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Q2(ii):
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution :
Given:
1. Five years ago, Nuri's age was thrice the age of Sonu.
2. Ten years later, Nuri's age will be twice the age of Sonu.
To Find:
The present ages of Nuri and Sonu.
Step 1: Defining Variables
Let the present age of Nuri be $x$ years.
Let the present age of Sonu be $y$ years.
Step 2: Formulating the Equations
Condition 1: Five years ago
Nuri's age was $(x - 5)$ and Sonu's age was $(y - 5)$.
According to the problem: $(x - 5) = 3(y - 5)$
$x - 5 = 3y - 15$
$x - 3y = -15 + 5$
$x - 3y = -10$ --- (Equation 1)
Condition 2: Ten years later
Nuri's age will be $(x + 10)$ and Sonu's age will be $(y + 10)$.
According to the problem: $(x + 10) = 2(y + 10)$
$x + 10 = 2y + 20$
$x - 2y = 20 - 10$
$x - 2y = 10$ --- (Equation 2)
Step 3: Solving by Elimination Method
We have the system of equations:
(1) $x - 3y = -10$
(2) $x - 2y = 10$
To eliminate $x$, subtract Equation (1) from Equation (2):
$(x - 2y) - (x - 3y) = 10 - (-10)$
$x - 2y - x + 3y = 10 + 10$
$y = 20$
Step 4: Finding the value of $x$
Substitute $y = 20$ into Equation (2):
$x - 2(20) = 10$
$x - 40 = 10$
$x = 10 + 40$
$x = 50$
Step 5: Verification
Five years ago: Nuri was $50 - 5 = 45$, Sonu was $20 - 5 = 15$. Since $45 = 3 \times 15$, the first condition is satisfied.
Ten years later: Nuri will be $50 + 10 = 60$, Sonu will be $20 + 10 = 30$. Since $60 = 2 \times 30$, the second condition is satisfied.
Final Answer: The present age of Nuri is 50 years and the present age of Sonu is 20 years.
More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.3
- Q1(i): Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x – 3y = 4
- Q1(ii): Solve the following pair of linear equations by the elimination method and the substitution method : (ii) 3x + 4y = 10 and 2x – 2y = 2
- Q1(iii): Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
- Q1(iv): Solve the following pair of linear equations by the elimination method and the substitution method : (iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$
- Q2(i): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?
- Q2(iii): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
- Q2(iv): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw ` 2000. She asked the cashier to give her ` 50 and ` 100 notes only. Meena got 25 notes in all. Find how many notes of ` 50 and ` 100 she received.
- Q2(v): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
CBSE Solutions for Class 10 Mathematics Pair of linear equations in two variable
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