10th Maths 3.1
NCERT Class 10th solution of Exercise 3.2
NCERT Class 10th solution of Exercise 3.3
NCERT Class 10th solution of Exercise 3.4
NCERT Class 10th solution of Exercise 3.5
NCERT Class 10th solution of Exercise 3.6
Exercise 3.1
Q1. Aftab tells her 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''(Is not this interesting?) Represent this situation algebraically and graphically.
Sol. :
Let the present age of Aftab =x years,
the present age of his daughter =y years,
According to question
(x-7)=7(y-7)
x-7=7y-49
x-7y+42=0
and after 3 years
(x+3)=3(y+3)
x+3=3y+9
x-3y-6=0
For graphical representation,
x-7y+42=0
x=7y-42
when x=0
0=7y-42
7y=42
y=6
When x=7
7=7y-42
7y=42+7
7y=49
y=7
|
X |
0 |
7 |
|
y |
6 |
7 |
and
x-3y-6=0
when x=0
0-3x-6=0
3x=-6
x=-2
When
x=6
6-3y-6=0
-3y=0
y=0
|
X |
0 |
6 |
|
y |
-2 |
0 |
Answer :
Algebraically the two situations can be represented as follow x-7y+42=0,x-3y-6=0
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Q2. The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Sol. :
Let the cost price of each bat x,
and the cost price of each ball y.
According to question
3x+6y=3900
x+2y=1300
and
x+3y=1300
For geometrically (graphically) representation
x+2y=1300
x=1300-2y
When x=300
300=1300-2y
2y=1300-300
y=10002
y=500
When x=500
500=1300-2y
2y=1300-500
2y=800
y=8002
y=400
|
X |
300 |
500 |
|
y |
500 |
400 |
and
x+3y=1300
x=1300-3y
When x=400
400=1300-3y
3y=1300-400
3y=900
y=9003
y=300
When x=100
100=1300-3y
3y=1300-100
3y=1200
y=12003
y=400
|
X |
400 |
100 |
|
y |
300 |
400 |
Algebraically the two situations can be represented as follow x+2y=1300,x+3y=1300.
Q3 The cost of 2 kg of apples on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Sol. :
Let the cost price of 1kg of apples is x,
and the cost price of 1kg of grapes is y.
According to question
2x+y=160
and
4x+2y=300
2x+y=150
when x=50
2(50)+y=160`
100+y=160
y=160-100
y=60
when x=60
2(60)+y=160
120+y=160
y=160-120
y=40
|
X |
50 |
60 |
|
y |
60 |
40 |
2x+y=150
when x=50
2(50)+y=150
100+y=150
y=150-100
y=50
when x=60
2(60)+y=150
120+y=150
y=150-120
y=30
|
X |
50 |
60 |
|
y |
50 |
30 |
Algebraically the two situations can be represented as follow 2x+y=150,2x+y=160.
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