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Subject: 

Fine feathers Predicate: don’t make fine birds

Answer is: boyish

Clay and his companion dug the underground tunnel to enter the bank and steal the gold without breaking open the doors of the bank.

Mr. Spaulding and Mr. Ross came out of the square hole.

1. the way to the Strand

2. closed for the weekend

3. narrow passage

4. one of the main banks of London

1. Holmes and others go to the bank.

2. Clay and Ross enter the cellar to steal the gold.

Correct answer is (c) beside

Spaulding was spending a lot of time in the cellar of the pawnshop because he was making a tunnel to get into the strong room of the bank in the next street at the back of the shop.

The two criminals were none other than Mr. Vincent Spaulding alias Clay and Mr. Duncan Ross.

No, Clay was not successful in using his revolver because Mr. Holmes at once hit him on his wrist and the revolver fell on the floor.

Mr. Merryweather’s banks had borrowed a huge quantity of gold from the Bank of France and the boxes in the cellar were full of gold. Hence the criminals were interested in the cellar.

Holmes wanted to see the knees of Mr. Spaulding’s trousers. It was to see if thery were dirty.

We must always work diligently.

Mr. Wilson was a simpleton. He could be easily taken for a ride. He was hardworking and diligent.

Mr. Wilson described his assistant as small, stout, with no hair an his face and had a white splash of acid on his forehead.

The best thing to do when you know a crime is being committed or you are being cheated is to go to the lawful authority. I would suggest/ advice Mr. Wilson to go to the police and file a complaint against the people who had cheated him. It is better to do so rather than taking matters into your own hands and make the situation worse.

He was followed by us down a narrow passage.

We will go to the square where Mr. Wilson has his shop.

I feel Mr. Wilson made the right decision to take the help of Mr. Holmes about whom he had heard a lot.

diligently – adverb

four – adjective

A Family Drama:

After attending the reception cum dinner of their close friend’s daughter, at the famous five star hotel ‘The Oberoi’, Mr. Pai and his family were returning home discussing the evening. Mr. and Mrs Pai along with their two children Suhani and Soham had attended the function. It was quite late at night, and to add to it, it began to rain making the surroundings dark and scary.

Mr. Pai drove the car cautiously and reached their colony safely. Mr. Pai parked the car while the others waited for him at the entrance of their building. It was still raining and the watchman was nowhere in sight. The common light in the building was not on. The way leading to their house was in darkness. Somehow they gropped their way to the second floor, where they lived. As they reached their house, Mr. Pai felt that he heard footsteps and whispering coming from inside their house. He asked his wife to listen carefully and she too agreed that there was whispering. The conversation indicated robbers searching for valuables. There was no light in their house.

Mr. Pai was an army officer. He asked his family members to stand behind him and he opened the main door very slowly. He could hear conversation and movements too. He slowly went towards the bedroom from where the conversation was coming. The door was closed.

He took out his revolver and pushed the door open quickly only to understand that the conversation was from the T.V. serial ‘Crime Patrol’. He then remembered that he had been watching T.V. before leaving home and had forgotten to switch it off. The other family members who had followed him gave a huge sigh of relief.

We have,

x² + xy + xz + yz

= x (x + y) + z (x + y)

= (x + y) (x + z)

The half time break was over, the star players of 9A were brimming with confidence. They did an encore by catching hold of ace raiders Mohan, Ajinkya and Ravi. As they say, the best team wins. Since 9A displayed excellence in all areas of the game, they emerged the winners. This coupled with team spirit, their confidence, strategies, swiftness and suppleness helped them emerge the winners beating 9B – a team which had never been defeated in any game of Kabaddi.

With the spectators clapping and boosting the morale of both the teams, the match got off to a good start. Having won the toss 9A sent their raider, Govind. Catching the opponent unawares, he touched Suresh from 9B and in no time headed towards home thus scoring a point.

Ravi being a strong player, with the knowledge of the right techniques appeared relaxed as he was aware of his strengths. Mangesh was wrestled down by Ravi and his friend, but Mangesh proved his capability. Even as he was on the ground with the anti¬raiders catching hold of his waist and legs, he slowly inched towards his court.

The best efforts of the anti-raiders failed in pulling him back and Mangesh successfully touched the mid-line scoring three points and getting three players of the anti-raiders out from the game.

Now, 9B began to play by exercising caution. Mihir and one of their raider tasted early success as they managed to tag Sohan in 9A. The other few raids were futile. Now, it was the turn of Vivek, the

strategist. The remaining three players from 9B trapped Vivek. He slowly made his way back towards his court with the anti-raiders moving with him

When they caught Vivek’s arm and tried to pull him back, Vivek slipped his leg beyond the midline with the anti-raiders still hanging onto his arm. Thanks to Vivek’s efforts, the whole of 9B was declared out.

As given that,

25x² – 10x + 1 – 36y²

= (5x)² – 2(5x) + 1 – (6y)²

= (5x – 1)² – (6y)²

= (5x – 6y – 1) (5x + 6y – 1)

The rules of the game of Kabaddi and the qualities required to excel in this game are title two aspects discussed in this extract.

Courts, teams, opponents, players Extract: In Kabaddi, two teams of seven members each, face each other on a flat rectangular court, divided by a midline. The game is usually played in two halves with a halftime break in between. After the break, the teams exchange their sides on the court.

To play the game, each team sends ‘raiders’ across the midline to the other team. The raider tries to ‘tag’ the opponents and run back to his side, all in one breath. To show that he hasn’t inhaled again, he has to chant ‘Kabaddi-Kabaddi’ all the time. If he has to inhale again, he is ‘out’.

The opponents try to catch the raider and stop him from going back to his half till he loses his breath, and has to inhale again. If the raider manages to go back successfully, all the persons he has tagged and all those who have touched him are declared out. For each player declared out, the opposite team scores a point. If all the seven players in a team are ‘out’, the opposite team gets bonus points – a Iona. The team with the maximum points wins the match.

The game of Kabaddi requires good health, muscular strength, strategic skills, a lot of practice and above all, great determination.

To play the game, each team sends ‘raiders’ across the midline to the other team.

Let the number of deer be x.

Number of deer grazing in the field = x/2

Number of deer playing nearby = 3/4 x Number of remaining deer

= 3/4 x (x - x/2) = 3/4 x x/2 = 3x/8

Number of deer drinking water from the pond = 9

x - (x/2 + 3x/8) = 9

x - (4x + 3x/8) = 9

x - 7x/8 = 9

x/8 = 9

Multiplying both sides by 8, we obtain

x = 72

Hence, the total number of deer in the herd is 72..

Let f (x) = x³ + 6x² + 11x + 6 be the given polynomial.

The constant term in f (x) is 6 and factors of 6 are +1,+2,+3 and +6

Putting x = - 1 in f (x) we have, 

f (-1) = (-1)³ + 6 (-1)² + 11 (-1) + 6 

= -1 + 6 – 11 + 6 

= 0 

Therefore, 

(x + 1) is a factor of f (x) 

Similarly, (x + 2) and (x + 3) are factors of f (x). 

Since, f (x) is a polynomial of degree 3. So, it cannot have more than three linear factors. 

Therefore, 

f (x) = k (x + 1) (x + 2) (x + 3) 

x³ + 6x² + 11x + 6 = k (x + 1) (x + 2) (x + 3) 

Putting x = 0, on both sides we get, 

0 + 0 + 0 + 6 = k (0 + 1) (0 + 2) (0 + 3) 

6 = 6k 

k = 1 

Putting k = 1 in f (x) = k (x + 1) (x + 2) (x + 3), we get 

f (x) = (x + 1) (x + 2) (x + 3) 

Hence, 

x³ + 6x² + 11x + 6 = (x + 1) (x + 2) (x + 3)

Let f(x) = x³ + 6x² + 11x + 6 

Step 1: Find the factors of constant term 

Here constant term = 6 

Factors of 6 are ±1, ±2, ±3, ±6 

Step 2: Find the factors of f(x)

Let x + 1 = 0 

⇒ x = -1 

Put the value of x in f(x) 

f(-1) = (−1)³ + 6(−1)² + 11(−1) + 6 

= -1 + 6 -11 + 6 

= 12 – 12 

= 0 

So, (x + 1) is the factor of f(x) 

Let x + 2 = 0 

⇒ x = -2 

Put the value of x in f(x) 

f(-2) = (−2)³ + 6(−2)² + 11(−2) + 6 

= - 8 + 24 – 22 + 6 

= 0 

So, (x + 2) is the factor of f(x) 

Let x + 3 = 0

⇒ x = -3 

Put the value of x in f(x) 

f(-3) = (−3)³ + 6(−3)2 + 11(−3) + 6 

= -27 + 54 – 33 + 6 

= 0

So, (x + 3) is the factor of f(x) 

Hence, f(x) = (x + 1)(x + 2)(x + 3).

We have,

f(x) = 3x⁴ + 17x³ + 9x² - 7x - 10 and g(x) = x + 5'

In order to find whether the polynomials g (x) = x – (-5) is a factor of f (x) or not, it is sufficient to show that f (-5) = 0

Now,

f(x) = 3x⁴ + 17x³ + 9x² - 7x - 10

f (-5) = 3 (-5)⁴ + 17 (-5)³ + 9 (-5)2 – 7 (-5) – 10

= 3 * 625 + 17 * (-125) + 9 * 25 + 35 – 10

= 1875 – 2125 + 225 + 35 – 10

= 0

Hence, 

g (x) is a factor of f (x).

Let, p (x) = ax³ + 3x² - 3 and q (x) = 2x 3-5x+a be the given polynomials.

Now,

R₁ = Remainder when p (x) is divided by (x – 4)

= p (4)

= a (4)³ + 3 (4)² – 3 [Therefore, p (x) = ax³ + 3x² - 3]

= 64a + 48 – 3

R₁ = 64a + 45

And,

R₂ = Remainder when q (x) is divided by (x – 4)

= q (4)

= 2 (4)³ – 5 (4) + a [Therefore, q (x) = 2x³ - 5x + a]

= 128 – 20 + a

R₂ = 108 + a

(i) Given condition is,

R₁ = R₂ 

64a + 45 = 108 + a 

63a – 63 = 0 

63a = 63 

a = 1

(ii) Given condition is R₁ + R₂ = 0

64a + 45 + 108 + a = 0

65a + 153 = 0

65a = -153

a = \(\frac{-153}{65}\)

(iii) Given condition is 2R₁ – R₂ = 0

2 (64a + 45) – (108 + a) = 0

128a + 90 – 108 – a

127a – 18 = 0

127a = 18

a = \(\frac{18}{127}\)

There are 30 boys and 20 girls. A teacher can select any boy as a class monitor from 30 boys in 30 different ways and he can select any girl as a class monitor from 20 girls in 20 different ways.

∴ by the fundamental principle of addition,

the total number of ways a teacher can select a class monitor

= 30 + 20 = 50

Hence, there are 50 different ways to select a class monitor.

Context: George was very much upset on the strange doings of his younger brother Tom. He told the writer that Tom married a woman of his mother’s age. After a few days she died and left him everything she had. Saying this George Ramsay was so angry with Tom’s activity that he beat his hands on the table. 

Explanation: George told the writer that Tom had married a very old lady. Few days after the marriage she died and Tom inherited all her property. Tom had become richer than George. At this the writer burst into laughter. He thought that George would never forgive him for this behaviour. But the writer had no grudge against Tom. He enjoyed rich dinners with Tom many times. Tom did not still leave his habit of borrowing money but it was never more than a sovereign.

Reference: These lines have been taken from the essay ‘The Ant and the Grasshopper’written by W. S. Maugham. 

[ N.B. : The above reference will be used for all explanations of this lesson. ] 

Context : In this lesson the writer recalls the old fable of ‘The Ant and the Grasshopper.’ The ant worked hard throughout summer and stored food for winter. On the other hand the grasshopper only sang and did nothing. In the same way George worked hard and earned money. But his brother Tom borrowed money and spent it on his luxuries. 

Explanation : The writer had learnt many stories. The story ‘The Ant and the Grasshopper’ was one of them. It teaches a great moral that industry is rewarded and giddiness punished. The laborious ant was happy in winter while the lazy grasshopper suffered starvation. In the same way a man who is sincere and laborious is always happy and contented. But a man who idles away his time repents in the end.

Over-fishing may lead to stripping the fisheries. Then man will lose a rich source of protein. The decimation of forests will harm ecology. Moreover, several species of life that live in forest will face extinction.

Let a₂ – a₁ = a₃ – a₂ = ........ = a_n – a_{n – 1} = d (common difference)

Then,

  \(\frac{1}{a_1a_2}\) + \(\frac{1}{a_2a_3}\) + ... + \(\frac{1}{a_{n-1}a_n}\) = \(\frac{1}{d}\)[\(\frac{d}{a_1a_2}\) + \(\frac{d}{a_2a_3}\) + ... + \(\frac{d}{a_{n-1}a_n}\)]

= \(\frac{1}{d}\)[\(\frac{a_2-a_1}{a_1a_2}\) + \(\frac{a_3-a_2}{a_2a_3}\) + ... + \(\frac{a_n- a_{n-1}}{a_{n-1}a_n}\) ] = \(\frac{1}{d}\)[ \(\frac{1}{a_1} -\frac{1}{a_2}\) + \(\frac{1}{a_2} -\frac{1}{a_3}\)+ ... + \(\frac{1}{a_{n-1}} -\frac{1}{a_n}\)]

= \(\frac{1}{d}\)[\(\frac{1}{a_1} -\frac{1}{a_n}\)] =  \(\frac{1}{d}\)[ \(\frac{a_n-a_1}{a_1a_n}\)] 

= \(\frac{1}{d}\)[ \(\frac{(a_1 +(n-1)d) -a_1}{a_1a_n}\)] = \(\frac{n-1}{a_1a_n}\).

Let required five numbers be x₁, x₂, x₃, x₄, x₅ 

∴ 8, x₁, x₂, x₃, x₄, x₅, 26 are in A.P. 

⇒ First term = 8 and seventh term = 26 

⇒ a = 8 and a + 6d - 26 

⇒ 3 = 8 and 6d = 26 – a = 26 – 8 = 18 

⇒ a = 8 and d = 3 

Hence x₁= a + d = 8 + 3 = 11 

x₂ = a + 2d = 8 + 6 = 14 

x₃ = 17, x₄ = 20, x₅ = 23

Answer : (b) 6

Let a₁, a₂, a₃, ..... , a_n be the n arithmetic means between 3 and 17. 

Then, 3, a₁, a₂, a₃, ..... , a_n, 17 form an A.P.

Let d be the common difference of this A.P. 

Then, a₁ = 3 + d and a_n = 17 – d

Given, 

\(\frac{a_n}{a_1} \) = \(\frac{3}{1}\) ⇒ \(\frac{17-d}{3+d}\) = \(\frac{3}{1}\) 

⇒ 17 – d = 9 + 3d 

⇒ 4d = 8 ⇒ d = 2. 

Also 17 is the (n + 2)th term of the given A.P. 

∴ 17 = 3 + (n + 2 – 1)2 

⇒ 17 = 3 + (n + 1)2 

⇒ 14 = (n + 1)2 

⇒ n + 1 = 7 

⇒ n = 6.

Correct option (c) 10,000

Explanation:

50, a₁ , a₂ ,...........a_n , 200 are in AP ......... (1)

 Also, 50, h₁ , h₂ ,.........,h_n 200 are in H.P

⇒ 1/50, 1/h₁, 1/h₂ , .... 1/h_n, 1/200 are in AP

⇒ 1/200, 1/h₁, 1/h_{n - 1}, ....1/h₁, 1/50 are in AP

Multiply by 200×50 = 10,000

⇒ 50, 10,000/h_n , 10,000/h_{n - 1}, ....10,000/h₂, 10,000/h₁, 200 are in AP ... (2)

Now (1) and (2) are identical.

⇒ a₂ = 10,000/h_{n - 1} gives a₂ .h_{n – 1} = 10,000

By data, the installments paid by the man are 

100, 105, 110, ………………  which is in A.P.with 

a = 100 and d = 5 

Hence 30th installment = 30^th term 

= a + 29d =100 + 29(5) = 245

विषम आवृत्ति वितरण में माध्यिका माध्य और भूयिष्ठक के बीच प्राप्त होगा ।\( \bar x\) > M > M _O अथवा \( \bar x\)< M < M_O

सही विकल्प है (D) विषम आवृत्ति वितरण में भूयिष्ठक के दोनों ओर समान दूरी पर आये अवलोकन में आवृत्ति समान रीति से वितरीत हुई है ।

सममितियता का अभाव अर्थात् विषमता ।