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YASH INDUSTRIES LIMITED

Registered Office: 102, New MIDC, Usha Tower, Shahu Chowk, Mumbai – 400 031.

CIN : L40407 MH 2005 PLC710007

Tel. No.: 022-23252323

Fax No.: 022-23600445 Ref. 

No.: Y/MR-B/5/19-20

Website: www.yashindustrieslimited.com

Email: [email protected] 

Date: 16th October, 2019

Ms. Yukta Shroff 

715, Narayan Peth, 

Laxmi Road, 

Pune – 411 038.

Sub.: Issue of Bonus Shares

Dear Madam,

I am directed by the Board of Directors to inform you that in accordance with the resolution passed in the Extra-ordinary General Meeting of the company held on 14th October, 2019, shareholders have unanimously approved the recommendation of Board of Directors to issue Bonus Shares. Bonus Shares are issued in the ratio of 1:1, i.e. one additional equity share for every equity share held as on 13th October, 2019.

The Details of issue of Bonus Shares are as follows:

The Company has complied with the provisions for the issue of Bonus Shares.

The Bonus Shares issued will rank pari passu /equal with the existing equity shares.

Thanking you,

Yours faithfully,

For Yash Industries Limited

Sign

(Mr. S. R. Naik)

Company Secretary

1. Capitalisation of reserves is the use of corporate reserves to pay a bonus to the shareholders in the form of additional shares. It is distributed to equity shareholders in pre-determined ratio.

2. Bonus shares are fully paid up shares given by a company as a gift out of its accumulated profits or reserves to the existing equity share holders in proportion of shares held by them. It is given free of cost and also known as capitalization of reserves.

3. Dividend warrant is written order given by the company to its banker to pay amount mentioned in it to the shareholder whose name is specified therein. Dividend can be paid through Dividend warrant or by means of electronic mode ECS or NEFT etc. to those shareholders who are registered members i.e their names appear in the Register of members.

Equity share holders enjoy the benefit of bonus shares.

Correct option: (c) 2

#include<stdio.h>
#include<conio.h>
void main( )
{
int n, i;
long int fact = 1;
clrscr ();
printf(“\n Enter any number”);
scanf(“%d”, & n);
for (i=1; i < = n; i + +)
{
fact = fact *i;
}
printf (“The factorial of % d is % 1d”, n, fact);
getch ();
}

\(\over3\times2\) is a multiple of 11.

A number is divisible by 11 if and only if the difference between the sum of odd and even place digits is a multiple of 11.

i.e., Sum of even placed digits – Sum of odd placed digits = 0, 11, 22…

x – (3+2) = 0, 11, 22…

x – 5 is a multiple of 11

x – 5 = 0

∴ x = 5

(i) By placing bars, we obtain

27225 = bar2 bar(72) bar(25) 

Since there are three bars, the square root of 27225 will have three digits in it.

(ii) By placing the bars, we obtain

390625 = bar(39) bar(06) bar(25)

Since there are three bars, the square root of 390625 will have 3 digits in it.

(i) Given 196 We know that from the property of perfect square, the square of an even number is always even.So the given number 196 is even, therefore it is square of even number

(ii) Given 441 We know that from the property of perfect square, the square of an even number is always even.But the given number 441 is odd, therefore it is not a square of even number

(iii) Given 900 We know that from the property of perfect square, the square of an even number is always even.So the given number 900 is even, therefore it is square of even number

(iv) Given 625 We know that from the property of perfect square, the square of an even number is always even.But the given number 625 is odd, therefore it is not a square of even number

(v) Given 324 We know that from the property of perfect square, the square of an even number is always even.So the given number 324 is even, therefore it is square of even number

(i) By placing bars, we obtain 64 = bar(64)

Since there is only one bar, the square root of 64 will have only one digit in it.

(ii) By placing bars, we obtain 144 = bar1 bar(44)

Since there are two bars, the square root of 144 will have 2 digits in it.

(iii) By placing bars, we obtain 4489 = bar(44) bar(89)

Since there are two bars, the square root of 4489 will have 2 digits in it.

Only even numbers be the square of even numbers 

So, 256, 324, 1296, 5476, 373758 can be square of even numbers but 373758 is not a perfect square 

So, 256, 324, 1296, 5476 are numbers

(i) 4 × (length of side)

Perimeter of regular polygon = 28 cm 

Measurement of each side = 7 cm 

Let number of sides in regular polygon is n, then, 

Total length of all sides = perimeter of polygon 

⇒ n × 7 = 28 

⇒ 7n = 28 

= n = 28/7 = 4

∴ Regular polygon has 4 sides only.

Length of each side = 3.5 cm 

Total number of sides = 3 

∴ Each side of regular polygon is equal 

∴ Its perimeter = 3.5+ 3.5 + 3.5 = 10.5 cm

Regular pentagon has 5 sides each side of this is of length 3 cm. 

Perimeter of regular pentagon 

= 5 × 3 cm = 15 cm.

Let the number of stamps of 20p and 25p be x and y respectively. 

Then as per the question 

x + y = 47 …….(i) 

0.20x + 0.25y = 10 

4x + 5y = 200 …….(ii) 

From (i), we get 

y = 47 – x 

Now, substituting y = 47 – x in (ii), we have 

4x + 5(47 – x) = 200 

⇒ 4x – 5x + 235 = 200 

⇒ x = 235 – 200 = 35 

Putting x = 35 in (i), we get 

35 + y = 47 

⇒ y = 47 – 35 = 12 

Hence, the number of 20p stamps and 25p stamps are 35 and 12 respectively

Let the number of stamps of 20p and 25p be x and y respectively. 

Then as per the question 

x + y = 47 …….(i) 

0.20x + 0.25y = 10 

4x + 5y = 200 …….(ii) 

From (i), we get 

y = 47 – x 

Now, substituting y = 47 – x in (ii), we have 

4x + 5(47 – x) = 200 

⇒ 4x – 5x + 235 = 200 

⇒ x = 235 – 200 = 35 

Putting x = 35 in (i), we get 

35 + y = 47 

⇒ y = 47 – 35 = 12 

Hence, the number of 20p stamps and 25p stamps are 35 and 12 respectively. 

Given digits = 0, 1, 2, 3, 4, 5

Hence, 6-digit numbers formed by these digits = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

If zero comes in first, then numbers becomes of 5-digit number then the 5-digit number formed is = 5!

= 5 × 4 × 3 × 2 × 1 = 120

Hence, the numbers that do not have zero in the begining are = 720 – 120 = 600.

Let ‘a’ be the first term and ‘d’ be the common difference of the give A.P. 

Middle most term = (11 + 1/2)^th = 6^th term 

t_n = a + (n – 1)d 

t₆ = a + 5d 

a + 5d = 30 

S_n = (n/2) [2a + (n – 1)d] 

S₁₁ = (11/2) [2a + 10d] 

= (11/2) x 2 [a + 5d] 

= 11 x 30

= 330 

Sum of 11 terms = 330

(c) 425

6 balls consisting of at least two balls of each colour from 5 red and 6 white balls can be made in the following ways: 

(a) Selecting 2 red balls out of 5 red balls and 4 white balls out of 6, i.e., 

Number of ways = ⁵C₂ x ⁶C₄ = \(\frac{5\times4}{2}\times\frac{6\times5}{2}\)

= 10 × 15 = 150 

(b) Selecting 3 red balls out of 5 red balls and 3 white balls out of 6, i.e., 

Number of ways =  ⁵C₃ x ⁶C₃ = \(\frac{5\times4}{2}\times\frac{6\times5\times4}{3\times2}\) 

= 10 × 20 = 200 

(c) Selecting 4 red balls out of 5 red balls and 2 white balls out of 6, i.e., 

Number of ways =  ⁵C₃ x ⁶C₂ = \(5\times\frac{6\times5}{2}\) = 5 × 15 = 75 

∴ Total number of ways of selecting at least two balls of each colour = 150 + 200 + 75 = 425.

Let us consider, larger number = x and

Smaller number = y

As per the statement,

3x = 4y + 8 …….(1)

5y = 3x + 5 …(2)

Using Substitution method:

Substitute the value of 3x in (2), we get

5y = 4y + 8 + 5

5y – 4y = 13

or y = 13

From (1): 3x = 4 x 13 + 8 = 60

x = 20

Answer:

Larger number = 20

Smaller number = 13

There are 10 lamps in a hall.

The hall can be illuminated if at least one lamp is switched.

.'. Total number of ways = ¹⁰C₁ + ¹⁰C₂ + ¹⁰C₃ … + ¹⁰C₁₀

= 2¹⁰ – 1 = 1024 - 1 = 1023

Area of 15 square colour papers 

= 10² + 11² + 12² + … + 24²

= (1² + 2² + 3² + … + 24²) – (1² + 2² + 9²)

= (24 x 25 x 49)/6 - (9 x 10 x 19)/6 

= 4 x 25 x 49 – 3 x 5 x 19 

= 4900 – 285 

= 4615 

Area can be decorated is 4615 cm²

There are two white, three black and four red balls.

We have to draw 3 balls, out of these 9 balls in which at least one black ball is included.

So we have following possibilities:

.’. Number of selections = ³C₁ x ⁶C₂ + ³C₂ x ⁶C₃ + ³C₃ x ⁶C₀

= 3×15+ 3×6+1= 45+ 18 + 1= 64

(a) 92 ___ 389:-

Let a be placed in the blank.

sum of the digits at odd places = 9 + 3 + 2 = 14

Sum of the digits at even places = 8 + a = 9 = 17a

Difference = 17 + a – 14 = 3 + a

For a number to be divisible by 11, this difference

should be zero or a multiple of 11.

If 3 + a = 0, then

a = -3

However, It cannot be negative.

A closest multiple of 11, which is near to 3, has to be taken. It is 11 itself

3 + a = 11

a = 11 – 3

a = 8

Therefore, the required digits is 8

(b) 8__9484:-

Let a be placed in the blank.

Sum of the digits at odds places = 4 + 4 + a = 8 + a

Sum of the digits at even place = 8 + 9 + 8 = 25

Difference = 25 – 8 + a

= 17 – a

For a number to be divisible by 11, this difference should be zero or a multiple of 11.

If 17 – a = 0, then

a = 17

This is not possible

A multiple of 11 has to be taken. Taking 11 we obtained

17 – a = 11

a = 6

Therefore the required digit is 6.

1. A number which has only two factors is called a prime number

2. A numbers Which has more than two factors is called a composite number

3. 1 is neither prime number nor composite number

4. The smallest prime number is 2

5. The smallest composite number is 4

6. The smallest even number is 2

(a) __ 6724:-

Sum of the remaining digits = 6 + 7 + 2 + 4 = 19

To make the number divisible by 3, the sum of its digits should be divisible by 3.

The smallest multiple of 3 Which comes after 19 is 21.

∴ Smallest number = 21 – 19 = 2

Now 2 + 3 + 3 = 8

However 2 + 3 + 3 + 3 = 11

If we put 8, then the sum of the digits will be 27 and as 27 is divisible by 3 the number will also be divisible by 3.

Therefore the largest number is 8.

(b) 4765 ___ 2:-

Sum of the remaining digits = 24

To make the number divisible by 3, the sum of its digits should be divisible by 3. As 24 is already divisible by 3, the smallest number that can be place here is 0

Now, 0 + 3 = 3

3 + 3 = 6

3 + 3 + 3 = 9

However, 3 + 3 + 3 + 3 = 12

If we put 9 then the sum of the digits will be 33 and as 33 is divisible by 3, the number will also be divisible by 3.

Therefore, the largest number is 9.

If a number is divisible by 5 its units digit be either ‘0’ or 5.

∴ a) 438750, b) 179015, c) 125 d) 639210 are divisible by 5.

(a) 297144: – Since, the last digits of the number is 4, it is divisible by 2.

On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3.

As the number is divisible by both 2 and 3, it is divisible by 6.

(b) 1258:- Since the last digit of the number is 8 , it is divisible by 2.

On adding all the digits of the number, the sum obtained is 16, is not divisible by 3, the given number is also not divisible by 3.

As the number is not divisible by both 2 &3. it is not divisible by 6.

(c) 4335:- The last digit of the number is 5, Which is not divisible by 2. Therefore the given number is also not divisible by 2.

On adding all the digits of the number, the sum obtained is 15, Since 15 is divisible by 3. As the number is not divisible by both 2 & 3. It is not divisible by 6

(d) 61233:- The last digit of the number is 3, Which is not divisible by 2. Therefore, the given number is also not divisible by 2

On adding all the digits of the number the sum obtained is 15. Since 15 is divisible by 3, the given number is also divisible by 3.

As the number is not divisible by both 2 & 3. it is not divisible by 6.

(e) 901352: -Since, the last digit of the number is 2 , It is divisible by 2.

On adding all the digits of the number, the sum obtained is 20. Since 20 is not divisible by 3, the given number is also not divisible by 3.

As the number is not division by both 2 & 3. it is not divisible by 6.

(f) 438750:- Since, the last digits of the number is 0, it is divisible by 2.

On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3.

As the number is divisible by both 2 & 3, it is divisible by 6.

(g) 1790184:- Since, the last digit of the number is 4, it is divisible by 2.

On adding all the digit of the number the sum obtained is 30. Since 30 is divisible by 3, the given number is also divisible by 3.

As the number is divisible by both 2&3. it is divisible by 6.

(h) 12583:- Since, the last digit of the number is 3, it is not divisible by 2.

On adding all the digit of the number, the sum obtained is 19, Since 19 is not divisible by 3
The given number is also not divisible by 3

As the number is not divisible by both 2 & 3. It is not divisible by 6

(i) 639210:- Since, the last digit of the number is 0. it is divisible by 2

On adding all the digits of the number the sum obtained is 21, Since 21 is divisible by 3, the given number is also divisible by 3.

As the number is divisible by both 2 & 3, it is divisible by 6.

(j) 17852:- Since, the last digits of the number is 2. it is divisible by 2

On adding all the digits of the numbers, the sum obtained is 23. Since 23 is not divisible by 3, the given number is also not divisible by 3.

As the number is not divisible by both 2 & 3, It is not divisible by 6.

(a) 5445:- Sum of the given digits at odd places = 5 + 4 = 9

Sum of the given digits at even places = 4 + 5 = 9

Difference = 9 – 9 = 0

As the difference between the sum of the digits at odd places and sum of the digits at even place is 0.

Therefore, 5445 is divisible by 11.

(b) 10824:-

Sum of the given digits at odd places = 4 + 8 + 1 = 13

Sum of the given digits at even places = 2 + 0 = 2

Difference = 13 – 2 = 11

The difference between the sum of the digits at odd places and the sum of the digit at even places is 11. Which is divisible by 11. Therefore 10824 is divisible by 11.

(c) 7138965:-

Sum of the given digits at odds places = 5 + 9 + 3 + 7 = 24

Sum of the given digits at even places = 6 + 8 + 1 = 15.

Difference = 24 – 15 = 9.

The difference between the sum of the digits at odd places and the sum of digits at even place is 9,

Which is not divisible by 11.

∴ 7138965 is not divisible by 11.

(d) 70169308:-

Sum of the digits at odd places = 8 + 3 + 6 + 0 = 17

Sum of digits at even places = 0 + 9 + 1 + 7 = 17

Difference = 17 – 17 = 0

As the difference between the sum of the digits at odd places and the sum of the digits at even place is 0.

Therefore, 70169308 is divisible by 11.

(e) 10000001:-

Sum of the digits at odd places=1 1

Sum of the digits even place = 1

Difference = 1 – 1 = 0

As the difference the sum of the digits at odd places and the sum of the digits at even places is 0. therefore 10000001 is divisible by 11.

(f) 901153:-
Sum of the digits at odd places = 3 + 1 + 0 = 4.

Sum of the digits at even places = 5 + 1 + 9 = 15

Difference = 15 – 4 = 11

The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, Which is divisible by 11.

Therefore, 901153 is divisible by 11.

If a number is divisible by 2 then the units digit of the number be 0, 2, 4, 6, 8. 

∴ a) 2144  b) 1258 c) 4336 e) 1352 are divisible by ‘2’.

Answer is: (b) Jaipur

Pratap Singh Barahat.

Ramnarayan Chowdhary.

Answer is: (b) Alwar

Revolutionary movement in Rajasthan: 

1. The background of the revolutionary movement in Rajasthan: 

Rajasthan also inherited anti – British ideology of failure of 1857 revolution against British rule in India in inheritence. The nationwide opposition of partition of Bengal in 1905, the idea of aggressive nationalism propounded by Bal Gangadhar Tilak, Bipin Chandra Pal and Lala Lajpat Rai, Veer Savarkar’s concept of nationalism and the influence of Rash Behari Bose and Sachindra Nath Sanyal, all combined together to create the background of revolutionary movement in Rajasthan. 

2. The heroes of the Revolt: 

In Rajasthan, the expression of martial nationalism emerged in the form of revolutionary activities At that time, there were three groups of revolutionaries in Rajasthan:

(1) The leader of Jaipur group was Arjunlal Sethi. He established Vardhman Vidyalaya at Jaipur for training the revolutionaries. Later, he shifted it to Indore. Pratap Singh Barahat the son of Thakur Kesari Singh and his son – in – law Ishwar Das were the students of the college owned by Ashia Sethi.

For the charge of forming the conspiracy of Neemage murder case (20th March,1913) he was made a prisoner. After the imprisonment of seven and half years, he was released. He is regarded as one of the most important national leaders of Rajasthan.

(2) Under the leadership of Thakur Kesari Singh Barahat,the revolutionary group of Kota was active. He favoured cultural nationalism. The phrase Chetavani – ra – chungataya written by him influenced Maharana Udai Singh and he did not participate in Delhi Court hold in 1903.On the charage of murder of a rich merhant pyarelal of Jodhpur in Kota, he was awarded punishment of imprisonment for 20 years His Jagir in Shahpura (Deopura) was confiscated.

In 1919, he was released from jail. His entire family underwent severe torture and sufferings for the sake of freedom of the country. His brother Jorawar Singh played a uniqe role in revolutionary movement. His name is associated with Neemage murder case, 1913 and throwing bomb on Lord Hardinge (23thDecember, 1912).

He lived underground for 27 years. He died while living undergound in 1939. Thakur Kesari Singh’s son Pratap Singh was taken as a prisoner in Varanasi conspiracy. He was sentenced to five years imprisonment. He breathed his last while facing inhuman tortures in Bareilly Jail.

3. The planning for an armed revolution: 

The revolutionary leader Rash Behari Bose had sent Bhoop Singh (later known by the name of Vijay Singh Pathik) to Rajasthan for collecting, making and repairing arms for revolution. The date fixed for armed revolution was 21th February, 1915. On that date. Gopal Singh Kharwa and Bhoop Singh, along with their hundreds of fellow revolutionaries, had been waiting in the jungle near Kharwa for signal of start of the revolt.

But. on 19th February, the secret of the mission was out and all the revolutionaries were taken prisoners. Bhoop Singh and Rao Gopal Singh Kharva were kept prisoners in todgarh fort. After some days, Bhoop Singh escaped from the fort and the leadership of Bijolia Peasant movemant was entrusted to him under the name of Vijay Singh Pathik. Rao Gopal Singh also escaped from Todgarh fort. He was arrested in August. 1915 and his Jagir in Kharwa village was confiscated. In 1920, he was released.

Evaluation: 

The revolutionary movement kept alive the public feelings of opposing the British power. During the time of national movement, the Indians went on receiving inspiration of getting rid of the British slavery.

In Jaisalmer.

In Ajmer Foy Sagar lake situated.

In Pratapgarh.

(c) Pratap Singh Barahat

(b) Sagarmal Gopa

1. Land was surveyed and measured with a bamboo jarib joined together with iron rings instead of the hemp rope used earlier. 

2. The land was grouped into four categories depending on whether it was regularly or occasionally cultivated. It was classified further into good, middling and bad categories. 

3. The average produce and the average price over the past 10 years were calculated. The revenue was calculated on the basis of these averages. One third of the average produce was the king’s share. It could be paid in cash or kind, though cash was preferred.

In Bundi  Chaurasi Khambo ki Chhatri situated.

Initially, the style of architecture was Persian, for instance, Humayun’s Tomb. However, with the passage of time, Akbar’s architectural style became more and more Indian. Inspired by the palaces of Hindu rajas, Akbar’s later buildings reflect Rajput traditions. Fatehpur Sikri, Akbar’s new capital city, contains many interesting Rajput-style buildings made of sandstone. It is referred to as a dream in stone. Among the many fine buildings are the Diwan-i-Khas, Panch Mahal, Jodha Bai’s Palace and the Buland Darwaza.

Akbar realized that the support of the Rajputs was necessary to build a powerful empire. 

For this he adopted the following policy: 

1. Akbar treated the Rajputs with honour and equality and won their respect and loyalty. 

2. He married Rajput princesses to strengthen his ties with Rajputs. 

3. He appointed Rajputs on High posts in his court to win their friendship, loyalty and cooperation. 

4. He abolished the Jaziya and pilgrim tax levied on Hindus. 

5. Akbar did not annexed the kingdoms of the Rajput rulers but only asked them to recognize him as their overlord and pay him regular tribute. The result of adopting this policy was that Akbar won the loyalty and support of the Rajputs, Who formed the backbone of a strong and stable empire.

Akbar’s position was unstable and insecure, when he ascended the throne of Delhi because his empire was surrounded by enemies on all the sides. Adil Shah who was the nephew of Sher Shah was determined to capture Delhi and re-establish the Afghan rule. Even the Rajputs were waiting for suitable opportunity to throw the Mughals out of India.

Akbar took following steps to transform the Islamic state into a secular one and unite the diverse races in the country into a single nation based on equality: 

1. He abolished the Jaziya tax which was imposed by earlier Muslim rulers on all non-Muslim subjects and pilgrim tax imposed on Hindus visiting places of pilgrimage. 

2. He allowed Hindus to build temples, celebrate festivals and worship freely. He even allowed his Hindu wives to celebrate festivals like Holi and Diwali. 

3. He gave high posts to talented and competent Hindus in his court, examples are: Todar Mai, Birbal and Raja Bhagwan Das. 

4. He married Rajput princesses and allowed them freedom of worship. 5. He granted land to all persons irrespective of their religious faith.

Akbar abolished the jaziya (poll) tax and the pilgrim tax.