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These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1.
From the pair of linear equations in the following problems and find their solutions graphically.
(i) 10 students of Class X took part in mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 45. Find the cost of one pencil and one pen.
Solution:
(i) Let the number of boys participating in mathematics quiz is x and the number of girls participating in mathematics quiz is y.
Therefore the equations form in this situations are

(ii) Let the cost of 1 pencil is ₹ x
and the cost of 1 pen is ₹ y
Therefore the linear equations on the given situations are
5x + 7y = 50 … (i)
and 7x + 5y = 46 … (ii)

Again from equation (ii)

7x + 5y = 46
∴ x = \(\frac{46-5 y}{7}\)
Multiply equation. (1) by 7 and (ii) by 5 the get
35x + 49y = 350 … (i)
35x + 25y = 230 …(ii)
Substracting equation (ii) from equation (i) we get
24y = 120
y = 5
Putting the value of y in equation (j)
We get
5x + 35 = 50
5x = 15
x = 5
Cost of one pencil = ₹ 3
Cost of one pen = ₹ 5

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\) find out whether the lines representing the following pairs of linear equations intersect at a point, parallel or coincident.
(i) 5x – 4y + 8 = 0
7x + 6y-9 = 0

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
(i) We have,
5x – 4y + 8 = 0
and 7x + 6y – 9 = 0
Here, a₁ = 5, b₁ = – 4, c₁ = 8
and a₂ = 7, b₂ = 6 and c₂ = – 9
Therefore, \(\frac{5}{7} \neq \frac{-4}{6}\)
So, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
Therefore the following pair of linear equations intersect each other at point.

(ii) We have,
9x + 3y + 12 = 0
18x + 6y + 24 = 0
Here, a₁ = 9, b₁ = 3, c₁ = 12
and a₂ = 18, b₂ = 6 and c₂ = 24
Therefore, \(\frac{9}{18}=\frac{3}{6}=\frac{12}{24}=\frac{1}{2}\)
So, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Therefore the lines representing the given pair of linear equations are coincident.

(iii) We have given,
6x – 3y + 10 = 0
2x – y + 9 = 0
Here, a₁ = 6, b₁ = – 3, c₁ = 10
and a₂ = 2, b₂ = – 1 and c₂ = 9
Therefore,
\(\frac{6}{2}=\frac{-3}{-1} \neq \frac{10}{9}\)
So, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Therefore the lines representing the given pair of linear equations are coincident.

Question 3.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the following pair of linear equations are consistent or inconsistent.
(i) 3% + 2y = 5
2x – 3y = 7

(ii) 2x – 3y = 8
4x – 6y = 9

(iii) \(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 3 }\)y = 7
9x – 10y = 14

(iv) 5x – 3y = 11
– 10x + 6y = – 22

(v) \(\frac { 4 }{ 3 }\)x + 2y = 8
2x + 3y = 12
Solution:
(i) We have given,
3x + 2y = 5
2x – 3y = 7
We can also write
3x + 2y – 5 = 0
2x – 3y – 7 = 0
Here, a₁ = 3, b₁ = 2, c₁ = – 5
and a₂ = 2, b₂ = – 3 and c₂ = – 7
So, \(\frac { 3 }{ 2 }\) ≠ \(\frac { 2 }{ – 3 }\)
Therefore, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
So the given pair of linear equations are consistent.

(ii) We have given,
2x – 3y = 8
4x – 4y = 9
We can also write
2x – 3y – 8 = 0
4x – 6y – 3 = 0
Here, a₁ = 2, b₁ = – 3, c₁ = – 8
and a₂ = 4, b₂ = – 6 and c₂ = – 3
So, \(\frac{2}{4}=\frac{-3}{-6} \neq \frac{-8}{-9}\)
Therefore,
So, the given pair of linear equations are inconsistent.

(iii) We have given,
\(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 3 }\)y = 7
9x – 10y = 14
We can also write,
\(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 3 }\)y – 7 = 0
9x – 10y – 14 = 0
Here, a₁ = \(\frac { 3 }{ 2 }\), b₁ = \(\frac { 5 }{ 3 }\), c₁ = – 7
and a₂ = 9, b₂ = – 10 and c₂ = – 14

So, the given pair of linear equations consistent.

(iv) We have given,
5x – 3y = 11
– 10x + 6y = – 22
We can also write,
5x – 3y – 11 = 0
– 10x + 6y + 22 = 0
So, \(\frac { 5 }{ -10 }\) = \(\frac { – 3 }{ 6 }\) = \(\frac { -11 }{ 22 }\)
Here, a₁ = 5, b₁ = – 3, c₁ = – 11
and a₂ = – 10, b₂ = 6 and c₂ = 22
Therefore, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
So, the given pair of linear equations consistent.

(v) We have given
\(\frac { 4 }{ 3 }\)x + 2y = 8
2x + 3y = 12
We can also write,
\(\frac { 4 }{ 3 }\)x + 2y – 8 = 0
2x + 3y – 12 = 0
Here, a₁ = \(\frac { 4 }{ 3 }\), b₁ = 2, c₁ = 8
and a₂ = 2, b₂ = 3 and c₂ = – 12

So, the given pair of linear equations consistent.

Question 4.
Which of the following pairs of linear equations are consistent inconsistent ? If consistent, obtain the solution graphically:
(i) x + y = 5
2x + 2y = 10

(ii) x – y = 8
3x – 3y = 16

(iii) 2x + y – 6 = 0
4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0
4x – 4y – 5 = 0
Solution:
We have given
x + y = 5
2x + 2y = 10
We can also write,
x + y – 5 = 0
2x + 2y – 10 = 0
Here, a₁ = 1, b₁ = 1, c₁ = – 5
and a₂ = 2, b₂ = 2 and c₂ = – 10
So, \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) = \(\frac { – 5 }{ – 10 }\)
Therefore, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
So, the given pair of linear equations are consistent from equation (i) we have,

So the given pair of linear equations are inconsistent.

(iii) We have given
2x + y -6 = 0 … (i)
4x – 2y – 4 = 0 … (ii)
Here, a₁ = 2, b₁ = 1, c₁ = – 6
and a₂ = 4, b₂ = – 2 and c₂ = – 4
Therefore, \(\frac{2}{4} \neq \frac{1}{-2}\)
So, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
Therefore, the given linear equations are consistent From equation (i)

Therefore, the given linear equations are inconsistent.

Question 5.
Half the perimeter of rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the length of the rectangular garden by y
and the breadth of the rectangular garden by y
According to question
x = y + 4
x – y = 4 … (i)
Perimeter of rectangular garden = 72
2(x + y)= 72
2x + 2y = 72 … (ii)
Multiplying equation (i) and (ii) and adding both equations, we get
2x = 40
x = 20 m
Putting this value in equation (i) we get
20 – y = 4
y = 16 m
Hence the length of garden is 20 m and the width of garden is 16 m.

Question 6.
Given the linear equation 2x + 3y – 8 = 0 write another linear equaton in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) parallel lines, (iii) coincident lines.
Solution:
Given equation is 2x + 3y – 8 = 0 i.e., a₁ = 2, b₁ = 3, c₁ = 8
(i)

one such equation can be 5x + 4y + 1 = 0 (Try forming other equations. How many such equation can be ?)

One such equation can be 2x + 3y + 5 = 0 (Try forming other equations. How many such equation can be ?]

(iii) Try yourself.

Question 7.
Draw the graphs of the equation x- y +1 = of the vertices of the triangle formed by these line
Solution:

Verticles of the triangle are (-1, 0) (4, 0) and (2, 3).

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Question 1.
Solve the following pair of linear equations by the substitution method,

Solution:
(i) The pair of linear equation formed are
x + y = 14 … (i)
x – y = 14 … (ii)
We express x in the term of y from equation (ii) to get
x = y + 4
Now we substitute this value of x in equation (i) we get
(y + 4) + y = 14
2y = 14 – 4
y = 5
Putting this value of equation (ii)
x – 5 = 4
x = 9
∴ x = 9, y = 5

(ii) We have
s – t = 3 … (i)
\(\frac { s }{ 3 }\) + \(\frac { t }{ 2 }\) = 6 … (ii)
We express s in the terms of t from equation (i)
s = 3 + t
2t + 6 + 3t = 36
5t = 30
t = 6
Putting this value in equation (i)
s – 6 = 3
s = 9
∴ t = 6 and s = 9

(iii) We have
3x – y = 3 … (i)
9x – 3y = 9 … (ii)
From equation (i)
y = 3x – 3
Substituting this value in equation (ii) we get
9x – 3(3x – 3) = 9
9x – 9x + 9 = 0
9 = 9
The statement is true for all value of y
So y = 3x – 3
Where x can take any value i.e., infinite many solution.

(iv) We have
0.2x + 0.3y = 1.3 …(i)
0.4x + 0.5y = 2.3 …(ii)
From equation (i) we get
x = \(\frac{1.3-0.3 y}{0.2}\)
Substituting this value in equation (ii)
0.4\(\frac{(1.3-0.3 y)}{0.2}\) + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
– 0.1y = – 0.3
y = 3
Putting this value in equation (i) we get
∴ 0.2x + 0.3 x 3 = 1.3
0.2x = 1.3-0.9 0.2x = 0.4 x= 2
∴ x = 2 and y = 3

(v) We have
\(\sqrt{2x}\) + \(\sqrt{3y}\)y = 0 … (i)
\(\sqrt{3x}\) + \(\sqrt{8y}\)y = 0 … (ii)
From equation (i) we can get
x = \(\frac{-\sqrt{3} y}{\sqrt{2}}\)
Substituting this value in equation (ii)

Putting this value in equation (i) we get

Substituting this value in equation (i) we get

Putting this value if y if equation (i) we get

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
We have the two equations
2x + 3y = 11 … (i)
2x – 4y = – 24 … (ii)
From equation (ii)
2x = 4y – 24
x = 2y – 12
Substituting this value in equation (ii)
2(2y – 12) + 3y = 11
4y – 24 + 3y = 11
7y = 35
y = 5
Putting this value in equation (i)
2x + 15 – 11 x = – 2
Putting these value in
y = mx + 3
we get
5 = – 2m + 3
2 = – 2m
m = – 1
Hence the value of m = – 1 and x = – 2, y = 5

Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes 5/6, Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let 1st number be x and 2nd number be y.
Let x > y
1st condition :
x – y = 26
2nd condition :
x = 3y
Putting x = 3y in equation (i)
3y – y = 26 ⇒ 2y = 26 ⇒ y = 13
From (ii)
x = 3 x 13 = 39
∴ One number is 13 and the other number is 39.

(ii) Let one angle be x and its supplementary angle = y
Let x > y
1st Condition :
x + y = 180°
2nd Condition :
x – y = 18° ⇒ X = 18° + y
From equation (ii), putting the value ofx in equation (i),
18° + y + y = 180° ⇒ 18° + 2y = 180°
2y = 162° ⇒ y = 81°
From (ii) x = 18° + 81° = 99° ⇒ x = 99°
∴ One angle is 81° and another angle is 99°.

(iii) Let the cost of each bat be x
and the cost of each ball by y.
According to question
7x + 6y = 3800 … (i)
3x + 5y = 1750 … (ii)
From equation (i)
3\(\frac{(3800-6 y)}{7}\) + 5y = 1750
3(3800 – 6y) + 35y = 7 x 1750
17y = 12250 – 11400
y = 50
Putting this value in equation (i)
7x + 300 = 3800
7x = 3500
x = 500
Hence the rate of each but in ₹ 500 and the rate of each ball is ₹ 50

(iv) Let the fixed charge be x
and let the charge per km be y.
Then according to question
x + 10y = 105 … (i)
x + 15 y = 155 … (ii)
From equation (i)
x = 105 – 10y
Substituting this value in equation (ii)
105 – 10y + 15y = 155
5y = 50
y = ₹ 10 per km.
Putting this value in equation (i)
x + 100 = 105
x = ₹ 5
A person travelling distance of 25 km the charges will be.
= x + 25y
= 5 + 25 x 10
= ₹ 255

(v) Let the numerator of the fraction be x
and the denominator of the fraction be y.
Therefore according to question

(vi) Let the present age of Jacob be x
and the present age of his son be y
According to question
(x + 5) = 3(y + 5)
Again according to question five years ago
(x – 5) = 7(y – 5)
x – 7y = – 30
From equation (i) we get
x= 3y + 10
Substituting this value in equation (ii)
3y + 10 – 7y = – 30
– 4y = – 40
y = 10
Putting this value in equation (i)
x – 30 = 10
x = 40
Hence the present age of Jacob is 40 years and the present age of his son is 10 years.

These NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions and Equations Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Chemical Reactions and Equations NCERT Solutions for Class 10 Science Chapter 1

Class 10 Science Chapter 1 Chemical Reactions and Equations InText Questions and Answers

In-text Questions (Page 6)

Question 1.
Why should magnesium ribbon be cleaned before burning air ?
Answer:
A layer of magnesium oxide is already present on the surface of magnesium ribbon which does not allow burning of magnesium ribbon in air. So magnesium ribbon should be cleaned before burning in air.

Question 2.
Write the balanced equation for the following chemical reactions:
(i) Hydrogen + Chlorine → Hydrogen chloride
(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
(iii) Sodium + Water → Sodium hydroxide + Hydrogen
Answer:
(i) H₂(g) + Cl₂(g) → 2HCl (g)
(ii) 3 BaCl₂ + Al₂(SO₄)₃ → 3 BaSO₄ + 2AlCl₃
(iii) Na (s) + H₂O (l) → NaOH (aq) + H₂(g)

Question 3.
Write a balanced chemical equation with state symbols for the following reactions :
(i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride and water.
Answer:
(i) BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (↓) + 2NaCl (aq)
(ii) NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)

In-text Questions (Page 10)

Question 1.
A solution of a substance ‘X is used for white-washing:
(i) Name the substance ‘X’ and write its formula.
(ii) Write the reaction of the substance ‘X’ named in (i) above with water.
Answer:
(i) The substance ‘X’ is quick lime or calcium oxide.

Question 2.
Why is the amount of gas collected in one of the test tubes in activity 1.7 double of the amount collected in the other ? Name this gas.
Answer:
When electricity is passed through the water in presence of an acid., water is electroysed and produces hydrogen and oxygen gas.
\(2 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { Electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\)
It is clear from the above equation that 2 moles .water undergoes electrolysis and produce 2 moles of hydrogen and one mole of oxygen in the given set of conditions of temperature and pressure because the volume of a gas is directly proportional to its no. of moles so the amount of one gas is doubled than the other.
The double amount gas is hydrogen.

In-text Questions (Page 13)

Question 1.
Why does the colour of copper sulphate solution change when an iron nail is dipped in it?
Answer:
Iron is more reactive that copper so it displaces copper metal from copper sulphate solution and produces sulphate of iron.

Question 2.
Give an example of a double displacement reaction other than the one given in Activity 1.10.
Answer:

Question 3.
Identify the substances that are oxidised and the substances that are reduced in the
following reactions.
(i) Na(s) + O₂(g) → 2Na₂O(s)
(ii) CuO(s) + H₂ (g) → CU(s) + H₂O(l)
Answer:
(i) 2 Na (s) + O₂ (g) → 2 NO₂O (s)
Na (s) → Oxidised
O₂(g) → Reduced
(ii) CuO (s) + H₂(g) → Cu (S) + H₂O (l)
CuO (s) → Reduced
H₂ (g) → Oxidised

Class 10 Science Chapter 1 Chemical Reactions and Equations Textbook Questions and Answers

Page No. 14

Question 1.
Which of the statements about the reaction below are incorrect?
2PbO(s) + C(s) → 2Pb(s) + CO₂(g)
(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced.
(i) (a) and (b)
(ii) (a) and (c)
(iii) (d), (b) and (c)
(iv) all
Answer:
(i) a and b

Question 2.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
The above reaction is an example of a
(a) combination reaction.
(b) double displacement reaction.
(c) decomposition reaction.
(d) displacement reaction.
Answer:
(d) displacement reaction.

Question 3.
What happens when dilute hydrochloric add is added to iron fillings? Tick the correct answer.
(a) Hydrogen gas and iron chloride are produced.
(b) Chlorine gas and iron hydroxide are produced.
(c) No reaction takes place.
(d) Iron salt and water are produced.
Answer:
(a) Hydrogen gas and iron chloride are produced.

Question 4.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer:
A chemical equation is said to be balanced when it has the same no. of atoms of different elements in both the sides, i. e., reactant and products sides.
e.g. H₂ + O₂ → H₂O
The above equation is not a balanced equation because it does not have the equal no. of atoms of oxygen in both sides.
But 2H₂ + O₂ → 2 H₂O is a balanced equation because it does not have the equal no. atoms of different elements in both the sides.

A chemical equation should be balanced because during a chemical change the no. of atoms of each element remain same. In other words the law of concervation of mass i.ethe total mass of all the products of a chemical reaction has to equal to the total mass of all reactants. And it is only possible when a chemical equation is balanced.

Question 5.
Translate the following statements into chemical equations and then balance them.
(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas bums in air to give water and sulpur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Answer:
(a) H₂ (g) + N₂(g) → NH₃ (g)
Step I. Look more closely the number of atoms of different elements present in the unbalanced chemical equation.

Step II. It is often convenient to start with the compound that contains maximum number of atoms, whether a reactant or a product. So we select NH3 and the element hydrogen in it. There are three hydrogen atoms on the right and only two hydrogen atoms on the left.

Now the partly balanced equation becomes as follows :
3H₂ + N₂ → 2NH₃

Step III. But nitrogen in the above equation is automatically balanced. So the balanced equation will be
3 H₂ + N₂ → NH₃
The above equation balanced because it contains the equal no. of atoms of hydrogen and nitrogen on both sides.
(b) H₂S + O₂ → H₂O + H₂O
The above equation can be balanced by algebraic sum method.

Let ‘a’, ‘b’, ‘c’ and ‘d’ are the no. of molecules of ‘H₂S’, ‘O₂‘, ‘H₂O’ and ‘SO₂‘ respectively in the balanced chemical equation. So the above equation can be written as
a H₂S + b O₂ → c H₂O + d SO₂
Now in a balanced chemical equation.
L.H.S. R.H.S.
Hydrogen 2 a= 2c …(i)
Sulphur a = d …(ii)
Oxygen 2b = c + 2d ….(iii)
Let a = 1 …..(iv)

So from equations (i) and (iv)
2 × 1 = 2 c
⇒ c = 1
From equation (iii)
2b = c + 2d
2b = 1 + 2 × 1
⇒ 2b = 3
⇒ b = \(\frac {3}{2}\)
Now by putting the value of a, b, c and d in the given equation, we get
1 × H₂S + \(\frac {3}{2}\)O₂ → 1 × H₂O + 1 × SO₂
OR 2 H₂S + 3O₂ → 2 H₂O + 2SO₂

(c)

Step I:

Step II: Now start from Al₂ (SO₄)₃ and balance oxygen first. The simple ratio between the atoms of oxygen in reactants to the product will be 12 : 4 or 3 : 1. So multiplying Al₂ (SO₄)₃ by ‘1’ ‘BaCl₂‘ by ‘3’, we get
BaCl2 + Al2 (SO4)3 → AlCl3 + 3BaSO4

Step III: Now balance ‘Al’ be multiplying ‘2’ in the product side (i.e. in AlCl₃)
BaCl₂ + Al₂ (SO₄)₃ → 2 AlCl₃ + 3 BaSO₄

Step IV : Balance barium by multiplying ‘3’ in the reactant to BaCl₂.
3 BaCl₂ + Al₂ (SO₄)₃ → 2 AlCl₃ + 3 BaSO₄
‘Cl’ is automatically balanced in the above equation.

(d) K + H₂O → KOH + H₂
Step I.

Step II: Now select ‘KOH’ in the reactant side because it contains the maximum no. of elements.
K + 2H₂O → 2 KOH + H₂

Step III: Balance potassium metal in the above partially balanced equation by multiplying ‘2’ in the potassium i.e. reactant side.
2K + 2H₂O → 2KOH + H₂

Question 6.
Balance the following chemical equations.
(a) HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + H₂O
(b) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
(c) NaCl + AgNO₃ → AgCl + NaNO₃
(d) BaCl₂ + H₂SO₄ → BaSO₄ + HCl
Answer:
(a)
Step I.

Step II. Select Ca (NO₃)₂ because it contains the maximum no. of elements. To balance oxygen multiply Ca (OH)₂ by two, we get
HNO₃+ 2Ca (OH)₂ → Ca(NO₃)₂ + H₂O

Step III. Now balance calcium atoms by multiplying two in reactants sides. We get
HNO₃ + 2 Ca (OH)₂ → Ca (NO₃)₂ + H₂O

Step IV : Balance nitrogen in the above partially balanced equation, multiply HNO₃ by ‘4’. We get
4 HNO₃ + 2 Ca (OH)₂ → 2 Ca (NO₃)₂ + H₂O

Step V : Balance hydrogen by multiplying ‘4’ in the product side. We get,
HNO₃ + 2 Ca (OH)₂ → 2 Ca (NO₃)₂ + 4H₂O
This is a balanced chemical equation.

(b) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
Step I.

Step II: Balance sodium in the above equation by multiplying ‘2’ in NaOH.
2 NaOH + H₂SO₄ → Na₂SO₄ + H₂O

Step III: Balance hydrogen by multiplying ‘2’ in H₂O.
2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O
This is a balanced equation.

(c) NaCl + AgNO₃ → AgCl + NaNO₃
Step I.

The above equation contains equal no. of atoms of different elements in reactants and products sides.

(d) BaCl₂ + H₂SO₄ → BaSO₄ + HCl
Step I.

Step II : Now balance chlorine atoms in the above equation by multiplying ‘2’ in the product side to HCl
BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
This is a balanced chemical equation.

Question 7.
Write the balanced chemical equations for the following reactions.
(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
(b) Zinc + Silver nitrate → Zinc nitrate + Silver
(c) Aluminium + Copper chloride → Aluminium chloride + Copper
(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride.
Answer:
(a) Ca (OH₂) + CO₂ → CaCO₃ + H₂O
Step I.

The above equation is balanced.

(b) Zn + AgNO₃ → Zn (NO₃)₂ + Ag
Step I.

Step II: Balance oxygen by multiplying ‘2’ in AgNO₃, we get
Zn + 2 AgNO₃ → Zn (NO₃)₂ + Ag

Step III ; Now Balance ‘Ag’ metal by multiplying ‘2’ in product side.
Zn + 2 AgNO₃ → Zn (NO₃)₂ + 2Ag

(c) Al + CuCl₂ AlCl₃ + Au
Step I.

By multiplying ‘AlCl₃‘ with ‘2’ and ‘CuCl₂‘ with ‘3’, ‘Al’ with ‘2′ and ‘Cu’ with ‘3’ we get
2Al + 3 CuCl₂ → 2AlCl₃ + 3 Cu
This is a balanced equation

(d) BaCl₂ + K₂SO₄ → BaSO₄ + KCl
Step I.

Balance chlorine by multiplying ‘2’ in NaCl, we get
BaCl₂ + K₂SO₄ → BaSO₄+ 2KCl

Question 8.
Write the balanced chemical equation for the following reactions and identify the type of reaction in each case.
(a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
(b) Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g)
(c) Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g)
(d) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g)
Answer:
(a) KBr (aq) + Bal₂ (aq) → KI (aq) BaBr₂ (aq)
Step I.

Balance bromine and iodine in the above equation.
2 KBr (aq) + Bal₂ (aq) → 2 Kl (aq) + BaBr₂ (aq)
It is a double displacement reaction.
(b) ZnCO₃ (s) → ZnO (s) + CO₂ (g)
It is decomposition reaction.
(c) H₂(g) + Cl₂(g) → 2 HCl (g)
It is combination reaction.
(d) Mg (s) + 2 HCl (aq) → MgCl₂ (aq) + H₂ (g)
It is a displacement reaction.

Question 9.
What does one mean by exothermic and endothermic reactions? Give examples.
Answer:
Exothermic Reaction: A reaction in which heat energy is given out along with the product is called exothermic reaction.
C (s) O₂ (g) → CO₂ (g) + Heat energy (394 kJ)
N₂ (g) + 3H₂ → (g) 2 NH₃ (g) + Heat energy (92.4 kJ)

Endothermic Reaction : A reaction in which heat energy is absorbed along with the product is called endothermic reaction.
CaCO₃ (s) → CaO (s) CO₂ (g)-Heat energy
2HgO (s) → 2 Hg (l) + O₂ (g)-Heat energy

Question 10.
Why is respiration considered an exothermic reaction? Explain.
Answer:
It is a well known fact that animals and human being require energy to stay alive. We and animals get this energy from food. During oxidation or digestion, food is broken down into simple substances. Rice, potatoes and bread contains carbohydrates. These carbohydrates are broken down to form glucose. This glucose combines with oxygen in the cells of our body and liberates energy. This reaction is known as respiration.
C₆H₁₂O₆ (aq) + 6O₂ (aq) → 6 CO₂ (aq) 6H₂O (l) + energy
Because energy is released during respiration so it is an exothermic reaction.

Question 11.
Why are decomposition reactions called opposite of combination reactions? Write equations for these reactions.
Answer:
In decomposition reactions a single reactant breaks down to give two or more simpler products.

In combination reaction two or more than two reactants combine with each other and to give simple product.

So it is dear from above discussion that decomposition reactions are opposite of the combination reactions,

Question 12.
Write equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.
Answer:
In a decomposition reaction a single reactant breaks down to give two or more simpler products. These reactions need energy in different forms to proceed.

The decomposition of CaCO₃ (s) takes place by supplying energy in the form of heat.
\(\mathrm{CaCO}_{3}(s) \stackrel{\Delta}{\rightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(g)\)
In the presence of sunlight white silver chloride decomposes and turns grey because silver metal is formed.

Water decomposes into H, and O, by passing electricity in an electrolytic cell.
\(2 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { Electricity }}{\rightarrow} 2 \mathrm{H}_{2}(g) \mathrm{O}_{2}(g)\)

Question 13.
What is the difference between displacement and double displacement reactions ? Write equations for these reactions.
Answer:
Displacement Reaction : In a displacement reaction, more active element displaces or removes another element from its compound.
e.g. Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)

Double displacement reactions: In a double displacement reaction two different atoms or groups of atoms are displaced by other atoms or group of atoms.
e.g

Question 14.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.
Answer:

Question 15.
What do you mean by a precipitation reaction? Explain by giving examples.
Answer:
Precipitation reaction : A reaction in which precipitates are formed is called precipitation reaction.
e.g.

Question 16.
Explain the following in terms of gain or loss of oxygen with two examples each.
(a) Oxidation
(b) Reduction
Answer:
(a) Oxidations A process which involves the gain of oxygen is called oxidation. In oxidation reaction oxygen is added to the reactant.
e.g. 2 Mg (s) + O₂ (g) → 2 MgO (s)
2 H₂S (g) + O₂ (g) → 2 S (s) + 2 H₂O (l)

(b) Reduction: A process which involves the loss of oxygen is called reduction.
2 Pb (NO₃)₂ (S) → 2 PbO (s) + 2NO₂ (g) + \(\frac {1}{2}\)O₂ (g)
2 KNO₃ → 2 KNO₂ + O₂

Question 17.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.
Answer:
The shiny brown coloured element ‘X’ is copper.
When copper is heated in air it forms copper oxide which is black in colour.

Question 18.
Why do we apply paint on iron articles?
Answer:
When metals are exposed by air, moisture or other atmospheric gases, it results in the formation of compounds like sulphides, oxides, carbonates etc. on the surface of metals. This phenomenon is called corrosion.

To Protect iron articles form corrosion they are painted i.e., a thin layer of paint is applied on the iron articles so they do not interact within the atmospheric gases and no corrosion of iron articles is occured.

Question 19.
Oil and fat containing food items are flushed with nitrogen. Why?
Answer:
When fats and oils are oxidised, they become rancid and their smell and taste change. Usually special type of substances i.e., antioxidants are added to fatty foods to prevent oxidation. It is known as rancidity. To protect food from oxidation or to slow down oxidation process of food it is kept in the refrigerator on we flush food with nitrogen.

Question 20.
Explain the following terms with one example each.
(a) Corrosion (b) Rancidity
Answer:
(a) Corrosion : The process of slowly eating away of the metals when they are exposed by air, moisture or other atmospheric gases, resulting into the formation of compounds such as oxides, sulphides, carbonates etc. is called corrosion.

Iron is corroded when exposed by moisture or air. Corrosion of iron is also known as rusting. Rust is mainy hydrated iron (III) oxide i.e. Fe₂O₃. xH₂O.

Rusting weakness the structure of car bodies, bridges, iron railings and other iron articles. Rusting of iron is a serious problem. Every year a large amount of money is spent to replace damaged iron.

(b) Rancidity: When fats and oils are oxidised they become rancid and their smell and taste change. Generally special type of substances i.e., antioxidants are added to fatty foods to prevent oxidation. At home oxidation of food can be slowed down by keeping it in the refrigerator. Keeping and food in air-tight containers can also help.

Class 10 Science Chapter 1 Chemical Reactions and Equations Textbook Activities

Activity 1.1 (Page 1)

Caution : This activity needs teacher’s assitsance. It would be better if students wear eye

Fig : Burning of a magnesium ribbon in air and collection of magnesium oxide in watch glass

• Clean a magnesium ribbon about 2 cm long by rubbing with sand paper.
• Hold it with a pair tongs. Born it using a spirit lamp or burner and collect the ash so formed in a watch glass as shown in Fig. Burn a magnesium ribbon keeping it as far as possible from your eyes.

Question 1.
What do you observe ?
Answer:
Observation : What observe that magnesium ribbon burns with a dazzling white flame and changes into a white powder. This powedered material is magnesium oxide. It is formed due to the reaction between magnesium and oxygen present in air.

Activity 1.2 (Page 2)

• Take lead nitrate solution in a test tube.
• Add potassium iodide solution in this.

Question 1.
What do you observe ?
Answer:
Pb (NO₃)₂ (aq) + 2KI (aq) → PbI₂(ppt.) + 2KNO₃
When the solution of potassium iodide is added to the lead nitrate solution, yellow colour precipitate of lead iodide is formed. It represents a chemical change.

Activity 1.3 (Page 2)

• Take few zinc granules in a conical flask or test tube.
• Add dilute hydrochloric acid or sulphuric add to this (Figure).

Caution : Handle the acid with care.

Question 1.
What do you observe immediately above the zinc granules in the conical flask or test tube ?
Touch the conical flask or test tube. Is there any change in temperature.
Answer:
Zinc granules react with dil. Hydrochloric acid or sulphuric acid liberates hydrogen gas in the form of bubbles along with zinc chloride or zinc sulphate.
Yes, the test tube becomes hot, because it is an exothermic reaction.

Activity 1.4 (Page 6)

• Take a small amount of calcium oxide or quick lime in a beaker.
• Slowly add water to this.
• Touch the beaker as shown in Fig.

Question 1.
Do you feel any change in temperature.
Answer:
Calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide) releasing a large amount of heat.

The temperature of the beaker is raised because this is an exothermic reaction.

Activity 1.5 (Page 8)

• Take 2 gm ferrous sulphate crystals in a dry boiling tube.
• Note the colour of the ferrous sulphate crystals.
• Heat the boiling tube over the flame of a burner or spirit lamp as shown fig.

Question 1.
Observe the colour of the crystals after heating.

Answer:
The colour of ferrous sulphate crystals is green. After heating for sometime the green colour of ferrous sulphate has changed into reddish brown.

Activity 1.6 (Page 8)

• Take about 2 gm lead nitrate powder in a boiling tube.
• Hold the boiling tube with a pair of tongs and heat it over a flame as shown in Fig.

Question 1.
What do you observe ? Note down the change, if any.

Answer:
Observation : When lead nitrate powder is heated in a boiling tube, we observe brown fumes of a nitrogen dioxide (NO.) release. The following decomposition reaction takes place :

Activity 1.7 (Page 9)

• Take a plastic mug. Drill two holes at its base and fit rubber stoppers in there holes, Insert carbon electrodes in these rubber stoppers as shown in figure.
• Connect these electrodes to a 6 volt battery.
• Fill in the mug with water such that the electrodes are immersed. Add a few drops of dilute sulphuric acid to the water.
• Take two graduated test tubes filled with water and invert them over the two carbon electrodes.
• Switch on the current and leave the appratus undisturbed for some time.
• You will observe the formation of bubbles at both the electrodes.
• These bubbles displace water in the graduted tubes.

Question 1.
Is the volume of the gas collected the same in both the test tubes ?
Answer:

• Once the test tube are filled with the respective gases, remove them carefully.
• Test these gases one by one by bringing a burning candle close to the mouth of the test tubes. Caution :

This step must be performed carefully by the teacher.

Question 2.
What happens in each case ?
Which gas is presen in each test tube ?
Answer:
Observation : No, the volume of the gases in both test tubes are not same.
The two test tubes contain ‘H₂‘ and ‘O₂‘ respectively.
Hydrogen burns with a pale blue flame but oxygen is not combustible.

Activity 1.8 (Page 9)

• Take about 2 gm silver chloride on a china dish.
• What is its colour ?
• Place this china dish under sunlight for sometime. (Fig.)

Question 1.
Observe the colour of the silver chloride after some time.
Answer:
The colour of the silver is white. We observe that the white colour of silver chloride turns grey in sunlight. This is due to the decomposition of silver chloride and chlorine by light.
\(2 \mathrm{AgCl}(\mathrm{s}) \stackrel{\text { Sunlight }}{\longrightarrow} 2 \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g})\)

Activity 1.9 (Page 10)

• Take three iron nails and clean them by rubbing with sand paper.
• Take two test tubes marked as (A) and (B). In each test tube, take about 10 ml copper sulphate solution.
• Tie two iron nails with a thread and immerse them carefully in the copper sulphate solution in test tube B for about 20 minutes [Fig. (a)]. Keep one iron nail aside for comparison.

• After 20 minutes, take out the iron nails from the copper sulphate solution.
• Compare the intensity of the blue colour of copper sulphate solutions in test tubes (A) and (B), [Fig. (b)].
• Also, compare the colour of the iron nails dipped in the copper sulphate solution with the one kept aside [Fig. (b)].

Question 1.
Why does the iron nail become brownish in colour and the blue colour of copper sulphate solution fade?

Answer:
Observatons: The following chemical reaction takes place in this Activity :

In this reaction, iron has displaced or removed another element, copper, from copper sulphate solution. This reaction is known as displacement reaction.

Activity 1.10 (Page 11)

• Take about 3 ml of sodium sulphate solution in a test tube.
• In another test tube, take about 3 ml of barium chloride solution.
• Mix the two solutions (Fig.).

Question 1.
What do you observe?
Answer:
Observations : We observe that white precipitate of barium sulphate is formed. Any reaction that produces a precipitate can be called a precipation reaction.

This is a double displacement reaction,

Recall Activity, 1.2, where you. have mixed the solutions of lead (II) nitrate and potassium iodide.
(i) What was the colour of the precipitate formed? Can you name the compound precipitated?
(ii) Write the balanced chemical equation for this reaction.
(iii) is this also a double displacement reaction ?
Answer:
(i) Yellow colour precipitates of lead iodide are formed.

(iii) Yes, this is a double displacement reaction.

Activity 1.11 (Page 12)

• Heat a china dish containing about 1g copper powder (Fig.).

Question 1.
What do you observe?
Answer:

Observation : The surface of copper powder becomes coated with black copper(II) oxide. This is because oxygen is added to copper and copper oxide is formed.
\(2 \mathrm{Cu}+\mathrm{O}_{2} \stackrel{\text { Heat }}{\longrightarrow} 2 \mathrm{CuO}\)

Recall Activity 1.1, where a magnesium ribbon bums with a dazzling flame in air (oxygen) and changes into a white substance, magnesium oxide. Is magnesium being oxidised or reduced in this reaction?
Answer:
Magnesium is being oxidised.

Class 10 Science Chapter 1 Chemical Reactions and Equations Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What are the state symbols of solid, liquid and gas ?
Answer:
Solid (s), liquid (l) and gas (g).

Question 2.
What is the state symbol of aqueous state?
Answer:
Aqueous (aq).

Question 3.
What are the symbols used to represent gaseous product and precipitate product ?
Answer:
Gaseous Product (↑); Precipitate Product (↓)

Question 4.
What is the formula of quick lime ?
Answer:
CaO.

Question 5.
What is the formula of slaked lime ?
Answer:
Ca(OH)₂.

Short Answer Type Questions

Question 1.
Name the component oxidised, component reduced, oxidising agent and reducing agent in the following displacement reactions.
(i) Mg + CuSO₄ MgSO₄ + Cu
(ii) Cu + 2 AgNO₃ Cu (NO₃) → 2 + 2 Ag.
Answer:
(i) Mg : Oxidised, recuding agent.
CU₂ : Reduced, oxidising agent.
(ii) CU : Oxisided, redusing agent.
Ag+ : Reduced, oxidising agent.

Question 2.
What type of chemical reactions take place when:
(a) A magnesium wire is brunt in air ?
(b) Limestone is heated ?
(c) Proteins are converted into amino acids.
(d) Electricity is passed through water ?
(e) Ammonia and hydrogen chloride are mixed ?
Answer:
(a) Combination reaction.
(b) Decomposition reaction.
(c) Decomposition reaction.
(d) Decomposition reaction.
(e) Combination reaction.

Question 3.
In the reaction represented by the equation:
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
(a) Name the substance oxidised.
(b) Name the oxidising agent.
(c) Name the substance reduced.
(d) Name the reducing asgent.
Answer:
(a) HCl, (b) MnO₂, (iii) MnO₂, (iv) HCl

Long Answer Type Question

Question 1.
What is corrosion ? What are the factors which promote corrosion ? How it is prevented ?
Answer:
The process of slowly eating away of the metal due to attack of the atmospheric gases on the surface of the metal resulting into the formation of compounds such as oxides, sulphides, carbonates, sulphates, etc. is called corrosion.

Factors which promote corrosion :

• Reactivity of the metal: More active metals are corroded readily e.g. Iron corroded more than zinc.
• Impurities : Impure metal are corroded more than pure metals e.g. pure iron does not rust.
• Presence of air and moisture : Air and moisture increase the rate of corrosion. Gases like ‘SO₂‘ and ‘CO₂‘ in air accelerate corrosion. Iron does not rust in vacuum.
• Electrolytes: Presence of electrolytes like NaCl, HCl etc. accelerate the corrosion. For example, iron rusts faster in saline water than in pure water.
• Strains in metals: Bends, scratches, nicks and cuts in a metal increase the rate of corrosion.

Prevention : When a metal surface is not allowed to come in contact with moisture, oxygen and carbon dioxide, it stops corrosion.

The can be achieved by the following methods:
(i) The metal surface is coated with paint which keeps it out of contact with air, moisture etc. till the paint layer develops cracks.

(ii) By applying film of oil and grease on the surface of the iron tools and machinery, the rusting of iron can be prevented since it keeps the iron surface away from moisture, oxygen and carbondioxide.

Multiple Choice Questions

Question 1.
The colour of lead iodide is:
(a) Red
(b) Blue
(c) White
(d) Yellow
Answer:
(b) Blue

Question 2.
Magnesium bums in air to form:
(a) Magnesium oxide
(b) Magnesium trioxide
(c) Magnesium dioxide
(d) A mixture of magnesium trioxide
Answer:
(a) Magnesium oxide

Question 3.
The following observations (s) help (s) us to determine that a chemical reaction has taken place:
(a) Change in colour
(b) Evolution of gas
(c) Change in temperature
(d) All the above
Answer:
(d) All the above

Question 4.
Zinc reacts will dil. H₂SO₄ to evolve:
(a) Oxygen
(b) Nitrogen
(c) Sulphur dioxide
(d) Hydrogen
Answer:
(d) Hydrogen

Question 5.
The molecular formula of rust is:
(a) Fe₂O₃
(b) FeO. xH₂O
(c) Fe₂O₃. xH₂O
(d) FeO.
Answer:
(c) Fe₂O₃. xH₂O

These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

Solution:
(i) 28
(ii) 2
(iii) 46
(iv) 10
(v) 3.5

Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …, is
(A) 97
(B) 77
(C) -77
(D) -87

(ii) 11th term of the AP: -3, \(\frac { -1 }{ 2 }\) , 2, …, is
(A) 28
(B) 22
(C) -38
(D) -48
Solution:
(i) 10, 7, 4, …,
a = 10, d = 7 – 10 = – 3, n = 30
a_n = a + (n – 1)d
⇒ a₃₀ = a + (30 – 1) d = a + 29 d = 10 + 29 (-3) = 10 – 87 = – 77
Therefore, 30th term of the sequence 10, 7, 4, is – 77 i.e., (C) is the correct choice.

(ii) We have given the sequence,

Therefore, 11th term of the sequence is 22 i.e., (B) is the correct choice.

Question 3.
In the following APs, find the missing terms in the boxes:

Solution:

Question 4.
Which term of the AP: 3, 8, 13, 18, …, is 78?
Solution:
Given: 3, 8, 13, 18, ………,
a = 3, d = 8 – 3 = 5
Let nth term is 78
a_n = 78
a + (n – 1) d = 78
⇒ 3 + (n – 1) 5 = 78
⇒ (n – 1) 5 = 78 – 3
⇒ (n – 1) 5 = 75
⇒ n – 1 = 15
⇒ n = 15 + 1
⇒ n = 16
Hence, a₁₆ = 78

Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19, …, 205
(ii) 18, 15\(\frac { 1 }{ 2 }\), 13, …, -47
Solution:
(i) We have given the sequence 7,13,19,…. 205 Here, a = 7, d = 13-7 = 6 an = 205 Let n terms are in this AP.
∴ a_n = 205
a + (n – 1)d= 205 (∴ a_n = a + (n – 1) d)
or, 7 + (n – 1) x 6 = 205 (∴ a = 7 and d = 6)
or, (n – 1)6 = 205 – 7
or, (n – 1) = \(\frac { 198 }{ 6 }\)
∴ n = 33 + 1 = 34
Therefore, this sequence has 34 terms.

(ii) We have given the sequence

Question 6.
Check, whether -150 is a term of the AP: 11, 8, 5, 2, ….
Solution:
11, 8, 5, 2, …….
Here, a = 11, d = 8 – 11= -3, a_n = -150
a + (n – 1) d = an
⇒ 11 + (n – 1) (- 3) = -150
⇒ (n – 1) (- 3) = -150 – 11
⇒ -3 (n – 1) = -161
⇒ n – 1 = \(\frac { -161 }{ -3 }\)
⇒ n = \(\frac { 161 }{ 3 }\) + 1 = \(\frac { 164 }{ 3 }\) = 53\(\frac { 4 }{ 3 }\)
Which is not an integral number.
Hence, – 150 is not a term of the AP.

Question 7.
Find the 31^st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
a₁₁ = 38 and a₁₆ = 73
⇒ a₁₁ = a + (11 – 1) d ⇒ a + 10d = 38 ….. (i)
⇒ a₁₆ = a + (16 – 1 )d ⇒ a + 15d = 73 …(ii)
Subtracting eqn. (i) from (ii), we get
a + 15d – a – 10d = 73 – 38
⇒ 5d = 35
⇒ d = 1
From (i), a + 10 x 7 = 38
⇒ a = 38 – 70 = – 32
a₃₁ = a + (31 – 1) d = a + 30d = – 32 + 30 x 7 = – 32 + 210 = 178
Therefore, the 31th term of this AP is 178.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Given:
a₅₀ = 106
a₅₀ = a + (50 – 1) d
⇒ a + 49d = 106 … (i)
and a₃ = 12
⇒ a₃ = a + (3 – 1 )d
⇒ a + 2d = 12 … (ii)
Subtracting eqn. (ii) from (i), we get
a + 49d – a – 2d = 106 – 12
⇒ 47d = 94
⇒ d = \(\frac { 94 }{ 47 }\) = 2
a + 2d = 12
⇒ a + 2 x 2 = 12
⇒ a + 4 = 12
⇒ a = 12 – 4 = 8
a₂₉ = a + (29 – 1) d = a + 28d = 8 + 28 x 2 = 8 + 56 = 64
Therefore, the 29^th term of this AP is 64.

Question 9.
If the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Given: a₃ = 4 and a₉ = – 8
⇒ a₃ = a + (3 – 1 )d ⇒ a + 2d = 4 …(i)
a₉ = a + (9 – 1) d ⇒ a + 8d = -8 ….(ii)
Subtracting eqn. (i) from (ii), we get
a + 8d – a – 2d = -8 – 4
⇒ 6d = -12.
⇒ d = -2
Now,
a + 2d = 4
⇒ a + 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let a_n = 0
⇒ a + (n – 1) d = 0
⇒ 8 + (n – 1) (- 2) = 0
⇒ 8 = 2 (n – 1)
⇒ n – 1 = 4
⇒ n = 4 + 1 = 5
Therefore, 5th term of this AP is 0.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Given: a₁₇ – a₁₀ = 7
⇒ [a + (17 – 1 ) d] – [a + (10 – 1 ) d] = 7
⇒ (a + 16d) – (a + 9d) = 7
⇒ 7d = 7
⇒ d = 1
Therefore, the common difference of this AP is 1.

Question 11.
Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:
3, 15, 27, 39, …..
Here, a = 3, d = 15 – 3 = 12
Let a_n = 132 + a₅₄
⇒ a_n – a₅₄ = 132
⇒ [a + (n – 1) d] – [a + (54 – 1) d] = 132
⇒ a + nd – d – a – 53d = 132
⇒ 12n – 54d = 132
⇒ 12n – 54 x 12 = 132
⇒ (n – 54)12 = 132
⇒ n – 54 = 11
⇒ n = 11 + 54 = 65
Therefore, 65th term of this AP be 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let a and A be the first term of two APs and d be the common difference.
Given:
a₁₀₀ – A₁₀₀ = 100
⇒ a + 99d – A – 99d = 100
⇒ a – A = 100
⇒ a₁₀₀₀ – A₁₀₀₀ = a + 999d – A – 999d
⇒ a – A = 100
⇒ a₁₀₀₀ – A₁₀₀₀ = 100 [From equation (i)]
Therefore, difference of their 1000th term is 100.

Question 13.
How many three-digit numbers are divisible by 7?
Solution:
The three-digit numbers which are divisible by 7 are 105, 112, 119, ………., 994
Here, a = 105, d = 112 – 105 = 7 , a_n = 994
a + (n – 1) d = 994
⇒ 105 + (n – 1) 7 = 994
⇒ (n – 1) 7 = 994 – 105
⇒ 7 (n – 1) = 889
⇒ n – 1 = 127
⇒ n = 127 + 1 = 128
Therefore, the number of three digit numbers which are divisible by 7 is 128.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The multiples of 4 between 10 and 250 be 12, 16, 20, 24,…., 248
Here, a = 12, d = 16 – 12 = 4, a_n = 248
a_n = a + (n – 1) d
⇒ 248 = 12 + (n – 1) 4
⇒ 248 – 12 = (n – 1) 4
⇒ 236 = (n – 1) 4
⇒ 59 = n – 1
⇒ n = 59 + 1 = 60
Therefore, number of terms between 10 and 250 which is multiple of 4 is 60.

Question 15.
For what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?
Solution:
First AP
63, 65, 67,…
Here, a = 63, d = 65 – 63 = 2
a_n = a + (n – 1) d = 63 + (n – 1) 2 = 63 + 2n – 2 = 61 + 2n
Second AP
3, 10, 17, …
Here, a = 3, d = 10 – 3 = 7
a_n = a + (n – 1) d = 3 + (n – 1)7 = 3 + 7n – 7 = 7n – 4
Now, a_n = a_n
⇒ 61 + 2n = 7n – 4
⇒ 61 + 4 = 7n – 2n
⇒ 65 = 5n
⇒ n = 13
Therefore, the 13th term of both APs are same.

Question 16.
Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
Solution:
Given: a₃ = 16
⇒ a + (3 – 1)d = 16
⇒ a + 2d = 16
and a₇ – a₅ = 12
⇒ [a + (7 – 1 )d] – [a + (5 – 1 )d] = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
⇒ d = 6
Since a + 2d = 16
⇒ a + 2(6) = 16
⇒ a + 12 = 16
⇒ a = 16 – 12 = 4
a₁ = a = 4
a₂ = a₁ + d = a + d = 4 + 6 = 10
a₃ = a₂ + d = 10 + 6 = 16
a₄ = a₃ + d = 16 + 6 = 22
Thus, the required AP is a₁, a₂, a₃, a₄,…,
Therefore the required AP is 4, 10, 16, 22.

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
Given: AP is 3, 8, 13,…….. ,253
On reversing the given A.P., we have
253, 248, 243 ,………, 13, 8, 3.
Here, a = 253, d = 248 – 253 = -5
a₂₀ = a + (20 – 1)d = a + 19d = 253 + 19 (-5) = 253 – 95 = 158

Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
We have,
or a₄ + a₈ – 24
a + 3d + a + 7d = 24
or 2a + 10d = 24 … (i)
and a₆ + a₁₀
or a + 5d + a + 9d = 44
or 2a + 14d = 44 … (ii)
Subtracting equation (ii) from equation (i), we
2a + 10d – (2a + 14d) = 24 – 44
or 2a + 10d – 2a – 14d = – 20
or 10d – 14d = – 20
or – 4d = – 20
∴ d = \(\frac { -20 }{ -4 }\)
Putting the value of d in equation (i), we get
2a + 10d = 24
or 2a + 10 x 5 = 24
or 2a + 50 = 24
or 2a = 24 – 50
or 2 a = – 26
∴ a = \(\frac { -26 }{ 2 }\) = – 13
Therefore, the sequence can be,
– 13, – 13 + 5, – 13 + 2 x 5,… or -13, -8, -3,…
∴ The first three terms of this AP are -13, -8, and-3

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000 ?
Solution:
We have,
Starting salary of Subba Rao is ₹ 5000 and annual increment = ₹ 500
Therefore, the sequence of Subba Rao’s salary is 5000, 5200, 5400,…
We have, a + (n – 1) d = 7000
⇒ 5000 + (n – 1) 200 = 7000
⇒ (n – 1) 200 = 7000 – 5000
⇒ (n – 1) 200 = 2000
⇒ (n- 1) = 10
⇒ n = 11
⇒ 1995 + 11 = 2006
Therefore, after 11 years from 1995, salary of Subba Roa each ₹ 7000.
∴ 19995 + 11 = 2006
So, in year 2006, salary of Subba Rao will reach will ₹ 7000.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving become ₹ 20.75, find n.
Solution:
Ramkali savings in the first week of the year be ₹ 5.
and weekly saving increased by ₹ 1.75
Therefore, the sequence of weekly savings be 5, 6.75,8.50,…
Here, a = 5 and d = 6.75 – 5 = 1.75
Let n^th term of this sequence be 20.75
a_n = 20.75
a + (n – 1) d – 20.75
⇒ 5 + (n – 1) 1.75 = 20.75
⇒ (n – 1) x 1.75 = 20.75 – 5
⇒ (n – 1) 1.75 = 15.75
⇒ n – 1 = 9
⇒ n = 9 + 1
⇒ n = 10
Hence, in 10th week Ramkali’s saving will be ₹ 20.75.

These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4

Question 1.
Which term of the AP: 121, 117. 113, ….., is its first negative term?
Solution:
Given sequence is 121,117,113,…
It is an AP. with a = 121, d = 117 – 121 = – 4
Let nth term of this sequence be its first negative term.
So, a < 0
i.e., a + (n – 1 )d < 0
121 + (n – 1) – 4 < 0
121 – 4n + 4 < 0
125 < 4n ⇒ 4n > 125
⇒ n > \(\frac { 125 }{ 4 }\) = 31.25
Hence, 32^th term of the given sequence is its first negative term.

Question 2.
The sum of the third term and the seventh term of an AP is 6 and their product is 8. Fibd the sum of first sixteen terms of the AP?
Solution:
Third term of AP is a + (3 – 1 )d = a + 2d
Seventh term of AP is a + (7 – 1)d = a + 6d
We have,
Sum of third and seventh terms of AP
(a + 2d) + (a + 6 d) = 6
2a + 8d= 6
a + 4d= 3
⇒ a = 3 – 4d
Also given,
Product (a + 2d) (a + 6d)= 6
a² + 6ad + 2ad + 12d² = 8
a² + 8ad + 12d² = 8
Now put the value of a in equation, we get,
(3 – 4d)² + 8(3 – 4d) + 12d² = 8
⇒ a + 16d² – 24d + 24d – 32d² + 12d² = 8
a – 4d² = 8
⇒ 4d² = 1
⇒ d² = \(\frac { 1 }{ 4 }\)
So, d = ±\(\frac { 1 }{ 2 }\)
Taking d = ±\(\frac { 1 }{ 2 }\), Put the value of d in

Question 3.
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 at the top. If the top and the bottom rungs are \({ 2 }\frac { 1 }{ 2 } m\) apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = \(\frac { 250 }{ 25 }\)]

Solution:
Length of the rungs of a ladder decreases uniformly from bottom to top.
Distance between bottom to top
= 2\(\frac { 1 }{ 2 }\)m
= \(\frac { 5 }{ 2 }\) x 100cm = 250cm
Hence, total no. of rungs = \(\frac { 250 }{ 25 }\) = 10
Series is 45,45 + d, 45 + 2d,… 45 + 9d
First term = a = 45, last term = 45 + 9d
According to question,
45 + 9d = 25
45 – 25 = 9d
20 = 9d
d = \(\frac { -20 }{ 9 }\)
Therefore, length of the wood required for rungs is,

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
Solution:
The series of houses will be 1, 2, 3, 4, …… 49
S_x-1 = S₄₉ – S_x
Let x be the number
According to the given condition

Now, put the values of S_x-1, S₄₉ and (S_x-1 = S₄₉ – S_x)

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50  m long and built of solid concrete. Each step has a rise of \(\frac { 1 }{ 2 } m\) and a tread of \(\frac { 1 }{ 2 } m\). Calculate the total volume of concrete required to build the terrace.
Solution:
Volume of concrete required to build the first step.
= \(\frac { 1 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³
Volume of concrete required to build the second step
= \(\frac { 2 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³
Similarly, third step = \(\frac { 3 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³
and for 15th = \(\frac { 15 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³
So, total volume of concrete required to build 15 steps

Hence, the total volume of concrete required to build the terrace is 750 m³.

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2

Question 1.
In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

Question 2.
E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution:
We have to show that EF || QR.
If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
(i) PE = 3.9cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Ratio are not same, so EF is not parallel to QR.

(ii) PE = 4cm, QE = 4.5cm, PF = 8cm and RF = 9cm

Therefore, the ratio are same, so EF || QR.

(iii) PQ = 1.28cm, PR = 2.56cm, PE = 0.18cm and PF = 0.36cm

Ratio are same, so EF || QR.

Question 3.
In the given figure, if LM || CB and LN || CD, Prove that \(\frac { AM }{ AB }\) = \(\frac { AN }{ AD }\)

Solution:
Given a quadrilateral ABCD, in which LM || BC and LN || CD
In ∆ABC, LM || BC
\(\frac { AM }{ MB }\) = \(\frac { AL }{ LC }\) … (i)
[Basic proportionality theorem in fig.]
In ∆ADC, LN || CD
\(\frac { AN }{ AD }\) = \(\frac { AL }{ LC }\) … (2)
[Basis proportionality theorem]
From equations (1) and (2), we have

Question 4.
In the adjoining figure DE || AC and DF || AE. Prove that \(\frac { BF }{ FE }\) = \(\frac { BE }{ EC }\)

Solution:
Given a quadrilateral ABC, in which DF || AE and DE || AC
In ∆ABE, we have
\(\frac { BD }{ AD }\) = \(\frac { BE }{ FE }\) … (i)
[Basic proportionality theorem in fig.]
In ∆ABC, we have
\(\frac { BD }{ AD }\) = \(\frac { BE }{ EC }\) … (2)
[Basis proportionality theorem]
From equations (1) and (2), we have
\(\frac { BE }{ FE }\) = \(\frac { BE }{ EC }\)

Question 5.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.
Solution:
In the triangle PQD, ED || OQ.
So applying Thales theorem, we get,

Equation (iii) shows that EF intersects PQ and PR at E and F respectively in equal ratio. So, according to the converse of the basic proportionality theorem (or Thales theorem), EF || QR. Flence Proved.

Question 6.
In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Solution:
Given : O is any point inside ∆PQR. A is any point PO and B is any point on OQ and C is | any point on OR. AB || PQ and AC || PR.

To Prove: BC || QR
Proof: In ∆POQ, AB || PQ
∴ \(\frac { OA }{ AP }\) = \(\frac { OB }{ BQ }\) …(1) [Basic proportionality theorem]
In ∆POR, AC || PR
∴\(\frac { OA }{ AP }\) = \(\frac { OC }{ OR }\) …(2) [Basic proportionality theorem]
From equations (1) and (2)
\(\frac { OB }{ BQ }\) = \(\frac { OC }{ OR }\)
But, we know that
In ∆OQR,if \(\frac { OB }{ BQ }\) = \(\frac { OC }{ OR }\)
then we can say BC || QR.
[Converse of basic proportionality theorem]
Hence Proved.

Question 7.
Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
Given: A ∆ABC in which D is the mid-point of AB and DE || BC
To Prove : E is the mid-point of AC.

Proof : We have to prove that E is the mid-point of AC. Suppose, let E be not the mid-point of AC. Let E’ be the mid-point of AC. Join DE’.
Now. in ∆ABC, D is the mid-point of AB and E’ is the mid-point of AC. Therefore, by Theorem (6.1).
DE || BC … (1)
Also DE || BC … (2)
From (1) and (2), we find that two intersecting lines DE and DE’ are both parallel to line BC. This is a contradiction to the parallel line axiom.
So, our assumption is wrong. Hence E is the mid-point of AC.

Question 8.
Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
Given: A ΔABC in which D and E are mid-points of sides AB and AC respectively.
To Prove: DE || BC
Construction : Produce the line segement DE to F, such that DE = EF. Join FC.

Proof: In ΔAED and CEF, we have
AE = CE
[∵ E is the mid-point of AC]
∠AED = ∠CEF
[Vertically opposite angle]
or DE = EF
So by SAS congruence
ΔAED ≅ ΔCEF
⇒ AD = CF … (1)
[corresponding of congruent triangle]
and ∠ADE = ∠CEF … (2)
Now, D is the mid-point of AB,
AD = DB
DB = CF … (3) [From (1)]
Now, DF intersects AD and FC at D and F such that
∠ADE = ∠CFE [From (2)]
i.e., alternate interior angles are equal
∴ AD || FC
DB || CF … (4)
From (3) and (4), DBCF is a quadrilateral such that one pair of sides are equal and parallel.
∴ DBCF is a parallelogram
⇒ DF || BC and DC = BC [opposite side of parallelogram]
But, D, E, F are collinear and DE = EF
∴ DE || BC

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the O. Show that \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\)

Solution:

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) Show that ABCD is a trapezium.
Solution:
Construction: Draw OF || AB, meeting AD in F.
To Prove: Quadrilateral ABCD is a trapezium.
Proof: In ABD, we have OF || AB

Hence Proved.