Laxmi

Students can Download Maths Chapter 4 Information Processing Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Intext Questions

Exercise 4.1
Activity 1. (Text book Page no. 83)

Question 1.
Draw a Golden Spiral using Fibonacci squares.

Answer:

Activity – 2 (Text book Page no. 85)

Question 1.
Using the given Table I, find the pattern, Answer the following questions and colour the values in the given Table II. One is done for you.
Table I

1. Where are the even Fibonacci Numbers?
Colour both the term n and where F(n) is even in
Do you find any pattern?
Every Third Fibonacci number is a multiple of 2(even).
i.e. a multiple of F(3) or 2 = F(3).

2. Where there are Fibonacci numbers which are multiple of 3?
Colour both the term n and where F(n) is multiple of 3 in red.
Write down the pattern you find
Every 4th Fibonacci number is a multiple of 3.
i.e. a multiple of F(4) or 3 = F(4).

3. What about the multiple of 5?
Colour both the term n and where F(n) is multiple of 5 in 1
Write down the pattern you find.
Every 5th Fibonacci number is a multiple of F(5) or 5 = F(5)
i.e. a multiple of F(4) or 3 = F(4)

4. What about the multiple of 8?
Colour both the term n where F(n) is multiple of 8 in green.
Write down the pattern you find.
Every 6th Fibonacci number i.e. a multiple of F(6) or 8 = F(6)

Table II

Solution:

R – Red
Y – Yellow
B – Blue
G – Green

Exercise 4.3
Try This (Text book Page no. 94)

Question 1.
Convert cipher code written in the school panel board as given in figure below.

Solution:
Cipher code given.
EAPLKEAPUSG UG NWP AJWYP NYEJKXG KOYAPUWNG, SWEFYPAPUWNG WX AVCWXUPLEG; UP UG AJWYP YNBKXGPANBUNC – QUVVAE FAYV PLYXGPWN
Mathematics is not about numbers, equations, computations or algorithms, it is about understanding. – William Paul Thurston

Activity 3. (Text book Page No. 95)

Code 1: Pigpen

Question 1.
Fill in the blank boxes and decode

The Pigpen code looks like meaningless writing, but it is quite easy to catch on to. Each letter is represented by the part of the “Pigpen” that surrounds it.

The first code uses the following key. To complete the code, you need to work out how to use the key to decode the message.

Solution:

To decode

Try These (Text book Page no. 97)

Question 1.
Use Pigpen Cipher code and write the code for your name ………. chapter names

(i) LIFE MATHEMATICS
(ii) ALGEBRA
(iii) GEOMETRY
(iv) INFORMATION PROCESSING

Solution:
For Pigpen, we should first construct the Criss Cross pattern & then fill with alphabets.
Step 1:

Step 2:
Fill the above with all the alphabets from A to Z.

Step 3:
So, the secret code is got by the outline of each letter in the above patterns. So for

Similarly for all the others. Therefore Pigpen code is

Hence if your name is RAM, the corresponding pattern is

Let us now write the given words with Pigpen code.

Question 2.
Decode the following Shifting and Substituting secret codes given below. Which one is a easier for you?

Solution:
Decode using shifting method. Let us shift & see by trial & error method.
Let us substitute ‘b’ for ‘a’, V for ‘b’ & so on.

Message: M N S G H M F H R H L O N R R H A K D
Decoded text: NOTHING IS IMPOSSIBLE
Decode using substituting method:

We find that Pigpen code has been used, by substituting for the symbols from the table that we constructed earlier, we go the message.
NOTHING IS IMPOSSIBLE

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.5

Question 1.
Write the following decimal numbers in the place value table.
(i) 247.36
(ii) 132.105
Solution:
(i) 247.36

(ii) 132.105

Question 2.
Write each of the following as decimal number.
(i) 300 + 5 + \(\frac { 7 }{ 10 } \) + \(\frac { 9 }{ 100 } \) + \(\frac { 2 }{ 100 } \)
(ii) 1000 + 400 + 30 + 2 + \(\frac { 6 }{ 10 } \) + \(\frac { 7 }{ 100 } \)
Solution:
(i) 300 + 5 + \(\frac { 7 }{ 10 } \) + \(\frac { 9 }{ 100 } \) + \(\frac { 2 }{ 100 } \) = 305.792
(ii) 1000 + 400 + 30 + 2 + \(\frac { 6 }{ 10 } \) + \(\frac { 7 }{ 100 } \) = 1432.67

Question 3.
Which is greater?
(i) 0.888 (or) 0.28
(ii) 23.914 (or) 23.915
Solution:
(i) 0.888 (or) 0.28
The whole number parts is equal for both the numbers.
Comparing the digits in the tenths place we get, 8 > 2.
0.888 > 0.28 ∴ 0.888 is greater.

(ii) 23.914 or 23.915
The whole number part is equal in both the numbers.
Also the tenth place and hundredths place are also equal.
∴ Comparing the thousandths place, we get 5 > 4.
23.915 > 23.914 ∴ 23.915 is greater.

Question 4.
In a 25 m swimming competition, the time taken by 5 swimmers A, B, C, D and E are 15.7 seconds, 15.68 seconds, 15.6 seconds, 15.74 seconds and 15.67 seconds respectively. Identify the winner.
Solution:
The winner is one who took less time for swimming 25 m.
Comparing the time taken by A, B, C, D, E the whole number part is equal for all participants.
Comparing digit in tenths place we get 6 < 7.
∴ Comparing 15.68, 15.6, 15.67, that is comparing the digits in hundredths place we get 15.60 < 15.67 < 15.68
One who took 15.6 seconds is the winner. ∴ C is the winner.

Question 5.
Convert the following decimal numbers into fractions
(i) 23.4
(ii) 46.301
Solution:
(i) 23.4 = \(\frac { 234 }{ 10 } \) = \(\frac{234 \div 2}{10 \div 2}\) = \(\frac { 117 }{ 5 } \)
(ii) 46.301 = \(\frac { 46301 }{ 1000 } \)

Question 6.
Express the following in kilometres using decimals,
(i) 256 m
(ii) 4567 m
Solution:
1 m = \(\frac { 1 }{ 1000 } \) km = 0.001 Km
(i) 256 m = \(\frac { 256 }{ 1000 } \) km = 0.256 km
(ii) 4567 m = \(\frac { 4567 }{ 1000 } \) km = 4.567 km

Question 7.
There are 26 boys and 24 girls in a class. Express the fractions of boys and girls as decimal numbers.
Solution:
Boys = 26; Girls = 24; Total = 50
Fraction of boys = \(\frac { 26 }{ 50 } \) = \(\frac{26 \times 2}{50 \times 2}\) = \(\frac { 52 }{ 100 } \) = 0.52
Fraction of girls = \(\frac { 24 }{ 50 } \) = \(\frac{24 \times 2}{50 \times 2}\) = \(\frac { 48 }{ 100 } \) = 0.48

Challenge Problems

Question 8.
Write the following amount using decimals.
(i) 809 rupees 99 paise
(ii) 147 rupees 70 paise
Solution:
100 paise = 1 rupee; 1 paise = \(\frac { 1 }{ 100 } \) rupee

(i) 809 rupees 99 paise = 809 rupees + \(\frac { 99 }{ 100 } \) rupees
= 809 + 0.99 rupees = ₹ 809.99

(ii) 147 rupees 70 paise = 147 rupees + \(\frac { 70 }{ 100 } \) rupees
= 147 rupees + 0.70 rupees = ₹ 147.70

Question 9.
Express the following in metres using decimals.
(i) 1328 cm
(ii) 419 cm
Solution:
100 cm = 1 m; 1 cm = \(\frac { 1 }{ 100 } \) m
(i) 1328 cm = \(\frac { 1328 }{ 100 } \) m = 13.28 m
(ii) 419 cm = \(\frac { 419 }{ 100 } \) m = 4.19 m

Question 10.
Express the following using decimal notation.
(i) 8 m 30 cm in metres
(ii) 24 km 200 m in kilometres
Solution:
(i) 8 m 30 cm in metres
8 m + \(\frac { 30 }{ 100 } \) m = 8 m + 0.30 m = 8.30 m

(ii) 24 km 200 m in kilometres
24 km + \(\frac { 200 }{ 1000 } \) km = 24 km + 0.200 km = 24.200 km

Question 11.
Write the following fractions as decimal numbers.
(i) \(\frac { 23 }{ 10000 } \)
(ii) \(\frac { 421 }{ 100 } \)
(iii) \(\frac { 37 }{ 10 } \)
Solution:
(i) \(\frac { 23 }{ 10000 } \) = 0.0023
(ii) \(\frac { 421 }{ 100 } \) = 4.21
(iii) \(\frac { 37 }{ 10 } \) = 3.7

Question 12.
Convert the following decimals into fractions and reduce them to the lowest form,
(i) 2.125
(ii) 0.0005
Solution:
(i) 2.125 = \(\frac { 2125 }{ 1000 } \) = \(\frac{2125 \div 25}{1000 \div 25}\) = \(\frac { 85 }{ 40 } \) = \(\frac{85 \div 5}{40 \div 5}\) = \(\frac { 17 }{ 8 } \)

(ii) 0.0005 = \(\frac { 5 }{ 1000 } \) = \(\frac{5 \div 5}{10000 \div 5}\) = \(\frac { 1 }{ 2000 } \)

Question 13.
Represent the decimal numbers 0.07 and 0.7 on a number line.
Solution:

0.07 lies between 0.0 and 0.1
The unit space between 0 and 0.1 is divided into 10 equal parts and 7th part is taken. Also 0.7 lies between 0 and 1.
The unit space between 0 and 1 is divided into 10 equal parts, and the 7th part is taken.

Question 14.
Write the following decimal numbers in words.
(i) 4.9
(ii) 220.0
(iii) 0.7
(iv) 86.3
Solution:
(i) 4.9 = Four and nine tenths
(ii) 220.0 = Two hundred and twenty
(iii) 0.7 = Seven tenths
(iv) 86.3 = Eighty six and three tenths.

Question 15.
Between which two whole numbers the given numbers lie?
(i) 0.2
(ii) 3.4
(iii) 3.9
(iv) 2.7
(v) 1.7
(vi) 1.3
Solution:
(i) 0.2 lies between 0 and 1.
(ii) 3.4 lies between 3 and 4.
(iii) 3.9 lies between 3 and 4.
(iv) 2.7 lies between 2 and 3.
(v) 1.7 lies between 1 and 2.
(vi) 1.3 lies between 1 and 2.

Question 16.
By how much is \(\frac { 9 }{ 10 } \) km less than 1 km. Express the same in decimal form.
Solution:
Given measures are 1 km and \(\frac { 9 }{ 10 } \) km. i.e., 1 km and 0.9 km.
Difference = 1.0 – 0.9 = 0.1 km.

Students can Download Maths Chapter 4 Information Processing Ex 4.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.1

Question 1.
Choose the correct answer:

Question (i)
What is the eleventh Fibonacci number?
(a) 55
(b) 77
(c) 89
(d) 144
Answer:
(c) 89
Hint:

∴ 11th Fibonacci number is 89

Question (ii)
If F(n) is a Fibonacci number and n = 8, which of the following is true?
(a) F(8) = F(9) + F(10)
(b) F(8) = F(7) + F(6)
(c) F(8) = F(10) x F(9)
(d) F(8) = F(7) – F(6)
Answer:
(b) F(8) = F(7) + F(6)
Hint:
Given F(n) is a Fibonacci number & n = 8
F(8) = F(7) + F (6) as any term in Fibonacci series is the sum of preceding 2 terms

Question (iii)
Every 3^rd number of the Fibonacci sequence is a multiple of ………..
(a) 2
(b) 3
(c) 5
(d) 8
Answer:
(a) 2
Hint:
Every 3rd number in Fibonacci sequence is a multiple,of 2

Question (iv)
Every ……….. number of the Fibonacci sequence is a multiple of 8
(a) 2^nd
(b) 4^th
(c) 6^th
(d) 8^th
Answer:
(c) 6^th

Question (v)
The difference between the 18th and 17th Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
Answer:
(d) 987
Hint:
F(18) = F(17) + F (16)
F(18) – F(17) = F(16) = F(15) + F(14) = 610-377 = 987

Question 2.
In the given graph sheet draw and colour how the Fibonacci number pattern makes a spiral in the Golden Rectangle.

Steps:
As per the instruction given, draw squares of side equal to each term of Fibonacci number – starting with 1 x 1 square, 2 x 2 square, 3 x 3 square and so on.
For spiral:
Starting from 1st square, join opposite diagonal points in a curve in all the squares to form a spiral.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.4

Question I.
Construct the following parallelograms with the given measurements and find their area.

Question 1.
ARTS, AR = 6 cm, RT = 5 cm and ∠ART = 70°.
Solution:

Given:
In the Parallelogram ARTS,
AR = 6 cm
RT = 5 cm and ∠ART = 70°

Construction:
Steps:

1. Draw a line segment AR = 6 cm.
2. Make an angle ∠ART = 70° at R on AR
3. With R as centre, draw an arc of radius 5 cm cutting RX at T
4. Draw a line TY parallel to AR through T.
5. With T as centre, draw an arc of radius 6 cm cutting TY at S. Join AS
6. ARTS is the required parallelogram.

Calculation of area:
Area of the parallelogram ARTS = b x h sq. units
= 6 x 4.7 = 28.2 sq. cm

Question 2.
BANK, BA = 7 cm, BK = 5.6 cm and ∠KBA = 85°.
Solution:

Given:
In the parallelogram BANK, BA =7 cm, BK = 5.6 cm, and ∠KBA = 85°

Construction:
Steps:

1. Draw a line segment KB = 5.6 cm.
2. Make an angle ∠KBA = 85° at B
3. Draw an arc of radius 7 cm with B as centre on BX.
4. Draw a line AY parallel to KB.
5. With A as centre, draw an arc of radius 5.6 cm cutting AY at N. Join KN
6. BANK is the required parallelogram.

Calculation of area:
Area of the Parallelogram BANK = b x h sq. units
= 5.6 x 7 = 39.2 sq. cm

Question 3.
CAMP, CA = 6 cm, AP = 8 cm and CP = 5.5 cm.
Solution:

Given:
In the parallelogram CAMP,
CA = 6 cm
AP = 8 cm, and CP = 5.5cm

Construction:
Steps:

1. Draw a line segment CA = 6 cm.
2. With C as centre, draw an arc of length 5.5 cm
3. With A as centre, draw an arc of length 8 cm
4. Mark the intersecting point of these two arcs as P
5. Draw a line PX parallel to CA
6. With P as centre draw an arc of radius 6 cm cutting PX at M. Join AM
7. CAMP is the required parallelogram.

Calculation of area:
Area of the Parallelogram CAMP = b x h sq. units
= 6 x 5.5 = 33 sq. cm

Question 4.
DRUM, DR = 7 cm, RU = 5.5 cm and DU = 8 cm.
Solution:

Given:
In the parallelogram DRUM,
DR = 7 cm,
RU = 5.5 cm, and DU = 8 cm

Construction:
Steps:

1. Draw a line segment DR = 7 cm.
2. With D and R as centres, draw arcs of radii 8 cm and 5.5 cm.
3. Mark the intersecting point of these arcs as U. Join DU and RU
4. Draw a line UX parallel to DR through U.
5. With U as centre draw an arc of radius 7 cm cutting UX at M. Join DM
6. DRUM is the required parallelogram.

Calculation of area:
Area of the Parallelogram DRUM = b x h sq. units
= 7 x 5.4 = 37.8 sq. cm

Question 5.
EARN, ER = 10 cm, AN = 7 cm and ∠EOA 110° where \(\overline { ER } \) and \(\overline { AN } \) intersect at O.
Solution:

Given:
In the parallelogram EARN,
ER= 10 cm
AN = 7 cm, and ∠EOA = 110°

Construction:
Steps:

1. Draw a line segment PX. Mark a {Joint O on PX
2. Make an angle ∠EOA = 110° on PX at O
3. Draw arcs of radius 3.5 cm with O as centre on either side of PX. Cutting YZ on A and N
4. With A as centre, draw an arc of radius 10 cm, cutting PX at E. Join AE
5. Draw a line parallel to AE at N cutting PX at R. Join EN and AR
6. EARN is the required parallelogram

Calculation of area:
Area of the Parallelogram EARN = b x h sq. units
= 10 x 5.5 = 55 sq. cm

Question 6.
FAIR, FI = 8 cm, AR = 6 cm and ∠IOR = 80° where \(\overline { FI } \) and \(\overline { AR } \) intersect at O.
Solution:

Given:
In the parallelogram FAIR,
FI = 8cm
AR = 6 cm and ∠IOR = 80°, where \(\overline { FI } \) and \(\overline { AR } \) intersect at O

Construction:
Steps:

1. Draw a line segment \(\overline { FI } \) = 8 cm
2. Mark O the midpoint of \(\overline { FI } \)
3. Draw a line \(\overline { XY } \) through O which makes ∠IOX = 80°
4. With O as centre and 3 cm as radius draw two arcs on XY on either sides of \(\overline { FI } \). Let the arcs cut \(\overline { OX } \) at R and \(\overline { OY } \) at A
5. Join \(\overline { IR } \), \(\overline { FR } \), \(\overline { FA } \) and \(\overline { AI } \)
6. FAIR is the required parallelogram

Calculation of area:
Area of the Parallelogram FAIR = b x h sq. units
= 5.4 x 4.3 = 23.22 sq. cm

Question 7.
GAIN, GA = 7.5 cm, GI = 9 cm and ∠GAI – 100°
Solution:

Given:
In the parallelogram GAIN,
GA = 7.5 cm
GI = 9 cm and ∠GAI= 100°

Construction:
Steps:

1. Draw a line segment GA = 7.5 cm.
2. Make an angle GAI = 100° at A.
3. With G as centre, draw an arc of radius 9 cm cutting AX at I. Join GI.
4. Draw a line IY parallel to GA through I.
5. With I as centre, draw an arc of radius T.5 cm on IY cutting at N. Join GN
6. GAIN is the required parallelogram.

Calculation of area:
Area of Parallelogram GAIN = b x h sq.units
= 7.5 x 3.9 = 29.25 sq. cm

Question 8.
HERB, HE = 6 cm, ∠EHB = 60° and EB = 7 cm.
Solution:

Given:
In the parallelogram HERB,
HE = 6 cm, ∠EHB = 60° and
EB = 7 cm

Construction:
Steps:

1. Draw a line segment HE = 6 cm.
2. Make an angle ∠BHE = 60° at H.
3. With E as centre, draw an arc of radius 7 cm cutting HX at B.
4. Draw a line BY parallel to HE through B.
5. Draw an arc of radius 6 cm with B as centre, cutting BY at R. Join ER.
6. HERB is the required parallelogram.

Calculation of area:
Area of the Parallelogram HERB = b x h sq. units
= 6 x 6.7 = 40.2 sq. cm

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Additional Questions

Additional Questions and Answers

Exercise 5.1

Question 1.
Find the relationship between x and y if

Solution:
The relationship between x and y is y = -5x

Question 2.
Find the relationship between x and y if

Solution:
The relationship between x andy is y = 3x + 2

Exercise 5.2

Question 1.
Find the sum of the elements of Pascal’s triangle and find the relationship between the numbers obtained.
Solution:
Pascal’s triangle with sum of elements is given by

The numbers formed are 1, 2, 4, 8, 16, 32, 64, …………..
They can be written as 2⁰, 2¹, 2², 2³, 2⁴, 2⁵, 2⁶, …………..
So they are the powers of base 2
The relationship is given by 2^x-1 if x represents the row.

Exercise 5.3

Question 1.
If the following numbers are taken from Pascal’s triangle find the missing numbers.

1. 9, 1, _____ = 45, 1, 11
2. 1, 6, _____ = 1, 4, 15
3. 21, 8, 1 = _____, 6, 28

Solutions:

1. 55
2. 10
3. 1

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Tamilnadu Samacheer Kalvi 9th English Solutions Poem Chapter 7 The Stick-together families

The Stick-together families Warm Up:

At the heart of life lie the relationships you have with other people: with family, classmates and friends close-by and far away. All relationships are based on some commonly accepted values like respect, honesty, consideration and commitment. Think about all the important relationships in your life and complete the table given below.

The Stick-together families Textual Questions

A. Based on your understanding of the poem, answer the questions in a sentence or two.

1. “The gladdest people living are the wholesome folks who make A circle at the fireside that no power but death can break. ”

Question (a).
Who are the gladdest people living?
Answer:
The gladdest people are those who live together as wholesome folks.

Question (b).
Where do they gather?
Answer:
They gather at the fireside.

Question (c).
What can break their unity?
Answer:
Only death can break their unity.

2. “And the finest of conventions ever held beneath the sun Are the little family gatherings when the busy day is done. ”

Question (a).
When do they have their family gatherings?
Answer:
They have their family gatherings at the end of the busy day.

Question (b).
Where do they have their family conventions?
Answer:
The family conventions are held beneath the sun.

Question (c).
What does the poet mean by ‘finest conventions’?
Answer:
The finest conventions means a family get together.

3. “There are rich folk, there are poor folk, who imagine they are wise,
And they’re very quick to shatter all the little family ties. ”

Question (a).
What do the rich and poor folk imagine themselves to be?
Answer:
The rich and the poor folks imagine themselves to be wise.

Question (b).
What do they do to their families?
Answer:
They are quick to shatter their little family ties.

Question (c).
Whom does ‘they’ refer to?
Answer:
They refers to the rich and poor people who do not consider the little family ties as valuable and shatter them.

4. “There are some who seem to fancy that for gladness they must roam,
That for smiles that are the brightest they must wander far from home”

Question (a).
Why do they roam?
Answer:
They roam to attain gladness.

Question (b).
According to them, when do they get bright smiles?
Answer:
They get bright smiles when they wander far from home.

5. “But the gladdest sort of people, when the busy day is done,
Are the brothers and the sisters who together share their fun.”

Question (a).
Who are the gladdest people?
Answer:
The gladdest people are those brothers and sisters who share their fun.

Question (b).
When do they share their fun?
Answer:
They share their fun when the busy day is done.

Question (c).
What does ‘who’ refer to?
Answer:
Who refers to the brothers and sisters.

6. “It’s the stick-together family that wins the joys of earth,

That hears the sweetest music and that finds the finest mirth;”

Question (a).
Who wins the joys of the earth?
Answer:
The stick-together families wins the joys of earth.

Question (b).
How do they find their joy?
Answer:
They find their joys by being together and hearing the sweetest music

Question (c).
What does the poet mean by ‘stick-together family’?
Answer:
The stick-together family means those families who spend time together (Joint families) and share their fun and sorrows and can be only separated after death.

Additional Questions

1. “ The stick-together families are happier by far
Than the brothers and the sisters who take separate highways are.
The gladdest people living are the wholesome folks who make
A circle at the fireside that no power but death can break.
And the finest of conventions ever held beneath the sun
Are the little family gatherings when the busyday is done. ”

Question (a).
Give the rhyme scheme of the above lines.
Answer:
‘aabbcc’ is the rhyme scheme of the above lines.

Question (b).
Pick out the rhyming words.
Answer:
‘far and are’, ‘make and break’ and ‘sun and done’ are the rhyming words.

Question (c).
What does the poet mean by ‘day is done’?
Answer:
The poet means that the day has come to an end.

Question (d).
What can break the circle at the fireside?
Answer:
Death can break a circle at the fireside.

Question (e).
What do the stick-together families do when the busy day is done?
Answer:
The stick-together families have little family gatherings when the busy day is done.

Question (f).
Mention the figure of speech in the last line.
Answer:
The figure of speech in the last line is alliteration, (day is done)

2. “There are rich folk, there are poor folk, who imagine they are wise,
And they’re very quick to shatter all the little family ties.”

Question (a).
What does the poet mean by the term, ‘family ties’?
Answer:
The poet by the term ‘family ties’ means the family bonds.

Question (b).
Pick out the rhyming words.
Answer:
The rhyming words are wise and ties.

Question (c).
Who shatters the family ties?
Answer:
The rich and the poor folk who think they are intelligent are quick to shatter the family ties.

3. “Each goes searching after pleasure in his own selected way,
Each with strangers likes to wander, and with strangers likes to play.”

Question (a).
Identify the figure of speech in the first line.
Answer:
Alliteration is the figure of speech in the first line.

Question (b).
Write the alliterated words in the first line.
Answer:
The alliterated words are searching and selected in the first line.

4. “There are some who seem to fancy that for gladness they must roam,
That for smiles that are the brightest they must wander far from home.
That the strange friend is the true friend, and they travel far astray
They waste their lives in striving for a joy that’s far away,
But the gladdest sort of people, when the busy day is done,
Are the brothers and the sisters who together share their fun. ”

Question (a).
Mention the rhyme scheme?
Answer:
‘aabbcc’ is the rhyme scheme.

Question (b).
Give the rhyming words for away and home.
Answer:
The rhyming words for away and home are astray and roam respectively.

Question (c).
What is meant by ‘astray’?
Answer:
Being lost in this world due to bad ways is the meaning of astray.

Question (d).
How do they waste their lives?
Answer:
They waste their lives in striving for a joy that cannot be easily got.

5. “And, O weary, wandering brother, if contentment you would win,
Come you back unto the fireside and be comrade with your kin.”

Question (a).
What are the alliterated words in the first line.
Answer:
The alliterated words are weary, wandering.

Question (b).
Who is a comrade?
Answer:
A comrade is a friend.

Question (c).
What should the weary wandering brother do?
Answer:
The weary wandering brother should come back to the fireside and be friends with the kith and kin.

Question (d).
What do you mean by ‘kin’?
Answer:
Kin means relatives.

B. Based on the understanding of the poem, fill in the blanks using the words and phrases given below to make a meaningful summary of the poem.

The poet brings out the difference in the attitudes of children living in joint family and nuclear family. The (1) …………………., are the happiest of all. Where as the (2) …………………… of nuclear families take (3) ………………………. The gladdest people are the children from (4) …………………….. who circle near the fireside. No power other than death can break them. The (5) …………………. imagine themselves to be wise and in the process they (6) ……………………. ties. Each of them goes searching for pleasure in their own selected way. They harvest only (7) …………. and find empty joy. But the wisest among them are the children of the stick-together families. When the busy day is done, they together (8) ……………… The stick-together family wins (9) ………………. The old house shelters all the (10) …………………… The poet invites wandering brothers to come and join the stick-together families in their fireside and have fun.
Answers

1. stick-together families
2. brothers and sisters
3. separate ways
4. joint family
5. rich and the poor folk
6. shatter their family
7. bitterness
8. share their fun
9. the joy of earth
10. charm of life

C. Answer the following questions in about 80-100 words.

Question 1.
The stick-together families are the happiest of all. Explain.
Answer:
The stick-together families are the happiest of all. There is so much fun living together in joint families. There is a lot of excitement in joint families. The gladdest people living are the good folks who create a circle for themselves. They share their joys and sorrows. The finest of all large formal gatherings is the little family gatherings. The wisest children are in the stick-together families. Some travel far away to seek joy but the happiest kind of people, are the ones who share their fun in stick-together families. It is sure that the stick-together family wins the joys of the earth.

Question 2.
Bring out the difference between the children of the joint family and nuclear family.
Answer:
The joint families are definitely happier than the members in the nuclear family who take separate ways. The joint family meets at the end of the day while the nuclear family meets occasionally. Members of the nuclear family go searching after pleasure in their own selected way but the joint family loves to travel together. People in nuclear families waste their lives in searching for a joy that’s far away. But the happiest kind of people in joint families share their fun together. The children in the stick-together families wins the joys of the earth whereas children in joint families travel far away seeking pleasure in vain.

Additional Questions:

Question 1.
What do those who think wise of themselves do?
Answer:
The rich and poor folk think wise of themselves and quickly destroy all the little family bonds. They do not like to mingle with one another and prefer to travel and seek pleasure. Every person goes searching after pleasure in his/her own selected way with strangers who likes to wander and play. On the contrary, they reap unpleasantness and find an empty joy. However, the children who are the wisest are surely those who live in the stick-together kind in the joint families and not those who are in nuclear families who think they are wise

Question 2.
Why do some people roam and what happens as they journey along?
Answer:
There are some who seem to like the thought of roaming around to seek happiness. They strongly feel that the brightest smile can be seen only when they travel far away from home. Such people also think that a strange friend is the best friend who will be true. However, when they travel far away and get lost in their path, they do waste their lives in searching for a joy that’s far away. Therefore, they really lack in happiness and lose the joy searching for gladness.

Appreciate The Poem

D. Answer the following

Question 1.
There are rich folk, there are poor folk, who imagine they are wise,…
Pick out the words in alliteration.
Answer:
‘There’ and ‘they’ are the words that alliterate.

Question 2.
Mention the rhyme scheme of the poem.
Answer:
The rhyme scheme of the poem is ‘aabbcc’ for all the stanzas.

Listening Activity

E. Listen to the passage and fill in the blanks with appropriate answer.

(For listening to the passage refer to our website www.fullcircleeducation.in) Family is where we all belong to and from where our identity comes from. A person is valued based on his family and upbringing. Family is a bond, a long lasting relationship that holds a bond with each other. There are many values that one has to learn to get the family bonding in the right manner. Bonding does not happen overnight. It forms with every second, every minute that you spend with your loved ones. The understanding, the acceptance, the belonging and the security all enclosed together is how a family bond is formed.

A close family bond is like a safe harbour, where we feel secure and where we trust that we have someone always there to whom we could turn to when we need them the most. It is through a family that we learn the values of love, trust, hope, belief, cultures, morals, traditions and every little matter that concerns to us. A strong foundation for any individual comes from being with a supportive family.

Family is one among the greatest gifts that we get from God. To have parents, who support us, teach us values in life, and gives us a strong foundation in character, teach us the importance of love and being loved, trust to be there for one another and many other morals that could be obtained only from a family. A gift not only with lovable parents, but siblings who care and love us beyond themselves. We cannot buy or demand all these things in life, as we are being given to understand their importance.

To be part of a happy family, one should always thank God for the blessing we have in lives, as having a family who cares and loves us is the greatest blessing that any person could get in life.

1. A person is valued based on his …………………. .
2. ………………….. does not happen overnight.
3. A close family bond is like a ……………………… .
4. A strong foundation for any individual comes from being with a …………………… .
5. A gift not only with …………….. but …………………….. who care and love us beyond themselves.

Answers

1. family and upbringing
2. Bonding
3. safe harbour
4. supportive family
5. lovable parents, siblings

Speaking:

F. “The building actually rests on the well laid out foundation and hence is strong and still.” How can this be related to a family? Discuss with your partner and share your views in the class.

The building actually rests on the well laid out foundation. Therefore it is strong and still. If the foundation is strong the building is strong. Likewise, if we lay a strong foundation for the children, then they will grow strong and still. For such a foundation to be strong, as parents and teachers, we must nurture them with virtues like prayer, faith, courage and other values.

You need a really solid foundation of friends and family to keep you where you need ‘ to be. If the mason is able and his materials are good, the house built will be strong and still. Just like the firm foundation with good material, the parents and teachers too should build a child on a strong foundation with good values like discipline, responsible citizenship, empathy, sympathy, sacrifice and so on.

It is surely the family which provides a child with roots, much- needed structure, and unconditional love. Families also provide their children with a happy home – a place where a child is always safe and welcome.

Writing:

G. Write a four-line poem with rhyming words describing your family.

I live in a lovely family off our Ladies two with Gentlemen two Missing days with Grandpa too Hope to hear the same from you.

The Stick-together families About The Poet:

Edgar Albert Guest born on 20th August 1881, was an English-born American poet, popular in the first half of the 20th century, commonly known as the People’s Poet due to his inspirational and optimistic view of everyday life. Elis first poem appeared on 11th December 1898. For 40 years, Guest was widely read throughout North America. Guest has penned around 11,000 poems that has appeared in 300 newspapers and collected in more than 20 books. Even today it occasionally appears in periodicals such as Reader’s Digest.

The Stick-together families Summary:

The poet envisions the joint families who are much happier than the brothers and sisters who take their separate ways. The gladdest people are those who circle around a fireside that no power but death can break. The finest conventions beneath the sun is the family gatherings after a busy day. Rich or poor folks, who think they are wise, break their family ties and wander away searching for joy in strangers and material things. But in fact, the stick together families alone enjoy the true joys of the earth, enjoy the sweetest music and finds mirth. An old home shelters the charm that life can give, and a happiest spot to live. Hence the poet asks people who go their separate ways to come together and comrade with their kin.

The Stick-together families Glossary:
Textual:
astray – away from the correct path or correct way of doing something
comrade – a friend
conventions – a large formal meeting of people who have a similar interest
mirth – laughter, humour or happiness
shatter – to break suddenly into very small pieces

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Intext Questions

Activity (Text book Page No. 91)

Question 1.
Observe the pattern given below. Continue the pattern for three more steps.

Let, ‘x’ be the number of steps and ‘y’ be the number of match sticks. Tabulate the values of ‘x’ and ‘y’ and verify the relationship y = 7x + 5.
Solution:
Three more patterns are

From the table y = 7x + 5 is verified.

Try These (Text book Page No. 92)

Question 1.
In the given figure, let Y denote the number of steps and y denote its area. Find the relationship between x and y by tabulation.

Solution:
Let x denote the number of steps and y denote the area.
In the first shape let x = 1 and the area be 1 cm²
when x = 2 : Area = 22 = 4 cm²
whenx = 3 : Area = 32 = 9 cm² and so on.
Tabulating the values of x and y

From the table:
x = 1 ⇒ y = 1²
x = 2 ⇒ y = 2²
x = 4 ⇒ y = 4²
x = 5 ⇒ y = 5²
Hence the relationship between x and y is y = x²

Question 2.
In the figure, let x denotes the number of steps and y denotes the number of matchsticks used. Find the relationship between Y and y by tabulation.

Solution:
Let x denote the number of steps and y denote the number of matchsticks used.
In step 1, x = 1 ⇒ y = number of mathsticks used is 1
In step 2, x = 2 ⇒ y = number of mathsticks used is 4
In step 3, x = 3 ⇒ y = number of mathsticks used is 7 and so on.
The values of v andy are tabulated as

x = 1 ⇒ y = 1 = 3(1) – 2
x = 2 ⇒ y = 4 = 3(2) – 2
x = 3 ⇒ y = 7 = 3(3) – 2
x = 4 ⇒ y = 10 = 3(4) – 2
From the table y = 3x – 2

Question 3.
Observe the table given below.

Find the relationship between x and y. What will be the value of y, when x = 8.
Solution:
Soil When x = – 2 y = 2 (-2) = -4
When x = -1 y = 2 (-1) = -2
When x = 0 y = 2 (0) = 0
When x =1 y = 2(1) = 2
When x = 2 y = 2 (2) = 4
When x = 8 y = 2 (8) = 16.
Also y = 2x is the relation between x and y.

Activity (Text book Page No. 93 & 94)

Question 1.
Complete the following Pascal’s Triangle by observing the number pattern.

Solution:
Pascal’s triangle is given by

Question 2.
Observe the above completed and r find the sequence that you see in it and complete them.

(i) 1,2, 3, 4, 5, 6, 7.
(ii) 1, 3, _____, _____, _____, _____.
(iii) 1, _____, _____, _____, _____,
(iv) _____, _____, _____, _____.
Solution:
(i) 1,2, 3, 4, 5, 6, 7.
(ii) 1,3,6,10,15,21.
(iii) 1,4,10,20,35.
(iv) 1,5,15,35.

Question 3.
Observe the sequence of numbers obtained in the 3rd and 4th slanting rows of Pascal’s Triangle and find the difference between the consecutive numbers and complete the table given below.

Solution:

Try These (Text book Page No. 96)

Question 1.
Observe the pattern of numbers given in the slanting rows earlier and complete the Pascal’s Triangle.

Solution:

Question 2.
Complete the given Pascal’s Triangle. Find the common property of the numbers filled by you. Can you relate this pattern with the pattern discussed in situation 2. Discuss.

Solution:

Common Properties:
The numbers filled by me are even numbers. Also they make triangular shape.
Yes, this pattern and the pattern given in situation 2 have the same properties.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.3

Miscellaneous Practice Problems

Question 1.
Choose the correct relationship between x and y for the given table.

(i) y = x + 4
(ii) y = x + 5
(iii) y = x + 6
(iv) y = x + 7
Answer:
(iii) y = x + 6

Question 2.
Find the triangular numbers from the Pascal’s Triangle and colour them.

Solution:
Triangular numbers are numbers the objects of which can be arranged in the form of equilateral triangle.
Example : 1, 3, 6, 10, 15,…
From Pascal’s Triangle, the triangular numbers are

Question 3.
Write the first five numbers in the third slanting row of the Pascal’s Triangle and find their squares. What do you infer?
Solution:

Numbers in the 3rd slanding row are 1, 3, 6, 10, 15, 21,….
The squares are 1², 3², 6², 10². 15², 21²,…. = 1, 9, 36, 100, 225, 441,…

From the above table we can conclude that the squares of the triangular numbers are the sum of cubes of natural numbers.

Challenge Problems

Question 4.
Tabulate and find the relationship between the variables (x and y) for the following patterns.

Solution:
(i) Let the number of steps be x and the number of shapes be y.
Tabulating the values of x and y

From the table
x = 1 ⇒ y = 1 = 12
x = 2 ⇒ y = 4 = 22
x = 3 ⇒ y = 9 = 32
x = 4 ⇒ y = 16 = 42
Hence the relationship between x and y is y = x2.

(ii) Let the number of steps be x and the number of shapes be y.
Tabulating the values of x and y

From the table x = 1 ⇒ y = 1 = 1
x = 2 ⇒ y = 2 + 1 = 3
x = 3 ⇒ y = 3 + 2 = 5
x = 4 ⇒ y = 4 + 3 = 7
x = 5 ⇒ y = 5 + 4 = 9
Hence the relationship between x and y is y = 2x-1.

Question 5.
Verify whether the following hexogonal shapes form a part of the Pascal’s Triangle.

Solution:
In Pascal’s Triangle product of the 3 alternate numbers given around the hexagon is equal to the product of remaining three numbers.

1 × 13 × 66 = 11 × 1 × 78 = 858
∴ It form a part of Pascal’s Triangle.

5 × 21 × 20 = 10 × 6 × 35 = 2100
∴ It form a part of Pascal’s Triangle

8 × 45 × 84 = 28 × 9 × 120 = 30240
∴ It form a part of Pascal’s Triangle

56 × 210 × 126 = 70 × 84 × 252 = 1481760
∴ It form a part of Pascal’s Triangle

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.2

Question 1.
Complete the Pascal’s Triangle.

Solution:

Question 2.
The following hexagonal shapes are taken from Pascal’s Triangle. Fill in the missing numbers.

Solution:

Question 3.
Complete the Pascal’s Triangle by taking the numbers 1,2,6,20 as line of symmetry.

Solution:
Corresponding numbers are equal about the line of symmetry.

Objective Type Questions

Question 1.
The elements along the sixth row of the Pascal’s Triangle is
(i) 1,5,10,5,1
(ii) 1,5,5,1
(iii) 1,5,5,10,5,5,1
(iv) 1,5,10,10,5,1
Answer:
(iv) 1,5,10,10,5,1

Question 2.
The difference between the consecutive terms of the fifth slanting row containing four elements of a Pascal’s Triangle is
(i) 3,6,10,…
(ii) 4,10,20,…
(iii) 1,4,10,…
(iv) 1,3,6,…
Answer:
(ii) 4,10,20,…

Question 3.
What is the sum of the elements of ninth row in the Pascal’s Triangle?
(i) 128
(ii) 254
(iii) 256
(iv) 126
Answer:
(iii) 256

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.1

Question 1.
Match the given patterns of shapes with the appropriate number pattern and its generalization.


Solution:
(i) (d)
(ii) (a)
(iii) (c)
(iv) (c)
(v) (b)

Objective Type Questions

Question 2.
Identify the correct relationship between x andy from the given table.

(i) y = 4x
(ii) y = x + 4
(iii) y = 4
(iv) y = 4 × 4
Answer:
(i) y = 4x

Question 3.
Identify the correct relationship between x and y from the given table.

(i) y = -2x
(ii) y = +2x
(iii) y = +3x
(iv) y = -3x
Answer:
(iv) y = -3x