Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

Students can download Maths Chapter 2 Real Numbers Ex 2.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.2

Therefore, one Decimal is equal to four hundred and thirty five decimal point six Square Feet (sq ft) in Online.

Question 1.
Express the following rational numbers into decimal and state the kind of decimal expression.
(i) \(\frac{2}{7}\)
(ii) -5\(\frac{3}{11}\)
(iii) \(\frac{22}{3}\)
(iv) \(\frac{327}{200}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 1
(i) \(\frac{2}{7}\) = 0.2857142….
= 0.\(\overline {285714}\)
Non-terminating and recurring decimal expansion.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

(ii) -5\(\frac{3}{11}\) = -5 + 0.272 = -5.272……..
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 2
= -5.\(\overline {27}\)
Non-terminating and recurring decimal expansion.

(iii) \(\frac{22}{3}\) = 7.333……..
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 3
= 7.\(\overline {3}\)
Non-terminating and recurring decimal expansion.

(iv) \(\frac{327}{200}\) = \(\frac{327}{2×100}\)
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 4
= \(\frac{3.27}{2}\)
= 1.635
Terminating decimal expansion.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

How to convert 1/32 to decimal form? … In the fraction 1/32, 1 is the numerator and 32 is the denominator, the fraction bar means “divided by”.

Question 2.
Express \(\frac{1}{13}\) in decimal form. Find the length of the period of decimals.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 5
\(\frac{1}{13}\) = 0.07692307
= 0.\(\overline {076923}\)
Length of the period of decimal is 6.

Question 3.
Express the rational number \(\frac{1}{33}\) in recurring decimal form by using the recurring decimal expansion of \(\frac{1}{11}\). Hence write \(\frac{71}{33}\) in recurring decimal form.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 6
\(\frac{1}{11}\) = 0.0909……… = 0.\(\overline {09}\)
∴ \(\frac{1}{33}\) = \(\frac{1}{3}\) × \(\frac{1}{11}\)
= \(\frac{1}{3}\) × 0.0909 ……..
= 0.0303 …… = 0.\(\overline {03}\)
\(\frac{71}{33}\) = 2\(\frac{5}{33}\) = 2 + \(\frac{5}{33}\) = 2 + 5 × \(\frac{1}{33}\)
= 2 + 5 × 0.\(\overline {03}\)
2 + (5 × 0.030303 ……..)
2 + 0.151515 ………
2+ 0.\(\overline {15}\)
2.\(\overline {15}\)

That’s literally all there is to it! 1/2 as a decimal is 0.5.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

Question 4.
Express the following decimal expression into rational numbers.
(i) 0.24
Solution:
Let x = 0.242424 ………. →(1)
100 x = 24.2424 ……… →(2)
(2) – (1) ⇒ 100 x – x = 24.2424 ……….. (-)
 0.2424 ……..
99 x = 24.0000
x = \(\frac{24}{99}\)
(or)
\(\frac{8}{33}\)

(ii) 2.327
Solution:
Let x = 2.327327327 ………. →(1)
1000 x = 2327.327327 ……… →(2)
(2) – (1) ⇒ 1000 x – x = 2327.327327 ……….. (-)
  2.327327 ……..
999 x = 2325.000
x = \(\frac{2325}{999}\)
(or)
\(\frac{775}{333}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

(iii) – 5.132
Solution:
– 5.132 = -5 + \(\frac{1}{10}\) + \(\frac{3}{100}\) + \(\frac{2}{1000}\)
= \(\frac{-5000 + 100 +30 + 2}{1000}\) = \(\frac{-4868}{1000}\)
(or)
\(\frac{-1217}{250}\)

(iv) 3.17
Solution:
Let x = 3.1777 ………. →(1)
10 x = 31.777 ……… →(2)
100 x = 317.77 …….. →(3)
(3) – (2) ⇒ 100 x – 10 x = 317.77 ……….. (-)
 31.777 ……..
90 x = 286.000
x = \(\frac{286}{90}\)
(or)
\(\frac{143}{45}\)

(v) 17.215
Solution:
Let x = 17.2151515 ………. →(1)
10 x = 172.151515 ……… →(2)
100 x = 17215.1515 …….. →(3)
(3) – (2) ⇒ 1000 x – 10 x = 17215.1515 ……….. (-)
 17215.1515 ……..
990 x = 17043
x = \(\frac{17043}{990}\)
(or)
\(\frac{5681}{330}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

(vi) -21.2137
Solution:
Let x = -21.213777 ………. →(1)
1000 x = -21213.777 ……… →(2)
100 x = -212137.77 …….. →(3)
(3) – (2) ⇒ 10000 x – 1000 x = -21213.777 ……….. (-)
-21213.777 ……..
9000 x = -190924
x = \(\frac{-190924}{9000}\)
(or)
\(\frac{-47731}{2250}\)

That’s literally all there is to it! 11/16 as a decimal is 0.6875.

Question 5.
Without actual division, find which of the following rational numbers have terminating decimal expression.
(i) \(\frac{7}{128}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 7
\(\frac{7}{128}\) = \(\frac{7}{2^{7}}\)
∴ \(\frac{7}{128}\) has terminating decimal expression.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2

(ii) \(\frac{21}{15}\)
Solution:
\(\frac{21}{15}\) = \(\frac{7}{5}\) = \(\frac{7}{5^1}\)
\(\frac{21}{15}\) has terminating decimal expression.

(iii) 4\(\frac{9}{35}\)
Solution:
4\(\frac{9}{35}\) = \(\frac{149}{35}\)
4\(\frac{149}{5×7}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ 4\(\frac{9}{35}\) has non-terminating recurring decimal expression.

(iv) \(\frac{219}{2200}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.2 8
\(\frac{219}{2200}\) = \(\frac{219}{2^{3} × 5^{2} × 11}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ \(\frac{219}{2200}\) has non-terminating recurring decimal expression.

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 1 Numbers Ex 1.4

Students can download Maths Chapter 1 Numbers Ex 1.4 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.4

Question 1.
Fill in the blanks.
(i) The nearest 100 of 843 is _____
(ii) The nearest 1000 of 756 is ______
(iii) The nearest 10,000 of 85654 is ______
Solution:
(i) 800.
The digit in tens place is 4 < 5
(ii) 1000.
The digit in hundred places is 7 ≥ 5
(iii) 90,000.
The digit in a thousand places is 5 ≥ 5

 Samacheer Kalvi 6th Maths Guide Term 1 Chapter 1 Set Language Ex 1.4

Question 2.
Say True or False
(i) 8567 is rounded off as 8600 to the nearest 10.
(ii) 139 is rounded off as 100 to the nearest 100.
(iii) 1,70,51,972 is rounded off as 1,70,00,000 to the nearest lakh.
Solution:
(i) False
Hint: In one’s place the digit is 7 ≥ 5. So 8580
(ii) True
Hint: In tens place, we have 3 < 5. So 100
(iii) False
Hint: In ten thousand places the digit is 5 ≥ 5. So 1,71,000,000

Question 3.
Round off the following to the given nearest place.
(i) 4,065; hundred
(ii) 44,555; thousand
(iii) 86,943; ten thousand
(iv) 50,81,739; lakh
(v) 33,75,98,482; ten crore
Solution:
(i) 4100
(ii) 45,000
(iii) 90,000
(iv) 51,00,000
(v) 30,00,00,000

 Samacheer Kalvi 6th Maths Guide Term 1 Chapter 1 Set Language Ex 1.4

Question 4.
Estimate the sum of 157826 and 32469 rounded off to the nearest ten thousand.
Solution:
= 157826 + 32469
= 190295
When rounded off to nearest ten thousand = 1,90,000

Decimal numbers can be rounded to the nearest hundredth by observing the digit in the thousandths place.

Question 5.
Estimate by rounding off each number to the nearest hundred.
(i) 8074 + 4178
(ii) 1768977 + 130589
Solution:
(i) 8074 + 4178 = 12,252
When rounded off to nearest hundred 12,300

(ii) 1768977 + 130589 = 18,99,566
When rounded off to nearest hundred = 18,99,600

 Samacheer Kalvi 6th Maths Guide Term 1 Chapter 1 Set Language Ex 1.4

Question 6.
The population of a city was 43,43,645 in the year 2001 and 46,81,087 in the year 2011. Estimate the increase in population by rounding off to the nearest thousands.
Solution:
Population in 2001 = 43,43,645
Population in 2011 = 46,81,087
Increase in population = 46,81,087 – 43,43,645 = 3,37,442
When rounded off to the nearest thousand = 3,37,000

 Samacheer Kalvi 6th Maths Guide Term 1 Chapter 1 Set Language Ex 1.4

Objective Type Questions

Question 7.
The number which on rounding off to the nearest thousand gives 11000 is
(a) 10345
(b) 10855
(c) 11799
(d) 10056
Solution:
(b) 10855

Question 8.
The estimation to the nearest hundredth of 76812 is
(a) 77000
(b) 76000
(c) 76800
(d) 76900
Solution:
(c) 76800
In tens place the digit is 1 < 5, So 76800

 Samacheer Kalvi 6th Maths Guide Term 1 Chapter 1 Set Language Ex 1.4

Question 9.
The number 9785764 is rounded off to the nearest lakh as
(a) 9800000
(b) 9786000
(c) 9795600
(d) 9795000
Solution:
(a) 9800000

Question 10.
The estimated difference of 167826 and 2765 rounded off to the nearest thousand is
(a) 180000
(b) 165000
(c) 140000
(d) 155000
Solution:
(b) 165000

 Samacheer Kalvi 6th Maths Guide Term 1 Chapter 1 Set Language Ex 1.4

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Multiple Choice Questions

Question 1.
The area of triangle formed by the points (-5, 0), (0, – 5) and (5, 0) is …………..
(1) 0 sq.units
(2) 25 sq.units
(3) 5 sq.units
(4) none of these
Answer:
(2) 25 sq.units Hint.
Hint:
Area of the ∆
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 1

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place.

Question 2.
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is …………
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
Answer:
(1) x = 10
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 2

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 3.
The straight line given by the equation x = 11 is …………….
(1) parallel to X axis
(2) parallel to Y axis
(3) passing through the origin
(4) passing through the point (0,11)
Answer:
(2) parallel to Y axis

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 4.
If (5,7), (3,p) and (6,6) are collinear, then the value of p is ……………
(1) 3
(2) 6
(3) 9
(4) 12
Answer:
(3) 9
Hint:
Since the three points are collinear. Area of a triangle is 0
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 3
5p + 18 + 42 – (21 + 6p + 30) = 0
5p + 60 – (51 + 6p) = 0
5p + 60 – 51 – 6p = 0
-p + 9 = 0
-p = -9
p = 9

Question 5.
The point of intersection of 3x – y = 4 and x + 7 = 8 is ……………
(1) (5,3)
(2) (2,4)
(3) (3,5)
(4) (4, 4)
Answer:
(3) (3, 5)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 4
Substitute the value of x = 3 in (2)
3 + 7 = 8
y = 8 – 3 = 5
The point of intersection is (3, 5)

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 6.
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 } \). The value of ‘a’ is …………….
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2
Hint:
Slope of a line = \(\frac { 1 }{ 8 } \)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 5

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (- 8, 8) is ………..
(1) -1
(2) 1
(3) \(\frac { 1 }{ 3 } \)
(4) -8
Answer:
(2) 1
Hint:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-0 }{ -8-0 } \) = \(\frac { 8 }{ -8 } \) = -1
Slope of the Perpendicular = 1

Question 8.
If slope of the line PQ is \(\frac{1}{\sqrt{3}}\) then slope of the perpendicular bisector of PQ is …………..
(1) \(\sqrt { 3 }\)
(2) –\(\sqrt { 3 }\)
(3) \(\frac{1}{\sqrt{3}}\)
(4) 0
Answer:
(2) –\(\sqrt { 3 }\)
Hint:
Slope of a line = \(\frac{1}{\sqrt{3}}\)
Slope of the ⊥r bisector = –\(\sqrt { 3 }\)

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 9.
If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is ……………
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
Answer:
(1) 8x + 5y = 40
Hint:
Let the point A be (0, 8) and B (5, 0)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 6
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 7

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 10.
The equation of a line passing through the origin and perpendicular to the line lx -3y + 4 = 0 is
(1) 7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0
Hint:
Slope of the line of 7x – 3y + 4 = 0
= \(\frac { -7 }{ -3 } \) = \(\frac { 7 }{ 3 } \)
Slope of its ⊥r = \(\frac { -3 }{ 7 } \)
The line passes through (0,0)
Equation of a line is
y – y1 = m(x – x1)
y – 0 = \(\frac { -3 }{ 7 } \) (x – 0)
y = \(\frac { -3 }{ 7 } \) x ⇒ 7y = -3x
3x + 7y = 0

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 11.
Consider four straight lines
(i) l1 : 3y = 4x + 5
(ii) l2 : 4y = 3x – 1
(iii) l3 : 4y + 3x = 7
(iv) l4 : 4x + 3y = 2
Which of the following statement is true?
(1) l1 and l2 are perpendicular
(2) l2 and l4 are parallel
(3) l2 and l4 are perpendicular
(4) l2 and l3 are parallel
Answer:
(3) l2 and l4 are perpendicular
Hint:
Slope of l1 = \(\frac { 4 }{ 3 } \); Slope of l2 = \(\frac { 3 }{ 4 } \)
Slope of l3 = – \(\frac { 3 }{ 4 } \); Slope of l4 = –\(\frac { 4 }{ 3 } \)
(1) l1 × l2 = \(\frac { 4 }{ 3 } \) × \(\frac { 3 }{ 4 } \) = 1 …….False
(2) l1 = \(\frac { 4 }{ 3 } \); l4 = – \(\frac { 4 }{ 3 } \) not parallel ………False
(3) l2 × l4 = \(\frac { 3 }{ 4 } \) × – \(\frac { 4 }{ 3 } \) = -1 …….True
(4) l2 = \(\frac { 3 }{ 4 } \); l3 = – \(\frac { 3 }{ 4 } \) not parallel ………False

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 12.
A straight line has equation 87 = 4x + 21. Which of the following is true …………………….
(1) The slope is 0.5 and the y intercept is 2.6
(2) The slope is 5 and the y intercept is 1.6
(3) The slope is 0.5 and they intercept is 1.6
(4) The slope is 5 and the y intercept is 2.6
Answer:
(1) The slope is 0.5 and they intercept is 2.6
Hint:
8y = 4x + 21
y = \(\frac { 4 }{ 8 } \) x + \(\frac { 21 }{ 8 } \)
= \(\frac { 1 }{ 2 } \) x + \(\frac { 21 }{ 8 } \)
\(\frac { 1 }{ 2 } \) = 0.5
\(\frac { 21 }{ 8 } \) = 2.625
Slope = \(\frac { 1 }{ 2 } \) = 0.5
y intercept = \(\frac { 21 }{ 8 } \) = 2.6

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 13.
When proving that a quadrilateral is a trapezium, it is necessary to show
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
Solution:
(2) Two parallel and two non-parallel sides.

Question 14.
When proving that a quadrilateral is a parallelogram by using slopes you must find …………………
(1) The slopes of two sides
(2) The slopes of two pair of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
Answer:
(2) The slopes of two pair of opposite sides

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 15.
(2,1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
Solution:
(2) x + y = 3; 3x + y = 7

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Students can download Maths Chapter 1 Numbers Ex 1.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.2

Question 1.
Fill in the blanks
(i) The HCF of 45 and 75 is …….
(ii) The HCF of two successive even numbers is …….
(iii) If the LCM of 3 and 9 is 9, then their HCF is ………
(iv) The LCM of 26, 39 and 52 is ……..
(v) The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is ………
Solution:
(i) 15
(ii) 2
(iii) 3
(iv) 156
(v) 3

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

The Least common multiple (LCM) of 15 and 20 is 60.

Question 2.
Say True or False
(i) The numbers 57 and 69 are co-primes.
(ii) The HCF of 17 and 18 is 1.
(iii) The LCM of two successive numbers is the product of the numbers.
(iv) The LCM of two co-primes is the sum of the numbers.
(v) The HCF of two numbers is always a factor of their LCM.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

So, now the prime factorization of 84 with the upside-down division method is 2 × 2 × 3 × 7.

Question 3.
Find the HCF of each set of numbers using the prime factorisation method.
(i) 18, 24
(ii) 51, 85
(iii) 61, 76
(iv) 84, 120
(v) 27, 45, 81
(vi) 45, 55, 95
Solution:
(i) 18, 24
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 1
HCF = 2 × 3 = 6

(ii) 51, 85
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 2
HCF = 17

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

(iii) 61, 76
61 = 1 × 61
For 61, 76, the common factor is 1
HCF = 1

(iv) 84, 120
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 3
HCF = 2 × 2 × 3 = 12

(v) 27, 45, 81
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 4
HCF = 3 × 3 = 9

(vi) 45, 55, 95
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 5
Factors of 55 = 5 × 11
Factors of 45 = 3 × 3 × 5
Factors of 95 = 5 × 19
HCF = 5

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 4.
Find the LCM of each set of numbers using the prime factorisation method.
(i) 6, 9
(ii) 8, 12
(iii) 10, 15
(iv) 14, 42
(v) 30, 40, 60
(vi) 15, 25, 75
Solution:
(i) 6, 9
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 6
Factors of 6 = 2 × 3
Factors of 9 = 3 × 3
LCM of 6 and 9 = 3 × 2 × 3 = 18

(ii) 8, 12
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 7

(iii) 10, 15
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 8
Factors of 10 = 2 × 5
Factors of 15 = 3 × 5
LCM of 10 and 15 = 5 × 2 × 3 = 30

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

(iv) 14, 42
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 9
Factors of 14 = 2 × 7
Factors of 42 = 2 × 3 × 7
LCM of 14 and 42 = 7 × 2 × 3 = 42

(v) 30, 40, 60
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 10
Factors of 30 = 2 × 3 × 5
Factors of 40 = 2 × 2 × 2 × 5
Factors of 60 = 2 × 2 × 3 × 5
LCM of 30, 40, 60 = 2 × 5 × 3 × 2 × 2 = 120

(vi) 15, 25, 75
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 11
Factors of 15 = 3 × 5
Factors of 25 =5 × 5
Factors of 75 = 3 × 5 × 5
LCM of 15, 25, 75 = 5 × 3 × 5 = 75

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 5.
Find the HCF and the LCM of the numbers 154, 198, 286
Solution:
HCF
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 12
Factors of 154 = 2 × 7 × 11
Factors of 198 = 2 × 3 × 3 × 11
Factors of 286 = 2 × 13 × 11
HCF of 154, 198, 286 = 11 × 2 = 22
LCM
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 13
LCM = 2 × 11 × 7 × 9 × 13
= 22 × 63 × 13
= 18018

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 6.
What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?
Solution:
This is an HCF related problem. So, we need to find the HCF of 80, 100, 120
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 14
80 = 2 × 2 × 2 × 2 × 5
100 = 2 × 2 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF of 80, 100 and 120 = 2 × 2 × 5 = 20
Volume of the vessel = 20 litres

Question 7.
The traffic lights at three different road junctions change after every 40 seconds, 60 seconds, and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?
Solution:
This is an LCM related problem. So, we need to find the LCM of 40, 60, 72
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 15
LCM of 40, 60 and 72
= 2 × 2 × 3 × 5 × 2 × 1 × 1 × 3
= 360 seconds
= 6 minutes
The traffic lights will change simultane¬ously again at 8 : 06 am.

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 8.
The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible.
Solution:
Product of two numbers = LCM × HCF
x × y = 210 × 14
x × y = 2940
(12, 245), (20, 147)
Two pairs are possible.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 16

Question 9.
The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers is 36, then find the other number.
Solution:
36x = 6 × 12 × 12
⇒ \(x=\frac{6 \times 12 \times 12}{36}=24\)

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Objective Type Questions

Question 10.
Which of the following pairs is co-prime?
(a) 51, 63
(b) 52, 91
(c) 71, 81
(d) 81, 99
Solution:
(c) 71, 81

Question 11.
The greatest four-digit number which is exactly divisible by 8, 9, and 12 is
(a) 9999
(b) 9996
(c) 9696
(d) 9936
Solution:
(d) 9936

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 12.
The HCF of two numbers is 2 and their LCM is 154. If the difference between the numbers is 8, then the sum is
(a) 26
(b) 36
(c) 46
(d) 56
Solution:
(b) 36

Question 13.
Which of the following cannot be the HCF of two numbers whose LCM is 120?
(a) 60
(b) 40
(c) 80
(d) 30
Solution:
(c) 80

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 1.
Fill in the blanks of the given equivalent ratios.
(i) 3 : 5 = 9 : ……
(ii) 4 : 5 = …… : 10
(iii) 6 : …… = 1 : 2
Solution:
(i) 15
Hint: \(\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}\)
(ii) 8
Hint: \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\)
(iii) 12
Hint: \(\frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12}\)

Question 2.
Complete the table.
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2 1
Solution:
(i) 1 feet = 12 inches
3 feet = 3 × 12 inches = 36 inches
72 inches = 6 × 12 inches = 6 feet

(ii) 1 week = 7 days
2 weeks = 2 × 7 days = 14 days
63 days = 9 × 7 days = 9 weeks

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 3.
Say True or False.
(i) 5 : 7 is equivalent to 21 : 15
(ii) If 40 is divided in the ratio 3 : 2, then the larger part is 24
Solution:
(i) False
Hint: \(\frac{21}{15}=\frac{7}{5}=7: 5\)
(ii) True
Hint: \(\frac{3}{5} \times 40=24\)

Handy LCM of two or more numbers Calculator displays LCM of 365, 838, 862 in a fraction of seconds i.e.

Question 4.
Give two equivalent ratios for each of the following.
(i) 3 : 2
(ii) 1 : 6
(iii) 5 : 4
Solution:
(i) 3 : 2
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2 2
3 : 2 = 6 : 4 = 9 : 6

(ii) 1 : 6
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2 3
1 : 6 = 2 : 12 = 3 : 18

(iii) 5 : 4
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2 4
5 : 4 = 10 : 8 = 15 : 12

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 5.
Which of the two ratios is larger?
(i) 4 : 5 or 8 : 15
(ii) 3 : 4 or 7 : 8
(iii) 1 : 2 or 2 : 1
Solution:
(i) 4 : 5 (or) 8 : 15
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2 5
4 : 5 > 8 : 15

(ii) 3 : 4 (or) 7 : 8
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2 6
7 : 8 > 3 : 4

(iii) 1 : 2 (or) 2 : 1
1 : 2 = \(\frac{1}{2}\)
2 : 1 = \(\frac{2}{1}\)
= 2
\(\frac{2}{1}\) > \(\frac{1}{2}\)
2 : 1 > 1 : 2

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 6.
Divide the numbers given below in the required ratio.
(i) 20 in the ratio 3 : 2
(ii) 27 in the ratio 4 : 5
(iii) 40 in the ratio 6 : 14.
Solution:
(i) Ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
5 parts = 20
1 part = \(\frac{20}{5}\) = 4
3 parts = 3 × 4 = 12
2 parts = 2 × 4 = 8
20 can be divided in the form as 12, 8.

(ii) Ratio = 4 : 5
Sum of the ratio = 4 + 5 = 9
9 parts = 27
1 part = \(\frac{27}{9}\) = 3
4 parts = 4 × 3 = 12
5 parts = 5 × 3 =15
27 can be divided in the form as 12, 15.

(iii) 40 in the ratio 6 : 14
Ratio = 6 : 14
Sum of the ratio = 6 + 14 = 20
20 parts = 40
1 part = \(\frac{40}{20}\) = 2
6 parts = 2 × 6 = 12
14 parts = 2 × 14 = 28
40 can be divided in the form as 12, 28.

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 7.
In a family, the amount spent in a month for buying Provisions and Vegetables are in the ratio 3 : 2. If the allotted amount is Rs 4000, then what will be the amount spent for
(i) Provisions and
(ii) Vegetables?
Solution:
Allotted amount = Rs 4000
Ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
5 parts = Rs 4000
1 part = Rs \(\frac{4000}{5}\) = Rs 800
Provisions : Vegetables = 3 : 2
3 parts = 3 × Rs 800 = Rs 2400
2 parts = 2 × Rs 800 = Rs 1600
Amount spent for provisions = Rs 2400
Amount spent for vegetables = Rs 1600

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 8.
A line segment 63 cm long is to be divided into two parts in the ratio 3 : 4. Find the length of each part.
Solution:
Total length = 63 cm Ratio = 3 : 4
Sum of the ratio = 3 + 4 = 7
7 parts = 63 cm
1 part = \(\frac{63}{7}\) = 9 cm
3 parts = 3 × 9 cm = 27 cm
4 parts = 4 × 9 cm = 36 cm
∴ 63 cm can be divided into the parts as 27 cm and 36 cm.

Objective Type Questions

Question 9.
If 2 : 3 and 4 : …… are equivalent ratios, then the missing term is
(a) 6
(b) 2
(c) 4
(d) 3
Solution:
(a) 6

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 10.
An equivalent ratio of 4 : 7 is
(a) 1 : 3
(b) 8 : 15
(c) 14 : 8
(d) 12 : 21
Solution:
(d) 12 : 21

Question 11.
Which is not an equivalent ratio of \(\frac{16}{24}\)?
(a) \(\frac{6}{9}\)
(b) \(\frac{12}{18}\)
(c) \(\frac{10}{15}\)
(d) \(\frac{20}{28}\)
Solution:
(d) \(\frac{20}{28}\)

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 12.
If Rs 1600 is divided
(a) Rs 480
(b) Rs 800
(c) Rs 1000
(d) Rs 200
Solution:
(c) Rs 1000

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1

Question 1.
Find all positive integers which when divided by 3 leaves remainder 2.
Answer:
All the positive integers when divided by 3 leaves remainder 2
By Euclid’s division lemma
a = bq + r, 0 < r < b
a = 3q + r where 0 < q < 3
a leaves remainder 2 when divided by 3
∴ The positive integers are 2, 5, 8, 11,…

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

Question 2.
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over.
Solution:
Using Euclid’s division algorithm,
a = 21q + r, we get 532 = 21 × 25 + 7.
The remainder is 7.
No. of completed rows = 25, left over flower pots = 7 pots.

Question 3.
Prove that the product of two consecutive positive integers is divisible by 2.
Answer:
Let n – 1 and n be two consecutive positive integers, then the product is n (n – 1)
n(n – 1) = n2 – n
We know that any positive integer is of the form 2q or 2q + 1 for same integer q

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

Case 1:
when n = 2 q
n2 – n = (2q)2 – 2q
= 4q2 – 2q
= 2q (2q – 1)
= 2 [q (2q – 1)]
n2 – n = 2 r
r = q(2q – 1)
Hence n2 – n. divisible by 2 for every positive integer.

Case 2:
when n = 2q + 1
n2 – n = (2q + 1 )2 – (2q + 1 )
= (2q + 1) [2q + 1 – 1]
= 2q (2q + 1)
n2 – n = 2r
r = q (2q + 1)
n2 – n divisible by 2 for every positive integer.

Question 4.
When the positive integers be a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a + b + c is divisible by 13.
Solution:
Let the positive integers be a, b, and c.
a = 13 q + 9
b = 13q + 1
c = 13 q + 10
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
= 13 (3q + 2)
which is divisible by 13.

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

Question 5.
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Answer:
Let the integer be ” x ”
The square of its integer is “x2
Let x be an even integer
x = 2q + 0
x2 = 4q2
When x is an odd integer
x = 2k + 1
x2 = (2k + 1)2
= 4k2 + 4k + 1
= 4k (k + 1) + 1
= 4q + 1 [q = k(k + 1)]
It is divisible by 4
Hence it is proved

The Prime Factors of 84 are: 2, 2, 3, 7. How many Prime Factors of 84.

Question 6.
Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
(ii) 867 and 255
(iii) 10224 and 9648
(iv) 84, 90 and 120
Solution:
To find the H.C.F. of 340 and 412. Using Euclid’s division algorithm.
We get 412 = 340 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm
340 = 72 × 4 + 52
The remainder 52 ≠ 0.
Again applying Euclid’s division algorithm
72 = 52 × 1 + 20
The remainder 20 ≠ 0.
Again applying Euclid’s division algorithm,
52 = 20 × 2 + 12
The remainder 12 ≠ 0.
Again applying Euclid’s division algorithm.
20 = 12 × 1 + 8
The remainder 8 ≠ 0.
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0.
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero.
Therefore H.C.F. of 340 and 412 is 4.

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

(ii) To find the H.C.F. of 867 and 255, using Euclid’s division algorithm.
867 = 255 × 3 + 102
The remainder 102 ≠ 0.
Again using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0.
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero.
Therefore the H.C.F. of 867 and 255 is 51.

(iii) To find H.C.F. 10224 and 9648. Using Euclid’s division algorithm.
10224 = 9648 × 1 + 576
The remainder 576 ≠ 0.
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
Remainder 432 ≠ 0.
Again applying Euclid’s division algorithm
576 = 432 × 1 + 144
Remainder 144 ≠ 0.
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is zero.
There H.C.F. of 10224 and 9648 is 144.

(iv) To find H.C.F. of 84, 90 and 120.
Using Euclid’s division algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0.
Again using Euclid’s division algorithm
84 = 6 × 14 + 0
The remainder is zero.
∴ The H.C.F. of 84 and 90 is 6. To find the H.C.F. of 6 and 120 using Euclid’s division algorithm.
120 = 6 × 20 + 0
The remainder is zero.
Therefore H.C.F. of 120 and 6 is 6
∴ H.C.F. of 84, 90 and 120 is 6.

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

Question 7.
Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Answer:
Find the HCF of ( 1230 – 12) and (1926- 12)
i.e HCF of 1218 and 1914
By Euclid’s division algorithm
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0
By Euclid’s division algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0
Again by Euclid’s division algorithm
696 = 522 × 1 + 174
The remainder 174 ≠ 0 Again by Euclid’s division algorithm
522 = 174 × 3 + 0
The remainder is zero
∴ HCF of 1218 and 1914 is 174
The largest value is 174

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

Question 8.
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Solution:
Applying Euclid’s divison lemma to 32 and 60, we get
60 = 32 × 1 + 28 ……………. (i)
The remainder is 28 ≠ 0.
Again applying division lemma
32 = 28 × 1 + 4 ……………. (ii)
The remainder 4 ≠ 0.
Again applying division lemma
28 = 4 × 7 + 0 ………….. (iii)
The remainder zero.
∴ H.C.F. of 32 and 60 is 4.
From (ii), we get
32 = 28 × 1 + 4
⇒ 4 = 32 – 28 × 1
⇒ 4 = 32 – (60 – 32 × 1) × 1
⇒ 4 = 32 – 60 + 32
⇒ 4 = 32 × 2+(-1) × 60
∴ x = 2 and y = -1

Question 9.
A positive integer, when divided by 88, gives the remainder 61. What will be the remainder when the same number is divided by 11?
Answer:
Let the positive integer be “x”
x = 88 × y + 61 (a = pq + r)
since 88 is a multiple of 11
61 = 11 × 5 + 6
∴ The remainder is 6

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.1

Question 10.
Prove that two consecutive positive integers are always coprime.
Solution:
Let the numbers be I, I + 1:
They are co-prime if only +ve integer that divides both is 1.
I is given to be +ve integer.
So I = 1, 2, 3, ….
∴ One is odd and the other one is even. Hence H.C.F. of the two consecutive numbers is 1. Hence the result.

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Students can download 10th Science Chapter 18 Heredity Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity

Samacheer Kalvi 10th Science Heredity Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1.
According to Mendel alleles have the following character:
(a) Pair of genes
(b) Responsible for character
(c) Production of gametes
(d) Recessive factors
Answer:
(b) Responsible for character

Question 2.
9 : 3 : 3 : 1 ratio is due to ______.
(a) Segregation
(b) Crossing over
(c) Independent assortment
(d) Recessiveness.
Answer:
(c) Independent assortment

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Question 3.
The region of the chromosome where the spindle fibres get attached during cell division:
(a) Chromomere
(b) Centrosome
(c) Centromere
(d) Chromonema
Answer:
(c) Centromere

Question 4.
The centromere is found at the centre of the ______ chromosome.
(a) Telocentric
(b) Metacentric
(c) Sub – metacentric
(d) Acrocentric.
Answer:
(b) Metacentric

Question 5.
The units form the backbone of the DNA.
(a) 5 carbon sugar
(b) Phosphate
(c) Nitrogenous bases
(d) Sugar phosphate
Answer:
(d) Sugar phosphate

Question 6.
Okazaki fragments are joined together by ______.
(a) Helicase
(b) DNA polymerase
(c) RNA primer
(d) DNA ligase.
Answer:
(d) DNA ligase.

Question 7.
The number of chromosomes found in human beings are:
(a) 22 pairs of autosomes and 1 pair of allosomes.
(b) 22 autosomes and 1 allosome
(c) 46 autosomes
(d) 46 pairs autosomes and 1 pair of allosomes.
Answer:
(a) 22 pairs of autosomes and 1 pair of allosomes.

Question 8.
The loss of one or more chromosome in ploidy is called ______.
(a) Tetraploidy
(b) Aneuploidy
(c) Euploidy
(d) Polyploidy.
Answer:
(b) Aneuploidy

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

II. Fill in the blanks:

  1. The pairs of contrasting character (traits) of Mendel are called ……..
  2. Physical expression of a gene is called ………..
  3. The thin thread like structures found in the nucleus of each cell are called ……….
  4. DNA consists of two ……….. chains.
  5. An inheritable change in the amount or the structure of a gene or a chromosome is called ……….

Answer:

  1. alleles or allelomorphs
  2. Phenotype
  3. Chromosomes
  4. Polynucleotide chain
  5. Mutation

III. Identify whether the statement are True or False. Correct the false statement.

  1. A typical Mendelian dihybrid ratio of F2 generation is 3 : 1.
  2. A recessive factor is altered by the presence of a dominant factor.
  3. Each gamete has only one allele of a gene.
  4. Hybrid is an offspring from a cross between genetically different parent.
  5. Some of the chromosomes have an elongated knob-like appendages known as telomere.
  6. New nucleotides are added and new complementary strand of DNA is formed with the help of enzyme DNA polymerase.
  7. Down’s syndrome is the genetic condition with 45 chromosomes.

Answer:

  1. False – A typical Mendelian dihybrid ratio of F2 generation is 9 : 3 : 3 : 1
  2. True
  3. True
  4. True
  5. False – Some of the chromosomes have an elongated knob-like appendages known as satellite
  6. True
  7. True

IV. Match the following:

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity 1
Answer:
A. (iii)
B. (iv)
C. (v)
D. (i)
E. (ii)

V. Answer in a sentence.

Question 1.
What is a cross in which inheritance of two pairs of contrasting characters are studied?
Answer:
Dihybrid cross is a cross in which inheritance of two pairs of contrasting characters.

Question 2.
Name the conditions when both the aisles are identical?
Answer:
Homozygous alleles.

Question 3.
A garden pea plant produces axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant trait.
Answer:
The dominant trait is axial white flower.

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Question 4.
What is the name given to the segments of DNA, which are responsible for the inheritance of a particular character?
Answer:
Genes are the segments of DNA, which are responsible for the inheritance of a particular character.

Question 5.
Name the bond which binds the nucleotides in a DNA.
Answer:
Hydrogen bond binds the nucleotides in a DNA.

VI. Short Answer Questions.

Question 1.
Why did Mendel select pea plant for his experiments?
Answer:

  1. It is naturally self-pollinating and so is very easy to raise pure breeding individuals.
  2. It has a short life span as it is an annual and so it was possible to follow several generations.
  3. It is easy to cross-pollinate.
  4. It has deeply defined contrasting characters.
  5. The flowers are bisexual.

Question 2.
What do you understand by the term phenotype and genotype?
Answer:

  • The external expression of a particular trait is known as the phenotype.
  • The genetic expression of an organism is a genotype.

Question 3.
What are allosomes?
Answer:
Allosomes are chromosomes which are responsible for determining the sex of an individual. They are also called as sex chromosomes or hetero-chromosomes. There are two types of sex chromosomes, X and Y- chromosomes.

Question 4.
What are the Okazaki fragments?
Answer:
The short segments of DNA are called Okazaki fragments.

Question 5.
Why is euploidy considered to be advantageous to both plants and animals?
Answer:
Euploidy is the condition in which individual bears more than the usual number of diploid (2n) chromosome. It is used in plant breeding and horticulture. It has economic significance by the production of large sized flowers and fruits. It plays a significant role in the evolution of new species.

Question 6.
A pure tall plant (TT) is crossed with the pure dwarf plant (tt), what would be the F1 and F2 generations? Explain.
Answer:
In the F1 generation, all are tall plants. (Genotype all are Tt and phenotype all are tall).
In F2 generation, genotype three tall and one dwarf. [TT : Tt : tt = 1 : 2 : 1] phenotype.
Tall : dwarf 3 : 1 [TT : Tt : Tt : tt].

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Question 7.
Explain the structure of a chromosome.
Answer:
The chromosomes are thin, long and thread like structures consisting of two identical strands called sister chromatids. They are held together by the centromere. Each chromatid is made up of spirally coiled thin structure called chromonema. The chromonema has number of bead-like structures along its length which are called chromomeres.

Question 8.
Label the parts of the DNA in the diagram given below. Explain the structure briefly.
Answer:
(i) DNA molecule consists of two polynucleotide chains.
(ii) These chains form a double helix structure with two strands which run anti-parallel to one another.
(iii) Nitrogenous bases in the centre are linked to sugar-phosphate units which form the backbone of the DNA.
(iv) Pairing between the nitrogenous bases is very specific and is always between purine and pyrimidine linked by hydrogen bonds.
Samacheer Kalvi 10th Science Guide Chapter 18 Heredity 2

VII. Long Answer Questions

Question 1.
Explain with an example of the inheritance of dihybrid cross. How is it different from a monohybrid cross?
Answer:
Dihybrid cross involves the inheritance of two pairs of contrasting characteristics (or contrasting traits) at the same time. The two pairs of contrasting characteristics chosen by Mendel were shape and color of seeds: round-yellow seeds and wrinkled-green seeds.
Mendel crossed pea plants having round-yellow seeds with pea plants having wrinkled-green seeds. Mendel made the following observations:

(i) Mendel first crossed pure breeding pea plants having round-yellow seeds with pure breeding pea plants having wrinkled-green seeds and found that only round-yellow seeds were produced in the first generation (F1). No wrinkled-green seeds were obtained in the F1 generation. From this it was concluded that round shape and yellow color of the seeds were dominant traits over the wrinkled shape and green color of the seeds.

(ii) When the hybrids of F1 generation pea plants having round-yellow seeds were cross-bred by self pollination, then four types of seeds having different combinations of shape and color were obtained in second generation or F2 generation. They were round yellow, round-green, wrinkled-yellow and wrinkled-green seeds.
The ratio of each phenotype (or appearance)of seeds in the F2 generation is 9 : 3 : 3 : 1. This is known as the Dihybrid ratio.
Samacheer Kalvi 10th Science Guide Chapter 18 Heredity 3
Monohybrid cross is a genetic cross that involves a single pair of gene or trait. In this parents differ by single trait. Eg: Height.
Dihybrid cross is a genetic cross that involves two pairs of genes, which are responsible for two trait. In this, parents have two different independent trait. Eg: flower colour, stem length.

Question 2.
How is the structure of DNA organised? What is the biological significance of DNA?
Answer:
DNA is the genetic material of almost all the organisms. One of the active functions of DNA is to make its copies which are transmitted to the daughter cells. Replication is the process by which DNA makes exact copies of itself. Replication is the basis of like and takes place during the inter phase stage.

During replication of DNA, two complementary strand of DNA unwind and separate from one end in a zipper like fashion. The enzyme helicase unwinds the two strands of the DNA. The enzyme called topoisomerase separates the double helix above the replication fork and removes the twists formed during the unwinding process. For the synthesis of new DNA, two things are required. One is RNA primer and enzyme primase. The DNA polymerase moves along the newly formed RNA primer nucleotides, which leads to the elongation of DNA. In the other strand DNA is synthesized in small fragments called Okazaki fragments.

These fragments are linked by the enzyme called ligase. In the resulting DNA, one of the strand is parental and the other is the newer strands which is formed discontinuously.

Significance of DNA:
(i) It is responsible for the transmission of hereditary information from one generation to next generation.
(ii) It contains information required for the formation of proteins.
(iii) It controls the developmental process and life activities of an organism.
Samacheer Kalvi 10th Science Guide Chapter 18 Heredity 4

Question 3.
The sex of the new born child is a matter of chance and neither of the parents may be considered responsible for it. What would be the possible fusion of gametes to determine the sex of the child?
Answer:
Sex determination is a chance of probability as to which category of sperm fuses with the egg. If the egg (X) is fused by the X-bearing sperm an XX individual (female) is produced. If the egg (X) is fused by the Y-bearing sperm an XY individual (male) is produced. The sperm, produced by the father, determines the sex of the child. The mother is not responsible in determining the sex of the child.

Now let’s see how the chromosomes take part in this formation. Fertilization of the egg (22+X) with a sperm (22+X) will produce a female child (44+XX). while fertilization of the egg (22+X) with a sperm (22+Y) will give rise to a male child (44+XY).

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

VIII. Higher Order Thinking Skills: (HOTS)

Question 1.
Flowers of the garden pea are bisexual and self-pollinated. Therefore, it is difficult to perform hybridization experiment by crossing a particular pistil with the specific pollen grains. How Mendel made it possible in his monohybrid and dihybrid crosses?
Answer:
He worked on nearly 10,000 pea plants of 34 different varieties. He had chosen 7 pairs of contrasting characters. As the pea plants, are self-pollinating it is easy to raise pure breeding individuals. It is easy to cross-pollinate. It has contrasting characters. So Mendel made it possible in his monohybrid and dihybrid crosses.

Question 2.
Pure-bred tall pea plants are first crossed with pure-bred dwarf pea plants. The pea plants obtained in F1 generation are then cross-bred to produce F2 generation of pea plants.
(a) What do the plants of F1 generation look like?
(b) What is the ratio of tall plants to dwarf plants in F2 generation?
(c) Which type of plants were missing in F1 generation but reappeared in F2 generation?
Answer:
(a) plants will be tall
(b) 3 : 1
(c) Tall heterozygous (Tt)

Question 3.
Kavitha gave birth to a female baby. Her family members say that she can give birth to only female babies because of her family history. Is the statement given by her family members true? Justify your answer.
Answer:
The statement given by her family members were not true. It is not hereditary or family history. The sex determination mainly depends on which category of sperm fuses with the egg. If the egg [X] is fused by the X – bearing sperm, an XX individual (female) is produced. If the egg [X] is fused by the Y – bearing sperm an XY individual (male) is produced.

IX. Value-Based Questions

Question 1.
Under which conditions does the law of independent assortment hold good and why?
Answer:
During meiosis, chromosomes assort randomly into gametes, such that the segregation of alleles of one gene is independent of alleles of another gene. This is stated in Mendel’s Second Law and is known as the law of independent assortment.

Samacheer Kalvi 10th Science Heredity Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
Exchange of genetic material take place in:
(a) vegetative reproduction
(b) asexual reproduction
(c) sexual reproduction
(d) budding
Answer:
(c) sexual reproduction

Question 2.
In human, the number of chromosomes in each cell is _______
(a) 22 pairs
(b) 21 pairs
(c) 23 pairs
(d) 20 pairs
Answer:
(c) 23 pairs

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Question 3.
If a round green seeded pea plant (RRYY) is crossed with wrinkled, yellow seeded pea plant,(rrYY) the seeds produced in F1 generation are:
(a) round and yellow
(b) round and green
(c) wrinkled and green
(d) wrinkled and yellow
Answer:
(a) round and yellow

Question 4.

In the new complementary strand of DNA, in one strand, the daughter strand is synthesized, as a continuous strand called ______
(a) Lagging strand
(b) Parent strand
(c) RNA primer
(d) Leading strand
Answer:
(d) Leading strand

Question 5.
A zygote which has an X-chromosome inherited from the father will develop into a:
(a) boy
(b) girl
(c) x- chromosome does not determine the sex of a child
(d) either boy or girl
Answer:
(b) girl

Question 6.
In pea, a pure tall plant (TT) is crossed to a short plant (tt). The ratio of pure tall plants to short plants in F2 is:
(a) 1 : 3
(b) 3 : 1
(c) 1 : 1
(d) 2 : 1
Answer:
(b) 3 : 1

Question 7.
The number of pairs of sex chromosomes in the zygote of human is:
(a) one
(b) two
(c) three
(d) four
Answer:
(a) one

Question 8.
Pure breeding varieties are otherwise called as:
(a) dominant
(b) recessive
(c) wild type
(d) mixed type
Answer:
(c) wild type

Question 9.
The genotype of a character is influenced by factors called:
(a) chromosome
(b) nucleus
(c) cytoplasm
(d) genes
Answer:
(d) genes

Find the Factors of 18 by Factoring Calculator … Factoring Calculator calculates the factors and factor pairs of positive integers.

Question 10.
Monosomy is represented by:
(a) 2n + 1
(b) 2n – 1
(c) 2n + 2
(d) 2n – 2
Answer:
(b) 2n – 1

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Question 11.
The term chromosome was introduced by:
(a) Bridges
(b) Waldeyer
(c) Balboni
(d) Flemming
Answer:
(b) Waldeyer

Question 12.
Diagrammatic representation of Karyotype of a species is:
(a) Idiogram
(b) Albinism
(c) Karyo tyning
(d) Heredity
Answer:
(a) Idiogram

II. Fill in the blanks:

1. The Genotypic ratio of Monohybrid cross is ………….
2. ……… is a graphical representation to calculate the probability of all possible genotype of off spring in a genetic cross.
3. The gene is present at a specific position on the chromosome called ……….
4. The end of the chromosome is called ………
5. The chromosomes with satellites are called as ………..
6. ……… act as a aging clock in every cell.
7. Nitrogen base + sugar = …………
8. The two strands of DNA open and separate at the point forming …………
9. Nullisomy is represented by ……….
10. The gametes produced by the organisms contain a single set of chromosomes is ……….
Answer:
1. 1 : 2 : 1
2. Punnet square
3. Locus
4. Telomere
5. Sat-chromosome
6. Telomeres
7. Nucleoside
8. Replication fork
9. 2n – 2
10. haploid (n)

III. Match the following:

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity 5
Answer:
A. (v)
B. (i)
C. (ii)
D. (iii)
E. (iv)

IV. State whether True or false, If false write the correct statement:

  1. The daughter strand synthesized in DNA is called logging strand.
  2. The centromere is found near the centre of the chromosome in sub metacentric.
  3. Primary construction in chromosome is called as nucleolar organizer.
  4. T.H. Morgan was awarded Nobel prize for determining the role of chromosome in heredity.
  5. Adenine links Thymine with three hydrogen bonds.

Answer:

  1. False – The daughter strand synthesized in DNA is called leading strand
  2. True
  3. False – Primary construction in chromosome is called as secondary construction.
  4. True
  5. False – Adenine links Thymine with two hydrogen bonds.

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

V. Answer in a word or sentence:

Question 1.
What is heredity?
Answer:
Heredity is transmission of characters from one generation to the next generation.

Question 2.
What is Alleles or Allelomorphs?
Answer:
The factors making up a pair of contrasting characters are called alleles or allelomorphs.

Question 3.
Define variation.
Answer:
Differences shown by the individuals of the same species and also by the offspring of the same parents.

Question 4.
What is Terminus?
Answer:
The replication fork of DNA, of the two sides, meet at a site called terminus.

Question 5.
Write the expanded form of DNA.
Answer:
Deoxyribo nucleic acid

Question 6.
What is the satellite?
Answer:
Some of the chromosomes have an elongated knob-like appendage at one end of the chromosome known as the satellite.

Question 7.
How many types of nitrogenous bases are present in DNA? Name them.
Answer:
There are two types of nitrogenous bases in DNA. They are purines (Adenine and Guanine) pyrimidines (Cytosine and Thymine).

Question 8.
Why is DNA called polynucleotide?
Answer:
DNA is a large molecule consisting of millions of nucleotides. Hence it is called as polynucleotide.

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Question 9.
Name two purine nitrogenous bases found in a DNA molecule.
Answer:
Adenine and Guanine

Question 10.
What are the three chemically essential parts of nucleotides containing a DNA?
Answer:
Nitrogenous base, pentose sugar and phosphate.

Question 11.
What are autosomes?
Answer:
Autosomes are chromosomes that contain genes which determine the somatic characters.

Question 12.
How is the sex of a new born determines in humans?
Answer:
The sperm produced by the father determines the sex of the child.

Question 13.
Define genetics.
Answer:
The branch of biology that deals with the genes genetic variation and heredity of living organisms is called genetics.

Question 14.
Define mutation.
Answer:
Mutation is an inheritable sudden change in the genetic material (DNA) of an organism.

Question 15.
Name the types of chromosomes based on the position of centromere.
Answer:
Based on the position of centromere, the chromosomes are classified as Telocentric, Aerocentric, submeta centric and meta centric.

VI. Short Answer Questions

Question 1.
What are chromosomes made up of?
Answer:
Chromosomes are made up of DNA, RNA, chromosomal proteins (histones and non-histones) and certain metallic ions. These proteins provide structural support to the chromosome.

Question 2.
What is the mechanism behind the expression of a particular trait? Explain.
Answer:
The factor for each character or trait remain independent and maintain their identity in the gametes. The factors are independent to each other and pass to the offspring through gametes.

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

Question 3.
If a pure tall pea plant is crossed with a pure dwarf plant, then in the first generation only tall plant appears.
(a) What happens to the traits of the dwarf plant?
(b) In the second generation, the dwarf trait reappears? Why?
Answer:
(a) The tall plant (dominant) mask the expression of the dwarf plant.
(b) When F1 hybrids are self crossed, the two entities separate and then unite independently forming tall and dwarf plant.

Question 4.
Explain the types of chromosome-based on function.
Answer:
Based on function, the chromosomes are classified into:

  1. Autosomes: Autosomes contain genes that determine the somatic (body) characters. Male and female have an equal number of autosomes.
  2. Allosomes: Allosomes are responsible for determining the sex of an individual. They are also called sex chromosomes or heterochromosomes. There are two types of sex chromosomes.

The human male has one X chromosome and one Y chromosome and human female have two X chromosomes.

VII. Long Question and Answer:

Question 1.
How are Mutation classified? Explain.
Answer:
Mutations are classified into two main types, namely chromosomal mutation and gene mutation.
Chromosomal mutation:
The sudden change in the structure or number of chromosomes is called a chromosomal mutation. This may result in
(i) Changes in the structure of chromosomes: Structural changes in the chromosomes usually occurs due to errors in cell division. Changes in the number and arrangement of genes take place as a result of deletion, duplication, inversion and translocation in chromosomes.

(ii) Changes in the number of chromosomes: They involve addition or deletion in the number of chromosomes present in a cell. This is called ploidy. There are two types of ploidy
(a) Euploidy (b) Aneuploidy.

Gene or point mutation: Gene mutation is the changes occurring in the nucleotide sequence of a gene. It involves substitution, deletion, insertion or inversion of a single or more than one nitrogenous base. Gene alteration results in abnormal protein formation in an organism.

Question 2.
What is a mutation? Explain the two types of mutation.
Answer:
The mutation is an inheritable sudden change in the genetic material (DNA) of an organism. Mutations are broadly classified into 1. Chromosomal mutation and 2. Gene mutation.

1. Chromosomal Mutation:
The sudden change in the structure or number of chromosomes is called chromosomal mutation. This result in
(a) Change in the structure of chromosomes: Structural changes occur due to errors in cell division. Changes in the number and arrangement of genes take place as a result of deletion, duplication, inversion and translocation in chromosomes.

(b) Changes in the number of chromosomes: They involve addition or deletion in the number of chromosomes present in a cell and is called ploidy. The two types of ploidy are:

(i) Euploidy: It is the condition, in which the individual bears more than the usual number. If an individual has three haploid sets of chromosomes, the condition is called triploidy [3n]. Triploid plants and animals are sterile. If an individual has four haploid sets of chromosomes, the condition is called tetraploidy [4n], Tetraploid plants often result in increased fruit and flower size.

(ii) Aneuploidy:
It is the loss or gain of one or more chromosomes in a set. It is of three types:

  • Monosomy [2n – 1]
  • Trisomy [2n + 1]
  • Nullisomy [2n – 2]

(iii) Down’s syndrome:
It is one of the commonly known aneuploid condition, in man. It is a genetic condition, in which there is an extra copy of chromosome 21 (Trisomy 21). It is associated with mental retardation, delayed development, behavioural problems, weak muscle tone, vision and hearing disability are some of the conditions seen in children.

2. Gene or point mutation:
Gene mutation is the changes occurring in the nucleotide sequence of a gene. It involves substitution, deletion, insertion or inversion of a single or more than one nitrogenous base. Gene alteration results in abnormal protein formation.

Question 3.
Write a note on down’s syndrome.
Answer:
Down’s syndrome: This condition was first identified by a doctor named Langdon Down in 1866. It is a genetic condition in which there is an extra copy of chromosome 21 (Trisomy 21). It is associated with mental retardation, delayed development, behavioural problems, weak muscle tone, vision and hearing disability are some of the conditions seen in these children.

Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

VIII. Higher Order Thinking Skills: (HOTS)

Question 1.
In a plant gene ‘A’ is responsible for tallness and its recessibe allele ‘a’ for dwarfness and ‘B’ is responsible for red colour to recessive allele ‘b’ for white flower colour. A tall and red flowered plant with genotype AaBb crossed with dwarf and red flowers (aaBb). What is the percentage of dwarf white flowered off spring of above cross?
Answer:
12.5 %

Question 2.
In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this data, the frequency of allele A in the population is:
Answer:
0.6

Question 3.
A tall true breeding garden pea plant is crossed with dwarf true breeding garden Pea plant. When the F1 plants were selfed the resulting genotypes were in the ratio of:
Answer:
1 : 2 : 1 :: Tall homozygous; Tall heterozygous Dwarf.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Computer Applications Guide Pdf Chapter 18 Electronic Data Interchange – EDI Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 18 Electronic Data Interchange – EDI

12th Computer Applications Guide Electronic Data Interchange – EDI Text Book Questions and Answers

Part I

Choose The Correct Answers

Question 1.
EDI stands for
a) Electronic Details Information
b) Electronic Data Information
c) Electronic Data Interchange
d) Electronic Details Interchange
Answer:
a) Electronic Details Information

Question 2.
Which of the following is an internationally recognized standard format for trade, transportation, insurance, banking and customs?
a) TSLFACT
b) SETFACT
c) FTPFACT
d) EDIFACT
Answer:
d) EDIFACT

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

What is the Greatest Common Factor of 12 and 18? … Greatest common factor (GCF) of 12 and 18 is 6.

Question 3.
Which is the first industry-specific EDI standard?
a) TDCC
b) VISA
c) Master
d) ANSI
Answer:
a) TDCC

Question 4.
UNSM stands for
a) Universal Natural Standard Message
b) Universal Notations for Simple Message
c) United Nations Standard Message
d) United Nations Service Message
Answer:
c) United Nations Standard Message

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 5.
Which of the following is a type of EDI?
a) Direct EDI
b) Indirect EDI
c) Collective EDI
d) Unique EDI
Answer:
a) Direct EDI

Question 6.
Who is called the father of EDI?
a) Charles Babbage
b) Ed Guilbert
c) Pascal
d) None of the above
Answer:
b) Ed Guilbert

Question 7.
EDI interchanges start with ……………. and end with ……………
a) UNA, UNZ
b) UNB, UNZ
c) UNA, UNT
d) UNB, UNT
Answer:
b) UNB, UNZ

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 8.
EDIFACT stands for
a) EDI for Admissible Commercial Transport
b) EDI for Advisory Committee and transport
c) EDI for Administration, Commerce, and Transport
d) EDI for Admissible Commerce and Trade
Answer:
c) EDI for Administration, Commerce, and Transport

Question 9.
The versions of EDIFACT are also called as
a) Message types
b) Subsets
c) Directories
d) Folders
Answer:
c) Directories

Question 10.
Number of characters in a single EDIFACT messages
a) 5
b) 6
c) 4
d) 3
Answer:
b) 6

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Part II

Short Answers

Question 1.
Define EDI.
Answer:
The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically. It is transferred through a dedicated channel or – through the Internet in a predefined format without much human intervention.

Question 2.
List few types of business documents that are transmitted through EDI.
Answer:

  1. Delivery notes
  2. Invoices
  3. Purchase orders
  4. Advance ship notice
  5. Functional acknowledgments etc.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 3.
What are the 4 major components of EDI?
Answer:
There are four major components of EDI. They are:

  1. Standard document format
  2. Translator and Mapper
  3. Communication software
  4. Communication network

Question 4.
What is meant by directories inEDIFACT?
Answer:

  • The versions of EDIFACT are also called as directories.
  • These EDIFACT directories will he revised twice a year.

Question 5.
Write a note on EDIFACT subsets.
Answer:
Due to the complexity, branch-specific subsets of EDIFACT have developed. These subsets of EDIFACT include only the functions relevant to specific user groups.
Example:

  • CEFIC – Chemical industry
  • EDIFURN – furniture industry
  • EDIGAS – gas business

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Part III

Explain In Brief Answer

Question 1.
Write a short note on EDI.

  • The Electronic Data Interchange (EDI)is the exchange of business documents between one trade partner and another electronically,
  • It is transferred through a dedicated channel or through the Internet in a predefined format without much human intervention,
  • It is used to transfer documents such as delivery notes, invoices, purchase orders, advance ship notices, functional acknowledgments, etc.

Question 2.
List the various layers of EDI.
Answer:
Electronic data interchange architecture specifies four different layers namely

  1. Semantic layer
  2. Standa, us translation layer
  3. Transport layer
  4. Physical layer

These EDI layers describe how data flows from one computer to another.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 3.
Write a note on UN/EDIFACT.
Answer:

  • United Nations / Electronic Data Interchange for Administration, Commerce, and Transport
  • (UN / EDIFACT) is an international EDI – a standard developed under the supervision of the United Nations.
  • In 1987, the UN / EDIFACT syntax rules were approved as ISO: IS09735 standard by the International Organization for Standardization.
  • EDIFACT includes a set of internationally agreed standards, catalogs, and guidelines for the electronic exchange of structured data between independent computer systems.

Question 4.
Write a note on the EDIFACT message.
Answer:

  • The basic standardization concept of EDIFACT is that there are uniform message types called United Nations Standard Message (UNSM).
  • In so-called subsets, the message types can be specified deeper in their characteristics depending on the sector.
  • The message types, all of which always have exactly one nickname consisting of six uppercase English alphabets.
  • The message begins with UNH and ends with UNT.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 5.
Write about EDIFACT separators
Answer:
EDIFACT has the following punctuation marks that are used as standard separators.
Character Uses

Character

Uses

Apostrophe (‘) segment terminator
Plus sign (+) segment tag and data element separator
Colon (;) component data element separator
Question mark (?) Release character
Period (.) decimal point

Part IV

Explain In Detail

Question 1.
Briefly explain various types of EDI.
Answer:
The types of EDI were constructed based on how EDI communication connections and the conversion were organized. Thus based on the medium used for transmitting EDI documents the following are the major EDI types.

  1. Direct EDI
  2. EDI via VAN
  3. EDI via-FTP/VPN, SFTP, FTPS
  4. Web EDI
  5. Mobile EDI
  6. Direct EDI/Point-to-Point

It is also called as Point-to-Point EDI. It establishes a direct connection between various business stakeholders and partners individually. This type of EDI suits to larger businesses with a lot of day to day business transactions.

EDI via VAN:
EDI via VAN (Value Added Network) is where EDI documents are transferred with the support of third-party network service providers. Many businesses prefer this network model to protect them from the updating ongoing complexities of network technologies.

EDI via FTP/VPN, SFTP, FTPS:
When protocols like FTP/VPN, SFTP, and FTPS are used for the exchange of EDI-based documents through the Internet or Intranet it is called EDI via FTP/VPN, SFTP, FTPS.

Web EDI:
Web-based EDI conducts EDI using a web browser via the Internet. Here the businesses are allowed to use any browser to transfer data to their business partners. Web-based EDI is easy and convenient for small and medium organizations.

Mobile EDI:
When smartphones or other such handheld devices are used to transfer EDI documents it is called mobile EDI. Mobile EDI applications considerably increase the speed of EDI transactions.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 2.
What are the advantages of EDI?
Answer:

  • EDI was developed to solve the problems inherent in paper-based transaction processing and in other forms of electronic communication.
  • Implementing an EDI system offers a company greater control over its supply chain and allow it to trade more effectively. It also increases productivity and promotes operational efficiency.

The following are the other advantages of EDI.

  • Improving service to end-users
  • Increasing productivity
  • Minimizing errors
  • Slashing response times
  • Automation of operations
  • Cutting costs
  • Integrating all business and trading partners

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 3.
Write about the structure of EDIFACT.
Answer:

  • EDIFACT is a hierarchical structure where the top level is referred to as an interchange, and lower levels contain multiple messages.
  • The messages consist of segments, which in turn consist of composites.
  • The final iteration is a data element.

Segment Tables

  • The segment table lists the message tags.
  • It contains the tags, tag names, requirements designator, and repetition field.
  • The requirement designator may be mandatory (M) or conditional (C).
  • The (M) denotes that the segment must appear at least once. The (C) denotes that the segment may be used if needed.
  • Example: CIO indicates repetitions of a segment or group between 0 and 10.

EDI Interchange

  • Interchange is also called an envelope.
  • The top-level of the EDIFACT structure is Interchange.
  • An interchange may contain multiple messages. It starts with UNB and ends with UNZ

EDIFACT message

  • The basic standardization concept of EDIFACT is that there are uniform message types called United Nations Standard Message (UNSM).
  • In so-called subsets, the message types can be specified deeper in their characteristics depending on the sector.
  • The message types, all of which always have exactly one nickname consisting of six uppercase English alphabets.
  • The message begins with UNH and ends with UNT

Service messages

  • To confirm/reject a message, CONTRL and APERAK messages are sent.
  • CONTRL- Syntax Check and Confirmation of Arrival of Message
  • APERAK – Technical error messages and acknowledgment

Data exchange

  • CREMUL – multiple credit advice
  • DELFOR- Delivery forecast
  • IFTMBC – Booking confirmation

EDIFACT Segment

  • It is the subset of messages.
  • A segment is a three-character alphanumeric code.
  • These segments are listed in segment tables.
  • Segments may contain one, or several related user data elements.

EDIFACT Elements

  • The elements are the piece of actual data.
  • These data elements may be either simple or composite.

EDI Separators
EDIFACT has the following punctuation marks that are used as standard separators.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

12th Computer Applications Guide Electronic Data Interchange – EDI Additional Important Questions and Answers

Part A

Choose The Correct Answers

Question 1.
……………………. is the exchange of business documents between one trade partner and another electronically.
(a) EDI
(b) UDI
(c) FDI
(d) DDI
Answer:
(a) EDI

Question 2.
First EDI standards were released by ………..
a) EDI
b) EFT
c) EDIA
d) TDCC
Answer:
d) TDCC

Question 3.
……………………. is a paperless trade.
(a) EDI
(b) XML
(c) EDIF
(d) EFT
Answer:
(a) EDI

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 4.
………… establishes a direct connection between various business stakeholders
and partners individually.
a) Direct EDI
b) EDI via VAN
c) Web EDI
d) Mobile EDI
Answer:
a) Direct EDI

Question 5.
Electronic data interchange architecture specifies ……………. different layers.
a) two
b) three
c) four
d) five
Answer:
c) four

Question 6.
TDCC was formed in the year …………………….
(a) 1964
(b) 1966
(c) 1968
(d) 1970
Answer:
(c) 1968

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 7.
In ……………… UN created the EDIFACT to assist with the global reach of technology in E-Commerce.
a)1985
b)1978
c)1974
d)1975
Answer:
a)1985

Question 8.
Expand EDIA
(a) Electronic Data Interchange Authority
(b) Electronic Data Information Association
(c) Electronic Data Interchange Association
(d) Electronic Device Interface Amplifier
Answer:
(c) Electronic Data Interchange Association

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 9.
Which of the following is for the exchange of EDI-based documents through the Internet?
a) FTP/VPN
b) SFTP
c) FTPS
d) All of the above
Answer:
d) All of the above

Question 10.
EDIA has become …………………….. committee.
(a) ANSIXI2
(b) ANSIXI3
(c) ANSIXI4
(d) ANSIX15
Answer:
(a) ANSIXI2

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Fill In The Blanks:

1. ……….. was developed to solve the problems inherent in paper-based transaction processing.
Answer:
EDT

2. ………….. is also called as Point-to-Point EDI.
Answer:
Direct EDT

3. Interchange is also called…………..
Answer:
Envelope

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

4. EDT is ……………… Trade.
Answer:
Paperless

5. EFT is …………….. Payment
Answer:
Paperless

6. ………… is “the computer-to-computer interchange of strictly formatted messages.
Answer:
EDI

7. …………….. EDI is easy and convenient for small and medium organizations.
Answer:
Web-based

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

8. The …………. is the most critical part of the entire EDI.
Answer:
standard

Abbreviations

  1. EDI – Electronic Data Interchange
  2. EFT – Electronic Transfer
  3. TDCC – Transportation Data Coordinating Committee
  4. EDIA – Electronic Data Interchange Association
  5. ANSI – American National Standards Institute
  6. VAN – Value Added Network
  7. ANSI ASC – American National Standards Institute Accredited Standard Committee
  8. GTDI – Guideline for Trade Data Interchange
  9. UN/ECE/ – United -Nations Economic Commission for Europe
  10. UN/EDIFACT -United Nations / Electronic Data Interchange for Administration, Commerce, and Transport
  11. UNSM -United Nations Standard Message

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Assertion And Reason
Question 1.
Assertion (A): According to the National Institute of Standards and Technology, EDI is the computer-to-computer interchange of strictly formatted messages that represent documents other than monetary instruments.
Reason(R): The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically.
a) Both (A) and (R) are correct and (R) ¡s the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)

Question 2.
Assertion(A): EFT is “Paperless Trade”
Reason(R): The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
d) (A) is false and (R) is true

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 3,
Assertion (A): United Nations / Electronic Data Interchange for Administration, Commerce, and Transport (UN / EDIFACT) is an international EDI – a standard developed under the supervision of the United Nations.
Reason(R): In 1985, the UN / EDIFACT syntax rules were approved as ISO: IS09735 standard by the International Organization for Standardization.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
c) (A) is true and (R) is false

Question 4.
Assertion (A): The segment table lists the message tags.
Reason(R): It contains the tags, tag names, requirements designator, and repatriation field.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 5.
Assertion (A): The top level of EDIFACT structure is Interchange.
Reason(R): Interchange is also called an envelope. An interchange may contain multiple messages. It starts with UNB and ends with UNZ
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Short Answer Questions

Question 1.
Who is the father of EDI?
Answer:
Ed Guilbert is called the father of EDI

Question 2.
What is Paperless trade?
Answer:
The exchange of business documents between one trade partner and another electronically is called Paperless trade.

Question 3.
What is Paperless Payment?
Answer:
Transfer of money from one bank account to another, via computer-based systems, is known as Paperless payment

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 4.
What is another name of Direct EDI?
Answer:
Another name of Direct EDI is Point-to-Point EDI.

Question 5.
How many alphabets require for EDI messages?
Answer:
Every EDI message requires six uppercase English Alphabets

Match The Following:

1. EDI – Booking confirmation
2. EFT – Paperless Trade
3. EDIFACT – Envelope
4. Interchange – Delivery forecast
5. CEFIC – Directories
6. EDIFURN – Chemical industry
7. EDIGAS – Technical error
8. CONTRL – Multiple credit advice
9. APERAK – Furniture industry
10. CREMUL – Arrival of Message
11. DELFOR – Gas business
12. IFTMBC – Paperless Payment

Answers
1. Paperless Trade
2. Paperless Payment
3. Directories
4. Envelope
5. Chemical industry
6. Furniture industry
7. Gas business
8. Arrival of Message
9. Technical error
10. Multiple credit advice
11. Delivery forecast
12. Booking confirmation

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Find The Odd One On The Following

1. (a) Deliver/ Notes
(b) Invoices
(c) Advance Ship Notice
(d) EDIFACT
Answer:
(d) EDIFACT

2. (a) EDIFACT
(b) XML
(c) CDMA
(d) ANSI ASCX12
Answer:
(c) CDMA

3. (a) Direct EDI
(b) InDirectEDI
(c) Web EDI
(d) Mobile EDI
Answer:
(b) InDirectEDI

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

4. (a) FTP/VPN
(b) HTTP
(c) SFTPP
(d) FTPS
Answer:
(b) HTTP

5. (a) Dial-Up Line
(b) I way
(c) point to point
(d) Internet
Answer:
(c) point to point

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

6. (a) Email
(b) MIME
(c) HTTP
(d) ANSI X12
Answer:
(d) ANSI X12

7. (a) Transport Layer
(b) Semantic Layer
(c) Application Layer
(d) physical Layer
Answer:
(c) Application Layer

8. (a) Standards
(b) Catalogs
(c) TDCC
(d) guidelines
Answer:
(c) TDCC

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

9. (a) CREMUL
(b) DELFOR
(c) APERAK
(d) IFTMBC
Answer:
(c) APERAK

10. (a) Segment Terminator
(b) : – component data
(c) ? – data element separator
(d). – decimal point
Answer:
(c) ? – data element separator

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Important Years To Remember:

1975 First EDI standards were released by TDCC
1977 Drafting and using an EDI project begin
1978 TDCC is renamed as Electronic Data Interchange Association (EDIA)
1979 ANSI ASC developed ANSI X12
1985 UN created the EDIFACT
1986 UN/EDIFACT is officially proposed
1987 UN / EDIFACT syntax rules were approved

Part B

Short Answers

Question 1.
What is VAN?
Answer:
A value-added network is a company, that is based on its own network, offering EDI services to other businesses. A value-added network acts as an intermediary between trading partners. The principal operations of value-added networks are the allocation of access rights and providing high data security.

Question 2.
What are the types of EDI?
Answer:

  1. Direct EDI
  2. EDI via VAN
  3. EDI via FTP/VPN, SFTP, FTPS
  4. Web EDI
  5. Mobile EDI

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 3.
Write a short note on the Segment Table?
Answer:
Segment Tables:
The segment table lists the message tags. It contains the tags, tag names, requirements designator, and repetitation field. The requirement designator may be mandatory (M) or conditional (C). The (M) denotes that the segment must appear atleast once. The (C) denotes that the segment may be used if needed.

Question 4.
Mention some International accepted EDI Standards.
Answer:

  • EDIFACT
  • XML
  • ANSI
  • ASC XI2,

Part C

Brief Answers

Question 1.
Write a short note on EDIFACT Structure.
Answer:

  • EDIFACT is a hierarchical structure where the top level is referred to as an interchange, and lower levels contain multiple messages.
  • The messages consist of segments, which in turn consist of composites.
  • The final iteration is a data element.

Question 2.
What is EDI interchange?
Answer:

  • The top-level of the EDIFACT structure is Interchange.
  • An interchange may contain multiple messages.
  • It starts with UNB and ends with UNZ

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 3.
What is the EDI segment?
Answer:

  • A segment is a three-character alphanumeric code.
  • These segments are listed in segment tables.
  • Segments may contain one, or several related user data elements.

Question 4.
Write a note on EDI Interchange?
Answer:
EDI Interchange:
Interchange is also called an envelope. The top-level of the EDIFACT structure is Interchange. An interchange may contain multiple messages. It starts with UNB and ends with UNZ.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Part D

Detailed Answers

Question 1.
Explain EDI standards?
Answer:
EDI Standards:

  • The standard is the most critical part of the entire EDI. Since EDI is the data transmission and information exchange in the form of an agreed message format, it is important to develop a unified EDI standard.
  • The EDI standard is mainly divided into the following aspects: basic standards, code-standards, message standards, document standards, management standards, application standards, communication standards, and security standards.
  • The first industry-specific EDI standard was the TDCC published by the Transportation Data Coordinating Committee in 1975.
  • Then other industries started developing unique standards based on their individual needs. E.g. WINS in the warehousing industry.
  • Since the application of EDI has become more mature, the target of trading operations is often not limited to a single industry.
  • In 1979, the American National Standards Institute Accredited Standard Committee (ANSI ASC) developed a wider range of EDI standards called ANSI XI2.
  • On the other hand, the European region has also developed an integrated EDI standard. Known as GTDI (Guideline for Trade Data Interchange).
  • ANSI X12 and GTDI have become the two regional EDI standards in North America and Europe respectively.
  • After the development of the two major regional EDI standards and a few years after the trial, the two standards began to integrate and conduct research and development of common EDI standards.
  • Subsequently, the United Nations Economic Commission for Europe (UN/ECE/WP.4) hosted the task of the development of international EDI standards. In 1986, UN/EDIFACT is officially proposed. The most widely used EDI message standards are the United Nations EDIFACT and the ANSI X12.

Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Question 2.
Draw the structure of the UN/EDIFACT message.
Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI 1

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11

Students can download Maths Chapter 3 Algebra Ex 3.11 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.11

Completing the Square Calculator is an online tool that helps to complete the square of a quadratic equation and calculate its roots.

Question 1.
Solve the following quadratic equations by completing the square method
(i) 9x2 – 12x + 4 = 0
Answer:
9x2 – 12x + 4 = 0
x2 – \(\frac { 12x }{ 9 } \) + \(\frac { 4 }{ 9 } \) = 0 (Divided by 9)
x2 – \(\frac { 4x }{ 3 } \) = \(\frac { -4 }{ 9 } \)
Add [\(\frac { 1 }{ 2 } \) (\(\frac { 4 }{ 3 } \))]2 on both sides
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11 1
The solution is \(\frac { 2 }{ 3 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11

(ii) \(\frac { 5x+7 }{ x-1 } \) = 3x + 2
Answer:
(3x + 2) (x – 1) = 5x + 7
3x2 – 3x + 2x – 2 = 5x + 7 ⇒ 3x2 – x – 5x – 2 – 7 = 0
3x2 – 6x – 9 = 0 ⇒ x2 – 2x – 3 = 0 (divided by 3)
x2 – 2x = 3
Adding (\(\frac { 1 }{ 2 } \) × 2)2 on both sides
x2 – 2x + 1 = 3 + 1
(x – 1)2 = 4 ⇒ x – 1 = \(\sqrt { 4 }\)
x – 1 = ±2
x – 1 = 2 or x – 1 = -2
x = 3 or x = -1
The solution set is -1 and 3

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11

Question 2.
Solve the following quadratic equations by formula method
(i) 2x2 – 5x + 2 = 0
Answer:
a = 2, b = -5, c = 2
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11 2
The solution set is \(\frac { 1 }{ 2 } \) and 2

(ii) \(\sqrt { 2 }\) f2 – 6 f + 3 \(\sqrt { 2 }\) = 0
Answer:
Here a = \(\sqrt { 2 }\), b = -6 and c = 3\(\sqrt { 2 }\)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11 3
The solution set is \(\frac{3+\sqrt{3}}{\sqrt{2}}\) and \(\frac{3-\sqrt{3}}{\sqrt{2}}\)

(iii) 3y2 – 20y – 23 = 0
Answer:
a = 3, b = -20, c = -23
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11 4
The solution set is -1 and \(\frac { 23 }{ 3 } \)

(iv) 36y2 – 12ay + (a2 – b2) = 0
Answer:
Here a = 36, b = -12a, c = a2 – b2
Samacheer Kalvi 10th <img class=
The solution set is \(\frac { (a+b) }{ 6 } \) and \(\frac { (a-b) }{ 6 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11

Question 3.
A ball rolls down a slope and travels a distance d = t2 – 0.75t feet in t seconds. Find the time when the distance travelled by the ball is 11.25 feet.
Answer:
Distance = t2 – 0.75t
11.25 = t2 – 0.75t
Multiply by 100
1125 = 100t2 – 75t
100t2 – 75t – 1125 = 0 (Divided by 25)
4t2 – 3t – 45 = 0
a = 4,
b = -3,
c = -45
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11 7
(time will not be negative)
The required time = \(\frac { 15 }{ 4 } \) seconds
= 3 \(\frac { 3 }{ 4 } \) second or 3.75 seconds

Tamil Nadu 11th English Model Question Paper 2

Students can Download Tamil Nadu 11th English Model Question Paper 2 Pdf, Tamil Nadu 11th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

We have provided The Lost Child MCQ Questions for Class 9 English Chapter 1 with Answers to help students understand the concept very well.

TN State Board 11th English Model Question Paper 2

General Instructions:

  1. The question paper comprises of four parts.
  2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
  3. All questions of Part I, II, III and IV are to be attempted separately.
  4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
    These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
  5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
  6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
  7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3:00 Hours
Maximum Marks: 90

PART – I

I. Answer all the questions. [20 × 1 = 20]
Choose the correct synonym for the underlined words from the options given.

Question 1.
The face shone bright through the delicate shroud.
(a) linen
(b) veil
(c) fabric
(d) saree
Answer:
(b) veil

Question 2.
The basket of apples rolled across the concourse.
(a) carriage
(b) cavalry
(c) courtyard
(d) candour
Answer:
(c) courtyard

Tamil Nadu 11th English Model Question Paper 2

Question 3.
The ordinary man seldom forgets things.
(a) often
(b) rarely
(c) frequently
(d) random
Answer:
(b) rarely

Choose the correct antonyms for the underlined words from the options given.

Question 4.
We were so amused with the quick response from the three year old.
(a) annoyed
(b) pleased
(c) happy
(d) entertained
Answer:
(a) annoyed

Question 5.
I cherished the moist imprint as the last sign of physical presence.
(a) dry
(b) clammy
(c) sultry
(d) soggy
Answer:
(a) dry

Question 6.
Her lips moved in inaudible prayer.
(a) indistinct
(b) muted
(c) audible
(d) altered
Answer:
(c) audible

Question 7.
Choose the unclipped form of “lunch”.
(a) lunchtime
(b) luncher
(c) lunchent
(d) luncheon
Answer:
(d) luncheon

Question 8.
Choose the right definition for the given term “fratricide”.
(a) The fear of the future
(b) One who makes major changes in the education system
(c) Killing small babies
(d) The murder of your sibling
Answer:
(d) The murder of your sibling

Question 9.
Choose the meaning of the idiom ‘Add insult to injury’.
(a) Hear from the authoritative source
(b) Everything about the case
(c) To worsen an unfavourable situation
(d) To see that two agree on something
Answer:
(c) To worsen an unfavourable situation

Tamil Nadu 11th English Model Question Paper 2

Question 10.
Choose the meaning of the foreign word in the sentence.
I saw an old lady wearing a babushka walking down the street.
(a) gown or night dress
(b) scarf or head covering
(c) rain coat
(d) baboons dress
Answer:
(b) scarf or head covering

Question 11.
Choose the word from the options given to form a compound word with “ran”.
(a) down
(b) floor
(c) sack
(d) town
Answer:
(c) sack

Question 12.
Form a new word by adding a suitable suffix to the root word, “nourish”.
(a) ly
(b) ile
(c) ment
(d) ness
Answer:
(c) ment

Question 13.
Choose the expanded form of “CPWD”.
(a) Centralised Public Works Department
(b) Central Public Works Director
(c) Central Public Works Department
(d) Central Private Works Department
Answer:
(c) Central Public Works Department

Question 14.
The correct syllabification of the word “circumstance” is………..
(a) cir-cum-stan-ces
(b) cir-cum-stance
(c) cir-cu-m-stance
(d) circ-um-stance
Answer:
(b) cir-cum-stance

Question 15.
The fear of being in the dark is known as…………
(a) Nyctophobia
(b) Topophobia
(c) Chronophobia
(d) Acronymania
Answer:
(a) Nyctophobia

Question 16.
Fill in the blank with the suitable preposition.
Aijun saw the train moving ……….. the lady with a cell phone on the track.
(a) for
(b) beneath
(c) towards
(d) from
Answer:
(c) towards

Question 17.
Add a suitable question tag to the following statement.
I haven’t answered your questions, ……….. ?
(d) haven’t I
(b) have I
(c) will I
(d) shan’t I
Answer:
(b) have I

Tamil Nadu 11th English Model Question Paper 2

Question 18.
Substitute the underlined word with the appropriate polite alternative.
John is jobless at the moment so I don’t think he can afford to come on holiday with us.
(a) firing
(b) hiring
(c) after jobs
(d) between jobs
Answer:
(d) between jobs

Question 19.
Substitute the phrasal verb in the sentence with a single word.
Monisha takes after her dad.
(a) follows
(b) receives
(c) resembles
(d) remembers
Answer:
(c) resembles

Question 20.
Fill in the blank with a suitable relative pronoun
This is the stream ……….. was contaminated with plastics.
(a) which
(b) that
(c) what
(d) whose
Answer:
(a) which

PART – II

II. Answer any seven of the following: [7 × 2 = 14]
(i) Read the following sets of poetic lines and answer any four of the following. [4 × 2 = 8]

Question 21.
“Cocktail face with all their conforming smiles
Like a fixed portrait smile”
(a) What is meant by ‘conforming smiles’?
(b) Mention the figure of speech employed in the first line.
Answer:
(a) The conforming smile symbolizes the artificial and stiff smile meant only for appearances or occasions.
(b) Metaphor

Question 22.
“I am just glad as glad can be
That I am not them, that they are not me.
With all my heart I do admire
Athletes who sweat for fun or hire”
(a) Who does he admire? Why?
(b) Pick out the rhyming words.
Answer:
(a) The poet admires athletes who play games and sweat for fun and money.
(b) ‘be, me’ and ‘admire, hire’ are the rhyming words.

Question 23.
“He’s outwardly respectable (They say he cheats at cards)
And his footprints are not found in any file of Scotland Yard’s”
(a) Identify the poem and the poet.
(b) Whose footprints are not found in any file of Scotland Yard’s?
Answer:
(a) The poem is ‘Macavity – The Mystery Cat’ written by T.S. Eliot.
(b) Macavity’s footprints are not found in any file of Scotland Yard’s.

Tamil Nadu 11th English Model Question Paper 2

Question 24.
“Let’s choose executors and talk of wills.
And yet not so – for what can we bequeath
Save our deposed bodies to the ground?”
(a) What do you mean by ‘deposed bodies’?
(b) Why should they choose executors and talk about the wills?
Answer:
(a) It means dead bodies.
(b) As the king’s death is nearing the king wants to talk about executors and wills.

Question 25.
“Our pride springs from the way we live.”
(a) Under normal circumstances what makes one feel proud?
(b) What is unique about the pride mentioned above?
Answer:
(а) One’s material wealth, high social standing and popularity makes one feel proud.
(b) The sense of pride springs from the way the people live their lives and not from positions or possessions.

Question 26.
“And much it grieved my heart to think
What Man has made of Man.”
(а) What grieves the poet?
(b) What is the significance of the second line?
Answer:
(a) Man’s greed to exploit natural resources and man’s moving away from nature gives ‘grief’ to the poet.
(b) The poet is unhappy with unnatural aspects of industrial revolution, the misery caused by wars, greedy and aggressive behaviour causing suffering in humans.

(ii) Do as directed (any three) [3 × 2 = 6]

Question 27.
Rewrite the following dialogue in reported form.
Paul : Do you know that Mrs. Kalpana was awarded the Best Teacher Trophy this year by the Rotary club?
Shanmugam : Is that so? I am glad. She is a deserving teacher.
Answer:
Paul asked Shanmugam if he knew that Mrs. Kalpana was awarded the Best Teacher Trophy that year by the Rotary club. Shanmugam enquired if that was so and that he was glad about it as she was a deserving teacher.

Question 28.
Rewrite the following sentence in its passive form.
He buys a portrait.
Answer:
A portrait is bought by him.

Question 29.
Priya did not start early. She was late to school. (Combine using ‘If’)
Answer:
If Priya had started early she wouldn’t have been late to school.

Tamil Nadu 11th English Model Question Paper 2

Question 30.
Convert the following complex sentence into a simple sentence.
This is the place where the meeting will be held.
Answer:
This is the venue of the meeting.

PART – III

III. Answer any seven of the following: [7 × 3 = 21]
(i) Explain any two of the following with Reference to the Context: [2 × 3 = 6]

Question 31.
There was a time indeed.
They used to shake hands with their hearts
But that’s gone, son
Answer:
Reference: This line is from the poem “Once upon a time” written by Gabriel Okara.

Context: The poet speaks about the falsity concealed behind smiles and the lack of innocence of childhood.

Explanation: The poet, Okara observes a marked change in the altitude of Africans. Those who were once so genuine, warm and sincere, have now suddenly turned cold and hostile towards him. He realizes that the early values like sincerity, good-naturedness, simplicity, whole-heartedness, hospitality, friendliness, originality and uniqueness have now drastically changed. The earlier warmth has gone.

Question 32.
We are proud of the position we
Hold; humble as we are
Answer:
Reference: These lines are from the poem “Everest is not the Only Peak” written by Kulothungan.

Context: The poet admits that he is proud of people’s humble positions because their pride springs not from positions or possessions but the way they live.

Explanation: The poet just doesn’t bother the height of the peak one reaches. It could even be a hillock. Their life knows no bending. What matters is how one reaches that spot. If merit and competence have paved the way for their success and positions, however humble they are, the poet admires them.

Question 33.
“For God’s sake let us sit upon the ground
And tell sad stories of the death of kings:”
Answer:
Reference: These lines are from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.

Context: This poem speaks of the vanity of life and how Death is the ultimate conqueror.

Explanation: King Richard started feeling distressed about his impending death. He realises his possessions will be reduced to a patch of land. His will bequeathing his wealth to his son will be treated like dust. He recalls how kings get slain in battlefield and cries on losing his belongings. The king feels he is also an ordinary mortal deceived by the jester ‘death’. He also needs to taste grief and needs the support of friends during distress.

Tamil Nadu 11th English Model Question Paper 2

(ii) Answer any two of the following questions briefly: [2 x 3 = 6]

Question 34.
What was Mary Kom’s first impression about America?
Answer:
America was cold and beautiful. What little she saw was very pleasing to her eyes. Americans were enormously nice too. She felt that this would be the place and event that would change her life.

Question 35.
What is the difference between a physical and mental tight corner?
Answer:
Physical tight comers are those situations which threaten the life of An individual. Mental tight comers are worries for which no solution is in sight. It upsets the individuals and confounds them.

Question 36.
How does Arignar Anna highlight the duties and responsibilities of graduates to the society?
Answer:
The graduates must acquire the means of a decent living. But it should be the only objective. As their education is funded by the tax from poor people, they have on obligation to pay back to the society if not in cash in terms of service. They must bring light into the dark alleys, sunshine into dingy places, solace into the affiliated hope unto the despondent and a new life into every one.

(iii) Answer any three of the following: [3 × 3 = 9]

Question 37.
Study the pie-chart and answer the questions that follow:
Tamil Nadu 11th English Model Question Paper 2 1
Questions.

  1. What percent of body weight constitutes of skin?
  2. Forty percent of the body weight constitutes of ………..
  3. Bones take up ……… percent of the body weight.

Answer:

  1. Ten percent of the body weight constitutes of skin.
  2. Hormones and enzymes
  3. Twenty

Question 38.
Build a dialogue of minimum three exchanges between a two friends.
Answer:
Mani : What’s wrong, Tarun? You look terrible!
Tarun : My car slid into a tree, because the roads were slippery.
Mani : Slippery roads and speed don’t mix Tarun. You should be careful.
Tarun : 1 know. But I have one more problem. I didn’t have my driver’s license with me.
Mani : Why were you driving without your license?
Tarun : Well, 1 lost my wallet some days ago, while I was travelling in the bus to work.

Tamil Nadu 11th English Model Question Paper 2

Question 39.
Describe the process of Assembling a piece of furniture.
Answer:

  1. Arrange the pieces of furniture to be assembled on the floor neatly.
  2. Follow the instructions given in the manual.
  3. Take inventory of all of the parts and pieces of your new furniture before you start building it.
  4. Keep aside the tools you required to fix the furniture.
  5. Reread all instructions and double check your handiwork before you proceed to the next step in the instruction manual.
  6. Screw in all the parts as seen in the manual tightly. Check for any loose contacts.
  7. Many furniture companies have videos and FAQ’s on their sites that are useful.
  8. Now the furniture is ready for use.

Question 40.
Complete the proverbs using the words given below.
(a) Appearances can be ……….. (funny, deceptive, tricky)
(b) Better ……….. than never, (late, soon, forget)
(c) Don’t ………. the hand that feeds you. (admire, thank, bite)
Answer:
(a) deceptive
(b) late
(c) bite

PART – IV

IV. Answer the following: [7 x 5 = 35]

Question 41.
“….But, when it’s my own – well, I think hysterics are fully justified’ – How?
Answer:
The author had planned to go to England with all his family members. He arrived at the Logan airport at Boston. When they were checking in, he suddenly remembered that he forgot to use his frequent flier card (British Airways). He also remembered how he had left it in a bag. He tried to open the bag. The zip was jammed. He tried to open it by force. After several attempts, it gave away spilling all the contents in a sprawling corridor in the airport. He ignored the flying documents, silver coins and even passport.

He worried about the tobacco box which was rolling away crazily disgorging its content on the way. He cried “My Tobacco” remembering how expensive it would be to buy tobacco for his pipe in England. Just then he realized that he was bleeding profusely. He had made a gash on his finger while trying to open the zip of his bag by force. He cried hysterically on seeing his own blood, “My finger” My finger”. In general, he was not comfortable flowing other’s blood. But when it came to spilling his own blood “hysterics” was really justified.

[OR]

How do Universities mould students apart from imparting academic education to them?
Answer:
Universities mould students by providing various opportunities to develop their soft skills and to develop values which would contribute to the process of nation building. They enable graduates to develop patience and perseverance. They help them develop faith in their own inherent ability to shoulder responsibilities. They are oriented to become citizens of democracy and repay to the society quality services which would reform the lives of the poor people.

They develop true spirit of democracy among young graduates. They enable appreciation of others point of view. The graduates are also provided opportunities to adjust with difference through amicable discussions. The universities, apart from imparting education mould the students’ character and personality too.

Tamil Nadu 11th English Model Question Paper 2

Question 42.
Give reasons to prove that the future generations remember easily the Victor more than the Vanquished with relevant references from King Richard’s speech.
Answer:
Unusually future generations remember victors. But there are rare instances of just rulers falling due to the conspiracy and greed of an aggressor. On such occasions, future generations remember the vanquished. A Shiva devotee king was very generous. His enemies entered his kingdom under the guise of Shiva devotees in saffron clothes and slew the king and captured his kingdom. Alexander, King Richard was a just ruler. He was loved by his subjects and loyal nobles.

He was defeated by his rebellious cousin simply because he wanted to be a king. When Richard was thinking about the welfare of his subjects, Bolingbroke was secretly raising an army to dethrone him. People who are mad after power resort to unjust means. So, British subjects respected and loved the vanquished but were helpless. Defeated Porus had fought so valiantly and wanted to be treated with respect befitting a king. Alexander himself respected him and returned his kingdom and sealed a life time friendship with him. From King Richard’s speech one understands that he was good at heart but in the strategy of war, he was not good.

Like a crooked end of a straight walking stick, a ruler has to have some secret deals with neighbouring countries to be protected during crisis. Bolingbroke turned out to be a more assertive and Shrewd king. But people would remember a just and noble person more even if defeated.

[OR]

When humanity fails to live in harmony with Nature, its effects are felt around the world. Justify.
Answer:
Man, the worst predator, kills for no reason. Man has to protect forests and live in harmony with nature. Instead man is callous. He kills elephants for their tusks, Rhinoceros for their horn, and polar bears for their fur. Huge trees, in Rainforests, which have been protecting lives of many species and insects, are being felled for timber and industrialization. Due to the increase in the denudation of forests, global warming has increased. Water levels in the ocean is increasing.

Heat waves are threatening the lives of people. Polar ice is melting. Scientists fear that if this persists, there will be hostility caused by water-sharing. Like Karnataka and Tamil Nadu, there will be political unrest and community conflicts demanding share in drinking water and water for irrigation purposes. In South Africa, zero water day is fast approaching. The scarcity of portable water is going to be a huge humanitarian crisis. As we have failed to protect the national resources, carbon foot print is expanding to alarming levels.

Delhi experiences difficulty as planes struggle to land or take off dufe to thick smog in and around Delhi. As toxic waste is released by Sterlyte and other industries people in Thoothukudi are becoming victims of cancer and other lung related disorders. Atomic power plants also retain potential hazards like radio-activity. Thus humanity’s failure to live in harmony with nature is threatening to wipe out human race.

Question 43.
Write a paragraph (150 words) by developing the following hints.
Miss Meadows – upset – remains gloomy – in class – taxes the students – sing sad – the girls sense her change – Basil – She thinks of the letter – called by headmistress – telegram – happy and returns to the class with vigour and good cheer.
Answer:
The Singing Lesson, written by Katherine Mansfield, is all about a surprising d&y of a music teacher’s life. Miss Meadows, a music teacher, receives a letter from her fiance which states quite plainly that Basil, her fiance, isn’t ready to marry her and feels that the marriage would fill him with disgust. Naturally she’s filled with despair, anger and sadness. Her usual calm and cheery demeanor turns gloomy and angry that day and this change doesn’t go unnoticed by her students.

During the lesson she’s rather harsh with her students. She tells them that today they would be practising a lament. Then she tells them that they must feel the despair, the pain and the sorrow in order to perform the piece perfectly. During the lesson she’s informed by another colleague that Basil, her fiance, has sent a telegram for her. Her first thought is that Basil has committed suicide! Yes, you read that right. It’s because the school has a rule; telegram can be sent to the workers during working hours only in case of death or emergency situation. But in the telegram Basil had asked her to ignore the first letter and that he had bought the hat-stand which they had been thinking of lately. In short, the marriage is happening.

The content of the telegram definitely lights up her mood. She returns and continues her class, now practicing a cheerful song, singing with expressions, more loudly and cheerfully than any of her students.

Tamil Nadu 11th English Model Question Paper 2

[OR]

Jack and Jill – call their house – a little nest – like birds make their nests – all collected free of cost – Jack and Jill – made their nest – right from villa – bought in instalment – take years to own.
Answer:
The Never Never Nest is a comic one-act play about a young couple who make full use of the buy-now-pay-later system. Jack and Jill were a young married couple who had a small baby. One day Aunt Jane visited them and was surprised to find that even though Jack’s salary was not high, they lived in a beautiful house with all comforts. She began to wonder whether, as a wedding gift she had given them 2000 pounds instead of 20 pounds. Otherwise how did Jack and Jill buy all these things? Then Jane understood that though they had everything, nothing really belonged to them. They bought everything on instalments.

Only a steering wheel of the car, a wheel and two cylinders had been paid for. The total amount to be paid towards instalments was more than their earnings. Aunt Jane was shocked at the way Jack and Jill ran their family. Before she left, she gave ten pounds to Jill and told them to make at least one article completely theirs, using that money. While Jack went with aunt Jane to the bus stop, Jill sent the money to Dr. Martin. Jack came back and said that he wanted to pay .two months instalments on the car using the aunt’s gift. But Jill said that by paying this money to Dr. Martin, their baby would become completely theirs.

Question 44.
Write a summary or Make notes of the following passage.
The interior maintenance of a house reflect the personality of the people who live in it. Attractive home furnishings set the stage for pleasant living. A home should have unity within each room and throughout the house. Each room should, harmonize with each other. The colour and styling of each room, particularly, should fit into the colour and styling of the rooms which run out of it. However, furnishings and surroundings expressive of just the right note of restfulness, or elegant simplicity are not often assembled by accident.

Most of the home decorators plan extensively by trying colour schemes, finding ingenious ways to make the best of what you have. They shop around to search out the right purchases at prices you can afford to pay. There is a keen pleasure in striving for the perfect result, and great satisfaction in achieving it.

A successful house and successful rooms will depend upon the proper relationship of each element used in it to the others and to the whole. Therefore, in selecting each piece it is well to consider the background, the usage, the ‘draperies, the floor covering, the upholstering materials, the woods, shapes, colour scheme, and the “feeling” you prefer for the room.

Work and plan to enjoy your house. Limit the expenditures of time, effort and money to the extent of your abilities. Elegance and delicate things may be a drain you can afford only in a limited way. If you can’t afford outside help, select a house and furnishings that require less care. Plan your activities so that tumult and upset are limited to a few rooms-an activity room or a bedroom, or a comer of the dining room.

You’ll get more pleasure out of a house if you have a hobby connected with it – collecting antiques or glass, gardening or indoor flower growing ceramics, art, cooking, decorating, flower arrangements, etc. And you’ll get more satisfaction and a great deal of help from studying household activities.

You can select a pleasing combination of colours from a wallpaper, a fabric, a flower or scene, or even a picture in a magazine. It is a good idea to make up a colour scheme. Let one colour predominate. Limit a colour scheme to two or three colours, with white or gray tones.
Answer:
Summary
No. of words given in the original passage: 369
No. of words to be written in the summary: 369/3 = 123 ±5
Rough Draft
The maintenance of the house reflects the personality of the people who live in that. So the distinctive decoration is as important as one attire in good clothes. A unity in the home can only be seen if the rooms in the house have a degree of harmony, colour and styling. Furniture is a working strategy for the pleasant living. If there is an expression of oneself then one will have a mental satisfaction every time one enter one’s home.

To attain such satisfaction one need to pore over plants, try colour schemes, window shopping to search the best thing for one’s home. Most of the home decorators plan extensively By trying colour schemes, finding ingenious ways to make the best of what you have. They shop around to search out the right purchases at prices you can afford to pay. There is a keen pleasure in striving for the perfect result, and great satisfaction in achieving it.

Fair Draft
Interior design of One’s Home
The interior furnishings of’ a house reflect the personality of the people who live in it. It is as important as one dresses in good clothes. A unity in the home can only be seen if the rooms in the house have a degree of harmony, colour and styling. Furniture also enhances one’s perception on pleasant living. One will have a mental satisfaction every time one enter one’s home.

To attain such satisfaction one need to plan extensively, try colour schemes, window shopping to search the best thing for one’s home within their budget. Home decorators helps one decide on these matters. One can get more pleasure out of a house if they have a hobby connected with it like collecting antiques, gardening, art, cooking, decorating, flower arrangements, etc.

No. of words in the summary: 129

Tamil Nadu 11th English Model Question Paper 2

[OR]

Notes
Title: Interior design of One’s Home
Answer:
Home reflects:

  • personality of house-owner
  • unity & harmony bet. rooms
  • colour & styling sh’d be uniform

Elements of decoration:

  • selection of colour schemes
  • draperies, rugs, upholstery, woods

Plan to enjoy the House:

  • limit time, effort & money
  • select furnish’gs which require little care
  • hobby connected with house-great pleasure.

Choice of Colours:

  • one colour sh’d predominate
  • calm colours for restfulness; intense for liveliness
  • colours sh’d harmonise with furniture, draperies, carpets

Abbreviations used: bet. – between; sh’d – should; fumish’gs – furnishings;

Question 45.
Read the following advertisement and prepare a resume/bio-data/CV considering yourself fulfilling the conditions specified.
[Write XXXX for your name and YYYY for your address]
Wanted
Computer Operator – Diploma holder with computer knowledge, fluency in English and good communication skills, Minimum 3 Years Experience.
Apply with your bio-data to : Post Box No : 545
C/o. The Hindu
Trichy- 620001.
Answer:

25.09.XXXX

From
XXXX
YYYY

To
Post box No. 545
C/o The Hindu
Trichy – 620001

Sir,
Sub: Applying for the post of Computer Operator – Reg.
I hereby apply for the post of Computer Operator vacant in your esteemed concern. I have the necessary qualification. My particulars are furnished below for your kind consideration. Bio-data

Name: XXXX
Father’s Name : Mr. R. Karthick
Address: YYYY
Qualification : B.Sc. Computer Science, 1st class, Madurai Kamaraj University
Technical Qualification : Tally, C++, PGDCA
Experience : Seven years of service in Aircel
Age: 28
Languages known : Tamil, English
Joining date : Can join immediately
Reference : My previous employer
Mr. Raj (9876543210)
I look forward to receiving your call letter. I shall offer my services to the best of my superiors’ satisfaction sir.

Yours sincerely
XXXX
Address on the Envelope
To
Post box No. 545
C/o. The Hindu
Trichy-620001

Tamil Nadu 11th English Model Question Paper 2

[OR]

Write an essay in about 150 words on ‘Cyber safety’.
Answer:
Cyber safety
Every child needs to be taught the basics of cyber safety. All of us are aware of the fact that ‘Blue Whale’ game cost lives of many young ones across the globe. Children who are befriended through social websites reveal personal information unwittingly and are exploited by persons who have access to their personal details. Children must be advised to refrain from sharing things with total strangers. Even adults are exploited through social websites and their budding lives are at stake. So, students must not evince keen interest in making friends with strangers online. If children do not do anything that is shameful to admit to parents, cyber crimes will be reduced to minimum.

Question 46.
Read the following sentences, spot the errors and rewrite the sentences correctly.
(a) There is nothing much selfish you can do than come to work sick.
(b) Elimination for child labour is undoubtedly one of the biggest challenge of our country.
(c) Today democracy is often assume to be a liberal form of governance.
(d) In the traditional sense prayer means communicating on God Almighty.
(e) Some of them have been converted into museums but libraries.
Answer:
(a) There is nothing more selfish you can do than come to work sick.
(b) Elimination of child labour is undoubtedly one of the biggest challenge of our country.
(c) Today democracy is often assumed to be a liberal form of governance.
(d) In the traditional sense prayer means communicating with God Almighty.
(e) Some of them have been converted into museums and libraries.

[OR]

Fill in the blanks appropriately.
(a) The Police tried ……….. to information from the boot leggar who used to sell ………. liqour. (illicit/ elicit)
(b) All citizens ………. obey the laws of the land. (Fill in with a modal verb)
(c) We ……….. go grocery shopping, (use semi-modal)
(d) ………. he is rich, he lives in a small house. (Use a suitable link word)
Answer:
(a) elicit, illicit
(b) must
(c) need to
(d) Although

Tamil Nadu 11th English Model Question Paper 2

Question 47.
Identify each of the following sentences with the fields given below.
(a) The board has decided to give the shareholders a dividend of 25 percent.
(b) A salaried employee in the highest slab pays income tax at 33.66 percent.
(c) For programming, people use the binary system.
(d) Every plant organ has a definite form and structure and performs certain specific functions.
(e) “My goal is winning a Grand Slam”, says Sania Mirza.
(Botany, Sports, Taxation, Business, Computer)
Answer:
(a) Business
(b) Taxation
(c) Computer
(d) Botany
(e) Sports

[OR]

Read the following passage carefully and answer the questions that follow.
Once upon a time a frog croaked in Bingle Bog all the night beginning from dusk to dawn. All the creatures hated his loud and unpleasant voice but still they did not have any other option. The voice came out from the sumac tree where every night the frog sang till morning.

He was so determined and also shameless that neither stones, prayers or sticks nor the insults or complaints could divert him from singing. One night, a nightingale started casting her melody in the moonlight to which both the frog and the other creatures were left dumbstruck. The whole bog remained, rapt and admired her voice and applauded her when she ended. The frog was obviously jealous of his rival and had finally decided to eliminate her.

So, the next night when the nightingale was again preparing to sing, the frog’s croak disturbed her. On being asked about himself by the nightingale he answered that he owned the sumac tree and he had been known for his splendid voice. Also he said that he had written a number of songs for the Bog Trumpet. The nightingale asked him whether he liked her song or not.

The frog said that the song wasn’t bad but too long and it lacked some force. The nightingale was greatly impressed that such a critic had discussed her song. She said that she was happy that the song was her own creation. To this the frog said that she needed a proper training to obtain a strong voice otherwise she would remain a beginner only. He also said that he would train her but would charge some fee.

Now, the nightingale was flushed with confidence and was a huge sensation, attracting animals from miles away and the frog with a great accuracy charged all of them admission fee. The frog began her vocal training despite of the bad and rainy weather where even the nightingale had first refused to sing. But the frog forced her to sing for six hours continuously till she was shivering and her voice had become rough and unclear. But, somehow her neck got clear the next day and she was able to sing again collecting a breathless crowd including rich ladies kings queens etc. To all this, the frog had both sweet and bitter feelings. Sweet because he was earning lots of money and bitter because of jealously as his rival was earning name and fame.

Every day, the frog scolded her to practice even longer finding out her little mistakes like nervousness not laying more trills and frills etc. He reminded her that she still owed him sixty shillings and that’s why the crowd should increase. But the condition of nightingale was getting worse. Her tired and uninspired song could no longer attract the crowd. She could not resist this as she had become used to applause and thus had become miserable too.

The heartless frog scolded her even then calling her a brainless bird. She trembled, puffed up, burst a vein and died. The frog said that he had tried to teach her but she was foolish, nervous and tensed and moreover much prone to influence. Then, once again the frog’s fog horn started blearing unrivalled in the bog.

The moral of the poem is that being inspired and influenced by someone much unknown and strange is indeed a foolish work. The nightingale could have very well, judged that how could the frog with such a harsh voice be music maestro and she had to suffer for her misjudgement. Many people in the human society also try to take advantage of the innocence or ignorance of the people.
Questions.

  1. How do you know that the frog was of a determined nature?
  2. A bog is a
  3. What did the nightingale become?
  4. Pick out one word from the passage which means ‘genius’.
  5. What is the moral of the story?

Answer:

  1. He was so determined and also shameless that neither stones, prayers or sticks nor the insults or complaints could divert him from singing.
  2. quagmire.
  3. The nightingale became a huge sensation.
  4. Maestro’ is the word which means genius.
  5. The moral of the story is that being inspired and influenced by someone much unknown and strange is indeed a foolish work.

Tamil Nadu 11th English Model Question Paper 2