NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.7

NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.7 are provided here to assist students with their Class 10 exam preparation. These solutions are created by subject experts to help students understand how to solve word problems involving linear equations in two variables using various algebraic methods like substitution, elimination, and cross-multiplication.

Exercise 3.7 focuses on applying real-world scenarios to form linear equations and then solving them using appropriate methods. These problems are designed to enhance students' skills in interpreting and solving equations that represent relationships between two variables. By practicing these problems, students will develop a strong understanding of how to model real-life situations mathematically.

NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.7

Pair of Linear Equations in Two Variables

Medium

Q.

$\begin{array}{l}\text{Form the pair of linear equations for the following}\\ \text{problems and find their solution by substitution method.}\\ \text{ (i) The difference betrween two numbers is 26 and one }\\ \text{ number is three times the other. Find them.}\\ \text{ (ii) The larger of two supplementary angles exceeds}\\ \text{ the smaller by 18 degrees. Find them.}\\ \text{ (iii) The coach of a cricket team buys a 7 bats and 6 balls}\\ \text{ for }₹\text{3800. Later she buys the 3 bats and 5 balls}\\ \text{ for }₹\text{1750. Find the cost of each bat and each ball.}\\ \text{ (iv) The taxi charges in a city consist of a fixed charge }\\ \text{ together with the charge for the distance covered.}\\ \text{ For\hspace{0.17em}\hspace{0.17em}a distance of 10 km, the charge paid is }₹\text{105 and}\\ \text{ for a journey of 15 km, the charge paid is 155. What }\\ \text{ are the\hspace{0.17em}\hspace{0.17em}fixed charges and the charge per km? How }\\ \text{ much does\hspace{0.17em}\hspace{0.17em}a person have to pay for travelling a }\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}distance of 25 km?}\\ \text{ (v) A fraction becomes }\frac{9}{11}\text{, if 2 is added to both the }\\ \text{ numerator and the denominator. If, 3 is added to}\\ \text{ both the numerator and the denominator it}\\ \text{ becomes }\frac{5}{6}.\text{ Find the fraction.}\\ \text{ (vi) Five years hence, the age of Jacob will be three times}\\ \text{ that of his son. Five years ago, Jacob's age was seven}\\ \text{ times that of his son. What are their present ages?}\end{array}$

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Pair of Linear Equations in Two Variables

Medium

Q.

$\begin{array}{l}\text{One says, "Give me a hundred, friend! I shall then}\\ \text{become twice as rich as you". The other replies, }\\ \text{"If you give me ten, I shall be six times as rich as}\\ \text{you". Tell me what is the amount of their}\\ \text{(respective) capital? }\\ \text{[From the Bijaganita of Bhaskara II]}\\ \text{[Hint: }\mathrm{x}+100=2(\mathrm{y}-100),\mathrm{y}+10=6(\mathrm{x}-10)]\end{array}$

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Pair of Linear Equations in Two Variables

Medium

Q.

$\begin{array}{l}\text{A train covered a certain distance at a uniform}\\ \text{speed. If the train would have been 10 km/h faster,}\\ \text{it would have taken 2 hours less than the scheduled}\\ \text{time. And, if the train were slower by 10 km/h; it}\\ \text{would have taken 3 hours more than the scheduled}\\ \text{time. Find the distance covered by the train.}\end{array}$

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Pair of Linear Equations in Two Variables

Medium

Q.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

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Pair of Linear Equations in Two Variables

Easy

Q.

$\begin{array}{l}\text{In a ΔABC, }\angle \text{C=3 }\angle \text{B=2}\left(\angle \text{A+}\angle \text{B}\right)\text{. Find the}\\ \text{three angles.}\end{array}$

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Pair of Linear Equations in Two Variables

Medium

Q.

Draw the graphs of the equations 5x – y = 5 and  3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis

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Pair of Linear Equations in Two Variables

Medium

Q.

ABCD is a cyclic quadrilateral. Find its all angles.

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Pair of Linear Equations in Two Variables

Medium

Q.

A father's age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.

[CBSE - 2019]

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These solutions are aligned with the latest CBSE syllabus and provide step-by-step guidance, making it easier for students to practice solving linear equations and excel in exams.

Q1: 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.

Solution:
Let the two numbers be $x$ and $y$, where $x$ is the larger number.

1. From the given information, we have:

   • $x - y = 26$

   • $x = 3y$ (since one number is three times the other)

Substitute $x = 3y$ into the first equation:

$$3y - y = 26 \quad \Rightarrow \quad 2y = 26 \quad \Rightarrow \quad y = 13$$

Substitute $y = 13$ into $x = 3y$:

$$x = 3 \times 13 = 39$$

Thus, the two numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:
Let the two angles be $x$ and $y$, where $x > y$.

1. Since the angles are supplementary, we know:

   • $x + y = 180$

   • $x = y + 18$ (since the larger angle exceeds the smaller by 18 degrees)

Substitute $x = y + 18$ into $x + y = 180$:

$$(y + 18) + y = 180 \quad \Rightarrow \quad 2y + 18 = 180 \quad \Rightarrow \quad 2y = 162 \quad \Rightarrow \quad y = 81$$

Substitute $y = 81$ into $x = y + 18$:

$$x = 81 + 18 = 99$$

Thus, the two angles are 99° and 81°.

(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.

Solution:
Let the cost of each bat be $x$ and the cost of each ball be $y$.

1. From the given information, we have:

   • $7x + 6y = 3800$

   • $3x + 5y = 1750$

Now, solve the system of equations. Multiply the second equation by 2 to align the coefficients of $x$:

$$6x + 10y = 3500$$

Now subtract the first equation from this:

$$(6x + 10y) - (7x + 6y) = 3500 - 3800$$ $$-x + 4y = -300 \quad \Rightarrow \quad x = 4y + 300$$

Substitute this back into one of the original equations to find $x$ and $y$.

(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:
Let the fixed charge be $x$ and the charge per km be $y$.

1. From the given information, we have:

   • $10x + 10y = 105$

   • $15x + 15y = 155$

Simplify and solve this system of equations to find $x$ and $y$. Then substitute the values of $x$ and $y$ into the equation for a 25 km journey:

$$25x + 25y$$

(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.

Solution:
Let the fraction be $\frac{x}{y}$.

1. From the given conditions, we have:

   • $\frac{x + 2}{y + 2} = \frac{9}{11}$

   • $\frac{x + 3}{y + 3} = \frac{5}{6}$

Now solve this system of equations for $x$ and $y$.

(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:
Let the present age of Jacob be $x$ and the present age of his son be $y$.

1. From the given information, we have:

   • $x + 5 = 3(y + 5)$

   • $x - 5 = 7(y - 5)$

Now solve this system of equations to find $x$ and $y$.

FAQs: Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.7

Q1. What is the focus of Exercise 3.7?
Answer:
Exercise 3.7 focuses on solving word problems that involve pair of linear equations in two variables. The problems require forming linear equations from given scenarios and solving them using methods like substitution, elimination, or cross-multiplication.

Q2. How do I approach word problems in this exercise?
Answer:
To solve word problems:

1. Read the problem carefully and identify the unknowns (variables).

2. Translate the problem into linear equations in two variables.

3. Use the appropriate method (substitution, elimination, or cross-multiplication) to solve the equations.

4. Interpret the solution and ensure it makes sense in the context of the problem.

Q3. What are the methods used to solve the problems in this exercise?
Answer:
The methods used are:

• Substitution Method: Solving one equation for one variable and substituting it in the other equation.

• Elimination Method: Adding or subtracting the equations to eliminate one variable.

• Cross-multiplication Method: A method used when equations are in standard form (Ax + By = C).

Q4. How do I choose the right method to solve a problem?
Answer:

• Substitution Method is useful when one equation can be easily solved for one variable.

• Elimination Method is preferred when the coefficients of one variable are easy to manipulate.

• Cross-multiplication Method is most efficient when the equations are in the form

  $\frac{A}{B} = \frac{C}{D}$BA​=DC​.

Q5. How do NCERT Solutions help in exam preparation?
Answer:
These solutions offer step-by-step explanations for solving problems, which help students understand the concepts clearly and apply them efficiently during exams. By practicing these solutions, students will improve their problem-solving abilities and be better prepared for board exams.