – (6x – 3)]

Solution:
3x + [4x – (6x – 3)]
= 3x + [4x – 6x + 3]
= 3x + 4x – 6x + 3
= 3x + 4x – 6x + 3
= 7x – 6x + 3= x + 3

Question 22.
5b – {6a + (8 – b – a)}

Solution:
5b- {6a + 8- 6-a}
= 5b – 6a – 8 + b + a
= -6a + a + 5b +b – 8
= -5a + 6b-8

Question 23.
2x-[5y- (3x -y) + x]

Solution:
2x – [5y- (3x – y) + x]
= 2x – {5y – 3x +y + x}
= 2x – 5y + 3x -y – x
= 2x + 3x – x – 5y – y
= 4x – 6y

Question 24.
6a – 3 (a + b – 2)

Solution:
6a – 3 (a + b – 2)
= 6a – 3a – 3b + 6
= 3a -3b + 6

Question 25.
8 [m + 2n-p – 7 (2m -n + 3p)]

Solution:
8 [m + 2n-p -1 (2m – n + 3p)]
8 [m + 2n-p- 14m + 7n-21p]
= 8m+ 16n -8p- 112m + 56n – 168p
= 8m – 112m + 16n + 56n -8p – 168p
= -104m + 72n – 176p

Question 26.
{9 – (4p – 6q)} – {3q – (5p – 10)}

Solution:
{9 – {4p – 6q)} – {3q – (5p – 10)}
{9 – 4p + 6q} – {3q -5p+ 10}
= 9 – 4p + 6q – 3q + 5p – 10
= 9 – 4p + 5p + 6q – 3q – 10
= p + 3q – 1

Question 27.
2 [a – 3 {a + 5 {a – 2) + 7}]

Solution:
2 [a – 3 {a + 5 {a – 2) + 7}]
= 2 [a- 3 {a + 5a- 10 + 7}]
= 2 [a -3a- 15a + 30 -21]
= 2a-6a- 30a + 60-42
= 2a- 36a + 60-42
= -34a + 18

Question 28.
5a – [6a – {9a – (10a – \(\overline { 4a-3a }\)  )}]

Solution:
5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – (10a – 4a + 3a)}]
= 5a – [6a – {9a – 10a + 4a – 3a}]
= 5a- [6a – 9a + 10a – 4a + 3a]
= 5a – 6a + 9a – 10a + 4a – 3a
= 5a + 9a + 4a – 6a – 10a – 3a
= 18a – 19a = – a

Question 29.
9x + 5 – [4x – {3x – 2 (4x – 3)}]

Solution:
9x + 5 – [4x – {3x – 2 (4x – 3)}]
= 9x + 5 – [4x – {3x – 8x + 6}]
= 9x + 5 – [4x – 3x + 8x – 6]
= 9x + 5-4x + 3x-8x + 6
= 9x + 3x-4x-8x + 5 + 6
= 12x- 12x+ 11 = 11

Question 30.
(x + y – z)x + (z + x – y)y – (x + y – z)z

Solution:
(x + y – z)x + (z + x -y )y – (x + y -z)z
= x² + xy – zx + yz + xy -y² – zx – yz + z²
= x² -y² + z² + 2xy – 2zx

Question 31.
-1 [a-3 {b -4 (a-b-8) + 4a} + 10]

Solution:
– 1 [a – 3 {b – 4(a – b – 8) + 4a} + 10]
= -1 [a-3 {b-4{a-b-8) + 4a} + 10]
= -1[a-3 {b-4a + Ab +32 + 4a} + 10]
= -1 [a-3b+ 12a- 126-96- 12a + 10]
= -a + 3b – 12a + 12b + 96 + 12a – 10
= -a-12a + 12a+ 3b+ 12b-96-10
= – a + 15b – 106

Question 32.

Solution:

Question 33.
10 – {4a – (7 – \(\overline { a-5 }\)) – (5a – \(\overline { 1+a }\))}

Solution:

10 – {4a – (7 – \(\overline { a-5 }\)) – (5a – \(\overline { 1+a }\))}
= 10 – {4a – (7 – a + 5) – (5a – 1 – a)}
= 10- {4a -(12 -a) -(4a- 1)}
= 10 – {4a – 12 + a- 4a + 1}
= 10 – 4a + 12 – a + 4a- 1
= 10 + 12 – 1 – 4a – a + 4a
= 21 -a

Question 34.
7a- [8a- (11a-(12a- \(\overline { 6a-5a }\))}]

Solution:
7a – [8a – {1 la – (12a –\(\overline { 6a-5a }\))}]
= 7a-[8a-{11a-(12a-6a + 5a)}]
= 7a -[8a -{11a -(17a -6a)}]
= 7a- [8a- {11a-(11a)}]
= 7a- [8a- {11a- 11a}]
= 7a – 8a = -a

Question 35.

Solution:

Question 36.
x-(3y- \(\overline { 4z-3x }\) +2z- \(\overline { 5y-7x }\))

Solution:
x-(3y- \(\overline { 4z-3x }\) +2z- \(\overline { 5y-7x }\))
= x – (3y – 4z + 3x  + 2z -5y + 7x)
= x-(-2y-2z+10x)
= x + 2y + 2z- 10x
= -9x + 2y + 2z

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

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Recognition of solids Exercise 18 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Identify the nets which can be used to form cubes

Solution:
Nets for a cube are (ii) , (iii) and (v).

Question 2.
Draw at least three different nets for making cube.
Solution:

Question 3.
The figure, given below, shows shadows of some 3D objects, when seen under the lamp of an overhead projector :

In each case, name the object.
Solution:

Question 4.
Using Euler’s formula, find the values of a, b, c and d.

Solution:

(i) a + 6 – 12 = 2 ⇒ a = 2 – 6 + 12 = 14 – 6 = 8
(ii) b + 5- 9 = 2 ⇒6 = 2 + 9- 5 = 6
(iii) 20+ 12 — c = 2 ⇒32 – c = 2 ⇒ c = 32-2 ⇒ c = 30
(iv) 6 + d-12=2 ⇒ d – 6 = 2 ⇒ d = 2 + 6 = 8

Question 5.
Dice are cubes with dot or dots on each face. Opposite faces of a die always have a total of seven on them.
Below are given two nets to make dice (cube), the numbers inserted in each square indicate the number of dots in it.

Insert suitable numbers in each blank so that numbers in opposite faces of the die have a total of seven dots.
Solution:

Question 6.
The following figures represent nets of some solids. Name the solids

Solution:
The given nets are of the solid as given below :
(i) Cube
(ii) Cuboid

Question 7.
Draw a map of your class room using proper scale and symbols for different objects.
Solution:

Question 8.
Draw a map of your school compound using proper scale and symbols for various features like play ground, main building, garden, etc.
Solution:

Question 9.
In the map of India, the distance between two cities is 13.8 cm.
Taking scale : 1 cm = 12 km, find the actual distance between these two cities.
Solution:
The scale for a map is given to be 1 cm = 12 km
The distance between these two cities = 13.8 cm on the map
∴ Actual distance between these two cities
= 12 x 13.8 km = 165.6 km