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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?
Solution :
Given:
1. The taxi charges consist of a fixed charge and a charge per kilometer.
2. For a distance of $10\text{ km}$, the total charge is $₹105$.
3. For a distance of $15\text{ km}$, the total charge is $₹155$.
To Find:
1. The fixed charge ($x$).
2. The charge per kilometer ($y$).
3. The total charge for a distance of $25\text{ km}$.
Step 1: Formulating the Linear Equations
Let the fixed charge be $x$ (in ₹) and the charge per kilometer be $y$ (in ₹/km).
Based on the problem statement, the total charge is given by the formula: $\text{Total Charge} = \text{Fixed Charge} + (\text{Distance} \times \text{Charge per km})$.
For the first condition ($10\text{ km}$ for $₹105$):
$x + 10y = 105$ --- (Equation 1)
For the second condition ($15\text{ km}$ for $₹155$):
$x + 15y = 155$ --- (Equation 2)
Step 2: Solving by Substitution Method
From Equation 1, express $x$ in terms of $y$:
$x = 105 - 10y$ --- (Equation 3)
Substitute the value of $x$ from Equation 3 into Equation 2:
$(105 - 10y) + 15y = 155$ [Substituting $x$]
$105 + 5y = 155$ [Combining like terms: $-10y + 15y = 5y$]
$5y = 155 - 105$ [Subtracting 105 from both sides]
$5y = 50$
$y = \frac{50}{5}$ [Dividing both sides by 5]
$y = 10$
Now, substitute $y = 10$ back into Equation 3 to find $x$:
$x = 105 - 10(10)$
$x = 105 - 100$
$x = 5$
Step 3: Calculating the Charge for 25 km
The total charge for $25\text{ km}$ is given by the expression: $x + 25y$.
Substitute $x = 5$ and $y = 10$ into the expression:
$\text{Total Charge} = 5 + 25(10)$
$\text{Total Charge} = 5 + 250$
$\text{Total Charge} = 255$
Final Answer:
The fixed charge is ₹5, the charge per km is ₹10, and the total charge for travelling 25 km is ₹255.
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(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.
- 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?
CBSE Solutions for Class 10 Mathematics Pair of linear equations in two variable
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