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.

9/20 as a decimal is 0.45

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 3 Chapter 1 Fractions Ex 1.1

Students can download Maths Chapter 1 Fractions Ex 1.1 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 3 Chapter 1 Fractions Ex 1.1

Question 1.
Fill in the blanks:
(i) 7\(\frac{3}{4}\) + 6\(\frac{1}{2}\) ………..
(ii) The sum of a whole number and a proper fraction is called ……….
(iii) 5\(\frac{1}{3}\) – 3\(\frac{1}{2}\) ………..
(iv) 8 ÷ \(\frac{1}{2}\) ………..
(v) The number which has its own reciprocal is ……….
Solution:
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 1

(ii) Mixed Fraction

(iii) 5\(\frac{1}{3}\) – 3\(\frac{1}{2}\)
= \(\frac{16}{3}\) – \(\frac{7}{2}\) = \(\frac{32-21}{6}\) = \(\frac{11}{6}\)
= 1\(\frac{5}{6}\)

(iv) 8 ÷ \(\frac{1}{2}\)
= 8 × \(\frac{2}{1}\)
= 16

(v) 1

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

The first step would be to write the mix number 223 as an improper fraction.

Question 2.
Say True or False
(i) 3\(\frac{1}{2}\) can be written as 3 + \(\frac{1}{2}\).
(ii) The sum of any two proper fractions is always an improper fraction.
(iii) The mixed fraction of \(\frac{13}{4}\) is 3\(\frac{1}{4}\).
(iv) The reciprocal of an improper fraction is always a proper fraction.
(v) 3\(\frac{1}{4}\) × 3\(\frac{1}{4}\) = 9\(\frac{1}{16}\)
Solution:
(i) True
(ii) False
(iii) True
(iv) True
(v) False

Question 3.
Answer the following:
(i) Find the sum of \(\frac{1}{7}\) and \(\frac{3}{9}\).
(ii) What is the total of 3\(\frac{1}{3}\) and 4\(\frac{1}{6}\)?
(iii) Simplify: 1\(\frac{3}{5}\)+5\(\frac{4}{7}\).
(iv) Find the difference between \(\frac{8}{9}\) and \(\frac{2}{7}\)
(v) Subtract: 1\(\frac{3}{5}\) from 2\(\frac{1}{3}\).
(vi) Simplify: 7\(\frac{2}{7}\) – 3\(\frac{4}{21}\)
Solution:
(i) \(\frac{1}{7}\) + \(\frac{3}{9}\)
= \(\frac{9+21}{63}\) = \(\frac{30}{63}\) = \(\frac{10}{21}\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 2
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 3

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 4.
Convert mixed fractions into improper fractions and vice versa:
(i) 3\(\frac{7}{18}\)
(ii) \(\frac{99}{7}\)
(iii) \(\frac{47}{6}\)
(iv) 12\(\frac{1}{9}\)
Solution:
(i) 3\(\frac{7}{18}\) = \(\frac{61}{18}\)
(ii) \(\frac{99}{7}\) = 14\(\frac{1}{7}\)
(iii) \(\frac{47}{6}\) = 7\(\frac{5}{6}\)
(iv) 12\(\frac{1}{9}\) = \(\frac{109}{9}\)

Question 5.
Multiply the following:
(i) \(\frac{2}{3}\) × 6
(ii) 8\(\frac{1}{3}\) × 5
(iii) \(\frac{3}{8}\) × \(\frac{4}{5}\)
(iv) 3\(\frac{5}{7}\) × 1\(\frac{1}{13}\)
Solution:
(i) \(\frac{2}{3}\) × 6 = 4

(ii) 8\(\frac{1}{3}\) × 5
= \(\frac{25}{3}\) × 5
= \(\frac{125}{3}\)
= 41\(\frac{2}{3}\)

(iii) \(\frac{3}{8}\) × \(\frac{4}{5}\)
= \(\frac{12}{40}\) = \(\frac{3}{10}\)

(iv) 3\(\frac{5}{7}\) × 1\(\frac{1}{13}\)
= \(\frac{26}{7}\) × \(\frac{14}{13}\)
= \(\frac{4}{1}\)
= 4

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 6.
Divide the following
(i) \(\frac{3}{7}\) ÷ 4
(ii) \(\frac{4}{3}\) ÷ \(\frac{5}{9}\)
(iii) 4\(\frac{1}{5}\) ÷ 3\(\frac{3}{4}\)
(iv) 9\(\frac{2}{3}\) ÷ 1\(\frac{2}{3}\)
Solution:
(i) \(\frac{3}{7}\) ÷ 4
= \(\frac{3}{7}\) × \(\frac{1}{4}\) = \(\frac{3}{28}\)

(ii) \(\frac{4}{3}\) ÷ \(\frac{5}{9}\)
= \(\frac{4}{3}\) × \(\frac{9}{5}\)
= \(\frac{12}{5}\)
= 2\(\frac{2}{5}\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 4

Question 7.
Gowri purchased 3\(\frac{1}{2}\) kg of tomatoes, \(\frac{3}{4}\) kg of brinjal and 1\(\frac{1}{4}\) kg of onion. What is the total weight of the vegetables she bought?
Solution:
Total weight of vegetables bought
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 5

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 8.
An oil tin contains 3\(\frac{3}{4}\) litres of oil of which 2\(\frac{1}{2}\) litres of oil is used. How much oil is left over?
Solution:
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 6

Question 9.
Nilavan can walk 4\(\frac{1}{2}\) km in an hour. How much distance will he cover in 3\(\frac{1}{2}\) hours?
Solution:
Distance walked in an hour = 4\(\frac{1}{2}\) km
Distance covered in 3\(\frac{1}{2}\)
Hours
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 7

Question 10.
Ravi bought a curtain of length 15\(\frac{3}{4}\) m. If he cut the curtain into small pieces each of length 2\(\frac{1}{4}\) m, then how many small curtains will he get?
Solution:
Total length = 15\(\frac{3}{4}\)
Length of the small pieces = 2\(\frac{1}{4}\)
Small curtains obtained
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 8
= 7

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Objective Type Questions

Question 11.
Which of the following statement is incorrect?
(a) \(\frac{1}{2}\) > \(\frac{1}{3}\)
(b) \(\frac{7}{8}\) > \(\frac{6}{7}\)
(c) \(\frac{8}{9}\) < \(\frac{9}{10}\)
(d) \(\frac{10}{11}\) > \(\frac{9}{10}\)
Solution:
(d) \(\frac{10}{11}\) > \(\frac{9}{10}\)

Question 12.
The difference between \(\frac{3}{7}\) and \(\frac{2}{9}\) is
(a) \(\frac{13}{63}\)
(b) \(\frac{1}{9}\)
(c) \(\frac{1}{7}\)
(d) \(\frac{9}{16}\)
Solution:
(a) \(\frac{13}{63}\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 13.
The reciprocal of \(\frac{53}{17}\) is
(a) \(\frac{53}{17}\)
(b) 5\(\frac{3}{17}\)
(c) \(\frac{17}{53}\)
(d) 3\(\frac{5}{17}\)
Solution:
(c) \(\frac{17}{53}\)

Question 14.
If \(\frac{6}{7}=\frac{\mathbf{A}}{49}\), then the value of A is
(a) 42
(b) 36
(c) 25
(d) 48
Solution:
(a) 42
Hint: \(\frac{6 \times 49}{7}=42\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 15.
Pugazh has been given four choices for his pocket money by his father. Which of the choices should he take in order to get the maximum money?
(a) \(\frac{2}{3}\) of Rs 150
(b) \(\frac{3}{5}\) of Rs 150
(c) \(\frac{4}{5}\) of Rs 150
(d) \(\frac{1}{5}\) of Rs 150
Solution:
(c) \(\frac{4}{5}\) of Rs 150

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 9th Maths Guide Chapter 3 Algebra Ex 3.8

Students can download Maths Chapter 3 Algebra Ex 3.8 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 3 Algebra Ex 3.8

Synthetic division calculator. This calculator divides two polynomials using synthetic division.

Question 1.
Factorise each of the following polynomials using synthetic division:
(i) x3 – 3x² – 10x + 24
Solution:
p(x) – x3 – 3x² – 10x + 24
p(1) = 13 – 3(1)² – 10(1) + 24
= 1 – 3 – 10 + 24
= 25 – 13
≠ 0
x – 1 is not a factor

p(-1) = (-1)3 – 3(-1)² – 10(-1) + 24
= – 1 – 3(1) + 10 + 24
= -1 – 3 + 10 + 24
= 34 – 4
= 30
≠ 0
x + 1 is not a factor

p(2) = 23 – 3(2)² – 10(2) + 24
= 8 – 3(4) – 20 + 24
= 8 – 12 – 20 + 24
= 32 – 32
= 0
∴ x – 2 is a factor
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 1
x² – x – 12 = x² – 4x + 3x – 12
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 2
= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
∴ The factors of x3 – 3x² – 10x + 24 = (x – 2) (x – 4) (x + 3)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(ii) 2x3 – 3x² – 3x + 2
Solution:
p(x) = 2x3 – 3x² – 3x + 2
P(1) = 2(1)3 – 3(1)² – 3(1) + 2
= 2 – 3 – 3 + 2
= 2 – 6
= -4
≠ 0
x – 1 is not a factor

P(-1) = 2(-1)3 – 3(-1)² – 3(-1) + 2
= -2 – 3 + 3 + 2
= 5 – 5
= 0
∴ x + 1 is a factor
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 3
2x² – 5x + 2 = 2x² – 4x – x + 2
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 4
= 2x(x – 2) – 1 (x – 2)
= (x – 2) (2x – 1)
∴ The factors of 2x3 – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(iii) – 7x + 3 + 4x3
Solution:
p(x) = – 7x + 3 + 4x3
= 4x3 – 7x + 3
P(1) = 4(1)3 – 7(1) + 3
4 – 7 + 3
= 7 – 7
= 0
∴ x – 1 is a factor
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 5
4x² + 4x – 3 = 4x² + 6x – 2x – 3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 6
= 2x(2x + 3) – 1 (2x + 3)
= (2x + 3) (2x – 1)
∴ The factors of – 7x + 3 + 4x3 = (x – 1) (2x + 3) (2x – 1)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(iv) x3 + x² – 14x – 24
Solution:
p(x) = x3 + x² – 14x – 24
p(1) = (1)3 + (1)2 – 14 (1) – 24
= 1 + 1 – 14 – 24
= -36
≠ 0
x + 1 is not a factor.

p(-1) = (-1)3 + (-1)² – 14(-1) – 24
= -1 + 1 + 14 – 24
= 15 – 25
≠ 0
x – 1 is not a factor.

p(2) = (-2)3 + (-2)2 – 14 (-2) – 24
= -8 + 4 + 28 – 24
= 32 – 32
= 0
∴ x + 2 is a factor
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 7
x² – x – 12 = x² – 4x + 3x – 12
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 8
= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
This (x + 2) (x + 3) (x – 4) are the factors.
x3 + x2 – 14x – 24 = (x + 2) (x + 3) (x – 4)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(v) x3 – 7x + 6
Solution:
p(x) = x3 – 7x + 6
P( 1) = 13 – 7(1) + 6
= 1 – 7 + 6
= 7 – 7
= 0
∴ x – 1 is a factor
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 9
x² + x – 6 = x² + 3x – 2x – 6
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 10
= x(x + 3) – 2 (x + 3)
= (x + 3) (x – 2)
This (x – 1) (x – 2) (x + 3) are factors.
∴ x3 – 7x + 6 = (x – 1) (x – 2) (x + 3)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

(vi) x3 – 10x² – x + 10
p(x) = x3 – 10x2 – x + 10
= 1 – 10 – 1 + 10
= 11 – 11
= 0
∴ x – 1 is a factor
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 11
x2 – 9x – 10 = x2 – 10x + x – 10
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8 12
= x(x – 10) + 1 (x – 10)
= (x – 10) (x + 1)
This (x – 1) (x + 1) (x – 10) are the factors.
∴ x3 – 10x2 – x + 10 = (x – 1) (x – 10) (x + 1)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.8

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