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Exercise 8A
Solve:
1. 8x+3=27+2x
Solution:
Given 8x+3=27+2x
By transposing the above equation we can write as
8x2x=273
6x=24
By cross multiplication
x=24/6=4
2. 5x+7=2x8
Solution:
Given 5x+7=2x8,
By transposing the above equation we can write as
5x2x=78
3x=15
By cross multiplication
x=15/3=5
3. 2z1=14z
Solution:
Given 2z1=14z,
By transposing the above equation we can write as
2z+z=14+1
3z=15
By cross multiplication
z=15/3=5
4. 9x+5=4(x2)+8
Solution:
Given 9x+5=4(x2) + 8,
By transposing the above equation we can write as
9x+5=4x8+8
9x4x=5
By cross multiplication
5x=5
x=5/5=1
5. (7y)/5=y4
Solution:
Given (7y)/5=y4,
By cross multiplying the above equation we can write as
7y=5(y4)
7y=5y20
Again by transposing
7y5y=20
2y=20
y=20/2=10
6. 3x+2/3=2x+1
Solution:
Given 3x+2/3=2x+1
By cross multiplying the above equation we can write as
3x+2=3(2x+1)
3x+2=6x+3
Again by transposing
3x6x=32
3x=1
x=1/3
7. 15(y4)2(y9)+5(y+6)=0
Solution:
Given 15(y4)2(y9) + 5(y+6)=0
Now by rearranging we get
15y602y+18+5y+30=0
18y12=0
By transposing the above equation we can write as
18y=12
Again by cross multiplication
y=12/18
y=6/9=2/3
8. 3(5x7)2(9x11)=4(8x13)17
Solution:
Given 3(5x7)2(9x11) = 4(8x13)17
Now by rearranging we get
15x2118x+22=32x5217
3x+1=32x69
By transposing the above equation we can write as
35x=70
Again by cross multiplication
x=70/35
x=2
9. (x5)/2(x3)/5=1/2
Solution:
Given (x5)/2(x3)/5=1/2
Now by taking L.C.M for 5 and 2 is 10
(5(x5)2(x3))/10=1/2
By cross multiplying the above equation we can write as
(5x252x+6)=10/2
3x19=5
Again by transposing
3x=19+5=24
x=24/3=8
10. (3t2)/4(2t+3)/3=2/3t
Solution:
Given (3t2)/4(2t+3)/32/3=t
Now by taking L.C.M for 4, 3 and 3 is 12
(3(3t2)4(2t+3)4(2))/12=t
By transposing the above equation we can write as
(9t68t128)/12=t
Again by cross multiplication
9t – 6 – 8t – 12 – 8 = – 12t
t – 18 – 8 = 12t
t – 26 = – 12t
Again by transposing
t + 12t = 26
13t = 26
t=26/13
t=2
11. (2x+7)/5(3x+11)/2=(2x+8)/35
Solution:
Given (2x+7)/5(3x+11)/2 = (2x+8)/35
Now by taking L.C.M for 3 and 1 is 3
(2x+7)/5(3x+11)/2 =(2x+815)/3
LCM of 5, 2 and 3 is 30
30[(2x+7)/5] – 30[(3x+11)/2] = 30[(2x7)/3]
6(2x+7) – 15(3x+11) = 10(2x7)
By further calculation
12x+4245x165=20x70
12x45x+42165=20x70
33x123=20x70
By transposing the above equation we can write as
33x20x=12370
Again by transposing
x=53/53
x=1
12. (5x4)/6=4x+1(3x+10)/2
Solution:
Given (5x4)/6=4x+1(3x+10)/2
Now by taking L.C.M for 1 and 2 is 2
5x4/6 = [2(4x+1) 3x10]/2
By further calculation we get
5x4/6=[8x+23x10]/2
(5x4)/6=(5x8)/2
By cross multiplication
2(5x4)=6(5x8)
10x8=30x48
So we get
10x30x=48+8
20x=40
Again by transposing,
x=40/20=2
13. 5x(1/3)(x+1)=6(x+(1/30))
Solution:
Given 5x(1/3) (x+1) = 6(x + (1/30))
Taking L.C.M on both sides, we get
(15x(x+1))/3=6(30x+1)/30
By cross multiplication,
10(14x1) = 6(30x+1)
140x180x=6+10
40x=16
x=2/5
14. 4(2(z4))/2=1/2(2z+5)
Solution:
Given 4(2(z4))/2=1/2(2z+5)
Now by taking L.C.M of 1 and 3 is 3
122(z4)/3 = (2z+5)/2
By cross multiplication we get
2(122z+8) = 3(2z+5)
404z=6z+15
10z=25
z=25/10=5/2
Exercise 8B
1. Two numbers are in the ratio 8:3. If the sum of the numbers is 143, find the numbers.
Solution:
Given two numbers are in the ratio 8:3
So let the numbers be 8x and 3x
According to the question we can write as 8x+3x=143
Which implies 11x=143
Again by transposing x=143/11
Therefore x=13
So the number are 8x=8(13) =104 and 3x=3(3)=39
2. 2/3 of a number is 20 less than the original number. Find the number.
Solution:
Let the original number be x.
According to the question we can write as (2/3)x+20=x
On rearranging x(2/3)x=20
Now taking the L.C.M of 1 and 3 is 3
(3x2x)/3=20
x/3=20
Again by transposing x=6o
So the original number is 60
3. Fourfifths of a number is 10 more than twothirds of the number. Find the number.
Solution:
Let the number required is x
According to the question we can write as
(4/5) x10 = (2/3) x
On rearranging,
(4/5) x(2/3) x = 10
Now taking the L.C.M of 3 and 5 is 15
(12x10x)/15=10
On cross multiplication,
2x=150
x=150/2=75
So the required number is 75
4. Twentyfour is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. Find each part.
Solution:
Let the two parts of the number be x and (24x)
According to the question we can write as 7x+5(24x)=146
7x+1205x=146
2x=146120
2x=26
x=26/2
x=13
And also (24x)=2413=11
So the parts are 13 and 11
5. Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.
Solution:
Let the number be x.
According to the question we can write as
(1/5) x+5 = (1/4) x – 5
On rearranging
(1/5) x â€“ (1/4) x = 55
(1/5) x â€“ (1/4)x =10
By taking L.C.M we get
(4x5x)/20=10
Again by transposing
x= 200
6. Three numbers are in the ratio of 4:5:6. If the sum of the largest and smallest equals the sum of the third and 55, find the numbers.
Solution:
Let the numbers be 4x, 5x and 6x
According to the question we can write as
6x+4x=5x+55
On rearranging we can write as
10x â€“ 5x = 55
5x=55
x=55/5=11
So the numbers are 44, 55 and 66
7. If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.
Solution:
Let the number be x.
According to the question we can write as
10+4x=5x5
On rearranging
4x5x=510
x=15
So x=15
8. Two numbers are such that the ratio between them is 3:5. If each is increased by 10, the ratio between the new numbers so formed is 5:7. Find the original number.
Solution:
Let the numbers be 3x and 5x.
According to the question we can write as
(3x+10) / (5x+10) = (5/7)
On cross multiplying we get
7(3x+10) = 5(5x+10)
21x+70=25x+50
On rearranging or transposing
7050=25x21x
4x=20
x=20/4=5
So the numbers are 15 and 25
9. Find three consecutive odd numbers whose sum is 147.
Solution:
Let the three consecutive numbers be (2x+1), (2x+3) and (2x+5)
According to the question we can write as 2x+1+2x+3+2x+5=147
On simplifying we get 6x+9=147
On rearranging we get
6x=1479
6x=138
x=138/6=23
So the numbers are (2x+1)=47
(2x+3)=49
(2x+5)=51
10. Find the three consecutive even number whose sum is 234.
Solution:
Let the three consecutive numbers be 2x, (2x+2) and (2x+4)
According to the question we can write as 2x+2x+2+2x+4=234
On simplifying we get 6x+6=234
On rearranging we get
6x=2346
6x=228
x=228/6=38
So the numbers are (2x) = 76
(2x+2)=78
(2x+4)=80
11. The sum of the digits of a two digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.
Solution:
Let the digits be x and y, so the number will be = (10x+y), on reversing the digits, the new number will be = (10y+x)
According to the question we can write as x + y=12 and also we can write as 10y+x10xy=54
Which implies 9y9x=54
yx=54/9
yx=6
y=6+x
Now on substituting this in x + y=12 we get
x+6+x=12
2x+6=12
2x=126
x=6/2=3
Now y=6+x=6+3=9
So the number is 39
To check: digit sum=3+9=12
Reversing the digit numbers becomes 93 and 9339=54
Hence verified.
12. The digit in the tens place of a two digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less than the original number. Find the original number.
Solution:
Let the digits be x and y, so the number will be = (10x+y), on reversing the digits, the new number will be = (10y+x)
According to the question we can write as x =3y and also we can write as 10y+x+36=10x+y
Which implies 10yy+36=10xx
9y+36=9x
Now substituting the value of x
9y+36=27y
18y=36
y=2 and x=6
Now y=6+x=6+3=9
So the number is 62
To check: digit sum=6+2=8
Reversing the digit numbers becomes 26 and 26+36=62
Hence verified.
13. The denominator of a rational number is greater than its numerator by 7. If the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2. Find the original number.
Solution:
Let the rational number be x/y
According to the question we can write as y=x+7
Which implies x=y7 and (x+17)/(y6) = 2
Now substituting the value of x in above equation we get
(y7+17)/(y6) = 2
Now by rearranging,
(y7+17)=2(y6)
y+10=2y12
y=22
But x=y7=227=15
So the number is 15/22
Exercise 8C
Select the correct answer in each of the following:
1. If 2x3=x+2, then x=?
(a)1 (b) 3 (c)5 (d) 7
Solution:
(c)5
Explanation:
Given 2x3=x+2
By transposing we can write as 2xx=2+3
Therefore x=5
2. If 5x+(7/2)=(3/2)x14, then x=?
(a)5 (b) 5 (c)6 (d) 6
Solution:
(b) 5
Explanation:
Given 5x+(7/2)=(3/2)x14
On rearranging we get
5x+(3/2)x=(7/2)14
Taking L.C.M we get
(10x3x)/2=(287)/2
7x=35
x=5
3. If z=4/5(z+10), then z=?
(a) 40 (b) 20 (c) 10 (d) 60
Solution:
(a) 40
Explanation:
Given z=4/5(z+10)
By cross multiplying
z(4/5)z=40/5
By taking L.C.M
z/5 =40/5
z=40
4. If 3m=5m(8/5), then m=?
(a) 2/5 (b) 3/5 (c) 4/5 (d) 1/5
Solution: (c) 4/5
Explanation:
By cross multiplying 5m3m=8/5
2m=8/5
m=4/5
5. If (5t3) = 3t5, then t=?
(a) 1 (b) 1 (c) 2 (d) 2
Solution:
(b) 1
Explanation:
By transposition 5t3t=5+3
2t=2
t=1
6. If 2y+(5/3)=(26/3)y, then y=?
(a) 1 (b) 2/3 (c) 6/5 (d) 7/3
Solution:
(d) 7/3
Explanation:
Given 2y+(5/3)=(26/3)y
On rearranging, 2y+y=(265)/3
3y=7
y=7/3
7. If (6x+1)/3+1=(x3)/6, then x=?
(a) 1 (b) 1 (c) 3 (d) 3
Solution:
(b) 1
Explanation:
Given (6x+1)/3+1=(x3)/6,
On rearranging, 12xx=38
x=1
8. If n/23n/4+5n/6=21, then n=?
(a) 30 (b) 42 (c) 36 (d) 28
Solution:
(c) 36
Explanation:
Given n/23n/4+5n/6=21
On rearranging and taking L.C.M,
(6n_9n+10n)/12=21
7n=21(12)
n=36
9. If (x+1)/(2x+3)=3/8, then x=?
(a) 1/4 (b) 1/3 (c) 1/6 (d)1/2
Solution:
(d) 1/2
Explanation:
Given (x+1)/(2x+3)=3/8,
On rearranging, 8(x+1)=3(2x+3)
8x6x=98
2x=1
x=1/2
RS Aggarwal Solutions for Class 8 Maths Chapter 8 Linear Equations
Chapter 8, Linear Equations, contains 3 exercises. RS Aggarwal Solutions given here contains the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.


 Definition of Equation



 Rules for solving linear Equations



 Definition of Transposing



 Applications of Linear Equations

Also, Access RS Aggarwal Solutions for Class 8 Maths Chapter 8 Linear Equations Exercises
Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 8 – Linear Equations
The RS Aggarwal Solutions for Class 8 Maths Chapter 8 â€“ Linear Equation deals with the definition of an equation, rules for solving linear equation, how to do transposing and also about cross multiplication. By using these rules we can find the unknown present in the given equation. It is helpful to solve verbally given problems in terms of equations. It is also helpful in solving problems related to geometry.