DIGITAL TEXT BOOK

1. A TEXT BOOK OF
MATHEMATICS
GRADE VIII
MATHEMATICS DIGITAL
TEXT BOOK
CLASS VIII
Shareena. Y
B.Ed. Mathematics
Reg.No. 18014350013

2. CONTENT
Chapter 1
Equal Triangles
1. Sides and Angles
2. One side and two angles
3. Bisector of an Angle
Chapter 2
Equations
1. Addition and Subtraction
2. Multiplication and Division
3. Different Changes
4. Algebraic Method

3. A TEXT BOOK OF
MATHEMATICS
GRADE VIII
MATHEMATICS DIGITAL
TEXT BOOK
CLASS VIII
Shareena. Y
B.Ed. Mathematics
Reg.No. 18014350013

4. CONTENT
Chapter 1
Equal Triangles
1. Sides and Angles
2. One side and two angles
3. Bisector of an Angle
Chapter 2
Equations
1. Addition and Subtraction
2. Multiplication and Division
3. Different Changes
4. Algebraic Method

5. Chapter 1
Equal Triangles
Sides and Angles
We know how to draw a triangle if the lengths of the
sides are given. We can draw triangle with sides 4 centimetres,
5 centimetres, 7 centimetres in various method.
We can draw with the 4 centimetres side as base.
Like this, with the 5 centimetres side as base.
with the 7 centimetres side as base.
Consider the triangles given below.
Since sides are equal, angles must also be equal. That is, each
angle of ABC is equal to some angle of PQR.
∠A is the largest angle in ABC
∠R is the largest angle in PQR
So, ∠A = ∠R
∠C is the smallest angle in ABC.
∠Q is the smallest angle in PQR.
So, ∠C = ∠Q
∠B is the medium sized angle in ABC.
∠P is the medium sized angle in PQR.
So, ∠B=∠P
In other words,
The longest side of ABC is BC, and the angle opposite
this side is ∠A, which is the largest angle.
The longest side of PQR is PQ and the angle opposite
this side is ∠R, which is the largest angle.
If the sides of a triangle are equal to the sides of another
triangle, then the angles of the triangles are also equal.
Equal triangles 1 Equal triangles 2

6. Chapter 1
Equal Triangles
Sides and Angles
We know how to draw a triangle if the lengths of the
sides are given. We can draw triangle with sides 4 centimetres,
5 centimetres, 7 centimetres in various method.
We can draw with the 4 centimetres side as base.
Like this, with the 5 centimetres side as base.
with the 7 centimetres side as base.
Consider the triangles given below.
Since sides are equal, angles must also be equal. That is, each
angle of ABC is equal to some angle of PQR.
∠A is the largest angle in ABC
∠R is the largest angle in PQR
So, ∠A = ∠R
∠C is the smallest angle in ABC.
∠Q is the smallest angle in PQR.
So, ∠C = ∠Q
∠B is the medium sized angle in ABC.
∠P is the medium sized angle in PQR.
So, ∠B=∠P
In other words,
The longest side of ABC is BC, and the angle opposite
this side is ∠A, which is the largest angle.
The longest side of PQR is PQ and the angle opposite
this side is ∠R, which is the largest angle.
If the sides of a triangle are equal to the sides of another
triangle, then the angles of the triangles are also equal.
Equal triangles 1 Equal triangles 2

7. So, ∠A=∠R
The smallest side of ABC is AB, and the angle opposite this
side is ∠C, which is the smallest angle.
The smallest side of  PQR is PR, and the angle
opposite this side is ∠Q, which is the smallest angle.
So, ∠C=∠Q
The medium sized side of ABC is AC, and the angle
opposite this side is ∠B, which is the medium sized angle.
Like this, the medium sized side of PQR is QR, and
the angle opposite this side is ∠P, which is the medium sized
angle.
So, ∠B=∠P
If the sides of a triangle are equal to the sides of an-
other triangle, then the angles opposite to the equal sides of
these triangles are equal.
Draw triangle ABC with sides 4 cnetimetres, 5 centimetres, 6
centimetres.
Draw the same triangle below AB, with right and left flipped.
The sides AC and BC of ABC are equal to the sides BD and
AD of ABD.
The third side of both triangle is AB.
Since the legths of all three sides are equal, the angles must
also be equal.
That is, CAB =DBA, CBA=DAB
CAB and DBA are alternate angles made by the line AB
meeting the pair AC, BD of lines. Since the angles are equal,
the lines AC and BD are parallel.
Similarly, CBA and DAB are alternate angles made by
the line AB meeting the pair BC, AD of lines. Since the angles
are equal, the lines BC and AD are parallel.
That is, ACBD is a parallelogram.
Q : Draw a parallelogram of sides 5 centimetres, 6
centimetres and one diogonal 8 cnetimetres.
Draw a triangle of sides 5 centimetres, 6 centimetres and
8 centimetres. (Take 8 centimetres side as base.)
Draw the same triangle below AB, with right and left flipped.
ACBD is a parallelogram.
Equal triangles 3 Equal triangles 4

8. So, ∠A=∠R
The smallest side of ABC is AB, and the angle opposite this
side is ∠C, which is the smallest angle.
The smallest side of  PQR is PR, and the angle
opposite this side is ∠Q, which is the smallest angle.
So, ∠C=∠Q
The medium sized side of ABC is AC, and the angle
opposite this side is ∠B, which is the medium sized angle.
Like this, the medium sized side of PQR is QR, and
the angle opposite this side is ∠P, which is the medium sized
angle.
So, ∠B=∠P
If the sides of a triangle are equal to the sides of an-
other triangle, then the angles opposite to the equal sides of
these triangles are equal.
Draw triangle ABC with sides 4 cnetimetres, 5 centimetres, 6
centimetres.
Draw the same triangle below AB, with right and left flipped.
The sides AC and BC of ABC are equal to the sides BD and
AD of ABD.
The third side of both triangle is AB.
Since the legths of all three sides are equal, the angles must
also be equal.
That is, CAB =DBA, CBA=DAB
CAB and DBA are alternate angles made by the line AB
meeting the pair AC, BD of lines. Since the angles are equal,
the lines AC and BD are parallel.
Similarly, CBA and DAB are alternate angles made by
the line AB meeting the pair BC, AD of lines. Since the angles
are equal, the lines BC and AD are parallel.
That is, ACBD is a parallelogram.
Q : Draw a parallelogram of sides 5 centimetres, 6
centimetres and one diogonal 8 cnetimetres.
Draw a triangle of sides 5 centimetres, 6 centimetres and
8 centimetres. (Take 8 centimetres side as base.)
Draw the same triangle below AB, with right and left flipped.
ACBD is a parallelogram.
Equal triangles 3 Equal triangles 4

9. One Sides and Two Angle
We can draw triangle in various methods with the length
of one sides 8 centimetres and the angles at its ends are 400
and 600
.
Third angle of all these triangles is 600
.
(Sum of angles in a triangle is 1800
.
Since the sum of all three angles in a triangle is 1800
, if any
two angles of a tringle are equal to any two angles of another
triangle, then the third angles are also equal.
In the figure, ABCD is a parallelogram
Drawing the diagonal AC, we can split it into two triangles.
In both ABC and ADC, one side is AC.
∠CAB and ∠DCA are alternated angles made by the line AC
meeting the pair AB, CD of parallel lines.
So ∠CAB=∠DCA
∠ACB and ∠DAC are alternate angles made by the line AC
meeting the pair AD, BC of parallel lines.
So ∠ACB =∠DAC
Thus in ABC and ADC, the side AC and the angles at its
ends are equal. So, the sides opposite to equal angles are also
equal.
That is, AB=CD, AD=BC.
Equal triangles 5 Equal triangles 6
If one side of a triangle and the angles at its ends are
equal to one side of another triangle and the angles at its
ends, then the third angles are also equal and the sides
opposite equal angles are equal.

10. One Sides and Two Angle
We can draw triangle in various methods with the length
of one sides 8 centimetres and the angles at its ends are 400
and 600
.
Third angle of all these triangles is 600
.
(Sum of angles in a triangle is 1800
.
Since the sum of all three angles in a triangle is 1800
, if any
two angles of a tringle are equal to any two angles of another
triangle, then the third angles are also equal.
In the figure, ABCD is a parallelogram
Drawing the diagonal AC, we can split it into two triangles.
In both ABC and ADC, one side is AC.
∠CAB and ∠DCA are alternated angles made by the line AC
meeting the pair AB, CD of parallel lines.
So ∠CAB=∠DCA
∠ACB and ∠DAC are alternate angles made by the line AC
meeting the pair AD, BC of parallel lines.
So ∠ACB =∠DAC
Thus in ABC and ADC, the side AC and the angles at its
ends are equal. So, the sides opposite to equal angles are also
equal.
That is, AB=CD, AD=BC.
Equal triangles 5 Equal triangles 6
If one side of a triangle and the angles at its ends are
equal to one side of another triangle and the angles at its
ends, then the third angles are also equal and the sides
opposite equal angles are equal.

11. In any parallelogram, opposite sides are equal.
Draw the second diagonal BD and the dioganals
intersecting at P.
The sides AB and CD of APB and CPD are equal. Also
∠CAB=∠DCA
That is, ∠PAB=∠PCD
∠PBA and ∠PDC are alternate angles made by the line BD
meeting the pair AB, CD of parallel lines.
So, ∠PBA =∠PDC
Thus in APB and CPD, the sides AB and CD are equal;
and the angles at their ends are also equal. So the sides oppo-
site equal angles are also equal.
That is, AP=CP, BP=DP
In other words, P is the mid point of both the diagonalsAC and
BD.
In any parallelogram, the point of intersection of the
diagonals is the midpoint of both.
In other words,
In any parallelogram, the diagonals bisect each other.
i. AB=RQ, BC=QP, AC=RP
ii. LM=YZ, MN=XZ, LN=XY
(∠N=1800
-(300
+800
)=700
∠Z=1800
-(700
+300
)=800
)
Q : In the figure, AP and BQ equal and parallel lines are
drawn at the ends of the line AB. The point of intersection
of PQ and AB is marked as M.
i. Are the sides of AMP equal to the sides of BMQ?
Why?
In AMP and BMQ,
∠PAM=∠QBM
(Alternate angles made by the line AB meeting the pair and
BQ of parallel lines.)
∠APM=∠BQM
(Alternate angles made by the line PQ meeting the pair AP and
BQ of parallel lines.)
AP=BQ
If one side of AMP and the angles at its ends are equal to
one side of BMQ and the angles at its ends, then the three
sides of AMP are equal to three sides of BMQ.
Equal triangles 7 Equal triangles 8

12. In any parallelogram, opposite sides are equal.
Draw the second diagonal BD and the dioganals
intersecting at P.
The sides AB and CD of APB and CPD are equal. Also
∠CAB=∠DCA
That is, ∠PAB=∠PCD
∠PBA and ∠PDC are alternate angles made by the line BD
meeting the pair AB, CD of parallel lines.
So, ∠PBA =∠PDC
Thus in APB and CPD, the sides AB and CD are equal;
and the angles at their ends are also equal. So the sides oppo-
site equal angles are also equal.
That is, AP=CP, BP=DP
In other words, P is the mid point of both the diagonalsAC and
BD.
In any parallelogram, the point of intersection of the
diagonals is the midpoint of both.
In other words,
In any parallelogram, the diagonals bisect each other.
i. AB=RQ, BC=QP, AC=RP
ii. LM=YZ, MN=XZ, LN=XY
(∠N=1800
-(300
+800
)=700
∠Z=1800
-(700
+300
)=800
)
Q : In the figure, AP and BQ equal and parallel lines are
drawn at the ends of the line AB. The point of intersection
of PQ and AB is marked as M.
i. Are the sides of AMP equal to the sides of BMQ?
Why?
In AMP and BMQ,
∠PAM=∠QBM
(Alternate angles made by the line AB meeting the pair and
BQ of parallel lines.)
∠APM=∠BQM
(Alternate angles made by the line PQ meeting the pair AP and
BQ of parallel lines.)
AP=BQ
If one side of AMP and the angles at its ends are equal to
one side of BMQ and the angles at its ends, then the three
sides of AMP are equal to three sides of BMQ.
Equal triangles 7 Equal triangles 8

13. ii. What is the special about the position of M on AB?
Since AM=BM, M is the midpoint of AB.
Isosceles Triangles
In this triangle two sies are equal.
If two sides of a triangle are equal, the angles opposite
these sides are also equal.
Eg: Let’s check whether, if two angles of a triangle are
equal, then the sides opposite to the equal angles are equal.
Draw ABC and draw a perpendicular from C to AB.
In APC and BPC,
∠A=∠B, ∠APC=∠BPC=900
Since the sum of all three angles in a triangle is 1800
, if any
two angles of a triangle are equal to any two angles of another
triangle, then the third angles are also equal.
So, ∠ACP=∠BCP
If two angles of a triangle are equal, then the sides
opposite to the equal angles are equal.
A triangle with two sides equal, is called an isosceles triangle.
or
Triangles with two angles equal are also isosceles.
The triangle with all three sides equal, is called an equilateral
trianle.
In ABC, since AC=BC, ∠B=∠A.
Q : One angle of an isosceles triangle is 900
. What are the
other two angles?
Two angles of an isosceles triangle are equal. 900
cannot
be the measure of equal angles. Therefore, sum of the mea-
sures of two equal angles is 900
(i.e., 1800
-900
)
i.e. each angle is equal to 450
(i.e., 900
+2)
Therefore, other angles are 450
, 450
.
Bisector of an Angle
First draw anb isosceles triangle including this angle like
this.
Equal triangles 9 Equal triangles 10

14. ii. What is the special about the position of M on AB?
Since AM=BM, M is the midpoint of AB.
Isosceles Triangles
In this triangle two sies are equal.
If two sides of a triangle are equal, the angles opposite
these sides are also equal.
Eg: Let’s check whether, if two angles of a triangle are
equal, then the sides opposite to the equal angles are equal.
Draw ABC and draw a perpendicular from C to AB.
In APC and BPC,
∠A=∠B, ∠APC=∠BPC=900
Since the sum of all three angles in a triangle is 1800
, if any
two angles of a triangle are equal to any two angles of another
triangle, then the third angles are also equal.
So, ∠ACP=∠BCP
If two angles of a triangle are equal, then the sides
opposite to the equal angles are equal.
A triangle with two sides equal, is called an isosceles triangle.
or
Triangles with two angles equal are also isosceles.
The triangle with all three sides equal, is called an equilateral
trianle.
In ABC, since AC=BC, ∠B=∠A.
Q : One angle of an isosceles triangle is 900
. What are the
other two angles?
Two angles of an isosceles triangle are equal. 900
cannot
be the measure of equal angles. Therefore, sum of the mea-
sures of two equal angles is 900
(i.e., 1800
-900
)
i.e. each angle is equal to 450
(i.e., 900
+2)
Therefore, other angles are 450
, 450
.
Bisector of an Angle
First draw anb isosceles triangle including this angle like
this.
Equal triangles 9 Equal triangles 10

15. Now we need only draw the perpendicular bisector of the side
PQ of PBQ.
Q : Draw an angle of 750
and draw its bisector.
* * * * * * * * * * * * * * * * * * *
Chapter 2
Equations
Addition and Subtraction
Q : “Six more marks and I would’ve got full hundred
marks in the math test”, Rajan was sad. How much
mark did he actually get?
Actual mark for Rajan = 100 - 6 = 94
Q : Gopalan bought a bunch of bananas. 7 of them were
rotten which he threw away. Now there are 46. How
many bananas were there in the bunch?
Number of bananas in the bunch = 46 + 7 = 53
Multiplication and Division
Q : A number multiplied by 12 gives 756. What is the
number?
Number x 12 = 756
Number = 756 ÷ 12 = 63
Q : A number divided by 21 gives 756. What is the
number?
Number ÷ 21 = 756
Number = 756 x 21 = 15876
Equal triangles 11 Equations 12

16. Now we need only draw the perpendicular bisector of the side
PQ of PBQ.
Q : Draw an angle of 750
and draw its bisector.
* * * * * * * * * * * * * * * * * * *
Chapter 2
Equations
Addition and Subtraction
Q : “Six more marks and I would’ve got full hundred
marks in the math test”, Rajan was sad. How much
mark did he actually get?
Actual mark for Rajan = 100 - 6 = 94
Q : Gopalan bought a bunch of bananas. 7 of them were
rotten which he threw away. Now there are 46. How
many bananas were there in the bunch?
Number of bananas in the bunch = 46 + 7 = 53
Multiplication and Division
Q : A number multiplied by 12 gives 756. What is the
number?
Number x 12 = 756
Number = 756 ÷ 12 = 63
Q : A number divided by 21 gives 756. What is the
number?
Number ÷ 21 = 756
Number = 756 x 21 = 15876
Equal triangles 11 Equations 12

17. Different changes
Q : When a number is tripled and then two added, it
became 50. What is the number?
Three times of number = 50 - 2=48
Number = 48 ÷ 3=16
Q : Anita and her friends bought pens. For five pens
bought together, they got a discount of three rupees
and it cost them 32 rupees. Had they bought the pens
separately, how much would each have to spend?
Actual cost of five pen = 32 + 3 = 35 rupees
Cost of one pen when they bought the pens separately
= 35 ÷ 5 = 7 rupees
Algebraic Method
Q : The price of a chair and a table together is 4500
rupees. The price of the table is 1000 rupees more than
that of the chair. What is the price of each?
Let x be the price of a chair
Since price of the table is 1000 rupees more than that of
the chair, the price of table = x + 1000 rupees
If x+(x+1000) = 4500, then find x.
2x+1000 = 4500
That is, 1000 is added to two times of a number, it
becomes 4500.
2x=4500 -1000 = 3500
Then, x = 3500 ÷ 2 = 1750
Price of a chair = 1750 rupees
Price of a table = x + 1000 = 1750 + 1000
= 2750 rupees
Q : i. The sum of three consecutive natural numbers is 36.
What are the numbers?
ii. The sum of three consecutive even numbers is 36.
What are the numbers?
iii. Can the sum of three consecutive odd numbers be
36? Why?
iv. The sun if three consecutive odd numbers is 33.
What are the numbers?
Addition and subtraction
If x + a = b, then x = b - a
If x - a = b, then x = b + a
Multiplication and Division
If ax = b(a ≠ 0, then x =
b
a
If
x
a
= b, then x = ab
Equations 13 Equations 14

18. Different changes
Q : When a number is tripled and then two added, it
became 50. What is the number?
Three times of number = 50 - 2=48
Number = 48 ÷ 3=16
Q : Anita and her friends bought pens. For five pens
bought together, they got a discount of three rupees
and it cost them 32 rupees. Had they bought the pens
separately, how much would each have to spend?
Actual cost of five pen = 32 + 3 = 35 rupees
Cost of one pen when they bought the pens separately
= 35 ÷ 5 = 7 rupees
Algebraic Method
Q : The price of a chair and a table together is 4500
rupees. The price of the table is 1000 rupees more than
that of the chair. What is the price of each?
Let x be the price of a chair
Since price of the table is 1000 rupees more than that of
the chair, the price of table = x + 1000 rupees
If x+(x+1000) = 4500, then find x.
2x+1000 = 4500
That is, 1000 is added to two times of a number, it
becomes 4500.
2x=4500 -1000 = 3500
Then, x = 3500 ÷ 2 = 1750
Price of a chair = 1750 rupees
Price of a table = x + 1000 = 1750 + 1000
= 2750 rupees
Q : i. The sum of three consecutive natural numbers is 36.
What are the numbers?
ii. The sum of three consecutive even numbers is 36.
What are the numbers?
iii. Can the sum of three consecutive odd numbers be
36? Why?
iv. The sun if three consecutive odd numbers is 33.
What are the numbers?
Addition and subtraction
If x + a = b, then x = b - a
If x - a = b, then x = b + a
Multiplication and Division
If ax = b(a ≠ 0, then x =
b
a
If
x
a
= b, then x = ab
Equations 13 Equations 14

19. v. The sum of three consecutive natural numbers is
33. What are the numbers?
i. Let three consecutive natural numbers are x, x+1, x+2
Sum of three consecutive natural numbers = 36
That is, x + x + 1 + x + 2 = 36
3x + 3 = 36
3x = 36 - 3 = 33
x =
33
3
= 11
Numbers = 11,12,13
ii. Let three consecutive even numbers are x, x + 2, x + 4.
Sum of three consecutive even numbers = 36
That is, x + x + 2 + x + 4 = 36
3x + 6 = 36
3x = 36 - 6 = 30
x =
30
3
= 10
Even Numbers = 10,12,14
iii. Sum of three consecutive odd numbers is an odd number.
There fore, sum of three consecutive odd numbers can
not be 36.
iv. Let three consecutive odd numbers are x,x+2, x+4.
Sum of three consecutive odd numbers = 33
That is, x + x + 2 + x + 4 = 33
3x + 6=33
3x = 33 - 6 = 27
x =
27
3
= 9
Odd numbers = 9,11,13
v. Let three consecutive natural numbers are x, x + 1, x + 2
Sum of three consecutive natural numbers = 33
x + x + 1 + x + 2 = 33
3x + 3=33
3x = 33 - 3 = 30
x =30
3
= 10
Natural numbers = 10,11,12
Q : i. In a calendar, a square of four numbers is marked.
The sum of the numbers is 80. What are the
numbers?
ii. A square of nine numbers is marked in a calendar.
The sum of all these numbers is 90. What are the
numbers?
i. Let the numbers in a square of four numbers marked in
the calendar are x, x + 1, x + 7, x + 8
x + x + 1 + x + 7 + x + 8 = 80
4x + 16 = 80
4x = 80 - 16 = 64
Equations 15 Equations 16
x x+1
x+7 x+8

20. v. The sum of three consecutive natural numbers is
33. What are the numbers?
i. Let three consecutive natural numbers are x, x+1, x+2
Sum of three consecutive natural numbers = 36
That is, x + x + 1 + x + 2 = 36
3x + 3 = 36
3x = 36 - 3 = 33
x =
33
3
= 11
Numbers = 11,12,13
ii. Let three consecutive even numbers are x, x + 2, x + 4.
Sum of three consecutive even numbers = 36
That is, x + x + 2 + x + 4 = 36
3x + 6 = 36
3x = 36 - 6 = 30
x =
30
3
= 10
Even Numbers = 10,12,14
iii. Sum of three consecutive odd numbers is an odd number.
There fore, sum of three consecutive odd numbers can
not be 36.
iv. Let three consecutive odd numbers are x,x+2, x+4.
Sum of three consecutive odd numbers = 33
That is, x + x + 2 + x + 4 = 33
3x + 6=33
3x = 33 - 6 = 27
x =
27
3
= 9
Odd numbers = 9,11,13
v. Let three consecutive natural numbers are x, x + 1, x + 2
Sum of three consecutive natural numbers = 33
x + x + 1 + x + 2 = 33
3x + 3=33
3x = 33 - 3 = 30
x =30
3
= 10
Natural numbers = 10,11,12
Q : i. In a calendar, a square of four numbers is marked.
The sum of the numbers is 80. What are the
numbers?
ii. A square of nine numbers is marked in a calendar.
The sum of all these numbers is 90. What are the
numbers?
i. Let the numbers in a square of four numbers marked in
the calendar are x, x + 1, x + 7, x + 8
x + x + 1 + x + 7 + x + 8 = 80
4x + 16 = 80
4x = 80 - 16 = 64
Equations 15 Equations 16
x x+1
x+7 x+8

21. x =
64
4
= 16
Numbers in the calandar = 16,17, 23, 24
ii. Let the numbers in a square of nine numbers marked in
the calandar are x, x + 1, x + 2, x + 7, x + 8, x + 9, x + 14,
x + 15, x+16
x + x + 1 + x + 2 + x + 7 + x + 8 + x + 9 + x + 14 + x + 15 +
x+16 = 90
That is, 9x + 72 = 90
9x = 90 - 72 = 18
x =
18
9
= 2
Numbers in the calandar = 2, 3, 4, 9, 10, 11, 16, 17, 18
Q : A class has the same number of girls and boys. Only
eight boys were absent on a particular day and then
the number of girls was double the number of boys.
What is the number of boys and girls?
Let the number of boys and girls each be x.
If eight boys were absent,
number of boys = x - 8
number of girls = x
Equations 17 Equations 18
Number of girls = 2 x number of boys
∴ x = 2(x - 8)
That is, x = 2x - 16
x - 2x = 2x - 16 - 2x
Subtract 2x from both sides
-x = -16
∴ x = 16
Number of boys = 16
Number of girls = 16
Q : Ajayan is ten years older than Vijayan. Next year,
Ajayan’s age would be double that of Vijayan. What
are their age now?
Let x be the present age of Vijayan. Then present
age of Ajayan = x+10
After one year
x + 11 = 2(x+1)
x + 11 = 2x+2
x - 2x = 2 - 11
-x = -9
x = 9
Age of Vijayan = 9, Age of Ajayan = 19.
x x+1 x+2
x+7 x+8 x+9
x+14 x+15 x+16
Present age Age in next year
Vijayan x x + 1
Ajayan x + 10 x + 11

22. x =
64
4
= 16
Numbers in the calandar = 16,17, 23, 24
ii. Let the numbers in a square of nine numbers marked in
the calandar are x, x + 1, x + 2, x + 7, x + 8, x + 9, x + 14,
x + 15, x+16
x + x + 1 + x + 2 + x + 7 + x + 8 + x + 9 + x + 14 + x + 15 +
x+16 = 90
That is, 9x + 72 = 90
9x = 90 - 72 = 18
x =
18
9
= 2
Numbers in the calandar = 2, 3, 4, 9, 10, 11, 16, 17, 18
Q : A class has the same number of girls and boys. Only
eight boys were absent on a particular day and then
the number of girls was double the number of boys.
What is the number of boys and girls?
Let the number of boys and girls each be x.
If eight boys were absent,
number of boys = x - 8
number of girls = x
Equations 17 Equations 18
Number of girls = 2 x number of boys
∴ x = 2(x - 8)
That is, x = 2x - 16
x - 2x = 2x - 16 - 2x
Subtract 2x from both sides
-x = -16
∴ x = 16
Number of boys = 16
Number of girls = 16
Q : Ajayan is ten years older than Vijayan. Next year,
Ajayan’s age would be double that of Vijayan. What
are their age now?
Let x be the present age of Vijayan. Then present
age of Ajayan = x+10
After one year
x + 11 = 2(x+1)
x + 11 = 2x+2
x - 2x = 2 - 11
-x = -9
x = 9
Age of Vijayan = 9, Age of Ajayan = 19.
x x+1 x+2
x+7 x+8 x+9
x+14 x+15 x+16
Present age Age in next year
Vijayan x x + 1
Ajayan x + 10 x + 11

23. MATHEMATICS
Grade VIII