Exercise 3.2 (NCERT) – Pair of Linear Equations in Two Varibales, Class 10
| Chapter 3 | Exercise 3.2 | Class 10 |
|---|
Below are the Quick links for all questions of Exercise 3.2, Pair of Linear Equations in Two Variables, Class 10. Click a link to view the solution of corresponding question.
Below are the solutions of every question of exercise 3.2 of chapter 3, Pair of Linear Equations in Two Variables, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.
Question 1
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
Answer: x = 9, y = 5
(ii) s – t = 3
s/3 + t/2 = 6
Answer: s = 9, t = 6
(iii) 3x – y = 3
9x – 3y = 9
Conclusion:
Since 9 = 9 is a true identity, hence the lines are coincident.
In other words, both equations represent the same line.
The system is consistent and has infinitely many solutions
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
From equation (1)
Answer: x = 2, y = 3
(v) √2x + √3y = 0
√3x – √8y = 0
From equation (1)
Substitute the value of x in equation (2)
Substitute the value of y in expression:
Answer: x = 0, y = 0
(vi) 3x/2 – 5y/3 = -2
x/3 + y/2 = 13/6
Let’s simplify equation (1)
Let’s simplify equation (2) as well
From equation (4)
Substitute the value of x in equation (3)
Substitute the value of y in expression
Answer: x = 2, y = 3
Question 2
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
From equation (1)
Substitute the value of x in equation (2)
Substitute the value of y in expression
Answer: x = -2, y = 5
Let us solve the Part 2
Answer: m = -1
Question 3
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.
Let the two numbers be x and y
Condition 1:
Difference between the two numbers = 26
=> x – y = 26 … (1)
Condition 2:
One number is three times the other
=> x = 3y … (2)
Substitute the value of x from (2) into (1)
Substitute the value of y into eq (2)
x = 3 (13) = 39
Answer: x = 39, y = 13
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let the two angles be x (larger) and y (smaller)
Condition 1:
Angles are supplementary
=> x + y = 180° … (1)
Condition 2:
Difference is 18°
=> x – y = 18° … (2)
Let’s apply substitution method
x = y + 18° … (from 2)
Substitute this value of x into (1)
Substitute y = 81 into (2)
x = y + 18
x = 81 + 18 = 99
Answer: 99°, 81°
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Let the cost of one bat be x and one ball be y
Condition 1:
Cost of 7 bats and 6 balls = 3800
=> 7x + 6y = 3800 … (1)
Condition 1:
Cost of 3 bats and 5 balls = 1750
=> 3x + 5y = 1750 … (2)
Let’s apply substitution method
From equation (2)
Substitute the value of x into equation (1)
Substitute the value of y into the expression for x
Answer: Cost of 1 bat = Rs. 500, Cost of 1 Ball = Rs. 50
(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 Rs. 105 and for a journey of 15 km, the charge paid is Rs. 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?
Let the fixed charge be x
& the charge per km be y
Conditions:
For 10 km, total charge = Rs. 105
For 15 km, total charge = Rs. 155
From equation (1)
Substitute the value of x into equation (2)
Substitute the value of y into equation (3)
Answer: Fixed Charges = Rs. 5, Charges per KM = Rs. 10
Part 2 – Total Fare to travel 25 KM
Answer: Total Fare = Rs. 255
(v) A fraction becomes 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 5/6. Find the fraction.
Condition 1:
Adding 2 to numerator and denominator
Condition 2:
Adding 3 to numerator and denominator
From equation (2)
Substitute the value of x into equation (1)
Substitute the value of y into equation (3)
(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?
Let present age of Jacob = x
& present age of Jacob’s son = y
| After 5 Years | 5 Years Ago | |
|---|---|---|
|
Jacob’s Age |
x + 5 |
x – 5 |
|
Age of Jacob’s son |
y + 5 |
y – 5 |
|
Condition |
x + 5 = 3 (y + 5) … (1) |
x – 5 = 7 (y – 5) … (2) |
Let us simplify equation (1) and (2)
From equation (3)
Substitute the value of x into equation (4)
Substitute the value of y into equation (5)
Answer: Jacob’s Age = 40 Years, Son’s Age = 10 Years