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Q3(vi):
Form the pair of linear equations for the following problems and find their solution by substitution method. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution :
Given:
1. Five years hence (in the future), Jacob's age will be three times that of his son.
2. Five years ago (in the past), Jacob's age was seven times that of his son.
To Find:
The present ages of Jacob and his son.
Step 1: Defining Variables
Let the present age of Jacob be $x$ years.
Let the present age of his son be $y$ years.
Step 2: Formulating the Equations
Case 1: Five years hence
Jacob's age will be $(x + 5)$ and his son's age will be $(y + 5)$.
According to the problem: $(x + 5) = 3(y + 5)$
$x + 5 = 3y + 15$
$x - 3y = 10$ --- (Equation 1)
Case 2: Five years ago
Jacob's age was $(x - 5)$ and his son's age was $(y - 5)$.
According to the problem: $(x - 5) = 7(y - 5)$
$x - 5 = 7y - 35$
$x - 7y = -30$ --- (Equation 2)
Step 3: Solving by Substitution Method
From Equation 1, express $x$ in terms of $y$:
$x = 3y + 10$ --- (Equation 3)
Substitute the value of $x$ from Equation 3 into Equation 2:
$(3y + 10) - 7y = -30$
$3y - 7y + 10 = -30$
$-4y = -30 - 10$
$-4y = -40$
$y = \frac{-40}{-4}$
$y = 10$
Step 4: Finding the value of $x$
Substitute $y = 10$ into Equation 3:
$x = 3(10) + 10$
$x = 30 + 10$
$x = 40$
Step 5: Verification
Five years hence: Jacob will be $45$, son will be $15$. $45 = 3 \times 15$ (Correct).
Five years ago: Jacob was $35$, son was $5$. $35 = 7 \times 5$ (Correct).
Final Answer: The present age of Jacob is 40 years and the present age of his son is 10 years.
More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.2
- Q1(i): Solve the following pair of linear equations by the substitution method. (i) x + y = 14; x – y = 4
- Q1(ii): Solve the following pair of linear equations by the substitution method. (ii) s – t = 3; $\frac{s}{3} + \frac{t}{2} = 6$
- Q1(iii): Solve the following pair of linear equations by the substitution method. (iii) 3x – y = 3; 9x – 3y = 9
- Q1(iv): Solve the following pair of linear equations by the substitution method. (iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
- Q1(v): Solve the following pair of linear equations by the substitution method. (v) $\sqrt{2}x + \sqrt{3}y = 0$; $\sqrt{3}x - \sqrt{8}y = 0$
- Q1(vi): Solve the following pair of linear equations by the substitution method. (vi) $\frac{3x}{2} - \frac{5y}{3} = -2$; $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
- Q2: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘$m$’ for which y = mx + 3.
- Q3(i): Form the pair of linear equations for the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them.
- Q3(ii): Form the pair of linear equations for the following problems and find their solution by substitution method. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
- Q3(iii): Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball.
- Q3(iv): Form the pair of linear equations for the following problems and find their solution by substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
- Q3(v): Form the pair of linear equations for the following problems and find their solution by substitution method. (v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.
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