Class 10 Maths Chapter 3 Practice Question Answers - Pair of Linear Equations in Two Variables

Sol:
Let the cost of 1 kg of apples be ‘Rs. x’.
And, let the cost of 1 kg of grapes be ‘Rs. y’.
According to the question, the algebraic representation is
2x + y = 160
And 4x + 2y = 300
For, 2x + y = 160 or y = 160 − 2x, the solution table is;

For 4x + 2y = 300 or y = (300 – 4x)/ 2, the solution table is;

Sol:
Given, half the perimeter of a rectangular garden = 36 m
so, 2(l + b)/2 = 36
(l + b) = 36 ……….(1)
Given, the length is 4 m more than its width.
Let width = x
And length = x + 4
Substituting this in eq(1), we get;
x + x + 4 = 36
2x + 4 = 36
2x = 32
x = 16
Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

Sol:
(i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0
and 2x – 3y = 7 or 2x – 3y – 7 = 0
Comparing the above equations with a1x + b1y + c1=0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ = -7
a₁/a₂ = 3/2, b₁/b₂ = 2/-3, c₁/c₂ = -5/-7 = 5/7

Since, a₁/a₂≠b₁/b₂ the lines intersect each other at a point and have only one possible solution.
Hence, the equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9
Therefore,
a₁ = 2, b₁ = -3, c₁ = -8
a₂ = 4, b₂ = -6, c₂ = -9
a1/a₂ = 2/4 = 1/2, b₁/b₂ = -3/-6 = 1/2, c₁/c₂ = -8/-9 = 8/9
Since, a₁/a₂=b₁/b₂≠c₁/c₂

Therefore, the lines are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

Sol:
(i) Given,
x + y = 14 and x – y = 4 are the two equations.
From 1st equation, we get,
x = 14 – y
Now, put the value of x in second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the value of x;
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.

(ii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From 1st equation, we get,
x = (3 + y)/3
Now, substitute the value of x in the given second equation to get,
9[(3 + y)/3] – 3y = 9
⇒ 3(3+y) – 3y = 9
⇒ 9 + 3y – 3y = 9
⇒ 9 = 9
Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

Sol:
2x + 3y = 11…………………………..(i)
2x – 4y = -24………………………… (ii)
From equation (i), we get;
x = (11 – 3y)/2 ……….…………………………..(iii)
Putting the value of x in equation (ii), we get
2[(11 – 3y)/2] – 4y = −24
11 – 3y – 4y = -24
-7y = -35
y = 5……………………………………..(iv)
Putting the value of y in equation (iii), we get;
x = (11 – 15)/2 = -4/2 = −2
Hence, x = -2, y = 5
Also,
y = mx + 3
5 = -2m +3
-2m = 2
m = -1
Therefore, the value of m is -1.

Sol:
Let the cost of a bat be x and the cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (i)
3x + 5y = 1750 ………………. (ii)
From (i), we get;
y = (3800 – 7x)/6 …………………… (iii)
Substituting (iii) in (ii). we get,
3x + 5[(3800 – 7x)/6] = 1750
⇒3x + (9500/3) – (35x/6) = 1750
3x – (35x/6) = 1750 – (9500/3)
(18x – 35x)/6 = (5250 – 9500)/3
⇒-17x/6 = -4250/3
⇒-17x = -8500
x = 500
Putting the value of x in (iii), we get;
y = (3800 – 7 × 500)/6 = 300/6 = 50
Hence, the cost of a bat is Rs 500 and the cost of a ball is Rs 50.

Sol:
Let the fraction be x/y.
According to the question,
(x + 2)/(y + 2) = 9/11
11x + 22 = 9y + 18
11x – 9y = -4 …………….. (1)
(x + 3)/(y + 3) = 5/6
6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)
From (1), we get
x = (-4 + 9y)/11 …………….. (3)
Substituting the value of x in (2), we get
6[(-4 + 9y)/11] – 5y = -3
-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)
Substituting the value of y in (3), we get
x = (-4 + 81)/11 = 77/11 = 7
Hence, the fraction is 7/9.

Sol: 
Let us assume, the present age of Nuri be x.
And the present age of Sonu is y.
According to the given condition, we can write as;
x – 5 = 3(y – 5)
x – 3y = -10…………………………………..(1)
Now,
x + 10 = 2(y +10)
x – 2y = 10…………………………………….(2)
Subtract eq. 1 from 2, to get,
y = 20 ………………………………………….(3)
Substituting the value of y in eq.1, we get,
x – 3(20) = -10
x – 60 = -10
x = 50
Therefore,
The age of Nuri is 50 years
The age of Sonu is 20 years.

Sol:
Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs. B.
According to the information given,
A + 4B = 27 …………………………………….…………………………. (i)
A + 2B = 21 ……………………………………………………………….. (ii)
When equation (ii) is subtracted from equation (i) we get,
2B = 6
B = 3 …………………………………………………………………………(iii)
Substituting B = 3 in equation (i) we get,
A + 12 = 27
A = 15
Hence, the fixed charge is Rs. 15.
And the Additional charge per day is Rs. 3.

Sol:
8x + 5y = 9 …………………..(1)
3x + 2y = 4 ……………….….(2)
From equation (2) we get;
x = (4 – 2y) / 3 ……………………. (3)
Substituting this value in equation 1, we get
8[(4 – 2y)/3] + 5y = 9
32 – 16y + 15y = 27
-y = -5
y = 5 ……………………………….(4)
Substituting this value in equation (2), we get
3x + 10 = 4
3x = -6
x = -2
Thus, x = -2 and y = 5.

Sol:
Let us consider,
Speed of boat is still water = x km/hr
Speed of current = y km/hr
Now, speed of Ritu, during,
Downstream = x + y km/hr
Upstream = x – y km/hr
As per the question given,
2(x + y) = 20
Or x + y = 10……………………….(1)
And, 2(x – y) = 4
Or x – y = 2………………………(2)
Adding both the eq.1 and 2, we get,
2x = 12
x = 6
Putting the value of x in eq.1, we get,
y = 4
Therefore,
Speed of Ritu is still water = 6 km/hr
Speed of current = 4 km/hr

Sol: 

The given condition for no solution = a1/a2 = b1/b2 ≠ c1/c2 (parallel lines)

(i) Yes.

Given that the pair of the equations are,

2x+4y – 3 = 0 as well as 6x + 12y – 6 = 0

On comparing the equations with ax+ by +c = 0;

We have,

a1 = 2, b1 = 4, c1 = – 3;

a2 = 6, b2 = 12, c2 = – 6;

a1 /a2 = 2/6 = 1/3

b1 /b2 = 4/12 = 1/3

c1 /c2 = – 3/ – 6 = ½

Where, a1/a2 = b1/b2 ≠ c1/c2, that is parallel lines

Therefore, the given pair of the linear equations has no solution. 

(ii) No.

Given that the pair of the equations,

x = 2y or x – 2y = 0

y = 2x or 2x – y = 0;

On comparing the equations with ax+ by +c = 0;

We have,,

a1 = 1, b1 = – 2, c1 = 0;

a2 = 2, b2 = – 1, c2 = 0;

a1 /a2 = ½

b1 /b2 = -2/-1 = 2

Where, a1/a2 ≠ b1/b2.

Therefore, the given pair of the linear equations has a unique solution. 

(iii) No.

Given that the pair of the equations,

3x + y – 3 = 0

2x + 2/3 y = 2

On comparing the equations with ax+ by +c = 0;

We have,

a1 = 3, b1 = 1, c1 = – 3;

a2 = 2, b2 = 2/3, c2 = – 2;

a1 /a2 = 2/6 = 3/2

b1 /b2 = 4/12 = 3/2

c1 /c2 = – 3/-2 = 3/2

Where, a1/a2 = b1/b2 = c1/c2, that is coincident lines

Sol:
Given,
x + 2y = 5
3x + ky + 15 = 0
Also, given that the pair of equations has a unique solution.
Comparing the given equations with standard form,
a₁ = 1, b₁ = 2, c₁ = -5
a₂ = 3, b₂ = k, c₂ = 15
Condition for unique solution is:
a₁/a₂ ≠ b₁/b₂
1/3 ≠ 2/k
k ≠ (2)(3)
k ≠ 6
Thus, for all real values of k except 6, the given pair of equations has a unique solution.

Sol: 

The given pair of the linear equations are as follows:

x + 2y = 1 ……(1)

(a-b)x + (a + b)y = a + b – 2 …..(2)

When we compare with ax + by = c = 0 we have,

a1 = 1, b1 = 2, c1 = – 1

a2 = (a – b), b2 = (a + b), c2 = – (a + b – 2)

a1 /a2 = 1/(a-b)

b1 /b2 = 2/(a+b)

c1 /c2 = 1/(a+b-2)

For infinitely many solutions, the pair of the given linear equations will,

a1/a2 = b1/b2=c1/c2 (here, coincident lines)

Thus, 1/(a-b) = 2/ (a+b) = 1/(a+b-2)

Taking the given first two parts,

1/(a-b) = 2/ (a+b)

a + b = 2(a – b)

a = 3b …(iii)

Taking the given last two parts,

2/ (a+b) = 1/(a+b-2)

2(a + b – 2) = (a + b)

a + b = 4 …(iv)

Also, putting the value of a from Eq. (iii) in Eq. (iv), we have,

3b + b = 4

4b = 4

b = 1

Putting the value for b in Eq. (iii), we get

a = 3

Thus, the values for (a,b) = (3,1) satisfies all the given parts. Thus, the required values of a and b are 3 and 1 respectively, and the given pair of the linear equations has infinitely many solutions.

Sol: 
Given,
2x + 3y = 8….(i)
4x + 6y = 7….(ii)
Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal.
4x + 6y = 16….(iii)
4x + 6y = 7….(iv)
Subtracting (iv) from (iii),
4x + 6y – 4x – 6y = 16 – 7
0 = 9, it is not possible
Therefore, the pair of equations has no solution.

Sol:  Given equations:

12x + 13y = 2

13x + 12y = 136

Let 1x = m and 1y = n.

The equations become:

m2 + n3 = 2

m3 + n2 = 136

Multiply the equations to remove fractions:

3m + 2n = 12

2m + 3n = 13

Using elimination method:

m = 2, n = 3

Substitute back:

1x = 2, so x = 12

1y = 3, so y = 13

Final Answer: x = 12, y = 13