## RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A

NCERT Maths Solutions for Ex 4.1 class 10 Quadratic Equations is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.

## RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A.

Other Exercises

Question 1.
Solution:
We know that a second degree of equation is called a quadratic equation. Therefore,

It is not a quadratic equation. It is of degree 5.
So, (i), (ii), (iii), (iv), (vi) and (ix) are quadratic equations.

Question 2.
Solution:
3x² + 2x – 1
= 3x² + 3x – x – 1
= 3x (x + 1) – 1 (x + 1)
= (x + 1) (3x – 1)
Either, x + 1 = 0 ⇒ x = -1
or 3x – 1 =0
⇒ 3x = 1
⇒ x = $$\frac { 1 }{ 3 }$$
Hence, (-1) and $$\frac { 1 }{ 3 }$$ are its roots.

Question 3.
Solution:

Question 4.
Solution:

Solve each of the following quadratic equations.

Question 5.
Solution:
Given : (2x – 3)(3x + 1) = 0
Either 2x – 3 = 0, then 2x = 3 ⇒ x = $$\frac { 3 }{ 2 }$$
or 3x + 1 = 0, then 3x = -1 ⇒ x = $$\frac { -1 }{ 3 }$$
x = $$\frac { 3 }{ 2 }$$ , $$\frac { -1 }{ 3 }$$

Question 6.
Solution:
4×2 + 5x = 0 ⇒ x (4x + 5) = 0
Either x = 0
or 4x + 5 = 0, then 4x = -5 ⇒ x = $$\frac { -5 }{ 4 }$$
x = $$\frac { -5 }{ 4 }$$ or 0

Question 7.
Solution:
3x² – 243 = 0
x² – 81 =0 (Dividing by 3)
⇒ (x)² – (9)² = 0
⇒ (x + 9) (x – 9) = 0
Either, x + 9 = 0, then x = -9
or x – 9 = 0, then x = 9
Hence, x = 9 or -9

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
4x² – 9x = 100
4x² – 9x – 100 = 0

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:
√3 x² + 10x + 7√3 = 0

Question 22.
Solution:

Question 23.
Solution:
3√7 x² + 4x + √7 = 0

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:
3x² – 2√6x + 2 = 0

Question 27.
Solution:

Question 28.
Solution:

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:

Question 32.
Solution:

Question 33.
Solution:

Question 34.
Solution:

Question 35.
Solution:

Question 36.
Solution:

Question 37.
Solution:

Question 38.
Solution:

Question 39.
Solution:

Question 40.
Solution:

Question 41.
Solution:

Question 42.
Solution:

Question 43.
Solution:

Question 44.
Solution:

Question 45.
Solution:

Question 46.
Solution:

Question 47.
Solution:

Question 48.
Solution:

Question 49.
Solution:

Question 50.
Solution:

Question 51.
Solution:

Question 52.
Solution:

Question 53.
Solution:

Question 54.
Solution:

Question 55.
Solution:

Question 56.
Solution:

Question 57.
Solution:

Question 58.
Solution:

Question 59.
Solution:

⇒ x = -2
Roots, x = -2

Question 60.
Solution:

Question 61.
Solution:

Question 62.
Solution:

Question 63.
Solution:

Question 64.
Solution:

Question 65.
Solution:

Question 66.
Solution:

Question 67.
Solution:

Question 68.
Solution:

Question 69.
Solution:

Question 70.
Solution:

Question 71.
Solution:

Question 72.
Solution:

Question 73.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A are helpful to complete your math homework.

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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D

NCERT Maths Solutions for Chapter 3 Ex 3.4 Class 10 acts as the best resource during your learning and helps you score well in your board exams.

## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

### RS Aggarwal Solutions Class 10 Chapter 3

Show that each of the following systems of equations has a unique solution and solve it:
Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

This system has a unique solution.
From (ii), x = 2 + 2y
Substituting the value of x in (i),
2(2 + 2y) + 3y = 18
=> 4 + 4y + 3y = 18
=> 7y = 18 – 4 = 14
=> y = 2
and x = 2 + 2 x 2 = 2 + 4 = 6
x = 6, y = 2

Find the value of k for which each of the following systems of equations has a unique solution:
Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:
Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

=> k (k – 6) = 0
Either k = 0, which is not true, or k – 6 = 0, then k = 6
k = 6

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Find the value of k for which each of the following systems of equations has no solution:
Question 27.
Solution:

Question 28.
Solution:
kx + 3y = 3
12x + ky = 6

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D are helpful to complete your math homework.

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## RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A

NCERT Maths Solutions for Ex 2.1 class 10 Polynomials is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.

## RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A

Other Exercises

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
Question 1.
Solution:

Question 2.
Solution:
x² – 2x – 8
Let f(x) = x² – 2x – 8

Question 3.
Solution:

Question 4.
Solution:
4x² – 4x – 3

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
Zeros of a quadratic polynomial are 2, -6

Question 14.
Solution:

Question 15.
Solution:
Sum of zeros = 8
Product of zeros = 12
Quadratic equation will be x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 8x + 12 = 0
=> x² – 6x – 2x + 12 = 0
=> x (x – 6) – 2 (x – 6) = 0
=> (x – 6) (x – 2) = 0
Either x – 6 = 0, then x = 6
or x – 2 = 0, then x = 2
Zeros are 6, 2
and quadratic polynomial is x² – 8x + 12

Question 16.
Solution:
Sum of zeros = 0
and product of zeros = -1
x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 0x – 1 = 0
=> x² – 1= 0
(x + 1)(x – 1) = 0
Either x + 1 = 0, then x = -1 or x – 1 =0, then x = 1
Zeros are 1, -1
and quadratic polynomial is x² – 1

Question 17.
Solution:
Sum of zeros = $$\frac { 5 }{ 2 }$$
Product of zeros = 1
x² – (Sum of zeros) x + Product of zeros = 0

and quadratic polynomial is 2x² – 5x + 2

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:
One zero of the given polynomial is $$\frac { 2 }{ 3 }$$

=> (x + 3) (x + 3) = 0
x = -3, -3
Hence, other zeros are -3, -3

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2A are helpful to complete your math homework.

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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B

Are you looking for the best Maths NCERT Solutions Chapter 3 Ex 3.2 Class 10? Then, grab them from our page and ace up your preparation for CBSE Class 10 Exams.

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B

Solve for x and y:
Question 1.
Solution:
x + y = 3 …..(i)
4x – 3y = 26 …(ii)
From (i), x = 3 – y
Substituting the value of x in (ii),
4(3 – y) – 3y = 26
=> 12 – 4y – 3y = 26
-7y = 26 – 12 = 14
y = -2
x = 3 – y = 3 – (-2) = 3 + 2 = 5
Hence, x = 5, y = -2

Question 2.
Solution:

Question 3.
Solution:
2x + 3y= 0 ……..(i)
3x + 4y = 5 …….(ii)

Question 4.
Solution:
2x – 3y = 13 ……(i)
7x – 2y = 20 ……….(ii)

Question 5.
Solution:

=> x = -2, y = -5

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Hence, x = $$\frac { 3 }{ 2 }$$ , y = $$\frac { -2 }{ 3 }$$

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Multiplying each term by 10
4x + 3y = 17
0.7x – 2y = 8
Multiply (i) by 2 and (ii) by 3,
8x + 6y = 34
21x – 6y = 24
29x = 58
x = 2
From (i) 4 x 2 + 3y = 17
=> 8 + 3y = 17
=> 3y = 17 – 8 = 9
y = 3
x = 2, y = 3

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
6x + 5y = 7x + 3y + 1
=> 6x + 5y – 7x – 3y = 1
=> -x + 2y = 1
=> 2y – x = 1 …(i)
7x + 3y + 1 = 2(x + 6y – 1)
7x + 3y + 1 = 2x + 12y – 2
=> 7x + 3y – 2x – 12y = -2 – 1
=> 5x – 9y = -3 …..(ii)
From (i), x = 2y – 1
Substituting the value of x in (ii),
5(2y – 1) – 9y = -3
=> 10y – 5 – 9y = -3
=> y = -3 + 5
=> y = +2
x = 2y – 1 = 2 x 2 – 1 = 4 – 1 = 3
x = 3, y = 2

Question 17.
Solution:

and x = y – 4 = 6 – 4 = 2
x = 2, y = 6

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

x = 3, y = -1

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

x = 3, y = 2

Question 28.
Solution:

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:

Question 32.
Solution:
71x + 37y = 253
37x + 71y = 287
108x + 108y = 540
x + y = 5 ………(i) (Dividing by 108)
and subtracting,
34x – 34y = -34
x – y = -1 ……..(ii) (Dividing by 34)
2x = 4 => x = 2
and subtracting,
2y = 6 => y = 3
Hence, x = 2, y = 3

Question 33.
Solution:
217x + 131y = 913 …(i)
131x + 217y = 827 …..(ii)
348x + 348y = 1740
x + 7 = 5 …..(iii) (Dividing by 348)
and subtracting,
86x – 86y = 86
x – y = 1 …(iv) (Dividing by 86)
2x = 6 => x = 3
and subtracting,
2y = 4 => y = 2
x = 3, y = 2

Question 34.
Solution:
23x – 29y = 98 ……(i)
29x – 23y = 110 ……(ii)
52x – 52y = 208
x – y = 4 ……(iii) (Dividing by 52)
and subtracting
-6x – 6y = -12
x + y = 2 …..(iv)
2x = 6 => x = 3
Subtracting (iii) from (iv)
2y = -2 => y = -1
x = 3, y = -1

Question 35.
Solution:

x = 1 and y = 2

Question 36.
Solution:

Question 37.
Solution:

Question 38.
Solution:

Question 39.
Solution:

Question 40.
Solution:

Question 41.
Solution:

Question 42.
Solution:

Question 43.
Solution:

Question 44.
Solution:

a = $$\frac { 1 }{ 2 }$$ , b = $$\frac { 1 }{ 3 }$$

Question 45.
Solution:

Question 46.
Solution:

Question 47.
Solution:

Question 48.
Solution:

Question 49.
Solution:
a²x + b²y = c² ……(i)
b²x + a²y = d² …….(ii)
Multiply (i) by a² and (ii) by b²,

Question 50.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B are helpful to complete your math homework.

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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E

NCERT Maths Solutions for Ex 3.5 class 10 Linear equations is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.

## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Question 1.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
According to the conditions,
5x + 4y = ₹ 5600 …(i)
4x + 3y = ₹ 4340 …(ii)

x = -560
and from (i)
5 x 560 + 4y = 5600
2800 + 4y = 5600
⇒ 4y = 5600 – 2800
⇒ 4y = 2800
⇒ y = 700
Cost of one chair = ₹ 560
and cost of one table = ₹ 700

Question 2.
Solution:
Let the cost of one spoon = ₹ x and cost of one fork = ₹ y
According to the conditions,
23x + 17y = 1770 …(i)
17x + 23y = 1830 …(ii)
40x + 40y = 3600
Dividing by 40,
x + y = 90 …(iii)
and subtracting,
6x – 6y = -60
Dividing by 6,
x – y = -10 …(iv)
2x = 80 ⇒ x = 40
and subtracting,
2y = 100 ⇒ y = 50
Cost of one spoon = ₹ 40
and cost of one fork = ₹ 50

Question 3.
Solution:
Let number of 25-paisa coins = x
and number 50-paisa coins = y
Total number of coins = 50
and total amount = ₹ 19.50 = 1950 paisa
x + y = 50 …(i)
25x + 50y = 1950
⇒ x + 2y = 78 …(ii)
Subtracting (i) from (ii), y = 28
x = 50 – y = 50 – 28 = 22
Number of 25-paisa coins = 22
and 50-paisa coins = 28

Question 4.
Solution:
Sum of two numbers = 137
and difference = 43
Let first number = x
and second number = y
x + y = 137 …..(i)
x – y = 43 ……(ii)
2x = 180 ⇒ x = 90
and subtracting,
2y = 94
y = 47
First number = 90
and second number = 47

Question 5.
Solution:
Let first number = x
and second number = y
According to the conditions,
2x + 3y = 92 …(i)
4x – 7y = 2 …(ii)
Multiply (i) by 2 and (ii) by 1
4x + 6y = 184 …..(iii)
4x – 7y = 2 …….(iv)
Subtracting (iii) from (iv),
13y = 182
y = 14
From (i), 2x + 3y = 92
2x + 3 x 14 = 92
⇒ 2x + 42 = 92
⇒ 2x = 92 – 42 = 50
⇒ x = 25
First number = 25
Second number = 14

Question 6.
Solution:
Let first number = x
and second number = y
According to the conditions,
3x + y=142 …(i)
4x – y = 138 …(ii)
7x = 280
⇒ x = 40
and from (i)
3 x 40 + y = 142
⇒ 120 + y = 142
⇒ y = 142 – 120 = 22
First number = 40,
second number = 22

Question 7.
Solution:
Let first greater number = x
and second smaller number = y
According to the conditions,
2x – 45 = y …(i)
2y – 21 = x …(ii)
Substituting the value of y in (ii),
2 (2x – 45) – 21 = x
⇒ 4x – 90 – 21 = x
⇒ 4x – x = 111
⇒ 3x = 111
⇒ x = 37
From (i),
y = 2 x 37 – 45 = 74 – 45 = 29
The numbers are 37, 29

Question 8.
Solution:
Let larger number = x
and smaller number = y
According to the conditions,
3x = 4 x y + 8 ⇒ 3x = 4y + 8 …….(i)
5y = x x 3 + 5 ⇒ 5y = 3x + 5 …(ii)
Substitute the value of 3x in (ii),
5y = 4y + 8 + 5
⇒ 5y – 4y = 13
⇒ y = 13
and 3x = 4 x 13 + 8 = 60
⇒ x = 20
Larger number = 20
and smaller number = 13

Question 9.
Solution:
Let first number = x and
second number = y
According to the conditions,

⇒ 11x – 44 = 5(2x + 2) – 20
⇒ 11x – 44 = 10x + 10 – 20
⇒ 11x – 10x = 10 – 20 + 44
⇒ x = 34
and y = 2 x 34 + 2 = 68 + 2 = 70
Numbers are 34 and 70

Question 10.
Solution:
Let first number = x
and second number (smaller) = y
According to the conditions,
x – y = 14
and x² – y² = 448
⇒ (x + y) (x – y) = 448
⇒ (x + y) x 14 = 448
⇒ x + y = 32 ……(i)
and x – y = 14 ……(ii)
2x = 46 ⇒ x = 23
and subtracting (i) and (ii),
2y = 18 ⇒ y = 9
Numbers are 23, 9

Question 11.
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Number = x + 10y
By interchanging the digits,
Ones digit = y
and tens digit = x
Number = y + 10x
According to the conditions,
x + y = 12 ………. (i)
y + 10x = x + 10y + 18
⇒ y + 10x – x – 10y = 18
⇒ 9x – 9y = 18
⇒ x – y = 2 …(ii) (Dividing by 9)
2x = 14 ⇒ x = 7
and subtracting,
2y = 10 ⇒ y = 5
Number = 7 + 10 x 5 = 7 + 50 = 57

Question 12.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After reversing the digits,
Ones digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 27 = y + 10x
⇒ y + 10x – x – 10y = -27
⇒ 9x – 9y = -27
⇒ x – y = -3 …(i)
and 7 (x + y) = x + 10y
7x + 7y = x+ 10y
⇒ 7x – x = 10y – 7y
⇒ 6x = 3y
⇒ 2x = y …(ii)
Substituting the value of y in (i)
x – 2x = -3
⇒ -x = -3
⇒ x = 3
y = 2x = 2 x 3 = 6
Number = x + 10y = 3 + 10 x 6 = 3 + 60 = 63

Question 13.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(i)
and x + y= 15 …(ii)
2x = 16
x = 8
and subtracting,
2y = 14
⇒ y = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 14.
Solution:
Let one’s digit of the two digit number = x
and ten’s digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
Now, according to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and 4(x + y) + 3 = x + 10y
4x + 4y + 3 = x + 10y
⇒ 4x + 4y – x – 10y = -3
3x – 6y = -3
⇒ x – 2y = -1 ……..(ii)
Subtracting,
y = 3
and x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5
Number = x + 10y = 5 + 10 x 3 = 5 + 30 = 35

Question 15.
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 9 = y + 10x
⇒ x + 10y – y – 10x = 9
⇒ -9x + 9y = 9
⇒x – y = -1 …(i) (Dividing by -9)

Question 16.
Solution:
Let the one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and xy = 35 …(ii)
Now, (x + y)² = (x – y)² + 4xy = (2)² + 4 x 35 = 4 + 140 = 144 = (12)²
⇒ (x + y) = 12 …(iii)
Subtracting (i) from (iii), we get

Question 17.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits One’s digit = y
Ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y – 63 = y + 10x

Question 18.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + 10y + y + 10x = 121
⇒ 11x + 11y = 121
⇒ x + y = 11 …(i)
x – y = 3 …(ii)
2x = 14 ⇒ x = 7
Subtracting,
2y = 8 ⇒ y = 4
Number = 7 + 10 x 4 = 7 + 40 = 47
or 4 + 10 x 7 = 4 + 70 = 74

Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,
x + y = 8 …(i)

Question 20.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,
$$\frac { x + 2 }{ y }$$ = $$\frac { 1 }{ 2 }$$
$$\frac { x }{ y – 1 }$$ = $$\frac { 1 }{ 3 }$$
⇒ 2x + 4 = y …(i)
3x = y – 1 …(ii)
⇒ 3x = 2x + 4 – 1
⇒ 3x = 2x + 3
⇒ 3x – 2x = 3
⇒ x = 3
and y = 2x + 4 = 2 x 3 + 4 = 6 + 4 = 10
Fraction = $$\frac { 3 }{ 10 }$$

Question 21.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,
y – x = 11
y = 11 + x …(i)

Question 22.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions,

Question 23.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,
x + y = 4 + 2x
⇒ y = 4 + x …(i)

Fraction = $$\frac { x }{ y }$$ = $$\frac { 5 }{ 9 }$$

Question 24.
Solution:
Let first number = x
and second number = y
According to the conditions,
x + y = 16

Question 25.
Solution:
Let in classroom A, the number of students = x
and in classroom B = y
According to the conditions,
x – 10 = y + 10
⇒ x – y = 10 + 10 = 20
⇒ x – y = 20 …(i)
and x + 20 = 2 (y – 20)
⇒ x + 20 = 2y – 40
⇒ x – 2y = -(40 + 20) = -60
x – 2y = -60 …(ii)
Subtracting, y = 80
and x – y = 20
⇒ x – 80 = 20
⇒ x = 20 + 80 = 100
Number of students in classroom A = 100 and in B = 80

Question 26.
Solution:
Let fixed charges = ₹ x
and other charges = ₹ y per km
According to the conditions,
For 80 km,
x + 80y = ₹ 1330 …(i)
and x + 90y = ₹ 1490 …(ii)
Subtracting (i) from (ii),
10y = 160 ⇒ y = 16
and from (i)
x + 80 x 16 = 1330
⇒ x + 1280 = 1330
⇒ x = 1330 – 1280 = 50
Fixed charges = ₹ 50
and rate per km = ₹ 16

Question 27.
Solution:
Let fixed charges of the hostel = ₹ x
and other charges per day = ₹ y
According to the conditions,
x + 25y = 4500 ……..(i)
x + 30y = 5200 ……(ii)
Subtracting (i) from (ii),
5y = 700
y = 140
and from (i),
x + 25 x 140 = 4500
⇒ x + 3500 = 4500
⇒ x = 4500 – 3500 = 1000
Fixed charges = ₹ 1000
and per day charges = ₹ 140

Question 28.
Solution:
Let first investment = ₹ x
and second investment = ₹ y
Rate of interest = 10% p.a. for first kind and 8% per second
Interest is for the first investment = ₹ 1350
and for the second = ₹ 1350 – ₹45 = ₹ 1305
According to the conditions,

Question 29.
Solution:
Ratio in the income of A and B = 5 : 4
Let A’s income = ₹ 5x and
B’s income = ₹ 4x
and ratio in their expenditures = 7 : 5
Let A’s expenditure = 7y
and B’s expenditure = 5y
According to the conditions,
5x – 7y = 9000 …(i)
and 4x – 5y = 9000 …(ii)
Multiply (i) by 5 and (ii) by 7,
25x – 35y = 45000
28x – 37y = 63000
Subtracting, we get
3x = 18000
⇒ x = 6000
A’s income = 5x = 5 x 6000 = ₹ 30000
and B’s income = 4x = 4 x 6000 = ₹ 24000

Question 30.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
In first case,
Profit on chair = 25%
and on table = 10%
and selling price = ₹ 1520
According to the conditions,

Question 31.
Solution:
Distance between two stations A and B = 70 km
Let speed of first car (starting from A) = x km/hr
and speed of second car = y km/hr

According to the conditions,
7x – 7y = 70
⇒ x – y = 10 …(i)
and x + y = 70 …(ii)
2x = 80 ⇒ x = 40
Subtracting (i) and (ii),
2y = 60 ⇒ y = 30
Speed of car A = 40 km/hr
and speed of car B = 30 km/hr

Question 32.
Solution:
Let uniform speed of the train = x km/hr
and time taken = y hours
Distance = x x y = xy km
Case I:
Speed = (x + 5) km/hr
and Time = (y – 3) hours
Distance = (x + 5) (y – 3)
(x + 5) (y – 3) = xy
⇒ xy – 3x + 5y – 15 = xy
-3x + 5y = 15 …(i)
Case II:
Speed = (x – 4) km/hr
and Time = (y + 3) hours
Distance = (x – 4) (y + 3)
(x – 4) (y + 3) = xy
⇒ xy + 3x – 4y – 12 = xy
3x – 4y = 12 …(ii)
y = 27
and from (i),
-3x + 5 x 27 = 15
⇒ -3x + 135 = 15
⇒ -3x = 15 – 135 = -120
⇒ x = 40
Speed of the train = 40 km/hr
and distance = 27 x 40 = 1080 km

Question 33.
Solution:
Let the speed of the train = x km/hr
and speed of taxi = y km/hr
According to the conditions,

Question 34.
Solution:
Distance between stations A and B = 160 km
Let the speed of the car starts from A = x km/hr
and speed of car starts from B = y km/hr
8x – 8y = 160
⇒ x – y = 20 …(i)
and 2x + 2y = 160
⇒ x + y = 80 …(ii)
2x = 100 ⇒ x = 50
and subtracting,
2y = 60 ⇒ y = 30
Speed of car starting from A = 50 km/hr
and from B = 30 km/hr

Question 35.
Solution:
Distance = 8 km
Let speed of sailor in still water = x km/hr
and speed of water = y km/hr
According to the conditions,

Question 36.
Solution:
Let speed of a boat = x km/hr
and speed of stream = y km/hr
According to the conditions,

Question 37.
Solution:
Let a man can do a work in x days
His 1 day’s work = $$\frac { 1 }{ x }$$
and a boy can do a work in y days
His 1 day’s work = $$\frac { 1 }{ y }$$
According to the conditions,

Question 38.
Solution:
Let length of a room = x m
and area = xy m²
According to the conditions,
x = y + 3 …(i)
(x + 3) (y – 2) = xy
xy – 2x + 3y – 6 = xy
-2x + 3y = 6 …(ii)
-2 (y + 3) + 3y = 6 [From (i)]
-2y – 6 + 3y = 6
⇒ y = 6 + 6 = 12
x = y + 3 = 12 + 3 = 15 …(ii)
Length of room = 15 m

Question 39.
Solution:
Let length of a rectangle = x m
Then area = x x y = xy m²
According to the conditions,
(x – 5) (y + 3) = xy – 8
⇒ xy + 3x – 5y – 15 = xy – 8
⇒ 3x – 5y = -8 + 15 = 7 …..(i)
and (x + 3) (y + 2) = xy + 74
⇒ xy + 2x + 3y + 6 = xy + 74
⇒ 2x + 3y = 74 – 6 = 68 …(ii)
Multiply (i) by 3 and (ii) by 5

Question 40.
Solution:
Let length of a rectangle = x m
Then area = xy m²
According to the conditions,
(x + 3) (y – 4) = xy – 67
⇒ xy – 4x + 3y – 12 = xy – 67
⇒ -4x + 3y = -67 + 12 = -55
⇒ 4x – 3y = 55 …(i)
and (x – 1) (y + 4) = xy + 89
⇒ xy + 4x – y – 4 = xy + 89
⇒ 4x – y = 89 + 4 = 93 ….(ii)
⇒ y = 4x – 93
Substituting the value of y in (i),
4x – 3(4x – 93) = 55
⇒ 4x – 12x + 279 = 55
⇒ -8x = 55 – 279 = -224
⇒ x = 28
and y = 4x – 93 = 4 x 28 – 93 = 112 – 93 = 19
Length of rectangle = 28 m

Question 41.
Solution:
Let reservation charges = ₹ x
and cost of full ticket from Mumbai to Delhi
According to the conditions,
x + y = 4150 …(i)
2x + $$\frac { 3 }{ 2 }$$ y = 6255
⇒ 4x + 3y = 12510 …(ii)
From (i), x = 4150 – y
Substituting the value of x in (ii),
4 (4150 – y) + 3y = 12510
⇒ 16600 – 4y + 3y = 12510
-y = 12510 – 16600
-y = -4090
⇒ y = 4090
and x = 4150 – y = 4150 – 4090 = 60
Reservation charges = ₹ 60
and cost of 1 ticket = ₹ 4090

Question 42.
Solution:
Let present age of a man = x years
and age of a son = y years
5 year’s hence,
Man’s age = x + 5 years
and son’s age = y + 5 years
x + 5 = 3 (y + 5) = 3y + 15
⇒ x – 3y = 15 – 5 = 10
x = 10 + 3y …(i)
and 5 years ago,
Man’s age = x – 5 years
and son’s age = y – 5 years
x – 5 = 7 (y – 5) = 7y – 35
x = 7y – 35 + 5 = 7y – 30 …(ii)
From (i) and (ii),
10 + 3y = 7y – 30
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
⇒ y = 10
and x = 10 + 3y = 10 + 3 x 10 = 10 + 30 = 40
Present age of a man = 40 years
and of son’s age = 10 years

Question 43.
Solution:
Let present age of a man = x years
and age of his son = y years
2 years ago,
Man’s age = x – 2 years
Son’s age = y – 2 years
x – 2 = 5 (y – 2)
⇒ x – 2 = 5y – 10
x = 5y – 10 + 2 = 5y – 8 …(i)
2 years later,
Man’s age = x + 2 years
and son’s age = y + 2 years
x + 2 = 3(y + 2) + 8
x + 2 = 3y + 6 + 8
⇒ x = 3y + 6 + 8 – 2 = 3y + 12 …(ii)
From (i) and (ii),
5y – 8 = 3y + 12
⇒ 5y – 3y = 12 + 8
⇒ 2y = 20
⇒ y = 10
and x = 5y – 8 = 5 x 10 – 8 = 50 – 8 = 42
Present age of man = 42 years
and age of son = 10 years

Question 44.
Solution:
Let age of father = x years
and age of his son = y years
According to the conditions,
2y + x = 10 …(i)
2x + y = 95 …(ii)
From (i),
x = 70 – 2y
Substituting the value of x in (ii),
2 (70 – 2y) + y = 95
⇒ 140 – 4y + y = 95
⇒ -3y = 95 – 140 = -45
⇒ -3y = -45
⇒ y = 15
and x = 70 – 2y = 70 – 2 x 15 = 70 – 30 = 40
Age of father = 40 years
and age of his son = 15 years

Question 45.
Solution:
Let present age of a woman = x years
and age of her daughter = y years
According to the conditions,
x = 3y + 3 …(i)
3 years hence,
Age of woman = x + 3 years
and age of her daughter = y + 3 years
x + 3 = 2 (y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒x = 2y + 16 – 3 = 2y + 13 …(ii)
From (i),
3y + 3 = 2y + 13
⇒ 3y – 2y = 13 – 3
⇒ y = 10
and x = 3y + 3 = 3 x 10 + 3 = 30 + 3 = 33
Present age of woman = 33 years
and age of her daughter = 10 years

Question 46.
Solution:
Let cost price of tea set = ₹ x
and of lemon set = ₹ y
According to the conditions,

Question 47.
Solution:
Let fixed charges = ₹ x (for first three days)
and then additional charges for each day = ₹ y
According to the conditions,
Mona paid ₹ 27 for 7 dyas
x + (7 – 3) x y = 27
⇒ x + 4y = 27
and Tanvy paid ₹ 21 for 5 days
x + (5 – 3) y = 21
⇒ x + 2y = 21 …(ii)
Subtracting,
2y = 6 ⇒ y = 3
But x + 2y = 21
⇒ x + 2 x 3 = 21
⇒ x + 6 = 21
⇒ x = 21 – 6 = 15
Fixed charges = ₹ 15
and additional charges per day = ₹ 3

Question 48.
Solution:
Let x litres of 50% solution be mixed with y litres of 25% solution, then
x + y = 10 …(i)

Subtracting (i) from (ii),
x = 6
and x + y = 10
⇒ 6 + y = 10
⇒ y = 10 – 6 = 4
50% solution = 6 litres
and 25% solution = 4 litres

Question 49.
Solution:
Let x g of 18 carat be mixed with y g of 12 carat gold to get 120 g of 16 carat gold, then
x + y = 120 …(i)
Now, gold % in 18-carat gold = $$\frac { 18 }{ 24 }$$ x 100 = 75%

⇒ 3x + 2y = 320 …(ii)
From (i),
x = 120 – y
Substituting the value of x in (ii),
3 (120 – y) + 2y = 320
⇒ 360 – 3y + 2y = 320
⇒ -y = 320 – 360
⇒ -y = -40
⇒ y = 40
and 40 + x = 120
⇒ x = 120 – 40 = 80
Hence, 18 carat gold = 80 g
and 12-carat gold = 40 g

Question 50.
Solution:
Let x litres of 90% pure solution be mixed withy litres of 97% pure solution to get 21 litres of 95% pure solution. Then,
x + y = 21 …(i)

⇒ 90x + 97y = 1995
From (i), x = 21 – y
Substituting the value of x in (ii),
90 (21 – y) + 97y = 1995
⇒ 1890 – 90y + 97y = 1995
⇒ 7y = 1995 – 1890 = 105
⇒ y =15
and x = 21 – y = 21 – 15 = 6
90% pure solution = 6 litres
and 97% pure solution = 15 litres

Question 51.
Solution:
Let larger supplementary angle = x°
and smaller angle = y°
According to the conditions,
x + y = 180° …(i)
x = y + 18° …(ii)
From (i),
y + 18° + y = 180°
⇒ 2y = 180° – 18° = 162°
⇒ 2y = 162°
⇒ y = 81°
and x= 180°- 81° = 99°
Hence, angles are 99° and 81°

Question 52.
Solution:
In ∆ABC,
∠A = x, ∠B = (3x – 2)°, ∠C = y°, ∠C – ∠B = 9°

Question 53.
Solution:

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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C.

### RS Aggarwal Solutions Class 10 Chapter 3

Solve each of the following systems of equations by using the method of cross multiplication:
Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

x = 15, y= 5

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself.

### RS Aggarwal Solutions Class 10 Chapter 3

MCQ
Question 1.
Solution:
(a)

Question 2.
Solution:
(d)

Question 3.
Solution:
(a)

Question 4.
Solution:
(d)

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:
Let first number = x
and second number = y
According to the conditions, x – y = 26 …(i)
and x = 3y …..(ii)
From (i),
3y – y = 26
⇒ 2y = 26
⇒ y = 13
and x = 3 x 13 = 39
Numbers are 39 and 13

Question 11.
Solution:
23x + 29y = 98 …..(i)
29x + 23y = 110 …..(ii)
Adding, we get 52x + 52y = 208
x + y = 4 …..(iii) (Dividing by 52)
and subtracting,
-6x + 6y = -12
x – y = 2. …..(iv) (Dividing by -6)
2x = 6 ⇒ x = 3
Subtracting,
2x = 2 ⇒ y = 1
Hence, x = 3, y = 1

Question 12.
Solution:

x = 1, y = $$\frac { 3 }{ 2 }$$

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to the condition,
5x + 7y = 195 …(i)
7x + 5y= 153 …(ii)
12x + 12y = 348
x + y = 29 ….(iii) (Dividing by 12)
and subtracting,
-2x + 2y = 42
-x + y = 21 …..(iv) (Dividing by -2)
2y = 50 ⇒ y = 25
and from (iv),
-x + 25 = 21 ⇒ -x = 21 – 25 = -4
x = 4
Cost of one pencil = ₹ 4
and cost of one pen = ₹ 25

Question 16.
Solution:
2x – 3y = 1, 4x – 3y + 1 = 0
2x – 3y = 1
2x = 1 + 3y
x = $$\frac { 1 + 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below

Now plot the points (2, 1), (5, 3) and (-1, -1) on the graph and join them to get a line.
Similarly,
4x – 3y + 1 = 0
⇒ 4x = 3y – 1
⇒ x = $$\frac { 3y – 1 }{ 4 }$$

Now plot the points (-1, -1), (-4, -5) and (2, 3) on the graph and join them to get another line which intersects the first line at the point (-1, -1).
Hence, x = -1, y = -1

Question 17.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Now, ∠A = 4x° + 20°, ∠B = 3x° – 5°, ∠C = 4y° and ∠D = 7y° + 5°
But ∠A + ∠C = 180°
4x + 20° + 4y° = 180°
⇒ 4x + 4y = 180° – 20 = 160°
x + y = 40° …(i) (Dividing by 4)
and ∠B + ∠D = 180°
⇒ 3x – 5 + 7y + 5 = 180°
⇒ 3x + 7y = 180° …(ii)
From (i), x = 40° – y
Substituting the value of x in (ii),
3(40° – y) + 7y = 180°
⇒ 120° – 3y + 7y = 180°
⇒ 4y = 180°- 120° = 60°
y = 15°
and x = 40° – y = 40° – 15° = 25°
∠A = 4x + 20 = 4 x 25 + 20 = 100 + 20= 120°
∠B = 3x – 5 = 3 x 25 – 5 = 75 – 5 = 70°
∠C = 4y = 4 x 15 = 60°
∠D = 7y + 5 = 7 x 15 + 5 = 105 + 5 = 110°

Question 18.
Solution:

Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Fraction = $$\frac { x }{ y }$$
According to the conditions,

Question 20.
Solution:

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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS.

### RS Aggarwal Solutions Class 10 Chapter 3

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(c)

Question 2.
Solution:
(c)

Question 3.
Solution:
(a)

Question 4.
Solution:
(d)

Question 5.
Solution:
(a)

Question 6.
Solution:
(b)

Question 7.
Solution:
(c) 4x + 6y = 3xy, 8x + 9y = 5xy
Dividing each term by xy,

Question 8.
Solution:
(a)

Question 9.
Solution:
(c)

Question 10.
Solution:
(b)

Question 11.
Solution:
(d)

Question 12.
Solution:
(b)

Question 13.
Solution:
(a)

Question 14.
Solution:
(d)

Question 15.
Solution:
(d)

Question 16.
Solution:
(d)

Question 17.
Solution:
(d)

Question 18.
Solution:
(d) The system of equations is consistent then their graph lines will be either intersecting or coincident.

Question 19.
Solution:
(a) The pair of lines of equation is inconsistent then the system will not have no solution i.e., their lines will be parallel.

Question 20.
Solution:
(b)

Question 21.
Solution:
(b) ABCD is a cyclic quadrilateral
∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°
∠A + ∠C = 180°
Now, x + y + 10°+ x + y – 30° = 180°
⇒ 2x + 2y – 20 = 180°
⇒ 2x + 2y = 180° + 20° = 200°
⇒ x + y = 100° …(i)
and ∠B + ∠D = 180°
⇒ y + 20° + x + y = 180°
⇒ x + 2y = 180° – 20° = 160° …(ii)
Subtracting,
-y = -60° ⇒ y = 60°
and x + 60° = 100°
⇒ x = 100° – 60° = 40°
Now, ∠B = y + 20° = 60° + 20° = 80°

Question 22.
Solution:
(d) Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + y = 15 …(i)
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(ii)
2x = 16 ⇒ x = 8
and x + y = 15
⇒ 8 + y = 15
⇒ y = 15 – 8 = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 23.
Solution:
(b) Let the numerator of a fractions = x
and denominator = y

Question 24.
Solution:
(d) Let present age of man = x years
and age of his son = y years
5 years hence,
Age of man = (x + 5) years
and age of son = y + 5 years
(x + 5) = 3 (y + 5)
⇒ x + 5 = 3y + 15
x = 3y + 15 – 5
x = 3y + 10 ……(i)
and 5 years earlier
Age of man = x – 5 years
and age of son = y – 5 years
x – 5 = 7 (y – 5)
x – 5 = 7y – 35
⇒ x = 7y – 35 + 5
x = 7y – 30 ……….(ii)
From (i) and (ii),
7y – 30 = 3y + 10
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
y = 10
x = 3y + 10 = 3 x 10 + 10 = 30 + 10 = 40
Present age of father = 40 years

Question 25.
Solution:
(b)

Question 26.
Solution:
(c)

Question 27.
Solution:
(a)

The system has infinitely many solutions.
The lines are coincident.

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## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F.

### RS Aggarwal Solutions Class 10 Chapter 3

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Let first, number = x
and second number = y
x – y = 5

Question 8.
Solution:
Let cost of one pen = ₹ x
and cost of one pencil = ₹ y
According to the conditions,
5x + 8y = 120 …(i)
8x + 5y = 153 …(ii)
13x + 13y = 273
x + y = 21 …(iii) (Dividing by 13)
and subtracting (i) from (ii),
3x – 3y = 33
⇒ x – y = 11 …….(iv) (Dividing by 3)
2x = 32 ⇒ x = 16
Subtracting,
2y = 10 ⇒ y = 5
Cost of 1 pen = ₹ 16
and cost of 1 pencil = ₹ 5

Question 9.
Solution:
Let first number = x
and second number = y
According to the conditions,

and x + y = 80
⇒ x + 15 = 80
x = 80 – 15 = 65
Numbers are : 65, 15

Question 10.
Solution:
Let one’s digit of a two digits number = x
and ten’s digit = y
Number = x + 10y
By reversing its digits One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + y = 10 …(i)
x + 10y – 18 = y + 10x
x+ 10y – y – 10x = 18
⇒ -9x + 9y = 18
⇒ x – y = -2 (Dividing by -9) …..(ii)
2x = 8 ⇒ x = 4
and by subtracting,
2y = 12 ⇒ y = 6
Number = x + 10y = 4 + 10 x 6 = 4 + 60 = 64

Question 11.
Solution:
Let number of stamps of 20p = x
and stamps of 25 p = y
According to the conditions,
x + y = 47 …..(i)
20x + 25y = 1000
4x + 5y = 200 …(ii)
From (i), x = 47 – y
Substituting the value of x in (ii),
4 (47 – y) + 5y = 200
188 – 4y + 5y = 200
⇒ y = 200 – 188 = 12
and x + y = 47
⇒ x + 12 = 47
⇒ x = 47 – 12 = 35
Hence, number of stamps of 20 p = 35
and number of stamps of 25 p = 12

Question 12.
Solution:
Let number of hens = x
and number of cows = y
According to the conditions,
x + y = 48 …..(i)
x x 2 + y x 4 = 140
⇒ 2x + 4y = 140
⇒ x + 2y = 70 ……(ii)
Subtracting (i) from (ii),
y = 22
and x + y = 48
⇒ x + 22 = 48
⇒ x = 48 – 22 = 26
Number of hens = 26
and number of cows = 22

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
12x + 17y = 53 …(i)
17x + 12y = 63 …(ii)
Adding, 29x + 29y = 116
Dividing by 29,
x + y = 4 …(iii)
Subtracting,
-5x + 5y = -10
⇒ x – y = 2 …(iv) (Dividing by -5)
2x = 6 ⇒ x = 3
Subtracting,
2y = 2 ⇒ y = 1
x = 3, y = 1
x + y = 3 + 1 = 4

Question 16.
Solution:

Question 17.
Solution:
kx – y = 2
6x – 2y = 3

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A.

### RS Aggarwal Solutions Class 10 Chapter 3

Solve each of the following systems of equations graphically.
Question 1.
Solution:
2x + 3y = 2 …..(i)
x – 2y = 8 …(ii)
From Eq. (i),
⇒ 2x = 2 – 3y
⇒x = $$\frac { 2 – 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (1, 0), (-2, 2) and (4, -2) on the graph and join them to get a line.
Similarly x – 2y = 8 ⇒ x = 8 + 2y

Now plot the points (6, -1), (4, -2) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at point(4, -2).
x = 4, y = -2

Question 2.
Solution:
3x + 2y = 4
⇒ 3x = 4 – 2y
⇒ x = $$\frac { 4 – 2y }{ 3 }$$
Giving some different values to y, we get corresponding values of x as given below:

Now plot the points (0, 2), (2, -1) and (4, -4) on the graph and join them to get a line. Similarly,
2x – 3y = 7
2x = 3y+ 7
x = $$\frac { 3y+ 7 }{ 2 }$$

Now plot the points on the graph and join them to get another line.
We see that these two lines intersect each other at point (2, -1).
x = 2, y = -1

Question 3.
Solution:
2x + 3y = 8
⇒ 2x = 8 – 3y
x = $$\frac { 8 – 3y }{ 2 }$$
Now, giving some different values to y, we get corresponding values of x as given below

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
x – 2y + 3 = 0
x = 2y – 3

Now, plot the points (-3, 0), (-1, 1) and (1, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (1, 2).
x = 1, y = 2

Question 4.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = $$\frac { 5y – 4 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-2, 0) (3, 2) and (8,4) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
⇒ y = 8 – 2x

Plot the points (1, 6), (2, 1) and (3, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (3, 2).
x = 3, y = 2

Question 5.
Solution:
3x + 2y = 12
⇒ 3x = 12 – 2y
x = $$\frac { 12 – 2y }{ 2 }$$
Giving some different values to y, we get corresponding the values of x as given below:

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
5x – 2y = 4
⇒ 5x = 4 + 2 y
⇒ x = $$\frac { 4 + 2 y }{ 5 }$$

Plot the points (0, -2), (2, 3) and (4, 8) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, 3).
x = 2, y = 3

Question 6.
Solution:
3x + y + 1 = 0 ⇒y = -3x – 1
Giving some different values to x, we get corresponding values of y as given below:

Now, plot the points (0, -1), (-1, 2) and (-2, 5) on the graph and join them to get a line.
Similarly,
2x – 3y + 8 = 0
2x = 3y – 8
x = $$\frac { 3y – 8 }{ 2 }$$

Now, plot the points (-4, 0), (-1, 2) and (2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the points (-1, 2).
x = -1, y = 2

Question 7.
Solution:
2x + 3y + 5 = 0
2x = -3y – 5
x = $$\frac { -3y – 5 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-4, 1), (-1, -1) and (2, -3) on the graph and join them to get a line. Similarly,
3x – 2y – 12 = 0
⇒ 3x = 2y + 12
x = $$\frac { 2y + 12 }{ 3 }$$

Plot the points (4, 0), (0, -6) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -3).
x = 2, y = -3

Question 8.
Solution:
2x – 3y + 13 = 0
⇒ 2x = 3y – 13
⇒ x = $$\frac { 3y – 13 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-5, 1), (-2, 3) and (1, 5) on the graph and join them to get a line.
Similarly,
3x – 2y + 12 = 0
3x = 2y – 12
x = $$\frac { 2y – 12 }{ 3 }$$

Now, plot the points (-4, 0), (-2, 3) and (0, 6) on the graph and join them to get another line.
We see that these lines intersect each other at the point (-2, 3).
x = -2, y = 3

Question 9.
Solution:
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
⇒ x = $$\frac { 4 – 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0) (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
3x – y + 5 = 0
⇒ -y = -5 – 3x
⇒ y = 5 + 3x

Plot the points (0, 5), (-1, 2) and (-2, -1) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (-1, 2).
x = -1, y = 2

Question 10.
Solution:
x + 2y + 2 = 0
⇒ x = – (2y + 2)
Giving some different values to y, we get corresponding the values of x as given below:

Plot the points (-2,0), (0, -1) and (2, -2) on the graph and join them to get a line.
Similarly,
3x + 2y – 2 = 0
⇒ 3x = 2 – 2y
x = $$\frac { 2 – 2y }{ 3 }$$

Plot the points (0, 1), (2, -2) and (-2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -2).
x = 2, y = -2

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:
Question 11.
Solution:
x – y + 3 = 0
⇒ x = y – 3
Giving some different value to y, we get corresponding values of x as given below:

Plot the points (-3, 0), (-1, 2) and (0, 3) on the graph and join them to get a line.
Similarly,
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
x = $$\frac { 4 – 3y }{ 2 }$$

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get another line.
We see that these line intersect each other at (-1, 2) and x-axis at A (-3, 0) and D (2, 0).

Area of ∆BAD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AD x BL
= $$\frac { 1 }{ 2 }$$ x 5 x 2 = 5 sq.units

Question 12.
Solution:
2x – 3y + 4 = 0
⇒ 2x = 3y – 4
x = $$\frac { 3y – 4 }{ 2 }$$
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-2, 0), (1, 2) and (4, 4) on the graph and join them to get a line.
Similarly,
x + 2y – 5 = 0
x = 5 – 2y

Now, plot the points (5, 0), (3, 1) and (1, 2) on the graph and join them to get another line.
We see that there two lines intersect each other at the point B (1, 2) and intersect x- axis at A (-2, 0) and D (5, 0) respectively.

Now, area of ∆BAD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AD x BL
= $$\frac { 1 }{ 2 }$$ x 7 x 2
= 7 sq. units

Question 13.
Solution:
4x – 3y + 4 = 0
⇒ 4x = 3y – 4
x = $$\frac { 3y – 4 }{ 4 }$$
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-1, 0), (2, 4) and (-4, -4) on the graph and join them, to get a line.
Similarly,
4x + 3y – 20 = 0
4x = 20 – 3y
x = $$\frac { 20 – 3y }{ 4 }$$

Plot the points (5, 0), (2, 4) and (-1, 8) on the graph and join them to get a line.

We see that these two lines intersect each other at the point B (2, 4) and intersect x-axis at A (-1, 0) and D (5, 0).
Area ∆BAD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AD x BL
= $$\frac { 1 }{ 2 }$$ x 6 x 4 = 12 sq. units

Question 14.
Solution:
x – y + 1 = 0 ⇒ x = y – 1
Giving some different values toy, we get the values of x as given below:

Plot the points (-1, 0), (0, 1) and (1, 2) on the graph and join them to get a line.
Similarly,
3x + 2y – 12 = 0
⇒ 3x = 12 – 2y
x = $$\frac { 12 – 2y }{ 3 }$$

Plot the points (4, 0), (2, 3) and (0, 6) on the graph and join them to get another line.

We see that these two lines intersect each ohter at the point E (2, 3) and intersect x- axis at A (-1, 0) and D (4, 0).
Area of ∆EAD = $$\frac { 1 }{ 2 }$$ x base x altitude 1
= $$\frac { 1 }{ 2 }$$ x AD x EL
= $$\frac { 1 }{ 2 }$$ x 5 x 3
= $$\frac { 15 }{ 2 }$$
= 7.5 sq. units

Question 15.
Solution:
x – 2y + 2 = 0 ⇒ x = 2y – 2
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (0, 1), (2, 2) on the graph and join them to get a line.
Similarly,
2x + y – 6 = 0
⇒ y = 6 – 2x

Plot the points on the graph and join them to get a line.

We see that these two lines intersect each other at the point C (2, 2) and intersect the x-axis at A (-2, 0) and F (3, 0).
Now, area of ∆CAF = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AF x CL
= $$\frac { 1 }{ 2 }$$ x 5 x 2 = 5 sq. units

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:
Question 16.
Solution:
2x – 3y + 6 = 0
⇒ 2x = 3y – 6
x = $$\frac { 3y – 6 }{ 2 }$$
Giving some different values to y, we get corresponding values of x, as given below:

Now, plot the points (-3, 0), (0, 2), (3, 4) on the graph and join them to get a line.
Similarly,
2x + 3y – 18 = 0
2x = 18 – 3y
x = $$\frac { 18 – 3y }{ 2 }$$

Plot the points (0, 6), (3, 4) and (6, 2) on the graph and join them to get a line.
We see that these two lines intersect each other at C (3, 4) and intersect y-axis at B (0, 2) and D (0, 6).

Now, area of ∆CBD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x BD x CL
= $$\frac { 1 }{ 2 }$$ x 4 x 3 sq. units = 6 sq. units

Question 17.
Solution:
4x – y – 4 = 0
⇒ 4x = y + 4
x = $$\frac { y + 4 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (1, 0), (0, -4) and (2, 4) on the graph and join them to get a line.
Similarly,
3x + 2y – 14 = 0
2y = 14 – 3x
y = $$\frac { 14 – 3x }{ 2 }$$

Now, plot the points (0, 7), (2, 4) and (4, 1) on the graph and join them to get another line.
We see that these two lines intersect each other at C (2, 4) and y-axis at B (0, -4) and D (0, 7) respectively.

Area of ∆CBD = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x BD x CL
= $$\frac { 1 }{ 2 }$$ x 11 x 2 sq. units = 11 sq. units

Question 18.
Solution:
x – y – 5 = 0 ⇒ x = y + 5
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (5, 0), (0, -5) and (1, -4) on the graph and join these to get a line.
Similarly,
3x + 5y – 15 = 0
3x = 15 – 5y
x = $$\frac { 15 – 5y }{ 3 }$$

Plot the points (5, 0), (0, 3) and (-5, 6) on the graph and join them to get a line.
We see that these two lines intersect each other at A (5, 0) and y-axis at B (0, -5) and E (0, 3) respectively.

Now area of ∆ABE = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x BE x AL
= $$\frac { 1 }{ 2 }$$ x 8 x 5 sq. units = 20 sq. units

Question 19.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = $$\frac { 5y – 4 }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (3, 2) and (-7, -2) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
2x = 8 – y
x = $$\frac { 8 – y }{ 2 }$$

Plot the points (4, 0), (3, 2) and (2, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at B (3, 2) and y-axis at G (0, 1) and H (0, 8) respectively.

Now area of ∆EGH = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x GH x EL
= $$\frac { 1 }{ 2 }$$ x 7 x 3 sq. units
= $$\frac { 21 }{ 2 }$$ = 10.5 sq. units

Question 20.
Solution:
5x – y – 7 = 0 ⇒ y = 5x – 7
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, -7), (1, -2), (2, 3) on the graph and join them to get a line.
Similarly,
x – y + 1 = 0 ⇒ x = y – 1

Plot the points (-1, 0), (1, 2) and (2, 3) on the graph and join them to get a line.

We see that these two lines intersect each other at C (2, 3) and intersect y-axis at A (0, -7) and F (0, 1) respectively.
Now, area of ∆CAF = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x AF x CL
= $$\frac { 1 }{ 2 }$$ x 8 x 2 = 8 sq.units

Question 21.
Solution:
2x – 3y – 12 ⇒ 2x = 12 + 3y
x = $$\frac { 12 + 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (3, -2) and (0, -4) on the graph and join them to get a line.
Similarly,
x + 3y = 6 ⇒ x = 6 – 3y

Plot the points (6, 0), (0, 2) and (-6, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at the points A (6, 0) and intersect the y-axis at C (0, -4) and E (0, 2) respectively.

Now area of ∆ACE = $$\frac { 1 }{ 2 }$$ x base x altitude
= $$\frac { 1 }{ 2 }$$ x CE x AO
= $$\frac { 1 }{ 2 }$$ x 6 x 6 sq. units = 18 sq.units

Show graphically that each of the following given systems of equations has infinitely many solutions:
Question 22.
Solution:
2x + 3y = 6
2x = 6 – 3y
x = $$\frac { 6 – 3y }{ 2 }$$
Giving some different values to y, we get the corresponding val ues of x as given below:

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get a line.
Similarly,
4x + 6y = 12
4x = 12 – 6y
x = $$\frac { 12 – 6y }{ 2 }$$

on the graph and join them to get a line.
We see that all the points lie on the same straight line.
This system has infinite many solutions.

Question 23.
Solution:
3x – y = 5
⇒ y = 3x – 5
Giving some different values to x, we get corresponding values of y as shown below:

Now, plot the points (0, -5), (1, -2), (2, 1) on the graph and join them to get a line:
Similarly,
6x – 2y = 10
⇒ 6x = 10 + 2y
x = $$\frac { 10 + 2y }{ 6 }$$

Plot the points (2, 1), (4, 7), (3, 4) on the graph and join them to get another line.
We see that these lines coincide each other.
This system has infinite many solutions.

Question 24.
Solution:
2x + y = 6
y = 6 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 6), (2, 2), (4, -2) on the graph and join them to get a line.
Similarly,
6x + 3y = 18
⇒ 6x = 18 – 3y
⇒ x = $$\frac { 18 – 3y }{ 2 }$$

Now, plot the points (3, 0), (1, 4) and (5, -4) on the graph and join them to get another line.
We see that these two lines coincide each other.
This system has infinitely many solutions

Question 25.
Solution:
x – 2y = 5 ⇒ x = 5 + 2y
Giving some different values to y, we get corresponding values of x as shown below:

Plot the points (5, 0), (3, -1), (1, -2) on the graph and join them to get a line.
similarly,
3x – 6y = 15
⇒ 3x = 15 + 6y
x = $$\frac { 15 + 6y }{ 3 }$$

Plot the points (7, 1), (-1, -3) and (-3, -4) on the graph and join them to get another line.
We see that all the points lie on the same line.
Lines coincide each other.
Hence, the system has infinite many solutions.

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:
Question 26.
Solution:
x – 2y = 6 ⇒ x = 6 + 2y
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (4, -1) and (0, -3) on the graph and join them to get a line.
Similarly,
3x – 6y = 0
⇒ 3x = 6y
⇒ x = 2y

Plot the points (0, 0), (2, 1), (4, 2) on the graph and join them to get a line.

We see that these two lines are parallel i.e., do not intersect each other.
This system has no solution.

Question 27.
Solution:
2x + 3y = 4
⇒ 2x = 4 – 3y
⇒ x = $$\frac { 4 – 3y }{ 2 }$$
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
4x + 6y = 12 ⇒ 4x = 12 – 6y
x = $$\frac { 12 – 6y }{ 4 }$$

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get another line.
We see that two lines are parallel i.e., these do not intersect each other at any point.
Therefore the system has no solution.

Question 28.
Solution:
2x + y = 6, y = 6 – 2x
Giving some different values to x, we get corresponding values ofy as given below:

Plot the points (0, 6), (1, 4) and (3, 0) on the graph and join them to get a line.
Similarly,
6x + 3y = 20 ⇒ 6x = 20 – 3y

We see that these two lines are parallel and do not intersect each other.
Therefore this system has no solution.

Question 29.
Solution:
2x + y = 2 ⇒ y = 2 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 2), (1, 0) and (-2, 6) on the graph and join them to get a line.
Similarly,
2x + y = 6 ⇒ y = 6 – 2x

Plot the points (0, 6), (2, 2) and (3, 0) on the graph and join them to get another line.
ABFD is the trapezium whose vertices are A (0, 2), B (1, 0), F (3, 0), D (0, 6).

Area of trapezium ABFD = Area ∆DOF – Area ∆AOB
= $$\frac { 1 }{ 2 }$$ (DO x OF) – $$\frac { 1 }{ 2 }$$ (AO x OB)
= $$\frac { 1 }{ 2 }$$ (6 x 3) – $$\frac { 1 }{ 2 }$$ (2 x 1) sq.units
= 9 – 1 = 8 sq.units

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself.

Other Exercises

MCQ
Question 1.
Solution:
In the given figure,
PT is the tangent and PQ is the chord of the circle with centre O.
∠OPT = 50°

OP is radius and PT is the tangent.
OP ⊥ PT ⇒ ∠OPT = 90°
∠OPQ + ∠QPT = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 40°
In ∆OPQ,
∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (40° + 40°) = 180° – 80°
= 100° (b)

Question 2.
Solution:
Angle between two radii of a circle = 130°

Then ∠APB = 180° – ∠AOB
= 180°- 130° = 50° (c)

Question 3.
Solution:
In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O
∠APB = 80°

OA is radius of the circle and AP is the tangent
OA ⊥ AP ⇒ ∠OAP = 90°
OP bisects ∠APB,
∠APO = $$\frac { 1 }{ 2 }$$ x 80 = 40°
∠POA = 180° – (90° + 40°)
= 180° – 130° = 50° (b)

Question 4.
Solution:
In the given figure, AD and AE are tangents to the circle with centre O.
BC is the tangent at F which meets AD at C and AE at B
AE = 5 cm

AE and AD are the tangents to the circle
AE = AD = 5 cm
Tangents from an external point drawn to the circle are equal
CD = CF and BE = BF
Now, perimeter of ∆ABC = AB + AC + BC
= AB + AC + BF + CF (BE = BF and CF = CD)
= AB + AC + BE + CD
= AB + BE + AC + CD
= 5 + 5 = 10 cm (b)

Question 5.
Solution:
In the given figure, a quadrilateral ABCD is circumscribed a circle touching its sides at P, Q, R and S respectively.
AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm

A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD …(i)
Now, AP and AS are tangents to the circle
AP = AS = 5 cm …(ii)
Similarly, CQ = CR = 3 cm
BP = BQ = x – 5 = 4
BQ = BC – CQ = 7 – 3 = 4 cm
x – 5 = 4
⇒ x = 4 + 5 = 9cm

Question 6.
Solution:
In the given figure, PA and PB are the tangents drawn from P to the circle.
OA and OB are the radii of the circle and AP and BP are the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
∠A + ∠B = 90° + 90° = 180°
But these are opposite angles of a quadrilateral
A, O, B, P are concyclic

Question 7.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O from an external point P.
∠PBA = 65°,
To find : ∠OAB and ∠APB
In ∆APB
AP = BP (Tangents from P to the circle)
∠PAB = ∠PBA = 65°
∠APB = 180° – (∠PAB + ∠PBA)
= 180° – (65° + 65°) = 180° – 130° = 50°
OA is radius and AP is tangent
OA ⊥ AP
∠OAP = 90°
∠OAB = ∠OAP – ∠PAB = 90° – 65° = 25°
Hence, ∠OAB = 25° and ∠APB = 50°

Question 8.
Solution:
Given : In the figure,
BC and BD are the tangents drawn from B
to the circle with centre O.
∠CBD = 120°
To prove : OB = 2BC
Construction : Join OB.
Proof: OB bisects ∠CBD (OC = OD and BC = BD)

Question 9.
Solution:
(i) A line intersecting a circle in two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinitely many tangents.

Question 10.
Solution:
Given : In a circle, from an external point P, PA and PB are the tangents drawn to the circle with centre O.
To prove : PA = PB
Construction : Join OA, OB and OP.
Proof : OA and OB are the radii of the circle and AP and BP are tangents.

OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
PA = PB (c.p.c.t.)
Hence proved.

Question 11.
Solution:
Given : In a circle with centre O and AB is its diameter.
From A and B, PQ and RS are the tangents drawn to the circle

To prove : PQ || RS
Proof : OA is radius and PAQ is the tangent
OA ⊥ PQ
∠PAO = 90° …(i)
Similarly, OB is the radius and RBS is tangent
∠OBS = 90° …(ii)
From (i) and (ii)
∠PAO = ∠OBS
But there are alternate angles
PQ || RS

Question 12.
Solution:
Given : In the given figure,
In ∆ABC,
AB = AC.
A circle is inscribed the triangle which touches it at D, E and F
To prove : BE = CE
Proof: AD and AF are the tangents drawn from A to the circle
But, AB = AC
AB – AD = AC -AF
⇒ BD = CF
But BD = BE and CF = CE (tangent drawn to the circle)
But BD = CF
BE = CE
Hence proved.

Question 13.
Solution:
Given : In a circle from an external point P, PA and PB are the tangents to the circle
OP, OA and OB are joined.

To prove: ∠POA = ∠POB
Proof: OA and OB are the radii of the circle and PA and PB are the tangents to the circle
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
∠POA = ∠POB (c.p.c.t.)
Hence proved.

Question 14.
Solution:
Given : A circle with centre O, PA and PB are the tangents drawn from A and B which meets at P.
AB is chord of the circle
To prove : ∠PAB = ∠PBA
Construction : Join OA, OB and OP

Proof: OA is radius and AP is tangent
OA ⊥ AP ⇒ ∠OAP = 90°
Similarly, OB ⊥ BP ⇒ ∠OBP = 90°
In ∆OAB, OA = OB (radii of the circle)
∠OAB = ∠OBA
⇒ ∠OAP – ∠OAB = ∠OBP – ∠OBA
⇒ ∠PAB = ∠PBA
Hence proved.

Question 15.
Solution:
Given : A parallelogram ABCD is circumscribed a circle.
To prove : ABCD is a rhombus.
Proof: In a parallelogram ABCD.
Opposite sides are equal and parallel.

AB = CD and AD = BC
Tangents drawn from an external point of a circle to the circle are equal.
AP = AS BP = BQ
CQ = CR and DR = DS
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC [AB = CD and AD = BC]
⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC
But AB = CD and BC = AD
AB = BC = CD = AD
Hence || gm ABCD is a rhombus.

Question 16.
Solution:
Given : O is the centre of two concentric circles with radii 5 cm and 3 cm respectively.
AB is the chord of the larger circle which touches the smaller circle at P.

OP and OA are joined.
To find : Length of AB
Proof: OP is the radius of the smaller circle and touches the smaller circle at P
OP ⊥ AB and also bisects AB at P
AP = PB = $$\frac { 1 }{ 2 }$$ AB
Now, in right ∆OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
Hence AB = 2 x AP = 2 x 4 = 8 cm

Question 17.
Solution:
In the figure, quad. ABCD is circumscribed about a circle which touches its sides at P, Q, R and S respectively

To prove : AB + CD = AD + BC
Proof: Tangents drawn from an external point to a circle are equal
AP = AS
BP = BQ
CR = CQ
DR = DS
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence AB + CD = AD + BC

Question 18.
Solution:
Given : A quad. ABCD circumscribe a circle with centre O and touches at P, Q, R and S respectively
OA, OB, OC and OD are joined forming angles AOB, BOC, COD and DOA
To prove : ∠AOB + ∠COD = 180°
and ∠BOC + ∠AOD = 180°
Construction : Join OP, OQ, OR and OS

Proof: In right ∆AOP and ∆AOS,
Side OP = OS (radii of the same circle)
Hyp. OA = OA (common)
∆AOP = ∆AOS (RHS axiom)
∠1 = ∠2 (c.p.c.t.)
Similarly, we can prove that
∠4 = ∠3
∠5 = ∠6
∠8 = ∠7
∠1 + ∠4 + ∠5 + ∠8 = ∠2 + ∠3 + ∠6 + ∠7
⇒ (∠1 + ∠8) + (∠4 + ∠5) = (∠2 + ∠3) + (∠6 + ∠7)
⇒ ∠AOB + ∠COD = ∠AOD + ∠BOC
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (angles at a point)
∠AOB + ∠COD = ∠AOD + ∠BOC = 180°
Hence proved

Question 19.
Solution:
Given : From an external point P, PA and PB are the tangents drawn to the circle,
OA and OB are joined.
To prove : ∠APB + ∠AOB = 180°
Construction : Join OP.

Proof : Now, in ∆POA and ∆PBO,
OP = OP (common)
PA = PB (Tangents from P to the circle)
OA = OB (Radii of the same circle)
∆POA = ∆PBO (SSS axiom)
∠APO = ∠BPO (c.p.c.t.)
and ∠AOP = ∠BOP (c.p.c.t.)
OA and OB are the radii and PA and PB are the tangents
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = 90° and ∠OBP = 90°
In ∆POA,
∠OAP = 90°
∠APO + ∠AOP = 90°
Similarly, ∠BPO + ∠BOP = 90°
(∠APO + ∠BPO) + (∠AOP + ∠BOP) = 90° + 90°
⇒ ∠APB + ∠AOB = 180°.
Hence proved.

Question 20.
Solution:
Given : PQ is chord of a circle with centre O.
TP and TQ are tangents to the circle
Radius of the circle = 10 cm
i.e. OP = OQ = 10 cm and PQ = 16 cm

To find : The length of TP.
OT bisects the chord PQ at M at right angle.
PM = MQ = $$\frac { 16 }{ 2 }$$ = 8 cm
In right ∆PMO,
OP² = PM² + MO² (Pythagoras Theorem)
⇒ (10)² = (8)² + MO²
⇒ 100 = 64 + MO²
⇒ MO² = 100 – 64 = 36 = (6)²
MO = 6 cm
Let TP = x and TM = y
In right ∆TPM,
TP² = TM² + PM²
⇒ x² = y² + 8²
⇒ x² = y² + 64 …(i)
and in right ∆TPM
OT² = TP² + OP²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y + 36 = y2 + 64 + 100 {From (i)}
⇒ 12y = 64 + 100 – 36 = 128

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself are helpful to complete your math homework.

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