Algebra Chapter 12 Equations
Equations Introduction
In class VII you have learned about the formation of linear equations of one variable and their solutions. Formation of equations, some useful information about equations, and rules of solving an equation have been discussed in detail in the chapter of ‘equation1 in the book for class VII. In this chapter, our aim is to deal with a little harder problems and also to discuss some special techniques for solving algebraic equations involving one variable.
Class 8 Maths Solutions Wbbse
Method of Transportation
In an equation, usually, there are some terms on the left-hand side of the ‘=’ sign and also there are some terms on the right-hand side of the sign. If any term is taken from the left-hand side to the right-hand side or from the right-hand side to the left-hand side then a change of sign of the respective term takes place. This is called the method of transportation.
For example, if x + a = b then x = b – a, and if x + c = d then x + c – d = 0
Cross multiplication
If x/a = b/c then cx =ab. This process of multiplication of the numerator of one side of an equation by the denominator of the other side is called cross multiplication.
Method Of Alternends
If a/b = c/d then a/c = b/d.
This is called the method of alternates.
Method Of Break Term
In order to solve the equation of the form \(\frac{p}{x+a}+\frac{q}{x+b}=\frac{p+q}{x+c}\) it is convenient to proceed by writing the equation as
\(\frac{p}{x+a}+\frac{q}{x+b}=\frac{p}{x+c}+\frac{q}{x+c}\) Then is known as the method of break term.
For example, to solve the equation, \(\frac{2}{x+1}+\frac{3}{x+2}=\frac{5}{x+3}\)
write the equation as: 2 / x+ 1 + 3 / x + 2 = 2/ x + 3 + 3 / x + 3.
Read And Learn More WBBSE Solutions For Class 8 Maths
Method of division
In some cases, each fraction of an equation is written in such a way that, a part of the numerator becomes divisible by the denominator.
For example, to solve the equation,
\(\frac{4 x+5}{2 x-3}+\frac{3 x+8}{x-5}=\frac{15 x}{3 x-1}\)WBBSE Class 8 Equations Notes
we may proceed as follows:
\(\frac{2(2 x-3)+11}{2 x-3}+\frac{3(x-5)+23}{x-5}=\frac{5(3 x-1)+5}{3 x-1}\)or, \(2+\frac{11}{2 x-3}+3+\frac{23}{x-5}=5+\frac{5}{3 x-1}\)
or, \(\frac{11}{2 x-3}+\frac{23}{x-5}=\frac{5}{3 x-1}\)
Algebra Chapter 12 Equations Some examples of equations
Example 1
Solve the equation : 5(x – 2) +7(x – 3) = 2(x + 3) + 3.
Solution :
Given 5(x – 2) +7(x – 3) = 2(x + 3) + 3.
5(x – 2) + 7(x – 3) = 2(x + 3) + 3
or, 5x – 10 + 7x – 21 = 2x + 6 + 3
or, 12x – 31 = 2x + 9
or, 12x – 2x = 31 + 9
or, 10x = 40
or, x = 40/10
= 4
The required solution is x = 4.
Example 2
Solve: (x – 2)(x – 3) = x2 – 44.
Solution:
Given (x – 2)(x – 3) = x2 – 44
or, x2 – 3x -2x + 6 = x2 – 44
or, x2 – x2 – 5x = -44 – 6
or, -5x = -50
or, x = -50 / -5
= 10
The required solution is x = 10.
Understanding Linear Equations in One Variable
Example 3
Solve: (x -+ 1) + x(x – 3) = 2x2 – 8.
Solution:
Given
(x -+ 1) + x(x – 3) = 2x2 – 8.
x(x + 1) + x(x – 3) = 2x2 – 8
or, x2 + x + x2 – 3x = 2x2 – 8
or, 2x2 – 2x2 – 2x = -8
or, -2x = -8
or, x = -8 / -2
= 4
The required solution is x = 4.
Example 4
Solve: \(\frac{x+1}{5}+x=\frac{2 x+7}{5}+2\)
Solution:
Given
\(\frac{x+1}{5}+x=\frac{2 x+7}{5}+2\).
\(\frac{x+1}{5}+x=\frac{2 x+7}{5}+2\) \(\text { or, } \frac{x+1+5 x}{5}=\frac{2 x+7+10}{5}\) \(\text { or, } 6 x+1=2 x+17 \text { or, } 6 x-2 x=17-1\) \(\text { or, } 4 x=16\) \(\text { or, } x=\frac{16}{4}=4\)Example 5
Solve: \(\frac{2x+1}{7}+4=\frac{5-x}{3}\)
Solution:
Given \(\frac{2x+1}{7}+4=\frac{5-x}{3}\)
\(\frac{2 x+1}{7}+4=\frac{5-x}{3}\)or, \(\frac{2 x+1+28}{7}=\frac{5-x}{3} \text { or, } \frac{2 x+29}{7}=\frac{5-x}{3} \text { or, } 6 x+87=35-7 x \text { or, } 6 x+7 x=35-87\)
or, 13x = -52
or, \(x=\frac{-52}{13}=-4\)
The required solution is x = -4.
Step-by-Step Guide to Solving Equations
Example 6
Solve: \(\frac{x+12}{6}-x=6 \frac{1}{2}-\frac{x}{12} \ldots\)
Solution:
Given \(\frac{x+12}{6}-x=6 \frac{1}{2}-\frac{x}{12} \ldots\).
\(\frac{x+12}{6}-x=6 \frac{1}{2}-\frac{x}{12}\)or, \(\frac{x+12}{6}-x=\frac{13}{2}-\frac{x}{12}\)
or, \(\frac{x+12-6 x}{6}=\frac{78-x}{12}\)
or, \(\frac{12-5 x}{6}=\frac{78-x}{12} \text { or, } 12-5 x=\frac{78-x}{2}\)
or, 78 – x = 24 – 10x or, 10x – x = 24 – 78
or, 9x = -54
or, \(x=\frac{-54}{9}=-6\)
The required solution is x = -6
Example 7
Solve: \(\frac{x-3 a}{b}+\frac{x-3 b}{a}+\frac{x-9 a-9 b}{a+b}=0 .\)
Solution:
Given \(\frac{x-3 a}{b}+\frac{x-3 b}{a}+\frac{x-9 a-9 b}{a+b}=0 .\)
\(\frac{x-3 a}{b}+\frac{x-3 b}{a}+\frac{x-9 a-9 b}{a+b}=0\)or, \(\frac{x-3 a}{b}-3+\frac{x-3 b}{a}-3+\frac{x-9 a-9 b}{a+b}+6=0\)
or, \(\frac{x-3 a-3 b}{b}+\frac{x-3 b-3 a}{a}\)
\(+\frac{x-9 a-9 b+6 a+6 b}{a+b}=0\)or, \(\frac{x-3 a-3 b}{b}+\frac{x-3 a-3 b}{a}+\frac{x-3 a-3 b}{a+b}=0\)
or, \((x-3 a-3 b)\left(\frac{1}{b}+\frac{1}{a}+\frac{1}{a+b}\right)=0 \text {. }\)
Now, the left-hand side is the product of two expressions. Its value will be equal to zero if at least one of them is zero. Clearly, the second factor cannot be equal to zero because nothing is known about the values of a, b, and c.
∴ x – 3a – 3b = 0
or, x = 3a +3b
The required solution is x = 3a +3b.
Common Mistakes in Solving Equations
Example 8
Solve: \(\frac{3}{5}(x-4)-\frac{1}{3}(2 x-9)=\frac{1}{4}(x-1)-2\)
Solution:
Given
\(\frac{3}{5}(x-4)-\frac{1}{3}(2 x-9)=\frac{1}{4}(x-1)-2\)Multiplying both sides by 60, which is the L.C.M of 5, 3, and 4.
36(x – 4) – 20(2x – 9) = 15(x – 1) – 120
or, 36x – 144 – 40x + 180 = 15x -15 -120
or, – 4x + 36 = 15x -135
or, 15x + 4x = 135 + 36
or, 19x = 171
or, x = 171 /19
= 9
The required solution is x = 9.
Example 9
Solve: \(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)
Solution:
Given \(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)
\(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)or, \(\frac{14(x+1)-(5 x+9)}{28}=\frac{x+6+105-7(x-12)}{21}\)
or, \(\frac{14 x+14-5 x-9}{28}=\frac{x+111-7 x+84}{21}\)
or, \(\frac{9 x+5}{28}=\frac{-6 x+195}{21}\)
or, \(\frac{9 x+5}{4}=\frac{-6 x+195}{3}\)
or, 27x + 15 = -24x + 780
or, 27x + 24x = 780 – 15
or, 51x = 765
or, \(x=\frac{765}{51}=15\)
The required solution is x = 15.
Example 10
Find the value of x from equation, \(\frac{7 x+5}{15}+\frac{2 x-5}{5}=\frac{2 x+4}{6}+\frac{5 x-2}{12}\)
Solution:
Given:-
\(\frac{7 x+5}{15}+\frac{2 x-5}{5}=\frac{2 x+4}{6}+\frac{5 x-2}{12}\) \(\frac{7 x+5}{15}+\frac{2 x-5}{5}=\frac{2 x+4}{6}+\frac{5 x-2}{12}\)or, \(\frac{7 x+5+6 x-15}{15}=\frac{4 x+8+5 x-2}{12}\)
or,\(\frac{7 x+5+6 x-15}{15}=\frac{4 x+8+5 x-2}{12}\)
or, 45x + 30 = 52x – 40
or, 45x – 52x = -40 -30
or, -7x = -70
or, \(x=\frac{-70}{-7}=10\)
Value of x is 10.
Practice Problems on Algebraic Equations
Example 11
Solve: \(\frac{x^2+2 x+3}{x^2+3 x+5}=\frac{x+2}{x+3}\)
Solution:
Given:-
\(\frac{x^2+2 x+3}{x^2+3 x+5}=\frac{x+2}{x+3}\) \(\frac{x^2+2 x+3}{x^2+3 x+5}=\frac{x+2}{x+3}\)or, \(\frac{x^2+2 x+3}{x+2}=\frac{x^2+3 x+5}{x+3}\)
[By the method of alternends]
or, \(\frac{x(x+2)+3}{x+2}=\frac{x(x+3)+5}{x+3}\)
or, \(x+\frac{3}{x+2}=x+\frac{5}{x+3}\)
or, \(\frac{3}{x+2}=\frac{5}{x+3}\)
or, 5x + 10 = 3x + 9 or, 5x – 3x = 9 – 10
or, 2x = -1 or, x = \(-\frac{1}{2}\)
The required solution x = \( -\frac{1}{2}\)
Example 12
Solve: \(\frac{2 x+1}{5}+\frac{7 x+2}{3}=\frac{x+3}{10}+\frac{2 x+1}{21}\)
Solution:
Given:-
\(\frac{2 x+1}{5}+\frac{7 x+2}{3}=\frac{x+3}{10}+\frac{2 x+1}{21}\) \(\frac{2 x+1}{5}+\frac{7 x+2}{3}=\frac{x+3}{10}+\frac{2 x-1}{21}\)or, \(\frac{2 x+1}{5}-\frac{x+3}{10}=\frac{2 x-1}{21}-\frac{7 x+2}{3}\)
[By transportation]
or, \(\frac{4 x+2-x-3}{10}=\frac{2 x-1-49 x-14}{21}\)
or, \(\frac{3 x-1}{10}=\frac{-47 x-15}{21}\)
or, 63x – 21 = -470x – 150
or, 470x + 63x = 21 – 150
or, 533x = -129
or, x = \(-\frac{129}{533}\)
The required solution is x = \(-\frac{129}{533}\)
Example 13
Solve: \(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{1}{x-1} .\)
Solution:
Given:-
\(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{1}{x-1} .\) \(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{1}{x-1}\)or, \(\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}=\frac{1}{x-1}\)
or, \(-\frac{1}{x-1}+\frac{1}{x-3}=\frac{1}{x-1}\)
or, \(\frac{1}{x-3}=\frac{2}{x-1} \text { or, } 2 x-6=x-1\)
or, 2x – x = 6 – 1 or, x = 5
The required solution is x = 5.
Short Notes on Types of Equations
Example 14
Solve: \(\frac{3}{x-1}+\frac{2}{x-2}=\frac{5}{x-3}\)
Solution:
Given:-
\(\frac{3}{x-1}+\frac{2}{x-2}=\frac{5}{x-3}\) \(\frac{3}{x-1}+\frac{2}{x-2}=\frac{5}{x-3}\)or, \(\frac{3}{x-1}+\frac{2}{x-2}=\frac{3+2}{x-3}\)
or, \(\frac{3}{x-1}+\frac{2}{x-2}=\frac{3}{x-3}+\frac{2}{x-3}\)
or, \(\frac{3}{x-1}-\frac{3}{x-3}=\frac{2}{x-3}-\frac{2}{x-2}\)
or, \(3\left(\frac{1}{x-1}-\frac{1}{x-3}\right)=2\left(\frac{1}{x-3}-\frac{1}{x-2}\right)\)
or, \(\text { 3. } \frac{-2}{(x-1)(x-3)}=2 \cdot \frac{1}{(x-3)(x-2)}\)
or, \(\frac{-3}{(x-1)(x-3)}=\frac{1}{(x-3)(x-2)}\)
Now, multiplying both sides by (x – 3) we get,
\(\frac{-3}{x-1}=\frac{1}{x-2}\)or, x – 1 = -3x + 6 or, x + 3x = 6 + 1
or, 4x = 7
or, \(x=\frac{7}{4}=1 \frac{3}{4}\)
The required equation is \(1 \frac{3}{4}\)
Example 15
Solve: \(\frac{1}{x-2}+\frac{1}{x-6}=\frac{1}{x-3}+\frac{1}{x-5}\)
Solution:
Given :-
\(\frac{1}{x-2}+\frac{1}{x-6}=\frac{1}{x-3}+\frac{1}{x-5}\) \(\frac{1}{x-2}+\frac{1}{x-6}=\frac{1}{x-3}+\frac{1}{x-5}\)or, \(\frac{1}{x-2}-\frac{1}{x-3}=\frac{1}{x-5}-\frac{1}{x-6}\)
or, \(\frac{x-3-x+2}{(x-2)(x-3)}=\frac{x-6-x+5}{(x-5)(x-6)}\)
or, \(\frac{-1}{(x-2)(x-3)}=\frac{-1}{(x-5)(x-6)}\)
or, (x – 5)(x – 6) = (x – 2)(x – 3)
or, x^2-11 x+30=x^2-5 x+6
or, -11x + 5x = 6 – 30
or, -6x = -24
or, x = \(\frac{-24}{-6}=4\)
The required solution is x = 4.
Important Definitions Related to Algebraic Equations
Example 16
Find the value of x from the relation \(\frac{a}{a-x}+\frac{b}{b-x}=\frac{a+b}{a+b-x} .\)
Solution:
Given:-
\(\frac{a}{a-x}+\frac{b}{b-x}=\frac{a+b}{a+b-x} .\)From the given relation,
\(\frac{a}{a-x}+\frac{b}{b-x}=\frac{a}{a+b-x}+\frac{b}{a+b-x}\)or, \(\frac{a}{a-x}-\frac{a}{a+b-x}=\frac{b}{a+b-x}-\frac{b}{b-x}\)
or, \(\frac{a^2+a b-a x-a^2+a x}{(a-x)(a+b-x)}=\frac{b^2-b x-a b-b^2+b x}{(a+b-x)(b-x)}\)
or, \(\frac{a b}{(a-x)(a+b-x)}=\frac{-a b}{(a+b-x)(b-x)}\)
or, \(\frac{1}{(a-x)(a+b-x)}=\frac{-1}{(a+b-x)(b-x)}\)
or, \(\frac{1}{a-x}=\frac{-1}{b-x}\)
[Multiplying both sides by a + b – x].
or, -a + x = b – x
or, x + x = a +b
or, 2x = a+ b
or, x = \(\frac{a+b}{2}\)
The required value of x is \(\frac{a+b}{2}\).
Example 17
Solve: [latex]\begin{aligned}
\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)} & +\frac{1}{(x+3)(x+4)} \\
& =\frac{1}{(x+1)(x+6)}
\end{aligned}[/latex]
Solution:
From the given equation,
\(\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}=\frac{1}{(x+1)(x+6)}\)or, \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}=\frac{1}{(x+1)(x+6)}\)
or, \(\frac{1}{x+1}-\frac{1}{x+4}=\frac{1}{(x+1)(x+6)}\)
or, \(\frac{x+4-x-1}{(x+1)(x+4)}=\frac{1}{(x+1)(x+6)}\)
or, \(\frac{3}{(x+1)(x+4)}=\frac{1}{(x+1)(x+6)}\)
or, \(\frac{3}{x+4}=\frac{1}{x+6}\)
[Multiplying both sides by (x + 1)]
or, 3x + 18 = x + 4
or, 3x – x = 4 – 18
or, 2x = -14
or, x = \(-\frac{14}{2}\) = -7
The required solution is x = -7.
Conceptual Questions on Solving Equations
Example 18
Solve: \(\frac{2 x+1}{2 x-1}+\frac{3 x+1}{3 x-2}=\frac{3 x+2}{3 x-1}+\frac{6 x+1}{6 x-5}\)
Solution:
From the given equation,
\(\frac{2 x+1}{2 x-1}-1+\frac{3 x+1}{3 x-2}-1=\frac{3 x+2}{3 x-1}-1+\frac{6 x+1}{6 x-5}-1\)or, \(\frac{2 x+1-2 x+1}{2 x-1}+\frac{3 x+1-3 x+2}{3 x-2}\)
= \(\frac{3 x+2-3 x+1}{3 x-1}+\frac{6 x+1-6 x+5}{6 x-5}\)
or, \(\frac{2}{2 x-1}+\frac{3}{3 x-2}=\frac{3}{3 x-1}+\frac{6}{6 x-5}\)
or, \(\frac{6 x-4+6 x-3}{(2 x-1)(3 x-2)}=\frac{18 x-15+18 x-6}{(3 x-1)(6 x-5)}\)
or, \(\frac{12 x-7}{6 x^2-7 x+2}=\frac{3(12 x-7)}{18 x^2-21 x+5}\)
or, \((12 x-7)\left(18 x^2-21 x+5\right)=(12 x-7)\left(18 x^2-21 x+6\right)\)
or, \((12 x-7)\left(18 x^2-21 x+5-18 x^2+21 x-6\right)=0\)
or, (12x-7)(-1) = 0 or, 12x – 7 = 0
or, 12x = 7 or, x = \(\frac{7}{12}\)
The required solution is x = \(\frac{7}{12}\).
Example 19
Solve: \(\frac{x-b c}{b+c}+\frac{x-c a}{c+a}+\frac{x-a b}{a+b}=a+b+c\)
Solution:
From the given equation,
\(\frac{x-b c}{b+c}-a+\frac{x-c a}{c+a}-b+\frac{x-a b}{a+b}-c=0\)or, \(\frac{x-b c-a b-a c}{b+c}+\frac{x-c a-b c-a b}{c+a}+\frac{x-a b-a c-b c}{a+b}=0\)
or, \((x-a b-b c-c a)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)=0\)
Now, the left-hand side is the product of two quantities. Its value will be equal to zero if at least one of them is zero. Clearly, the second factor cannot be equal to zero because nothing is known about the values of a, b, and c.
Hence, x – ab – be – ca = 0
or, x = ab + bc + ca
The required solution is x = ab + bc + ca.
Example 20
Solve: \(\frac{x+a^2+2 c^2}{b+c}+\frac{x+b^2+2 a^2}{c+a}+\frac{x+c^2+2 b^2}{a+b}=0 .\)
Solution:
From the given equation,
\(\begin{array}{r}\frac{x+a^2+2 c^2}{b+c}+(b-c)+\frac{x+b^2+2 a^2}{c+a}+(c-a) \\
+\frac{x+c^2+2 b^2}{a+b}+(a-b)=0
\end{array}\)
or, \(\frac{x+a^2+2 c^2+b^2-c^2}{b+c}+\frac{x+b^2+2 a^2+c^2-a^2}{c+a}\)
\(+\frac{x+c^2+2 b^2+a^2-b^2}{a+b}=0\)or, \(\frac{x+a^2+b^2+c^2}{b+c}+\frac{x+a^2+b^2+c^2}{c+a}\)
\(+\frac{x+a^2+b^2+c^2}{a+b}=0\)or, \(\left(x+a^2+b^2+c^2\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)=0 .\)
Now, the left-hand side is the product of two quantities. Its value will be equal to zero if at least one of them is zero’. Clearly, the second factor cannot be equal to zero because nothing is known about the values of a, b, and c.
Hence, x + a2 + b2 + c2 = 0
or, x = – (a2 + b2 + c2)
The required solution is x = – (a2+ b2 + c2)
Example 21
The sum of the digits of a number of two digits is 10. if the digit in the units’ place is 4 times the digit in the tens’ place, find the number.
Solution:
Let, the digit in the tens place be x.
Then the digit in the units place = 4x
∴ The number = 10 x x + 4x
= 10x + 4x
= 14x.
Again, according to the question,
x + 4x = 10
or, 5x = 10
or, x = 10/5
= 2
Hence, the number = 14 x 2
= 28
The number is 28.
Examples of Real-Life Applications of Equations
Example 22
A number consisting of two digits is such that its digit in the tens’ place is twice that in the units’ place. If the digits are reversed, the number thus formed is 27 less than the original one. Find the number.
Solution :
Let, the digit in the units’ place be x.
Then the digit in the tens’ place = 2x.
∴ The number = 10 x 2x + x = 20x + x= 21x
If the digits are reversed, the number obtained
= 10 x x + 2x
= 12x
According to the question,
21x – 12x = 27
or, 9x = 27
or, x = 27/9
= 3
Hence, the number
= 21 x 3
= 63
The number is 63.
Example 23
The denominator of a fraction is greater than its numerator by 2. The fraction becomes 1/2 when 3 is subtracted from both the numerator and the denominator. Find the fraction.
Solution:
Let, the numerator of the fraction be x.
Then its denominator = x + 2.
Hence, the fraction = \(\frac{x}{x+2}\)
According to the question,
\(\frac{x-3}{x+2-3}=\frac{1}{2}\)or, \(\frac{x-3}{x-1}=\frac{1}{2} \text { or, } 2 x-6=x-1\)
or, 2x – x = 6 – 1 or, x = 5
Hence, the required fraction is \(\frac{5}{7}\)
The required fraction is \(\frac{5}{7}\).
Example 24
The denominator of a fraction is 2 more than twice the numerator keeping the denominator unaltered then the fraction becomes 7/12. Find the fraction.
Solution:
Let, the numerator be x.
Then the denominator = 2x + 2
∴ The fraction = \(\frac{x}{2 x+2}\)
According to the question,
\(\frac{x+4}{2 x+2}=\frac{7}{12} \text { or, } 14 x+14=12 x+48\)or, 14x – 12x = 48 – 14
or, 2x = 34
or, x = \(\frac{34}{2}=17\)
∴Numerator = 17
and denominator = 2 x 17 + 2 = 36
∴ The required fraction = \(\frac{17}{36}\)
The fraction is \(\frac{17}{36}\)
Example 25
₹ 624 was distributed among A, B, C, and D, If A would receive ₹ 2 more, B would receive ₹ 6 less, C would receive 5 times his money and D would receive 1/4th of his money, then they would receive equal money. How much money did each of them receive?
Solution:
Given
₹ 624 was distributed among A, B, C, and D, If A would receive ₹ 2 more, B would receive ₹ 6 less, C would receive 5 times his money and D would receive 1/4th of his money, then they would receive equal money.
Let, the equal money be ₹ x,
Then A received ₹( x – 2),
B received ₹ ( x + 6),
C received ₹ x/5 ,
D received ₹ 4x.
According to the question,
x – 2 + x + 6 + x/5 + 4x = 624
or, 6x + x/5 = 624 – 4
or, 31x / 5 = 620
or, x = 620 x 5/31 = 100
∴ A received ₹(100-2)
= ₹98
B received ₹ (100+6)
= ₹ 106
C received ₹ 100/5
= ₹ 20
D received ₹ 4 x 100
= ₹ 400.
∴ A received v 98,
B received ₹ 106,
C received ₹ 20,
D received ₹ 400.