Ques 1: Show that x = 4 is a solution of the equation:
Solution: Substituting x = 4 in 



Since, LHS = RHS
∴ x = 4 is a solution of the given equation.
Ques 2: Solve
Solution: We have
LCM of 3, 5, 2 and 4 is 60.
∴ The given equation can be expressed as:

⇒ 
⇒ 
⇒ 
⇒ 
Thus, x = 27/10 is the required solution.
Ques 3: Solve for
Solution: We have 
⇒ 
⇒ 
⇒ 3x - 4 + 44 - 4x - 3 = 2x + 4
⇒ 3x - 4x - 2x = 4 + 3 - 44 + 4
⇒ 3x - 6x = 11 - 44
⇒ -3x = -33 ⇒ x = 11
Ques 4: Solve for
Solution: We have
By cross multiplication, we get:
(2 + x)(7 - x) = (5 - x)(4 + x)
⇒ 2(7 - x) + x(7 - x) = 5(4 + x) - x(4 + x)
⇒ 14 - 2x + 7x - x2 = 20 + 5x - 4x - x2
⇒ -x2 + x2 - 2x + 7x - 5x + 4x = 20 - 14
⇒ -7x + 7x + 4x = 6
⇒ 4x = 6 ⇒ x = 6/4 or 3/2
Thus, the solution of the given equation is x = 3/2
Ques 5: A number is such that it is as much greater than 65 as it is less than 91. Find the number.
Solution: Let the number be x.
Since, we have [The number] - 65 = 91 - [The number]
⇒ x - 65 = 91 - x
⇒ x + x = 91 + 65
⇒ 2x = 156
⇒ x = 156/2 = 78
Thus, the required number is 78.
Ques 6: The numerator of a fraction is 2 less than the denominator. If 1 is added to its denominator, it becomes 1/2. Find the fraction.
Solution: Let the denominator of the fraction be x.
∴ Numerator = x - 2
The fraction =
Since it becomes 1/2
When 1 is added to its denominator.
i.e
By cross multiplication, we have
2(x - 2) = x + 1
⇒ 2x - 4 = x + 1
⇒ 2x - x = 1 + 4
⇒ x = 5
⇒ Fraction =
Ques 7: After 24 years I shall be 3 times as old as I was 4 years ago. Find my present age.
Solution: Let my present age be x years.
∴ After 24 years, my age will be (x + 24) years.
4 years ago, my age was (x - 4) years.
According to the given condition, we have
(x + 24) = 3(x - 4)
⇒ x + 24 = 3x - 12
⇒ x - 3x = -12 - 24
⇒ -2x = -36
⇒ x = -36/-2 = 18
Thus, my present age is 18 years.
Ques 8: If the sum of two numbers is 30 and their ratio is 2 : 3, then find the numbers.
Solution: Let one of the numbers be x.
∴ The other number = (30 - x)
According to the condition, we have
[∵ The ratio number is 2 : 3]
⇒ 3x = 2(30 - x) [By cross multiplication]
⇒ 3x = 60 - 2x
⇒ 3x + 2x = 60
⇒ 5x = 60
⇒ x = 60/5 = 12
∴ 30 - x = 30 - 12 = 18
Thus, the required numbers are 12 and 18.
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