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=1000/2`
`y=500`
When `x=500`
`500=1300-2y`
`2y=1300-500`
`2y=800`
`y=800/2`
`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=900/3`
`y=300`
When `x=100`
`100=1300-3y`
`3y=1300-100`
`3y=1200`
`y=1200/3`
`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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