Category: Class 8

Students can Download Maths Chapter 3 Geometry Ex 3.1 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 3 Chapter 3 Geometry Ex 3.1

Question 1.
Fill in the blanks:

1. The altitudes of a triangle intersect at………..
2. The medians of a triangle cross each other at………..
3. The meeting point of the angle bisectors of a triangle is………..
4. The perpendicular bisectors of the sides a triangle meet at………..
5. The centroid of a triangle divides each medians in the ratio………..

Solution:

1. Orthocentre
2. Centroid
3. Incentre
4. Circumcentre
5. 2 : 1

Question 2.
Say True or False:
(i) In any triangle the Centroid and the Incentre are located inside the triangle.
(ii) The centroid, orthocentre, and incentre of a triangle are collinear.
(iii) The incentre is equidistant from all the vertices of a triangle.
Solution:
(i) True
(ii) True
(iii) False

Question 3.
(a) Where does the circumcentre lie in the case of
(i) An acute angled triangle.
Solution:
Inside the triangle.

(ii) An obtuse-angled triangle.
Solution:
Exterior of the triangle.

(iii) A right angled triangle.
Solution:
On the hypotenuse.

(b) Where does the orthocentre lie in the case of
(i) An acute-angled triangle.
Solution:
Interior of the triangle.

(ii) An obtuse-angled triangle.
Solution:
Exterior of the triangle.

(iii) A right angled triangle.
Solution:
On the vertex containing 90°.

Question 4.
Fill in the blanks:
In the triangle ABC,

(i) The angle bisector is……….
(ii) The altitude is………..
(iii) The median is…………
Solution:
(i) BE
(ii) AD
(iii) CF

Question 5.
In right triangle ABC, what is the length of altitude drawn from the vertex A to BC?

Solution:
In this right angled triangle ΔABC, length of the altitude drawn from vertex A is the leg AB itself. By Pythagoras theorem.
AC² = AB² + BC²
13² = AB² + 12²
169 = AB² + 144
AB² = 169 – 144 = 25
AB² = 52
AB = 5cm

Question 6.
In triangle XYZ, YM is the angle bisector of ∠Y and ∠Y is 100°. Find ∠XYM and ∠ZYM.

Solution:
Given YM is the angle bisector of ∠Y.
also ∠Y = 100°
Angle -bisector divides the angle into two congruent angles.
∠XYM = ∠ZYM
∠Y = ∠XYM + ∠ZYM
100° = ∠XYM + ∠ZYM [∴ ∠XYM = ∠ZYM]
2 ∠XYM = 100°
∠XYM = \(\frac{1}{2}\) (100°)
∠XYM = 50°
∴ ∠ZYM = 50°

Question 7.
In triangle PQR, PS is a median and QS = 3.5 cm, then find QR?

Solution:
Given PS is the median and QS = 3.5 cm
Median is the line drawn from a vertex to the midpoint of the opposite side.
∴ QS = RS
so QS = RS = 3.5 cm
∴ QR = QS +SR = 3.5 + 3.5 = 7 cm
QR = 7 cm

Question 8.
In triangle ABC, line is a perpendicular bisector of BC. If BC = 12 cm, SM = 8 cm, find CS.

Solution:
Given l₁ is the perpendicular bisector of BC.
∴ ∠SMC = 90° and BM = MC
BC = 12 cm
⇒ BM + MC = 12 cm
MC + MC = 12 cm [∴ BM = MC]
2MC = 12
MC = \(\frac{12}{2}\)
MC = 6 cm
Given SM = 8 cm
By Pythagoras theorem SC² = SM² + MC²
SC² = 8² + 6²
SC² = 64 + 36
CS² = 100
CS² = 10²
CS = 10 cm

Students can Download Maths Chapter 2 Life Mathematics Ex 2.2 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 3 Chapter 2 Life Mathematics Ex 2.2

Miscellaneous and Practice Problems

Question 1.
5 boys or 3 girls can do a science project in 40 days. How long will it take for 15 boys and 6 girls to do the same project?
Solution:
Let B and G denote Boys and Girls respectively.
Given 5B = 3G ⇒ 1B = \(\frac{3}{5}\)G
now 15B + 6G = 15 × \(\frac{3}{5}\) G + 6G = 9G + 6G = 15G
If 3 girls can do the project in 40 days then 15 girls can do it in
3G × 40 ÷ 15G = 3G × 40 × \(\frac{1}{15G}\) = \(\frac{40}{5}\)
= 8 days.
∴ 15 boys and 6 girls can complete the project in 8 days.

Question 2.
If 32 men working 12 hours a day can do a work in 15 days, how many men working 10 hours a day can do double that work in 24 days?
Solution:
Let the required number of men be x.

Let P₁= 32, H₁ = 12, D₁ = 12, W₁ = 1
P₂ = x, H₂ = 10, D₂ = 24, W₂ = 1

x = 24 persons
To complete the same work 24 men needed.
To complete double the work 24 × 2 = 48 men are required.

Question 3.
Amutha can weave a saree in 18 days. Anjali is twice as good a weaver as Amutha. If both of them weave together, in how many days can they complete weaving the saree?
Solution:
Amutha can weave a saree in 18 days. Anjali is twice as good as Amutha.
ie. If Amutha weave for 2 days, Anjali do the same work in 1 day.
If Anjali weave the saree she will take
\(\frac{18}{2}\) = 9 days
Hence time taken by them together = \(\frac{ab}{a + b}\) days = \(\frac{18 × 9}{18 + 9}\) = \(\frac{18 × 9}{27}\) = 6 days

In 6 days they complete weaving the saree.

Question 4.
A, B and C can complete a work in 5 days. If A and C can complete the same work in 7½ days and A alone in 15 days then, in how many days can B and G finish the work?
Solution:
A + B + C complete the work in 5 days.
∴ (A + B + C)’s 1 day work = \(\frac{1}{5}\)
(A + C) complete the work in 7 \(\frac{1}{2}\) days = \(\frac{15}{2}\) days
∴ (A + C)’s 1 day work = \(\frac{1}{\frac{15}{2}}\) = \(\frac{2}{15}\)
∴ B’s 1 day work = (A + B + C)’s 1 day work – (A + C)’s 1 day work.
\(\frac{1}{5}\) – \(\frac{2}{15}\) = \(\frac{3}{15}\) – \(\frac{2}{15}\)
C’s 1 day work = (A + C)’s 1 day work – A’s 1 day work
= \(\frac{2}{5}\) – \(\frac{1}{15}\) = \(\frac{1}{15}\)
Now (A + C)’s 1 day work = B’s 1 day work + C’s 1 day work
= \(\frac{1}{15}\) + \(\frac{1}{15}\) = \(\frac{2}{15}\)
∴ (B + C) can complete the work in \(\frac{1}{\frac{2}{15}}\) days. = \(\frac{15}{2}\) days = 7\(\frac{1}{2}\) days
∴B and C finish the work in 7\(\frac{1}{2}\) days.

Question 5.
P and Q can do a piece of work in 12 days and 15 days respectively. P started the work alone and then, after 3 days Q joined him till the work was completed. How long did the work last?
Solution:
p can do a piece of work in 12 days.
∴ p’s 1 day work = \(\frac{1}{12}\)
p’s 1 day work = 3 × \(\frac{1}{12}\) = \(\frac{3}{12}\)
Q can do a piece of work in 15 days.
∴ Q’s 1 day work = \(\frac{1}{15}\)
Remaining work after 3 days = 1 – \(\frac{3}{12}\) = \(\frac{9}{12}\)
(P + Q)’s 1 day work = \(\frac{1}{12}\) + \(\frac{1}{15}\) = \(\frac{5}{60}\) + \(\frac{4}{60}\) = \(\frac{9}{60}\)
Number of days required to finish the remaining work
= \(\frac{Remaining work}{(P + Q)’s 1 day work}\) = \(\frac{\frac{9}{12}}{\frac{9}{60}}\) = \(\frac{9}{12}\) × \(\frac{60}{9}\) = 5
Remaining work lasts for 5 days. Total work lasts for 3 + 5 = 8 days.

Challenging Problems

Question 6.
A camp had provisions for 490 soldiers for 65 days. After 15 days, more soldiers arrived and the remaining provisions lasted for 35 days. How many soldiers joined the camp?
Solution:

Now as the soldiers increases food last for less days.
∴ It is inverse proportion.
The proportion is (490 + x): 490 : : 50 : 35
Product of the extremes = Product of the means
(490 + x) × 35 = 490 × 50
(490 + x) = \(\frac{490 × 50}{35}\)

x = 700 – 490
x = 210
∴ 210 soldiers joined the camp.

Question 7.
A small – scale company undertakes an agreement to produce 540 motor pumps in 150 days and employs 40 men for the work. After 75 days, the company could make only 180 motor pumps. How many more men should the company employ so that the work is completed on time as per the agreement?
Solution:
Let the number of men to be appointed more be x.

To produce more pumps more men required
∴ It is direct variation.
∴ The multiplying factor is \(\frac{360}{180}\)
More days means less employees needed.
∴ It is Indirect proportion.
∴ The multiplying factor is \(\frac{75}{75}\)
Now 40 + x = 40 × \(\frac{360}{180}\) × \(\frac{75}{75}\)
40 + x = 80
x = 80 – 40
x = 40
40 more man should be employed to complete the work on time as per the agreement.

Question 8.
A can do a work in 45 days. He works at it for 15 days and then, B alone finishes the remaining work in 24 days. Find the time taken to complete 80% of the work if they work together.
Solution:
A can do a work in 45 days.
A’s 1 day work = \(\frac{1}{45}\)
∴ A’s 15 days work = 15 × \(\frac{1}{45}\) = \(\frac{1}{3}\)
Remaining work = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
B alone completes the remaining \(\frac{2}{3}\) work in 24 days
∴ B completes the whole work in \(\frac{24}{\frac{2}{3}}\) days = 24 × \(\frac{3}{2}\) = 36 days.
∴ B’s 1 day’s work = \(\frac{1}{36}\)
∴ (A + B)’s together complete the work in \(\frac{ab}{a + b}\) days = \(\frac{45 × 36}{45 + 36}\) = \(\frac{45 × 36}{81}\)

Whole wok will be completed by (A + B) in = 20 days.
∴ 80% of the work will be completed in \(\frac{80 × 20}{100}\) = 16 days.

Question 9.
P alone can do \(\frac{1}{2}\) of a work in 6 days and Q alone can do \(\frac{2}{3}\) of the same work in 4 days. In how many days working together, will they finish \(\frac{3}{4}\) of the work?
Solution:
\(\frac{1}{2}\) of the work is done by P in 6 days
∴ Full work is done by P in \(\frac{6}{\frac{1}{2}}\) = 6 × 2 = 12 days
\(\frac{2}{3}\) of work done by Q in 4 days.
∴ Full work done by Q in \(\frac{4}{\frac{2}{3}}\) = 4 × \(\frac{3}{2}\) = 6 days
(P + Q) will finish the whole work in \(\frac{ab}{a + b}\) days = \(\frac{12 × 6}{12 + 6}\) = \(\frac{12 × 6}{18}\) = 4 days
(P + Q) will finish \(\frac{3}{4}\) of the work in 4 × \(\frac{3}{4}\) = 3 days.

Question 10.
X alone can do a piece of work in 6 days and Y alone in 8 days. X and Y undertook the work for Rs 4800. With the help of Z, they completed the work in 3 days. How much is Z’s share?
Solution:
X can do the work in 6 days.
X’s 1 day work = \(\frac{1}{6}\)
X’s share for 1 day = \(\frac{1}{6}\) × 48000 = Rs 800
X’s share for 3 days = 3 × 800 = 2400
Y can complete the work in 8 days.
Y’s 1 day work = \(\frac{1}{8}\)
Y’s 1 day share = \(\frac{1}{8}\) × 4800 = 600
Y’s 3 days share = 600 × 3 = 1800
(X + Y)’s 3 days share = 2400 + 1800 = 4200
Remaining money is Z’s share
∴ Z’s share = 4800 – 4200 = 600

Students can Download Maths Chapter 1 Numbers Ex 1.5 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 3 Chapter 1 Numbers Ex 1.5

Miscellaneous and Practice Problems

Question 1.
A square carpet covers an area of 1024 m² of a big hall. It is placed in the middle of the hall. What is the length of a side of the carpet?
Solution:
Area of the carpet = 1024 m²
side × side = 1024 m²
(side)² = 1024 m²

(side)² = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2²
= (2 × 2 × 2 × 2 × 2)²
(side)² = 32²
side = 32
Length of a side of the carpet = 32 m

Question 2.
There is a large square portrait of a leader that covers an area of 4489 cm². If each side has a 2 cm liner, what would be its area?
Solution:
Area of the square = 4489 cm²
(side)² = 4489 cm²

(side)² = 67 x 67
side² = 67²
Length of a side = 67

Length of a side Length of a side with liner = 67 + 2 + 2 cm = 71 cm
Area of the larger square = 71 x 71 cm²
= 5041 cm²
Area of the liner = Area of big square-Area of small square
= (5041 – 4489) cm² = 552 cm²

Question 3.
2401 plants are planted in a garden such that each contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Given number of plants in a row = Number of rows.
Number of rows × number of plants in a row = Total plants
Total plants = 2401

= 7 × 7 × 7 × 7 = 7² × 7²
= 49 × 49
∴ number of rows = 49
number of plants in a row = 49

Question 4.
If \(\sqrt[3]{1906624} \times \sqrt{x}\) = 3100, find x.
Solution:

\(\sqrt{x}\) = 25
Squaring on both sides \((\sqrt{x})^{2}\) = 25²
x = 625

Question 5.
If (625)^x = 15625, find x² and x³
Solution:
(625)^x = 15625

(5 x 5 x 5 x 5)^x = 5 x 5 x 5 x 5 x 5 x 5
(5⁴)^x = 5⁶
5^4x = 5⁶
5^4x = 5⁶
Comparing the powers of 5 both sides
4x = 6
x = \(\frac{6}{4}\)
x = \(\frac{3}{2}\)

Question 6.
If 2^m-1 + 2^m+1 = 640, then find ‘m’
Solution:
Given 2^m-1 + 2^m+1 = 640
2^m-1 + 2^m+1 = 128 + 512 [consecutive powers of 2]
2^m-1 + 2^m+1 = 2⁷+ 2⁹ [powers of 2: 2, 4, 8, 16, 32, 64, 128, 256, 512, …..]
m – 1 = 7
m = 7 + 1
m = 8

Question 7.
Simplify \(\frac{16 \times 10^{2} \times 64}{4^{2} \times 2^{4}}\)
Solution:

\(\frac{16 \times 10^{2} \times 64}{4^{2} \times 2^{4}}\) = \(\frac{2^{4} \times 10^{2} \times 2^{6}}{\left(2^{2}\right)^{2} \times 2^{4}}=\frac{2^{4+6} \times 10^{2}}{2^{4} \times 2^{4}}=\frac{2^{10} \times 100}{2^{8}}\)
= 2^10-8 × 100= 2² × 100 = 400

Question 8.
Give the answer in scientific notation:
A human heart beats at an average of 80 beats per minute. How many times does it beat in
(i) an hour?
(ii) a day?
(iii) a year?
(iv) 100 years?
Solution:
Heart beat per minute = 80 beats
(i) an hour One hour = 60 minutes
Heart beat in an hour = 60 x 80 = 4800 = 4.8 x 10³

(ii) In a day
One day = 24 hours = 24 x 60 minutes
∴ Heart beat in one day = 24 x 60 x 80 = 24 x 4800 = 115200 = 1.152 x 10⁵

(iii) a year
One year = 365 days = 365 x 24 hours = 365 x 24 x 60 minutes
∴ Heart beats in a year = 365 x 24 x 60 x 80 = 42048000 = 4.2048 x 10⁷

(iv) 100 years
Heart beats in one year = 4.2048 x 10⁷
Heart beats in 100 years = 4.2048 x 10⁷ x 100 = 4.2048 x 10⁷ x 10²
= 4.2048 x 10⁹

Challenging Problems

Question 9.
A greeting card has an area 90 cm². Between what two whole numbers is the length of its side?
Solution:
Area of the greeting card = 90 cm²
(side)² = 90 cm²

(side)² = 2 x 5 x 3 x 3 = 2 x 5 x 3²
\(\sqrt{({side})^{2}}=\sqrt{2 \times 5 \times 3^{2}}\)

side = 3 \(\sqrt{2 × 5}\)
side = \(\sqrt{10}\) cm
side = 3 × 3.2 cm
side = 9.6 cm
∴ Side lies between the whole numbers 9 and 10.

Question 10.
225 square shaped mosaic tiles, each of area 1 square decimetre exactly cover a square shaped verandah. How long is each side of the square shaped verandah?
Solution:
Area of one tile = 1 sq. decimeter
Area of 225 tiles = 225 sq.decimeter
225 square tiles exactly covers the square shaped verandah.
∴ Area of 225 tiles = Area of the verandah
Area of the verandah = 225 sq.decimeter
side x side = 15 x 15 sq.decimeter
side = 15 decimeters
Length of each side of verandah = 15 decimeters.

Question 11.
A group of 1536 cadets wanted to have a parade forming a square design. Is it possible? If it is not possible how many more cadets would be required?
Solution:
Number of cadets to form square design

1536 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
The numbers 2 and 3 are unpaired
It is impossible to have the parade forming square design with 1536 cadets.

39 x 39 = 1521
Also 40 x 40 = 1600
∴ We have to add (1600 – 1536) = 64 to make 1536 a perfect square.
∴ 64 more cadets would be required to form the square design.

Question 12.
Find the decimal fraction which when multiplied by itself gives 176.252176.
Solution:
We will find the square root of 176.252176

176.252176 = 13.276 x 13.276
∴ The required number is 13.276

Question 13.
Evaluate \(\sqrt{286225}\) and use it to compute \(\sqrt{2862.25}\) + \(\sqrt{28.6225}\)
Solution:

Question 14.
The speed of light in glass is about 2 x 10⁸ m/sec. Use the formula, time = \(\frac{distence}{speed}\) to find the time (in hours) for a pulse of light to travel 7200 km in glass.
Solution:

Required time = 10^-5 hours

Question 15.
Simplify : (3.769 x 10⁵) + (4.21 x 10⁵)
Solution:

(3.769 x 10⁵) + (4.21 x 10⁵) = 3,76,900 + 4,21,00
= 7,97,900 = 7.979 x 10⁵

Question 16.
Order the following from the least to the greatest: 1625, 8100, 3500, 4400, 2600
Solution:
16²⁵ = (2⁴)²⁵ = 2¹⁰⁰
8¹⁰⁰ = (2³)¹⁰⁰ = 2³⁰⁰
4⁴⁰⁰ = (2²)⁴⁰⁰ = 2⁸⁰⁰
2⁶⁰⁰ = 2⁶⁰⁰
Comparing the powers we have, 2¹⁰⁰ < 2³⁰⁰ < 2⁶⁰⁰< 2⁸⁰⁰
∴ The required order : 16²⁵, 8¹⁰⁰, 2⁶⁰⁰, 3⁵⁰⁰, 4⁴⁰⁰

Students can Download Maths Chapter 1 Numbers Ex 1.3 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 3 Chapter 1 Numbers Ex 1.3

Question 1.
Fill in the blanks:
(i) The ones digits in the cube of 73 is……….
(ii) The maximum number of digits in the cube of a two digit number is……….
(iii) The cube root of 540 × 50 is………..
(iv) The cube root of 0.000004913 is………..
(v) The number to be added to 3333 to make it a perfect cube is………….
Solution:
(i) 7
(ii) 6
(iii) 30
(iv) 0.017
(v) 42

Question 2.
Say True or False:
(i) The cube of 0.0012 is 0.000001728.
(ii) The cube root of 250047 is 63.
(iii) 79570 is not a perfect cube.
Solution:
(i) false
(ii) true
(iii) true

Question 3.
Show that 1944 is not a perfect cube.
Solution:

1994 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2³ × 3³ × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.

Question 4.
Find the cube root of 24 × 36 × 80 × 25.
Solution:

Question 5.
Find the smallest number by which 10985 should be divided so that the quotient is a perfect cube.
Solution:

We have 10985 = 5 × 13 × 13 × 13
= 5 × 13 × 13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.

Question 6.
Find two smallest perfect square numbers which when multiplied together gives a perfect cube number.
Solution:
Consider the numbers 22 and 42
The numbers are 4 and 16.
Their product 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16

Question 7.
If the cube of a squared number is 729, find the square root of that number.
Solution:
729 = 3 × 3 × 3 × 3 × 3 × 3
(729)^1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]

we have to find out √3, √3 = 1.732

Question 8.
What is the square root of cube root of 46656?
Solution:
We have to find out \(\sqrt{(\sqrt[3]{46656})}\)
First we will find \(\sqrt[3]{46656}\)

\(\sqrt[3]{46656}\) = (2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3)^1/3
\(\sqrt[3]{46656}\) = 2 × 2 × 3 × 3
\(\sqrt[3]{46656}\) = 2² × 3² = 36
Now \(\sqrt{(\sqrt[3]{46656})}\) = \(\sqrt{36}\) = \(\sqrt{2^2×3^2 }\) = 2 × 3 = 6
∴ The required is 6.

Question 9.
Find the cube root of 729 and 6859 by prime factorisation.
Solution:

(i) \(\sqrt[3]{729}\) = \(\sqrt[3]{3 × 3 × 3 × 3 × 3 × 3}\)
= 3 × 3
\(\sqrt[3]{729}\) = 9

(ii) \(\sqrt[3]{6859}\) = \(\sqrt[3]{19 × 19 × 19 }\)
\(\sqrt[3]{6859}\) = 19

Question 10.
Find the smallest number by which 200 should be multiplied to make it a perfect cube.
Solution:
We find 200 = 2 × 2 × 2 × 5 × 5
Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.

200 × 5 = (2 × 2 × 2 × 5 × 5) × 5
1000 = 2 × 2 × 2 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.

Students can Download Maths Chapter 2 Life Mathematics Ex 2.1 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 3 Chapter 2 Life Mathematics Ex 2.1

Question 1.
Fill in the blanks:
(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job together is………..days.
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in………..days.
(iii) A can do a work in 24 days. A and B together can finish the work in 6 days. Then B alone can finish the work in…………days.
(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in………..days.
(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for Rs 200000. The amount that A will get is………
Solution:
(i) 2 days
(ii) 5
(iii) 8
(iv) 25
(v) Rs 1,20,000

Question 2.
210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Solution:
Let the required number of men be x.

More working hours ⇒ less men required.
∴ It is inverse proportion.
∴ Multiplying factor is \(\frac{12}{14}\)
Also more number of days ⇒ less men
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{18}{20}\)
∴ x = 210 × \(\frac{12}{14}\) × \(\frac{18}{20}\)= 162 men

162 men are required.

Question 3.
A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?
Solution:
Let the required number of cement bags be x.

Number of days more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{18}{12}\)
Number of machines more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{24}{36}\)
∴ x = 7000 × \(\frac{18}{12}\) × \(\frac{24}{36}\)

x = 7000 cement bags
7000 cement bags can be made.

Question 4.
A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 hours more a day?
Solution:
Let the required number of days be x.

To produce more soaps more days required.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{14400}{9600}\)
If more hours spend, less days required.
∴ It is indirect proportion.
∴ Multiplying factor = \(\frac{15}{18}\)
∴ x = 6 × \(\frac{14400}{9600}\) × \(\frac{15}{18}\)

x = \(\frac{15}{2}\)
\(\frac{15}{2}\) days will be needed.

Question 5.
If 6 container lorries transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?
Solution:
Let the number of lorries required more = x.

As the goods are more ⇒ More lorries are needed to transport.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{180}{135}\)
Again if more days ⇒ less number of lorries enough.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{5}{4}\)
∴ 6 + x = 6 × \(\frac{180}{135}\) × \(\frac{5}{4}\)

6 + x = 10
x = 10 – 6
x = 4
∴ 4 more lorries are required.

Question 6.
A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?
Solution:
Time taken by A to complete the work =12 hrs.
∴ A’s 1 hr work = \(\frac{1}{12}\)…………(1)
(B + C) complete the work in 3 hrs.
∴ (B + C)’s 1 hour work = \(\frac{1}{3}\)…………(2)
(1) + (2) ⇒
∴ (A + B + C)’s 1 hour work = \(\frac{1}{12}\) + \(\frac{1}{3}\) = \(\frac{1+4}{12}\) = \(\frac{5}{12}\)
Now (A + C) complete the work in 6 hrs.
∴(A + C)’s 1 hour work = \(\frac{1}{6}\)
∴ B’s 1 hour work = (A + B + C)’s 1 hour work – (A + C)’s 1 hour work
= \(\frac{5}{12}\) – \(\frac{1}{6}\) = \(\frac{5-2}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
∴ B alone take 4 days to complete the work.

Question 7.
A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?
Solution:
(A + B) complete the work in 12 days.
∴ (A + B)’s 1 day work = \(\frac{1}{12}\)……….(1)
(B + C) complete the work in 15 days
∴ (B + C)’s 1 day work = \(\frac{1}{15}\)……….(2)
(A + C) complete the work in 20 days
∴ (A + C)’s 1 day work = \(\frac{1}{20}\)……….(3)
Now (1) + (2) + (3) ⇒
[(A + B) + (B + C) + (A + C)]’s 1 day work = \(\frac{1}{12}\) + \(\frac{1}{15}\) + \(\frac{1}{20}\)

(2A + 2B + 2C)’s 1 day work = \(\frac{5}{60}\) + \(\frac{4}{60}\) + \(\frac{3}{60}\)
2(A + B + C)’s 1 day work = \(\frac{5+4+3}{60}\)
(A + B + C)’s 1 day work = \(\frac{12}{60×2}\)
(A + B + C)’s 1 day work = \(\frac{1}{10}\)
Now A’s 1 day’s work = (A + B + C)’s 1 day work – (B + C)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{15}\) = \(\frac{3}{30}\) – \(\frac{2}{30}\) = \(\frac{1}{30}\)
∴ A takes 30 days to complete the work.
B’s 1 day work = (A + B + C)’s 1 day’s work – (A + C)’s 1 day’s work
= \(\frac{1}{10}\) – \(\frac{1}{20}\) = \(\frac{6}{60}\) – \(\frac{3}{60}\)
\(\frac{6-3}{60}\) = \(\frac{3}{60}\) = \(\frac{1}{20}\)
B takes 20 days to complete the work.
C’s 1 day work = (A + B + C)’s 1 day work – (A + B)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{12}\) = \(\frac{6}{60}\) – \(\frac{5}{60}\) = \(\frac{6-5}{60}\) = \(\frac{1}{60}\)
∴ C takes 60 days to complete the work.

Question 8.
Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 more minutes than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Solution:
Time taken by A to fit a chair = 15 minutes
Time taken by B = 3 minutes more than A
= 15 + 3 = 18 minutes
∴ A’s 1 minute work = \(\frac{1}{15}\)
B’s 1 minute work = \(\frac{1}{18}\)
(A + B)’s 1 minutes work = \(\frac{1}{15}\) + \(\frac{1}{18}\)

\(\frac{12}{180}\) + \(\frac{22}{180}\) = \(\frac{22}{180}\) = \(\frac{11}{90}\)
∴ Time taken by (A + B) to fit a chair
= \(\frac{1}{\frac{11}{90}}\) = \(\frac{90}{11}\) minutes
∴ Time taken by (A + B) to fit 22 chairs
= \(\frac{90}{11}\) × 22 = 180 minutes
= \(\frac{180}{60}\) = 3 hours

Question 9.
A man takes 10 days to finish a job where as a woman takes 6 days to finish the same job. Together they worked for 3 days and then the woman left. In how many days will the man complete the remaining job?
Solution:
Man can finish the work in 10 days and women can finish the same work in 6 days.
∴ Man’s 1 day work = \(\frac{1}{10}\)
Woman’s 1 day work = \(\frac{1}{6}\)
(Man + Woman)s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{6}\) = \(\frac{6}{60}\) + \(\frac{10}{60}\) = \(\frac{16}{60}\)
(Man + Woman)s 3 days work

In 3 days \(\frac{4}{5}\) th of the whole work is completed.
Remaining work = 1 – \(\frac{4}{5}\) = \(\frac{5}{5}\) – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Complete work is done by the man by 10 days
∴ \(\frac{1}{5}\) of the work is done by man in \(\frac{1}{5}\) × 10 = 2 days.

Question 10.
A is thrice as fast as B. If B can do a piece of work in 24 days then, find the number of days they will take to complete the work together.
Solution:
If B does the work in 3 days, A will do it in 1 day.
B complete the work in 24 days.
∴ A complete the same work in \(\frac{24}{3}\) = 8 days.
∴ (A + B) complete the work in \(\frac{ab}{a+b}\) days = \(\frac{24×8}{24+8}\) days = \(\frac{24×8}{32}\)days = 6 days

They together complete the work in 6 days.

Students can Download Maths Chapter 1 Life Mathematics 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 1 Life Mathematics Intext Questions

Exercise 1.1
Try This (Text book Page no. 1)

Question 1.
Find the indicated percentage value of the given numbers
Solution:

Think (Text book page no. 6)

Question 1.
An increase from 200 to 600 ¡s clearly a 200%increase. Isn’t it? (check!). With a lot of pride, the traffic police commiuêener of a city reported that the accidents had decreased by 200% ¡n one year. He cameupwith this number stating that the accidents had gone down from 600 last year to 2this year. Is the decrease from 600 to 200, the same 200% as above? Justify.
Solution:
Increase from original value 200 to 600

Decrease from original value 600 to 200

here original value is 600
% decrease = \(\frac{600-200}{600}\) x 100 = \(\frac{400}{600}\) x 100 = 66.67% decrease
Increase from 200 → 600 and % decrease from 600 → 200 are not the same

Try This (Text book page no. 7)

Question 1.
What percent of a day is 10 hours?
Solution:
In a day, there are 24 hours
10 hrs out of 24 hrs is \(\frac{10}{24}\)
As a percentage, we need to multiply by 100
∴ Percentage = \(\frac{10}{24}\) x 100 = 41.67%

Question 2.
Divide ₹ 350 among P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.
Solution:
Let R get x, Q gets 50% of what R gets
∴ Q gets = \(\frac{50}{100}\) × x = \(\frac{x}{2}\)
P gets 50% of what Q gets
∴ P gets = \(\frac{50}{100}\) x \(\frac{x}{2}\) = \(\frac{x}{4}\)
Since 350 is divided among the three
∴ 350 = x + \(\frac{x}{2}\) + \(\frac{x}{4}\)
350 = \(\frac{4x+2x+x}{4}\) = \(\frac{7x}{4}\) = 350
x = \(\frac{350×4}{7}\) = 200
Q gets = \(\frac{x}{2}\) = \(\frac{200}{2}\) = 100,
P gets = \(\frac{x}{4}\) = \(\frac{200}{4}\) = 50
∴ P = 50
Q = 100
R = 200

Exercise 1.2
Think (Text book Page No. 13)

Question 1.
A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?
Solution:
Let cost price of marker board be 100
CP = 100 Marks it 15% above CP
∴ Marked price = MP = \(\frac{15}{100}\) x CP + CP = \(\frac{15}{100}\) 100 + 100 = 15 + 100 = 115
Discount % = 15%

∴ He sells it 97.75 which is less than his cost price. Therefore he loses
Loss = 97.75 – 100 = – 2.25

Try This (Text book Page No. 14)

Question 1.
By selling 5 articles, a man gains the cost price of 1 article. Find his gain percentage.
Solution:
Let cost price of article be C.P. Let S.P of 1 articles be SP by selling 5 articles at SP he makes a gain of cost price of one article.
Gain on 1 article = SP – CP; ⇒ Gain% = \(\frac{SP-CP}{CP}\) x 100
Gain on 2 articles = 2 x (SP – CP)
Gain on 5 articles = 5 x (SP – CP)
Given than gain on 5 articles is CP of 1 article
∴ 5(SP – CP) = CP
\(\frac{SP-CP}{CP}\) = \(\frac{1}{5}\)
Gain percentage \(\frac{SP-CP}{CP}\) x 100 = \(\frac{1}{5}\) x 100 % = 20%

Question 2.
By selling 8 articles, a shopkeeper gains the selling price of 3 articles. Find his gain percentage.
Solution:
Let cost price of 1 article be CP. Let selling price of 1 article be SP.
Gain on 1 article = SP – CP
Gain on 8 articles = 8 x (SP – CP)
Given that gain on 8 articles in selling price of 3 articles
8(SP – CP) = 3 x SP
∴ \(\frac{SP-CP}{CP}\) = \(\frac{3}{8}\)
∴ \(\frac{SP}{SP-CP}\) = \(\frac{8}{3}\)
Subtracting 1 on both sides
\(\frac{SP}{SP-CP}\) – 1 = \(\frac{8}{3}\) – 1 = \(\frac{SP-(SP-CP)}{SP-CP}\)= \(\frac{8-3}{3}\)
\(\frac{SP-SP-CP}{SP-CP}\) = \(\frac{5}{3}\) ⇒ \(\frac{CP}{SP-CP}\) = \(\frac{5}{3}\)
\(\frac{SP-CP}{CP}\) = \(\frac{3}{5}\) (taking reciprocals on both sides)
Gain% \(\frac{SP-CP}{CP}\) x 100 = \(\frac{3}{5}\) x 100 = 3 x 20 = 60%

Question 3.
If the C.P of 20 articles is equal to the S.P of 15 articles, find the profit or loss percentage.
Solution:
Given CP of 20 article = SP of 15 articles
∴ SP of 15 articles = CP of 20 articles
∴ SP of 1 article = \(\frac{1}{15}\) x CP of 20 articles
SP = \(\frac{1}{15}\) x 20 CP = \(\frac{20}{15}\) CP = \(\frac{4}{3}\) CP
∴ SP = \(\frac{4}{3}\) CP ⇒ SP is greater than CP
It is a profit.

Exercise 1.3
Try This (Text book Page No. 23)

Question 1.
Find the principal which gives ₹ 420 as C.I @ 20% p.a compounded half yearly for one year.
Solution:
CI = ₹420
Rate = ₹ 20% p.a
Principle = ₹ [required to find] Time period (n) = 1 year.
However, let us value of r to be 20% p.a so for half yearly, r is \(\frac{20}{2}\) = 10%
Formula for Amount (A) when compounded half yearly is

Question 2.
The price of a laptop depreciates @ 4% p.a. If its present price is ₹ 24,000, find its price after 3 years.
Solution:
Let original price of laptop be ‘P’, Rate of depreciation is 4% p.a,
Present price is ₹ 24,000 (D).
Formula for depreciation is

Price after 3 years from now is
P(1 – \(\frac{4}{100}\))ⁿ⁺³ = ? ⇒ (1 – \(\frac{4}{100}\))ⁿ (1 – \(\frac{4}{100}\))³
From (1)
24,000 x (1 – \(\frac{4}{100}\))³
24,000 x \(\frac{96}{100}\) x \(\frac{96}{100}\) x \(\frac{96}{100}\) = 21233.66

Actvity I (text book Page No. 23)

Question 1.
Mukunthan invests 30,000/- for 3 months in a bank which gives C.I at the rate of 12% compounded monthly. A private company offers his S.l at the rate of 12% p.a What is the difference in the interests received by Mukunthan? Do by traditional method and verify your answer by calculator.
Solution:
Principal = 30,000
Time period = 3 months
In Bank rate of interest for CI = 12% compounded monthly
∴ A = (1 + \(\frac{r}{100}\))ⁿ = 30,000(1 + \(\frac{12}{100}\))³
30,000 x \(\frac{112}{100}\) x \(\frac{112}{100}\) x \(\frac{112}{100}\) = 42147.84
∴ CI = A – P = 42147.84 – 30,000
CI = 12147.84
In private company,
Rate of single Interest SI = 12% p.a
So, for 3 months, i.e \(\frac{3}{12}\) = \(\frac{1}{4}\) year,

∴ Difference in interest = CI – SI = 12,147.84 – 900 = 11247.84

Read More: BHARTIARTL Pivot Calculator

Students can Download Maths Chapter 1 Numbers Additional 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 3 Chapter 1 Numbers Additional Questions

Question 1.
Find the least number by which 1100 must be multiplied so that the product becomes a perfect square. Also, in each case find the square root of the perfect square so obtained.
Solution:
We find 1100 = 2 × 2 × 5 × 5 × 11 =2² × 5² × 11
∴ The prime factor 11 has no pair.
∴ If we multiply 1100 by 11, then the product becomes a perfect square.
∴ New number = 1100 × 11 = 12100

12100 = 2² × 5² × 11²
\(\sqrt{12100}\) = 2 × 5 × 11 = 110
Square root of the new number = 110

Question 2.
Find the square of 509 using (a + b)² = a² + 2ab + b²
Solution:
509² = (500 + 9)² = 500² + 2 x 500 x 9 + 9²
= 250000 + 9000 + 81
509² = 259081

Question 3.
Find the sum without adding
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Given sum is the sum of first 12 odd natural numbers.
Sum of first n odd natural numbers is n².
∴ 1 +3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12² = 144

Question 4.
Write a Pythagorean triplet whose one number is 110
Solution:
Here let 2m = 110
m = 55

m² – 1 = 55² – 1 = 3025 – 1 = 3024
m² + 1 = 55² + 1 = 3025 + 1 = 3026
∴ Pythagorean triplet is 110, 3024, 3026.

Question 5.
Find the square root of 10 \(\frac{2}{3}\) correct to three places of decimal.
Solution:
10 \(\frac{2}{3}\) = 10.6666…….

\(\sqrt{10 \frac{2}{3}}=3.2659 \Rightarrow \sqrt{10 \frac{2}{3}}\) = 3.266 correct to three places of decimal.

Question 6.
Find the square root of 0.053361
Solution:

\(\sqrt{0.053361}\) = 0.231

Question 7.
Three numbers are in the ratio 2:3:4. The sum of their cubes is 33957. Find the numbers.
Solution:
Let the numbers be 2x, 3x and 4x
(2x)³ + (3x)³ + (4x)³ = 33957
8x³ + 27x³ + 64x³ = 33957
99x³ = 33957
x³ = \(\frac{33957}{99}\)
x³ = 343
x³ = 7 × 7 × 7
x³ = 7³
x = 7
∴ The numbers are 2x = 2 × 7 = 14
3x = 3 × 7 = 21
4x = 4 × 7 = 28

Question 8.
The volume of a cube is 9261000 m³. Find the side of the cube?
Solution:
Volume of the cube = side x side x side
side x side x side = 9261000
side = \(\sqrt[3]{9261×1000}\) = \(\sqrt[3]{9261}\) × \(\sqrt[3]{1000}\)
= \(\sqrt[3]{3^{3}×7^{3}}\) × \(\sqrt[3]{10×10×10}\) = 3 × 7 × 10 = 210
∴ Side of the cube = 210 m

Question 9.
If the diameters of the sun and Earth are 1.4 × 109 m and 1.275 × 107 m respectively. compare these two.
Solution:

So the diameter of the Sun is about 100 times the diameter of the Earth.

Question 10.
The size of a red blood cell is 0.000007 m and that of a plant, cell is 0.00001275 m. Compare these two.
Solution:

∴ RBC size if half the size of a plant cell.

Students can Download Science Term 3 Chapter 9 Visual Communication Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 Science Solutions Term 3 Chapter 9 Visual Communication

Samacheer Kalvi 8th Science Visual Communication Text Book Exercise

I. Choose the best answer:

Question 1.
The Keyboard shortcut is used to copy the selected text
(a) Ctrl + C
(b) Ctrl + V
(c) Ctrl + X
(d) Ctrl + A
Answer:
(a) Ctrl + C

Question 2.
The Keyboard shortcut is used to cut the selected text
(a) Ctrl + C
(b) Ctrl + V
(c) Ctrl + X
(d) Ctrl + A
Answer:
(c) Ctrl + X

Question 3.
If the ruler is not displayed in the screen, option is clicked.
(a) View → ruler
(b) view → task
(c) File → save
(d) Edit → paste
Answer:
(a) View → ruler

Question 4.
How many types of page orientation are there in Libre office Writer?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 5.
The menu used to save the document is –
(a) File → open
(b) file → print
(c) file → save
(d) Edit → close
Answer:
(c) file → save

II. Answer briefly:

Question 1.
What is the use for Text document software?
Answer:

1. A text file is used to store standard and structured textual data or information that is human readable.
2. It is defined in several different formate including the most popular ASCII for cross platform usage and ANSI for windows – based operating platforms.

Question 2.
What is selecting text?
Answer:
Selecting is the process of highlighting text or picking an object. For example, a user may select text to copy, cut or move that text to an alternate location or select a file they want to view.

Question 3.
How to close a document?
Answer:
Close the current document by selecting File → Close command on the menu bar or click the Close icon if it is visible on the Standard tool bar.

Question 4.
What is right alignment?
Answer:
Right alignment is text or page formatting that aligns text along the right side of a page or containing element.

Question 5.
How to open an existing document?
Answer:
To open an existing document, do any one of the following methods:

1. Click the Open File button on the menu bar.
2. Choose File → Open command from the menu bar.
3. Press CTRL+O keys on the keyboard. Each of the above method will show the Open dialog box. Choose the file and click the Open button.

Samacheer Kalvi 8th Science Visual Communication Additional Questions

I. Choose the correct answer:

Question 1.
………….. is a powerful and free office suite, used by millions of people.
(a) LibreOffice
(b) Microsoft window
(c) JAVA
(d) HTML
Answer:
(a) LibreOffice

Question 2.
………….. can create and edit forms, views and relations.
(a) Calc
(b) Impress
(c) Base
(d) Math
Answer:
(c) Base

Question 3.
………….. is the LibreOffice formula or equation editor.
(a) Impress
(b) Drawing
(c) Base
(d) Math
Answer:
(d) Math

Question 4.
The menu is used to print the document.
(a) File → open
(b) File → print
(c) File → save
(d) File → close
Answer:
(b) File → print

Question 5.
A ………….. is a set of characters and numbers in a certain style.
(a) Font
(b) Bullets
(c) Underline
(d) Paragraph
Answer:
(a) Font

Question 6.
………….. alignment refers to the appearance of the left and right sides of the Paragraph.
(a) Right
(b) Letf
(c) Paragraph
(d) None
Answer:
(c) Paragraph

Question 7.
How many types of alignment can be selected in LibreOffice?
(a) two
(b) three
(c) four
(d) five
Answer:
(c) four

Question 8.
A ………….. orientation means a horizontal display.
(a) Landscape
(b) Portrait
(c) Both (a) and (b)
(d) None of these
Answer:
(a) Landscape

Question 9.
A page is shorter in height but wider in width
(a) Landscape
(c) Both (a) and (b)
(b) Portrait
(d) None of these
Answer:
(b) Portrait

II. Answer the following question:

Question 1.
What is drawing?
Answer:
Draw is a vector drawing tool that can produce everything from simple diagrams or flowcharts to 3D artwork.

Question 2.
How can you create a new document?
Answer:
To create a new document, do any one of the following methods

1. Click the New Document button on the menu bar.
2. Choose File → New command from the menu bar.
3. Press CTRL+N keys on the keyboard.

Question 3.
How can you print a document?
Answer:
To print a document or selected pages follow the steps given below:

1. Open the document to be printed.
2. Choose File → Print command on the menu bar. The Print dialog box will open. Select the Options like print range, Number of copies, Printer name etc. See that printer is switched on and the paper is available in the printer tray.
3. Click OK.

Question 4.
What is the difference between cut and copy?
Answer:
The main difference between Cut and Copy is that cut removes the selected data from its original position while copy creates a duplicate of the original content.

Question 5.
What is font?
Answer:
A font is a set of characters and numbers in a certain style. Each font looks different from other fonts.

Question 6.
What is paragraph alignment?
Answer:
Paragraph alignment refers to the appearance of the left and right sides of the paragraph.

III. Paragraph Questions:

Question 1.
What are the components of LibreOffice?(any five)
Answer:
LibreOffice includes the following components.

Text Document:
Writer is a featurerich tool for creating letters, books, reports, newsletters, brochures, and other documents.

Calc (spreadsheet):
Calc has all of the advanced analysis, charting, and decision making features expected from a high-end spreadsheet. It includes over 300 functions for financial, statistical and mathematical operations, among others.

Impress (presentations):
Impress provides all the common multimedia presentation tools, such as special effects, animation, and drawing tools.

Base (database):
Base provides tools for day-to-day database work within a simple interface. It can create and edit forms, reports, queries, tables, views, and relations, so that managing a relational database is much the same as in other popular database applications.

Math (formula editor):
Math is the LibreOffice formula or equation editor. You can use it to create complex equations that include symbols or characters not available in standard font sets.

Question 2.
How can you selecting the text?
Answer:
For selecting text, the mouse or the keyboard can be used.
Selecting Text with Mouse – Following steps are to be followed:

1. Insertion point is moved to the start of the text to be selected.
2. The left mouse button should be clicked, held down and dragged across the text to be selected.
3. When the intended text is selected, the mouse button should be released.

Selecting Text with Keyboard – Following are the steps to be followed:

1. Insertion point is moved to the start of the text to be selected.
2. The Shift key is pressed down and the movement keys are used to highlight the required text.
3. When the Shift key is released, the text is selected.

Question 3.
List the steps of moving the text.
Answer:
The selected text can be easily cut and pasted in the required location. Following steps are to be followed.

1. The text to be moved to a new location is selected.
2. Edit → Cut is selected or in the tool bar is selected to cut the selected text.
3. Insertion point is moved to the place where the text is to be pasted.
4. Edit → Paste is selected or in the tool bar is selected to paste the text in the new location. The text can also be pasted in this way to another or another type of document.

The following keyboard shortcuts can be used to move text.

• Ctrl + X → to Cut,
• Ctrl + V → to Paste

Question 4.
How can you change the margins?
Answer:
If the user is not having the exact value for the margins then the Ruler option on the View menu can be used to change the margins.
Following steps are used in this method:

1. If the ruler is not displayed in the screen, View → Ruler option is clicked.
2. The gray area of the ruler indicates the margin’s top area.
3. The mouse pointer is then moved in between the gray and white area of the ruler.
4. When the pointer is in the right spot, it changes into a line with arrows on both sides.
5. The margin guide is dragged to a new location.

Students can Download Maths Chapter 2 Life Mathematics 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 3 Chapter 2 Life Mathematics Intext Questions

Exercise 2.1
Try These (Text book Page No. 33)

Classify the given examples as direct or inverse proportion:
(i) Weight of pulses to their cost.
Solution:
As weight increases cost also increases.
∴ Weight and cost are direct proportion.

(ii) Distance travelled by bus to the price of ticket.
Solution:
As the distance increases price to travel also increases.
∴ Distance and price are direct proportion.

(iii) Speed of the athelete to cover a certain distance.
Solution:
As the speed increases, the time to cover the distance become less.
So speed and time are in indirect proportion.

(iv) Number of workers employed to complete a construction in a specified time.
Solution:
As the number of workers increases, the amount of work become less, so they are in indirect proportion.

(v) Volume of water flown through a pipe to its pressure.
Solution:
As the pressure increases, volume also increases.
∴ They are direct proportions.

(vi) Area of a circle to its radius.
Solution:
If the radius of the circle increases its area also increases.
∴ Area and radius of circles are direct proportion.

Use the concept of direct and inverse proportions and try to answer the following questions:
Question 1.
A student can type 21 pages in 15 minutes. At the same rate, how long will it take the student to type 84 pages?
Solution:
Direct proportion
No. of minutes = x
k = \(\frac{21}{15}\)
\(\frac{21}{15}\) = \(\frac{84}{x}\)

Question 2.
The weight of an iron pipe varies directly with its length. If 8 feet of an iron pipe weighs 3.2 kg, find the proportionality constant k and determine the weight of a 36 feet iron pipe.
Solution:

Weight of 36 feet iron pipe = x
\(\frac{36}{x}\) = 2.5

Question 3.
A car covers a distance of 765 km in 51 litres of petrol. How much distance would it cover in 30 litres of petrol?
Solution:
Direct proportion
k = \(\frac{51}{765}\)
Distance cover = x km
\(\frac{30}{x}\) = \(\frac{51}{765}\)

Question 4.
If x and y vary inversely and x = 24 when y = 8, find x when y = 12.
Solution:
k = xy = 24 × 8 = 192
∴ 12 × x = 192

Question 5.
If 35 women can do a piece of work in 16 days, in how many days will 28 women do the same work?
Solution:
Inverse proportion
No. of days = x
k = 35 × 16
∴ 28 × x = 35 × 16

Question 6.
A farmer has food for 14 cows which can last for 39 days. How long would the food last, if 7 more cows join his cattle?
Solution:
Inverse variation
k = xy = 14 × 39
No. of cow = 14 + 7 = 21
No. of days = x
21 × x = 14 × 39

Question 7.
Identify the type of proportion and fill in the blank boxes:

Solution:
Direct proportion
\(\frac{x}{y}\) = k = \(\frac{1}{20}\)
(i) x = 2; y = ?
\(\frac{2}{y}\) = \(\frac{1}{20}\) ⇒ y = 2 × 20 = 40

(ii) x = ?; y = 60
\(\frac{x}{60}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{60}{20}\) = 3

(iii) x = 4; y = ?
\(\frac{4}{y}\) = \(\frac{1}{20}\) ⇒ y = 80

(iv) x = 4; y = ?
\(\frac{8}{y}\) = \(\frac{1}{20}\) ⇒ y = 20 × 8 = 160

(v) x = ?; y = 180
\(\frac{x}{180}\) = \(\frac{1}{20}\)
x = \(\frac{180}{20}\) = 9

(vi) x = 12; y = ?
\(\frac{12}{y}\) = \(\frac{1}{20}\)
y = 12 × 20 = 240

(vii) x = ?; y = 360
\(\frac{x}{360}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{360}{20}\) = 18

(viii) x = 24; y = ?
\(\frac{24}{y}\) = \(\frac{1}{20}\) ⇒ y = 24 × 20 = 480

Question 8.
Identify the type of proportion and fill in the blank boxes:

Solution:
Inverse proportion
k = xy = 1 × 144 = 144
(i) x = 2; y = ?
2y = 144
y = 72

(ii) X = ?; y = 48
48x = 144
x = \(\frac{144}{48}\) = 3

(iii) x = 4; y = ?
4y = 144
y = \(\frac{144}{4}\) = 36

(iv) x = 8; y = ?
8 y = 144
y = \(\frac{144}{8}\) = 18

(v) x = ?; y = 16
16x = 144
y = \(\frac{144}{16}\) = 9

(vi) x = 12; y = ?
12y = 144
y = \(\frac{144}{12}\) = 12

(vii) x = ?; y = 9
9x = 144
x = \(\frac{144}{9}\) = 16

(viii) x = 24; y = ?
24y = 144
y = \(\frac{144}{24}\) = 6

Try These (Text book Page No. 38)

Question 1.
When x = 5 and y = 5 find k, if x and y vary directly.
Solution:
If x and y vary directly then \(\frac{x}{y}\) = k
Here x = 5; y = 5
∴ k = \(\frac{5}{5}\)
k = 1

Question 2.
When x and y vary inversely, find the constant of variation when x = 64 and y = 0.75
Solution:
Given
x =64, y = 0.75
and also given x and y vary inversely.
∴ xy = k. the constant of variation.
∴ Constant = 64 × 0.75
Constant of variation = 48

Think (Text book Page No. 38)

(i) When x and y are in direct proportion and if y is doubled, then what happens to x?
Solution:
If x and y are in direct proportion \(\frac{x}{y}\) = k, constant.
if y is doubled, then \(\frac{x}{2}\) must be equal to k. So x also to be doubled.

(ii) if \(\frac{x}{y-x}\) = \(\frac{6}{7}\) What is \(\frac{x}{y}\)?
Solution:
if \(\frac{x}{y-x}\) = \(\frac{6}{7}\)
\(\frac{y-x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) – \(\frac{x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + \(\frac{x}{x}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + 1
\(\frac{y}{x}\) = \(\frac{7+6}{6}\)
\(\frac{y}{x}\) = \(\frac{13}{6}\)
\(\frac{x}{y}\) = \(\frac{6}{13}\)

Try These (Text book Page No. 40)

Identify the different variations present in the following questions:
Question 1.
24 men can make 48 articles in 12 days. Then, 6 men can make …………. articles in 6 days.
Solution:
Let the required no. of articles be x

(i) Mens and days are Indirect variables.
(ii) Men and Articles are direct variables
(iii) Days and articles are also direct variables using formula.
Let P₁ = 24, D₁ = 12, W₁ = 48
P₂ = 6, D₂ = 6, W₂ = x

Question 2.
15 workers can lay a road of length 4 km In 4 hours. Then, …………. workers can lay a road of length 8 km in 8 hours.
Solution:
Let the required number of workers be x

Length and workers are direct variable as more length need more workers.
The proportion is 4 : 8 : : 15 : x ……….(1)
Hours and workers are indirect variables as more working hours need less men.
∴ The proportion is 8 : 4 : : 15 : x ………..(2)
Combining (1) and (2)

Product of the extremes = product of the means
4 × 8 × x = 8 × 4 × 15
x = \(\frac{8×4×15}{4×8}\)
x = 15 workers

Question 3.
25 women working 12 hours a day can complete a work in 36 days. Then, 20 women must ……….. work hours to complete the same work in 30 days.
Solution:
Let the required hours be x

As women increases hours to work decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{25}{20}\)
As days increases hours needed become less
∴ It is also an indirect variation.
∴ Multiplying factor is \(\frac{36}{30}\)
∴ x = 12 × \(\frac{25}{20}\) × \(\frac{36}{30}\)
x = 18 hours

Question 4.
In a camp, there are 420 kg of rice sufficient for 98 persons for 45 days. The number of days that 60 kg of rice will last for 42 persons is…………
Solution:
Let the required number of days be x.

If amount of rice is more it will last for more days.
∴ It is Direct Proportion.
∴ Multiplying factor is \(\frac{60}{420}\)
If men increases number of days the rice lasts decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{98}{42}\)
x = 45 × \(\frac{60}{420}\) × \(\frac{98}{42}\)

x = 15 days

Try These (Text book Page No. 44)

Question 1.
Vikram can do one-third of work in p days. He can do \(\frac{3}{4}\)th of work in ………… days.
Solution:
\(\frac{1}{3}\) of the work will be done in p days
∴ Full work will be completed in 3p days
\(\frac{3}{4}\)th of the work will be done in = 3p × \(\frac{3}{4}\) = \(\frac{9}{4}\)p = 2\(\frac{1}{4}\)p days

Question 2.
If m persons can complete a work in n days, then 4m persons can complete the same work in ……….. days and \(\frac{m}{4}\) persons can complete the same work in…….. days
Solution:
Given m persons complete a work in n days
(i) Then work measured in terms of Man days = mn
4 m men do the work it will be completed in \(\frac{mn}{4m}\) days = \(\frac{n}{4}\) days.
(ii) \(\frac{m}{4}\) persons can complete the same work in \(\frac{mn}{\frac{m}{4}}\) days = \(\frac{4mn}{m}\) = 4n days

Students can Download Science Term 3 Chapter 8 Conservation of Plants and Animals Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 Science Solutions Term 3 Chapter 8 Conservation of Plants and Animals

Samacheer Kalvi 8th Science Conservation of Plants and Animals Text Book Exercise

I. Choose the best answer:

Question 1.
The plants found in a particular area are known as …………….
(a) fauna
(b) flora
(c) endemic
(d) rare
Answer:
(c) endemic

Question 2.
Deforestation means …………….
(a) cleaning of forest
(b) to grow plants
(c) to look after plants
(d) None of these
Answer:
(d) None of these

Question 3.
The Red Data Book gives a list of …………….
(a) endemic species
(b) extinct species
(c) natural species
(d) None of these
Answer:
(d) None of these

Question 4.
Insitu conservation is …………….
(a) off site conservation
(b) on site conservation
(c) Both a and b
(d) None of these
Answer:
(b) on site conservation

Question 5.
Wildlife Protection Act was implemented in …………….
(a) 1986
(b) 1972
(c) 1973
(d) 1971
Answer:
(b) 1972

II. Fill in the blanks:

1. WWF stands for …………….
2. The animal found in a particular area is known as …………….
3. Red Data Book is maintained by …………….
4. Mudhumalai Wildlife Sanctuary is located in ……………. district.
5. ……………. is observed as ‘World Wildlife day’.

Answer:

1. World Wildlife Fund
2. endemic
3. International Union for conservation of Nature
4. Nilgiris
5. March 3

III. Match the following:

1. Gir national park – Madhya Pradesh
2. Sundarabans National Park – Uttarangal
3. Indira Gandhi National Park – West Bengal
4. Corbett National Park – Gujarat
5. Kanha National Park – Tamil Nadu

Answer:

1. Gir national park – Gujarat
2. Sundarabans National Park – West Bengal
3. Indira Gandhi National Park – Tamil Nadu
4. Corbett National Park – Uttarangal
5. Kanha National Park – Madhya Pradesh

IV. Answer very briefly:

Question 1.
What is global warming?
Answer:
Gases methane and carbon dioxide accumulating in the atmosphere and trap the heat energy inside the atmosphere leading to increase in temperature is called Global warming.

Question 2.
What is known as extent species?
Answer:
Species which no longer exist on Earth are called extinct species.

Question 3.
Give few example for extinct species.
Answer:
Dinosaurs and Dodo are examples of extinct species.

Question 4.
Name two endangered animals.
Answer:

1. Snow Leo pard
2. Asiatic Lion

Question 5.
What is ICBN?
Answer:
International Code of Botanical Nomenclature. These are a set of International rules proposed by botanists to ensure a stable, universal and uniform system of naming plants.

V. Answer briefly:

Question 1.
What is biosphere reserve?
Answer:

1. Biosphere is a protected area where human population also forms the part of the system.
2. The area of these places will be around 5000 square kilometers. They conserve the eco system, species and genetic resources. These areas are set up mainly for economic development.

Question 2.
What is tissue culture?
Answer:
It is a technique of growing plant cells, tissues, organs, seeds or other plant parts in a sterile environment on a nutrient medium.

Question 3.
What is endangered species? Give two examples.
Answer:

1. An endangered species is an animal or a plant that is considered to be at the risk of extinction.
2. It means that there are only few of them left on the Earth and soon they might extinct.
3. Snow leopard, Bengal tiger, Asiatic lion, Purple frog and Indian giant squirrel are some of the endangered animals in India.

Question 4.
Write the advantages of the Red Data Book.
Answer:

1. It helps to evaluate the population of a particular species.
2. The data given in this book can be used to evaluate the species at the global level.
3. The risk of a species becoming globally extinct can be estimated with the help of this book.
4. It provides guidelines for implementing protective measures for endangered species.

Question 5.
Mention four main reasons for the conservation of forests.
Answer:

1. According to WWF (World Wildlife Fund) there has been 60% decrease in the size of population of animals, birds, fish, reptiles and amphibians over the past 40 years.
2. In order to leave something for the future generation, we need to conserve it now.
3. Conservation is the protection, preservation, management of wildlife and natural resource such as forest and water.
4. Conservation of biodiversity helps us to protect, maintain and recover endangered animals and plant species.

Question 6.
What do you understand by the term bio-magnification?
Answer:

1. Bio – magnification is the increase in contaminated substances due to the intoxicating environment.
2. The contaminants might be heavy metals such as mercury, arsenic, and pesticides such as polychlorinated biphenyls and DDT (Dichloro Diphenyl Trichloro ethane).
3. These substances are taken up by the organisms through the food they consume.
4. When the organisms in the higher food chain feed on the organisms in the lower food chain containing these toxins, these toxins get accumulated in the higher organisms.

Question 7.
What is PBR?
Answer:
1. Peoples Biodiversity Register is a document which contains comprehensive information on locally available bio-resources including landscape and demography of a particular area or village.

2. Bio-resources mean plants, animals and microorganisms or parts thereof, their genetic material and by-products with actual or potential use or value.

3. A Biodiversity Management Committee is set up in each local body which prepares the People’s Biodiversity Registers with the guidance and technical support of National Biodiversity Authority and the State Biodiversity Boards.

4. Preparation of this register promotes conservation, preservation of habitats and breed of animals and gathering of knowledge relating to biological diversity.

5. The register entails a complete documentation of biodiversity in the area related to the plant, food source, wildlife, medicinal source, traditional knowledge etc.

VI. Answer in detail:

Question 1.
What is Deforestation? Explain the causes and effects of Deforestation.
Answer:
Destruction of forests in order to make the land available for different uses is known as deforestation.
Causes of Deforestation:

• Fires and floods are the natural causes for deforestation.
• Human activities are responsible for deforestation include agricultural expansion, cattle breeding, illegal logging, mining, oil extraction, dam construction and infrastructure development.

1. Agricultural Expansion:

• With increasing population, there is an overgrowing demand for food production.
• Hence, large amount of trees are chopped down for crops and for cattle grazing.

2. Urbanization:

• Increase in population needs the expansion of cities.
• More land is needed to establish housing and settlement.
• Requirements like construction of roads, development of houses, mineral exploitation and expansion of industries also arise due to urbanisation.
• Forests are destroyed to meet all these needs.

3. Mining:

• Mining of coal, diamond and gold require a large amount of forest land.
• Large number of trees are cut down to clear the forest area. The waste that comes out from mining pollutes the environment and affects the nearby plants.

4. Construction of dams:
To provide water supply to the increasing population, large size dams are constructed. Hence, a great extend of forest area is being cleared.

5. Timber Production:

• Wood-based industries like paper, match-sticks, furniture need a substantial amount of wood supply.
• Wood is the most commonly used fuel, thus, a large number of trees are being cut down for fuel supplies.
• Illegal wood cutting is the main reason for the destruction of some valuable plants.

6. Forest fire:

• Forest fire be caused by humans, accidents or natural factors.
• Forest fires wipe out thousands of acres of forest land each year all over the world. This has tremendous effects on biodiversity and the economy as well.

7. Cyclones:
Cyclones destroy the trees on a massive scale.

Effects of Deforestation:
1. Extinction of species:
Deforestation has resulted in the loss of many wonderful species of plants and animals and many are on the verge of extinction.

2. Soil Erosion:

• When the trees are cut down, soils are exposed to the Sun’s heat.
• Extreme temperature of the summer dries up the moisture and makes the nutrients to evaporate, ft also affects the bacteria that helps in the breakdown of organic matter.

3. Water cycle:
When trees are cut down, the amount of water vapour released decreases for transpiration and hence there is a decrease in the rainfall. ‘

4. Floods:
When the trees are cut down, the flow of water is disrupted and it leads to flooding.

Question 2.
Discuss the advantages of Insitu and exsitu conservation.
Answer:
Advantages of In-situ conservation:

1. Species can be adapted to their habitat.
2. Species can interact with each other.
3. Natural habitat is maintained.
4. It is less expensive and easy to manage.
5. Interests of indigenous people are protected.

Advantages of Exquisite conservation:

1. It prevents the decline of species.
2. Endangered animals can be breeded in these ways.
3. Threatened species are breeded and released in natural environment.
4. It is useful for conducting research and scientific work.

Question 3.
Write about the types of conservation.
Answer:
Conservation is the protection, preservation, management of wildlife and natural resource such as forest and water.

Types:
In-situ conservation:
It is conservation of living resources within the natural ecosystem in which they occur.

1. National Parks:

• It is an area which is strictly reserved for the betterment of the wildlife.
• Here, activities like forestry, grazing or cultivation are not permitted.
• Eg – Guindy National Park in Chennai district.

2. Wildlife sanctuaries:

• A sanctuary is a protected area reserved for the conservation of animals only.
• Human activities like harvesting of timber, collection of forest products and private ownership rights are allowed here.
• Controlled interference like tourist activity is also allowed.

3. Biosphere reserves:

• It is a protected area where human population also forms the part of the system.
• The area of these places will be around 5000 square kilometers.
• They conserve the eco system, species and genetic resources.
• Eg – These areas are set up mainly for economic development.

Ex-situ Conservation:
It is the conservation of wildlife outside their habitat. Establishing zoos and botanical gardens, conservation of genes, seedling and tissue.

1. Botanical gardens:

• It is a place where flowers, fruits and vegetables are grown.
• These places provide a healthy and calm environment.

2. Zoological parks:

• Zoological parks are the areas where wild animals are conserved.
• In India there are about 800 zoological parks.

3. Tissue Culture:
It is a technique of growing plant cells, tissues, organs, seeds or other plant parts in a sterile environment on a nutrient medium.

4. Seed bank:
The seed bank preserves dried seeds by storing them in a very low temperature.

5. Cryo Bank:
It is a technique by which a seed or embryo is preserved at a very low temperature.

Question 4.
Write a note on Blue Cross.
Answer:
1. Blue Cross is a registered animal welfare charity in the United Kingdom, founded in 1897 as Our Dumb Friends League.

2. The vision of this charity is that every pet will enjoy a healthy life in a happy home.

3. The charity provides support for pet owners who cannot afford private veterinary treatment, helps to find homes for unwanted animals, and educates the public in the responsibilities of animal ownership.

4. Captain V. Sundaram founded the Blue Cross of India, the largest animal welfare organization of Asia in Chennai in the year 1959.

5. He was an Indian pilot and animal welfare activist. Now, Blue Cross of India the country’s largest animal welfare organizations and it runs several animal welfare events like pet adaptation and animal right awareness.

6. Blue Cross of India has received several international and national awards.

7. This organization is entirely looked after by volunteers.

8. The main office is located at Guindy, Chennai, with all amenities like hospitals, shelters, ambulance services and animal birth controls, etc.

9. Activities of the organization include, providing shelters, re-homing, adoption, animal birth control, maintaining hospitals and mobile dispensary and providing ambulance services.

VII. HOT:

Question 1.
Is it possible to find Dinosaurs today? Why?
Answer:

1. No. It is not possible to find Dinosaurs today since they have become extinct.
2. The reasons may be shortage of food, space or climate changes.

Question 2.
Animals are affected by Deforestation. How?
Answer:

1. Deforestation involves destruction of forests by man to make the lend available for different uses.
2. When forests are destroyed, the wild animals living there are left homeless.
3. They start moving out in search of shelter.
4. Since animals start moving out the food chains are also affected and animals cannot find food also.
5. They also get isolated from their groups and may be killed by accidents or hunted.

Question 3.
Why did the numbers of Tiger and Black buck decrease?
Answer:

1. The number of Black buck and Tigers have decreased due to excessive hunting, Deforestation and habitat degradation.
2. Tigers are also facing problems of prey depletion since the number of deers and other herbivores are decreasing due to disappearance of forests.

Samacheer Kalvi 8th Science Conservation of Plants and Animals Additional Question

I. Choose the correct answer:

Question 1.
…………… is not a green house gas.
(a) Oxygen
(b) Carbon dioxide
(c) Nitro-us oxide
(d) Methane
Answer:
(a) Oxygen

Question 2.
Chipko Movement was started in ……………
(a) 1980
(b) 1970
(c) 1960
(d) 1953
Answer:
(b) 1970

Question 3.
Each year …………… is celebrated as ‘World Biodiversity Day’.
(a) April 20
(b) May 22
(c) December 8
(d) October 12
Answer:
(b) May 22

Question 4.
…………… is not an endangered animal.
(a) Nilgiri Tahr
(b) Asiatic Lion
(c) Snow leopard
(d) Dodo duck
Answer:
(d) Dodo duck

Question 5.
Yeoman Butterfly has been declared as state butterfly of ……………
(a) Manipur
(b) Nagaland
(c) Tamil Nadu
(d) West Bengal
Answer:
(c) Tamil Nadu

II. Fill in the blanks:

1. The founder of Chipko movement was ……………
2. In a Cryo bank, the seeds are preserved in ……………
3. The variety of life forms is called …………….
4. Replanting of trees is called ……………
5. The details of endangered species can be viewed in ……………
6. Blue cross of India was established in …………… in India.
7. …………… has led to destruction of coral seeds.
8. …………… Biosphere reserve is located in Tamil Nadu.
9. National park is an example for …………… conservation.
10. World Wild life Day is celebrated on …………….

Answer:

1. Sunderlal Bahuguna
2. liquid nitrogen
3. Biodiversity
4. Reforestation
5. Red Data Book
6. Chennai
7. Biomagnification
8. Nilgiri
9. In-situ
10. March 3rd

III. Very short answer:

Question 1.
What is Reforestation?
Answer:
Reforestation is the natural or intentional replanting of the existing forests that have been destroyed through deforestation.

Question 2.
Differentiate Afforestation and Reforestation.
Answer:
Afforestation:

1. Trees are planted in new areas where there was no forest cover.
2. One sapling is planted to get one tree.
3. It is practiced to bring more area under forest

Reafforestation:

1. It is practiced in areas where forests have been destroyed.
2. Two saplings are planted to replace every felled tree
3. It is practiced to avoid deforestation.

Question 3.
Name two endangered plant species?
Answer:
Malabar Lily and Rafflesia flower.

Question 4.
What are the functions of CPCSEA?
Answer:

1. Approval of animal house facilities.
2. Permission for conducting experiments involving usage of animals.
3. CPCSEA stands for ‘The Committee for the Purpose of Control and Supervision of Experiments on Animals.
4. It is a statutory committee set up under the Preservation of Cruelty to Animals Act,1960.

Question 5.
The Red Data Book is maintained by which organisation?
Answer:

1. Red Data Book is maintained by the International Union for Conservation of Nature.
2. It is an international organization working in the field of nature conservation and sustainable use of natural resources.

IV. Long Answer:

Question 1.
List some measure to save endangered species.
Answer:

1. Animal species are endangered mainly because of hunting and poaching. If it is controlled, there can be a significant change in the number of endangered animals.
2. Controlling pollution can have a positive impact on animals, fish and birds all over the world.
3. By consuming less pollutants, we can protect the ecosystems.
4. Animals often mistake plastic for food and hence plastics harm and cause endangerment of many species. Limiting the amount of plastic and recycling it can save the endangered animals.
5. Recycling things and buying Eco friendly products will preserve the environment resources and hence the animals.
6. Pesticides and chemicals which cause damage to the environment should be avoided.
7. Planting native trees will provide food to the animals.

Question 2.
What are causes for endangerment of species?
Answer:

1. Loss of habitat.
2. Trees that provide food and shelter to so many species are destroyed due to human intervention.

Over hunting and poaching:
Large number of animals is hunted for their horns, skin, teeth and many other valuable products.

3. Pollution:

• Number of animals are affected by pollutions like air pollution and water pollution.
• In the recent years more number of animals is affected by wastes in the form of plastic.

4. New habitat:

• Sometimes animals are taken by people to new habitat where they do not naturally live.
• Some, of them may extinct and some may survive.
• The new ones may also get attacked by the species already living there and cause their extinction.

5. Chemicals:

• We use pesticides and other chemicals to get rid of damaging insects, pests or weeds.
• But they can also poison desired plants and animals if we do not use them correctly.

6. Diseases:
Diseases due to various unknown reasons may affect the animals and make them extinct.

7. Natural calamities:
Animals may also be destroyed due to natural disasters like flood and fire.

Question 3.
Explain the significance of Afforestation.
Answer:

1. Afforestation helps the wild animals and even humans to have shelter and to find their food source.
2. Through afforestation we can increase the supply of oxygen. Trees planted can increase the water vapour in the atmosphere to get the rainfall.
3. By planting trees the amount of carbon dioxide in the atmosphere can be reduced and thus the effects of air pollution, green house gases and global warming can be controlled.
4. Afforestation enables us to avoid desertification of land.
5. Barren lands experience strong winds and it causes soil erosion. Top soil is washed away during rainfall. Afforestation helps to grow more trees so that they can hold the top soil along with the nutrients.
6. Creating forests provides us fodder, fruits, firewood and many other resources.
7. Industries need specific type of trees. Afforestation helps us to grow a particular type of trees.