Samacheer Class 7 Maths - Algebra Intext Questions

Area of each box = xy
Totally 12 boxes
∴ Total area = 12 × xy = 12xy
Also multiplying the length 4x and breadth 3y
We have area of the rectangle = 4x × 3y = 12xy
(ii) Area of each small box = x 2
Total number of boxes = 3
∴ Total area = 3x 2
Also length of the rectangle = 3x
breadth of the rectangle = x
Area of the rectangle = length × breadth
= 3x × x
= 3x 2
(iii) Area of each small box is ay, by, cy
∴ Total area = ay + by + cy = y (a + b + c)
Area of the rectangle = length × breadth
= (a + b + c) y
(iv) Area of each small square = x 2
There are 4 small squares
∴ Total area of the given square = 4x 2
Also side of the big square = 2x
∴ Area of the square = (2x) 2 = 4x 2
(v) Area of each small rectangle = xy
There are 9 such rectangles
∴ Total area = 9xy
Area of big rectangle = lenght × breath
= 3x × 3y = 9xy

Let a = 52
b = 4(a – b) 2 = a 2 – 2ab + b 2 = 52 2 – 2 (52) (4) + 4 2
= 2704 – 416 + 16 = 2304
Think (Text book Page No. 60)

(i) x 2 + 5x + 4 = x2 + (1 + 4)x + (1 × 4)
Which is of the form x 2 + (a + b) x + ab
= (x + a) (x + b)
x 2 + (1 + 4)x + (1 × 4) = (x + 1) (x + 4)
∴ x 2 + 5x + 4 = (x + 1) (x + 4)
(ii) x 2 – 5x + 4 = x 2 + ((-1) + (- 4))x + (-1) (- 4)
Which is of the form x 2 + (a + b) x + ab
= (x + a) (x + b)
x 2 + ((-1) + 4))x + ((-1)(-4)) = (x + (-4)) = (x – 1) (x – 4)
x 2 – 5x + 4 = (x – 1) (x – 4))Exercise 3.2
Try These (Text book Page No. 63)

1. x > 25,000, where x is Ramesh’s Salary per month.
2. y < 5, where y is the maximum number of persons the left can carry.
3. z > 100, where z is the number of days when the exhibition is there.Think (Text book Page No. 65)

Let x be the age and y be the height then
40 < x < 45 and 160 < y < 170
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(i) 24ab 2 c 2 = 2 × 2 × 2 × 3 × a × b × b × c × c
(ii) 36 x 3 y 2 z = 2 × 2 × 3 × 3 × x × x × x × y × y × z
(iii) 56 mn 2 p 2 = 2 × 2 × 2 × 7 × m × n × n × p × p

(i) (x + 3) (x + 7)
Let a = 3; b = 7, then
(x + 3) (x + 7) is of the form x 2 + x (a + b) + ab
(x + 3) (x + 7) = x 2 + x (3 + 7) + (3 × 7) = x 2 + 10x + 21
(ii) (6a + 9) (6a – 5)
Substituting x = 6a ; a = 9 and b = -5
In (x + a) (x + b) = x 2 + x (a + b) + ab, we get
(6a + 9)(6a – 5) = (6a) 2 + 6a (9 + (-5)) + (9 × (-5))
6 2 a 2 + 6a (4) + (-45) = 36a 2 + 24a – 45
(6a + 9) (6a – 5) = 36a 2 + 24a – 45
(iii) (4x + 3y) (4x + 5y)
Substituting x = 4x ; a = 3y and b = 5y in
(x + a) (x + b) = x 2 + x (a + b) + ab, we get
(4x + 3y) (4x – 5y) = (4x) 2 + 4x (3y + 5y) + (3y) (5y)
= 4 2 x 2 + 4x (8y) + 15y 2 = 16x 2 + 32xy + 15y 2
(4x + 3y) (4x + 5y) = 16x 2 + 32xy + 15y 2
(iv) (8 + pq) (pq + 7)
Substituting x = pq ; a = 8 and b = 7 in
(x + a) (x + b) = x 2 + x (a + b) + ab, we get
(pq + 8) (pq + 7) = (pq) 2 + pq (8 + 7) + (8) (7)
= p 2 q 2 + pq (15) + 56
(8 + pq) (pq + 7) = p 2 q 2 + 15pq + 56

(i) (2x + 5) 2
Comparing (2x + 5) 2 with (a + b) 2 we have a = 2x and b = 5
a = 2x and b = 5,
(a + b) 2 = a 2 + 2ab + b 2
(2x + 5) 2 = (2x) 2 + 2(2x) (5) + 5 2 = 2 2 x 2 + 20x + 25
= 2 2 x 2 + 20x + 25
(2x + 5) 2 = 4x 2 + 20x + 25
(ii) (b – 7) 2
Comparing (b – 7) 2 with (a – b) 2 we have a = b and b = 7
(a – b) 2 = a 2 – 2ab + b 2
(b – 7) 2 = b 2 – 2(b) (7) + 7 2
(b – 7) 2 = b 2 – 14b + 49
(iii) (mn + 3p) 2
Comparing (mn + 3p) 2 with (a + b) 2 we have
(a + b) 2 = a 2 + 2ab + b 2
(mn + 3p) 2 = (mn) 2 + 2(mn) (3p) + (3p) 2
(mn + 3p) 2 = m 2 n 2 + 6mnp + 9p 2
(iv) (xyz – 1) 2
Comparing (xyz – 1) 2 with (a – b) 2 we have = a + xyz and b = 1
a = xyz and b = 1
(a – b) 2 = a 2 – 2ab + b 2
(xyz – 1) 2 = (xyz) 2 – 2 (xyz) (1) + 1 2
(xyz -1) 2 = x 2 y 2 z 2 – 2 xyz + 1

(i) (p + 2) (p – 2)
Substituting a = p ; b = 2 in the identity (a + b) (a – b) = a 2 – b 2 , we get
(p + 2) (p – 2) = p 2 – 2 2
(ii) (1 + 3b)(3b – 1)
(1 + 3b) (3b -1) can be written as (3b + 1) (3b – 1)
Substituting a = 36 and b = 1 in the identity
(a + b) (a – b) = a 2 – b 2 , we get
(3b + 1)(3b – 1) = (3b) 2 – 1 2 = 3 2 × b 2 – 1 2
(3b + 1) (3b – 1) = 9b 2 – 1 2
(iii) (4 – mn) (mn + 4)
(4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn)
Substituting a = 4 and b = mn is
(a + b) (a – b) = a 2 – b 2 , we get
(4 + mn) (4 – mn) = 4 2 – (mn) 2 = 16 – m 2 n 2
(iv) (6x + 7y) (6x – 7y)
Substituting a = 6x and b = 7y in
(a + b) (a – b) = a 2 – b 2 , We get
(6x + 7y) (6x – 7y) = (6x) 2 – (7y) 2 = 6 2 x 2 – 7 2 y 2
(6x + 7y) (6x – 7y) = (6x) 2 – (7y) 2 = 6 2 x 2 – 7 2 y 2
(6x + 7y) (6x – 7y) = 36x 2 – 49y 2

51 2
= (50 + 1) 2
Taking a = 50 and b = 1 we get
(a + b) 2 = a 2 + 2ab + b 2
(50 + 1) 2 = 50 2 + 2 (50) (1) + 1 2 = 2500 + 100 + 1
51 2 = 2601
(ii) 103 2
103 2 = (100 + 3) 2
Taking a = 100 and b = 3
(a + b) 2 = a 2 + 2ab + b 2 becomes
(100 + 3) 2 = 100 2 + 2 (100) (3) + 3 2 = 10000 + 600 + 9
103 2 = 10609
(iii) 998 2
998 2 = (1000 – 2) 2
Taking a = 1000 and b = 2
(a – b) 2 = a 2 + 2ab + b 2 becomes
(1000 – 2) 2 = 1000 2 – 2 (1000) (2) + 2 2
= 1000000 – 4000 + 4
998 2 = 10,04,004
(iv) 47 2
47 2 = (50 – 3) 2
Taking a = 50 and b = 3
(a – b) 2 = a 2 – 2ab + b 2 becomes
(50 – 3) 2 = 50 2 – 2 (50) (3) + 3 2
= 2500 – 300 + 9 = 2200 + 9
47 2 = 2209
(v) 297 × 303
297 × 303 = (300 – 3) (300 + 3)
Taking a = 300 and b = 3, then
(a + b) (a – b) = a 2 – b 2 becomes
(300 + 3) (300 – 3) = 300 2 – 3 2
303 × 297 = 90000 – 9
297 × 303 = 89,991(vi) 990 × 1010
990 × 1010 = (1000 – 10) (1000 + 10)
Taking a = 1000 and b = 10, then
(a – b) (a + b) = a 2 – b 2 becomes
(1000 – 10) (1000 + 10) = 1000 2 – 10 2
990 × 1010 = 1000000 – 100
990 × 1010 = 999900
(vii) 51 × 52
= (50 + 1) (50 + 1)
Taking x = 50, a = 1 and b = 2
then (x + a) (x + b) = x 2 + (a + b) x + ab becomes
(50 + 1) (50 + 2) = 50 2 + (1 + 2) 50 + (1 × 2)
2500 + (3) 50 + 2 = 2500 + 150 + 2
51 × 52 = 2652

We have (a + b) 2 = a 2 + 2ab + b 2
(a + b) 2 = a 2 + b 2 + 2ab
given a + b = 0 and ab = 18
10 2 = = a 2 + b 2 + 2(18)
100 = = a 2 + b 2 + 36
100 – 36 = a 2 + b 2
a 2 + b 2 = 64

(i) z 2 – 16
z 2 – 16 = z 2 – 4 2
We have a 2 – b 2 = (a + b) (a – b)
let a = z and b = 4,
z 2 – 4 2 = (z + 4) (z – 4)
(ii) 9 – 4y 2
9 – 4y 2 = 3 2 – 2 2 y 2 = 3 2 – (2y) 2
let a = 3 and b = 2y, then
a 2 – b 2 = (a + b) (a – b)
∴ 3 2 – (2y) 2 = (3 + 2y) (3 – 2y)
9 – 4y 2 = (3 + 2y) (3 – 2y)
(iii) 25a 2 – 49b 2
25a2 – 49b2 = 52 – a2 – 72 = (5a)2 – (7b)2
let A = 5a and B = 7b
A 2 B 2
(5a) 2 – (7b) 2 = (5a + 7b) (5a – 7b)
(iv) x 4 – y 4
Let x 4 – y 4 = (x 2 ) 2 – (y 2 ) 2
We have a 2 – b 2 = (a + b) (a – b)
(x 2 ) 2 – (y 2 ) 2 = (x 2 + y 2 ) (x 2 – y 2 )
x 4 – y 4 = (x 2 + y 2 ) (x 2 – y 2 )
Again we have x 2 – y 2 = (x + y) (x – y)
∴ x 4 – y 4 = (x 2 + y 2 ) (x + y) (x – y)

(i) (x – 3)(x – 3)
Hint: x 2 – 6x + 9 = x 2 – 2(x) (3) + 3 2
a 2 – 2ab + b 2 – (a- b) 2 = (x – 3) 2 = (x – 3) (x – 3)

(i) x < 7, where x is a natural number.
Since the solution belongs to the set of natural numbers, that are less than or equal to 7, we take the values of x as 1,2, 3, 4, 5, 6 and 7.
(ii) x – 6 < 1, where x is a natural number.
x – 6 < 1 Adding 6 on the both the sides x – 6 + 6 < 1 + 6
x < 7
Since the solutions belongs to the set of natural numbers that are less than 7, we take the values of x as 1,2, 3, 4, 5 and 6
(iii) 2a + 3 < 13, where a is a whole number.
2a + 3 < 13
Subtracting 3 from both the sides 2a + 3 – 3 < 13 – 3
2a < 10
Dividing both the side by 2. \(\frac { 2a }{ 2 } \) < \(\frac { 10 }{ 2 } \)
a < 5
Since the solutions belongs to the set of whole numbers that are less than or equal to 5 we take the values of a as 0, 1, 2, 3, 4 and 5(iv) 6x – 7 > 35, where x is an integer.
6x – 7 > 35 Adding 7 on both the sides
6x – 7 + 7 > 35 + 7
6x > 42
Dividing both the sides by 6 we get \(\frac { 6x }{ 6 } \) > \(\frac { 42 }{ 6 } \)
x > 7
Since the solution belongs to the set of integers that are greater than or equal to 7, we take the values of x as 7, 8, 9, 10…
(v) 4x – 9 > -33, where x is a negative integer.
4x – 9 > – 33 + 9 Adding 9 both the sides
4x – 9 + 9 > -33 + 9
4x > – 24
Dividing both the sides by 4
\(\frac { 4x }{ 4 } \) > \(\frac { -24 }{ 4 } \)
x > -6
Since the solution belongs to a negative integer that are greater than -6, we take values of u as -5, -4, -3, -2 and -1

(i) k > -5, k is an integer.
Since the solution belongs to the set of integers, the solution is -4, -3, -2, -1, 0,… It’s graph on number line is shown below.(ii) -7 < y, y is a negative integer.
-7 < y
Since the solution set belongs to the set of negative integers, the solution is
-7, -6, -5, -4, -3, -2, -1.
Its graph on the number line is shown below(iii) -4 < x < 8, x is a natural number.
-4 < x < 8
Since the solution belongs to the set of natural numbers, the solution is
1, 2, 3, 4, 5, 6, 7 and 8.
Its graph on number line is shown below(iv) 3m – 5 < 2m + 1, m is an integer.
3m – 5 < 2m + 1
Subtracting 1 on both the sides
3m – 5 – 1 < 2m + 1 + 1
3m – 6 < 2m
Subtracting 2m on both the sides 3m- 6 – 2m < 2m -2m
m – 6 < 0
Adding 6 on both the sides m – 6 + 6 < 0 + 6
m < 6
Since the solution belongs to the set of integers, the solution is
6, 5, 4, 3, 2, 1, 0,-1,…
Its graph on number line is shown below

Given the artist can spend any amount between ₹ 80 to ₹ 200
Let the number of packets of brush he can buy be x
Given cost of 1 brush = ₹ 5
Cost of 1 packet brush (6 brushes) = ₹ 5 × 6 = ₹ 30
∴ Cost of x packets of brushes = 30 x
∴ The inequation becomes 80 < 30x < 200
Dividing throughout by 30 we get \(\frac { 80 }{ 30 } \) < \(\frac { 30x }{ 30 } \) < \(\frac { 200 }{ 30 } \)
\(\frac { 8 }{ 3 } \) < x < \(\frac { 20 }{ 3 } \) ;
2 \(\frac { 2 }{ 3 } \) < x < 6 \(\frac { 2 }{ 3 } \)
brush packets cannot get in fractions.
∴ The artist can buy 3 < x < 6 packets of brushes,
or x = 3, 4, 5 and 6 packets of brushes.Objective Type Questions

(i) 1 and 2
Hint: 5x + 5 < 15
5x < 15 – 5 = 10
x < \(\frac { 10 }{ 5 } \) = 2

(iii) 6
Hint:
Price of 1 pen = ₹ 8
Price of 1 pack = 10 × 8 = 80
Number of packs Swetha can buy = x
80x < 500
8x < 50
x < \(\frac { 50 }{ 8 } \) = 6.25
x is a natural number x = 1, 2, 3, 4, 5, 6

(i) (4.9) 2
(4.9) 2 = (5 – 0.1) 2
Substituting a = 5 and b = 0.1 in
(a – b) 2 = a 2 – 2ab + b 2 , we have
(5 – 0.1) 2 = 5 2 – 2(5) (0.1) + (0.1) 2
(4.9) 2 = 25 – 1 + 0.01 = 24 + 0.01
(4.9) 2 = 24.01
(ii) (100.1) 2
(100.1) 2 = (100 + 0.1) 2
Substituting a = 100 and b = 0.1 in
(a + b) 2 = a 2 + 2ab + b 2 , we have
(100 + 0.1) 2 = (100) 2 + 2(100) (0.1) + (0.1) 2
(100.1) 2 = 10000 + 20 + 0.01
(100.1) 2 = 10020.01(iii) (1.9) × (2.1)
(1.9) × (2.1) = (2 – 0.1) × (2 + 0.1)
Substituting a = 100 and b = 0.1 in
(a – b) (a + b) = a 2 – b 2 we have
(2 – 0.1) (2 + 0.1) = 2 2 – (0.1) 2
(1.9) × (2.1) = 4 – 0.01
(9.9) (2.1) = 3.99

(i) (3p + q) (3p + r)
Substitute x = 3p,a = q and b = r in
(x + a) (x + b) = x 2 + x(a + b) + ab
(3p + q)(3p + r) = (3p) 2 + 3p (q + r) + (q × r)
= 3 2 p 2 + 3p (q + r) + qr
(3p + q)(3p + r) = 9p 2 + 3p(q + r) + qr
(ii) (3p + q) (3p – q)
Substitute a = 3p and b = q in
(a + b) (a – b) = a 2 – b 2 , we have
(3p + q) (3p – q) = (3p) 2 – q 2 = 32 p 2 – q 2
(3P + q) (3p – q) = 9p 2 – q 2

Given X = a 2 – 1
Y = I – b 2
X + Y = (a 2 – 1) + (1 – b 2 )
= a 2 – 1 + 1 – b 2
We know the identity that a 2 – b 2 = (a + b) (a – b)
∴ X + Y = (a + b) (a – b)

We know that (a – b) (a + b) = a 2 – b 2
Put a = x and b = y in the identity (1) then
(x – y) (x + y) = x 2 – y 2
Now (x – y) (x + y)(x 2 + y 2 ) = (x 2 – y 2 ) (x 2 + y 2 )
Again put a = x 2 and b = y 2 in (1)
We have (x 2 – y 2 ) (x 2 + y 2 ) = (x 2 ) 2 – (y 2 ) 2 = x 4 – y 4
So (x – y) (x + y) (x 2 + y 2 ) = x 4 – y 4

Let the side of the lawn = a m
then side Of big square = (a + 2(2)) m
= (a + 4)mArea of the path – Area Of large square – Area of smaller square
136 = (a + 4)2 – a2
136 = a 2 + (2 × a × 4) + 4 2 – a 2
136 = a 2 + 8a + 16 – a 2
136 = 8a + 16
136 = 8 (a + 2)
Dividing by 8
17 = a + 2
Subtracting 2 on both sides
17 – 3 = a + 2 – 2
15 = a
∴ side of small square = 15 m
Area of square = (side × side) Sq. units
∴ Area of the lawn = (15 × 15)m 2 = 225 m 2
∴ Area of the lawn = 225 m 2

(i) 4n + 7 > 3n + 10, n is an integer.
4n + 7 – 3n > 3n + 10 – 3n
n(4 – 3) + 7 > 3n + 10 – 3n
n (4 – 3) + 7 > n (3 – 3) + 10
n + 7 > 10
Subtracting 7 on both sides
n + 7 – 7 > 10 – 7
n > 3
Since the solution is an integer and is greater than or equal to 3, the solution will be 3,
4, 5, 6, 7, …..
n = 3, 4, 5, 6,7, ….
(ii) 6 (x + 6) > 5 (x – 3), x is a whole number.
6x + 36 > 5x – 15
Subtracting 5x on both sides
6x + 36 – 5x > 5x – 15 – 5x
x (6 – 5) + 36 > x(5 – 5) – 15
x + 36 > -15
Subtracting 36 on both sides
x + 36 – 36 > -15 -36
x > -51
The solution is a whole number and which is greater than or equal to -51
∴ The solution is 0, 1, 2, 3, 4,…
x = 0,1,2, 3,4,…(iii) -13 < 5x + 2 < 32, x is an integer.
Subtracting throughout by 2
-13 – 2 < 5x + 2 – 2 < 32 – 2
-15 < 5x < 30
Dividing throughout by 5
\(\frac { -15 }{ 5 } \) < \(\frac { 5x }{ 5 } \) < \(\frac { 30 }{ 5 } \)
– 3 < x < 6
∴ Since the solution is an integer between -3 and 6 both inclusive, we have the solution
as -3, -2, -1,0, 1,2, 3, 4, 5, 6.
i.e. x = -3, -2, 0, 1, 2, 3,4, 5 and 6.
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