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Brilliant’s Composite Mathematics Class 7 Solutions Chapter 6 Linear Equations in One Variable
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Brilliant’s Composite Mathematics Class 7 Math Book, Chapter 6, Linear Equations in One Variable. Here students can easily find step by step solutions of all the problems for Linear Equations in One Variable, Exercise 6.1 and 6.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Linear Equations in One Variable Exercise 6.1 Solution
Question no – (1)
Solution :
Given, 4x + 3/5 = 5
or, 20 + 3/5 = 5
= 20x + 3 = 15 – 3
= x 12/20
= 2x = 15
∴ x = 3/5
Question no – (2)
Solution :
Given, 2x + 5 = 11
= x 11 – 5/2
= 6/2
= 3
Question no – (3)
Solution :
Given, x/3 – 5/2 = 6
= 2x – 15/6 = 6
= 2x = 36 + 15
= x = 51/2
Question no – (4)
Solution :
Given, 3x – 7/8 = 2/3
= 3x = 2/3 + 7/8
= 16 + 21/24
= 37/24
= x = 37/24×3
= 37/72
Question no – (5)
Solution :
Given, 0.6x + 4/5 = 0.28x + 1.16
or, 0.6 – 0.28x
= 1.16 – 4/5
= 1.16 – 0.8
= 0.36
or, 0.32x = 0.36
= x = 0.36/0.32
= 9/8
Question no – (6)
Solution :
Given, 0.6x + 0.2x = 0.4x + 8
= 0.8x – 0.4x = 8
= 0.4x = 8
= x = 8/0.4
Question no – (7)
Solution :
Given, 5x – 6x + 2x = 28 + 3x
= 5x-6x+2x-3x = 28
= -x –x = 28
= – 2x = 28
= x = -28/2
= – 14
Question no – (8)
Solution :
Given, 3x – 2 [2x + 3] = 2 [x + 3]
Or, 6x2 + 9x – 4x – 6
= 2x + 6
Or, 6x + 5x – 2x
= 6 + 6 = 36
Or, 6x2 + 3x = 36
Or, 6x2 + 3x – 36 = 0
Or, (2x2 + x – 12) = 0
Or, 2x2 + x – 12 = 0
Question no – (9)
Solution :
Given, x – 30% of x = 35
Or, x – 30/100 × x = 35
Or, x (1 – 3/10) = 35
Or, (10 – 3/10) = 35
Or, x (7/10) = 35
Or, x = 35 × 10/7 = 50
Question no – (10)
Solution :
Given, 1/2 [x + 5] – 1/3 [x – 2] = 4
Or, x/2 + 5/2 – x/3 + 2/3 = 4
Or, x/2 – x/3 = 4 – 5/2 – 2/3
= 24 – 15 – 4/6
= 24 – 19/6
Or, 3x – 2x/6 = 5/6
Or, x = 5
Question no – (11)
Solution :
Given, 5x – 3 = 2x + 9
Or, 5x – 2x
= 9 + 3 = 12
= 3x = 12
= x = 12/5
= 4
Question no – (12)
Solution :
Given, x/2 + x/3 = 3 + 3x/4
x/2 + x/3 = 3 + 3x/4
Or, x/2 + x/3 – 3x/4 = 3
Or, x (1/2 + 1/3. – 3/4) = 3
Or, x (6 + 2 -9/12) = 3
Or, x (- 1/12) = 3
= x = – (3 × 12)
= – 36
Question no – (13)
Solution :
Given, 2[x-2]/3 + 3x+1/4 = x + 3/2 + 2
Or, 2x-4/3 + 3x+1/4 – (x + 3/2) = 2
Or, 2x/3 – 4/3 + 3x/4 + 1/4 – x/2 – 3/2 = 2
Or, 2x/3 + 3x/4 – x/2
= 2 + 4/3 –1/4 + 3/2
Or, 78x + 9x – 6x/12
= 24 + 16 – 13 + 18/12
Or, 11x = 55
Or, x = 55/11
= 5
Question no – (14)
Solution :
Given, y – 1.32= 0.09y – 1.19
or, y + 0.09y = 1.32 – 1.19
or, 1.09y = 0.13
or, y = 0.13/1.09
= 0.119
Question no – (15)
Solution :
Given, 4[3p + 2] + 6[7 p-4] – 4p = 2 (p – 8) + 5 (6p – 1)
or, 12p + 8 + 42p – 24 – 4p
= 2p – 16 + 30p – 5
or, 12p + 42p – 4p – 2p – 3op
= 24 – 8 – 16 – 5
or, 18p = – 5
or, p = -5/18
Question no – (16)
Solution :
Given, a – a – 1/2 = 1 – a -2/3
or, a – a/2 + 1/2 = 1 – a/3 + 2/3
or, a – a/2 + a/3 = 1 + 2/3 – 1/2
or, a (1 – 1/2 + 1/3) = (1 + 2/3 – 1/2)
or, a (6 – 3 + 2/6) = (6 + 4 – 3/6)
or, 5a = 7
or, a = 7/5
Question no – (17)
Solution :
Given, 3a-2/3 + 2a+3/2 = a +7/6
or, 6a – 4 + 6a + 9/6 = 6a + 7/6
or, 12a + 5 = 6a + 7
or, 12a – 6a = 7 – 5 = 2
or, 6a = 2
or, a = 2/6
= 1/3
Question no – (18)
Solution :
Given, x – [2x – 3x – 4/7] = 4x-27/3 – 3
or, x – [14x – 3x+4/7] = 4x – 27 – 90/ 3
= 4x – 36/3
or, x – [11x + 4/7] = 4x – 36/3
or, 7x – 11x – 4/7 = 4x – 48/3
or, – 4x – 4/7 = 4x – 48/3
or, – 12x – 12 = 28x – 336
or, 28x + 12x = 336 – 12 = 324
or, 40x = 324
or, x = 324/40
Question no – (19)
Solution :
Given, 2p – 1/3 – 6p – 2/5 = 1/3
or, 10p – 5 – 18p + 6/15 = 1/3
or, -8p + 1/5 = 1
or, -8p + 1 = 5
or, – 8p = 5 – 1 = 4
or, p = – 4/8 = – 1/2
Question no – (20)
Solution :
Given, 3x – 4/12 + [11 – x/3 – 1/4] = x + 2/6
3x – 4/12 + [44 – 4x – 3/12] = x + 2/6
or, 3x – 4/12 + 41 – 4x/12 = x + 2/6
or, 3x – 4 + 41 – 4x/12 = x + 2/6
or, -x + 37/2 = x + 2 or, -x + 37 = 2x + 4
= 2x + x = 37 – 4 = 33
or, 3x = 33
or, x = 33/3
= 11
Linear Equations in One Variable Exercise 6.2 Solution
Question no – (1)
Solution :
Let, 1st consecutive number = x
2nd According to questions –
x + (x + 1) = 53
or, 2x + 1 = 53
or, 2x = 53 – 1 = 52
or, x = 52/2 = 26
1st number = 26
2nd number = 26 + 1
= 27
Therefore, two consecutive natural numbers are 26 and 27
Question no – (2)
Solution :
Let, The number = x
Now, according to the questions,
x + 13 = 25
or, x = 25 – 13
= x = 12
Therefore, the number will be 12.
Question no – (3)
Solution :
Let, the number = x
Now, as per the given question,
5 + 7x = 68
or, 7x = 68 – 5 = 63
or, x = 63/7
= x = 9
Therefore, the required number will be 9
Question no – (4)
Solution :
Let, the number = x
Now, according to the questions,
13 – x = 24
or, -x = 24 – 13 = 11
or, x = -11
Hence, the number will be -11
Question no – (5)
Solution :
Let, are number = x
Another number = y
= x + y = 45 – (i)
= x – y = 9 – (ii)
Now, subtracting (i) + (ii)
x + y = 45
x – y = 9
——————————
2x = 54
or, x = 54/2
= 27
∴ x = 27, y = 18
∴ y = 45 – x
= 45 – 27
= 18
Question no – (6)
Solution :
Let, the number = x
Now, according to the questions
3x + 3 = 7
or, 3x = 7 – 3 = 4
or, x = 4/3
Thus, the number will be 4/3
Question no – (7)
Solution :
Let, odd integers = (2x + 1), (2x + 3), (2x + 5)
Now, as per the given question,
= (2x + 1) + (2x + 3) + (2x + 5) = 87
or, 6x + 9 = 87
or, 6x = 87 – 9 = 87
or, 6x = 87 – 9 = 78
or, x = 78/6 = 13
∴ Odd integers are,
= (2 × 13+ 1) = 27
= (2 × 13 + 3) = 29
= (2 × 13 + 5) = 31
Question no – (8)
Solution :
Let, 1st integers are = x
another integers are = (x + 1), (x + 2), (x + 3), (x + 4)
As per the given question,
= x + (x + 1) + (x + 2) + (x + 3) = 266
or, 4x + 6 = 266
or, 4x = 266 – 6 = 260
or, x = 260/4 = 65
∴ Consecutive integers are,
= 65, (65 + 1) = 66
= (65 + 2) = 67
= (65 + 3) = 68
Question no – (9)
Solution :
Let, 3rd side of triangle = x
Another two side of triangle = (3x – 4)
According to questions – x + (3x – 4) + (3x – 4) = 55
or, x + 3x – 4 + 3x – 4 = 55
or, 7x – 80 = 55
or, 7x = 55 + 8 = 63
or, x = 63/7
One side = 9 cm
Another two sides = (9 × 3 – 4)
= 27 – 4
= 23 cm 4
Question no – (10)
Solution :
Let, no of boys = x
= x of girls = 2x/5
Now, according to the question,
x + 2/5 = x = 35
or, 5x + 2x/5 = 35
or, 7x = 35 × 5
or, x = 25 × 5/7 = 25
∴ Number of boys = 25
Number of girls
= 2/5 × 25
= 10
Therefore, the number of boys in the class is 25 and girls 10.
Question no – (11)
Solution :
Let, the number = x
= x/3 – 1/4 (x + 1) = 1
or, x/3 – (x + 1/4) = 1
or, 4x – 3x – 13/12 = 1
or, x – 3 = 12
or, x = 12 + 3
= 15
Therefore, the numbers are 12 and 15
Question no – (12)
Solution :
Let, the numerator = x
Denominator = x + 7
According to questions,
= x/x + 7 = x + 2/(x + 7) + 9
or, x/x + 7 = x + 2/x + 16
or, x2 + 16x = x2 + 7x + 2x + 14 = 9x + 14
or, 16x – 9x = 14
or, x = 14/7
= 2
Number = 2/2 + 7
= 2/9
Therefore, the number will be 2/9
Question no – (13)
Solution :
Let, the number = x
According to questions –
= 6 (x – 1/2) = 21
or, 6x – 6/2 = 21
or, 3 (2x – 1) = 21
or, (2x – 1) = 21/3 = 7
or, x = 8/2
= x = 4
Therefore, the number will be 4
Question no – (14)
Solution :
Let, numerator = x
Denominator = x + 3
According to questions,
= x/x + 3 = x + 8/(x + 3) + 14
or, x/x+3 = x + 8/x + 17
or, x2 + 17x = x2 + 8x + 3x + 24 = 11x + 24
or, 17x – 11x = 24
or, 6x = 24
or, x = 24/6
∴ Numerator = 4
∴ Denominator = 4 + 3 = 7
Question no – (15)
Solution :
Let, numerator = x
denominator = y
According to questions,
= x – 3 = y
or, x – y = 3 – (i)
According to questions,
= y + x/y – x = 11/3 – (ii)
or, 3y + 3x = 11y – 11x
or, 11y – 3y = 3x + 11x = 14x
or, 8y = 14x
or, x/y
Therefore, the number will be 4/7
Question no – (16)
Solution :
Let, the numerator = x
Denominator = y
According to questions,
x + 4 = y
According to questions,
= x +1/y + 1 =1/2
or, 2x + 2 = y + 1
or, 2x + 2 = (x + 4) + 1
or, 2x + 2 = x + 4 + 1 = x + 5
or, 2x – x = 5 – 2 = 3
or, x = 3
∴ y = 3 + 4 = 7
Therefore, the number will be 3/7
Question no – (17)
Solution :
Let, breadth = b
length = l = 3b
According to questions
= 2(3b + b) = 200
or, 2 × 4b = 200
or, b = 200/2 ×4 = 25
∴ breadth = 25 m
∴ length = 25 × 3
= 75 m
Hence, the dimensions of the garden will be 75 m.
Question no – (18)
Solution :
Let, Breadth = b
Length = l = 2b
According to questions,
= 2 (b + 2b) = 450
or, 2 × 3b = 450
or, b = 450/2 × 3 = 75
∴ b = 75 cm
∴ l = 2 × 75 = 150 cm
Therefore, the dimension of the plot will be length 75 cm and breadth 150 cm.
Question no – (19)
Solution :
Let, breadth = b
Length = l = (3b + 6)
According to questions,
= 2 [(b + (3b + 6)] = 148
or, 2 (4b + 6) = 148
or, 8b + 12 = 148
or, 8b = 148 – 12 = 136
or, b = 136/8 = 17 m
∴ l = (3 × 17 + 6)
= 51 + 6
= 57 m
Therefore, the dimension of the rectangle will be length 57 m and breadth 17 m.
Question no – (20)
Solution :
Let, breadth = b
length = l = b + 5
According to the question,
2 [b + (b + 5)] = 142
or, 2 (2b + 5)] = 142
or, 4b + 10 = 142
or, 4b = 142 – 10 = 132
or, b = 132/4 = 33
∴ b = 33 m
∴ l = 33 + 5
= 38 m
Therefore, the dimension of the plot are length 38 m and breadth 33 m.
Question no – (21)
Solution :
Let, one digit = x
Another digit = y
The number = 10x + y
The reversed = 10y + x
x + y = 9
According to questions,
= (10x + y) + 27 = 10y + x
or, (10x + 4) – (10y + x) = – 27
or, 10x + y – 10y – x = – 27
= 9x – 9y = – 27
= 9 (x – y) = – 27
or, x – y = 27/9 = – 3
or, x – (9 – x) = – 3
or, x – 9 + x = – 3
or, 2x = 9 – 3 = 6
or, x = 6/2 = 3
Question no – (22)
Solution :
Let, one digit = x
another digit = y
The number 10x + y
the reversed = 10y + x
According to questions –
= (10x + y) + 54 = 10y + x
or, 10x + y – 10y – x = – 54
or, 9x – 9y = – 54
or, 9 (x – y) = – 54
or, x – 8 + x = – 54/9 = – 6
or, 2x = 8 – 6 = 2
or, 2/2 = 1
∴ y = 8 – 1 = 7
∴ The number = 10 × 1 + 7 = 17
Question no – (23)
Solution :
Let, one angle = x°
another angle = (90 – x)
According to questions,
= x – (90 – x) = 18
or, x – 90 + x = 18
or, 2x = 18 + 90 = 108
or, x = 108/2 = 54°
∴ 2 angles are = (i) 54° and (ii) (90 – 54°) = 36°
Question no – (24)
Solution :
Let, one angle = x°
supplementary angles = (180 – x)°
According to questions,
= x – (180 – x) 36
or, x – 180 + x = 36
or, x – 180 + x = 36
or, 2x = 36 + 180 = 216
or, x = 216/2 = 108°
Angles are = 108°
= (180 – 108)
= 72°
Question no – (25)
Solution :
Let, the common factor = x
According to question,
2x + 3x + 4x = 180°
or, x = 180/9 = 20°
∴ The 3 angles are
= (20 × 2) = 40°
= (20 × 3) = 60°
= (20 × 4) = 80°
Therefore, the angles are 40°, 60° and 80°
Question no – (26)
Solution :
Let, Kamal age = x
Pankaj age = x + 3
6 years ago
Kamal age – x – 6
Pankaj age = 4 (x – 6)
According to question,
= 4(x – 6) – (x – 6) = (x + 3) – x
or, 4x – 24 – + 6 = x + 3 – x
or, 3x – 18 + 3 = 21
or, x = 21/3 = 7
∴ Kamal age = 7
∴ Pankaj age = 7 + 3 = 10
Question no – (27)
Solution :
Let, the father’s age = x
Son’s age = y
According to questions,
x = y + 30
or, x – y = 30 – (i)
After 12 years father’s age = x + 12
son’s age = y + 12
According to questions,
= (x + 12) = 3 (y + 12)
or, x + 12 = 3y + 36
or, x – 3y = 36 – 12 = 24
or, x – 3y = 24
x – y = 30
x – 3y = 24
———————————————–
2y = 6
or, y = 6/2 = 3
∴ The father’s age = 3 years
∴ Son’s age = 30 + 3 = 33 years
Question no – (28)
Solution :
5 years ago,
son’s age = x
Father’s age = 7x
Present age –
Son’s age = x + 5
Father’s age = 7x + 5
After 5 years,
Son’s age = x + 5 + 5 = x + 10
Father’s age = 7x + 5 + 5 = 7x + 10
According to questions,
= 7x + 10 3 (x + 10)
or, 7x + 10 = 3x + 30
or, 7x – 3x = 30 – 10 = 20
or, 4x = 20
or, x = 20/4 = 5
Present age of son = 5 + 5 = 10
∴ Present age of Father’s
= 7 × 5 + 5
= 40 years
Question no – (29)
Solution :
Let, Mamta’s age = x
After 12 years Mamta’s age = x + 2
According to questions,
x + 12 = 3x – 12
or, 3x – x = 12 + 12 = 24
or, 2x = 24
or, x = 24/2
= 12
Question no – (30)
Solution :
Let, present age of son1 = x
present age of son2 = y
present age of fathers = 3 (x + y)
24 years father of son1 = x + 24
24 years father of son2 = y + 24
24 years son fathers = 3 (x + y) + 24
According to questions,
= 3 (x + y) + 24 = (x + 24) + (y + 24)
or, 3x + 3y + 24 = x + 24 + y + 24
or, 3x + x + 3y – y = 24
or, 2x + 2y = 24
or, 2(x + y) = 24
or, x + y = 24/2 = 12
∴ Present age of father = 3 (x + y)
= 3 × 12
= 36 years
Question no – (31)
Solution :
Let, total property = x
Property for boy = x/3
Property for daughter = x/4
rest property for with = x – (x/3 + x/4)
= x – (4x + 3x/12) = x – 7x/12
= 12x – 7x/12 = 5x/12
∴ According to question,
= 5x/12 = 3200
or, x = 32000 × 12/5
= 76800
Therefore, the value of his property will be 76800 Rs.
Question no – (32)
Solution :
Let, no of 500 prizes = x
no of 100 prizes = y
x + y = 200 – (i)
According to questions –
= 500x + 100y = 50000
or, 100 (5x + y) = 50000
or, 5x + y = 500 – (ii)
5x + y = 500
x + y = 200
———————————————–
4x = 300
or, x = 300/4 = 75
∴ x = 75
y = 200 – 75 = 125
Question no – (34)
Solution :
Let, no of 50 paise coins = x
no of 25 paise coins = 2x
Value of 50 coins = 50 x
value of 25 paise coins = 25 × 2x = 50
According to question,
= 50x + 50x = 900
or, 100x = 900
or, x = 9
∴ Number of 50 paise coins = 9
∴ Number of 25 paise = 2 × 9 = 18
∴ Total coins = 9 + 18 = 27
Question no – (35)
Solution :
Let, no of 50 p coins = x
no of 1 paise coins = y
According to questions – x + y = 90 – (i)
again x + y/2 = 50 – (ii)
or, x + 0.50y = 50
∴ (i) and (ii)
x + y = 90
x + 0.50y = 50
———————————————–
0.50y = 40
or, y = 40/0.550 = 80
∴ x = 90 – 90 = 10
∴ x = 10, y = 80
Question no – (36)
Solution :
Quality of lead of 78 kg = 78 × 67/100
Let, lead need to add = x kg
∴ Let of lead = 2613 + x
let of alloy = 78 + x
According to question,
= 26.73 + x/78 + x
= 60/100
= 3/5
or, 234 + 3x = 130.65-5x
or, 5x – 3x = 234 – 130.65
or, 2x = 103.35
or, x = 103.35/2
= 51.67 kg
Question no – (37)
Solution :
Quality of lead of 50 kg = 50 × 60/100 = 300 kg
Let, lead need to added = x kg
∴ Let, of lead = (30 + x)
Let, of lead alloy = (50 + x)
Lead% of new allow = 30 + x/50 + x × 100
According to questions
= 30 + x/50 + x = 70/100
= 3/4
or, 120 + 4x = 150 + 3x
or, 4x – 3x = 150 – 120 = 30
or, x = 30 kg
Question no – (38)
Solution :
Let, no of day labourer work = x
Number of day labourer absent = 30 – x
Amount to days he worked = 100x
Amount to days cut for absence = 20 (30 – x)
= 600 – 20x
According to questions –
= 100x – (600 – 20x) = 2040
or, 100x – 600 + 20x = 2040
or, 120x = 2040 + 600 = 2640
or, x = 2040/120 = 22 days
∴ Day of absence = 30 – 22
= 8 days
Question no – (39)
Solution :
Let, no of goats = x;
Number of hens = y
We know that = x + y = 60 – (i)
∴ Total legs = legs of goats + legs to knows
= 4x + 2y
∴ Again 4x + 20y = 190
or, 2(2x + y) = 190
= 2x + x = 95 = 95 – (ii)
(ii) – (i)
2x + y = 95
x + y = 60
————————————-
x = 35
∴ y = 60 – 35 = 25
∴ Goats = 35
∴ Hens = 25
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