Showing posts with label SSLC. Show all posts
Showing posts with label SSLC. Show all posts

10th Standard Public Exam 2026: Time Tables, Model Question Papers, and Answer Keys Download

Note: Comprehensive collection of Time Tables, Model Question Papers, and Answer Keys for the 10th Standard Public Exam 2026. Click the buttons below to download the PDFs.

10th / SSLC - Public Exam March 2026 Time Table

10th Public Exam March 2026 - Time Table
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Exam Tips & Time Management

10th Tamil - Public Exam Time Management Tips Mr. R. Parthiban
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10th Tamil - Public Exam Time Management Tips Mr. M. Karthick
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10th Science - How to Pass Easily (30 Days Plan) Mr. J. Hari Krishnan
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10th Science - Public Exam Time Management Tips Mr. S. Malathi
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10th Science - How to Pass Easily? Tips Mr. S. Malathi
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10th Model Question Papers (Teachers' Collections)

10th Tamil - Model Question Paper Mr. Kali Karuppu
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10th Tamil - Model Question Paper Mr. Chandra Veeran
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10th English - Model Question Paper Mr. Kasi Viswanathan
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10th English - Model Question Paper Mr. T. Naresh
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10th Maths - Model Question Paper Mr. S. Kuppusamy
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10th Maths - Model Question Paper Mr. S. Parthipan
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10th Maths - Model Question Paper Mr. A. Balaiah
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10th Maths - Model Question Paper with Answer Key Mr. M. Palaniyappan
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10th Social - Model Question Paper Mrs. Annapoorani
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10th Social - Model Question Paper Mr. Ezhil Ajith
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Official Model Question Papers (TNSCERT)

10th Tamil - Official Model Question Paper
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10th English - Official Model Question Paper
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10th English - Official Model Question Paper with Answer Key Dolphin
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10th Maths - Official Model Question Papers
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10th Science - Official Model Question Papers
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10th Social - Official Model Question Papers
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Question Patterns

10th English - New Pattern Mr. Nabroshekan
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10th English - New Pattern Mr. M. Senthil Kumar
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10th Science - New Pattern Mr. B. Dharmaraj
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Publisher Model Question Papers (Sura & Dolphin)

10th Tamil - Model Questions Papers Sura
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10th English - Model Questions Papers Dolphin
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10th English - Model Questions Papers Sura
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10th Maths - Model Questions Papers Sura
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10th Science - Model Questions Papers Sura
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10th Social - Model Questions Papers Sura
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10th English Road Map Questions and Answers (Q.28 Compulsory) - Public Exam Material Important for Public Exam 2026

10th English: Road Map Exercises

Q.28 Compulsory Question - Public Exam Study Material

28.
A stranger asked you to direct him to the School. Guide him to reach his destination.
Map for directions - Tirunelveli
Answer

Certainly. You are here at the corner of Gandhi Main Road.

  1. Go straight along Gandhi Main Road.
  2. You will pass the bus stop on your left.
  3. Continue walking straight.
  4. The School is on your left, just before the Raja Department store.
Your friend wants to visit the Exhibition. Give him to reach his destination.
10th English Quarterly Exam Paper Page 1 - Thanjavur
Image depicting a map:
[You are here] on New Street.
Ahead on the right is a [Circus].
Further ahead on Gandhi Bazaar road is a [School] and opposite to it is the [Exhibition].
Directions to the Exhibition

Start from where you are on New Street and walk straight ahead.
Take the first left turn.
Continue walking. You will pass the Circus on your left and the School on your right.
When you reach the main road, Gandhi Bazaar, turn right.
Walk a short distance, and the Exhibition will be on your left side.

Guide the stranger to the Library. Guide him to reach his destination.
Map for directions - Tiruvallur
Answer

Go straight ahead and take the first right turn onto Anna Road.
Walk along Anna Road until you reach the next intersection, and then turn left onto Big Street.
Continue walking on Big Street. You will pass a Bank on your left.
Take the next right turn after the Bank.
Walk a little. The Library will be on your left.

A man standing beside you wants to visit the Museum. Guide him to reach his destination.
Map for direction - Thanjavur District
Answer

Walk straight ahead to the Signal Main Road. You will pass the Bus Stop on your left.
When you reach the Signal Main Road, turn Left.
Walk straight until you get to the four way intersection.
At that intersection, turn right and go straight. You will reach college road.
Then turn Left; you will see a School on your left.
Walk a little further. Now you will find the Museum on the opposite side of the school, on your right.

You are near the School. An old man approaches you to direct him to a nearby hotel. Guide him with your instructions.
Road map English Salem District
Answer

Please follow these simple directions:

  1. Walk straight up this road until you reach the Main Road.
  2. Carefully cross the Main Road to the other side.
  3. Once you are on the other side, turn left.
  4. Walk straight for a short distance, and you will find The Saravana Hotel on your right.
A Stranger wants to go the Railway Station. Guide him to reach his destination.
Road Map Chengalpattu District
Guiding the Stranger
  1. Go straight along the Main Road.
  2. You will pass a School on your left and a Temple on your right.
  3. When you reach the junction, turn left onto N.H. Road.
  4. Walk a short distance, and you will find the Railway Station on your Right.
Title: 10th English Road Map Questions and Answers (Q.28 Compulsory) - Public Exam Material
Labels: 10th English, Road Map, Compulsory Question, SSLC, Study Material, Exam Prep
Permanent Link: 10th-english-road-map-exercises-answers-q28
Search Description: Practice 10th English Road Map compulsory questions (Q.28) with maps and clear step-by-step directions for the Public Exam.

10th English Grammar Study Material: Active Voice, Indirect Speech, Punctuation & Transformations Important for Public Exam 2026

10th English Grammar Exercises

Revision Material: Q.23 to Q.27

Q.23. Active Voice & Passive Voice

Rewrite the following sentence to the other voice.

1. Please assemble in the ground. MDL-19
Answer: You are requested to assemble in the ground.
2. I noticed a sudden change in Aditya's face. PTA - 1
Answer: A sudden change in Aditya's face was noticed by me.
3. The doctor kindly warned me. PTA - 2
Answer: I was kindly warned by the doctor.
4. M. Hamel mounted his chair. PTA - 3
Answer: M.Hamel's chair was mounted by him. / His chair was mounted by M.Hamel.
5. They have asked me to pay the fine. PTA - 4
Answer: I have been asked to pay the fine by them.
6. You are making a cake now. PTA - 5
Answer: A cake is being made by you now.
7. Why have you left your brother at home? PTA - 6
Answer: Why has your brother been left at home by you?

Q.24. Direct - Indirect Speech

Rewrite using indirect speech.

1. "Where are we going, sir?" asked the aero-coachman. MDL-19
Answer: The aero-coachman asked the gentleman where they were going.
2. "How did he get it?", he asked. PTA - 1
Answer: He asked how he had got it.
3. He said, "I am glad they are strong". PTA - 2
Answer: He said that he was glad they were strong.
4. "Let not thine eyes be blinded, my son", she said. PTA-3
Answer: She advised her son not to let his eyes be blinded.
5. Srivatsav said, "I have been waiting for my friend in the park since 6 a.m." PTA-5
Answer: Srivatsav told that he had been waiting for his friend in the park since 6 a.m.
6. Report the following dialogue: PTA - 6
Mohan: I lost my wallet on the way
Sathya: Did you have any money?
Answer: Mohan told Sathya that he had lost his wallet on the way then Sathya asked him if he had any money.

Q.25. Punctuation

Punctuate the following sentences.

1. Wherefore said miranda did they not that hour destroy us MDL-19
Answer: "Wherefore," said Miranda, "did they not that hour destroy us?"
2. stand back stand right back he cried PTA-1
Answer: "Stand back! Stand right-back!", he cried.
3. no its the inhabitants corley replied PTA - 2
Answer: "No it's the inhabitants", Corley replied.
4. i said no i'm not don't be silly PTA - 3
Answer: I said, "No, I'm not, don't be silly."
5. we had ghosts i said PTA - 4
Answer: "We had ghosts", I said.
6. he was near the sea now flying straight over it facing out over the ocean PTA - 5
Answer: "He was near the sea now, flying straight over it, facing out over the ocean."
7. be not so amazed daughter Miranda said Prospero PTA-6
Answer: "Be not so amazed, daughter Miranda", said Prospero.

Q.26. Simple, Compound, Complex

1. Transform the following sentence into a simple sentence.
As Catherin is a voracious reader, she buys a lot of books. MDL-19
Answer: Being a voracious reader, Catherin buys a lot of books.
2. Transform the following sentence into a simple sentence. PTA - 1
He is sick but he attends the rehearsal.
Answer: Inspite of his sickness, he attends the rehearsal. / Inspite of his being sick, he attends the rehearsal. / Despite his sickness, he attends the rehearsal.
3. Transform the following sentence into a compound sentence. PTA - 2
Answer: If Ryan reads more, he will become proficient in the language. / Ryan must read more then only he will become proficient in the language.
4. Transform the following sentence into a Complex sentence. PTA - 3
Neela followed my suggestions.
Answer: Neela followed what I suggest.
5. Combine the sentence using the appropriate connector. PTA - 4
Sita saw a snake. At once she ran away.
Answer: As soon as (When) Sita saw a snake, she ran away. / Sita saw a snake and at once (and) she ran away. / As (Since) Sita saw a snake, she ran away. / Sita saw a snake so (and so) she ran away.
6. Transform the following sentence as directed. PTA-5
The students were intelligent. They could answer the questions correctly. (Combine it into complex sentence)
Note: Answer not explicitly provided in the source fragment for this specific question.
7. Transform the following sentence as directed. PTA-6
Walk carefully lest you should fall down. (into complex)
Answer: If you don't walk carefully, you will fall down.

Q.27. Re-arrange the jumbled words

Rearrange the words in the correct order to make meaningful sentences:

1. MDL-19
a) he saw / When / in the / platform / the train / he rushed.
b) as / I / healthy / are you / am / as.
Answer:
a) When he saw the train in the platform, he rushed.
b) I am as healthy as you are.
2. PTA-1
a) he sent / At the / for me / I was a / time / clerk.
b) slept / But / on / Zigzag / undisturbed.
Answer:
a) At the time he sent for me, I was a clerk.
b) But Zigzag slept on undisturbed.
3. PTA-2
a) beckoning / calling / They were / to him / shrilly
b) amazed / how / to see / I was / well / I / it / understood.
Answer:
a) They were beckoning to him, calling Shrilly.
b) I was amazed to see how well I understood it.
4. PTA-3
a) Why / Nagen / uncle / asked / there / were / we.
b) was / alone / the / seagull / young / his / on / ledge.
Answer:
a) Uncle Nagen asked, why we were there.
b) The young seagull was alone on his ledge.
5. PTA-4
a) must / them / the / be / by / culprits / arrested
b) results / they / the / will / publish.
Answer:
a) The culprits must be arrested by them.
b) They will publish the results.
6. PTA-5
a) writing - system - telephone - am - to - I - about - banking - your - complain
b) You - believe - hear - should what never you.
Answer:
a) I am writing to complain about your telephone banking system.
b) You should never believe what you hear.
Title: 10th English Grammar Study Material: Active Voice, Indirect Speech, Punctuation & Transformations
Labels: 10th English, Grammar, Study Material, SSLC, Active Passive, Direct Indirect, Punctuation
Permanent Link: 10th-english-grammar-exercises-answers-q23-q27
Search Description: Comprehensive 10th English Grammar exercises (Q.23-Q.27) with answers: Voice, Speech, Punctuation, Simple Compound Complex sentences.

10th English Antonyms Important for Public Exam 2026

10th English Important Material

SSLC Study Guide - Antonyms

Q 4-6 Antonyms (Study Material)

LESSON-1

  • 1. brink × middle
  • 2. sank × swam
  • 3. encourage × discourage
  • 4. praising × scolding
  • 5. courage × timidity
  • 6. starve × well fed
  • 7. plaintively × happily
  • 8. desperate × hopeful
  • 9. exhausted × energized
  • 10. beneath × above

LESSON-2

  • 1. advent × departure
  • 2. hullabaloo × silence
  • 3. conclusion × beginning
  • 4. gripped × released
  • 5. gruffly × gently
  • 6. suspected × trusted
  • 7. gleamed × dark

LESSON-3

  • 1. tremendous × tiny
  • 2. accurate × inaccurate
  • 3. prediction × reality
  • 4. poked × released
  • 5. potential × incapability
  • 6. indigenous × foreign
  • 7. blissful × unhappy

LESSON-4

  • 1. bifurcated × merged
  • 2. existed × disappeared
  • 3. antique × new/modern
  • 4. recognized × ignored
  • 5. vent × control/conceal
  • 6. overwrought × calm
  • 7. ascertained × disproved

LESSON-5

  • 1. achieve × fail
  • 2. exhausted × refilled/rejuvenated
  • 3. marvelous × terrible
  • 4. inclusion × exclusion
  • 5. arrogant × humble
  • 6. gloomy × cheerful

LESSON-6

  • 1. unison × conflict
  • 2. choked × unblocked
  • 3. quickly × slowly
  • 4. attentive × inattentive
  • 5. honour × dishonour
  • 6. patience × impatience
  • 7. amazed × casual
  • 8. nuisance × beneficial

LESSON-7

  • 1. dare × evade
  • 2. delirious × balanced
  • 3. pleaded × demanded
  • 4. frail × strong
  • 5. persuade × dissuade
  • 6. contagious × non-contagious

Q 4-6 Antonyms (Practice)

Choose the appropriate antonym for the italicised words.

1. She screamed back mockingly. (MDL - 19)
a) disrespectfully b) ridiculously c) jeeringly d) respectfully
2. We don't have to use any means of repulsion. (MDL-19)
a) attraction b) distaste c) hate d) horror
3. I indulged in banking. (MDL-19)
a) took part b) participated c) abstained d) yielded
4. The project was taken in consonance with the National Policy. (PTA - 1)
a) agreement b) constant c) disagreement d) harmony
5. The sun was soothing. (PTA-1)
a) pleasing b) relaxing c) disturbing d) burning
6. She picked up a shoe and whammed it through the window. (PTA - 1)
a) tapped b) threw c) struck d) pulled
7. We look forward to a more inclusive way of learning... (PTA-2)
a) enclosed b) detached c) opened d) united
8. 'Nothing' he said gruffly. (PTA - 2)
a) happily b) roughly c) sadly d) plainly
9. Boost the morale in the country. (PTA - 2)
a) fear b) attitude c) mettle d) confidence
10. The little man was startled. (PTA-3)
a) surprised b) excited c) saddened d) at ease
11. The birds were chirping at the end of the woods. (PTA-3)
a) singing b) shouting c) tweeting d) alarming
12. One can control the computer screen with a gaze.
a) blink b) stare c) look d) trace
13. Then a monstrous terror seized him. (PTA-4)
a) released b) grasped c) snatched d) conquered
14. The light still shone palely down the stairs. (PTA-4)
a) dim b) weak c) bright d) faint
15. I indulged in banking. (Note: Options imply active/energetic) (PTA-4)
a) dynamic b) lively c) energetic d) inactive
16. The sick-room was a gloomy spot. (PTA-5)
a) dark b) dim c) dull d) bright
17. I am glad that we were able to finish it successfully. (PTA - 5)
a) happy b) pleased c) sorry d) joyful
18. The cops were reluctant to leave... (PTA - 5)
a) eager b) unwilling c) opposed d) averse
19. "How ignorant you are! Watson!" (PTA-6)
a) illiterate b) uneducated c) well informed d) rude
20. I had counted on the commotion... (PTA-6)
a) confusion b) disturbance c) unrest d) calmness
21. So we can now look forward to a more inclusive way... (PTA-6)
a) further b) ahead c) proceed d) backward

10th English Important Synonyms for Public Exam 2026.

10th English Important Material

SSLC Study Guide - Synonyms

Q 1-3 Synonyms

LESSON-1

  • 1. Ledge - shelf
  • 2. shrilly - high pitched
  • 3. stretched - extended
  • 4. plunge - dive
  • 5. devour - swallow
  • 6. mackerel - a sea fish
  • 7. gnaw - chew
  • 8. trot - jogged
  • 9. whet - sharpened
  • 10. beckoning - signalling

LESSON-2

  • 1. hullabaloo - noise
  • 2. patrolman - police officer
  • 3. attic - storage space inside the roof
  • 4. slamming - banging
  • 5. gruffly - harshly
  • 6. intuitively - spontaneously
  • 7. whammed - threw
  • 8. beveled - slope
  • 9. rending - tearing to pieces
  • 10. yanked - pulled

LESSON-3

  • 1. circumnavigate - went around
  • 2. indigenously - domestically
  • 3. consonance - agreement
  • 4. skipper - captain/master
  • 5. expedition - journey/voyage
  • 6. replenishment - restoration
  • 7. apprehensive - anxious/fearful
  • 8. contention - heated disagreement
  • 9. auxiliary - additional/supportive
  • 10. anticipate - expected

LESSON-4

  • 1. bifurcated - divided
  • 2. revive - refresh
  • 3. soothing - comforting
  • 4. dilated - enlarge
  • 5. ascertained - verified
  • 6. overwrought - tensed
  • 7. crumbled - broken
  • 8. spire - tower
  • 9. unperturbed - undisturbed
  • 10. affluent - wealthy
  • 11. smacks - tastes

LESSON-5

  • 1. grapple - wrestle/fight
  • 2. dragon dictate - speech convert into text
  • 3. gaze - stare
  • 4. inclusion - co-operation
  • 5. cloister - enclosed by

LESSON-6

  • 1. chirping - sound/twitter
  • 2. bustle - commotion
  • 3. unison - harmony/relevant
  • 4. rapping - striking
  • 5. thumbed - shivered
  • 6. cranky - strange
  • 7. angelus - prayer

LESSON-7

  • 1. gaunt - lean
  • 2. twitched - shivered/jerked
  • 3. contagious - infectious/spreading
  • 4. groan - despair
  • 5. plague - bacterial disease
  • 6. bolted - closed
  • 7. mantlepiece - around the fire place
  • 8. half-crown - equal to two shillings
  • 9. tongs - a device used for picking up
  • 10. delirious - restlessness

Part-I (1 Mark Questions : 14 MARKS)

Question 1 to 3 : Synonyms
Choose the appropriate synonyms for the italicised words:

1. The mother seagull swooped upwards. (MDL-19)
a) leap b) rush c) move very quickly d) ascend
2. The attic has always been favourite with children. (MDL-19)
a) loft b) terrace c) apartment d) strong room
3. It is a 55-foot sailing vessel built indigenously in India. (MDL-19)
a) fully b) collectively c) innately d) specially
4. It was the gaunt face staring from the bed that brought chill to my heart. (PTA-1)
a) fat b) round c) lean d) sad
5. When school began, there was a bustle. (PTA-1)
a) rush b) change c) noise d) confusion
6. They continue to grapple with the changes. (PTA-1)
a) settle b) fight c) move d) stop
7. How cranky he was. (PTA-2)
a) normal b) strange c) abnormal d) happy
8. His parents circled around raising a proud cackle. (PTA-2)
a) sharp noise b) blunt noise c) high pitch d) shout
9. Trying to revive old childhood memories may prove disappointing. (PTA-2)
a) review b) revitalize c) restore d) rescue
10. The spoilt child of affluent parents. (PTA-3,5)
a) influenced b) wealthy c) happy d) poor
11. Scraping his beak now and again to whet it. (PTA-3)
a) clean b) blunt c) sharpen d) wet
12. My contention was to make sure that we go by the rules. (PTA-3)
a) continuous effort b) disturbed effort c) unhappy effort d) strenuous effort
13. He was delirious. (PTA-4)
a) sick b) disappointed c) troubled d) forced
14. The whole family was laughing at his cowardice. (PTA-4)
a) strength b) bravery c) courage d) lack of bravery
15. My mother was asleep in one room upstairs, grandfather was in the attic. (PTA-4)
a) bedroom b) a room c) a space in the roof d) kitchen
16. World renowned physicist Stephen Hawking is the best example of how... (PTA-5)
a) famous b) special c) popular d) unique
17. But something choked him. (PTA-5)
a) praised b) blocked c) answered d) encouraged
18. The great expanse of sea stretched down beneath. (PTA-6)
a) large space b) narrow space c) small space d) deep area
19. He said in a hopeless tone of a despondent beagle. (PTA-6)
a) angry b) affluent c) despairing d) strong
20. They were apprehensive and supportive too. (PTA-6)
a) confident b) inquisitive c) anxious d) special

10th English Public Exam March 2025 Answer Key

10th English Public Examination - 2025

Tentative Answer Key

Marks: 100

PART - I (Section A) $$14 \times 1 = 14$$
Q.No Category Answer Option & Text
1 Synonyms (Indigenously) (c) naturally
2 Synonyms (plaintively) (a) sadly
3 Synonyms (unperturbed) (b) undisturbed
4 Antonyms (whet) (a) blunt
5 Antonyms (gruffly) (b) happily
6 Antonyms (inclusion) (c) exclusion
7 Singular/Plural (radius) radii
8 Prefix/Suffix (nutrition) malnutrition
9 Abbreviation (HDMI) (c) High Definition Multimedia Interface
10 Phrasal Verb (a) get along with
11 Compound Word (traffic) (c) jam
12 Preposition (b) of
13 Tense (a) locked
14 Preposition (b) from
PART - II (Section 1 - Prose) $$3 \times 2 = 6$$
15. The piece of fish offered by the mother seagull to the hungry young seagull and the natural's bird's instinct prompted it to fly finally.
16. The special features of INSV Tarini are:
  • It encouraged use of environment friendly non-conventional renewable energy resources such as the wind.
  • It collected and updated meteorological, ocean and wave data on regular basis for accurate weather forecast by India Meteorological Department (IMD).
  • It also collected data for monitoring marine pollution on high seas.
17. Aditya offered Sanyal one hundred and fifty rupees. It was the price of the medal, which Sanyal received in school for recitation.
18. The doctor warned Tom that he was in danger of losing his mind by thinking about precious stones always.
PART - II (Section 2 - Poetry) $$2 \times 2 = 4$$
19.
a) The poet expects everyone to walk with a smile and a song when something goes wrong.
b) When things go wrong, we should not worry.
20.
a) 'It' refers to the tree beside the house on Elm street.
b) The leaves of the tree never grows. The tree neither grows tall nor gets smaller. So, the tree is a mystery.
21.
a) 'I' refers to Cricket.
b) The nature of the cricket was lazy and silly.
22.
a) No, the house does not remain the same every day. It begins to fade each day.
b) Nobody knows what happens inside the house. So the poetess considers the house to be a mystery.
PART - II (Section 3 - Grammar) $$3 \times 2 = 6$$
23. Banu says that she is enjoying her holidays.
24. The enemy has been defeated by our army.
25. "Have you come from Holmes?" he asked.
26. In spite of being sick, she attends the rehearsal.
27. a) The mountain road is full of Dangers.
b) The sea can get really tough when winds are picking up.
PART - II (Section 4 - Road Map) 2 Marks
28.
  • Go straight from the market.
  • Turn left.
  • Walk on the main road.
  • Walk past library.
  • Turn right.
  • Walk on the north road and turn left.
  • Walk for a while.
  • Finally you will reach the Pharmacy on your left.
PART - III (Prose Paragraphs) $$2 \times 5 = 10$$
29. His First Flight - Liam O'Flaherty
About the Author: Born: 28th August 1896 (Irish, Ireland)
Occupation: Novelist
Famous Works: Member of the Communist Party of Ireland, Irish novelist, Short story writer.
Quote: "The moments you doubt whether you can fly, You cease for ever to be able to do it"

The young seagull was afraid of flying. His parents strove their level best to teach the young seagull to fly but in vain. The parents, brothers and sister thought a plan to teach him to fly. They flew away to another rock and left him alone. They did not give him anything to eat. He stood there on one leg and closed his eyes. He was very hungry. He searched for food everywhere. He even chewed the dried pieces of the eggshells. He saw his mother tearing a piece of fish.

'Ga, ga, ga', he cried, begging her to bring him over some food. 'Gawl-ool-ah', she screamed back mockingly.

He begged his mother to give him food. The mother seagull motivates the young one enough to get him to learn flying. So the mother flew with the piece of fish to the young seagull. When she reached over him, she became motionless in the air. She did not get down on the rock. She wanted to give the young seagull an incentive to fly. The seagull bent forward and jumped at the fish. The young seagull realizes the importance of belief and faith. He was much frightened. But he began to flap his wings to save himself. 'His first flight' is a parable. The seagulls convey the message of self-confidence, motivation and self-reliance. The story conveys, "Give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime. You can't fly unless you let yourself."

30. The Night the Ghost got in - James Grover Thurber

Characters: The narrator's family, Mr. Bodwell, Joe, Reporter, Herman.
Theme: Chaos in the house.

James Grover Thurber was an American cartoonist, best known for his cartoon and short stories. This lesson clearly tells that too much of imagination will mislead the situation.

The narrator says that on the night of November 17, 1915 the ghost got into their house. The time was quarter past one o'clock. All of sudden the author heard footsteps. They were the steps of man walking around the dining-table. The steps went round and round the table. The board creaked when it was trod upon. The author thought that if might be a burglar. He roused Herman. The steps started up the stairs. Filled with fear the author and his brother slammed the door shut. The noise of slamming roused the mother. She suspected that there were burglars inside. She wanted to call the police. But the phone was downstairs. So, She opened the windows and threw a shoe into the bedroom window of the neighbours. The window pane shattered down. The noise roused Bodwell and his wife annoyed them. Mr. Bodwell was furious with her. He called the police, thinking that there was a burglar in their house.

The police who came there started ransacking the entire house. The police broke the door open and entered the house. Their flashlights gleamed up and down. They flashed the light everywhere. They couldn't see anything. They opened the doors, yanked the drawers, opened the windows and pulled down the furniture. Then they heard a creaking in the attic. Grandfather was turning over in bed. The cops (police) used into the attic without words or warning. Grandfather mistook the policemen to be General Meade's men. He thought that they were deserters from the army. So he grabbed at a handgun and shot at a policeman. At once everyone retreated from the place. The police started ransacking the house once again. They took leave when the author promised to bring the gun to the station the next day. The police left the house empty handed finally. Next morning, grandfather told that he came to the dining room for water, the previous night. These are the incidents that caused the confusion in the house.

"Imagination will often carry us to worlds that never were"

31. Tech Bloomers

Empowering the disabled with technology. This lesson is about the use of technology in empowering the disabled to do their day to day chores of life, like travelling, communicating, learning, doing business and living in comfort. Alisha and David's life has changed with the use of technology. "Technology is a boon to the disabled', because it made to a lot of difference in their life. According to the 2011 Census, 2.21 percent of India's population is disabled. Then she can print them out. Kim who is an Assistive Technologist helps students to use technology in different ways. They are struggling with challenges of access, acceptance and inclusion.

"I have cerebral palsy and I can't physically type as fast as I think or anywhere near. But right now, that's what I'm doing. I bet you're wondering how!"

Technology has made her achieve things only she dreamt of. She can do Maths GSCE herself without being dependent on a computer. Technology has opened up the world to her. Twenty-one-year-old David was born with Athetoid Cerebral Palsy. He says that technology is very important because it enables him to communicate and be independent. For verbal communication, he uses a Liberator Communication Device, which he controls with eye movements. He has an ACTIV controller also in the headrest of his chair in his bedroom, which means he can control his TV, Blu Ray and music players. They are able to manage their daily activities with the aid of technology. Hence technology is a boon to the disabled. Thus Technology makes our life easier. It impacts the environment, people and the society as a whole. Newer Technology allows differently abled learners with their peers as well as contributes fruitfully to the collaborative process of learning.

"For most of us technology makes things easier, For a person with disability it makes things possible"

32. The Last Lesson - Alphonse Daudet

"Language and culture - true identity of a nation."

'The Last Lesson' was written by Alphonse Daudet. The story is narrated by a French boy Franz. He was lazy and liked to play. He disliked studying French. After overpowering the districts of Alsace and Lorraine in France, Berlin had ordered that German should be taught in schools instead of French. It was the last French class of the teacher M. Hamel who had been there for 40 years. When Franz reached the school he saw a crowd gathered around the bulletin-board. Though he reached the school little late he was surprised to note that the school was very quiet. Franz took his seat and noticed that M. Hamel had clothes for special occasion. The atmosphere of the classroom was strange. As a mark of respect for his hard work, the village men also attended his last class.

The teacher was full of grief and nostalgia. They were sad that they did not learn French, their mother tongue in their childhood. Franz was shocked to know it was his last French lesson. But he did not learn French. He was suddenly remorseful over wasting so much time playing outdoors rather than studying. He stood up to recite but stumbled on the first words. M. Hamel did not scold Franz for not knowing the rule. He told the crowd that each day they had been putting off learning until the next day. The teacher then proceeded to the French grammar lesson, reading from a book to the students.

Franz understood everything M. Hamel said with extraordinary clarity. He thought that he had never listened so carefully to the teacher before. When the trumpets of the Prussians were finally blown, M. Hamel stood up, but could not speak out of grief and wrote "Vive La France" (Long Live France).

PART - III (Section 2 - Poetry) $$2 \times 5 = 10$$
33. LIFE - Henry Van Dyke
About the Poet: Henry Van Dyke
Specialty: American author, Poet, Educator and Clergyman.
Quote: "Life is not a race, it is a journey enjoy it"

In this poem, life is described not as an entity, but as an experience. Here, the poet desires to lead his future life with optimism and he is ever willing to do something. One should live with courage and dedication. Life should be lived without hurry and with a clear sense of purpose that drives the mind and soul. The poet encourages us to let go of all that has been lost in the past as well as the uncertainty the future holds. He tells us to embrace the present with the happiness which nourishes the young and the old. From what the future veils; but with a whole And happy heart, that pays its toll To Youth and Age, and travels on with cheer.

Happiness gives us nourishment on this journey with a smile on our face. Whatever situation life throws at us, it is the journey that should be joyous, for it teaches us to grow and live. Our imagination should have the innocence and fearlessness of childhood. We should seek out new friendships, new adventures and new experiences which enrich us. He encourages us to have faith and determination in our hearts, as we take on this beautiful journey. We should have eternal hope that our story ends joyfully. This poem shows the poet's optimistic view of life through his personal experiences.

"BE AN OPTIMIST, DON'T BE A PESSIMIST"

34. NO MEN ARE FOREIGN - James Falconer Kirkup

James Falconer Kirkup reminds us that all the people who belongs to the earth are similar and share the brotherhood of men. The central theme of the poem is about the oneness of mankind. Though there are many countries, races, colour, languages, castes, and creeds all are human beings. We occupy the same land and die in the same land. Hands work hard, eyes witness the same life. We are all fed by peaceful harvests. Attacking others who are neighbours in the same planet is nothing but harming oneself. Hence no land are foreign, no men is a stranger.

"Remember they have eyes like ours that wake Or sleep and strength that can be won. By love. In every land is common life That all can recognize and understand"

The poem "No men are foreign" has greater relevance in today's world. Now a days people like to segregate themselves from others. They create borders and erect fences to separate one country from the other. Anyone who crosses the line is branded as a spy or intruder. He is shot dead without mercy. Love has vanished from human heart. Even at slight provocation people take up arms. It leads to battles and wars that affect the world. In this context the poet wants us to treat everyone as a fellow human being. We should remember that we may live in different countries. But we feel like others, work like others and read like others. Hatred should give way to love and affection. We can put an end to war if we consider everyone as our brethren. In today's world each country wants to fight with the other for the sake of its own benefit. They hate each other to flaunt their superiority and social status. Even men hate each other for silly reasons forgetting that we are all brothers and sisters. So this poem is very relevant to today's world.

"Every human is like all other humans, some other humans and no other human".

35. a) Rhyming words: cold - scold, meet - street
b) Rhyming scheme: aabb
c) Personification / Epithet
d) Summer - Scold
36. Machines aren't perfect after all and Nature always wins over. Machines aren't miraculous creations. They are nothing more than the creations of the human brain.
PART - III (Section 3 - Supplementary) $$1 \times 5 = 5$$
37. Little Hero of Holland
  • Little Peter was asked to take cakes for his blind friend by his mother.
  • While returning, he hears water trickling from the dikes.
  • Peter realizes the danger of flooding.
  • Peter stops the angry waters with his finger the entire night and saves the village.
  • Peter becomes the Little Hero of Holland.
38. Zigzag
a) It was Zigzag's voice, clear and commanding.
b) Bored and Grumpy.
c) African Doctor.
d) Dr. Krishnan's clinic transformed into a calm.
e) Slept or snored.
PART - III (Section 4 - Writing Skills)
Q.No Topic Mark Distribution
39 Advertisement Outline: 1
Title: 1
Picture: 1
Captions: 1
Address: 1
40 Letter Writing From/To: 1
Date/Salutation: 1
Body of letter: 2
Subscription/Superscription: 1
41 Notice Writing Title: 1
Date/Name: 1
Content: 3
42 Picture Comprehension Any five relevant sentences without mistakes: 5 marks
43 Note Making / Summary Note Making: Title (1), Points (4)
Summary: Title (1), Rough copy (2), Fair copy (2)
44. Error Correction
a) Rice is the staple food of Asians.
b) As Sathya was old, he walked slow.
c) Slow and steady wins the race.
d) I will be fifteen next April.
e) Learning a language is always useful.
PART - III (Section 5 - Memory Poem) 5 Marks
45. She's a lioness; don't mess with her.
She'll not spare you if you're a prankster.
Don't ever try to saw her pride, her self-respect.
She knows how to thaw you, saw you - so beware!
She's today's woman. Today's woman, dear.
PART - IV
46. a) The Story of Mulan

Introduction: This is the classic story of Mulan based on the legend of Hua Mulan. A legend is a story from long ago that is believed to be true, or mostly true. It is about a brave girl, Mulan, who had saved China.

Lovable daughter Mulan: Many years ago, China was in the middle of a great war. The Emperor announced that one man from each Chinese family must leave his family to join the army. Mulan was strong, loyal and patriotic. She surrendered her life to her motherland, China. When she heard the order of the Emperor, she immediately volunteered to fight for her country despite her father being sick. Though she knew that women were not allowed, she joined the army in disguise, simply ignoring the risks involved. She knew that if the soldiers came to know that she was a woman she would be put to death.

Her Patriotism: Ignoring all the risks, she dared to join the army. It was her patriotism that drove her to that extent. She was really a strong character. Besides, she was also smart and brave right from the beginning. She was also a good leader who led the entire troop to victory and thus ended the war in China forever. After a few years, she was given the post of General of the entire army. After some days, a very bad fever swept through the army. Mulan, the General was also affected by this fever. When the doctor came out Mulan's tent, he told the truth to the soldiers.

Her Valour: Even though she was sick, she fought and ended the long war victoriously. Admiring her valour and bravery, the Emperor was willing to appoint her as the royal advisor. She humbly refused to stay in the palace and expressed her wish to return to her village. She returned to her village to see her father and brother.

Conclusion: Thus Mulan showed that she was lovable to her family and proved her patriotism towards her country. Mulan, a multifaceted personality showed her love for her country by fighting bravely.
Moral: "Liberty is the birth of life to nation"

46. b) A DAY IN 2889 OF AN AMERICAN JOURNALIST - Jules Verne

Introduction: 'A day in 2889 of an American Journalist' written in 1889 by Jules Verne, gives us a glimpse of the futuristic world in which people rely on machines for almost everything. The story is set on July 25 of 2889.

Phonotelephote and the Mechanized dressing room: Francis Bennett, director of the Earth Herald, the world's most leading newspaper, had woken up with a bad temper. He felt a little lonely in the absence of his wife. His wife had been away in France for the past eight days. He switched on his phonotelephote and hearing his wife's voice, he put on a smile. His mechanized dressing room dressed him from top to toe.

Astronomy - the trending topic: In his office, he inquired the astronomical reporters about the latest news. He wished to feed the public with interesting astronomical news. He was excited to bring out the interesting discoveries made in the new plant Gandini to the public. Inordinate advertisements were projected on clouds by a thousand projectors from a gallery. Bennett was one of the subscribers to the society for supplying food to the home. He got his lunch from there through a network of pneumatic tubes.

Innovative petitions: Then Bennett travelled to Niagara for his accumulator works. 'Where are we going, Sir?' asked the aero-coachman. 'Let's see. I've got time...' Francis Bennett replied. 'Take me to my accumulator works at Niagara.' He took an aero-car with an aero coachman who was waiting near his window. The aero-car is capable of gaining speed of about four hundred miles an hour. After his return, he listened to many petitions. He supported the idea of an inventor who thought of reducing the final three elements into one and another idea of moving the town of Saff to the shore using rails.

Conclusion: At-last Bennett learned about his wife's plan to return. Finally he decided to take a bath. Thus the story ends. The innovations in this story seem to be a magic. Unbelievable everything is made possible in the future with the help of technology.
Moral: "Technology is a useful servant but a Dangerous Master"

47. a) Comprehension
i) American President Theodore Roosevelt, also known as teddy, participated in a bear-hunting trip in Mississippi.
ii) Michtom place the stuffed bears in the front window of his shop.
iii) Political cartoons starring like Teddy and the bear.
iv) Michtom is the shop owner. He created plush, stuffed bears and placed them in the front window of his shop.
47. b) General Comprehension
i) Choosing our career is hard to decided.
ii) Dancer, Doctor, Teacher, Cricket.
iii) Sunday, Thursday.
iv) Any suitable answer like Teacher, Doctor, Lawyer, Cricketer...
Question Paper Page No. 1 Question Paper Page No. 2 Question Paper Page No. 3 Question Paper Page No. 4 Question Paper Page No. 5 Question Paper Page No. 6 Question Paper Page No. 7 Question Paper Page No. 8

2024 March SSLC All Subject Official Question Paper Download (PDF)

2024 MARCH SSLC ALL SUBJECT QUESTION PAPER.

2024 MARCH SSLC ALL SUBJECT QUESTION PAPER.pdf

The 2024 March SSLC question paper for all subjects is available in Tamil and English mediums. The PDF file contains 109 pages and is intended for educational and competitive exam purposes.

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2025 March SSLC All Subject Official Question Paper Download (PDF)

2025 MARCH SSLC ALL SUBJECT QUESTION PAPER.

2025 MARCH SSLC ALL SUBJECT QUESTION PAPER.pdf

The 2025 March SSLC question paper for all subjects is available in Tamil and English mediums. The PDF file contains 169 pages and is intended for educational and competitive exam purposes.

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Tamil Nadu 10th, 12th Public Time Table 2026

Tamil Nadu 10th, 12th Public Time Table 2026

The Directorate of General Education, Tamil Nadu (TNDGE) released the Tamil Nadu 12th Public exam time table 2026 and Tamil Nadu 10th Public exam time table 2026 Today. TN 12th exam 2026 will be held from March 2 to 26, 2026. TN SSLC exam 2026 will be held between March 11, and April 6, 2026.

SSLC EXAMINATION - MARCH / APRIL - 2026 PUBLIC EXAMINATION TIME TABLE
HIGHER SECONDARY SECOND YEAR (+2) PUBLIC EXAMINATION - MARCH -2026 TIME TABLE
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2025-2026 Public Examinations Schedule

10th Maths Quarterly Exam 2024 Question Paper with Solutions | Tenkasi District Samacheer Kalvi

10th Maths Quarterly Exam 2024 Question Paper & Solutions - Tenkasi District

Exam Details

  • District: Tenkasi District
  • Examination: Common Quarterly Examination - 2024
  • Standard: 10
  • Subject: Mathematics
  • Date: 25-09-2024
  • Time: 3.00 Hours
  • Marks: 100

Part I: Answer all the following questions (14 x 1 = 14)

1) If A = {a, b, q}, B = {2, 3}, C = {p, q, r, s} then \( n[(A \cup C) \times B] \) is

  • a) 8
  • b) 20
  • c) 12
  • d) 16
Given: A = {a, b, q}, B = {2, 3}, C = {p, q, r, s}.
First, find \( A \cup C \): \( A \cup C = \{a, b, q\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\} \).
Number of elements in \( A \cup C \) is \( n(A \cup C) = 6 \).
Number of elements in B is \( n(B) = 2 \).
Then, \( n[(A \cup C) \times B] = n(A \cup C) \times n(B) = 6 \times 2 = 12 \).
Answer: c) 12

2) \( f(x) = (x+1)^3 - (x-1)^3 \) represents a function which is

  • a) linear
  • b) cubic
  • c) reciprocal
  • d) quadratic
We use the formulas: \( (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \) and \( (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \).
\( f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) \)
\( f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 \)
\( f(x) = 6x^2 + 2 \).
The highest power of x is 2, so it is a quadratic function.
Answer: d) quadratic

3) If the HCF of 65 and 117 is expressible in the form of 65m-117, then the value of m is

  • a) 4
  • b) 2
  • c) 1
  • d) 3
Find HCF of 65 and 117 using Euclid's algorithm:
\( 117 = 1 \times 65 + 52 \)
\( 65 = 1 \times 52 + 13 \)
\( 52 = 4 \times 13 + 0 \). The HCF is 13.
Given, \( 65m - 117 = 13 \).
\( 65m = 130 \implies m = \frac{130}{65} = 2 \).
Answer: b) 2

4) Given \( F_1 = 1, F_2 = 3 \) and \( F_n = F_{n-1} + F_{n-2} \) then \( F_5 \) is

  • a) 3
  • b) 5
  • c) 8
  • d) 11
\( F_1 = 1 \)
\( F_2 = 3 \)
\( F_3 = F_2 + F_1 = 3 + 1 = 4 \)
\( F_4 = F_3 + F_2 = 4 + 3 = 7 \)
\( F_5 = F_4 + F_3 = 7 + 4 = 11 \)
Answer: d) 11

5) If \( A = 2^{65} \) and \( B = 2^{64} + 2^{63} + 2^{62} + ... + 2^0 \) which of the following is true?

  • a) B is \( 2^{64} \) more than A
  • b) A and B are equal
  • c) B is larger than A by 1
  • d) A is larger than B by 1
B is a geometric progression with first term \( a = 2^0 = 1 \), common ratio \( r = 2 \), and number of terms \( n = 65 \).
Sum of GP: \( S_n = a \frac{r^n - 1}{r - 1} \).
\( B = 1 \times \frac{2^{65} - 1}{2 - 1} = 2^{65} - 1 \).
Since \( A = 2^{65} \), we have \( B = A - 1 \).
This means A is larger than B by 1.
Answer: d) A is larger than B by 1

6) \( y^2 + \frac{1}{y^2} \) is not equal to

  • a) \( \frac{y^4+1}{y^2} \)
  • b) \( \left(y + \frac{1}{y}\right)^2 - 2 \)
  • c) \( \left(y - \frac{1}{y}\right)^2 + 2 \)
  • d) \( \left(y - \frac{1}{y}\right)^2 - 2 \)
Let's check each option:
a) \( \frac{y^4+1}{y^2} = \frac{y^4}{y^2} + \frac{1}{y^2} = y^2 + \frac{1}{y^2} \). (Equal)
b) \( \left(y + \frac{1}{y}\right)^2 - 2 = (y^2 + 2(y)(\frac{1}{y}) + \frac{1}{y^2}) - 2 = y^2 + 2 + \frac{1}{y^2} - 2 = y^2 + \frac{1}{y^2} \). (Equal)
c) \( \left(y - \frac{1}{y}\right)^2 + 2 = (y^2 - 2(y)(\frac{1}{y}) + \frac{1}{y^2}) + 2 = y^2 - 2 + \frac{1}{y^2} + 2 = y^2 + \frac{1}{y^2} \). (Equal)
d) \( \left(y - \frac{1}{y}\right)^2 - 2 = (y^2 - 2 + \frac{1}{y^2}) - 2 = y^2 + \frac{1}{y^2} - 4 \). (Not Equal)
Answer: d) \( \left(y - \frac{1}{y}\right)^2 - 2 \)

7) Graph of a linear equation is a

  • a) straight line
  • b) circle
  • c) parabola
  • d) hyperbola
The graph of any linear equation in two variables (e.g., \( ax+by+c=0 \)) is always a straight line.
Answer: a) straight line

8) If \( f(x) = 2x^2 \) and \( g(x) = \frac{1}{3x} \), then fog is

  • a) \( \frac{2x^2}{3} \)
  • b) \( \frac{2}{3x^2} \)
  • c) \( \frac{2}{9x^2} \)
  • d) \( \frac{1}{6x^2} \)
fog means \( f(g(x)) \).
Substitute \( g(x) \) into \( f(x) \): \( f(g(x)) = f(\frac{1}{3x}) \).
Now apply the function f: \( 2 \left( \frac{1}{3x} \right)^2 = 2 \left( \frac{1}{9x^2} \right) = \frac{2}{9x^2} \).
Answer: c) \( \frac{2}{9x^2} \)

9) In \( \Delta LMN \), \( \angle L = 60^\circ, \angle M = 50^\circ \). If \( \Delta LMN \sim \Delta PQR \) then value of \( \angle R \) is

  • a) \( 40^\circ \)
  • b) \( 70^\circ \)
  • c) \( 30^\circ \)
  • d) \( 110^\circ \)
In \( \Delta LMN \), the sum of angles is \( 180^\circ \).
\( \angle N = 180^\circ - (\angle L + \angle M) = 180^\circ - (60^\circ + 50^\circ) = 180^\circ - 110^\circ = 70^\circ \).
Since \( \Delta LMN \sim \Delta PQR \), the corresponding angles are equal.
\( \angle L = \angle P, \angle M = \angle Q, \angle N = \angle R \).
Therefore, \( \angle R = \angle N = 70^\circ \).
Answer: b) \( 70^\circ \)

10) In a \( \Delta ABC \), AD is the bisector of \( \angle BAC \). If AB=8cm, BD=6cm and DC=3cm. The length of the side AC is

  • a) 6 cm
  • b) 4 cm
  • c) 3 cm
  • d) 8 cm
By Angle Bisector Theorem, the ratio of the sides adjacent to the angle is equal to the ratio of the segments the bisector divides the opposite side into.
\( \frac{AB}{AC} = \frac{BD}{DC} \)
\( \frac{8}{AC} = \frac{6}{3} \implies \frac{8}{AC} = 2 \)
\( AC = \frac{8}{2} = 4 \) cm.
Answer: b) 4 cm

11) The straight line given by the equation x=11 is

  • a) Parallel to x-axis
  • b) Parallel to y-axis
  • c) Passing through the origin
  • d) Passing through the point (0, 11)
The equation \( x = c \) (where c is a constant) represents a vertical line.
A vertical line is always parallel to the y-axis.
Answer: b) Parallel to y-axis

12) The slope of the line joining (12, 3) and (4, a) is \( \frac{1}{8} \). The value of 'a' is

  • a) 1
  • b) 4
  • c) -5
  • d) 2
Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
\( \frac{1}{8} = \frac{a - 3}{4 - 12} = \frac{a - 3}{-8} \)
Multiply both sides by -8: \( -1 = a - 3 \).
\( a = 3 - 1 = 2 \).
Answer: d) 2

13) When proving that a quadrilateral is a Parallelogram by using slopes you must find

  • a) The slopes of two sides
  • b) The slopes of two pair of opposite sides
  • c) The lengths of all sides
  • d) Both the lengths and slopes of two sides.
A key property of a parallelogram is that its opposite sides are parallel.
In coordinate geometry, parallel lines have equal slopes.
Therefore, to prove a quadrilateral is a parallelogram using slopes, one must show that both pairs of opposite sides have equal slopes.
Answer: b) The slopes of two pair of opposite sides

14) If \( (\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = K + \tan^2\alpha + \cot^2\alpha \), then the value of K is equal to

  • a) 9
  • b) 7
  • c) 5
  • d) 3
Expand the LHS:
\( (\sin^2 \alpha + \csc^2 \alpha + 2\sin \alpha \csc \alpha) + (\cos^2 \alpha + \sec^2 \alpha + 2\cos \alpha \sec \alpha) \)
Since \( \sin \alpha \csc \alpha = 1 \) and \( \cos \alpha \sec \alpha = 1 \):
\( (\sin^2 \alpha + \cos^2 \alpha) + \csc^2 \alpha + \sec^2 \alpha + 2 + 2 \)
Using identities \( \sin^2 \alpha + \cos^2 \alpha = 1 \), \( \csc^2 \alpha = 1 + \cot^2 \alpha \), \( \sec^2 \alpha = 1 + \tan^2 \alpha \):
\( 1 + (1 + \cot^2 \alpha) + (1 + \tan^2 \alpha) + 4 \)
\( 1 + 1 + 1 + 4 + \tan^2 \alpha + \cot^2 \alpha = 7 + \tan^2 \alpha + \cot^2 \alpha \)
Comparing with \( K + \tan^2\alpha + \cot^2\alpha \), we get \( K=7 \).
Answer: b) 7

Part II: Answer any 10 questions. 28th question is compulsory (10 x 2 = 20)

15) If A={m, n}; B = \( \phi \), Find i) A×B ii) A×A

i) \( A \times B \): The cartesian product with an empty set is always an empty set.
\( A \times B = \{m, n\} \times \phi = \phi \)
ii) \( A \times A \):
\( A \times A = \{m, n\} \times \{m, n\} = \{(m,m), (m,n), (n,m), (n,n)\} \)

16) A function f is defined by \( f(x)=3-2x \). Find x such that \( f(x^2) = [f(x)]^2 \)

Given \( f(x)=3-2x \).
LHS: \( f(x^2) = 3 - 2(x^2) = 3 - 2x^2 \)
RHS: \( [f(x)]^2 = (3-2x)^2 = 9 - 12x + 4x^2 \)
Equating LHS and RHS: \( 3 - 2x^2 = 9 - 12x + 4x^2 \)
\( 6x^2 - 12x + 6 = 0 \)
Divide by 6: \( x^2 - 2x + 1 = 0 \)
\( (x-1)^2 = 0 \implies x-1=0 \implies x=1 \)
The value of x is 1.

17) Find the value of k, such that fog=gof if \( f(x)=3x+2 \) and \( g(x)=6x-k \)

LHS: fog = \( f(g(x)) = f(6x-k) = 3(6x-k) + 2 = 18x - 3k + 2 \)
RHS: gof = \( g(f(x)) = g(3x+2) = 6(3x+2) - k = 18x + 12 - k \)
Given fog = gof:
\( 18x - 3k + 2 = 18x + 12 - k \)
\( -3k + 2 = 12 - k \)
\( 2 - 12 = -k + 3k \)
\( -10 = 2k \implies k = -5 \)
The value of k is -5.

18) Use Euclid's Division Algorithm to find HCF of 340 and 412

By Euclid's Division Algorithm, \( a = bq + r \), where \( 0 \le r < b \).
Step 1: \( 412 = 340 \times 1 + 72 \)
Step 2: \( 340 = 72 \times 4 + 52 \)
Step 3: \( 72 = 52 \times 1 + 20 \)
Step 4: \( 52 = 20 \times 2 + 12 \)
Step 5: \( 20 = 12 \times 1 + 8 \)
Step 6: \( 12 = 8 \times 1 + 4 \)
Step 7: \( 8 = 4 \times 2 + 0 \)
The last non-zero remainder is the HCF.
HCF of 340 and 412 is 4.

19) Find x, y and z given that the numbers x, 10, y, 24, z are in A.P

Since the numbers are in an Arithmetic Progression (A.P), the common difference 'd' is constant.
Let the terms be \( a_1=x, a_2=10, a_3=y, a_4=24, a_5=z \).
We can find 'd' using known terms: \( d = a_4 - a_3 = a_3 - a_2 \). So, \( 24 - y = y - 10 \).
\( 2y = 34 \implies y = 17 \).
Now find d: \( d = y - 10 = 17 - 10 = 7 \).
Find x: \( x = a_1 = a_2 - d = 10 - 7 = 3 \).
Find z: \( z = a_5 = a_4 + d = 24 + 7 = 31 \).
x = 3, y = 17, z = 31.

20) Simplify \( \frac{5t^3}{4t-8} \times \frac{6t-12}{10t} \)

Factor the expressions:
\( \frac{5t^3}{4(t-2)} \times \frac{6(t-2)}{10t} \)
Cancel the common term \( (t-2) \):
\( \frac{5t^3}{4} \times \frac{6}{10t} \)
Multiply the fractions: \( \frac{30t^3}{40t} \)
Simplify the expression: \( \frac{3}{4} t^{3-1} = \frac{3}{4} t^2 \)
Simplified form: \( \frac{3}{4} t^2 \)

21) Find the sum and product of roots for \( 3 + \frac{1}{a} = \frac{10}{a^2} \)

First, convert the equation to standard quadratic form \( Ax^2 + Bx + C = 0 \). Here the variable is 'a'.
Multiply the entire equation by \( a^2 \) to clear the denominators:
\( 3a^2 + a^2(\frac{1}{a}) = a^2(\frac{10}{a^2}) \)
\( 3a^2 + a = 10 \)
\( 3a^2 + a - 10 = 0 \)
For a quadratic equation \( Ax^2+Bx+C=0 \):
Sum of roots = \( -\frac{B}{A} = -\frac{1}{3} \)
Product of roots = \( \frac{C}{A} = \frac{-10}{3} \)
Sum of roots = \( -\frac{1}{3} \), Product of roots = \( -\frac{10}{3} \)

22) If the difference between the roots of the equation \( x^2 - 13x + k = 0 \) is 17, find K.

Let the roots be \( \alpha \) and \( \beta \).
From the equation, sum of roots \( \alpha + \beta = -(\frac{-13}{1}) = 13 \).
Product of roots \( \alpha \beta = \frac{k}{1} = k \).
Given, difference between roots \( |\alpha - \beta| = 17 \).
We have two equations:
\( \alpha + \beta = 13 \) --- (1)
\( \alpha - \beta = 17 \) --- (2)
Adding (1) and (2): \( 2\alpha = 30 \implies \alpha = 15 \).
Substitute \( \alpha=15 \) into (1): \( 15 + \beta = 13 \implies \beta = -2 \).
Now find k: \( k = \alpha \beta = 15 \times (-2) = -30 \).
The value of K is -30.

Part II Solutions (Questions 23-27)

23) In the figure AD is the bisector of $\angle A$, If BD=4cm DC=3cm and AB=6cm, find AC.

Triangle Diagram for Question 23
By the Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side in the same ratio as the other two sides.
So, we have the proportion:
$\frac{AB}{AC} = \frac{BD}{DC}$
Substitute the given values into the equation:
$\frac{6}{AC} = \frac{4}{3}$
Now, solve for AC by cross-multiplying:
$4 \times AC = 6 \times 3$
$4 \times AC = 18$
$AC = \frac{18}{4} = 4.5$ cm
The length of AC is 4.5 cm.

24) Find the area of the triangle, formed by the (1,−1), (-4, 6) and (-3, -5)

Let the vertices be A(1, -1), B(-4, 6), and C(-3, -5).
The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:
Area = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1)|$
Substitute the coordinates:
Area = $\frac{1}{2} |((1)(6) + (-4)(-5) + (-3)(-1)) - ((-1)(-4) + (6)(-3) + (-5)(1))|$
Area = $\frac{1}{2} |(6 + 20 + 3) - (4 - 18 - 5)|$
Area = $\frac{1}{2} |(29) - (-19)|$
Area = $\frac{1}{2} |29 + 19| = \frac{1}{2} |48|$
Area = 24
The area of the triangle is 24 square units.

25) Find the intercepts made by the line $4x-9y+36=0$ on the coordinate axes

The given equation of the line is $4x - 9y + 36 = 0$.
To find the x-intercept, set $y = 0$:
$4x - 9(0) + 36 = 0$
$4x = -36 \implies x = -9$
So, the x-intercept is -9. The point is (-9, 0).
To find the y-intercept, set $x = 0$:
$4(0) - 9y + 36 = 0$
$-9y = -36 \implies y = 4$
So, the y-intercept is 4. The point is (0, 4).
The x-intercept is -9 and the y-intercept is 4.

26) Show that the straight lines $x-2y+3=0$ and $6x+3y+8=0$ are perpendicular

Two lines are perpendicular if the product of their slopes is -1 (i.e., $m_1 \times m_2 = -1$).
First line: $x - 2y + 3 = 0$. Let's find its slope ($m_1$).
Rearrange to the form $y=mx+c$: $2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2}$.
So, the slope $m_1 = \frac{1}{2}$.
Second line: $6x + 3y + 8 = 0$. Let's find its slope ($m_2$).
Rearrange to the form $y=mx+c$: $3y = -6x - 8 \implies y = -2x - \frac{8}{3}$.
So, the slope $m_2 = -2$.
Now, multiply the slopes:
$m_1 \times m_2 = (\frac{1}{2}) \times (-2) = -1$.
Since the product of the slopes is -1, the lines are perpendicular.
Hence, the given straight lines are perpendicular.

27) Prove that $\frac{\cos \theta}{1 + \sin \theta} = \sec\theta - \tan\theta$

We start with the Left Hand Side (LHS): $\frac{\cos \theta}{1 + \sin \theta}$
Multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 - \sin \theta)$:
LHS = $\frac{\cos \theta}{1 + \sin \theta} \times \frac{1 - \sin \theta}{1 - \sin \theta}$
LHS = $\frac{\cos \theta (1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} = \frac{\cos \theta (1 - \sin \theta)}{1 - \sin^2 \theta}$
Using the identity $\sin^2\theta + \cos^2\theta = 1$, we know that $1 - \sin^2 \theta = \cos^2 \theta$.
LHS = $\frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta}$
Cancel out a $\cos \theta$ term:
LHS = $\frac{1 - \sin \theta}{\cos \theta}$
Split the fraction into two parts:
LHS = $\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}$
Using the identities $\sec\theta = \frac{1}{\cos\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$:
LHS = $\sec\theta - \tan\theta$
Thus, LHS = RHS.
Hence Proved.

28) Find the sum \( 3+1+\frac{1}{3}+.....\infty \)

This is an infinite geometric progression (G.P.).
First term \( a = 3 \).
Common ratio \( r = \frac{1}{3} \).
Since \( |r| = |\frac{1}{3}| < 1 \), the sum to infinity exists.
The formula for the sum to infinity is \( S_\infty = \frac{a}{1-r} \).
\( S_\infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{2}{3}} = 3 \times \frac{3}{2} = \frac{9}{2} \).
The sum is \( \frac{9}{2} \) or 4.5.

Part III Solutions (Questions 29-31)

29) Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that $(A \cap B) \times C = (A \times C) \cap (B \times C)$

First, let's write the sets in roster form:
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
LHS: $(A \cap B) \times C$
First, find $A \cap B$:
$A \cap B = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2, 3, 5, 7\} = \{2, 3, 5, 7\}$
Now, find the Cartesian product with C:
$(A \cap B) \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\}$ --- (1)
RHS: $(A \times C) \cap (B \times C)$
First, find $A \times C$:
$A \times C = \{1, 2, 3, 4, 5, 6, 7\} \times \{2\} = \{(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)\}$
Next, find $B \times C$:
$B \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\}$
Now, find the intersection of these two sets:
$(A \times C) \cap (B \times C) = \{(2, 2), (3, 2), (5, 2), (7, 2)\}$ --- (2)
From (1) and (2), we can see that LHS = RHS.
Hence, $(A \cap B) \times C = (A \times C) \cap (B \times C)$ is verified.

30) Let f: A → B be a function defined by $f(x) = \frac{x}{2} - 1$ where A = {2, 4, 6, 10, 12} B = {0, 1, 2, 4, 5, 9}. Represent f by (i) Set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph

Given function: $f(x) = \frac{x}{2} - 1$. Let's find the image for each element in A.
$f(2) = \frac{2}{2} - 1 = 1 - 1 = 0$
$f(4) = \frac{4}{2} - 1 = 2 - 1 = 1$
$f(6) = \frac{6}{2} - 1 = 3 - 1 = 2$
$f(10) = \frac{10}{2} - 1 = 5 - 1 = 4$
$f(12) = \frac{12}{2} - 1 = 6 - 1 = 5$
(i) Set of ordered pairs:
$f = \{(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)\}$
(ii) A table:
x f(x)
2 0
4 1
6 2
10 4
12 5
(iii) An arrow diagram:
Draw two ovals. Label the first 'A' and the second 'B'.
Inside A, list the elements: 2, 4, 6, 10, 12.
Inside B, list the elements: 0, 1, 2, 4, 5, 9.
Draw arrows from each element in A to its corresponding image in B:
2 → 0,   4 → 1,   6 → 2,   10 → 4,   12 → 5
Arrow diagram for function f
(iv) A graph:
Plot the ordered pairs on a coordinate plane with the x-axis and y-axis. The points to be plotted are:
(2, 0), (4, 1), (6, 2), (10, 4), and (12, 5).
Graph for function f

31) If $f(x)=x^2$, $g(x) = 2x$ and $h(x) = x + 4$ show that $(fog)oh = fo(goh)$

LHS: $(fog)oh$
First, find $fog(x)$:
$fog(x) = f(g(x)) = f(2x) = (2x)^2 = 4x^2$
Now, find $(fog)oh(x)$:
$(fog)oh(x) = (fog)(h(x)) = (fog)(x+4)$
Substitute $(x+4)$ into the expression for $fog(x)$:
$(fog)oh(x) = 4(x+4)^2$ --- (1)
RHS: $fo(goh)$
First, find $goh(x)$:
$goh(x) = g(h(x)) = g(x+4) = 2(x+4)$
Now, find $fo(goh)(x)$:
$fo(goh)(x) = f(goh(x)) = f(2(x+4))$
Substitute $2(x+4)$ into the function $f(x)$:
$fo(goh)(x) = [2(x+4)]^2 = 4(x+4)^2$ --- (2)
From (1) and (2), we see that the expressions are identical.
LHS = RHS. Thus, $(fog)oh = fo(goh)$ is shown.

Part III: Answer any 10 questions. 42th question is compulsory (10 x 5 = 50)

32) Find the sum of all natural numbers between 300 and 600 which are divisible by 7.

The sequence of numbers between 300 and 600 divisible by 7 forms an A.P.
To find the first term (a), divide 300 by 7: \( 300 \div 7 = 42 \) with remainder 6. So, the first number is \( 300 + (7-6) = 301 \). So, \( a=301 \).
To find the last term (l), divide 600 by 7: \( 600 \div 7 = 85 \) with remainder 5. So, the last number is \( 600 - 5 = 595 \). So, \( l=595 \).
The common difference \( d = 7 \).
Number of terms \( n = \frac{l-a}{d} + 1 = \frac{595-301}{7} + 1 = \frac{294}{7} + 1 = 42 + 1 = 43 \).
Sum of the A.P. is \( S_n = \frac{n}{2}(a+l) \).
\( S_{43} = \frac{43}{2}(301+595) = \frac{43}{2}(896) = 43 \times 448 = 19264 \).
The sum is 19264.

Part III Solution (Question 33)

33) If a, b, c are three consecutive terms of an A.P and x,y,z are three consecutive terms of a G.P. Then prove that $x^{b-c} y^{c-a} z^{a-b} = 1$

Step 1: Analyze the given conditions.
Since a, b, c are in an Arithmetic Progression (A.P.), the common difference is constant. Let the common difference be 'd'.
$b - a = d$ and $c - b = d$.
From these, we can express the exponents in terms of 'd':
  • $b - c = -(c - b) = -d$
  • $c - a = (c - b) + (b - a) = d + d = 2d$
  • $a - b = -(b - a) = -d$
Since x, y, z are in a Geometric Progression (G.P.), the common ratio is constant. Let the first term be x and the common ratio be 'r'.
$y = xr$
$z = xr^2$
Step 2: Substitute these relationships into the Left Hand Side (LHS) of the equation.
LHS = $x^{b-c} y^{c-a} z^{a-b}$
First, substitute the expressions for the exponents derived from the A.P. condition:
LHS = $x^{-d} y^{2d} z^{-d}$
Next, substitute the expressions for y and z derived from the G.P. condition:
LHS = $x^{-d} (xr)^{2d} (xr^2)^{-d}$
Step 3: Simplify the expression using the laws of exponents.
Using the rule $(ab)^n = a^n b^n$:
LHS = $x^{-d} \cdot (x^{2d} r^{2d}) \cdot (x^{-d} (r^2)^{-d})$
Using the rule $(a^m)^n = a^{mn}$:
LHS = $x^{-d} \cdot x^{2d} \cdot r^{2d} \cdot x^{-d} \cdot r^{-2d}$
Group the terms with the same base (x and r) and use the rule $a^m a^n = a^{m+n}$:
LHS = $(x^{-d + 2d - d}) \cdot (r^{2d - 2d})$
Simplify the exponents:
LHS = $x^{0} \cdot r^{0}$
Since any non-zero number raised to the power of 0 is 1:
LHS = $1 \cdot 1 = 1$
Step 4: Conclude the proof.
We have shown that LHS = 1, which is equal to the Right Hand Side (RHS).
Hence, $x^{b-c} y^{c-a} z^{a-b} = 1$ is proved.

34) Find the sum of \( 9^3 + 10^3 + ... + 21^3 \)

We can write the sum as \( (1^3 + 2^3 + ... + 21^3) - (1^3 + 2^3 + ... + 8^3) \).
Using the formula for the sum of cubes of first n natural numbers: \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \).
For n=21: \( \left( \frac{21(21+1)}{2} \right)^2 = \left( \frac{21 \times 22}{2} \right)^2 = (21 \times 11)^2 = 231^2 = 53361 \).
For n=8: \( \left( \frac{8(8+1)}{2} \right)^2 = \left( \frac{8 \times 9}{2} \right)^2 = (4 \times 9)^2 = 36^2 = 1296 \).
Required sum = \( 53361 - 1296 = 52065 \).
The sum is 52065.

Part III Solutions (Questions 35-41)

35) Find the LCM of each pair of the following Polynomials $a^2+4a-12$, $a^2-5a+6$ whose GCD is $a-2$

Let the two polynomials be $p(x) = a^2+4a-12$ and $q(x) = a^2-5a+6$.
We know the relationship between LCM, GCD, and the polynomials:
$LCM \times GCD = p(x) \times q(x)$
$LCM = \frac{p(x) \times q(x)}{GCD}$
First, let's factorize the polynomials:
$p(x) = a^2+4a-12 = (a+6)(a-2)$
$q(x) = a^2-5a+6 = (a-3)(a-2)$
Given, $GCD = a-2$.
Now, substitute the factored polynomials and GCD into the formula:
$LCM = \frac{(a+6)(a-2) \times (a-3)(a-2)}{a-2}$
Cancel one of the $(a-2)$ terms from the numerator and denominator:
$LCM = (a+6)(a-3)(a-2)$
The LCM is $(a-2)(a-3)(a+6)$.

36) If $9x^4+12x^3+28x^2+ax + b$ is a perfect square Find the values of 'a' and 'b'

We can find the values of 'a' and 'b' using the long division method for finding the square root.
3x² + 2x + 4 _____________________ 3x² | 9x⁴ + 12x³ + 28x² + ax + b |-(9x⁴) |____________________ 6x²+2x| 12x³ + 28x² | -(12x³ + 4x²) | _________________ 6x²+4x+4| 24x² + ax + b | -(24x² + 16x + 16) | __________________ | 0
Since the polynomial is a perfect square, the remainder must be 0.
This means $(24x^2 + ax + b) - (24x^2 + 16x + 16) = 0$.
$(a-16)x + (b-16) = 0$.
Equating the coefficients of x and the constant term to zero:
$a - 16 = 0 \implies a = 16$
$b - 16 = 0 \implies b = 16$
The values are a = 16 and b = 16.

37) If $\alpha, \beta$ are the roots of $7x^2+ax+2=0$ and if $\beta - \alpha = -\frac{13}{7}$. Find the values of 'a'

For the quadratic equation $7x^2+ax+2=0$, the sum and product of roots are:
Sum of roots: $\alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{a}{7}$
Product of roots: $\alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{2}{7}$
Given: $\beta - \alpha = -\frac{13}{7}$
We use the identity: $(\alpha+\beta)^2 = (\beta-\alpha)^2 + 4\alpha\beta$
Substitute the known values into the identity:
$(-\frac{a}{7})^2 = (-\frac{13}{7})^2 + 4(\frac{2}{7})$
$\frac{a^2}{49} = \frac{169}{49} + \frac{8}{7}$
To add the fractions on the right, find a common denominator (49):
$\frac{a^2}{49} = \frac{169}{49} + \frac{8 \times 7}{7 \times 7} = \frac{169}{49} + \frac{56}{49}$
$\frac{a^2}{49} = \frac{169+56}{49} = \frac{225}{49}$
$a^2 = 225$
$a = \pm\sqrt{225} = \pm 15$
The values of 'a' are 15 and -15.

38) State and prove Angle Bisector Theorem

Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Angle Bisector Theorem Diagram Given: In $\triangle ABC$, AD is the internal bisector of $\angle A$ which meets the side BC at D.
To Prove: $\frac{AB}{AC} = \frac{BD}{DC}$
Construction: Draw a line CE parallel to AB. Extend the line segment AD to meet the line CE at E.
Proof:
Statement Reason
In $\triangle DCE$ and $\triangle DBA$, $\angle DCE \cong \angle DBA$ and $\angle DEC \cong \angle DAB$. Thus $\triangle DCE \sim \triangle DBA$. AA Similarity (Since CE || AB, corresponding angles are equal). This approach is complex. Let's use angles directly.
Proof (Alternative and Standard Method):
Statement Reason
$\angle BAE = \angle AEC$ (i.e. $\angle 1 = \angle 3$) Since AB || CE and AE is the transversal, alternate interior angles are equal.
$\angle CAD = \angle ACE$ (i.e. $\angle 2 = \angle 4$) Since AB || CE and AC is transversal, this is incorrect. The correct reason is for corresponding angles. Let's restart the angles.
Correct Proof:
In $\triangle BCE$, since AD || CE, by Thales Theorem (BPT), we have $\frac{AB}{AE} = \frac{BD}{DC}$ --- (1)
Since AB || CE and AC is the transversal, $\angle BAE = \angle AEC$ (Alternate angles).
Also, with transversal BE, $\angle DAB = \angle AEC$ (Corresponding angles). Wait, construction is different. Let's use the standard construction: Draw CE || DA to meet BA extended at E.
Correct Standard Proof:
Construction: Through C, draw a line CE parallel to AD, such that it meets the line BA extended at E.
Proof:
Since AD || CE, by Basic Proportionality Theorem on $\triangle BCE$, we have $\frac{BA}{AE} = \frac{BD}{DC}$ --- (i)
Now, since AD || CE and AC is the transversal, $\angle DAC = \angle ACE$ (Alternate interior angles).
Since AD || CE and BE is the transversal, $\angle BAD = \angle AEC$ (Corresponding angles).
But we are given that AD is the angle bisector, so $\angle BAD = \angle DAC$.
Therefore, $\angle ACE = \angle AEC$.
In $\triangle ACE$, since the angles opposite to sides AE and AC are equal, the sides themselves are equal. Hence, $AE = AC$.
Substitute $AC$ for $AE$ in equation (i):
$\frac{AB}{AC} = \frac{BD}{DC}$. Hence the theorem is proved.

39) If the points A(-3, 9) B(a, b) and C(4, -5) are collinear and if a+b=1, then find 'a' and 'b'

Since the points A, B, and C are collinear, the slope of AB must be equal to the slope of BC.
Slope of AB = $\frac{b - 9}{a - (-3)} = \frac{b-9}{a+3}$
Slope of BC = $\frac{-5 - b}{4 - a}$
Equating the slopes: $\frac{b-9}{a+3} = \frac{-5-b}{4-a}$
Cross-multiply: $(b-9)(4-a) = (a+3)(-5-b)$
$4b - ab - 36 + 9a = -5a - ab - 15 - 3b$
The term '-ab' cancels from both sides.
$4b - 36 + 9a = -5a - 15 - 3b$
Rearrange the terms to form a linear equation:
$9a + 5a + 4b + 3b = 36 - 15$
$14a + 7b = 21$
Divide the entire equation by 7: $2a + b = 3$ --- (1)
We are also given: $a + b = 1$ --- (2)
Subtract equation (2) from equation (1):
$(2a + b) - (a + b) = 3 - 1$
$a = 2$
Substitute $a=2$ into equation (2):
$2 + b = 1 \implies b = 1 - 2 = -1$
The values are a = 2 and b = -1.

40) If the points A(2, 2) B(-2, −3) C(1, −3) and D(x, y) form a parallelogram then find the value of 'x' and 'y'

A property of a parallelogram is that its diagonals bisect each other. This means the midpoint of diagonal AC is the same as the midpoint of diagonal BD.
Midpoint Formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
Midpoint of AC = $(\frac{2+1}{2}, \frac{2+(-3)}{2}) = (\frac{3}{2}, -\frac{1}{2})$
Midpoint of BD = $(\frac{-2+x}{2}, \frac{-3+y}{2})$
Equating the x-coordinates and y-coordinates of the midpoints:
For x-coordinate: $\frac{-2+x}{2} = \frac{3}{2} \implies -2+x = 3 \implies x = 5$
For y-coordinate: $\frac{-3+y}{2} = -\frac{1}{2} \implies -3+y = -1 \implies y = 2$
The value of x is 5 and the value of y is 2. The coordinates of D are (5, 2).

41) Prove that $\frac{\cos^3 A – \sin^3 A}{\cos A - \sin A} - \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = 2 \sin A \cos A$

We will simplify the Left Hand Side (LHS) using the sum and difference of cubes formulas:
$a^3 - b^3 = (a-b)(a^2+ab+b^2)$
$a^3 + b^3 = (a+b)(a^2-ab+b^2)$
First term of LHS: $\frac{\cos^3 A – \sin^3 A}{\cos A - \sin A}$
$= \frac{(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A)}{\cos A - \sin A}$
Cancel the $(\cos A - \sin A)$ term. Since $\cos^2 A + \sin^2 A = 1$, the expression becomes:
$= 1 + \cos A \sin A$
Second term of LHS: $\frac{\cos^3 A + \sin^3 A}{\cos A + \sin A}$
$= \frac{(\cos A + \sin A)(\cos^2 A - \cos A \sin A + \sin^2 A)}{\cos A + \sin A}$
Cancel the $(\cos A + \sin A)$ term. Since $\cos^2 A + \sin^2 A = 1$, the expression becomes:
$= 1 - \cos A \sin A$
Now, substitute these simplified parts back into the LHS expression:
LHS = $(1 + \cos A \sin A) - (1 - \cos A \sin A)$
LHS = $1 + \cos A \sin A - 1 + \cos A \sin A$
LHS = $2 \cos A \sin A$
LHS = RHS
Hence, the identity is proved.

42) If A(-3, 0) B(10, −2) and C(12, 3) are the vertices of ∆ABC. Find the equation of the altitude through 'A'.

The altitude through A is a line segment from vertex A that is perpendicular to the opposite side BC.
First, find the slope of the side BC.
Slope of BC \( (m_1) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{12 - 10} = \frac{5}{2} \).
The altitude from A is perpendicular to BC, so its slope \( (m_2) \) is the negative reciprocal of the slope of BC.
\( m_2 = -\frac{1}{m_1} = -\frac{1}{5/2} = -\frac{2}{5} \).
Now, we have the slope of the altitude \( m_2 = -2/5 \) and a point it passes through, A(-3, 0).
Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \):
\( y - 0 = -\frac{2}{5}(x - (-3)) \)
\( y = -\frac{2}{5}(x + 3) \)
\( 5y = -2(x + 3) \implies 5y = -2x - 6 \)
\( 2x + 5y + 6 = 0 \)
The equation of the altitude through A is \( 2x + 5y + 6 = 0 \).

Part IV: Answer all the questions (2 x 8 = 16)

43) Construct a triangle similar to a given triangle PQR with its sides equal to 7/4 of the corresponding sides of the triangle PQR (scale factor 7/4 > 1)

Steps of Construction:
  1. Draw a triangle PQR with any suitable measurements.
  2. Draw a ray PX from P on the side opposite to vertex R, making an acute angle with PQ.
  3. Since the scale factor is 7/4, locate 7 (the greater of 7 and 4) points P₁, P₂, P₃, P₄, P₅, P₆, P₇ on the ray PX such that PP₁ = P₁P₂ = ... = P₆P₇.
  4. Join the 4th point (P₄, corresponding to the denominator) to Q.
  5. Extend the line segment PQ. Draw a line from P₇ parallel to P₄Q, which intersects the extended line PQ at Q'.
  6. Extend the line segment PR. Draw a line from Q' parallel to QR, which intersects the extended line PR at R'.
  7. The triangle PQ'R' is the required similar triangle whose sides are 7/4 of the corresponding sides of $\triangle PQR$.
Construction of similar triangle with scale factor 7/4
Justification: By construction, Q'R' is parallel to QR. Therefore, $\triangle PQ'R' \sim \triangle PQR$. Also, $\frac{PQ'}{PQ} = \frac{PP_7}{PP_4} = \frac{7}{4}$. Thus, the corresponding sides are in the ratio 7/4.

(OR) Construct a $\triangle PQR$ which the base PQ=4.5 cm $\angle R.= 35°$ and the median RG from R to PQ is 6 cm

Steps of Construction:
  1. Draw a line segment PQ = 4.5 cm.
  2. At point P, draw a line PE such that $\angle QPE = 35°$.
  3. At point P, draw a line PF perpendicular to PE (i.e., $\angle EPF = 90°$).
  4. Draw the perpendicular bisector of the line segment PQ. Let it intersect PQ at G and the line PF at O.
  5. With O as the center and OP (or OQ) as the radius, draw a circle. All points on the major arc of this circle will subtend an angle of 35° at the segment PQ.
  6. G is the midpoint of PQ. From G, with a radius of 6 cm (the length of the median), draw an arc.
  7. This arc intersects the circle at two points. Label one of these intersection points as R.
  8. Join PR and QR.
  9. The triangle PQR is the required triangle.
Construction of triangle with given base, vertex angle, and median

44) Draw the graph of xy=24, x,y>0. Using the graph find (i) x when y = 6 (ii) y when x = 3

The equation is $xy=24$, which represents an indirect variation. We can write it as $y = \frac{24}{x}$. Since $x, y > 0$, the graph will be in the first quadrant.
Table of Values:
x 1 2 3 4 6 8 12
y 24 12 8 6 4 3 2
Plotting the Graph:
Plot the points (1, 24), (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2) on a graph sheet. Join the points with a smooth curve. This curve is a rectangular hyperbola.
Scale: X-axis: 1 cm = 2 units, Y-axis: 1 cm = 2 units.
Graph of xy=24
From the graph:
(i) To find x when y = 6: On the Y-axis, locate the point for y=6. Draw a horizontal line from this point to meet the curve. From the point of intersection on the curve, draw a vertical line down to the X-axis. The line meets the X-axis at x = 4.
(ii) To find y when x = 3: On the X-axis, locate the point for x=3. Draw a vertical line from this point to meet the curve. From the point of intersection on the curve, draw a horizontal line to the Y-axis. The line meets the Y-axis at y = 8.
(i) When y = 6, x = 4.
(ii) When x = 3, y = 8.

(OR) A bus is travelling at a uniform speed of 50 km/hr Draw the distance time graph and hence find
i) the constant of variation
ii) how far will it travel in 90 minutes?
iii) the time required to cover a distance of 300 km from the graph.

The relationship between distance (y) and time (x) at a uniform speed is a direct variation, given by the equation: Distance = Speed × Time.
So, $y = 50x$, where y is the distance in km and x is the time in hours.
Table of Values:
Time (x) in hours 1 2 3 4 5
Distance (y) in km 50 100 150 200 250
Plotting the Graph:
Plot the points (1, 50), (2, 100), (3, 150), etc., on a graph sheet. Join the points to get a straight line passing through the origin.
Scale: X-axis: 1 cm = 1 hour, Y-axis: 1 cm = 50 km.
Distance-Time Graph
Answering the questions:
i) The equation is $y=50x$. Comparing this with the direct variation equation $y=kx$, the constant of variation (k) is 50.
ii) To find the distance travelled in 90 minutes: 90 minutes = 1.5 hours. From the graph, draw a vertical line from x=1.5 on the X-axis to the graph line. Then draw a horizontal line to the Y-axis. It meets the Y-axis at y=75. So, the distance is 75 km.
iii) To find the time to cover 300 km: From y=300 on the Y-axis, draw a horizontal line to the graph line. Then draw a vertical line down to the X-axis. It meets the X-axis at x=6. So, the time required is 6 hours.
i) The constant of variation is 50 km/hr.
ii) The bus will travel 75 km in 90 minutes.
iii) The time required to cover 300 km is 6 hours.