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

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

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

Multiple Choice Questions

Question 1.
The area of triangle formed by the points (-5, 0), (0, – 5) and (5, 0) is …………..
(1) 0 sq.units
(2) 25 sq.units
(3) 5 sq.units
(4) none of these
(2) 25 sq.units Hint.
Hint:
Area of the ∆

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

Question 2.
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is …………
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
(1) x = 10
Hint:

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

Question 4.
If (5,7), (3,p) and (6,6) are collinear, then the value of p is ……………
(1) 3
(2) 6
(3) 9
(4) 12
(3) 9
Hint:
Since the three points are collinear. Area of a triangle is 0

5p + 18 + 42 – (21 + 6p + 30) = 0
5p + 60 – (51 + 6p) = 0
5p + 60 – 51 – 6p = 0
-p + 9 = 0
-p = -9
p = 9

Question 5.
The point of intersection of 3x – y = 4 and x + 7 = 8 is ……………
(1) (5,3)
(2) (2,4)
(3) (3,5)
(4) (4, 4)
(3) (3, 5)

Substitute the value of x = 3 in (2)
3 + 7 = 8
y = 8 – 3 = 5
The point of intersection is (3, 5)

Question 6.
The slope of the line joining (12, 3), (4, a) is $$\frac { 1 }{ 8 }$$. The value of ‘a’ is …………….
(1) 1
(2) 4
(3) -5
(4) 2
(4) 2
Hint:
Slope of a line = $$\frac { 1 }{ 8 }$$

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

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

Question 9.
If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is ……………
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
(1) 8x + 5y = 40
Hint:
Let the point A be (0, 8) and B (5, 0)

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

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

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

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

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

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

## Samacheer Kalvi 10th Maths Guide Book Back Answers Solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Samacheer Kalvi 10th Science Guide Chapter 18 Heredity

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

## Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity

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

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

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

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

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

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

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

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

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

II. Fill in the blanks:

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

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

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

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

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

IV. Match the following:

A. (iii)
B. (iv)
C. (v)
D. (i)
E. (ii)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(ii) When the hybrids of F1 generation pea plants having round-yellow seeds were cross-bred by self pollination, then four types of seeds having different combinations of shape and color were obtained in second generation or F2 generation. They were round yellow, round-green, wrinkled-yellow and wrinkled-green seeds.
The ratio of each phenotype (or appearance)of seeds in the F2 generation is 9 : 3 : 3 : 1. This is known as the Dihybrid ratio.

Monohybrid cross is a genetic cross that involves a single pair of gene or trait. In this parents differ by single trait. Eg: Height.
Dihybrid cross is a genetic cross that involves two pairs of genes, which are responsible for two trait. In this, parents have two different independent trait. Eg: flower colour, stem length.

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

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

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

Significance of DNA:
(i) It is responsible for the transmission of hereditary information from one generation to next generation.
(ii) It contains information required for the formation of proteins.
(iii) It controls the developmental process and life activities of an organism.

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

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

VIII. Higher Order Thinking Skills: (HOTS)

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

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

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

IX. Value-Based Questions

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

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

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

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

Question 4.

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

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

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

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

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

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

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

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

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

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

II. Fill in the blanks:

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

III. Match the following:

A. (v)
B. (i)
C. (ii)
D. (iii)
E. (iv)

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

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

1. False – The daughter strand synthesized in DNA is called leading strand
2. True
3. False – Primary construction in chromosome is called as secondary construction.
4. True

V. Answer in a word or sentence:

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

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

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

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

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

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

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

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

Question 9.
Name two purine nitrogenous bases found in a DNA molecule.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

VIII. Higher Order Thinking Skills: (HOTS)

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

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

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

## Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11

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

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

Question 1.
Solve the following quadratic equations by completing the square method
(i) 9x2 – 12x + 4 = 0
9x2 – 12x + 4 = 0
x2 – $$\frac { 12x }{ 9 }$$ + $$\frac { 4 }{ 9 }$$ = 0 (Divided by 9)
x2 – $$\frac { 4x }{ 3 }$$ = $$\frac { -4 }{ 9 }$$
Add [$$\frac { 1 }{ 2 }$$ ($$\frac { 4 }{ 3 }$$)]2 on both sides

The solution is $$\frac { 2 }{ 3 }$$

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

Question 2.
Solve the following quadratic equations by formula method
(i) 2x2 – 5x + 2 = 0
a = 2, b = -5, c = 2

The solution set is $$\frac { 1 }{ 2 }$$ and 2

(ii) $$\sqrt { 2 }$$ f2 – 6 f + 3 $$\sqrt { 2 }$$ = 0
Here a = $$\sqrt { 2 }$$, b = -6 and c = 3$$\sqrt { 2 }$$

The solution set is $$\frac{3+\sqrt{3}}{\sqrt{2}}$$ and $$\frac{3-\sqrt{3}}{\sqrt{2}}$$

(iii) 3y2 – 20y – 23 = 0
a = 3, b = -20, c = -23

The solution set is -1 and $$\frac { 23 }{ 3 }$$

(iv) 36y2 – 12ay + (a2 – b2) = 0
Here a = 36, b = -12a, c = a2 – b2

The solution set is $$\frac { (a+b) }{ 6 }$$ and $$\frac { (a-b) }{ 6 }$$

Question 3.
A ball rolls down a slope and travels a distance d = t2 – 0.75t feet in t seconds. Find the time when the distance travelled by the ball is 11.25 feet.
Distance = t2 – 0.75t
11.25 = t2 – 0.75t
Multiply by 100
1125 = 100t2 – 75t
100t2 – 75t – 1125 = 0 (Divided by 25)
4t2 – 3t – 45 = 0
a = 4,
b = -3,
c = -45

(time will not be negative)
The required time = $$\frac { 15 }{ 4 }$$ seconds
= 3 $$\frac { 3 }{ 4 }$$ second or 3.75 seconds

## Samacheer Kalvi 10th Maths Guide Chapter 7 Mensuration Additional Questions

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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Multiple Choice Questions:

Question 1.
The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to ______
(1) π cm2
(2) 2π cm2
(3) 3π cm2
(4) 2 cm2
(2) 2π cm2
Hint:
C.S.A of a cylinder = 2πrh sq. units = 2 × π × 1 × 1 cm2 = 2π cm2

Question 2.
The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to ______ sq. units.
(1) $$\frac{3}{2} \pi h$$
(2) $$\frac{2}{3} \pi h^{2}$$
(3) $$\frac{3}{2} \pi h^{2}$$
(4) $$\frac{2}{3} \pi h$$
(3) $$\frac{3}{2} \pi h^{2}$$
Hint:
T.S.A = 2πr(h + r)
[radius is half of the height]
= $$2 \pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)$$
= $$=\pi h\left(\frac{3 h}{2}\right)=\frac{3}{2} \pi h^{2}$$ sq. units

Question 3.
Base area of a right circular cylinder is 80 cm2. If its height is 5 cm, then the volume is equal to _______
(1) 400 cm3
(2) 16 cm3
(3) 200 cm3
(4) $$\frac{400}{3}$$ cm3
(1) 400 cm3
Hint:
Volume of a cylinder = πr2h cu. units
[Base area (πr2) = 80 cm2 = 80 × 5 cm3 = 400 cm3

Question 4.
If the total surface area of a solid right circular cylinder is 200π cm2 and its radius is 5 cm, then the sum of its height and radius is ______
(1) 20 cm
(2) 25 cm
(3) 30 cm
(4) 15 cm
(1) 20 cm
Hint:
T.S.A of a cylinder = 200π cm2
2πr (h + r) = 200π
2 × 5 (h + r) = 200
(h + r) = 20 cm

Question 5.
The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to ______
(1) πa2b sq.cm
(2) 2πab sq.cm
(3) 2π sq.cm
(4) 2 sq.cm
(2) 2πab sq.cm .
Hint:
C.S.A. of a cylinder = 2πrh sq. units = 2 × π × a × b sq. cm = 2πab sq. cm

Question 6.
Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm3, then the volume of the cone is equal to _______
(1) 1200 cm3
(2) 360 cm3
(3) 40 cm3
(4) 90 cm3
(3) 40 cm3
Hint:
Volume of the cone = $$\frac{1}{3}$$ × volume of the cylinder
= $$\frac{1}{3}$$ × 120 cm3
= 40 cm3

Question 7.
If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is
(1) 10 cm
(2) 20 cm
(3) 30 cm
(4) 96 cm
(1) 10 cm
Hint:
Slant height of a cone

Question 8.
If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ______
(1) 1200π cm2
(2) 600π cm2
(3) 300π cm2
(4) 600 cm2
(2) 600π cm2
Hint:
Circumference (2πr) = 120π cm
Slant height (l) = 10 cm;
Curved surface area of a cone = πrl sq. units
= $$\frac{120 \pi}{2}$$ × 10 cm2 = 600π cm2

Question 9.
If the volume and the base area of a right circular cone are 48π cm and 12π cm respectively, then the height of the cone is equal to ______
(1) 6 cm
(2) 8 cm
(3) 10 cm
(4) 12 cm
(4) 12 cm
Hint:
Volume of a cone = 48π cm3
[Base area (πr2) = 12π]
$$\frac{1}{3}$$ πr2h = 48π
$$\frac{1}{3}$$ × 12π × h = 48π
[Substitute πr2 = 12π]
h = $$\frac{48}{4}$$ = 12 cm

Question 10.
If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the volume of the cone is equal to _______
(1) 240 cm3
(2) 120 cm3
(3) 80 cm3
(4) 480 cm3
(3) 80 cm3
Hint:
Volume of a cone (V) = $$\frac{1}{3}$$ πr2h sq. units
Base area (πr2) = 48 sq. cm
V = $$\frac{1}{3}$$ × 48 × 5 = 80 cm3

Question 11.
The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1 respectively. Then their respective volumes are in the ratio _______
(1) 4 : 1
(2) 1 : 4
(3) 2 : 1
(4) 1 : 2
(3) 2 : 1
Hint:
h1 : h2 = 1 : 2
r1 : r2 = 2 : 1
Ratio of their volumes
= $$\frac{1}{3} \pi r_{1}^{2} h_{1}: \frac{1}{3} \pi r_{2}^{2} h_{2}$$
= 22 × 1 : 12 × 2 = 4 : 2 = 2 : 1

Question 12.
If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to ________
(1) 8π cm2
(2) 16 cm2
(3) 12π cm2
(4) 16π cm2
(4) 16π cm2
Hint:
C.S.A of a sphere = 4πr2 sq. units
= 4 × π × 22 cm2
= 16π cm2

Question 13.
The total surface area of a solid hemisphere of diameter 2 cm is equal to _______
(1) 12 cm2
(2) 12π cm2
(3) 4π cm2
(4) 3π cm2
(4) 3π cm2
Hint:
Radius of a hemisphere = $$\frac{2}{2}$$ = 1 cm
Total surface area of a hemisphere = 3πr2 sq. units = 3 × π × 12 cm2 = 3π cm2

Question 14.
If the volume of a sphere is $$\frac{9}{16} \pi$$ cu.cm, then its radius is ________
(1) $$\frac{4}{3}$$ cm
(2) $$\frac{3}{4}$$ cm
(3) $$\frac{3}{2}$$ cm
(4) $$\frac{2}{3}$$ cm
(2) $$\frac{3}{4}$$ cm
Hint:
Volume of the sphere = $$\frac{9}{16} \pi$$

Question 15.
The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio _______
(1) 81 : 625
(2) 729 : 15625
(3) 27 : 75
(4) 27 : 125
(4) 27 : 125
Hint:
Ratio of their surface area = 9 : 25

Question 16.
The total surface area of a solid hemisphere whose radius is a units, is equal to ________
(1) 2πa2 sq. units
(2) 3πa2 sq. units
(3) 3πa sq. units
(4) 3a2 sq. units
(2) 3πa2 sq. units
Hint:
T.S.A. of a solid hemisphere = 3πr2 sq. units
= 3 × π × a × a sq.units
= 3πa2 sq. units

Question 17.
If the surface area of a sphere is 100π cm2, then its radius is equal to ______
(1) 25 cm
(2) 100 cm
(3) 5 cm
(4) 10 cm
(3) 5 cm
Hint:
Surface area of a sphere = 100π cm2
4πr2 = 100π
r2 = 25
r = √25 = 5 cm

Question 18.
If the surface area of a sphere is 36π cm2, then the volume of the sphere is equal to _______
(1) 12π cm3
(2) 36π cm3
(3) 72π cm3
(4) 108π cm3
(2) 36π cm3
Hint:
Surface area of a sphere = 36π cm2
4πr2 = 36π
r2 = 9
r = 3 cm
Volume of a sphere = $$\frac{4}{3} \pi r^{3}$$ cu. units
= $$\frac{4}{3} \pi$$ × 3 × 3 × 3 cm3 = 36π cm3

Question 19.
If the total surface area of a solid hemisphere is 12π cm2 then its curved surface area is equal to ______
(1) 6π cm2
(2) 24π cm2
(3) 36π cm2
(4) 8π cm2
(4) 8π cm2
Hint:
T.S.A of a hemisphere = 12π cm2
3πr2 = 12π
r2 = 4
r = 2
Curved surface area of a hemisphere = 2πr2 = 2 × π × 4 = 8π cm2

Question 20.
If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio _____
(1) 1 : 8
(2) 2 : 1
(3) 1 : 2
(4) 8 : 1
(1) 1 : 8
Hint:
$$r_{1}=\frac{r_{2}}{2} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2} \Rightarrow r_{1}: r_{2}=1: 2$$

Question 1.
Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.
Given, Circumference of the base of a cylinder = 110 cm
2πr = 110 ……. (1)
Curved surface area = 4400 cm2
2πrh = 4400 cm2 ……. (2)
From (1) & (2), $$\frac{(2)}{(1)} \Rightarrow \frac{2 \pi r h}{2 \pi r}=\frac{4400}{110}=40 \mathrm{cm}$$
Height of the cylinder (h) = 40 cm
From (1), 2πr = 110
2 × $$\frac{22}{7}$$ × r = 110
r = $$\frac{35}{2}$$
We know that, diameter (d) = 2 × radius
d = 2 × $$\frac{35}{2}$$ = 35 cm
Diameter of the Circular cylinder = 35 cm

Question 2.
A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost of painting the lateral surface of the pillars at ₹ 20 per square metre.
Given, Radius of a cylinder (r) = 50 cm = 0.5 m
Height of a cylinder (h) = 3.5 m
Curved surface area of a pillar = 2πrh sq. units
Curved surface area of 12 pillars = 12 × 2πrh
= 12 × 2 × $$\frac{22}{7}$$ × 0.5 × 3.5 m2
= 132 sq. m.
Cost for painting the lateral surface of pillars per metre = ₹ 20
Cost of painting = 132 × ₹ 20 = ₹ 2640

Question 3.
The total surface area of a solid right circular cylinder is 231 cm2. Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.
Given, Total surface area of a cylinder (T.S. A) = 231 sq.cm
Curved surface area = $$\frac{2}{3}$$ × T.S.A = $$\frac{2}{3}$$ × 231 = 154 cm2
2πrh = 154 cm2 …… (1)
Total surface area = 231 cm2
2πr (h + r) = 231
2πrh + 2πr2= 231
154 + 2πr2 = 231 [from (1)]
2πr2 = 231 – 154 = 77

Radius of the cylinder = 3.5 cm
Height of the cylinder = 7 cm

Question 4.
The total surface area of a solid right circular cylinder is 1540 cm2. If the height is four times the radius of the base, then find the height of the cylinder.
Given, Let the radius of the cylinder be ‘r’
Height of a cylinder (h) = 4r (by given condition)
Total surface area = 1540 cm2
2πr(h + r) = 1540 cm2

Height of the cylinder = 4r = 4 × 7 = 28 cm

Question 5.
If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.
Given, In the figure, OAB is a cone and OC ⊥ AB
∠AOC = $$\frac{60^{\circ}}{2}$$ = 30°
In the right ∆OAC, tan 30° = $$\frac{\mathrm{AC}}{\mathrm{OC}}$$

Slant Height of the cone (l) = 15 × 2 = 30 cm

Question 6.
The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.

Given, Radius of a sector (r) = 21 cm
The angle of the sector (θ) = 180°
Let “R” be the radius of the cone.
Circumference of the base of a cone = Arc length of the sector

Radius of the cone (R) = 10.5 cm

Question 7.
If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.
Given, the Curved surface area of a solid hemisphere = 2772 cm2
2πr2 = 2772

Total surface area = 3πr2 sq. units
= 3 × $$\frac{22}{7}$$ × 21 × 21
= 4158 sq.cm
Aliter:
C.S.A of a hemisphere = 2772 cm2
2πr2 = 2772 cm2
πr2 = $$\frac{2772}{2}$$ = 1386 cm
T.S.A of a hemisphere = 3πr2 sq.units = 3 × 1386 cm2 = 4158 cm2

Question 8.
An inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of ₹ 5 per sq. m.
Given, Circumference of the dome = 17.6 m
2πr = 17.6
$$r=\frac{17.6 \times 7}{2 \times 22}=\frac{8.8 \times 7}{22}=2.8 \mathrm{m}$$
Curved surface area of the dome = 2πr2 sq. units
= 2 × $$\frac{22}{7}$$ × 2.8 × 2.8 m2
= 49.28 m2
Cost of painting for one sq.metre = ₹ 5
Cost of painting the curved surface = 49.28 × ₹ 5 = ₹ 246.40

Question 9.
Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.
Given, Height of a cylinder (h) = 4.5 cm
Volume of a solid cylinder = 62.37 cu. cm

Radius of a cylinder (r) = 2.1 cm

Question 10.
A rectangular sheet of metal foil with dimension 66 cm × 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.
After rolling the rectangular sheet into a cylinder

Volume of the cylinder = 4158 cm3

Question 11.
The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.
Given, Height of the wooden solid cone (h) = 12 m
Circumference of the base = 44 m
2πr = 44
r = $$\frac{44 \times 7}{2 \times 22}$$ = 7 m
Volume of the wooden solid = $$\frac{1}{3} \pi r^{2} h$$ cu. units
= $$\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \mathrm{m}^{3}$$
= 88 × 7
= 616 m3
Volume of the solid = 616 m3

Question 12.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.
Given, Edge of the cube = 14 cm
The largest circular cone is cut out from the cube.
Radius of the cone (r) = $$\frac{14}{2}$$ = 7 cm
Height of the cone (h) = 14 cm
Volume of a cone

Volume of a cone = 718.67 cm3

Question 13.
The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. (Take π = $$\frac{22}{7}$$)
Let r, R and w be the inner and outer radii and thickness of the hemispherical bowl respectively.

Given that r = 5 cm, w = 0.25 cm
R = r + w = 5 + 0.25 = 5.25 cm
Now, outer surface area of the bowl = 2πR2
= 2 × $$\frac{22}{7}$$ × 5.25 × 5.25
= 173.25 sq. cm
Thus, the outer surface area of the bowl = 173.25 sq. cm

Question 14.
Volume of a hollow sphere is $$\frac{11352}{7}$$ cm3. If the outer radius is 8 cm, find the inner radius of the sphere. (Take π = $$\frac{22}{7}$$)
Let R and r be the outer and inner radii of the hollow sphere respectively.
Let V be the volume of the hollow sphere.

Hence, the inner radius r = 5 cm

Question 15.
How many litres of water will a hemispherical tank whose diameter is 4.2 m?
Radius of the tank = $$\frac{4.2}{2}$$ = 2.1 m
Volume of the hemisphere
= $$\frac{2}{3} \pi r^{3}$$ cu.units
= $$\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{m}^{3}$$
= 19.404 m3
= 19.404 x 1000 lit
= 19,404 litres

Question 1.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

For cylindrical part:
Height (h) = 6 cm
Curved surface area = 2πrh = 2 × $$\frac{22}{7}$$ × 7 × 6 cm2 = 264 cm2
For hemispherical part:
Surface area (h) = 2πr2
= 2 × $$\frac{22}{7}$$ × 7 × 7 cm2
= 308 cm2
Total surface area = (264 + 308) = 572 cm2

Question 2.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Question 3.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

For cylinderical part:
Height (h) = 2.4 cm
Diameter (d) = 1.4 cm
Total surface area of the cylindrical part

For conical part:
Base area (r) = 0.7 cm
Height (h) = 2.4 cm

Question 4.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Diameter of the cylindrical well = 7 m
Radius of the cylinder (r) = $$\frac{7}{2}$$ m
Depth of the well (h) = 20 m
Volume = πr2h
= $$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{m}^{3}$$
= 22 × 7 × 5 m3
Volume of the earth taken out = 22 × 7 × 5 m3
Now this earth is spread out to form a cuboidal platform having
Length (l) = 22 m
Let ‘h’ be the height of the platform.
Volume of the platform = 22 × 14 × h m3
Volume of the platform = Volume of the earth taken out
22 × 14 × h = 22 × 7 × 5
$$h=\frac{22 \times 7 \times 5}{22 \times 14}=\frac{5}{2} \mathrm{m}=2.5 \mathrm{m}$$
Thus, the required height of the platform is 2.5 m.

Question 5.
The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum is 8 cm, find the whole surface area of the frustum. [Use π = 3.14]

Let the radii of circular ends are R and r [R > r]
Perimeter of circular ends are 207.24 cm and 169.56 cm
2πR = 207.24 cm

The whole surface area of the frustum = π [(R2 + r2) + (R + r) l]
Required whole surface area of the frustum
= 3.14 [332 + 272 + (33 + 27) × 10] cm2
= 3.14 [1089 + 729 + 600] cm2
= 3.14 [2418] cm2
= 7592.52 cm2

Question 6.
A cuboid-shaped slab of iron whose dimensions are 55 cm × 40 cm × 15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe. (Take π = $$\frac {22}{7}$$)

Let h1 be the length of the pipe
Let R and r be the outer and inner radii of the pipe respectively.
Iron slab:
Volume = lbh = 55 × 40 × 15 cm3
Iron pipe:
Outer diameter, 2R = 8 cm
Outer radius, R = 4 cm
Thickness, w = 1 cm
Inner radius, r = R – w = 4 – 1 = 3 cm
Now, the volume of the iron pipe = Volume of the iron slab

Time is taken by the pipe to empty half of the tank = 3 hours 12 minutes.

Question 7.
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4π cm. If the depth is 14 cm., then find its volume.
Given let the radius of the top of the frustum be “R” and the radius of the bottom of the frustum be “r”

Question 8.
A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.

Given, Total height of solid = 13.5 cm
Diameter of the cylinder (d) = 28 m
Height of a cylinder (h) = 3 m
Height of a conical portion = 13.5 – 3 = 10.5 m
From the diagram, Radius of a cone = Radius of a cylinder

## Samacheer Kalvi 10th English Guide Poem 6 No Men Are Foreign

Tamilnadu State Board New Syllabus Samacheer Kalvi 10th English Guide Pdf Poem 6 No Men Are Foreign Questions and Answers, Summary, Notes.

## Tamilnadu Samacheer Kalvi 10th English Solutions Poem 6 No Men Are Foreign

### 10th English Guide No Men Are Foreign Textbook Questions and Answers

A. Based on the understanding of the poem, read the following lines and answer the questions given below.

1. “Beneath all uniforms, a single body breathes
Like ours: the land our brothers walk upon
Is earth like this, in which we all shall lie”
(a) What is found beneath all uniforms?
(b) What is same for every one of us?
(c) Where are we all going to lie finally?
(d) What is the alliterated words in the 2nd line?
(e) What is the figure of speech in the 2nd and 3rd line?
(a) Human body is found beneath all uniforms.
(b) The earth we walk up on is the same for every one of us.
(c) We are all going to lie beneath the earth.
(d) beneath; body; breathes
(e) Simile

2. “They, too, aware of sun and air and water,
Are fed by peaceful harvests, by war’s long winter starv ’d.
(a) What is common for all of us? (or) What are they aware of?
(b) How are we fed?
(c) Mention the season referred here.
(a) The sun, air and water are common for all of us.
(b) We are fed by peaceful harvest.
(c) The winter season

3. Their hands are ours, and in their lines we read A labour not different from our own.
(a) Who does “their” refer to?
(b) What does the poet mean by lines we read?
(c) What does not differ?
(a) ‘Their’ refers to the other people of the world whom we consider as strange and foreign.’
(b) The poet by the words, Tines we read’ means that their destiny is similar to ours.The lines of their hands also show their capacity of doing hard work or labour.
(c) Labour does not differ.

No Men Are Foreign MCQ Questions Class 9 English with Answers.

4. “Let us remember, whenever we are told
To hate our brothers, it is ourselves
That we shall dispossess, betray, condemn ”
(a) Who tells us to hate our brothers?
(b) What happens when we hate our brothers?
(c) What do we do to ourselves?
(a) The evil rulers tell us to hate our brothers.
(b) When we hate our brothers, we hate ourselves.
(c) We dispose, betray and blame ourselves.

5. “Our hells of fire and dust outrage the innocence
Of air that is everywhere our own,
Remember, no men are foreign, and no countries Strange ”
(a) What outrages the innocence?
(b) Who are not foreign?
(c) What is not strange?
(d) Who defiles the earth?
(a) War, which is futile spoiling the very earth with hells of fire and dust outrages the innocence.
(b) Any human being who breathes the same air is not foreign.
(c) The world which becomes a more difficult place to live in and any country in this world is not strange.
(d) The men who fight with each other defile our earth.

1. Remember they have eyes like ours that wake Or sleep, and strength that can be won By love
(a) What do they have like ours?
(b) What do the eyes do?
(c) How can strength be won?
(a) They have eyes like ours.
(b) The eyes wake up or sleep.
(c) Strength can be won by love.

2. In every land is common life
That all can recognize and understand
(a) What is common in every land?
(b) What can all recognise and understand?
(a) Life is common in every land.
(b) All can recognise and understand that life is common in everyland.

3. Remember we who take arms against each other
It is the human earth that we defile
(a) Who defiles the earth?
(b) Whose earth is this?
(a) We who take arms against each other defile the earth.
(b) This is our earth. It is the human earth.

4. Or sleep, and strength that can be won.
Pick out the words that are in alliteration in this line.
The alliterated words are: Sleep, Strength

5. Remember, we who take arms against each other.
Write down the words that are in assonance here.
The words in assonance are : arm, against.

6. Beneath all uniforms, a single body breathes like ours;
(a) Who does all refer to?
(b) What does the poet denote?
(a) All refers to the people from the countries.
(b) The poet denotes universal brotherhood and equality.

7. Are fed by peaceful harvests, by war’s long winter starv’d
What is the poetic device employed here?
The poetic device employed here is transferred epithet’. It is used in the phrase – “winter starv’d”.

8. Our hells of fire and dust outrage the innocence.
(a) What is the figure of speech used here?
Metaphor is used here. ‘Hells of fire’ is a metaphor.

9. Remember, no men are strange, no countries foreign
Remember, they have eyes like ours that wake
Remember, we who take arms against each other
Remember, no men are foreign, and no countries strange
What is the figure of speech used in these lines?
The figure of speech used here is ‘repetition’.

B. Based on your understanding of the poem complete the following by choosing the appropriate words/phrases given in brackets:

This poem is about the ………………….. (1) …………………… of all men. The subject of the poem is the …………………. (2) …………………. race, despite of the difference in colour, caste, creed, religion, country, etc. All human beings are the same. We walk on the ……………… (3) ……………….. and w/-e will be buried under it. Each and every one of us is related to the other. We all are born the same and die in the same way. We may wear different uniforms like ………………….. (4) ………………… during wars the opposing side will also have the same …………………… (5) ……………….. like ours. We as human do the same labour with ……………….. (6) ………………… and look at the world with the …………………. (7) ………………… Waging war against others as they belong to a different country is like attacking our own selves. It is the ………………… (8) ………………….. we impair. We all share the same ……………………. (9) …………………….. We are similar to each other. So the poet concludes that we shouldn’t have wars as it is …………………… (10) ……………………. to fight against us.
(unity of human, dreams and aspirations, same land, our hands, unnatural, breathing body, same eyes, brotherhood, language, human-earth)

1. brotherhood
2. unity of human
3. same land
4. language
5. breathing body
6. our hands
7. same eyes
8. dreams and aspirations
9. human earth
10. unnatural

C. Based on your understanding of the poem answer the following questions in a paragraph of about 100-150 words.

Question 1.
What is the central theme of the poem “No Men are Foreign”.
James Kirkup gives a positive message of hope to mankind. In spite of obvious divisions and variances, all are united together by the common bond of civilization and mankind. For their entrusted interests, some selfish people divide lands and people. They collaborate to create hatred and divisions among people. The poet validates the statement that people living in different countries are essentially the same by proclaiming that ‘no men are strange and no men are foreign’.

That is the part of the title of the poem and it is the central theme too. Every single body breathes and functions in the same way as ours. Each one of us equally needs the sun, air and water. Human hands too are used for the similar purpose of labouring for livelihood. Even eyes perform similar purpose of sleeping and waking up. Love wins us all and we all identify its power.

In peace times, we all flourish and wars starve us. Hatred leads us astray and when we take up arms against each other, the entire earth is defiled and destroyed. Therefore, we all like peace which showers abundance and prosperity on us. Therefore, fundamentally we all are the same.

We should understand and try to recognise that the same soul runs through all the people. Let us work for the unity and affluence of all lands and all people. Let us not pollute and taint the earth which is ours. Hatred and narrow ideas pollute the minds of the people.

Conflicts and wars bring destruction and violence. We should remember that raising our arms against anyone means fighting against ourselves. The poet reminds us to remember, recognise and strengthen the common bond that unites mankind and humanity which is the main theme of this poem.

‘Sometimes one feels better speaking to a stranger than someone known.’

(OR)

Poem: No men are foreign
Poet: James Falconer Kirkup
Theme: Universal brotherhood

Human beings are same. We walk on the same land. We will be buried under it. We are related to each other. We all utilise the sunlight, air and water. We live by the food crops of the earth. All have their dreams and aspiration like us. We do the same labour with our hands. We look at the world with the same eyes. We hate ourselves, when we hate our brothers. We are similar to each other. We shouldn’t have war and fight against them.

(OR)

1. Human beings are same.
2. In the world, all are brothers.
3. We live and die in the same earth.
4. The sun, air, water are common to all.
5. We live by the earth food crops.
6. Land is common to all.
7. Love strengthens us.
8. During war, we hate our brothers.
9. So, learn to live in peace with all.

Question 2.
The poem “No Men are Foreign” has a greater relevance ¡n today’s world – Elucidate.
Introduction:
The poem “No men are foreign” has a greater relevance ¡n today’s world. Let us see what are the relevance in today’s world.

Theme of the poem:
The poem tells us about unity of human race, despite the differences ¡n colour, caste, creed, etc. The poet tells that there are a lot of conflicts and disasters in and out of every country.

Today’s world:
This has totally affected the world peace and harmony. We defile our earth by means of war. Enmity and hatred must be given to peace and harmony. The earth is full of fire and dust created by means of war. We should not hate our brothers.

Conclusion:
In such situation of considering the earth as the single living place for all, we shall live together, strengthened by love. Thus the earth will be a better place and there will be no more of fire and dust.

(OR)

The poem ‘No Men are Foreign’ has a greater relevance in today’s world. There are a lot of fights and disasters in and out of every countries. The wars are always there between countries for one matter or the other. This has totally affected the world peace and harmony. No one wants to be defeated by the other.

No one bothers about the damage caused to the countries and the loss of lives of lot of people. People pay more attention to the differences and shoot troubles always. As there is no love and spirit of brotherhood, they wage war every now and then. Enmity and hatred must be given to maintain peace and harmony.

Let us consider the earth as the single living place for all who live on it. In such a situation we shall live together, strengthened by love, admiring one another with better understanding. Thus the earth will be a paradise and there will be no more hell of fire and dust.

(OR)

1. This poem has greater relevance to the present world.
2. There is endless war everywhere.
3. There is no spirit of brotherhood.
4. We look at others as different creations.
5. But we are the same and live on the same earth.
6. We have fights and confusions because of our nationalism and racism.
7. We can overcome this if we think about the oneness of mankind.
8. Our inventions destroy us.
9. Only understanding each other can save us.
10. Love alone can bind us and bring out of our blind thoughts.
11. The earth will be our paradise if we give up the differences.
12. Let “No Men are Foreign” be our motto to promote us to peaceful life.

### No Men Are Foreign Summary of the poem

The poem ‘No men are foreign’ tells about that no people unity of the human race. The people of one country shouldn’t think of the people of other countries as strangers. The poet tells that they are all humans the same as we should not hate it, brothers. At last, the poet asks us to remember that no men are strange and no countries foreign.

Glossary:

Condemn – express complete disapproval
Labour – hard work
Betray – disloyal
Defile – damage the purity or appearance
Outrage – the extremely strong reaction of anger shock

## Samacheer Kalvi 10th Social Science Model Question Paper 3 English Medium

Students can Download Samacheer Kalvi 10th Social Science Model Question Paper 3 English Medium Pdf, Samacheer Kalvi 10th Social Science Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the questions in each part. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III, and IV are to be attempted separately.
4. Question numbers 1 to 14 in Part I are Multiple Choice Questions of one mark each.
These are to be answered by writing the correct answer along with the corresponding option code and the corresponding answer
5. Question numbers 15 to 28 in Part II are of two marks each. Any one question should be answered compulsorily.
6. Question numbers 29 to 42 in Part III are of five marks each. Any one question should be answered compulsorily.
7. Question numbers 43 to 44 in Part IV are of Eight marks each. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 100

Part – I

Answer all the questions. Choose the correct answer [14 × 1 = 14]

Question 1.
What were the three major empires shattered by the end of First World War?
(a) Germany, Austria-Hungary, and the Ottomans
(b) Germany, Austria-Hungary, and Russia
(c) Spain, Portugal and Italy
(d) Germany, Austria-Hungary, Italy
(a) Germany, Austria-Hungary, and the Ottomans

Question 2.
Which quickened the-process of liberation in South America?
(a) Support of US
(b) Napoleonic Invasion
(c) Simon Bolivar’s involvement
(d) French Revolution
(d) French Revolution

Question 3.
When was people’s Political consultative conference held in China?
(a) September 1959
(b) September 1948
(c) September 1954
(d) September 1949
(d) September 1949

Question 4.
Who were driven out of their homeland during the process of creation of zamins under permanent Settlement?
(a) Santhals
(b) Titu Mir
(c) Munda
(d) Kol.
(a) Santhals

Question 5.
Which among the following was declared as ‘Independence Day’?
(a) 26th January 1930
(b) 26th December 1929
(c) 16th June 1946
(d) 15th January 1947
(d) 15th January 1947

Question 6.
The extent of Himalayas in the east-west is about …………………
(a) 2,500 km
(b) 2,400 km
(c) 800 km
(d) 2,200 km
(a) 2,500 km

Question 7.
Western disturbances cause rainfall in …………………
(b) Kerala
(c) Punjab
(c) Punjab

Question 8.
The soil formed by the river are …………………
(a) Red soils
(b) Black soils
(c) Desert soils
(d) Alluvial soils
(d) Alluvial soils

Question 9.
Retreating monsoon wind pick up moisture from …………………
(a) Arabian sea
(b) Bay of Bengal
(c) Indian ocean
(d) Timor sea
(b) Bay of Bengal

Question 10.
Second staple food of the people of Tamil Nadu is …………………
(a) Pulses
(b) Millets
(c) Oilseeds
(d) Rice
(b) Millets

Question 11.
Which one of the following rights was described by Dr. B. R. Ambedkar as the heart and soul of the constitution?
(a) Right to freedom of religion
(b) Right to equality
(c) Right to constitutional remedies
(d) Right to property
(c) Right to constitutional remedies

Question 12.
Which of the following country is not the founder member of NAM?
(a) Yugoslavia
(b) Indonesia
(c) Egypt
(d) Pakistan
(d) Pakistan

Question 13.
………………… is the process of providing or obtaining the food necessary for health and growth.
(a) Health
(b) Nutrition
(c) Sanitation
(d) Security
(b) Nutrition

Question 14.
Gross value added at current prices for services sector is estimated at lakh crore
in 2018-19.
(a) 91.06
(b) 92.26
(c) 80.07
(d) 98.29
(b) 92.26

Part – II

Answer any 10 questions. Question No. 28 is compulsory. [10 x 2 = 20]

Question 15.
Access the role of Ayyankali in fighting for the cause of “Untouchables”.
(i) Ayyankali brought tremendous social changes especially in caste structure. The discrimination he faced as a child turned him into a leader of an anti-caste movement and who later fought for basic rights including access to public spaces and entry to schools.

(ii) Ayyankali challenged many caste conventions such as clothing style, he wore clothes associated with upper castes that were prohibited for lower castes.

Question 16.
How did Hitler get the support from the people of germany?
(i) Hitler was well aware of the discontent among the Germans. He used his oratorical skills to sway the common people and promised them to return the glorious military past of Germany. He founded the National Socialist Party, generally known as “the Nazis”.

(ii) The fundamental platform on which Hitler built his support was the notion of the racial superiority of the Germans as a pure, ‘Aryan’ race and a deep-seated hatred of the Jews. Hitler came to power in 1933 and ruled Germany for twelve long years.

Question 17.
Why was Heron dismissed from service?

• Colonel Heron was urged to deal with Puli Thevar as he continued to defy the authority of the company. Puli Thevar wielded much influence over the western Palayakkarars.
• Heron had to abandon the plan for want of cannon and of supplies and pay to soldiers. He retired to Madurai. He was then recalled and dismissed from service.

Question 18.
Why did Gandhi withdraw the Non- cooperation Movement?

• The Non-cooperation Movement started in 1920. It soon became a nation-wide movement because it got support of the people across the country. But in February 1922, a violent incident occurred at Chauri Chaura, a village near Gorakhpur in Uttar Pradesh.
• In this incident a procession of nationalists provoked by the police turned violent. The police finding themselves outnumbered shut themselves inside the police station.
• The mob burnt the police station in which 22 policemen lost their lives. The incident hurt Gandhiji too much and he immediately withdrew the movement.

Question 19.
What are “Jet Stream”?
In the upper layers of the atmosphere, there are strong westerly winds concentrated in a relatively narrow and shallow streams known as “Jet streams” They cause heavy rainfall in North-west India.

Question 20.
State any two characteristics of block cotton soil.

• This soil is rich in calcium carbonate, magnesium, potash, lime and iron but deficient in phosphorous. It is clayey and impermeable which has great capacity to retain moisture for a long time.
• It becomes sticky when wet but develops cracks during dry summer season. The soil is suited for dry farming due to its high moisture retentivity.

Question 21.
Name the tributaries of river Thamirabarani.
Karaiyar, Servalar, Manimuthar, Gadananathi, Pachaiyar, Chittar and Ramanathi are its main tributaries.

Question 22.
List out the three leads of the relations between the centre and the states.
There are –

• Legislative relations
• Financial relations.

Question 23.
(i) India is building the Kaladan Multi-Model Transit Transport, a road-river-port cargo transport project to link Kolkata to Sittwe in Myanmar.

(ii) A project aiming to connect Kolkata with Ho Chi Minh City on the South Sea for the formation of an economic zone will have a road pass through Myanmar, Cambodia and Vietnam and work on the first phase connecting Guwahati with Mandalay is currently underway.

Question 24.
What is national emergency?
National emergency is a situation beyond the ordinary. The President declares this emergency if he is satisfied that India’s security is threatened due to war, external aggression or armed rebellion or if there is an imminent danger or threat.

Question 25.
How the state of Jammu and Kashmir differ from the other states of India.

• The Constitution of India grants special status to Jammu and Kashmir among Indian States, and it is the only state in India to have a separate Constitution.
• The Directive Principles of the State Policy and Fundamental Duties of the Constitution are not applicable to the State of Jammu and Kashmir.
• Rights to property, which is denied as a Fundamental Right to rest of India is still guaranteed in Jammu and Kashmir.

Question 26.
what are the factors supporting to develop the Indian economy?
Factors supporting to develop the Indian economy :

• A fast growing population of working age.
• India has a strong legal system and many English language speakers
• Wage costs are low here.
• India’s economy has successfully developed highly advanced and attractive clusters of business in the technology space.

Question 27.
What is the main objective of WTO?
The main objective of WTO is to set and enforce rules for international trade and to provide a forum for negotiating and monitoring further trade liberalisations.

Question 28.
Name the important multipurpose projects of Tamil Nadu.
Mettur Dam, Amaravathi Dam, Papanasam Dam, Bhavani Sagar Dam.

Part – III

Answer any 10 questions. Question No. 42 is compulsory. [10 x 5 = 50]

Question 29.
Fill in the blanks:
(i) ………………… treaty signed on February 7, 1922 created the EU.
(ii) The riverine island of Srirangam is located between …………………and ………………… branches of Cauvery.
(iii) …………………was India’s policy in the face of the bipolar order of the cold war.
(iv) In the year ………………… National Food Security Act was passed by the Indian Parliament.
(v) Tamil Nadu ranks …………………in India with a share of over 20% in total road projects under
operation in the Public Private Partnership (PPP).
(i) The Maastricht
(ii) Northern, Southern
(iii) Non-Alignment
(iv) 2013
(v) Second

Question 30.
Match the following:

Question 31.
Match the following:

Question 32.
(a) Distinguish between
(i) North East Monsoon and South West Monsoon
(ii) Rabi and Kharif Crop Season
(a) (i) North East Monsoon and South East Monsoon:
North East Monsoon :

1. North east monsoon occurs between October and December.
2. It is Winter Monsoon.
3. The coromandal coast of Tamil Nadu gets, heavy rainfall from north east monsoon.

South West Monsoon :

1. Southwest Monsoon occurs between June to September.
2. It is Summer Monsoon.
3. The districts of Nilgiris, Kanyakumari, Salem, Coimbatore and Erode get rainfall.

(ii) Rabi and Kharif crop season:
Rabi Crop Season :

1. Rabi crops are sown in the beginning of the winter (i.e.) in November.
2. The crops are harvested in the beginning of summer (i.e) in March. days of November.
3. Major crops are wheat, tobacco, mustard, pulses, linseed and grains.

Kharif Crop Season :

1. Kharif crops are sown in the beginning of monsoon (i.e.) in June.
2. The crops are harvested in the early
3. Major crops are paddy, maize, cotton, millets, jute and sugarcane.

(b) Give reason: Alluvial soil is fertile
Alluvial soil are formed by the deposition of silt by the rivers. Alluvial soils are generally fertile as they are rich in minerals such as lime, potassium, magnesium, nitrogen and phosphoric acid. It is deficient in nitrogen and humus. It is porous and loamy.

Question 33.
Discuss the main causes of the First World War.
The causes of the First World War are given below:

• Formation of European alliances and counter alliances
• Emergence of violent forms of nationalism in countries like England, France and Germany
• Aggressive attitude of the German Emperor Kaiser Wilhelm II
• Hostility of France towards Germany
• Opportunity for imperial power politics in the Balkans
• The Balkans wars
• Immediate cause which included the assassination of Archduke Franz Ferdinand,nephew and heir to Franz Joseph, Emperor of Austria-Hungary, by Princip, a Bosnian Serb, on 28 June 1914.

Question 34.
Discuss the causes and consequences of the Revolt of 1857?
The Great Rebellion of 1857 is a unique example of resistance to the British authorities, in India. There were several reasons that triggered the Revolt:

(i) The annexation policy of British India created dissatisfaction among the native rulers. The British claimed themselves as paramount, exercising supreme authority. New territories were annexed on the grounds that the native rulers were corrupt, and inept.

(ii) The British annexed several territories such as Satara, Sambalpur, parts of Punjab, Jhansi and Nagpur through the Doctrine of Lapse. This also angered many Indian rulers.

(iii) Indian sepoys were upset with discrimination in salary and promotion. They were paid much less than their European counterparts. They felt humiliated and racially abused by their seniors.

Consequences:

• India was pronounced as one of the many crown colonies to be directly governed by the Parliament. This resulted in the transfer of power from the East India company to the British crown.
• Queen Victoria proclaimed to the Indian people that the British government would not interfere in traditional institutions and religious matters. It was promised that Indians would be absorbed in government services.
• There came significant changes in the Indian army. The number of Indians was reduced. Indians were restrained from holding important ranks and position.
• It was also decided that instead of recruiting soldiers from Rajputs, Brahmins and North Indian Muslims, more soldiers would be recruited from the Gorkhas, Sikhs and Pathans.

Question 35.
Bring out the characteristics of Intensive and Plantation farming.
Intensive farming:

• It involves various types of agriculture with higher levels of input, such as capital and labour, per unit of agricultural land area.
• It aims to maximize yields from available land through various means, such as heavy use of pesticides and chemical fertilizers. ,
• Intensive farming is practiced in Punjab, parts of Rajasthan, Uttar Pradesh, and Madhya Pradesh in India.

Plantation farming:

• It is a single crop farming, practised on a large area.
• Crops are mainly grown for the market.
• It is both labour intensive and capital intensive.
• It has an interface of agriculture and industry.
• Developed network of transport and communication connecting the plantation processing industries and markets play an important role in the development of plants. Example- tea, coffee, rubber, sugarcane, etc.

Question 36.
Write an account on river Cauvery.

• The main river of Tamil Nadu is Cauvery which originates at Talacauvery in the Brahmragiri hills of Kodagu (Coorg) district of Karnataka in the Western Ghats.
• About 416 km of its course falls in Tamil Nadu. It serves as the boundary between Karnataka
and Tamil Nadu for a distance of 64 km. A tributary called Bhavani joins Cauvery on the right bank about 45 km from the Mettur Reservoir.
• Thereafter, it takes easterly course to enter into the plains of Tamil Nadu. Cauvery and its distributaries in its lower course drain the districts of Nagapattinam, Thanjavur, Thivarur and Thiruchirapalli.
• The Cauvery, Kollidam and the vellar jointly drain central part of the Tamil Nadu. The head of the Cauvery delta is near the islands of Srirangam. Kollidam branches off from cauvery at Grand Anaicut, also called as Kallanai was built across the river Cauvery.
• After Kallanai, the river breaks into a large number of distributaries and forms a network all over the delta. The network of tributaries within the delta of cauvery in the coast is called as the ‘Garden of Southern India’.
• It merges into Bay of Bengal to the south of Cuddalore. Cauvery along with its tributaries Bhavani, Noyyal, Mayar and Amaravathi is the most important source of canal irrigation.

Question 37.
Write briefly on the Right to Constitutional Remedies.
(i) Our Constitution guarantees six Fundamental Rights to its citizens. It safeguards all these rights by granting us the Right to Constitutional Remedies. It is possible that the Government or private bodies may violate one of our Fundamental Rights.

(ii) Right to Constitutional Remedies protects us from such violations. It allows us to file a case against the Government or private bodies in the High Courts of the States and the Supreme Court of the India.

(iii) Both the Supreme Court and High Courts are empowered to issue five kinds of writs of habeas corpus, mandamus, prohibition two warrants and certiorari to protect the Fundamental Rights of the citizens. That is why the Supreme Court is called the “Guardian of the Constitutions”.

Question 38.
Briefly discuss the Functions of the State Legislature.
The powers and functions of the State Legislature are almost the same as that of the Parliament.

• The State Legislature can pass laws on all subjects mentioned in the State List as per the constitutions. It can also pass laws on concurrent subjects.
• The Legislature controls the finances of the State. The Lower House enjoys greater power than the Upper House in money matters. Money Bills can be introduced only in the Lower House of the Assembly.
• The Legislature controls the Executive. The council of Ministers is responsible to the Assembly. The ministers have to answer questions asked by the members of the Legislature.
• The Council cannot vote for grants.
• No new tax can be levied without the sanction and permission of the Assembly.

Question 39.
Multinational companies first started their activities in the extractive industries and controlled raw materials in the host countries during 1920s and then entered the manufacturing and service sectors after 1950s. Most of the MNCs at present belong to the four major exporting countries i.e., USA, UK, France and Germany. However, the largest is America. In 1971, the American Corporations held 52% of the total world stock of foreign direct inverstment.
Great Britain held 14.5% followed by France 5% and Federal Republic of Germany 4.4% and Japan 2.7%.

• MNCs produce the same quality of goods at lower cost and without transaction cost.
• They reduce prices and increase the purchasing power of consumers world wide.
• They are able to take advantage of tax variation.
• They spur job growth in the local economy.

Disdvantages of MNCs:

(i) They have led to the downfall of smaller, local business.

(ii) With more companies transferring offices and centering operations in other countries, jobs for the people living in developed countries are threatened.

(iii) MNCs often invest in developing countries where they can take advantage of cheaper labour. Some MNCs prefer to put up branches in these parts of the world where there are no stringent policies in labour and where people need jobs because these MNCs can demand for cheaper labour and lesser healthcare benefits.

Question 40.
Write a note on history of industrialisation in Tamil Nadu.

Industrialisation in the Colonial Period:

• The introduction of cotton cultivation in western and southern Tamil Nadu by the colonial government led to the emergence of a large-scale textile sector in these parts, which involved ginning, pressing, spinning and weaving operations.
• Introduction of railways also expanded the market for cotton yam and helped develop the sector.
• There was increase in trade during this period which led to industrial development. The two active ports in the region were Chennai and Tuticorin.
• In Western Tamil Nadu, the emergence of textiles industries also led to demand and starting of textile machinery industry in the region.

Post-Independence to early 1990s:

• After independence, several large enterprises were set up by both the central and state governments.
• The Integral Coach Factory in Chennai made railway coaches and the Bharat Heavy Electricals Limited (BHEL) in Tiruchirapalli manufactured boilers and turbines.
• Ashok Motors and Standard Motors together helped form an automobile cluster in the Chennai region.
• The 1970s and 1980s saw the setting up of emergence of powerloom weaving clusters in the Coimbatore region as well as expansion of cotton knitwear cluster in Tiruppur and home furnishings cluster in Karur.
• The Hosur industrial cluster is a successful case of how such policy efforts to promote industrial estates helped develop industries in a backward region.

Industrialisation in Tamil Nadu – Liberalization Phase:

• The final phase of industrialisation is the post-reforms period since the early 1990s.
• Because of trade liberalisation measures, exports of textiles, home furnishings and leather products began to grow rapidly.
• Efforts to attract investments led to entry of leading multinational firms (MNCs) into the state, especially in the automobile sector.
• Chennai region also emerged as a hub for electronics industry with MNCs such as Nokia, Foxconn, Samsung and Flextronics opening plants on the city’s outskirts.
• A significant share of these investments has come up in special economic zones in the districts bordering Chennai.
• The major industries are automobiles, autocomponents, light and heavy engineering, machinery, cotton, etc.
• This diffused process of industrialisation and corresponding urbanisation has paved the way for better rural-urban linkages in Tamil Nadu than in most other states.

Question 41.
Draw a time line for the following:
Write any five important events between 1905-1920

Question 42.
Mark the following places on the world map.
(i) Portugal
(ii) Spain
(iii) Morocco
(iv) France
(v) Great Britian

Part – IV

Answer both questions. [2 x 8 = 16]

Question 43.
(i) When did Germany attack Stalingrad?
(ii) What were the main manufactures of Stalingrad?
(iii) What was the name of the plan formulated by Hitler to attack Stalingrad?
(iv) What is the significance of the Battle of Stalingrad?
(i) In August 1942, Germany attacked Stalingrad.
(ii) The main manufactures of Stalingrad were armaments and tractors.
(iii) Fall Blau or Operation Blue
(iv) The people of Russia were grateful for Stalin’s conduct of the war. They regarded him as ‘a prodigy of patience, tenacity and vigilance, almost omnipresent, almost omniscient.’

(b) Political developments in South America.
(i) By which year did the whole of South America become free from European domination?
(ii) How many republics came into being from the Central America?
(iii) In which year was Cuba occupied by the USA?
(iv) What made oligarchic regimes unpopular in South America?
(b) Political developments in South America:
(i) By 1830 the whole of South America was free from European domination.
(ii) Five republics came into being from the Central America.
(iii) The USA occupied Cuba in the year 1898.
(iv) Economic growth, urbanisation and industrial growth in countries like Argentina, Chile, Brazil, and Mexico helped consolidate the hold of middle class and the emergence of militant working class oganisations. At the same time American power and wealth came to dominate Central and South America. These factors made olgarchic regimes unpopular in South America.

[OR]

(c) Gandhi and Mass nationalism.
(i) Which incident is considered a turning point in the life of Gandhi?
(ii) Name the works that influenced Gandhi?
(iii) How did Gandhi use satyagraha as a strategy in South Africa?
(iv) What do you know about the Champaran Satyagraha?
(c) Gandhi and Mass nationalism:
(i) On his journey from Durban to Pretoria, at the Pietermaritzburg railway station, he was physically thrown out of the first class compartment in which he was travelling despite having a first class ticket. This incident is considered a turning point in the life of Gandhi.
(ii) Tolstoy’s The Kingdom of God is Within You, Ruskin’s Unto This Last and Thoreau’s Civil Disobedience.
(iii) Gandhi developed satyagraha (truth-force) as a strategy, in which campaigners went on peaceful marches and presented themselves for arrest in protest against unjust laws.
(iv) The Champaran Satyagraha of 1916 was the first satyagraha movement inspired by Gandhi. It was a farmer’s uprising that took place in Champaran district of Bihar, India during the British colonial period.

(d) Periyar E. V. R.
(i) When did Periyar found Dravidar Kazhagam?
(ii) What were the Newspapers and Journals run by Periyar?
(iii) Why was Periyar known as Vaikom hero?
(iv) Which was the most important work of Periyar?
(d) Periyar E. V. R.
(i) Periyar found Dravidar Kazhagam in 1944.
(ii) The newspapers and journals started by Periyar were – Kudi Arasu, Revolt, Puratchi, Paguththarivu and Viduthalai.
(iii) In Vaikom, people protested against the practice of no access to the temples by the lower caste people. After the local leaders were arrested Periyar led the Temple Entry Movement and was imprisoned. So, people hailed him as Vaikom Virar or hero of Vaikom.
(iv) Right from 1929, when the Self-respect Conferences began to voice its concern over the plight of women, Periyar had been emphasising women’s right to divorce and property. Periyar’s most important work on this subject is Why the Woman is Enslaved.

Question 44.
Mark the following places on the given outline map of India.
(i) Himalayas
(ii) Nilgiris
(v) Deccan Plateau
(vi) Southwest Monsoon
(viii) Chennai to Mumbai air route

[OR]

Mark the following places are given outline map of Tamil Nadu:
(i) Coromandel coast
(iii) Nilgiri Hills
(iv) Nagapattinam
(v) Meenambakkam
(vi) Thamirabarani
(vii) Magnesite region (any one place)
(viii) Vaigai Dam

Map for Q. 42
(i) Portugal
(ii) Spain
(iii) Morocco
(iv) France
(v) Great Britain

Map for Q. 44
(i) Himalayas
(ii) Nilgiris
(v) Deccan Plateau
(vi) Southwest Monsoon
(viii) Chennai to Mumbai air route

Map for Q. 44
(i) Coromandel coast
(iii) Nilgiri Hills
(iv) Nagapattinam
(v) Meenambakkam
(vi) Thamirabarani
(vii) Magnesite region (any one place)
(viii) Vaigai Dam

## Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1

Students can download Maths Chapter 8 Statistics and Probability Ex 8.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.1

Question 1.
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
(i) Here the largest value (L) = 125
The smallest value (S) = 63
Range = L – S = 125 – 63 = 62

Question 2.
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Solution:
If the range = 36.8 and
the smallest value = 13.4 then
the largest value = L = R + S
= 36.8 + 13.4 = 50.2

Question 3.
Calculate the range of the following data.

Smallest value (S) = 400
Largest value (L) = 650
Range = L – S = 650 – 400 = 250

Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
The remaining number of pages to be completed is 60 – 32; 60 – 35; 60 – 37; 60 – 30; 60 – 33; 60 – 36; 60 – 35 and 60 – 37
The pages to be completed are, 28, 25, 23, 30, 27, 24, 25, and 23
Arrange in ascending order we get, 23, 23, 24, 25, 25, 27, 28 and 30

Free online sample standard deviation calculator and variance calculator with steps.

Question 5.
Find the variance and standard deviation of the wages of 9 workers given below:
₹ 310, ₹ 290, ₹ 320, ₹ 280, ₹ 300, ₹ 290, ₹ 320, ₹ 310, ₹ 280.
Arrange in ascending order we get,
280, 280, 290, 290, 300, 310, 310, 320 and 320
Assumed mean = 300

Variance = 222.222
Standard deviation = √Variance = √222.222 = 14.907 = 14.91
Variance = 222.22
Standard deviation = 14.91

Question 6.
A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock, 3 times at 3 o’clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Wall clock strikes the bell in 12 hours
1, 2, 3, 4, 5,… ,12
Wall clock strikes in a day (24 hours)
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Assumed mean = 14

The standard deviation of bell strike in a day is 6.9

Question 7.
Find the standard deviation of the first 21 natural numbers.
Here n = 21
The standard deviation of the first ‘n’ natural numbers,

The standard deviation of the first 21 natural numbers = 6.06

Question 8.
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Solution:
If the standard deviation of a data is 4.5 and each value of the data decreased by 5, the new standard deviation does not change and it is also 4.5.

Question 9.
If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
The standard deviation of the data = 3.6
Each value of the data is divided by 3
New standard deviation = $$\frac{3.6}{3}$$ = 1.2
New Variance = (1.2)2 = 1.44 [∴ Variance = (S.D)2]
New standard Deviation = 1.2
New variance = 1.44

Question 10.
The rainfall recorded in various places of five districts in a week are given below.

Find its standard deviation.
Assumed mean = 60

Question 11.
In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Assumed mean = 35

Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.

Assumed mean = 34.5

Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Assumed mean = 11

Standard deviation (σ) = 1.24

Question 14.
For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Number of candidates = 100
n = 100
Mean ($$\bar{x}$$) = 60
standard deviation (σ) = 15
Mean ($$\bar{x}$$) = $$\frac{\Sigma x}{n} \Rightarrow 60=\frac{\Sigma x}{100}$$
Σx = 6000
Correct total = 6000 + (45 – 40) + ( 72 – 27) = 6000 + 5 + 45 = 6050
Correct mean ($$\bar{x}$$) = $$\frac{6050}{100}$$ = 60.5
Given standard deviation = 15

Correct mean = 60.5
Correct standard deviation (σ) = 14. 61

Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8

## Samacheer Kalvi 10th Social Science Guide Book Back Answers Solutions

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