KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Students can Download Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Question 1.
Aftab tells his daughter, “Seven years ago, I was, seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Answer:
Let the present age of father be ‘x’
Let the age of daughter be ‘y’.
Before seven years, age of father is x – 7
Before seven years, age of daughter is y – 7.
Father was 7 times as old as daughter.
∴ x – 7 = 7 (y – 7)
x – 7 = 7y – 49
x – 7y = – 49 + 7
x – 7y = – 42
∴ x – 7y + 42 = 0 …………….. (i)
After three years, age of father is x + 3
After three years, age of daughter is y + 3
Three years from now will three times as old as father.
∴ x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y + 3 – 9 = 0
3 – 3y – 6 = 0 ………….. (ii)
∴ Algebraically linear equations:
x – 7y + 42 = 0
x – 3y – 6 =0
If we represent this linear equations through graph, we have

i) x – 7y + 42 = 0
– 7y = – x – 42
7y = x + 42
$\therefore \quad y=\frac{x+42}{7}$

x	0	7
$y=\frac{x+42}{7}$	6	7

ii) x – 3y – 6 = 0
– 3y = – x + 6
3y = x – 6
$\therefore \quad y=\frac{x-6}{3}$

x	0	+6
$y=\frac{x-6}{3}$	-2	0

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Answer:
Let the cost of each bat be Rs. ‘x’
Let the cost of each ball be Rs. ‘y’.
3x + 6y = 3900
∴ x + 2y = 1300 ………. (i)
Cost of another bat and ball be x + 3y = 1300 ………….. (ii)
∴ Algebriacally
x + 2y = 1300
x + 3y = 1300
And to represent through graph:
i) x + 2y = 1300
2y = 1300 – x
$\therefore \quad x=\frac{1300-x}{2}$

x	100	200
$y=\frac{1300-x}{2}$	600	500

ii) x + 3y = 1300
3y = 1300 – x
$\therefore \quad y=\frac{1300-x}{3}$

x	100	400
$y=\frac{1300-x}{3}$	400	900

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Answer:
Algebraically,
If Cost of each kg. of apple be Rs. ‘x’,
Cost of grapes be Rs. ‘y’. Then
2x + y = 160
4x + 2y = 300
To represent geometrically,

i) 2x + y = 160
y = – 2x + 160

x	20	80
y = 2x + 160	120	0

ii) 4x + 2y = 300
2x + y = 150
y = 150 – 2x

x	40	60
y = 150 – 2x	70	30