Category: Class 10

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
If the radii of the circular ends of a conical bucket which is 45 cm high are 28 cm and 7 cm, find the capacity of the bucket. (Use \(\pi=\frac{22}{7}\))
Solution:
Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm.
Capacity of the bucket = volume of the frustum

Question 2.
Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m × 16 m × 11 m.
Solution:
Volume of the cylindrical tank = Volume of the rectangle tank

Question 3.
What is the ratio of the volume of a cylinder, a cone, and a sphere. If each has the same diameter and same height?
Solution:

Question 4.
Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:

Question 5.
A spherical ball of iron has been melted and made into small balls. If the raidus of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made?
Solution:

Question 6.
A wooden article was made by scooping out a hemisphere from each end of a cylinder as shown in figure. If the height of the cylinder is 10cm and its base is of radius 3.5 cm find the total surface area of the article.

Solution:
Radius of the cylinder be r Height of the cylinder be h Total surface area of the article = CSA of cylinder + CSA of 2 hemispheres = 2πrh + 2πr² = 2πr (h + 2r)

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Unit Exercise 7

Question 1.
The barrel of a fountain-pen cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one fifth of a litre?
Solution:

Question 2.
A hemi-spherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litre per second. How much time will it take to empty the tank completely?
Solution:
Suppose the pipe takes x seconds to empty the tank. Then, volume of the water that flows out of the tank in x seconds = Volume of the hemispherical tank.
Volume of the water that flows out of the tank in x seconds.
= Volume of hemispherical shell of radius 175 cm.

Question 3.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Solution:
Radius of the base of cone = Radius of the hemisphere = r
Height of the cone = Radius of the hemisphere

Question 4.
An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion be 8cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Solution:
Slant height of the frustum

Question 5.
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
No. of coins required .

Question 6.
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm
and whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of solid cylinder.
Solution:
Volume of the solid cylinder = Volume of the hollow cylinder melted.

Question 7.
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m.
Solution:

Question 8.
A hemi-spherical hollow bowl has material of volume \(\frac{436 \pi}{3}\) cubic cm. Its external diameter is 14 cm. Find its thickness.
Solution:

Question 9.
The volume of a cone is \(1005 \frac{5}{7}\) cu. cm. The area of its base is \(201 \frac{1}{7}\) sq. cm. Find the slant height of the cone.
Solution:

Question 10.
A metallic sheet in the form of a sector of T a circle of radius 21 cm has central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Solution:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.5

Multiple choice questions.
Question 1.
The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is ______
(1) 60π cm²
(2) 68π cm²
(3) 120π cm²
(4) 136π cm²
Answer:
(4) 136π cm²
Hint:

Question 2.
If two solid hemispheres of same base radius r units are joined together along with their bases, then the curved surface area of this new solid is
(1) 4πr² sq. units
(2) 67πr² sq. units
(3) 3πr² sq. units
(4) 8πr² sq. units
Solution:
(1) 47πr² sq. units]

Question 3.
The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be __________
(1) 12 cm
(2) 10 cm
(3) 13 cm
(4) 5 cm
Answer:
(1) 12 cm
Hint:

Question 4.
If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is
(1) 1 : 2
(2) 1 : 4
(3) 1 : 6
(4) 1 : 8
Solution:
(2) 1 : 4

Question 5.
The total surface area of a cylinder whose radius is \(\frac{1}{3}\) of its height is

Solution:
(3) \(\frac{8 \pi h^{2}}{9}\) sq. units

Question 6.
In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is _______
(1) 560π cm³
(2) 1120π cm³
(3) 56π cm³
(4) 360π cm³
Answer:
(2) 1120π cm³
Hint:
R + r = 14 cm
w = 4 cm
h = 90 cm

Question 7.
If the radius of the base of a cone is tripled and the height is doubled then the volume is
(1) made 6 times
(2) made 18 times
(3) made 12 times
(4) unchanged
Solution:
(2) made 18 times
Hint:

Question 8.
The total surface area of a hemisphere is how many times the square of its radius ______
(1) π
(2) 4π
(3) 3π
(4) 2π
Answer:
(3) 3π
Hint:

Question 9.
A solid sphere of radius x cm is melted and cast into a shape of a solid cone of same radius. The height of the cone is
(1) 3x cm
(2) x cm
(3) 4x cm
(4) 2x cm
Solution:
(3) 4x cm
Hint:

Question 10.
A frustum of a right circular cone is of height 16 cm with radii of its ends as 8 cm and 20 cm. Then, the volume of the frustum is _______
(1) 3328π cm3
(2) 3228π cm3
(3) 3240π cm3
(4) 3340π cm3
Answer:
(1) 3328π cm3Hint:

Question 11.
A shuttlecock used for playing badminton has the shape of the combination of
(1) a cylinder and a sphere
(2) a hemisphere and a cone
(3) a sphere and a cone
(4) frustum of a cone and a hemisphere
Solution:
(4) frustum of a cone and a hemisphere

Question 12.
A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Solution:
Hint:

Question 13.
The volume (in cm³) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is

Solution:
(1) \(\frac{4}{3} \pi\)

Question 14.
The height and radius of the cone of which the frustum is a part are h₁ units and r₁ units respectively. Height of the frustum is h₂ units and the radius of the smaller base is r₂ units. If h₂: h₁ = 1 : 2 then r₂ : r₁ is
(1) 1 : 3
(2) 1 : 2
(3) 2 : 1
(4) 3 : 1
Solution:
(2) 1 : 2

Question 15.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is ____
(1) 1 : 2 : 3
(2) 2 : 1 : 3
(3) 1 : 3 : 2
(4) 3 : 1 : 2
Answer:
(4) 3 : 1 : 2
Hint:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Additional Questions

Question 1.
Let A = {1, 2, 3, 4} and B = {-1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let R = {(1, 3), (2, 6), (3,10), (4, 9)} ⊂ A × B be a relation. Show that R is a function and find its domain, co-domain and the range of R.
Answer:
Domain of R = {1, 2, 3, 4}
Co-domain of R = B = {-1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12}
Range of R= {3, 6, 10, 9}

Question 2.
Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f: A → B be a function given by f(x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow and (iv) a graph.
Solution:
A = {0, 1, 2, 3}, B = {1, 3, 5, 7, 9}
f(x) = 2x + 1
f(0) = 2(0) + 1 = 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

(iii) An arrow diagram

Question 3.
State whether the graph represent a function. Use vertical line test.

Solution:
It is not a function as the vertical line PQ cuts the graph at two points.

Question 4.
Let f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5, 3)} be a function from A = {-1, 0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Is this (i) an one-one function (ii) an onto function, (iii) both one- one and onto function?
Solution:
It is both one-one and onto function.

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

Question 5.
A function f: (-7,6) → R is defined as follows.

Find (i) 2f(-4) + 3 f(2)
(ii) f(-7) – f(-3)
Solution:

(i) 2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 × 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

(ii) f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 – 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) – f(-3) = 36 – 2 = 34

Question 6.
If A = {2,3, 5} and B = {1, 4} then find
(i) A × B
(ii) B × A
Answer:
A = {2, 3, 5}
B = {1, 4}

(i) A × B = {2,3,5} × {1,4}
= {(2, 1) (2, 4) (3, 1) (3, 4) (5,1) (5, 4)}.

(ii) B × A = {1,4} × {2,3,5}
= {(1,2) (1,3) (1,5) (4, 2) (4, 3) (4, 5)}

Question 7.
Let A = {5, 6, 7, 8};
B = {- 11, 4, 7, -10, -7, – 9, -13} and
f = {(x,y): y = 3 – 2x, x ∈ A, y ∈ B}.
(i) Write down the elements of f.
(ii) What is the co-domain?
(iii) What is the range?
(iv) Identify the type of function.
Answer:
Given, A = {5, 6, 7, 8},
B = {- 11,4, 7,-10,-7,-9,-13}
y = 3 – 2x
ie; f(x) = 3 – 2x
f(5) = 3 – 2 (5) = 3 – 10 = – 7
f(6) = 3 – 2 (6) = 3 – 12 = – 9
f(7) = 3 – 2(7) = 3 – 14 = – 11
f(8) = 3 – 2 (8) = 3 – 16 = – 13
(i) f = {(5, – 7), (6, – 9), (7, – 11), (8, – 13)}
(ii) Co-domain (B)
= {-11,4, 7,-10,-7,-9,-13} i
(iii) Range = {-7, – 9, -11,-13}
(iv) It is one-one function.

Question 8.
A function f: [1, 6] → R is defined as follows:

Find the value of (i) f(5)
(ii) f(3)
(iii) f(2) – f(4).
Solution:

(i) f(5) = 3x² – 10
= 3 (5²) – 10 = 75 – 10 = 65
(ii) f(3) = 2x – 1
= 2(3) – 1 = 6 – 1 = 5
(ii) f(2) – f(4)
f(2) = 2x – 1
= 2(2) – 1 = 3
f(4) = 3x² – 10
= 3(4²) – 10 = 38
∴ f(2) – f(4) = 3 – 38 = 35

Question 9.
The following table represents a function from A = {5, 6, 8, 10} to B = {19, 15, 9, 11}, where f(x) = 2x – 1. Find the values of a and b.
Solution:

A = {5, 6, 8, 10}, B = {19, 15, 9, 11}
f(x) = 2x – 1
f(5) = 2(5) – 1 = 9
f(8) = 2(8) – 1 = 15
∴ a = 9, b = 15

Question 10.
If R = {(a, -2), (-5, 6), (8, c), (d, -1)} represents the identity function, find the values of a,b,c and d.
Solution:
R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.6

Question 1.
If n(A × B) = 6 and A = {1, 3} then n(B) is
(1) 1
(2) 2
(3) 3
(4) 6
Answer:
(3) 3
Hint:
If n(A × B) = 6
A = {1, 1}, n(A) = 2
n(B) = 3

Question 2.
A = {a, b,p}, B = {2, 3}, C = {p, q, r, s)
then n[(A ∪ C) × B] is ………….
(1) 8
(2) 20
(3) 12
(4) 16
Answer:
(3) 12
Hint: A ∪ C = [a, b, p] ∪ [p, q, r, s]
= [a, b, p, q, r, s]
n (A ∪ C) = 6
n(B) = 2
∴ n [(A ∪ C)] × B] = 6 × 2 = 12

Question 3.
If A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8} then state which of the following statement is true.
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)
Answer:
(1) (A × C) ⊂ (B × D)]
Hint:
A = {1, 2}, B = {1, 2, 3, 4},
C = {5, 6}, D ={5, 6, 7, 8}
A × C ={(1,5), (1,6), (2, 5), (2, 6)}
B × D = {(1, 5),(1, 6),(1, 7),(1, 8),(2, 5),(2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8)}
∴ (A × C) ⊂ B × D it is true

Question 4.
If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is ………………….
(1) 3
(2) 2
(3) 4
(4) 8
Answer:
(2) 2
Hint: n(A) = 5
n(A × B) = 10
(consider 1024 as 10)
n(A) × n(B) = 10
5 × n(B) = 10
n(B) = \(\frac { 10 }{ 5 } \) = 2
n(B) = 2

Question 5.
The range of the relation R = {(x, x²)|x is a prime number less than 13} is
(1) {2, 3, 5, 7}
(2) {2, 3, 5, 7, 11}
(3) {4, 9, 25, 49, 121}
(4) {1, 4, 9, 25, 49, 121}
Answer:
(3) {4, 9, 25, 49, 121}]
Hint:
R = {(x, x²)/x is a prime number < 13}
The squares of 2, 3, 5, 7, 11 are
{4, 9, 25, 49, 121}

Question 6.
If the ordered pairs (a + 2,4) and (5, 2a + 6) are equal then (a, b) is ………
(1) (2, -2)
(2) (5, 1)
(3) (2, 3)
(4) (3, -2)
Answer:
(4) (3, -2)
Hint:

The value of a = 3 and b = -2

Question 7.
Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is
(1) mⁿ
(2) n^m
(3) 2^mn – 1
(4) 2^mn
Answer:
(4) 2^mn
Hint:
n(A) = m, n(B) = n
n(A × B) = 2^mn

Question 8.
If {(a, 8),(6, b)} represents an identity function, then the value of a and 6 are respectively
(1) (8,6)
(2) (8,8)
(3) (6,8)
(4) (6,6)
Answer:
(1) (8,6)
Hint: f = {{a, 8) (6, 6)}. In an identity function each one is the image of it self.
∴ a = 8, b = 6

Question 9.
Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f : A → B given by f = {(1, 4),(2, 8),(3, 9),(4, 10)} is a
(1) Many-one function
(2) Identity function
(3) One-to-one function
(4) Into function
Answer:
(3) One-to one function
Hint:
A = {1, 2, 3, 4), B = {4, 8, 9,10}

Question 10.
If f(x) = 2x² and g (x) = \(\frac{1}{3 x}\), Then fog is

Answer:
(3) \(\frac{2}{9 x^{2}}\)
Hint:
f(x) = 2x²
g(x) = \(\frac{1}{3 x}\)
fog = f(g(x)) = \(f\left(\frac{1}{3 x}\right)=2\left(\frac{1}{3 x}\right)^{2}\)
= 2 × \(\frac{1}{9 x^{2}}=\frac{2}{9 x^{2}}\)

Queston 11.
If f: A → B is a bijective function and if n(B) = 7 , then n(A) is equal to …………..
(1) 7
(2) 49
(3) 1
(4) 14
Answer:
(1) 7
Hint:
n(B) = 7
Since it is a bijective function, the function is one – one and also it is onto.
n(A) = n(B)
∴ n(A) = 7

Question 12.
Let f and g be two functions given by f = {(0, 1), (2, 0), (3, -4), (4, 2), (5, 7)} g = {(0, 2), (1, 0), (2, 4), (-4, 2), (7, 0)} then the range of fog is
(1) {0, 2, 3, 4, 5}
(2) {-4, 1, 0, 2, 7}
(3) {1, 2, 3, 4, 5}
(4) {0, 1, 2}
Answer:
(4) {0, 1, 2}
Hint:
gof = g(f(x))
fog = f(g(x))
= {(0, 2),(1, 0),(2, 4),(-4, 2),(7, 0)}
Range of fog = {0, 1, 2}

Question 13.
Let f (x) = \(\sqrt{1+x^{2}}\) then ………………..
(1) f(xy) = f(x) f(y)
(2) f(xy) > f(x).f(y)
(3) f(xy) < f(x). f(y)
(4) None of these
Answer:
(3) f(xy) < f(x) . f(y)

Question 14.
If g = {(1, 1),(2, 3),(3, 5),(4, 7)} is a function given by g(x) = αx + β then the values of α and β are
(1) (-1, 2)
(2) (2, -1)
(3) (-1, -2)
(4) (1, 2)
Answer:
(2) (2,-1)
Hint:
g(x) = αx + β
α = 2
β = -1
g(x) = 2x – 1
g(1) = 2(1) – 1 = 1
g(2) = 2(2) – 1 = 3
g(3) = 2(3) – 1 = 5
g(4) = 2(4) – 1 = 7

Question 15.
f(x) = (x + 1)³ – (x – 1)³ represents a function which is …………….
(1) linear
(2) cubic
(3) reciprocal
(4) quadratic
Answer:
(4) quadratic
Hint: f(x) = (x + 1)³ – (x – 1)³
[using a³ – b³ = (a – b)³ + 3 ab (a – b)]
= (x + 1 – x + 1)³ + 3(x + 1) (x – 1)
(x + 1 – x + 1)
= 8 + 3 (x² – 1)²
= 8 + 6 (x² – 1)
= 8 + 6x² – 6
= 6x² + 2
It is quadratic polynomial

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Additional Questions

Question 1.
Use Euclid’s algorithm to find the HCF of 4052 and 12756.
Solution:
Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get HCF
12576 = 4052 × 3 + 420.
Since the remainder 420 ≠ 0, we apply the division lemma to 4052
4052 = 420 × 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 × 1 + 148, 148 ≠ 0.
∴ Again by division lemma
272 = 148 × 1 + 124, here 124 ≠ 0.
∴ Again by division lemma
148 = 124 × 1 + 24, Here 24 ≠ 0.
∴ Again by division lemma
124 = 24 × 5 + 4, Here 4 ≠ 0.
∴ Again by division lemma
24 = 4 × 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
∴ The HCF of 12576 and 4052 is 4.

Question 2.
If the HCF of 65 and 117 is in the form (65m – 117) then find the value of m.
Answer:
By Euclid’s algorithm 117 > 65
117 = 65 × 1 + 52
52 = 13 × 4 × 0
65 = 52 × 1 + 13
H.C.F. of 65 and 117 is 13
65m – 117 = 13
65 m = 130
m = \(\frac { 130 }{ 65 } \) = 2
The value of m = 2

Question 3.
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution:
We have 6 = 2¹ × 3¹ and
20 = 2 × 2 × 5 = 2² × 5¹
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 × 2 × 3 × 5 = 60. As done in your earlier classes. Note that HCF (6, 20) = 2¹ = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 2² × 3¹ × 5¹ = 60.
= Product of the greatest power of each prime factor, involved in the numbers.

Question 4.
Prove that \(\sqrt { 3 }\) is irrational.
Answer:
Let us assume the opposite, (1) \(\sqrt { 3 }\) is irrational.
Hence \(\sqrt { 3 }\) = \(\frac { p }{ q } \)
Where p and q(q ≠ 0) are co-prime (no common factor other than 1)

Hence, 3 divides p²
So 3 divides p also …………….. (1)
Hence we can say
\(\frac { p }{ 3 } \) = c where c is some integer
p = 3c
Now we know that
3q² = p²
Putting = 3c
3q² = (3c)²
3q² = 9c²
q² = \(\frac { 1 }{ 3 } \) × 9c²
q² = 3c²
\(\frac{q^{2}}{3}\) = C²
Hence 3 divides q²
So, 3 divides q also ……………. (2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 }\) is irrational.

Question 5.
Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
(i) 4, 10, 16, 22, …
(ii) 1, -1,-3, -5,…
(iii) -2, 2, -2, 2, -2, …
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3,…
Solution:
(i) 4, 10, 16, 22, …….
We have a₂ – a₁ = 10 – 4 = 6
a₃ – a₂ = 16 – 10 = 6
a₄ – a₃ = 22 – 16 = 6
∴ It is an A.P. with common difference 6.
∴ The next two terms are, 28, 34

(ii) 1, -1, -3, -5
t₂ – t₁ = -1 – 1 = -2
t₃ – t₂ = -3 – (-1) = -2
t₄ – t₃ = -5 – (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

(iii) -2, 2,-2, 2,-2
t₂ – t₁ = 2-(-2) = 4
t₃ – t₂ = -2 -2 = -4
t₄ – t₃ = 2 – (-2) = 4
It is not an A.P.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3
t₂ – t₁ = 1 – 1 = 0
t₃ – t₂ = 1 – 1 = 0
t₄ – t₃ = 2 – 1 = 1
Here t₂ – t₁ ≠ t₃ – t₂
∴ It is not an A.P.

Question 6.
Find n so that the n^th terms of the following two A.P.’s are the same.
1, 7,13,19,… and 100, 95,90,…
Answer:
The given A.P. is 1, 7, 13, 19,….
a = 1, d = 7 – 1 = 6
t_n1 = a + (n – 1)d
t_n1 = 1 + (n – 1) 6
= 1 + 6n – 6 = 6n – 5 … (1)
The given A.P. is 100, 95, 90,….
a = 100, d = 95 – 100 = – 5
tn₂ = 100 + (n – 1) (-5)
= 100 – 5n + 5
= 105 – 5n …..(2)
Given that, t_n1 = t_n2
6n – 5 = 105 – 5n
6n + 5n = 105 + 5
11 n = 110
n = 10
∴ 10^th term are same for both the A.P’s.

Question 7.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 is the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Answer:
The number of rose plants in the 1st, 2nd, 3rd,… rows are
23, 21, 19,………….. 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = -2, l = 5.
As, a_n = a + (n – 1)d i.e. t_n = a + (n – 1)d
We have 5 = 23 + (n – 1)(-2)
i.e. -18 = (n – 1)(-2)
n = 10
∴ There are 10 rows in the flower bed.

Question 8.
Find the sum of the first 30 terms of an A.P. whose n^th term is 3 + 2n.
Answer:
Given,
t_n = 3 + 2n
t₁ = 3 + 2 (1) = 3 + 2 = 5
t₂ = 3 + 2 (2) = 3 + 4 = 7
t₃ = 3 + 2 (3) = 3 + 6 = 9
Here a = 5,d = 7 – 5 = 2, n = 30
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S₃₀ = \(\frac { 30 }{ 2 } \) [10 + 29(2)]
= 15 [10 + 58] = 15 × 68 = 1020
∴ Sum of first 30 terms = 1020

Question 9.
How many terms of the AP: 24, 21, 18, . must be taken so that their sum is 78?
Solution:
Here a = 24, d = 21 – 24 = -3, S_n = 78. We need to find n.
We know that,
S_n = \(\frac { n }{ 2 } \) (2a + (n – 1)d)
78 = \(\frac { n}{ 2 } \) (48 + 13(-3))
78 = \(\frac { n}{ 2 } \) (51 – 3n)
or 3n² – 51n + 156 = 0
n² – 17n + 52 = 0
(n – 4) (n – 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

Question 10.
The sum of first n terms of a certain series is given as 3n² – 2n. Show that the series is an arithmetic series.
Solution:
Given, S_n = 3n² – 2n
S₁ = 3 (1)² – 2(1)
= 3 – 2 = 1
ie; t₁ = 1 (∴ S₁ = t₁)
S₂ = 3(2)² – 2(2) = 12 – 4 = 8
ie; t₁ + t₂ = 8 (∴ S₂ = t₁ + t₂)
∴ t₂ = 8 – 1 = 7
S₃ = 3(3)² – 2(3) = 27 – 6 = 21
t₁ + t₂ + t₃ = 21 (∴ S₃ = t₁ + t₂ + t₃)
8 + t₃ = 21 (Substitute t₁ + t₂ = 8)
t₃ = 21 – 8 ⇒ t₃ = 13
∴ The series is 1,7,13, …………. and this series is an A.P. with common difference 6.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Unit Exercise 2

Question 1.
Prove that n² – n divisible by 2 for every positive integer n.
Answer:
To prove n² – n divisible by 2 for every positive integer n.
We know that any positive integer is of the form 2q or 2q + 1, for some integer q.
So, following cases arise:
Case I. When n = 2q.
In this case, we have
n² – n = (2q)² – 2q = 4q² – 2q = 2q(2q – 1)
⇒ n² – n = 2r where r = q(2q – 1)
⇒ n² – n is divisible by 2.

Case II. When n = 2q + 1.
In this case, we have
n² – n = (2q + 1)² – (2q + 1)
= (2q + 1)(2q + 1 – 1) = (2q + 1)2q
⇒ n² – n = 2r where r = q (2q + 1)
⇒ n² – n is divisible by 2.
Hence n² – n is divisible by 2 for every positive integer n.

Question 2.
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following (i) Capacity of a can
(ii) Number of cans of cow’s milk
(iii) Number of cans of buffalow’s milk.
Answer:
Cow’s milk = 175 litres
Buffalow’s milk = 105 litres
Find the H.C.F. of 175 and 105 using Euclid’s division method of factorisation method.

175 = 5 × 5 × 7
105 = 3 × 5 × 7
H.C.F. of 175 and 105 = 5 × 7 = 35
(i) The capacity of the milk can’s is 35 litres

(ii) Cows milk = 175 litres
Number of cans = \(\frac { 175 }{ 35 } \) = 5

(iii) Buffalow’s milk = 105 litres
Number of cans = \(\frac { 105 }{ 35 } \) = 3
(i) Capacity of one can = 35 litres
(ii) Number of can’s for cow’s milk= 5 litres
(iii) Number of can’s for Buffalow’s milk = 3 litres

Question 3.
When the positive integers a, b and c are divided by 13 the respective remainders are 9, 7 and 10. Find the remainder when a + 2b + 3c is divided by 13.
Answer:
Let the positive integers be a, b, and c.
a = 13q + 9
b = 13q + 7
c = 13q + 10
a + 2b + 3c = 13 q + 9 + 2(13q + 7) + 3(13q + 10)
= 13q + 9 + 269 + 14 + 39q + 30
= 78q + 53 = (13 × 6)q + 53
The remainder is 53.
But 53 = 13 × 4 + 1
∴ The remainder is 1

Question 4.
Show that 107 is of the form 4q + 3 for any integer q.
Answer:

107 = 4 x 26 + 3
This is in the form of a = bq + r
Hence it is proved.

Question 5.
If (m + 1)^th term of an A.P. is twice the (n + 1)^th term, then prove that (3m + 1)^th term is twice the (m + n + 1)^th term.
Solution:
t_n = a + (n – 1)d
t_m+1 = a + (m + 1 – 1)d
= a + md
t_n+1 = a + (n + 1 – 1)d
= a + nd
2(t_n+1) = 2(a + nd)
t_m+1 = 2t_n+1 …………… (1)
⇒ a + md = 2(a + nd)
2a + 2nd – a – md = 0
a + (2n – m)d = 0
t_(3m+1) = a + (3m + 1 – 1)d
= a + 3md
t_(m+n+1) = a + (m + n + 1 – 1)d
= a + (m + n)d
2(t_(m+n+1)) = 2(a + (m + n)d)
= 2a + 2md + 2nd
t_(3m+1) = 2t_(m+n+1) ………….. (2)
a + 3md = 2a + 2md + 2nd
2a + 2md + 2nd – a – 3md = 0
a – md + 2nd = 0
a + (2n – m)d = 0
∴ It is proved that t_(3m+1) = 2t_(m+n+1)

Question 6.
Find the 12^th term from the last term of the A.P -2, -4, -6, … -100.
Solution:

12^th term from the last = 39^th term from the beginning
∴ t₃₉ = a + 38d
= -2 + 38(-2)
= – 2 – 76
= – 78

Question 7.
Two A.P.’s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. Show that the difference between their 10^th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.
Answer:
Let the common difference for the 2 A.P be “d”
For the first A.P
a = 2, d = d, n = 10
t_n = a + (n – 1) d
t₁₀ = 2 + 9 d ….(1)
For the 2^nd A.P
a = 7, d = d n = 10
t₁₀ = 7 + (9)d
= 7 + 9d ….(2)
Difference between their 10th term ⇒ (1) – (2)
= 2 + 9d – (7 + 9d)
= 2 + 9d – 7 – 9d
= -5
For first A.P when n = 21, a = 2, d = d
t₂₁ = 2 + 20d …….(3)
For second A.P when n = 21, a = 7, d = d
t₂₁ = 7 + 20d …….(4)
Difference between the 21^st term ⇒ (3) – (4)
= 2 + 20d – (7 + 20d)
= 2 + 20d – 7 – 20d
= -5
Difference between their 10th term and 21^st term = -5
Hence it is proved.

Question 8.
A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?
Solution:
S₁₀ = ₹16500
a, a + d, a + 2d…
d = 100
n = 10
S_n = \(\frac { n }{ 2 } \)(2a+(n-1)d)
S₁₀ = 16500
S₁₀ = \(\frac { 10 }{ 2 } \)(2×a+9×100)
16500 = 5(2a+900)
16500 = 10a + 4500
10a = 16500 – 4500
10a = 12000
a = \(\frac { 12000 }{ 10 } \) = ₹1200
∴ He saved ₹1200 in the first year

Question 9.
Find the G.P. in which the 2nd term is \(\sqrt { 6 }\) and the 6th term is 9\(\sqrt { 6 }\).
Solution:

Question 10.
The value of a motorcycle depreciates at the rate of 15% per year. What will be the value of the motor cycle 3 year hence, which is now purchased for ₹45,000?
Answer:
Value of the motor cycyle = ₹ 45000
a = 45000
Depreciation = 15% of the cost value
= \(\frac { 15 }{ 100 } \) × 45000
= 15 × 450
= 6750
d = – 6750 (decrease it is depreciation
Value of the motor cycle lightning of the 2nd year = 45000 – 6750
= ₹ 38250
Depreciation for the 2nd year = \(\frac { 15 }{ 100 } \) × 38250
= ₹ 57370.50

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.10

Multiple choice questions
Question 1.
Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy.
(1) 1 < r < b
(2) 0 < r < b
(3) 0 < r < b
(4) 0 < r < b
Answer:
(3) 0 < r < b

Question 2.
Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are
(1) 0, 1, 8
(2) 1, 4, 8
(3) 0, 1, 3
(4) 1, 3, 5
Answer:
(1) 0,1,8
Hint:
Cube of any +ve integers 1³, 2³, 3³, 4³,. . .
1, 8, 27, 64, 125, 216 …
Remainders when 27, 64, 125 are divided by 9.

Question 3.
If the H.C.F of 65 and 117 is expressible in the form of 65m -117 , then the value of m is ………………….
(1) 4
(2) 2
(3) 1
(4) 3
Answer:
(2) 2
Hint:
117 = 3 × 3 × 13
65 = 5 × 13
H.C.F = 13
65m – 117 = 13 ⇒ 65m = 117 + 13 = 130
m = \(\frac { 130 }{ 65 } \) = 2
The value of m = 2

Question 4.
The sum of the exponents of the prime factors in the prime factorization of 1729 is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3
Hint:
1729 = 7¹ × 13¹ × 19¹

Question 5.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(1) 2025
(2) 5220
(3) 5025
(4) 2520
Answer:
(4) 2520
Hint:

∴ L.C.M. of 1,2,3,4,…,10 is 2 × 2 × 3 × 5 × 7 × 2 × 3 = 2520

Question 6.
7^4k ≡ …………………. (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(1) 1
Hint:
7^4k ≡ . . . . . (mod 100)
7^4k = (7⁴)^k ≡ ……….. (mod 100) (7⁴ – 2401)
The value is 1.

Question 7.
Given F₁ = 1, F₂ = 3 and Fn = F_n-1 + F_n-2 then
(1) 3
(2) 5
(3) 8
(4) 11
Answer:
(4) 11
Answer:
F₁ = 1, F₂ = 3
F_n = F_n-1 + F_n-2
F₅ = F_5-1 + F_5-2 = F₄ + F₃
= F₃ + F₂ + F₂ + F₁
= F₂ + F₁ + F₂ + F₂ + F₁
= 3 + 1 + 3 + 3 + 1 = 11

Question 8.
The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P …………..
(1) 4551
(2) 10091
(3) 7881
(4) 13531
Answer:
(3) 7881
Hint:
Here a = 1, d = 4
t_n = a + (n – 1) d = 1 + (n – 1) 4
= 1 + 4n – 4
= 4n – 3
4554
(i) 4n – 3 = 4551 ⇒ 4n = 4551 + 3 ⇒ n = \(\frac { 4554 }{ 4 } \) = 1138.5.
It is not a term of A.P.

(ii) 4n – 3 = 10091 ⇒ 4n = 10091 + 3 = 10094
n = \(\frac { 10094 }{ 4 } \) = 2523.5 it is a term of A.P.

(iii) 4n – 3 = 7881 ⇒ 4n = 7881 + 3
n = \(\frac { 7884 }{ 4 } \) = 1971.
∴ 7881 is a term of the A.P.

Question 9.
If 6 times of 6^th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P. is
(1) 0
(2) 6
(3) 7
(4) 13
Answer:
(1) 0
Hint:
6t₆ = 7t₇
6(a + 5d) = 7(a + 6d)
6a + 30d = 7a + 42d
7a + 42d – 6a – 30d = 0
a + 12d = 0 = t₁₃

Question 10.
An A.P consists of 31 terms. If its 16^th term is m, then the sum of all the terms of this A.P. is …………..
(1) 16m
(2) 62m
(3) 31m
(4) \(\frac { 31 }{ 2 } \)m
Answer:
(3) 31m
Hint:
M = 31
t₁₆ = m ⇒ a + 15d = m
S_n = \(\frac { n }{ 2 } \)[2a + (n – 1)d]
S_n = \(\frac { 31 }{ 2 } \)[2a + 30d]= \(\frac { 31 }{ 2 } \) × 2[a + 15d]
= 31 (m) = 31m

Question 11.
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
(1) 6
(2) 7
(3) 8
(4) 9
Answer:
(3) 8
Hint:

Question 12.
If A = 2⁶⁵ and B = 2⁶⁴ + 2⁶³ + 2⁶² + +2⁰ which of the following is true?
(1) B is 264 more than A
(2) A and B are equal
(3) B is larger than A by 1
(4) A is larger than B by 1
Answer:
(4) A is larger than B by 1
Hint:
A = 2⁶⁵
B = 2⁶⁴ + 2⁶³ + 2⁶² + … + 20
B = 2⁰ + 2¹ + 2² + … + 264
G.P = 1 + 2¹ + 2² + … + 2⁶⁴ it is a G.P
Here a = 1, r = 2, n = 65

A = 2⁶⁵, B = 2⁶⁵ – 1
∴ B is smaller.
A is larger than B by 1.

Question 13.
The next term of the sequence \(\frac { 3 }{ 16 } \),\(\frac { 1 }{ 8 } \),\(\frac { 1 }{ 12 } \),\(\frac { 1 }{ 18 } \), ….
(1) \(\frac { 1 }{ 24 } \)
(2) \(\frac { 1 }{ 27 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) \(\frac { 1 }{ 81 } \)
Answer:
(2) \(\frac { 1 }{ 27 } \)
Hint:

Question 14.
If the sequence t₁, t₂, t₃, …………are in A.P. then the sequence t₆, t₁₂, t₁₈, …… is ………….
(1) a Geometric progression
(2) an Arithmetic progression
(3) neither an Arithmetic progression nor a Geometric progression
(4) a constant sequence
Answer:
(2) an Arithmetic progression
Hint: t₁, t₂, t₃ …. are in A.P
t₆, t₁₂, t₁₈ …… is also an A.P. (6, 12, 18 …….. is an A.P.)

Question 15.
The value of (1³ + 2³ + 3³ + … + 15³) – (1 + 2 + 3 + … + 15) is
(1) 14400
(2) 14200
(3) 14280
(4) 14520
Answer:
(3) 14280
Hint:
\(\left(\frac{15 \times 16}{2}\right)^{2}-\frac{15 \times 16}{2}\) = (120)² – 120 = 14280

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Additional Questions

Question 1.
Solve the following system of linear equations in three variables. x + y + z = 6; 2x + 3y + 4z = 20;
3x + 2y + 5z = 22
Solution:
x + y + z = 6 ………….. (1)
2x + 3y + 4z = 20 ………… (2)
3x + 2y + 5z = 22 …………(3)

Sub. z = 3 in (5) ⇒ y – 2(3) = -4
y = 2
Sub. y = 2, z = 3 in (1), we get
x + 2 + 3 = 6
x = 1
x = 1, y = 2, z = 3

Question 2.
Using quadratic formula solve the following equations.
(i) p²x² + (p² – q²) x – q² = 0
(ii) 9x² – 9 (a + b)x + (2a² + 5ab + 2b²) = 0
Solution:
(i) p²x² + (p² – q²)x – q² = 0
Comparing this with ax² + bx + c = 0, we have
a = p²
b = p² – q²
c = -q²
Δ = b² – 4ac
= (p² – q²)-4 × p² × -q²
= (p² – q²)² + 4p² q²
= (p² + q²)² > 0
So, the given equation has real roots given by
\(\alpha=\frac{-b-\sqrt{\Delta}}{2 a}=\frac{-\left(p^{2}-q^{2}\right)+\left(p^{2}+q^{2}\right)}{2 p^{2}}=\frac{q^{2}}{p^{2}}\)
\(\beta=\frac{-b-\sqrt{\Delta}}{2 a}=\frac{-\left(p^{2}-q^{2}\right)-\left(p^{2}+q^{2}\right)}{2 p^{2}}\)
= -1

(ii) 9x² – 9(a + b)x + (2a² + 5ab + 2b²) = 0
Comparing this with ax² + bx + c = 0.
a =9
b = -9 (a + b)
c = (2a² + 5 ab + 2b²)
Δ = B² – 4AC
⇒ 81 (a + b)² – 36(2a² + 5ab + 2b²)
⇒ 9a² + 9b² – 18ab
⇒ 9(a – b)² > 0
∴ the roots are real and given by

Question 3.
Find the HCF of x³ + x² + x + 1 and x⁴ – 1.
Answer:
x³ + x² + x + 1 = x² (x + 1) + 1 (x + 1)
= (x + 1) (x² + 1)
x⁴-1 = (x²)² – 1
= (x² + 1) (x²– 1)
= (x² + 1) (x + 1) (x – 1)
H.C.F. = (x² + 1)(x + 1)

Question 4.
Prove that the equation x²(a² + b²) + 2x(ac + bd) + (c² + d²) = 0 has no real root if ad ≠ bc
Solution:
Δ = b² – 4ac
⇒ 4(ac + bd)² – 4(a² + b²)(c² + d²)
⇒ 4[(ac + bd)² – (a² + b²)(c² + d²)]
⇒ 4(a²c² + b²d² + 2acbd – a²c²b²c² – a²d² – b²d²]
⇒ 4[2acbd – a²d² – b²c²]
⇒ 4 [a²c² + b²c² – 2adbc]
⇒ -4[ad – bc]²
We have ad ≠ bc
∴ ad – bc > 0
⇒ (ad – bc)² > 0
⇒ -4(ad – bc)² < 0 ⇒ Δ < 0
Hence the given equation has no real roots.

Question 5.
Find the L.C.M of 2(x³ + x² – x – 1) and 3(x³ + 3x² – x – 3)
Answer:
2[x³ + x² – x – 1] = 2[x²(x+ 1)- 1 (x + 1)]
= 2(x + 1) (x² – 1)
= 2(x + 1) (x + 1) (x – 1)
= 2(x + 1)² (x – 1)
3[x³ + 3x² – x – 3] = 3[x²(x + 3) -1 (x + 3)]
= 3[(x + 3)(x² – 1)]
= 3(x + 3)(x + 1) (x – 1)
L.C.M. = 6(x + 1)²(x – 1) (x + 3)

Question 6.
A two digit number is such that the product of its digits is 12. When 36 is added to the number the digits interchange their places. Find the number.
Solution:
Let the ten’s digit of the number be x. It is given that the product of the digits is 12.
Unit’s digit = \(\frac{12}{x}\)
Number = 10x + \(\frac{12}{x}\)
If 36 is added to the number the digits interchange their places.

x = -6, 2.
But a number can never be (-ve). So, x = 2.
The number is 10 × 2 + \(\frac{12}{2}\) = 26

Question 7.
Seven years ago, Vanin’s age was five times the square of swati’s age. Three years hence Swati’s age will be two fifth of Varun’s age. Find their present ages.
Solution:
Seven years ago, let Swathi’s age be x years.
Seven years ago, let Varun’s age was 5x² years.
Swathi’s present age = x + 7 years
Varun’s present age = (5x² + 7) years
3 years hence, we have Swathi’s age = x + 7 + 3 years
= x + 10 years
Varun’s age = 5x² + 7 + 3 years
= 5x² + 10 years
It is given that 3 years hence Swathi’s age will be \(\frac{2}{5}\) of Varun’s age.
∴ x + 10 = \(\frac{2}{5}\) (5x² + 10)
⇒ x + 10 = 2x² + 4
⇒ 2x² – x – 6 = 0
⇒ 2x(x – 2) + 3(x – 2) = 0
⇒ (2x + 3)(x – 2) = 0
⇒ x – 2 = 0
⇒ x = 2 (∵ 2x + 3 ≠ 0 as x > 0)
Hence Swathi’s present age = (2 + 7) years
= 9 years
Varun’s present age = (5 × 2² + 7) years = 27 years

Question 8.
A chess board contains 64 equal squares and the area of each square is 6.25 cm². A border round the board is 2 cm wide find its side.
Solution:
Let the length of the side of the chess board be x cm. Then,

Area of 64 squares = (x – 4)²
(x – 4)² = 64 × 6.25
⇒ x² – 8x+ 16 = 400
⇒ x² – 8x-384 = 0
⇒ x² – 24x + 16x – 384 = 0
⇒ (x – 24)(x + 16) = 0
⇒ x = 24 cm.

Question 9.
Find two consecutive natural numbers whose product is 20.
Solution:
Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
x² + x – 20 = 0
(x + 5)(x – 4) = 0
x = -5, 4
∴ x = 4 (∵ x ≠ -5, x is natural number)
The next number = 4 + 1 = 5
Two consecutive numbers are 4, 5

Question 10.
A two digit number is such that the product of its digits is 18, when 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Let the tens digit be x. Then the units digits = \(\frac{18}{x}\)

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Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Paper 2 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.
Which country emerged as the strongest in East Asia towards the close of nineteenth century?
(a) China
(b) Japan
(c) Korea
(d) Mongolia
Answer:
(b) Japan

Question 2.
Who made peru as part of their dominions?
(a) English
(b) Spaniards
(c) Russian
(d) French
Answer:
(b) Spaniards

Question 3.
What was the name of the Samaj founded by Dayanand Saraswati?
(a) Arya Samaj
(b) Brahmo Samaj
(c) Prathana Samaj
(d) Adi Brahmo Samaj
Answer:
(a) Arya Samaj

Question 4.
Who had borrowed money from the East India Company to meet the expenses he had incurred during the Carnatic war?
(a) Velunachiyar
(b) Puli Thevar
(c) NawabofArcot
(d) Raja of Travancore
Answer:
(c) NawabofArcot

Question 5.
……………………..was the official newspaper of the Self-Respect Movement.
(c) Kudi Arasu
(b) Puratchi
(c) Viduthalai
(d) Paguththarivu
Answer:
(c) Kudi Arasu

Question 6.
The Southern most point of India is ……………..
(a) Andaman
(b) Kanyakumari
(c) Indira Point
(d) Kavaratti
Answer:
(c) Indira Point

Question 7
……………………is the highest gravity in India.
(a) Hirakud dam
(b) Bhakra Nangal dam
(c) Mettur dam
(d) Nagarjuna Sagar dam
Answer:
(b) Bhakra Nangal dam

Question 8.
The first Jute mill of India was established at ………………
(a) Gujarat
(b) Rajasthan
(c) Maharashtra
(d) Tamil Nadu
Answer:
(c) Maharashtra

Question 9.
Which of the following passes is not located in the Western Ghats of Tamil Nadu?
(a) Palghat
(b) Shencottah
(c) Bhorghat
(d) Achankoil
Answer:
(c) Bhorghat

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

Question 11.
The Indian constitution gives to its citizens………………………
(a) Double citizenship
(b) Single citizenship
(c) Single citizenship in some states and double in others .
(d) None of the above
Answer:
(b) Single citizenship

Question 12.
The Agreement signed by India and China in 1954 related …………..
(a) Trade and Commerce
(b) Restoration of normal relations
(c) Cultural exchange programmes
(d) The Five Principles of Co-existence
Answer:
(d) The Five Principles of Co-existence

Question 13.
Which one sector is highest employment in the GDP?
are an essential aspect of a nation’s development.
(a) Agricultural Sector
(b) Industrial Sector
(c) Service Sector
(d) None of the above
Answer:
(c) Service Sector

Question 14.
………………are an essential aspect of a nation’s development.
(a) Agriculture
(b) Industry
(c) Railway
(d) None of these
Answer:
(a) Agriculture

Part – II

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

Question 15.
Point out the essence of the Berlin Colonial Conference 1884-1885.
Answer:

• The Berlin Colonial Conference of 1884-85 had that Africa should be divided into spheres of influence of various colonial powers.
• The war between the British and Boers in South Africa, however, was in defiance of this
  resolution.

Question 16.
Write a note on Third World Countries.
Answer:

• The capitalist countries led by the US were politically designated as the First Worlds, while the communist states led by the Soviet Union came to be known as the Second World states, outside these two were called third World.
• During the Cold War, third World consisted of the developing world the former colonies of Africa, Asia, and Latin America.
• With the break up of the Soviet Union in 1991, and the process of globalisation, the term Third World has lost its relevance.

Question 17.
What do you mean by drain of wealth?
Answer:
The colonial economy was a continuous transfer of resources from India to Britain without any favourable returns back to India. This is called the drain to wealth.

Question 18.
Discuss the importance of Hindu Religious Endowment Act passed by the Justicite ministry ?
Answer:

• Tamil Nadu has a large number of temples. These temples commanded huge resources which were monopolised and exploited by the dominant caste in the society and led to mismanagement of public resources.
• The Justice Party introduced the Hindu Religious Endowment Act in 1926 and enabled any individual, irrespective of their caste affiliation, to become member of the temple committee and govern the resources of the religious institutions.

Question 19.
State the west flowing rivers of India.
Answer:

• The Narmada and the Tapi are the west flowing rivers of the Peninsular India:
• These rivers drain into the Arabian Sea.
• These rivers form estuaries on the west coast.
• These rivers are devoid of an large tributary system.
• Narmada river is the largest among the west flowing rivers of peninsular India. Its principal tributaries are Bushner, Halon, Heran, Banjar, Dudhi, Shakkar, Tawa, Bama and Kolar.
• Tapti river is one of only the three rivers in peninsular India that run from east to west. The others being the Narmada and the Mahi.
• The major tributaries are Vaki, Gomai, Arunavati, Aner, Nesu, Buray Panjhra and Bori. It outfall into the Arabian sea through the Gulf of Cambay.

Question 20.
Write a note on Pipeline network transport in India.
Answer:

• Pipelines are used for transporting crude oil, petroleum products and natural products and natural gas from oil fields to the refineries, factories and big thermal power plants.
• Pipelines are more reliable and considerably safer mode of transportation.
• The possibility of pilferage or product less on pipelines is almost negligible. The basic limitations of pipelines is that they are capital intensive mode of transportation.

Question 21.
How is coastal plain formed?
Answer:
It is formed by the rivers that flow towards east and drain in the Bay of Bengal.

Question 22.
What is MRTS?
Answer:
MRTS means mass Rapid Transport System and currently developing a Metro System, with its first underground stretch in operation since May 2017.

Question 23.
What is a Writ?
Answer:
A writ is an order or command issued by a Court in writing under its seal.

Question 24.
Write about India’s foreign policy.
Answer:
Nehru, India’s first Prime Minister, was opposed to the rivalry of the two superpowers (America and Russia). The aim of India’s foreign policy of that time was ‘world co-operation, world peace, end of colonial imperialism, racial equality and non-alignment’.

Question 25.
What is meant by Gross Domestic Product?
Answer:
The GDP is the market value of all the final goods and services produced in the country during a time period.

Question 26.
Write the types of globalization.
Answer:
There are three types of globalization –

• Archaic globalization
• Proto globalization and
• Modem globalization.

Question 27.
Why we pay tax to the government?
Answer:
The levying of taxes aims to raise revenue to fund governance or to alter prices in order to affect demand. States and their functional equivalents throughout history have used money provided by taxation to carry out many functions.

Some of these include expenditures on economic infrastructure (transportation, sanitation, public safety, education, healthcare cur systems, to name a few), military, scientific research, culture and the arts, public works and public insurance and the operation of government itself. A government’s ability to raise taxes is called its fiscal capacity.

Question 28.
Name the states that lead in the production of iron ore in India.
Answer:
Jharkhand, Odisha, Chattisgarh, Karnataka, Andhra Pradesh and Tamil Nadu.

Part – III

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

Question 29.
Fill in the blanks:
(i) Japan forced a war on China in the year ………………
(ii) The major tribal revolt took place in chota nagpur region was ……………
(iii) The Cauvery rises in hills of …………………. district in Karnataka.
(iv) Governor of the state government surrenders his resignation to
(v) A better economy introduce rapid development of the ……………….
Answers
(i) 1894
(ii) Kolrevolt
(iii) Coorg
(iv) President
(v) Capital market

Question 30.
Match the following:

Answer:

Question 31.
Match the following:

Answer:

Question 32.
(a) Distinguish between
(i) Himalayan Rivers and Peninsular Rivers.
(ii) Andaman and Nicobar Islands and Lakshadweep Islands.
Answer:
(a) (i) Himalayan Rivers and Peninsular Rivers

Himalayan Rivers:

• Originate from Himalayas.
• Long and wide Perennial in nature.
• Perennial in nature.
• Unsuitable for hydropower generation.
• Middle and lower courses are navigable.

Peninsular Rivers:

• Originate from Western Ghats.
• Short and arrow.
• Non Perennial in nature.
• Suitable for hydropower generation.
• Not useful for navigation.

(ii) Andaman and Nicobar Islands and Lakshadweep Islands
Andaman and Nicobar Islands :

• They are located in the Bay of Bengal.
• They are far off from India.
• Port Blair is the capital.
• They are about 572 islands.
• Only 38 are inhabited.

Lakshadweep Islands:

• They are located in the Arabian Sea.
• They are of coral origin
• Kavaratti is the administration headquaters.
• They are 27 islands here.
• Only 11 islands are inhabited.

(b) Give reason: The great Indian desert is called Marusthali.
Answer:
The Thar desert, also known as the Great Indian desert is a large arid region in the north western part of the Indian Subcontinent that covers an area of 2,00,000 Km2 and forms a natural boundary between India and Pakistan.

Marusthali means sand-dune. It covered eastern portion of the Great Indian Thar Desert in western Rajasthan. It extends over about 24,000 square miles north of the Luni River.

Question 33.
Attempt an essay on the Arab-Israeli wars of 1967 and 1973.
Answer:
(i) The formation of Palestinian Liberation Organisation (PLO) was never friendly to Israel. It came to be attached frequently by Palestinian guerrilla groups based in Syria, Lebanon and Jordan. Israel also made violent retaliation.

(ii) In November 1966, Israel attacked the village of Al-Sami in the Jordanian West Bank. The death toll in this attack was 18. In April 1967 Israel started air battle with Syria which resulted in the shooting down of six Syrian Mig fighter jet.

(iii) In his bid to show Egypt’s support for Syria, Nasser mobilised Egyptian forces in the Sinai, seeking the removal of UN emergency forces stationed there on May 18. On May 22, he closed the Gulf Aqaba to Israeli shifting.

(iv) On June 5, Israel stopped a sudden pre-emptive air strike that destroyed more than 90 percent of Egypt’s air force on the tarmac. A similar air assault was in capacitated the Syria air force. Within three ways, Israel captured the Gaza Strip and all the Senai Peninsula up to the East Bank of the Suez Canal.

(v) On June 7, the Israeli forces drove Jordanian forces out of East Jerusalem and most the West Bank. The War ended when the UN Security Council called for a ceasefire. Arab-Israel War of 1973: Egypt and Syria made a secret agreement in January 1973 to bring their armies under one command.

(vi) Hafez al-Assad, the President of Syria was keen on retrieving Golan Heights. As Assad was aware that his country’s weapons were dated, he offered the Israelis a peace deal of they would withdraw from Sinai. Israel rejected the offer.

(vii) Egypt and Syria then launched a sudden attack on the Yom Kippur religions holiday on 6 October 1973. Though Israel suffered heavy casualties, it could finally push back the Arab forces. But this time, due to UN intervention, Israel was forced to return to 1967 position. Arabs gained nothing out of this war too.

Question 34.
Estimate Periyar E.V.R’s decisive contribution to the social transformation of Tamil Nadu?
Answer:
Periyar was a great social reformer. His contribution to the social transformation of Tamil Nadu is really praise worthy.

• He launched ‘Temple entry’ movement to provide the lower caste people easy access to the temples.
• He also started the Self-Respect movement in 1925, with the determination that there ought to be ‘no God; no religion; no Gandhi; no Congress; and no Brahmins’.
• Periyar was critical of patriarchy. He condemned child marriage and the Devadasi system. Right from 1929, when the Self-Respect conference began to voice its concern over the plight of women, Periyar had been emphasising on women’s right to divorce and property.
• He advocated atheism as a mode of critique to deconstruct the established practices of faith, culture and custom. Periyar wanted religion to be replaced by rationalism.
• He welcomed equal rights for males and females in property, guardianship and adoption. He was strong champion of birth control and contraception.

Question 35.
What is urbanization? Explain its impacts.
Answer:
The process of society’s transformation from rural to urban is known as urbanization. The level of urbanization of a place is assessed based on the size of population of the towns and cities and the proportion of population engaged in non agricultural sectors. These two are closely linked to the process of industrialization and expansion of the secondary and tertiary sectors of economy.

Impacts of urbanization:

• Urbanization and population concentration go hand – in – hand and are closely related to each other. A rapid rate of urbanization in a society is taken as an indicator of its economic development.
• Urbanization is increasing rapidly in the developing countries including India.
• Rural to urban migration leads to population explosion in urban areas. Metropolitan cities like Mumbai, Kolkatta and Delhi have more population than that can accommodate.

The following are the major problem of urbanization in India:

• It creates urban sprawl.
• It makes overcrowding in urban centres.
• It leads to shortage of houses in urban areas.
• It leads to the formation of slums.
• It increases traffic congestion in cities.
• It creates water scarcity in cities.
• It creates drainage problems.
• It poses the problem of solid waste management.
• It increases the rate of crime.

Question 36.
Write about Road Safety Rules.
Answer:

• Aware of the road signals
• Stop, look and cross
• Listen and ensure whether a vehicle is approaching
• Don’t rush on roads
• Cross roads in pedestrian crossings
• Don’t stretch hands while driving vehicles
• Never cross road at bends and stay safe in a moving vehicle

Question 37.
What are the powers and functions of the Chief Minister?
Answer:
The powers and the functions of the Chief Minister are:

• The Chief Minister is the head of the Council of Ministers. He recommends the persons who can be appointed as ministers by the Governor. He allocates the portfolio among the ministers.
• He presides over the meetings of the council of Ministers and influences its decisions.
• The Chief Minister is the principal channel of communication between Governor and the Council of Ministers.
• He announces the Government policies on the floor of the House. He can introduce the Bills in the Legislative Assembly.
• For smooth functioning of the State and for good Centre – State relations, he has to develop a rapport with the Union Government.

Question 38.
Explain the role of taxation in economic development.
Answer:
The role of taxation in developing economics is as follows.

1. Resource mobilisation: Taxation enables the government to mobilise a substantial amount of revenue. The tax revenue is generated by imposing direct taxes such as personal income tax and corporate tax and indirect taxes such as customs duty, excise duty, etc.

2 Reduction in equalities of income: Taxation follows the principle of equity. The direct
taxes are progressive in nature. Also certain indirect taxes, such as taxes on luxury goods, is also progressive in nature.

3. Social welfare: Taxation generates social welfare. Social welfare is generated due to higher taxes on certain undesirable products like alcoholic products.

4. Foreign exchange: Taxation encourages exports and restricts imports, Generally developing countries and even the developed countries do not impose taxes on export items.

5. Regional development: Taxation plays an important role in regional development, Tax incentives such as tax holidays for setting up industries in backward regions, which induces business firms to set up industries in such regions.

6. Control of inflation: Taxation can be used as an instrument for controlling inflation. Through taxation the government can control inflation by reducing the tax on the commodities.

Question 39.
Write briefly the history of globalization.
Answer:
The historical background of globalization can be discussed in three steps –
(i) Archaic Globalization – It is an early form of globalization. It existed during the Hellenistic Age. It established a trade link between the Roman Empire, Parthian Empire and the Han Dynasty. The commercial links between these powers inspired the development of the Silk Road. The Islamic Golden Age was also an important early stage of globalization.

(ii) Proto Globalization – It was characterised by the rise of maritime European empires in the 16th and 17th centuries. In the 17th century, globalization became private business phenomenon like British East India Company founded in 1600, described as the first multinational company.

(iii) Modern Globalization – The 19th and 20th centuries witnessed the advent of modem globalization. Global trade and capital investment increased. Several multinational firms came into being.

Question 40.
What are the contributions of Industrialization to development?
Answer:

(i) As stated earlier, it is essential to produce inputs to other producers in an economy. Even agriculture requires inputs from industry such as fertilisers and tractors to increase productivity.

(ii) Second, a market exists for both producers and consumer goods. Even services like banking, transport and trade are dependent on production of industrial goods.

(iii) Third, by using modem methods of production, industries contribute to better productivity and hence lower cost of production of all goods produced. It therefore helps people to buy goods at a cheaper rate and help create demand for more products.

(iv) Fourth, through such expansion of production, industrialisation helps to absorb the labour force coming out of agriculture. Employment generation is therefore an important objective of industrialisation.

(v) Fifth, a related advantage of industrialisation is therefore technological change. Through use of modem techniques, industrialisation contributes to learning of such methods and their improvement. As a result labour productivity, ie, output per unit of labour input increases, which can help workers earn higher wages.

(vi) Sixth, expanding incomes lead to more demand for goods and services. If an economy is not able to produce enough to meet such demand, it has to rely on imports and therefore spend a lot of foreign exchange. If the economy does not earn enough from exporting, it will be difficult to meet the growing demand. Industrialisation therefore helps an economy to save and also generate foreign exchange through exports. .

Question 41.
Draw a time line for the following:
five important events between 1889-1935.
Answer:
Samacheer Kalvi 10th Social Science Model Question Paper 2 English Medium – 5

Question 42.
Mark the following places on the world map.
(i) Great Britain
(ii) Russia
(iii) Hawai Island
(iv) Germany
(v) Norway
Answer:

Part – IV

Answer both questions. [2 × 8 = 16]

Question 43.
(a) Japanese Aggression in South-east Asia
(i) Name the South-east Asian countries which fell to the Japanese.
(ii) Account for the setback of Allies in the Pacific region?
(iii) What is the significance of Battle of Midway?
(iv) What happened to the Indians living in Burma?
Answer:
(a) Japanese Aggression in South-east Asia:

(i) Guam, the Philippines, Hong Kong, Singapore, Malaya, the Dutch East Indies (Indonesia) and Burma.
(ii) The Allies faced many reverses in the Pacific region because of their inadequate preparations. The colonial rulers, especially the British, withdrew from their territories, leaving the local people to face the atrocities of the Japanese.
(iii) The US navy defeated the Japanese navy in the Battle of Midway.
(iv) Many Indians walked all the way from Burma to the Indian border, facing hardships. Many died of disease and exhaustion. Those who remained suffered under the Japanese.

(b) Aligarh Movement
(i) What is the main aim of this movement? ‘
(ii) Who is considered the soul of this movement?
(iii) Why were English books translated into Urudu?
(iv) Name the college which was later raised to the status of a University?
Answer:
(b) Aligarh Movement:

(i) The main aim of the Aligarh Movement was to persuade the Muslims to acquire modem knowledge and English language.
(ii) Sir Sayed Ahmed Khan is considered the soul of the Aligarh Movement.
(iii) Many English Books were translated into Urdu in order to enable the Muslims to accept the western science and take up government services.
(iv) Aligarh Muhammadan Anglo-Oriental College.

[OR]

Question 43.
(c) Factors leading to the rise of National Movement.
(i) How did the national leaders inspire the people?
(ii) When was the Vernacular Press Act passed?
(iii) What was the policy of the British?
(iv) How did the British consider the Indians?
Answer:
(c) Factors leading to the rise of National Movement:

(i) They inspired the people with the ideas of self-respect and self-confidence.
(ii) The Vernacular Press Act was passed in 1878.
(iii) The British followed the policy of “Divide and Rule”.
(iv) The British considered the Indians as inferior and uncivilized.

(d) Maraimalai Adigal
(i) Name the Sangam texts for which Maraimalai Adigal wrote commentaries.
(ii) Name the journal where he worked as a young man.
(iii) Why did he oppose imposition of Hindi?
(iv) Who were the key influences in Maraimalai Adigal’s life?
Answer:
(d) Maraimalai Adigal

(i) Pattinappalai and Mullaipattu.
(ii) Siddhanta Deepika
(iii) Adigal promoted the use of pure Tamil words and removal of the Sanskrit influences from Tamil language. He painted out that Tamil language would suffer with the introduction of Hindi.
(iv) His teachers P. Sundaram Pillai and Somasundara Nayagar were the key influences in Maraimalai Adigal’s life.

Question 44.
Mark the following places on the given outline map of India.
(i) Zaskar range
(ii) River Ganga
(iii) Baghelkhand Plateau
(iv) Coffee growing area
(v) Any one Iron ore production area
(vi) Bombay high
(vii) Andaman & Nicobar Island
(viii) Cotton growing area
Answer:

[OR]

Mark the following places on the given outline map of Tamil Nadu:
(i) Red soil area
(ii) Mangrove forest
(iii) Villupuram
(iv) One oil seeds area
(v) Udagamandalam
(vi) Kancheepuram
(vii) Gomukhi dam
(viii) River Tamirabarani
Answer:

Map for Q. 42
(i) Great Britain
(ii) Russia
(iii) Hawai Island
(iv) Germany
(v) Norway
Answer:

Map for Q. 44
(i) Zaskar range
(ii) River Ganga
(iii) Baghelkhand Plateau
(iv) Coffee growing area
(v) Any one Iron ore production area
(vi) Bombay high
(vii) Andaman & Nicobar Island
(viii) Cotton growing area
Answer:

Map for Q. 44
(i) Red soil area
(ii) Mangrove forest
(iii) Villupuram
(iv) One oil seeds area
(v) Udagamandalam
(vi) Kancheepuram
(vii) Gomukhi dam
(viii) River Tamirabarani
Answer: