Category: Class 8

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 எண்கள் Ex 1.3 Textbook Questions and Answers, Notes.

TN Board 8th Maths Solutions Chapter 1 எண்கள் Ex 1.3

கேள்வி 1.
\(\frac{-5}{7}\) மற்றும் \(\frac{8}{9}\) ஆகிய விகிதமுறு எண்களுக்குக் கூட்டல் மற்றும் பெருக்கலுக்கான அடைவுப் பண்பினைச் சரிபார்க்கவும்.
தீர்வு :
a = \(\frac{-5}{7}\), b = \(\frac{8}{9}\) என்க
a + b = \(\frac{-5}{7}+\frac{8}{9}=\frac{-45+56}{63}=\frac{11}{63}\) ∈ Q
ab = \(\frac{-5}{7} \times \frac{8}{9}=\frac{-40}{63} \mathrm{Q}\)

கேள்வி 2.
\(\frac{-10}{11}\) மற்றும் \(\frac{-8}{33}\) ஆகிய விகிதமுறு எண்களுக்குக் கூட்டல் மற்றும் பெருக்கலுக்கானப் பரிமாற்றுப் பண்பினைச் சரிபார்க்கவும்.
தீர்வு :
கூட்டல் பரிமாற்றுப் பண்பு
a + b = b + a
இடப்பக்கம் = a+b =
\(\left(\frac{-10}{11}\right)+\left(\frac{-8}{33}\right)=\frac{(-30)+(-8)}{33}=\frac{-38}{33}\)
வலப்பக்கம் = b+a= \(\left(\frac{-8}{33}\right)+\left(\frac{-10}{11}\right)=\frac{(-8)+(-30)}{33}=\frac{-38}{33}\)
பெருக்கல் பரிமாற்றப் பண்பு. இடப்பக்கம்
a x b = \(\left(\frac{-10}{11}\right) \times\left(\frac{-8}{33}\right)=\frac{80}{363}\)
வலப்பக்கம்
= b x a = \(\left(\frac{-8}{33}\right) \times\left(\frac{-10}{11}\right)=\frac{80}{363}\)
இடப்பக்கம் = வலப்பக்கம்

கேள்வி 3.
\(\frac{-7}{9}, \frac{5}{6}\) மற்றும் \(\frac{-4}{3}\) ஆகிய விகிதமுறு எண்களுக்குக் கூட்டல் மற்றும் பெருக்கலுக்கானச் சேர்ப்புப் பண்பினைச் சரிபார்க்கவும்.
தீர்வு :
a = \(\frac{-7}{9}\) b = \(\frac{5}{6}\) மற்றும் C = \(\frac{-4}{3}\)
கூட்டல் சேர்ப்பு பண்பு :
a + ( b + c ) = (a + b) + c

(1), (2) லிருந்து a+(b+c)=(a+b) + c நிரூபிக்கப்பட்டது.
இடப்பக்கம் = வலப்பக்கம்
பெருக்கல் சேர்ப்பு பண்பு :
a x (b x c) = (a x b) x c
இடப்பக்கம்

வலப்பக்கம்

(1), (2) லிருந்து
a x (b x c) = (a x b) x c நிரூபிக்கப்பட்டது இடப்பக்கம் = வலப்பக்கம்

கேள்வி 4.
விகிதமுறு எண்களுக்கான a x (b + c)= (axb) +(a x c) என்ற பங்கீட்டுப் பண்பினை a = \(\frac { -1 }{ 2 }\)
b = \(\frac { 2 }{ 3 }\), c = \(\frac { -5 }{ 6 }\) ஆகிய விகிதமுறு எண்களுக்கு சரிபார்க்கவும்.
தீர்வு :
பங்கீட்டுப் பண்பு
a x (b+c) = (a x b) + (a x c)
இடப்பக்கம் ⇒ b + c
\(\frac{2}{3}+\frac{(-5)}{6}=\frac{4+(-5)}{6}=\frac{-1}{6}\)
வலப்பக்கம் a x (b+c)
= \(\left(\frac{-1}{2}\right) \times\left(\frac{-1}{6}\right)=\frac{1}{12}\)
வலப்பக்கம்

இடப்பக்கம் = வலப்பக்கம்

கேள்வி 5.
கூட்டல் மற்றும் பெருக்கலுக்கான சமனிப் பண்பினை \(\frac { 15 }{ 9 }\) மற்றும் \(\frac { -18 }{ 25 }\) ஆகிய விகிதமுறு எண்களுக்குச் சரிபார்க்க வும்.
தீர்வு :
a = \(\frac { 15 }{ 9 }\), b = \(\frac { -18 }{ 25 }\)
கூட்டலுக்கான சமனிப் பண்பு
0 + a = a + 0 = a
இடப்பக்கம்
= 0 + a = 0 + \(\frac { 15 }{ 9 }\) = \(\frac { 15 }{ 9 }\) = a
வலப்பக்கம்
= a + 0 = \(\frac { 15 }{ 9 }\) + 0 = \(\frac { 15 }{ 9 }\) = a ……. (2)
(1), (2) லிருந்து
0 + a = a + 0 = a
0 + b = b + 0 = b
இடப்பக்கம்
= 0 + b = 0 + ( \(\left(\frac{-18}{25}\right)=\frac{-18}{25}\) ) = b ……………… (1)
வலப்பக்கம்
= b + 0 = \(\left(\frac{-18}{25}\right)+0=\frac{-18}{25}\) = b ……………. (2)
(1) & (2) லிருந்து 0 + b = b + 0 = b
பெருக்கலுக்கான சமனிப்பண்பு :
1 x a = a x 1 =a
இடப்பக்கம் = 1 x a = 1 x \(\frac{15}{19}=\frac{15}{19}\) = a ….. (1)
வலப்பக்கம் = a x 1 = \(\) = a ………(2)
(1) & (2) லிருந்து 1 x a = a x 1 = a
1 x b = b x 1 = b
இடப்பக்கம்
1 x b = 1 x \(\left(\frac{-18}{25}\right)=\frac{-18}{25}\) = b …………………(1)
வலப்பக்கம்
= b x 1 = \(\left(\frac{-18}{25}\right) \times 1=\frac{-18}{25}\) = b ……………….(2)
(1) & (2) லிருந்து
1 x b = b x 1 = b

கேள்வி 6.
கூட்டல் மற்றும் பெருக்கலுக்கான நேர்மாறு பண்பினை \(\frac { -7 }{ 17 }\) மற்றும் \(\frac { 17 }{ 27 }\) ஆகிய விகிதமுறு எண்களுக்குச் சரிபார்க்கவும்.
கூட்டலுக்கான நேர்மாறு பண்பு :
தீர்வு :
a + (-a) = 0 = (-a) + a
a = \(\frac { -7 }{ 17 }\) b = \(\frac { 17 }{ 27 }\)
இடப்பக்கம் = a + (-a)
\(\left(\frac{-7}{17}\right)+\left(\frac{7}{17}\right)=\frac{-7+7}{17}=\frac{0}{17}\) = 0 ….(1)

(1) & (2) லிருந்து a + (-a) = 0 = (-a) + a
(ii) இடப்பக்கம் = b + (-b)

(1) & (2) லிருந்து b + (-b) = 0 = (-b) + b
பெருக்கலுக்கான நேர்மாறு பண்பு :
a x \(\frac{1}{a}=\frac{1}{a}\) x a = 1
(i) இடப்பக்கம் =

(1) & (2) லிருந்து
a x \(\frac{1}{a}=\frac{1}{a}\) x a = 1

(ii) இடப்பக்கம் =


(1) & (2) லிருந்து
b x \(\frac{1}{b}=\frac{1}{b}\) x b = 1

கொள்குறிவகை வினாக்கள்

கேள்வி 7.
விகிதமுறு எண்களுக்கு, ………. என்ற எண்ணால் அடைவுப் பண்பானது வகுத்தலுக்கு உண்மையாகாது.
(அ) 1
(ஆ)-1
(இ) 0
(ஈ) 1/2
விடை :
(இ) 0

கேள்வி 8.
\(\frac{1}{2}-\left(\frac{3}{4}-\frac{5}{6}\right) \neq\left(\frac{1}{2}-\frac{3}{4}\right)-\frac{5}{6}\) என்பது, கழித்தலானது, விகிதமுறு எண்களின் ……… பண்பினை நிறைவு செய்யாது என்பதை விளக்குகிறது.
(அ) பரிமாற்று
(ஆ) அடைவு
(இ) பங்கீட்டு
(ஈ) சேர்ப்பு
விடை :
(ஈ) சேர்ப்பு

கேள்வி 9.
பின்வருவனவற்றுள் எது கூட்டலின் நேர்மாறுப் பண்பினை விளக்குகிறது?
\(\frac{1}{8}-\frac{1}{8}\) = 0
\(\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\)
\(\frac{1}{8}+0=\frac{1}{8}\)
\(\frac{1}{8}-0=\frac{1}{8}\)
விடை :
\(\frac{1}{8}-\frac{1}{8}\) = 0

கேள்வி 10.
\(\frac{3}{4} \times\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{3}{4} \times \frac{1}{2}-\frac{3}{4} \times \frac{1}{4}\) என்பது, பெருக்கலானது …………….. இன் மீது பங்கீடு செய்கிறது என்பதை
விளக்குகிறது.
(அ) கூட்டல்
(ஆ) கழித்தல்
(இ) பெருக்கல்
(ஈ) வகுத்தல்
விடை :
(ஆ) கழித்தல்

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 எண்கள் Ex 1.2 Textbook Questions and Answers, Notes.

TN Board 8th Maths Solutions Chapter 1 எண்கள் Ex 1.2

கேள்வி 1.
கோடிட்ட இடங்களை நிரப்புக :
(i) \(\frac{-5}{12}+\frac{7}{15}\) இன் மதிப்பு ……… ஆகும்
(ii) \(\frac{16}{-30}, \frac{-8}{15}\) இன் மதிப்பு ……….. ஆகும்
(iii) \(\frac{-18}{36}, \frac{-20}{44}\) இன் மதிப்பு ………… ஆகும்
(iv) …………… என்ற விகிதமுறு எண்ணிற்கு தலைகீழி கிடையாது.
(v) -1 இன் பெருக்கல் நேர்மாறு ………….. ஆகும்.
விடை :
(i) \(\frac{1}{20}\)
(ii) 1
(iii) 1
(iv) 0
(v) -1

கேள்வி 2.
சரியா? தவறா? எனக் கூறுக. –
(i) எல்லா விகிதமுறு எண்களும் ஒரு கூட்டல் தலைகீழியைப் பெற்றிருக்கும்.
(ii) 0 மற்றும் -1 ஆகியன அவற்றின் கூட்டல் நேர்மாறுகளுக்குச் சமமான விகிதமுறு எண்கள் ஆகும்.
(iii) \(\frac{-11}{-17}\) இன் கூட்டல் நேர்மாறு \(\frac{11}{17}\) ஆகும்
(iv) தன்னைத்தானே தலைகீழியாகக் கொண்ட விகிதமுறு எண் -1 ஆகும்.
(v) அனைத்து விகிதமுறு எண்களுக்கும் பெருக்கல் நேர்மாறு உண்டு.
விடை :
(i) சரி
(ii) தவறு
(iii) தவறு
(iv) சரி
(v) தவறு

கேள்வி 3.
கூடுதலைக் காண்க
(i) \(\frac{7}{5}+\frac{3}{5}\)
(ii) \(\frac{7}{5}+\frac{5}{7}\)
(iii) \(\frac{6}{5}+\left(\frac{-14}{15}\right)\)
(iv) \(-4 \frac{2}{3}+7 \frac{5}{12}\)
விடை :

கேள்வி 4.
\(\frac{-17}{11}\) இலிருந்து \(\frac{-8}{44}\) ஐக் கழிக்கவும்.
தீர்வு :

கேள்வி 5.
மதிப்பு காண்க.
(i) \(\frac{9}{132} \times \frac{-11}{3}\)
(ii) \(\frac{-7}{27} \times \frac{24}{-35}\)
தீர்வு :
(i) \(\frac{9}{132} \times \frac{-11}{3}\)
= \(\frac{-1}{4}\)
(ii) \(\frac{-7}{27} \times \frac{24}{-35}\)
= \(\frac{8}{45}\)

கேள்வி 6.
வகுக்கவும் (i) \(\frac{-21}{5}\) ஜ \(\frac{-7}{-10}\) ஆல்
(ii) \(\frac{-3}{-13}\) ஜ – 3 ஆல் (iii) -2 ஜ \(\frac{-6}{15}\)ஆல்
தீர்வு :

கேள்வி 7.
(i) a = \(\frac { 1 }{ 2 }\) , b = \(\frac { 2 }{ 3 }\), (ii) a = \(\frac { -3 }{ 5 }\) ,b = \(\frac { 2 }{ 15 }\) எனில் (a + b) = (a – b) ஐக் காண்க.
தீர்வு :
(i) a = \(\frac { 1 }{ 2 }\) , b = \(\frac { 2 }{ 3 }\)

(ii) a = \(\frac { -3 }{ 5 }\) ,b = \(\frac { 2 }{ 15 }\)

கேள்வி 8.
\(1 / 2+(3 / 2-2 / 5) \div 3 / 10 \times 3\) ஐச் சுருக்கி, அது 11 மற்றும் 12க்கு இடையில் அமைந்துள்ள ஒரு விகிதமுறு எண் என நிரூபிக்கவும்.
தீர்வு :
im 2
கொடுக்கப்பட்ட எண்ணை சுருக்கும் போது 11.5 கிடைப்பதால் அது 11 மற்றும் 12க்கு இடையில் அமையும்.

கேள்வி 9.
சுருக்குக.
(i) \(\left[\frac{11}{8} \times\left(\frac{-6}{33}\right)\right]+\left[\frac{1}{3}+\left(\frac{3}{5} \div \frac{9}{20}\right)\right]-\left[\frac{4}{7} \times \frac{-7}{5}\right]\)
தீர்வு :

(ii) \(\left[\frac{4}{3} \div\left(\frac{8}{-7}\right)\right]-\left[\frac{3}{4} \times \frac{4}{3}\right]+\left[\frac{4}{3} \times\left(\frac{-1}{4}\right)\right]

கேள்வி 10.
ஒரு மாணவர் ஓர் எண்ணை [latex]\frac { 4 }{ 3 }\) ஆல் வகுப்பதற்குப் பதிலாக \(\frac { 4 }{ 3 }\) ஆல் பெருக்கி சரியான விடையைக் காட்டிலும் 70ஐக் கூடுதலாகப் பெற்றார். அந்த எண்ணைக் காண்க. தீர்வு :

அந்த எண் 120

கொள்குறிவகை வினாக்கள்

கேள்வி 11.
\(\frac{3}{4}+\frac{5}{6}+\left(\frac{-7}{12}\right)\) இன் திட்ட வடிவம் …………………….. ஆகும்.
(அ) 1
(ஆ) \(\frac{-1}{2}\)
(இ) \(\frac{-1}{12}\)
(ஈ) \(\frac{1}{22}\)
விடை :
(அ) 1

கேள்வி 12.
\(\left(\frac{3}{4}-\frac{5}{8}\right)+\frac{1}{2}\) = ………………..
(அ) \(\frac{15}{64}\)
(ஆ) 1
(இ) \(\frac{5}{8}\)
(ஈ) \(\frac{1}{16}\)
விடை :
(இ) \(\frac{5}{8}\)

கேள்வி 13.
\(\frac{3}{4} \div\left(\frac{5}{8}+\frac{1}{2}\right)\) = …………………
(அ) \(\frac{13}{10}\)
(ஆ) \(\frac{2}{3}\)
(இ) \(\frac{3}{2}\)
(ஈ) \(\frac{5}{8}\)
விடை :
(ஆ) \(\frac{2}{3}\)

கேள்வி 14.
\(\frac{3}{4} \times\left(\frac{5}{8} \div \frac{1}{2}\right)\) = ………………………….
(அ) \(\frac{5}{8}\)
(ஆ) \(\frac{2}{3}\)
(இ) \(\frac{15}{32}\)
(ஈ) \(\frac{15}{16}\)
விடை :
(ஈ) \(\frac{15}{16}\)

கேள்வி 15.
இவற்றுள் எந்த விகிதமுறு எண்ணிற்கு கூட்டல் நேர்மாறு உள்ளது?
(அ) 7
(ஆ) \(\frac{-5}{7}\)
(இ) 10
(ஈ) இவை அனைத்திற்கும்
விடை :
(ஈ) இவை அனைத்திற்கும்

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 எண்கள் Ex 1.1 Textbook Questions and Answers, Notes.

TN Board 8th Maths Solutions Chapter 1 எண்கள் Ex 1.1

கேள்வி 1.
கோடிட்ட இடங்களை நிரப்புக :

i) \(\frac{-19}{5}\) ஆனது………….. மற்றும்…………….. என்ற முழுக்களுக்கிடையே இருக்கும்.
ii) \(\frac{15}{-4}\)என்ற விகிதமுறு எண்ணின் தசம வடிவம் …….. ஆகும்.
iii) \(\frac{-8}{3}\) மற்றும் \(\frac{8}{3}\) ஆகிய விகிதமுறு எண்கள்….. இலிருந்து சம தொலைவில் இருக்கும். iv) \(\frac{-15}{24}, \frac{20}{-32}, \frac{-25}{40}\) என்ற வரிசையின் அடுத்த விகிதமுறு எண் ………. ஆகும்.
v) \(\frac{+58}{-78}\) இன் திட்டவடிவம் ………. ஆகும்.
விடை :
(i) -4 மற்றும் – 3
(ii) -3.75
(iii) 0
(iv) \(\frac{30}{-48}\)
(v) \(\frac{-29}{39}\)

கேள்வி 2.
சரியா தவறா எனக் கூறுக :
i) 0 ஆனது மிகச் சிறிய விகிதமுறு எண் ஆகும்.
ii) \(\frac{-4}{5}\) ஆனது \(\frac{-3}{4}\) இன் இடதுபுறமாக உள்ளது.
iii) \(\frac{-19}{5}\) ஆனது \(\frac{15}{-4}\) ஐ விடப் பெரியது
iv) இரு விகிதமுறு எண்களின் சராசரியானது அவற்றிற்கிடையே அமையும்.
v) 10 மற்றும் 11 இக்கு இடையில் எண்ணிலடங்கா விகிதமுறு எண்கள் உள்ளன.
விடை :
i) தவறு
ii) சரி
iii) தவறு
iv) சரி
v) சரி

கேள்வி 3.
எண்கோட்டின் மீது கேள்விக் குறியிட்டுள்ள இடங்களில் அமைந்த விகிதமுறு எண்களைக் காண்க.

தீர்வு :

கேள்வி 4.
ஒர் எண்கோட்டின் மீது S, Y, N, C, R, A,T, I மற்றும் 0 ஆகிய புள்ளிகள் CN = NY=YS மற்றும் RA= AT=TI =I0 என்றுள்ளவாறு இருக்கின்றன. Y, N, A, T மற்றும் 1 ஆகிய எழுத்துக்களால் குறிக்கப்பெறும் விகிதமுறு எண்களைக் காண்க.

விடை :

Y = \(\frac { -5 }{ 3 }\)
N = \(\frac { -4 }{ 3 }\)
A = \(\frac { 9 }{ 4 }\)
T = \(\frac { 10 }{ 4 }\)
I = \(\frac { 11 }{ 4 }\)

கேள்வி 5.
ஓர் எண்கோட்டினை வரைந்து, அதன்மீது பின்வரும் விகிதமுறு எண்களைக் குறிக்கவும்.
(i) \(\frac { 9 }{ 4 }\)
(ii) \(\frac { -8 }{ 3 }\)
(iii) \(\frac { -17 }{ -5 }\)
(iv) \(\frac { 15 }{ -4 }\)
தீர்வு :
(i) \(\frac { 9 }{ 4 }\)

(ii) \(\frac { -8 }{ 3 }\)

(iii) \(\frac { -17 }{ -5 }\)

(iv) \(\frac { 15 }{ -4 }\)

கேள்வி 6.
பின்வரும் விகிதமுறு எண்களின் – தசம வடிவத்தை எழுதவும்.
(i) \(\frac { 1 }{ 11 }\)
(ii)\(\frac { 13 }{ 4 }\)
(iii) \(\frac { -18 }{ 7 }\)
(iv) 1\(\frac { 2 }{ 5 }\)
(v) -3\(\frac { 1 }{ 2 }\)
விடை :
(i) \(\frac { 1 }{ 11 }\) = 0.090909
(ii)\(\frac { 13 }{ 4 }\) = 3.25
(iii) \(\frac { -18 }{ 7 }\) = -2.5714285714
(iv) 1\(\frac { 2 }{ 5 }\) = \(\frac { 7 }{ 5 }\) = 1.4
(v) -3\(\frac { 1 }{ 2 }\) = \(\frac { -7 }{ 2 }\) = -3.5

கேள்வி 7.
கொடுக்கப்பட்ட விகிதமுறு எண்களுக்கு இடையில் ஏதேனும் ஐந்து விகிதமுறு எண்களைப் பட்டியலிடுக.
(i) -2 மற்றும் 0
(ii) \(\frac{-1}{2}\) மற்றும் \(\frac{-3}{5}\)
(iii) \(\frac{1}{4}\) மற்றும் \(\frac{7}{20}\)
(iv) \(\frac{-6}{4}\) மற்றும் \(\frac{-23}{10}\)
தீர்வு :
(i) -2 மற்றும் 0

(ii) \(\frac{-1}{2}\) மற்றும் \(\frac{-3}{5}\)

(iii) \(\frac{1}{4}\) மற்றும் \(\frac{7}{20}\)
4 மற்றும் 20 ன் மீ.சி.ம. 20 ஆகும்

(iv) \(\frac{-6}{4}\) மற்றும் \(\frac{-23}{10}\)

கேள்வி 8.
சராசரிகள் முறையைப் பயன்படுத்தி \(\frac{14}{5}\) மற்றும் \(\frac{16}{3}\) ஆகியவற்றுக்கு இடையே 2 விகிதமுறு எண்களை எழுதவும்.
தீர்வு :

கேள்வி 9.
பின்வரும் விகிதமுறு சோடிகளை ஒப்பிடுக.
(i) \(\frac{-11}{5}, \frac{-21}{8}\)
(ii) \(\frac{3}{-4}, \frac{-1}{2}\)
(iii) \(\frac{2}{3}, \frac{4}{5}\)
தீர்வு :
(i) \(\frac{-11}{5}, \frac{-21}{8}\)
பகுதி 5, 8 ன் மீ.சி.ம 40.
\(\frac{-11}{5} \times \frac{8}{8}=\frac{-88}{40}\)

(ii) \(\frac{3}{-4}, \frac{-1}{2}\)
பகுதி 4மற்றும் 2 ன் மீ.சி.ம 4.

(iii) \(\frac{2}{3}, \frac{4}{5}\)
பகுதி 3 மற்றும் 5 இன் மீ.சி.ம 15

கேள்வி 10.
பின்வரும் விகிதமுறு எண்களை ஏறுவரிசை மற்றும் இறங்கு வரிசையில் எழுதுக.
(i) \(\frac{-5}{12}, \frac{-11}{8}, \frac{-15}{24}, \frac{-7}{-9}, \frac{12}{36}\)
(ii) \(\frac{-17}{10}, \frac{-7}{5}, 0, \frac{-2}{4}, \frac{-19}{20}\)
தீர்வு :
(i) \(\frac{-5}{12}, \frac{-11}{8}, \frac{-15}{24}, \frac{-7}{-9}, \frac{12}{36}\)

ஏறு வரிசை :
\(\frac{-11}{8}<\frac{-5}{8}<\frac{-5}{12}<\frac{1}{3}<\frac{7}{9}\) இறங்கு வரிசை : \(\frac{7}{9}>\frac{1}{3}>\frac{-5}{12}>\frac{-5}{8}>\frac{-11}{8}\)

ii) \(\frac{-17}{10}, \frac{-7}{5}, 0, \frac{-2}{4}, \frac{-19}{20}\)
10, 5, 4, 20, ன் மீ.சி.ம 20.
\(\frac{-17}{10} \times \frac{2}{2}=\frac{-34}{20}\)

கொள்குறிவகை வினாக்கள்

கேள்வி 11.
\(\frac { 8 }{ 9 }\) கிடைக்க ……. என்ற எண்ணை \(\frac { -6 }{ 11 }\) இலிருந்து கழிக்க வேண்டும்
(ஆ) \(\frac { 34 }{ 99 }\)
(ஆ) \(\frac { -142 }{ 99 }\)
(இ) \(\frac { 142 }{ 99 }\)
(ஈ) \(\frac { -34 }{ 99 }\)
விடை :
(ஆ) \(\frac { -142 }{ 99 }\)

கேள்வி 12.
பின்வரும் சோடிகளில் எது சமான எண்களின் சோடியாகும்?
(ஆ) \(\frac{-20}{12}, \frac{5}{3}\)
(ஆ) \(\frac{16}{-30}, \frac{-8}{15}\)
(இ) \(\frac{-18}{36}, \frac{-20}{44}\)
(ஈ) \(\frac{7}{-5}, \frac{-5}{7}\)
விடை :
(ஆ) \(\frac{16}{-30}, \frac{-8}{15}\)

கேள்வி 13.
\(\frac { -5 }{ 4 }\) என்ற விகிதமுறு எண்ணானது ……….. ஆகியவற்றின் இடையில் அமையும்
(அ) 0 மற்றும் \(\frac { -5 }{ 4 }\)
(ஆ) -1 மற்றும் 0
(இ) -1 மற்றும் -2
(ஈ) -4 மற்றும் -5
விடை :
(இ) -1 மற்றும் – 2

கேள்வி 14.
பின்வரும் விகிதமுறு எண்க ளில், எது மிகப் பெரியது?
(ஆ) \(\frac { -17 }{ 24 }\)
(ஆ) \(\frac { -13 }{ 16 }\)
(இ) \(\frac { 7 }{ -8 }\)
(ஈ) \(\frac { -31 }{ 32 }\)
விடை :
(ஆ) \(\frac { -17 }{ 24 }\)

கேள்வி 15.
\(\frac { 112 }{ 528 }\) இன் எளிய வடிவில் உள்ள பகுதியின் இலக்கங்களின் கூடுதல்
(அ) 4
(ஆ) 5
(இ) 6
(ஈ) 7
விடை :
(இ) 6

Students can Download Maths Chapter 3 Algebra Ex 3.2 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 1 Chapter 3 Algebra Ex 3.2

Samacheerkalvi.Guru 8th Maths Question 1.
Fill in the blanks:

Solution:
(i) \(\frac{18 m^{4}\left(n^{8}\right)}{2 m^{(3)} n^{3}}\) = 9 mn⁵
(ii) \(\frac{l^{4} m^{5} n^{(7)}}{2 l m^{(3)} n^{6}}=\frac{l^{3} m^{2} n}{2}\)
(iii) \(\frac{42 a^{4} b^{5}\left(c^{2}\right)}{6(a)^{4}(b)^{2}}\) = (7)b³c²

Samacheer Guru 8th Maths Question 2.
Say True or False:
(i) 5x³y ÷ 4x² = 2xy
(ii) 7ab² ÷ 14ab = 2b²
Solution:
(i) True
(ii) False

Samacheer Kalvi 8th Standard Maths Question 3.
(i) 27y³ ÷ 3y
(ii) x³y² ÷ x²y
(iii) 45x³y²z⁴ ÷ (-15xyz)
(iv) (3xy)² ÷ 9xy
Solution:

Exercise 3.2 Class 8 Question 4.

Solution:

Class 8 Maths Chapter 3 Exercise 3.2 Solution Question 5.
Divide
(i) 32y² – 8yz by 2y
(ii) (4m² n³ + 16m⁴ n² – mn) by 2 mn
(iii) 10 (4x – 8y) by 5 (x – 2y)
(iv) 81 (9⁴q²r³ + 2p³q³r² – 5p²q²r²) by (3pqr)²
Solution:

Class 8 Maths Chapter 3 Notes Question 6.
Find Adirai’s percentage of marks who scored 25m³n²p out of 100m²np
Solution:
Total marks = 100 m²np
Adirai’s score = 25 m³n²p

Class 8 Maths Chapter 3 Exercise 3.2 Question 7.
Identify the error and correct them.
(i) 7y² – y² + 3y² = 10y²
(ii) 6xy + 3xy = 9x²y²
(iii) m (4m – 3) = 4m² – 3
(iv) (4n)² – 2n + 3 = 4n² – 2n + 3
(v) (x – 2) (x + 3) = x² – 6
Solution:
(i) 7y² – y² + 3y² = (7 – 1 + 3)y² = (6 + 3)y² = 9y²
(ii) 6xy + 3xy = (6 + 3)xy = 9xy
(iii) m (4m – 3) = m (4m) + m (-3) = 4m² – 3m
(iv) (4n)² – 2n + 3 = 16n² – 2n + 3
(v) (x – 2) (x + 3) = x (x + 3) – 2 (x + 3) = x (x) + (x) × 3 + (-2) (x) + (-2) (3)
= x² + 3x – 2x – 6 = x² + x – 6

Students can Download Social Science Geography Term 3 Chapter 3 Map Reading Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Social Science 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 Social Science Geography Solutions Term 3 Chapter 3 Map Reading

Samacheer Kalvi 8th Social Science Geography Map Reading Textual Evaluation

I. Choose the best answer:

Map Reading Samacheer Kalvi Question 1.
The subject which deals with map making process is-
(a) Demography
(b) Cartography
(c) Physiography
(d) Topography
Answer:
(b) Cartography

8th Social Map Reading Question 2.
A map that shows the physical features of an area is called.
(a) Cadastral map
(b) Relief map
(c) Climatic map
(d) Resource map
Answer:
(b) Relief map

Map Reading 8th Standard Question 3.
Shallow water bodies are represented by color.
(a) Yellow
(b) Brown
(c) Light blue
(d) Dark blue
Answer:
(c) Light blue

Map Reading Answers Question 4.
The maps which are known as plans are-
(a) Cadastral maps
(b) Topographical maps
(c) Isoline maps
(d) Transport maps
Answer:
(a) Cadastral maps

Map Reading Questions And Answers Question 5.
Actual distribution of population can be represented by-
(a) lines
(b) Shades
(c) Dots
(d) Contours
Answer:
(b) Shades

II. Fill in the Blanks:

1. The globe is the true representation of the ……………
2. A way of representing the spherical earth on a flat surface is ……………
3. A line that joins the points of equal elevation is ……………
4. Cadastral maps are usually maintained by ……………
5. …………… map is focused on a specific theme

Answer:

1. Earth
2. map projection
3. Isoline
4. Government
5. Thematic

III. Choose the option which matches the following correctly:

(a) Legend – 1. 45%
(b) North East – 2. brown colour
(c) Contour Line – 3. thematic map
(d) Cadastral map – 4. key of a map
(e) Choropleth – 5. taxation

(a) 3,5,1,4,2
(b) 4,1,2,5,3
(c) 2,5,1,3,4
(d) 5,2,4,1,3
Answer:
(b) 4,1, 2, 5, 3

IV. Match the statement with the reason and select the correct answer:
Map Reading Questions And Answers Pdf Question 1.
Statement: Small scale maps can show only major features
Reason: Due to lack of space ,it shows large areas like Continents and countries
(a) Statement is true but reason is wrong
(b) Statement is wrong and reason is correct
(c) Both the statement and reasons are correct
(d) Both the statement and reasons are wrong
Answer:
(c) Both the statement and reasons are correct

Samacheer Kalvi Guru 8th Social Question 2.
Statement: The conventional signs and symbols are the keys of map reading
Reason: These symbols give a lots of information in a limited area
(a) Both the statement and reasons are correct
(b) Statement is wrong and reason is correct
(c) Statement is true but reason is wrong
(d) Both the statement and reasons are wrong
Answer:
(a) Both the statement and reasons are correct

V. Answer the following in one or two sentence:

Samacheer Kalvi 8th Social Book Solutions Question 1.
Define “Map scale”
Answer:

1. Map scale refers to the relationship (or ratio) between distance on a map and the corresponding distance on the ground.
2. The map scale is stated in words i.e., 1cm to 1 km.

8th Standard Samacheer Kalvi Social Question 2.
What is a physical map?
Answer:
The map that shows the physical features of an area is usually called a Physical Map or a Relief Map.

8th Samacheer Kalvi Social Question 3.
Write a short note on map projection.
Answer:
A map projection is a way of representing the spherical earth on a flat surface of a map. The curved surface of the earth cannot be shown accurately on a map. So, cartographers use map projections while mapping the earth surface which would help them to reduce distortions.

8th Standard Map Question 4.
Name the Intermediate directions.
Answer:
The Intermediate directions are north east, north west, south east and south west.

8th Standard Social Science Map Question 5.
What are the uses of a cadastral map?
Answer:
Cadastral maps are useful for local administration such as the city survey, taxation, management of estates and to define property in legal documents.

VI. Differentiate:

Samacheer Kalvi 8th Social Book Solutions Guide Question 1.
Relief map and thematic map.
Answer:

1. Relief map:

• The map that shows the physical features of an area is usually called a Physical Map or a Relief Map.
• Their primary purpose is to show land forms like deserts, rivers, mountains, plains, plateaus etc.

2. Thematic map:

• A thematic map is a map that focuses on a specific theme or subject area.
• They show the subject such as physical phenomena like temperature variation, rainfall distribution and population density in an area.

Samacheer Kalvi Guru 8th Social Science Question 2.
Large Scale map and small scale map.
Answer:
1. Large Scale:

• The Large scale maps portray the information in detail than the small scale maps.
• For example physical map of India represents a small area of the earth but gives us more information.

2. Small Scale map:

• Small scale maps can show only major features omitting the minor ones due to lack of space.
• For example physical map of the world will show us only the major physical features in the world.

Question 3.
Globe and Map.
Answer:
1. Globe:

• Globe gives a three dimensional representation of the entire world.
• It is a miniature form of the earth (model of the earth).

2. Map:

• Map gives a two dimensional representation of certain regions or the entire world.
• It is a visual representation of an entire or a part of an area typically represented on a flat surface.

VII. Answer in a paragraph:

Question 1.
Explain the different types of scales in detail.
Answer:
Scales on maps can be represented in three different ways. They are:

1. Statement or Verbal scale
2. Representative Fraction (RF) or Ratio Scale
3. Graphical or Bar Scale

1. Statement or Verbal scale:

• In this method, the map scale is stated in words i.e., 1 cm to 1 km.
• It means 1 cm distance on the map corresponds to 1 km distance on the ground.
• Thus it is written on the map like 1 cm to 1 km.

2. Representative Fraction (RF) or Numerical Fraction or Ratio Scale:

• It shows the relationship between the map distance and the corresponding ground distance in the same units of length.
• R.F. is generally shown as a fraction.

3. Graphical or Bar Scale or Linear Scale:

• A graphic scale looks like a small ruler drawn at the bottom of the page.
• This line is line is divided and sub divided into lengths each of which represents a certain distance on the ground.
• This scale has added advantage for taking copies of maps as the measurement does not change.

Question 2.
Describe the Cadastral map and its importance.
Answer:
The Cadastral map:

1. Cadastral map refers to a map that shows the boundaries and ownership of land within a specified area.
2. These maps are sometimes known as plans.
3. They are useful for local administration such as the city survey, taxation, management of estates.
4. They are used to define property in legal documents.
5. They are maintained by the government and they are a matter of public record.

Importance of Cadastral maps:

1. Cadastral surveys document the boundaries of land ownership, by the production of documents, diagrams, sketches, plans, charts and maps.
2. They were originally used to ensure reliable facts for land valuation and taxation.

Question 3.
Write a paragraph about the conventional signs and symbols.
Answer:
The conventional signs and symbols:

1. Conventional signs are symbols used in maps to represent different features.
2. The symbols are explained in the key of the map.
3. These symbols give a lot information in a limited space.
4. With the use of these symbols, maps can be drawn easily and the concept of the map can be understood well. There is an International agreement regarding the use of certain symbols.
5. The symbols fall under this category are Called Conventional Symbols. Other category is called contextual symbols which are decided by the cartographers.

VIII. Students Activity:

1. Underline the map title.
2. Write N,S,W & E on the compass rose.
3. Label the road Kambar Street.
4. The rail track runs from Southwest to.
5. The park is situated on of the rail track.
6. Colour the school with red.
7. Colour the supermarket with brown.
8. Colour the restaurant with yellow.
9. Colour the house east of the railroad with orange.

Samacheer Kalvi 8th Social Science Geography Map Reading Additional Questions

I. Choose the correct answer:

Question 1.
Map reading is the act of-
(a) Interpreting geographical information
(b) Industrially geographical information
(c) Develops mental maps of the real world
(d) All of the above
Answer:
(d) All of the above

Question 2.
Some of the components of the map.
(a) Title and Scale
(b) Legend and Direction
(c) Source and map projection
(d) All of the above
Answer:
(d) All of the above

Question 3.
Physical map shows.
(a) Physical features
(b) Cities
(c) Industries
(d) Population
Answer:
(a) Physical features

Question 4.
Scale is represented by-
(a) Veíbal
(b) Representative Fraction
(c) Graphical
(d) All of the above
Answer:
(d) All of the above

Question 5.
Symbols in the key might be-
(a) pictures
(b) icons
(e) both a and b
(d) neither a nor b
Answer:
(c) both a and b

Question 6.
The source should normally be given-
(a) outside the frame of the map
(b) inside frame of the map
(c) at the top of the map
(d) at the bottom of the map
Answer:
(a) outside the frame of the map

Question 7.
Contextual symbols are decided by the-
(a) government
(b) countries
(c) cartographers
(d) scientists
Answer:
(c) cartographers

Question 8.
symbol represents.
(a) Bridge
(b) Baffle field
(e) Foot path
(d) Íain load
Answer:
(a) Bridge

II. Fill in the blanks:

1. A map ………….. is a representation of an entire or a part of an area.
2. Globe is the miniature form of the …………..
3. The art and science of map making in known as …………..
4. The content of the map is revealed by its …………..
5. A graphical scale looks like a ………….. drawn at the bottom of the page.
6. Locational information of an area is given by ………….. and …………..
7. Different features on the map are represented by …………..
8. Cadastral maps are matter of ………….. record.
9. Cadastre means …………..
10. General reference maps do not focus any …………..
11. Map showing the distribution of soil is a ………….. map.
12. The distribution of atmospheric pressure is represented by …………..

Answer:

1. Visual
2. earth
3. Cartography
4. Title
5. small ruler
6. Conventional signs
7. Conventional signs
8. public
9. Register of Territorial property
10. Specific theme
11. Qualitative
12. Isobar

III. Match the following:

1. Globe – a) Percapital income
2. Source – b) Temperature
3. Choropleth – c) High Concentration
4. Isotherms – d) Model of the Earth
5. Many dots – e) Bottom right

Answer:

1. d
2. e
3. a
4. b
5. c

IV. Match the statement with the reason and select the correct answer:

Question 1.
Statement: A map must indicate direction.
Reason : By means of an arrow printing to the north, the rest of the directions are easily located.
(a) Statement is true but reason is wrong
(b) Statement is wrong and reason is correct
(c) Both the statement and reasons are correct
(d) Both the statement and reasons are wrong
Answer:
(c) Both the statement and reasons are correct

Question 2.
Statement: A dot density map is a type of Thematic Map.
Reason : Each dot on a dot density map represents some amount of data.
(a) Statement is true but reason is wrong
(b) Statement and reason are correct
(c) Statement and reason are wrong
(d) Statement is wrong but reason is Correct.
Answer:
(b) Statement and reason are correct

V. Answer the following one or two sentences:

Question 1.
What is a map?
Answer:

1. A map is a visual representation of an entire or a part of an area, typically represented on a flat surface.
2. The work of a map is to illustrate specific and detailed features of a particular area, most frequently used to illustrate geography.

Question 2.
What is map reading?
Answer:
Map reading:

1. Map reading is the act of interpreting or understanding the geographic information portrayed on a map.
2. By map reading,the reader should be able to develop a mental map of the real-world information.
3. This is done by processing the symbolized information shown on maps.

Question 3.
Mention the components of a map.
Answer:
The basic components of a map are:

1. Title
2. Scale
3. Legend or key
4. Direction
5. Source
6. Map projection and location information and
7. Conventional signs and symbols.

Question 4.
What is meant by the Term Title of the map?
Answer:
Term Title of the map:

1. Title tells about the content of the map.
2. It is placed mostly at the top comer or at the bottom comer of the map.

Question 5.
Explain the term ‘Source’ of the map.
Answer:
The term ‘Source’ of the map:

1. All maps must show the source of the data used in the respective maps.
2. The source should normally be given outside the frame of the map on the bottom right.
3. On the bottom left the name of the author, publisher, place of publication and year of publication must be given.

Question 6.
What is the scale of a Cadastral Map?
Answer:
The scale of a Cadastral Map:

1. Cadastral maps commonly range from scales of 1:500 to 1:10,000.
2. Large scale diagrams or map shows more precise dimensions and features.
3. Exampl; Buildings, irrigation units, etc. are often prepared by cadastral surveys.

Question 7.
What is a Choropleth Mapping?
Answer:
A choropleth map is a Thematic map in which areas are shaded or patterned in proportion to the measurement of the statistical variable being displayed on the map, such as population density or per-capita income.

VI. Differentiate:

Question 1.
Isoline map and Dot Density Map.
Answer:
1. Isoline map:

• Isolines are lines drawn to link different places that share a common value.
• An Isoline is a line joining equal points. Eg; Isobars,Isotherms etc.

2. Dot Density Map:

• A dot-density map is a type of Thematic map that uses dots on the map to show the values of one or more numeric data fields.
• Each dot on a dot-density map represents some amount of data. In a dot-density map, areas with many dots indicate high concentration of values.

VII. Answer in a paragraph:

Question 1.
What are the uses of maps?
Answer:
Uses of maps:

1. To find the location of objects and places
2. To find the transportation routes
3. Maps showing strategic locations are useful for military
4. Serve as tourist guide
5. To find the spatial distribution of different phenomena
6. Display weather conditions
7. Highly helpful in learning geography
8. Represent the real world on a small scale

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

8th Maths Algebra Exercise 3.1 Question 1.
Multiply a monomial by a monomial.
(i) 6x, 4
(ii) -3x, 7y
(iii) -2m², (-5m)³
(iv) a³, – 4a²b
(v) 2p²q³, -9pq²
Solution:
(i) 6x × 4 = (6 × 4) (x) = 24x
(ii) -3x × 7y = (-3 × 7) (x × y) = -21xy
(iii) (-2m²) × (-5m)³ = -2m² × (-)³ (5³ (m)³) = -2m² × (-125m³)
= (-) × (-)(2 × 125)(m² × m³) = + 250m5 = 250 m
(iv) a³ × (-4a² b) = (-4) × (a³ × a²) × (b) = -4a⁵b
(v) (2p²q³) × (-9pq²) = (+) × (-) × (2 × 9) (p² × p(q³ × q²)) = -18p³q⁵

Maths 8th Samacheer Kalvi Question 2.
Complete the table

Solution:

8th Standard Maths 3.1 Solution Question 3.
Find the product of the terms.
(i) -2mn, (2m)², -3mn
(ii) 3x²y, -3xy³, x²y²
Solution:
(i) (-2mn) × (2m)² × (-3mn) = (-2mn) × 2² m² × (-3mn) = (-2mn) × 4m² × (-3mn)
= (-) (+)(-) (2 × 4 × 3) (m × m² × m) (n × n)
= + 24 m⁴ n²

(ii) (3²y) × (-31xy³) × (x²y²) = (+) × (-) × (+) × (3 × 3 × 1) (x² × x × x²) x (y × y³ × y²)
= -9x⁵y⁶

Samacheer Kalvi Guru 8th Maths Question 4.
If l = 4pq², b = -3p 2q, h = 2p³q³ then, find the value of 1 × b × h.
Solution:
Given l = 4pq²
b = -3p²q
h = 2p³q³
l × b × h = (4pq²) × (-3p² q) × (2p³q³)
= (+) (-) (+) (4 × 3 × 2) (p × p² × p³) (q² × q × q³)
= -24p⁶q⁶

Samacheer Kalvi.Guru 8th Maths Question 5.
Expand
(i) 5x (2y – 3)
(ii) -2p (5p² – 3p + 7)
(iii) 3mn (m³n³ – 5m²n + 7mn²)
(iv) x² (x + y + z) y² (x + y + z) + z² (x – y – z)
Solution:
(i) 5x(2y – 3) = (5x) (2y) – (5x) (3)
= (5 × 2) (x × y) – (5 × 3) x
= 10xy – 15x

(ii) -2p (5p² – 3p + 7) = (-2p) (5p²) + (-2p) (-3p) + (-2p) (7)
= [(-) (+) (2 × 5) (p × p²)] + [(-) (+) (2 × 3) (p × q)] + (-) (+) (2 × 7) p
= -10p³ + 6p² – 14p

(iii) 3mn(m³n³ – 5m²n + 7mn²)
= (3mn) (m³n³) + (3mn) (-5m²n) + (3mn)(7mn²)
= (3) (m × m³) (n × n³) + (+) (-) (3 × 5) (m × m²) (n × n) + (3 × 7) (m × m)(n × n²)
= 3m⁴n⁴ – 15m³ n² + 21m²n³

(iv) x² (x + y + z) + y² (x + y + z) + z² (x – y – z)
= (x² × x) + (x² × y) + (x² × z) + (y² × x) + (y² × y) + (y² × z) + (z² × x) + z² (-y) + z² (-z)
= x³ + x²y + x²z + xy² + y³ + y²z + xz² – yz² – z³
= x³ + y³ – z² + x²y + x²z + xy² + zy² + xz² – yz²

Samacheer Kalvi Guru Maths 8th Question 6.
Find the product of
(i) (2x + 3)(2x – 4)
(ii) (y² – 4) (2y² + 3y)
(iii) (m² – m) (5m²n² – n²)
(iv) 3(x – 5) × 2(x – 1)
Solution:
(2x) (2x – 4) + 3 (2x – 4) = (2x) (2x – 4) + 3 (2x – 4)
= (2x × 2x) – 4 (2x) + 3(2x) – 3 (4)
= 4x² – 8x + 6x – 12
= 4x² + (- 8 + 6)x – 12
= 4x² – 2x – 12

(ii) (y² -4) (2y² + 3y) = y² (2y² + 3y) – 4 (2y² + 37)
= y²(2y²) + y²(3y) – 4(2y²) -4 (3y)
= 2y⁴ + 3y³ – 8y² – 12y

(iii) (m² – n) (5m²n² – n²) = m² (5m²n² – n²) – n (5m²n² – n²)
= m² (5m²n²) + m² (-n²) – n (5m²n²) + (-) (-) n (n²)
= 5m⁴n² – m²n² – 5m²n³ + n³

(iv) 3(x – 5) × 2(x – 1) = (3 × 2) (x – 5) (x – 1)
= 6 × [x (x – 1) – 5 (x- 1)]
= 6 [x.x – x . 1 – 5.x + (-1) (-) 5 1]
= 6 [x² – x – 5x + 5] = 6 [x² + (-1 – 5)x + 5]
= 6 [x² – 6x + 5] = 6x² – 36x + 30

Samacheer Kalvi 8 Maths Question 7.
Find the missing term.
(i) 6xy – × ______ = -12x³y
(ii) ________ × (-15m²n³p) = 45m³n³p²
(iii) 2y(5x²y – ___ + 3 ___) = 10x²y² – 2xy + 6y³
Solution:
(i) 6xy – × (-2x²) = -12x³y
(ii) -3mp × (-15m²n³p) = 45m³n³p²
(iii) 2y(5x²y – x + 3 y²) = 10x²y² – 2xy + 6y³

Samacheer Kalvi 8th Maths Book Solutions Question 8.
Match the following

(A) iv, v, ii, i, iii
(B) v, iv, iii, ii, i
(C) iv, v, ii, iii, i
(D) iv, v, ii, iii, i
Solution:
(a) iv
(b) v
(c) ii
(d) iii
(e) i

Samacheer Kalvi 8 Maths Book Question 9.
A car moves at a uniform speed of (x + 30) km/hr. Find the distance covered by the car in (y + 2)hours. (Hint: distance = speed × time).
Solution:
Sppeed of the car = (x + 30) km / hr.
Time = (y + 2) hours
Distance = Speed × time = (x + 30) (y + 2) = x(y + 2) + 30 (y + 2) = x (y + 2) + 30 (y + 2)
= (x) (y) + (x) (2) + (30) (y) + (30) (2)
= xy + 2x + 30y + 60
Distance covered = (xy + 2x + 30y + 60) km

Objective Type Questions

Samacheer Kalvi Class 8 Maths Solutions Question 10.
The product of 7p³ and (2p²)² is
(A) 14p²
(B) 28p⁷
(C) 9p⁷
(D) 11p¹²
Solution:
(B) 28p⁷

Samacheerkalvi.Guru 8th Maths Question 11.
The missing terms in the product -3m³ n × 9(- -) = ____ m⁴n³ are
(A) mn², 27
(B) m²n, 27
(C) m²n², -27
(D) mn², -27
Solution
(A) mn² ,27

Samacheer Kalvi Guru 8th Maths Guide Question 12.
If the area of a square is 36x⁴y² then, its side is ______ .
(A) 6x⁴y²
(B) 8x²y²
(C) 6x²y
(D) -6x²y
Solution:
(C) 6x²y

Samacheer Kalvi 8th Maths Question 13.
If the area of a rectangle is 48m²n³ and whose length is 8mn2 then, its breadth is ____ .
(A) 6 mn
(B) 8m²n
(C) 7m²n²
(D) 6m²n²
Solution:
(A) 6mn

Maths Class 8 Samacheer Kalvi Question 14.
If the area of a rectangular land is (a² – b²) sq.units whose breadth is (a – b) then, its length is _____
(A) a – b
(B) a + b
(C) a² – b
(D) (a + b)²
Solution:
(B) a + b

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Tamilnadu Samacheer Kalvi 8th Social Science History Solutions Term 3 Chapter 2 Status of Women in India through the Ages

Samacheer Kalvi 8th Social Science Status of Women in India through the Ages Textual Evaluation

I. Choose the correct answer:

Status Of Women In India Through The Ages Question 1.
…………… society is constantly changing with additions, assimilation and omissions from within and outside.
(a) Human
(b) Animal
(c) Forest
(d) Nature
Answer:
(a) Human

Question 2.
The First woman doctor in India was
(a) Dharmambal
(b) Muthulakshmi Ammaiyar
(c) Moovalur Ramamirdham
(d) Panditha Ramabai
Answer:
(b) Muthulakshmi Ammaiyar

Question 3.
The practice of sat was abolished in.
(a) 1827
(b) 1828
(c) 1829
(d) 1830
Answer:
(c) 1829

Question 4.
B.M Malabar was a
(a) teacher
(b) doctor
(c) lawyer
(d) journalist
Answer:
(d) journalist

Question 5.
Which of the following was/were the reform movement(s)?
(a) Brahma Samaj
(b) Prarthana Samaj
(c) Arya Samaj
(d) all the above
Answer:
(d) all the above

Question 6.
The Bethune school was founded in by J.E.D. Bethune.
(a) 1848
(b) 1849
(c) 1850
(d) 1851
Answer:
(b) 1849

Question 7.
Which commission recommended to start primary schools for girls in 1882?
(a) Wood’s
(b) Welly
(c) Hunter
(d) Muddiness
Answer:
(c) Hunter

Question 8.
Sara’s child Marriage Bill fixing the minimum marriageable age for girls at.
(a) 11
(b) 12
(c) 13
(d) 14
Answer:
(d) 14

II. Fill in the Blanks:

1. ……………. society was setup by the Christian missionaries in 1819.
2. ……………. of Sivaganga fought bravely against the British.
3. Servants of India Society was started by ……………
4. ……………. was the one of the greatest social reformer of Tamil Nadu.
5. Kandukuri Veeresalingam published a journal called …………….

III. Match the following:

1. Theosophical society – Italian traveler
2. Sarada Sadan – Social evil
3. Wood’s Despatch – Annie Besant
4. Niccolo Conti – Pandita RamaBhai
5. Dowry – 1854

Answer:

1. Theosophical society — Annie Besant
2. Sarada Sadan — Pandita RamaBhai
3.  Wood’s Despatch — 1854
4. Niccolo Conti — Italian traveler
5. Dowry — Social evil

IV. State True or False:

1. Women were honoured in Rig Vedic period.
2. Devadasi system was a social evil.
3. Raja Rammohan Roy, was the pioneer of Indian social reform movement.
4. Reservation of 23 percent to women envisaged an improvement in the sociopolitical status of women.
5. The age of marriage was raised for boys and girls by the Sharda Act of 1930.

Answer:

1. True
2. True
3. True
4. False
5. True

V. Choose the correct statement:

Question 1.
Find out the correct pair:
(a) Women’s university – Prof. D.K. Karve
(b) Justice Ranade – Arya Samaj
(c) Widow Remarriage Act – 1855
(d) Rani Lakshmi Bhai – Delhi
Answer:
(a) Women’s university – Prof. D.K. Karve

Question 2.
Find the odd
(a) Child marriage devadasi system)
(b) sati
(c) devadasi system
(d) widow remarriage
Answer:
(d) widow remarriage

Question 3.
Consider the following Statements
i) Begum Hazarat Mahal, Rani Lakshmi Bhai led an armed revolt against the British
ii) Velunachiyar of Sivaganga, Tamil Nadu fought bravely against the British Which of the statement (s) given above is/or correct?
(a) i only
(b) ii only
(c) i and ii
(d) neither i nor ii
Answer:
(c) i and ii

Question 4.
Assertion (A): Raja Rammohan Roy is most remembered by all Indians
Reason (R): He wiped out the evil practice of Sati form the Indian Society
(a) A and R are wrong
(b) A is correct and R is Wrong
(c) A is correct and R explains A
(d) A is correct and R does not explain A
Answer:
(c) A is correct and R explains A

VI. Answer the following in one or two sentences:

Quesstion 1.
Name the prominent leaders who fought for the enlistment of women.
Answer:
During the British Raj, many socioreligious reformers like Raja Rammohan Roy, Dayananda Saraswathi, Keshab Chandra Sen, Iswara Chandra Vidya Sagar, Pandita Ramabai, Dr. Muthulakshmi, Jyoti rao phule, Periyar E.V.R, Dr. Dharmambal were the prominent leaders who fought for the upliftment of women.

Question 2.
List out some social evils.
Answer:
Sat, child marriages, female infanticide, Purdah system and slavery were some of the social evils that existed in Indian Society.

Question 3.
Who were the notable women during the medieval period?
Answer:
Some of the notable women during the medieval period were Razia sultana, Queen Durgavati, Chand bibi, Nurjahan, Jahan nara, Jijabai and Mira bai.

Question 4.
Mention the important women freedom fighters of India.
Answer:
Velunachiyar of Sivaganga, Begum Hazrat Mahal and Rani Lakshmi Bhai of Jhansi were important women freedom fighters of India.

Question 5.
Give a note on Sati.
Answer:

1. Sati was social evil that prevailed in Indian Society especially among the Rajputs,
2. The feudal society of the time encouraged “sati” which meant self-immolation of the widow on the funeral pyre of her husband.
3. Earlier it was a voluntary act but later by the relatives forced the widow to sit on the funeral pyre.

VII. Answer the following:

Question 1.
Trace the role of women in freedom struggle.
Answer:

• The spread of female education led to several other social reforms of great consequences, such as the abolition of purdah system, participation of women in the freedom struggle.
•  In the early anti-colonial Struggle women played major roles in various capacities,
• Velunachiyar of Sivaganga fought violently against the British and restored her rule in Sivaganga.
•  Begum Hazrat Mahal, Rani Lakshmi Bhai of Jhansi led an armed revolt of 1857* against the British.
• In the freedom struggle thousands of women came out of their homes, boycotted foreign goods, marched in processions, defied laws, received lathi charges and Courted jails.

Quesstion 2.
Explain the contribution of the Social Reformers for the eradication of social evils.
Answer:
The contribution of the social reformers for the eradication of social evils is listed below.
(a) Raja Ram Mohan Roy:

• Raja Rammohan Roy was, the pioneer of Indian social reform movement by abolishing sati in 1829 with the help of Lord William Bentinck.
• He also protested against the child marriage and female infanticide and favoured the remarriage of widows.

(b) Ishwar Chandra Vidhyasagar:

• Ishwar Chandra Vidhyasagar carried on the movement for female education, widow remarriage and abolition of polygamy in Bengal.
• He was instrumental to passing of the Hindu Widow Remarriage Act in 1856.
•  His son Narayanachandra set an example to others by marrying a widow of his choice.

(c) Kandukuri Veeresalingam:

• Veeresalingam Pantulu published a journal Viveka Vardhani.
• He opened his first girls’ school in 1874, and made widow remarriage and female education the key points of his programme for social reform.

(d) M.G. Ranade and B.M. Malabari:

• In Bombay presidency, M.G. Ranade and B.M. Malabari carried on the movement
  for the upliftment of women.
• In 1869, Ranade joined the Widow Remarriage Association and encouraged widow remarriage and female education.
• He opposed child marriage..
• In 1884, B.M. Malabari, a journalist, started a movement for the abolition of child marriage.

(e) Gopal Krishna Gokhale:
In 1905, Gopal Krishna Gokhale started the Servants of India Society to reform the society with various measures.

(f) Periyar E.V.R:
Periyar E.V.R. was one of the greatest social reformers of Tamil Nadu. He advocated upliftment of women education, widow remarriage and opposed child marriages.

(g) Women Reformers:

• In 1889, Pandita Ramabai opened Sarada Sadan for Hindu widows in Bombay.
• Dr. Annie Besant established Theosophical society to developed general social reform programme.
• Dr. S.Dharmambal showed great interest in implementing widow remarriage and women education.
• Moovalur Ramamirdham Ammaiyar opposed Devadasi system along with Dr. Muthulakshmi Ammaiyar.

Question 3.
Give a detailed account on the Impact of reform movement.
Answer:

• Significant advances were made in the field of emancipation of women.
• It created of national awakening among the masses.
• It created the feeling of sacrifice, service and rationalism.
• The practice of sati and infanticide were made illegal.
• It permitted widow remarriage.
• The following legislations have enhanced the status of women in matters of marriage adoption and inheritance.

Legislation Provisions:

• Bengal regulation of XXI, 1804 – Female infanticide was declared illegal
• Regulation of XVII, 1829 – Practice of sat was declared illegal
• Hindus Widow’s Remarriage Act, 1856 – It permitted widow remarriage
• The Native Marriage Act, 1872 – The Child Marriage was prohibited
• The Sharda Act, 1930 – The age of marriage was raised for boys and girls
• Devadasi abolition Act, 1947 – It abolished Devadasi system

Samacheer Kalvi 8th Social Science Status of Women in India through the Ages Additional Questions

I. Choose the correct answer:

Question 1.
Which of the following put limitations and restrictions on the liberty of women?
(a) New Social practices
(b) New Customs
(c) New Systems
(d) All of the above
Answer:
(d) All of the above

Question 2.
Who fought for the upliftment of women?
(a) Keshab Chandra Sen
(b) Jyoti rao Phule
(c) Pandit Rama Bai
(d) All of the above
Answer:
(d) All of the above

Question 3.
When was there a transitional development in the status of women restricting her role in the social life?
(a) In Indus Valley Civilization
(b) In Rig Vedic period
(c) In Later Vedic period
(d) None of the above
Answer:
(c) In Later Vedic period

Question 4.
The social evils which affected the position of women in the medieval period was / were ……………..
(a) Purdah system
(b) Slavery
(c) Female infanticide
(d) All of the above
Answer:
(d) All of the above

Question 5.
The Mughal ruler …………… attempted to abolish sati.
(a) Aurangazeb
(b) Jahangir
(c) Akbar
(d) Babar
Answer:
(c) Akbar

Question 6.
Female infanticide was particularly in vogue in ……………
(a) Rajputana
(b) North Western Provinces
(c) Punjab
(d) All of the above
Answer:
(d) All of the above

Question 7.
Sati was abolished during the time of …………….
(a) Lord William Bentinck
(b) Lord Cornwallis
(c) Lord Wellesley
(d) Lord Dalhousie
Answer:
(a) Lord William Bentinck

Question 8.
The person who was nominated to the Tamil Nadu Legislative Council in 1929 was ……………
(a) Moovalur Ramamirdham
(b) Dr. Muthulakshmi Ammaiyar
(c) Pandit Rama Bai
(d) Rajaji
Answer:
(b) Dr. Muthulakshmi Ammaiyar

Question 9.
The person who was responsible for the abolition of polygamy in Bengal was.
(a) Raja Ram
(b) Dayananda Saraswathi,
(c) Periyar E.V.R
(d) Vidya Sagar
Answer:
(d) Vidya Sagar

Question 10.
Ranade started the National Social Conference in.
(a) 1829
(b) 1853
(c) 1887
(d) 1882
Answer:
(c) 1887

Question 11.
Dr. Annie Besant came to India from
(a) U.S.A
(b) Europe
(c) AustraliaAfrica
(d) Africa
Answer:
(b) Europe

Quesstion 12.
Dr. S. Dharmambal was very much influenced by the ideas of
(a) Gokhale
(b) Periyar
(c) Rajaji
(d) None ofthe above
Answer:
(b) Periyar

Question 13.
The National Commission for women was set up by January
(a) 1947
(b) 1950
(c) 1992
(d) 2000
Answer:
(c) 1992

II. Fill in the blanks:

1. The position of women was not ……………. in all periods.
2. The widow remarriage Act was passed in ……………..
3. Among the Rajputs of Rajasthan ……………. was practiced.
4. Purdah system became popular as a result of ……………. invasion.
5. J.E.D. Bethune was the president of the council of education in ……………..
6.  In ……………. the women’s medical service did a lot of work in training mid-wives.
7. D.K. Karve established a number of female schools in ……………..
8. Lady Harding Medical College was started in ……………..
9. Bengal Regulatory Act XXI of 1795 banned the practice of …………….
10. Child marriage was prevalent among the …………….
11. Radhakanta Deb, and Bhawani Charan Banerji advocated the orthrodox Hindu opinion against the abolition of …………….
12. The Person who was instrumental in passing the “Devadasi abolition bill” was …………….
13. Narayanachandra was the son of …………….
14. Pandit Ramabai’s greatest legacy was her effort to educate …………….
15. Equal opportunity and equal pay for equal work in guaranteed by the Indian Constitution through …………….

Answer:

1. uniform
2. 1856
3. Jauhar
4. Muslim
5. Calcutta
6. 1914
7. poona
8. Delhi
9. Female Infanticide
10. tribes
11. Sati
12. Periyar E.V. Ramasamy
13. Iswar Chandra Vidya sagar
14. Widows
15. Article 14

III. Match the following:

1. Raja Rammohan Roy – a) Opened first girl’s school
2. Vidhyasagar – b) Women Reformer
3. Veeresalingam – c) Revolt of 1857
4. Tarabai Shinde – d) Abolition of Sati
5. Rani Laxmi Bai – e) Widow Remarriage

Answer:

1. d
2. e
3. a
4. b
5. c

IV. State True or False:

1. Women enjoyed equal rights among the later Vedic period.
2. The social reformers rightly realised that female education as an emancipating agent in eradicating social evils.
3. The position of wife was honoured during the ancient period of India.
4. The patriarchal system was flexible in later vedic period.
5. Sati was in practice particularly among the royal and upper strata of the society.
6. The major effect of national awakening in the nineteenth century was seen in the field of social reform.
7. Charles Wood’s despatch recommended female education in 1857.
8. The Indian Women’s University was started in 1919.
9. In 1846, the minimum marriageable age for a girl was only 14 years.
10. Niccolo Conti was an Italian Traveller.
11. Moovalur Ramamirdham fought for the emancipation of the Devadasi.
12. M.G. Ranade opposed child marriage.
13. Sarada Sadan was stated by Tarabai Shinde.
14. The Native Marriage Act was passed in 1872.
15. The National policy on Education of 1986 was passed for empowerment of women.

Answer:

1. False
2. True
3. True
4. False
5. True
6. True
7. False
8. False
9. False
10. True
11. True
12. True
13. False
14. True
15. True

V. Choose the correct statement:

Question 1.
Find out the correct pair.
(a) Gokhale – Servants of India Society
(b) Veereslingam – The Indian Women’s University
(c) Jagan Mohan Roy – Abolition of Sati
(d) Serampore missionaries – Female infanticide
Answer:
(a) Gokhale – Servants of India Society

Question 2.
Find the odd one out.
(a) Raja Ram
(b) Dayananda saraswathi
(c) Niccolo Conti
(d) Rama Bai
Answer:
(c) Niccolo conti

Question 3.
Consider the following Statements
i) Akbar made obligatory for the parents to obtain the approval of both the bride and the bridegroom before the marriage.
ii) According to Bengal regulation of XXI, 1804 Female infanticide was declared illegal. Which of the statement
given above is/or correct?
(a) i only
(b) ii only
(c) i and ii
(d) neither i nor ii
Answer:
(c) i and ii

Question 4.
Assertion (A): Veeresalingam Pantulu was the earliest champion in South India of women’s emancipation.
Reason (R) : Veeresalingam opened his first girls’ school in 1874 and made widow remarriage and female
education the key points of his programme for social reform.
(a) A and R are wrong
(b) A is correct and R is Wrong
(c) A is correct and R explains A
(d) A is correct and R does not explain A
Answer:
(c) A is correct and R explains A

VI. Answer the following in one or two sentences:

Question 1.
What was the positive of women in Rig Vedic period?
Answer:
During the Rig Vedic period the position of wife was honoured and women’s position was acknowledged,
especially in the performance of religious ceremonies.

Question 2.
How were women treated in the later Vedic period?
Answer:

• In the later vedic period women’s social and political freedom was restricted.
• Sati became popular
• The patriarchal system became rigid.

Question 3.
Explain the term Jauhar.

• The Jauhar refers to the practice of collective voluntary immolation by wives and daughters of defeated Rajput warriors.
• They followed it in order to avoid capture and dishonor.

Question 4.
Mention the type of education followed for women in medieval India.
Answer:

• Female education was informal
• Girls usually had their lessons from their parents in their childhood.
• The rich appointed tutors to teach their daughters at home.

Question 5.
Mention the main provisions of Indian Education Commission of 1882.
Answer:

• Indian Education Commission of 1882 recommended to start primary schools for girls.
• It also suggested teacher – training institutions to be established.
• Special scholarships and prizes were mentioned for girls.

Question 6.
Explain the Madras Devadasi Act of 1947.
Answer:

• The Madras Devadasi Act was a law that was enacted on 9th October 1947.
• It gave Devadasis the legal right to marry.
•  It made it illegal to dedicated girls to India temples.

Question 7.
How did Vidyasagar promote female education?
Answer:
To promote female education, Vidhyasagar founded several girls’ schools in the districts of Nadia, Midnapur, Hugh and Burdwan in Bengal.

Question 8.
What were the key points of Veeresalingam programme for social reform?
Answer:

• Veeresalingam published a journal Viveka Vardhani and opened his first girls’ school in 1874.
• Widow remarriage and female education the key points of his programme for social reform.

Question 9.
Name the institution started by M.G. Ranade in 1887.
Answer:

• In 1887, M.G. Ranade started the National Social Conference.
• It became a preeminent institution for social reform.

Question 10.
What are the measures taken by the servants of India society of reform the society.
Answer:

• In 1905 Gokhale started the Servant of India Society.
• It took up such social reform measures as primary education, female education and uplifitment of the depressed classes in society.

Question 11.
Who was Periyar E.V.R?
Answer:

• Periyar E.V.R. was one of the greatest social reformers of Tamil Nadu.
• He advocated women education, widow remarriage and inter-caste marriages.
• He opposed child marriages.

Question 12.
Write about Sarada Sadan.
Answer:

• • Sarada Sadan means Home of Learning. It was started by Pandita Ramabai in 1889.
  • It was started to educate Hindu widows in Bombay.
  • It was later shifted to Poona.

Question 13.
How did the government of Tamil Nadu recognise the service of Moovalur Ramamirdham Ammaiyar?
Answer:

• Moovalur Ramamirdham Ammaiyar raised her voice against Devadasi system along with Dr. Muthulakshmi Ammaiyar.
• In her memory, the government of Tamil Nadu has instituted the “Moovalur Ramamirdha Ammal Ninaivu Marriage assistance scheme”.
• It is a social welfare scheme to provide financial assistance to poor women.

Question 14.
Mention the three major National Women’s organisation.
Answer:

• Leading women realized the need of forming their own association in order to safeguard their interests.
• As a result three major natural women’s organisation namely
• Women’s India Association
• National Council of Women in India .
• The All India Women’s Conference were started

VII. Answer the following in detail:

Question 1.
What was the position of women in the Medieval period?
Answer:

• In the medieval period the position of women in society further deteriorated.
• Women suffered from many social evils such as sati, child marriages, Purdah system, female infanticide and slavery.
• Normally monogamy was in practice but among the rich polygamy was prevalent.
• Widow re-marriage was rare.
• Devadasi system was in practice in some parts of India.
• The Rajputs of Rajastan, the Jauhar was practiced.
•  The condition of widow became miserable during rhe medieval period.
• But People like Akbar attempted to abolish sati.
• There were exceptional women like Razia Sultana, Queen Durgavati, Chand bibi, Nurjahan and Mira bai.
• Women’s education was not completely ignored. Female education was informal. Girls usually had their lessons from their parents in their childhood.
• The rich appointed tutors to teach their daughters at home.
• The daughters of Rajput chiefs and Zamindars studied literature and philosophy.

Question 2.
Why were the three major national women’s organisation started?
Answer:

1. Most of the reform movements like Brahma Samaj, Prarthana Samaj and Arya Samaj were led by male
   reformers.
2. Women reformers like Pandita Ramabai, Rukhmabai and Tarabai Shinde tried to extent further.
3. In 1889, Pandita Ramabai opened Sarada Sadan in Bombay to educate Hindu widows.
4. Dr. Annie Besant established Theosophical society at Chennai developed general social reform programme.
5. Dr. S.Dharmambal showed great interest in implementing widow remarriage and women education.
6. Moovalur Ramamirdham Ammaiyar’ opposed Devadasi system along with Dr. Muthulakshmi Ammaiyar.
7. Thus women reformers also contributed a lot for winning their own rights
8.  As a result three major national women’s organizations were started. They were
   • Women’s India Association
   • National Council of Women in India
   • The All India Women’s Conference.

Question 3.
Write about women in independent India.
Answer:

• Women in India now participate in all activities such as education, politics, medical, culture, service sectors, science and technology.
• The constitution of India guarantees through Article 14 equal opportunity and equal pay for equal work.
• The National policy for empowerment of women was passed under the National Policy on Education in 1986.
• New programme was launched called Mahila Samakhya.
• It mainly focused on empowerment of women.
• Reservation of 33 percent to women, envisaged an improvement in the socio-political status of women.
• The National Commission for women was set up in January 1992.
• it reviews women related legislation and intervene in specific individual complaints of atrocities and denial of rights.

Students can Download Maths Chapter 3 Algebra Ex 3.4 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 1 Chapter 3 Algebra Ex 3.4

8th Standard Maths Exercise 3.4 Question 1.
Factorise the following by taking out the common factor
(i) 18xy – 12yz
(ii) 9x⁵y³ + 6x³y² – 18x²y
(iii) x(b – 2c) + y(b – 2c)
(iv) (ax + ay) + (bx + by)
(v) 2x²(4x – 1) – 4x + 1
(vi) 3y(x – 2)² – 2(2 – x)
(vii) 6xy – 4y² + 12xy – 2yzx
(viii) a³ – 3a² + a – 3
(ix) 3y³ – 48y
(x) ab² – bc² – ab + c²
Solution:
(i) 18xy – 12yz = (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)
Taking out the common factors 2, 3, y, we get
= 2 × 3 × y (3x – 2z) = 6y (3x- 2z)

i) 9x⁵y³ + 6x³y² – 18x²y = (3 × 3 × x² × x³ × y × y²) + (2 × 3 × x² × x × y × y)
Taking out the common factors 3, x², y, we get
= 3 × x² × y (3x³ y² + 2xy – 6)
= 3x²y (3x³ y² + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)
Taking out the binomial factor (b – 2c) from each term, we have
= (b – 2c)(x + y)

(iv) (ax + ay) + (bx + by)
Taking at ‘a’ from the first term and ‘b’ from the second term we have
(ax + ay) + (bx + by) = a (x + y) + b (x + y)
Now taking out the binomial factor (x + y) from each term
= (x + y)(a + b)

(v) 2x²(4x – 1) – 4x + 1
Taking out -1 from last two terms
2x² (4x – 1) – 4x + 1 = 2x² (4x – 1) – 1 (4x- 1)
Taking out the binomial factor 4x – 1, we get
= (4x – 1)(2x² – 1)

(vi) 3y(x – 2)² – 2(2 – x)
3y(x – 2)² – 2(2 – x) = 3y(x – 2)(x – 2)-2(-1) (x – 2) [∵ Taking out -1 from 2 – x]
= 3y (x – 2) (x – 2) + 2 (x – 2)
Taking out the binomial factor x – 2 from each term, we get
= (x – 2) [3y (x – 2) + 2]

(vii) 6xy – 4y² + 12xy – 2yzx
= 6xy + 12xy – 4y² – 2yzx [∵ Addition is commutative]
= (6 × x × y) + (2 × 6 × x × y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))
Taking out 6 × x × y from first two terms and (-1) × 2 × y from last two terms we get
= 6 × x × y(1 + 2) + (-1)(2)y[2y + zx]
= 6xy (3) -2y(2y + zx)
= (2 × 3 × 3 × x × y) – 2xy (2y + zx)
Taking out 2y from two terms
= 2y (9x – (2y + zx)) = 2y (9x – 2y – xz)

(viii) a² – 3a² + a – 3 = a² (a – 3) + 1 (a – 3) [∵ Grouping the terms suitably]
= (a – 3) (a² + 1)

(ix) 3y² – 48y = 3 × y × y² – 3 × 16 × y
Taking out 3 × y = 3y (y² – 16) = 3y (y² – 4²)
Comparing y² – 4² with a² – b²
a = y, b = 4
a² – b² = (a + b) (a – b)
y² – 4² = (y + 4) (y – 4)
∴ 3y (y² – 16) = 3y (y + 4) (y – 4)

(x) ab² – bc² – ab + c²
Grouping suitably
ab² – bc² – ab + c² = b ((ab – c²) – 1(ab – c²)
Taking out the binomial factor ab – c² = (ab – c²) (b – 1)

Samacheer Kalvi Guru 8th Maths Question 2.
Factorise the following expressions
(i) x² + 14x + 49
(ii) y² – 10y + 25
(iii) c² – 4c – 12
(iv) m² + m – 72
(v) 4x² – 8x + 3
Solution:
x² + 14x + 49 = x² + 14x + 72
Comparing with a² + 2ab + b² = (a + b)² we have a = x and b = 7
⇒ x² + 2(x) (7) + 7² = (x + 7)²
∴ x² + 14x + 49 = (x + 7)²

(ii) y² – 10y + 25 = y² – 10y + 5²
Comparing with a² – 2ab + b² = (a – b)² we get a = y ; b = 5
⇒ y² – 2(y) (5) + 5² = (y – 5)²
∴ y² – 10y + 25 = (y – 5)²

(iii) c² – 4c – 12
This is of the form ax² + bx + c
Where a = 1,b = – 4 c = – 12, x = c
Now the product ac = 1 × – 12 = – 12 and the sum b = -4

(iv) m² + m – 72

This is of the form ax² + bx + c
where a = 1, b = 1, c = -12
Product a × c = 1 × -72 = -72
Sum b = 1
The middle term m can be written as 9m – 8m
m² + m – 72 = m² + 9m – 8m – 12
= m (m + 9) – 8 (m + 9)
Taking out (m + 9)
= (m + 9) (m – 8)
∴ m² + m – 72 = (m + 9) (m – 8)

(v) 4x² – 8x + 3
This is of the form ax² + bx + c with a = 4 b = -8 c = 3
Product ac = 4 × 3 = 12
Sum b = -8

Samacheer Kalvi Guru Maths 8th Question 3.
Factorize the following expressions using a³ + b³ = (a + b)(a² – ab + b²) identity
(i) h³ + k²
(ii) 2a³ + 16
(iii) x³y³ + 27
(iv) 64m³ + n³
(v) r⁴ + 27p³r
Solution:
(i) h³ + k³
Comparing h³ + k³ with a³ + b³ = (a + b)(a² – ab + b²) we have a = h, b = k
∴ h³ + k³ = (h + k)(h² – hk + k²)

(ii) 2a² + 16
(2 × a³) + (2 × 8) = 2(a³ + 8) = 2 (a³ + 2³)
∴ 2a³ + 16 = 2 (a³ + 2³)
Comparing with a³ + b³ we have a = a and b = 2
a³ + b³ = (a + b)(a² – ab + b²)
2(a³ + 2³) = 2[(a + 2) (a² – (a) (2) + 2²)] = 2[(a + 2) (a² – 2a + 4)]
2a³ +16 = 2 (a + 2) (a² – 2a + 4)

(iii) x³ y³ + 27 = (xy)³ + 3³
Comparing with a³ + b³ we have a = xy ;b = 3
a³ + b³ = (a + b) (a² – ab + b²)
(xy)³ + 3³ = (xy + 3) ((xy)² – (xy) (3) + 3²) = (xy + 3) (x²y² – 3xy + 9)
∴ x³y³ + 27 = (xy + 3)(x²y² – 3xy + 9)

(iv) 64m³ + n³ = (4³m³) + n³
= (4m)³ + n³
Comparing this with a³ + b³ we have a = 4m; b = n
a³ + b³ = (a + b) (a² – ab + b²)
(4m)³ + n³ = (4m + n) [(4m)2 – (4m) (n) + n2]
= (4m + n) [4²m² – 4mn + n²]
= (4m + n) [ 16m² – 4mn + n²]
64m³ + n³ = (4m + n) (16m² – 4mn + n²)

(v) r³ + 27p³r = r (r³ + 27p³) = r [r³ + 3³p³]
Comparing r³ + (3p)³ with a³ + b³ we have a = r; b = 3p
a³ + b³ = (a + b)(a² – ab + b²)
r[r³ + (3p)³] = r[(r + 3p)(r² – r(3p) + (3p)²)]
= r[(r+ 3p) (r² – 3rp + 3²p²]
= r(r + 3p) (r² – 3rp + 9p²)
r⁴ + 27p³r = r(r + 3p) (r² – 3rp + 9p²)

11th Maths Exercise 3.4 Samacheer Kalvi Question 4.
Factorize the following expressions using a³ – b³ = (a – b)(a² + ab + b²) identity
(i) y³ – 27
(ii) 3b³ – 192c³
(iii) -16y³ + 2x³
(iv) x³ y³ – 7³
(v) c³ – 27b³ a³
Solution:
(i) y³ – 27 = y³ – 3³
Comparing this with a³ – b³ , we have a = y and b = 3
a³ – b³ = (a + b) (a² + ab + b² )
y³ – 3³ = (y – 3)(y² + (y) (3) + 3² ) = (y – 3) (y² + 3y + 9)
y³ – 27 = (y – 3) (y² + 3y + 9)

(ii) 3b³ + 192c³ = (3 × b³) – (3 × 4 × 4 × 4 × c³) = 3(b³ – 4³ c³)
= 3(b³ – (4c)³ )
Comparing b³ – (4c)³ with a³ – b³ we have a = b and b = 4c
a³ – b³ = (a – b) (a² + ab + b²)
3(b³ – (4c)³) = 3[(b – 4c)(b² + (b)(4c) + (4c)²)]
= 3[(b – 4c)(b² + 4bc + 4² c²)]
3b³ – 192c³ = 3 [(b – 4c) (b² + 4bc + 16c²)]

(iii) -16y³ + 2x³ = 2x³ – 16y³ [∵ Addition is commatative]
= 2(x³ – 8y³) = 2(x³ – 23y³)
= 2(x³ – (2y)³)
Comparing x³ – (2y)³ with a³ – b³ we have a = x and b = 2y
a³ – b³ = (a – b)(a² + ab + b²)
2[x³ – (2y)³] = 2[(x – 2y) (x² + (x) (2y) + (2y)²)]
= 2[(x – 2y) (x² + 2xy + 2² y²)]
-16y³ + 2x³ = 2[(x – 2y)(x² + 2xy + 4y²)]

(iv) x³y³ – 7³ = (xy)³ – 7³
Comparing with a³ – b³ we have a = xy and b = 7
a³ – b³ = (a – b)(a² + ab + b²)
(xy)³ – 7³ = (xy – 7) ((xy)² + (xy) (7) + 7²)
x³y³ – 7³ = (xy – 7) (x²y² + 7xy + 49)

(v) c³ – 27 b³ a³ = c³ – 3³b³ a³ = c³ – (3ba)³
Comparing this with a³ – b³ we have a = x and b = 3ba
a³ – b³ = (a – b)(a² + ab + b²)
∴ c³ – (3ba)³ = (c – 3ba) (c² + (c) (3ba) + (3ba)²)
= (c – 3ba) (c² + 3bac + 3² b²a²)
c³ – 27b³a³ = (c – 3ab) (c² + 3bac + 9a²b²)

Students can Download Maths Chapter 3 Algebra Ex 3.3 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 1 Chapter 3 Algebra Ex 3.3

8th Maths Algebra Exercise 3.3 Question 1.
Expand
(i) (3m + 5)²
(ii) (5p – 1)²
(iii) (2n – 1)(2n + 3)
(iv) 4p² – 25q²
Solution:
(i) (3m + 5)²
Comparing (3m + 5)² with (a + b)² we have a = 3m and b = 5
(a + b)² = a² + 2 ab + b²
(3m + 5)² = (3m)² + 2 (3m) (5) + 5²
= 3²m² + 30m + 25 = 9m² + 30m +25

(ii) (5p – 1)²
Comparing (5p – 1)² with (a – b)² we have a = 5p and b = 1
(a – b)² = a² – 2ab + b²
(5p – 1)² = (5p)² – 2 (5p) (1) + 1²
= 5²p² – 10p + 1 = 25p² – 10p + 1

(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3
(x + a) (x + b) = x² + (a + b)x + ab
(2n +(- 1)) (2n + 3) = (2n)² + (-1 + 3)2n + (-1) (3)
= 2²n² + 2 (2n) – 3 = 4n² + 4n – 3

(iv) 4p² – 25q² = (2p)² – (5q)²
Comparing (2p)² – (5q)² with a² – b² we have a = 2p and b = 5q
(a² – b²) = (a + b)(a – b) = (2p + 5q) (2p – 5q)

Maths Chapter 3 Exercise 3.3 Class 8 Question 2.
Expand
(i) (3 + m)³
(ii) (2a + 5)³
(iii) (3p + 4q)³
(iv) (52)³
(v) (104)³
Solution:
(i) (3 + m)³
Comparing (3 + m)³ with (a + b)³ we have a = 3; b = m
(a + b)³ = a² + 3a²b + 3 ab² + b³
(3 + m)³ = 3³ + 3(3)² (m) + 3 (3) m² + m³
= 27 + 27m + 9m² + m³ = m³ + 9 m² + 27m + 27

(ii) (2a + 5)³
Comparing (2a + 5)³ with (a + b)³ we have a = 2a, b = 5
(a + b)³ = a³ + 3a²b + 3ab² + b³ = (2a)³ + 3(2a)² 5 + 3 (2a) 5² + 5³
= 2³a³ + 3(2²a²) 5 + 6a (25) + 125
= 8a³+ 60a² + 150a + 125

(iii) (3p + 4q)³
Comparing (3p + 4q)³ with (a + b)³ we have a = 3p and b = 4q
(a + b) ³ = a³ + 3a²b + 3ab² + b³
(3p + 4q)³ = (3p)³ + 3(3p)² (4q) + 3(3p)(4q)² + (4q)³
= 3³p³ +3 (9p²) (4q) + 9p (16q²) + 43q³
= 27p³ + 108p²q + 144pq² + 64q³

(iv) (52)³ = (50 + 2)³
Comparing (50 + 2)³ with (a + b)³ we have a = 50 and b = 2
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(50 + 2)³ = 50³ + 3 (50)²2 + 3 (50)(2)² + 2³
52³ = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
52³ = 1,40,608

(v) (104)³ = (100 + 4)³
Comparing (100 + 4)³ with (a + b)³ we have a = 100 and b = 4
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(100 + 4)³ = (100)³ + 3 (100)² (4) + 3 (100) (4)² + (4)³
= 10,00,000 + 3(10000) 4 + 300 (16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

Samacheer Kalvi 8 Maths Question 3.
Expand
(i) (5 – x)³
(ii) (2x – 4y)³
(iii) (ab – c)³
(iv) (48)³
(v) (97xy)³
Solution:
(i) (5 – x)³
Comparing (5 – x)³ with (a – b)³ we have a = 5 and b = x
(a – b)³ = a³ – 3a²b + 3ab² – b³
(5 – x)³ = 5³ – 3 (5)² (x) + 3(5)(x²) – x³
= 125 – 3(25)(x) + 15x² – x³ = 125

(ii) (2x – 4y)³
Comparing (2x – 4y)³ with (a – b)³ we have a = 2x and b = 4y
(a – b)³ = a³ – 3a²b + 3ab³ – b³
(2x – 4y)³ = (2x)³ – 3(2x)² (4y) + 3(2x) (4y)² – (4y)³
= 2³x³ – 3(2²x²) (4y) + 3(2x) (4²y²) – (4³y³)
= 8x³ – 48x²y + 96xy² – 64y³

(iii) (ab – c)³
Comparing (ab – c)³ with (a – b)³ we have a = ab and b = c
(a – b)³ = a³ – 3a²b + 3ab² – b³
(ab – c)³ = (ab)³ – 3 (ab)² c + 3 ab (c)² – c³
= a³b³ – 3(a²b²) c + 3abc² – c³
= a³b³ – 3a²b² c + 3abc² – c³

(iv) (48)³ = (50 – 2)³
Comparing (50 – 2)³ with (a – b)³ we have a = 50 and b = 2
(a – b)³ = a³ – 3a²b + 3ab² – b³
(50 – 2)³ = (50)³ – 3(50)²(2) + 3 (50)(2)² – 2³
= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592
= 1,10,592

(v) (97xy)³
= 97³ x³ y³ = (100 – 3)³ x³y³
Comparing (100 – 3)³ with (a – b)³ we have a = 100, b = 3
(a – b)³ = a³ – 3a²b + 3ab² – b³
(100 – 3)³ = (100)³ – 3(100)² (3) + 3 (100)(3)² – 3³
97³ = 10,00,000 – 90000 + 2700 – 27
97³ = 910000 + 2673
97³ = 912673
97x³y³ = 912673x³y³

Samacheer Kalvi 8th Maths Book Question 4.
Simplify (i) (5y + 1)(5y + 2)(5y + 3)
(ii) (p – 2)(p + 1)(p – 4)
Solution:
(i) (5y + 1) (5y + 2) (5y + 3)
Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² (ab + bc + ca) x + abc
= (5y)³ + (1 + 2 + 3) (5y)² + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)
= 5³y³ + 6(5²y²) + (2 + 6 + 3)5y + 6
= 125³ + 150y² + 55y + 6

(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + be + ca) x + abc
= p³ + (-2 + 1 + (-4))p² + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)
= p³ + (-5 )p² + (-2 + (-4) + 8)p + 8
= p³ – 5p² + 2p + 8

Samacheer Kalvi Guru 8th Maths Question 5.
Find the volume of the cube whose side is (x + 1) cm.
Solution:
Given side of the cube = (x + 1) cm
Volume of the cube = (side)³ cubic units = (x + 1)³ cm³
We have (a + b)³ = (a³ + 3a²b + 3ab² + b³) cm³
(x + 1)³ = (x³ + 3x² (1) + 3x (1)² + 1³) cm³
Volume = (x³ + 3x² + 3x + 1) cm³

Samacheerkalvi.Guru 8th Maths Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3).
Solution:
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units³
= (x + 2) (x – 1) (x – 3) units³
We have (x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + bc+ ca)x + abc
∴ (x + 2)(x – 1) (x – 3) = x³ + (2 – 1 – 3)x² + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)
x³ – 2x² + (-2 + 3 – 6)x + 6
Volume = x³ – 2x² – 5x + 6 units³

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 6.1 வளம் பெருகுக Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 6.1 வளம் பெருகுக

கற்பவை கற்றபின்

Question 1.
உமது பகுதியில் நடைபெறும் ஏதேனும் ஒரு தொழிலின் பல செயல்களை வரிசைப்படுத்தி எழுதுக.
Answer:
மண்பாண்டத் தொழில் :
குளங்கள், ஆற்றங்கரைகள், வயல்வெளிகள் ஆகிய இடங்களிலிருந்து களிமண்ணை எடுத்து வருவர், பெரிய பள்ளம் தோண்டி அதில் களிமண்ணை நிரப்பி தண்ணீர் ஊற்றி ஒரு நாள் முழுவதும் ஊற வைப்பர். பிறகு அதனுடன் மெல்லிய மணல் சாம்பல் ஆகியவற்றைக் கலந்து பயன்படுத்துவார்கள். பிறகு பானை செய்யும் சக்கரத்தில் வைத்து வேண்டிய வடிவங்களில் அதை உருவாக்குவார்கள். உரிய வடிவம் வந்ததும் அடிப்பகுதியில் நூல் அல்லது ஊசியால் அறுத்து எடுத்து காய வைப்பர்.

ஓரளவுகாய்ந்ததும், தட்டுப்பலகை கொண்டுதட்டி பானையின் அடிப்பகுதியில் இருக்கும் ஓட்டையை மூடி பானையை முழுமையாக்குகின்றனர். பிறகு உருட்டுக்கல் கொண்டு தேய்த்து பானையைப் பளபளபாக்குகின்றனர். பிறகு வண்ணங்களையும், ஓவியங்களையும் தகுந்தாற்போல வரைகின்றனர்.

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
தோட்டத்தில் தம்பி ஊன்றிய ……………………. எல்லாம் முளைத்தன.
அ) சத்துகள்
ஆ) பித்துகள்
இ) முத்துகள்
ஈ) வித்துகள்
Answer:
ஈ) வித்துகள்

Question 2.
என் நண்பன் செய்த தொழிலில் அவனுக்கு …………………… பெருகிற்று.
அ) காரி
ஆ) ஓரி
இ) வாரி
ஈ) பாரி
Answer:
இ) வாரி

Question 3.
‘அக்களத்து’ என்ற சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………….
அ) அ + களத்து
ஆ) அக் + களத்து
இ) அக்க + அளத்து
ஈ) அம் + களத்து
Answer:
அ) அ + களத்து

Question 4.
கதிர் + ஈன என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ………………….
அ) கதிரென
ஆ) கதியீன
இ) கதிரீன
ஈ) கதிரின்ன
Answer:
இ) கதிரீன

குறுவினா

Question 1.
பயிர்கள் வாட்டமின்றி கிளைத்து வளரத் தேவையானது யாது?
Answer:
தகுந்த காலத்தில் பெய்யும் மழையே பயிர்கள் வாட்டமின்றி கிளைத்து வளரத் தேவையானது ஆகும்.

Question 2.
உழவர்கள் எப்போது ஆரவார ஒலி எழுப்புவர்?
Answer:
நெற்போரினை அடித்து நெல்லினைக் கொள்ளும் (எடுக்கும்) காலத்தில் உழவர்கள் ஆரவார ஒலி எழுப்புவர்.

சிறு வினா

Question 1.
உழவுத் தொழில் பற்றித் தகடூர் யாத்திரை கூறுவன யாவை?
Answer:

• சேரனின் நாட்டில் பெருகிய மழைநீரால் வருவாய் சிறந்து விளங்குகிறது.
• அகலமான நிலப்பகுதியில் விதைகள் குறைவின்றி முளை விடுகின்றன.
• முளைத்த விதைகள் செழிப்புடன் வளர தட்டுப்பாடின்றி மழை பொழிகின்றது.
• தகுந்த காலத்தில் மழை பொழிவதால் பயிர்கள் வாட்டம் இன்றி கிளைத்து வளர்கிறது. செழித்த பயிர்கள் பால் முற்றிக் கதிர்களைப் பெற்றிருக்கின்றன.
• அக்கதிர்கள் அறுவடை செய்யப் பெற்று ஏரினால் வளம் சிறக்கும் செல்வர்களின் களத்தில் நெற்போர் காவல் இல்லாமலே இருக்கின்றது.
• நெற்போரினை அடித்து நெல்லினைக் கொள்ளும் (எடுக்கும்) காலத்தில் உழவர்கள் எழுப்பும் ஆரவார ஒலியால் நாரை இனங்கள் அஞ்சித் தம் பெண் பறவைகளோடு பிரிந்து செல்லும் சிறப்புடைய, சேர மன்னரின் அகன்ற பெரிய நாடு புது வருவாயுடன் சிறந்து விளங்குகின்றது.

சிந்தனை வினா

Question 1.
உழவுத் தொழில் சிறக்க இன்றியமையாதனவாக நீங்கள் கருதுவன யாவை?
Answer:
உழவுத் தொழில் உயிர் தொழில்

நாகரீகம் என்ற பெயரில் இன்று யாரும் உழவுத் தொழில் செய்ய முன்வருவதில்லை. ஒவ்வொரு வீட்டிலும் உள்ள ஒருவர் கட்டாயம் உழவுத் தொழில் செய்தல் வேண்டும். உழவுத் தொழில், அரசுப் பணிகளில் ஒன்றாகச் சேர்க்கப்பட வேண்டும் உழவுத் தொழிலில் சிறந்து விளங்கும் உழவர்களுக்கு ஆண்டுதோறும் விருதுகளும் பரிசுத் தொகையும் கொடுக்க வேண்டும். இன்றைய இளைஞர்கள் வேலை விருப்பப் பட்டியலில் உழவுத்தொழிலைச் சேர்த்துக் கொள்ள வேண்டும். இவ்வாறு செய்தால் மட்டுமே உழவுத் தொழில் நிச்சயம் சிறக்கும்.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1. ம
ன்பதை காக்கும் மாபெரும் சிறப்பு …………………… க்கு உண்டு .
அ) மழை
ஆ) வயல்
இ) நிலம்
ஈ) உழவர்
Answer:
அ) மழை

Question 2.
தகடூர் யாத்திரை பாடல் பேசும் தொழில் ……………………..
அ) உழவுத் தொழில்
ஆ) நெய்தல் தொழில்
இ) மீன்பிடித் தொழில்
ஈ) மண்பாண்டத் தொழில்
Answer:
அ) உழவுத் தொழில்

Question 3.
புது வருவாய் என்னும் பொருளினைக் குறிக்கும் சொல் …………………
அ) வாரி
ஆ) எஞ்சாமை
இ) ஒட்டாது
ஈ) யாணர்
Answer:
ஈ) யாணர்

Question 4.
வளம் பெருக பாடல் ………………….. மன்னர் பற்றியது.
அ) சோழர்
ஆ) சேரர்
இ) பாண்டியர்
ஈ) பல்ல வர்
Answer:
ஆ) சேரர்

Question 5.
தர்மபுரியின் பழைய பெயர் ……………………..
அ) மாமண்டூர்
ஆ) வடுவூர்
இ) தகடூர்
ஈ) குரும்பூர்
Answer:
இ) தகடூர்

குறுவினா

Question 1.
சேரநாட்டில் வருவாய் சிறந்து விளங்கக் காரணம் யாது?
Answer:
பெருகிய மழை நீரால் சேர நாட்டின் வருவாய் சிறந்து விளங்குகிறது.

Question 2.
அறுவடை செய்யப்பட்ட கதிர்கள் எங்கு நிறைகின்றன?
Answer:
அறுவடை செய்யப்பட்ட கதிர்கள் ஏறினால் வளம் சிறக்கும் செல்வர்களின் களத்தில் வந்து நிறைகின்றன.

Question 3.
நாரை இனங்கள் பெண்பாற் பறவைகளோடு பிரிந்து செல்லக் காரணம் யாது?
Answer:
நெற்போரினை அடித்து நெல்லினைக் கொள்ளும் (எடுக்கும்) காலத்தில் உழவர்கள் எழுப்பும் ஆரவார ஒலியால், நாரை இனங்கள் தன் பெண்பாற் பறவைகளோடு பிரிந்து செல்கின்றன.

சொல்லும் பொருளும்

வாரி – வருவாய்
எஞ்சாமை – குறைவின்றி
முட்டாது – தட்டுப்பாடின்றி
ஒட்டாது – வாட்டம் இன்றி
வைகுக – தங்குக
ஓதை – ஓசை
வெரீஇ – அஞ்சி
யாணர் -புது வருவாய்