Q 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.

SOLUTION:

Soln. : At present : Let Aftab's age = x years
His daughter's age = y years
Seven years ago : Aftab's age = (x - 7)years
Daughter's age = (y - 7)years
According to the condition, [Aftab's age] = 7[His daughter's age]
⇒ [x - 7] = 7[y - 7] ⇒ x - 7 = 7y - 49
⇒ x - 7y -7 + 49 = 0 ⇒ x - 7y + 42 = 0 ..... (1)
After three years : Aftab's age = (x + 3) years
His daughter's age = (y + 3) years
According to the condition,
[Aftab's age] = 3[His daughter's age]
⇒ [x + 3] = 3[y + 3]
⇒ x + 3 = 3y + 9 ⇒ x - 3y + 3 - 9 = 0
⇒ x - 3y - 6 = 0 .... (2)
Graphical representation of equation (1) and (2) :
From equation (1), we have :


From equation (2), we have


The lines l₁ and l₂ intersect at (42, 12).

Q 2.

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.

SOLUTION:

Soln. : Let the cost of a bat = ₹ x and the cost of a ball = ₹ y
∴ Cost of 3 bats = ₹ 3x and
Cost of 6 balls = ₹ 6y
Again, cost of 1 bat = ₹ x and
Cost of 3 balls = ₹ 3y
Algebraic representation :
Cost of 3 bats + Cost of 6 balls = ₹ 3900
⇒ 3x + 6y = 3900 ⇒ x + 2y = 1300 ..... (1)
Also, cost of 1 bat + cost of 3 balls = ₹ 1300
⇒ x + 3y = 1300 ..... (2)

We can also see from the obtained figure that the straight lines representing the two equations intersect at (1300, 0).

Q 3.

The cost of 2 kg of apples and 1 kg of grapes 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.

SOLUTION:

Soln. : Let the cost of 1 kg of apples = ₹ x
Thus, (1) and (2) are the algebraic representations of the given situation.
Geometrical representation :
We have for equation (1),


and for equation (2),


And the cost of 1 kg of grapes = ₹ y
Algebraic representation :
2x + y = 160 ...... (1) and 4x + 2y = 300
⇒ 2x + y = 150 ...... (2)
Geometrical representation :
We have, for equation (1), l₁ : 2x + y = 160
⇒ y = 160 - 2x

From the equation (2), we have l₂ : 2x + y = 150 ⇒ y = 150 - 2x

Q 4.

Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

SOLUTION:

Soln. : (i) Let the number of boys = x
∴ Number of girls = y
∴ x + y = 10 .... (1)
∴ Number girls = [Number of boys] + 4
∴ y = x + 4 ⇒ x - y = - 4 ..... (2)
Now, from the equation (1), we have l 1 : y = 10 - x

The straight lines l₁ and l₂ are the graphical representations of the equations (1) and (2) respectively. The lines are parallel.
And from the equation (2), we have l₂ : y= x + 4


l₁ and l₂ intersects at the point (3, 7)
∴ The solution of the pair of linear equations is : x = 3, y = 7
∴ Required number of boys = 3 and required number of girls = 7
(ii) Let the cost of a pencil = ₹x and cost of a pen = ₹y
Since cost of 5 pencils + Cost of 7 pens = ₹ 50
⇒ 5x + 7y = 50 ........ (1)
Also cost of 7 pencils + cost of 5 pens = ₹ 46
⇒ 7x + 5y = 46 ........ (2)
Now, from equation (1), we have
These two straight lines intersects at (3, 5).
∴ Cost of a pencil = ₹ 3 and cost of a pen = ₹ 5.

Q 5.

On comparing the ratios find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident :
(i) 5 x - 4 y + 8 = 0, 7 x + 6y - 9 = 0
(ii) 9 x + 3y + 12 = 0, 18 x + 6y + 24 = 0
(iii) 6 x - 3 y + 10 = 0, 2 x - y + 9 = 0

SOLUTION:

Soln. : Comparing the given equations with
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, we have
(i) For, 5x - 4y + 8 = 0, 7x + 6y - 9 = 0

And from equation (2), we have

Plotting the points (10, 0), (3, 5) and (-4, 10), we get a straight line l₁ and plotting the points (8, -2), (3, 5) and (0, 9.2) we get a straight line l₂.
a₁= 5, b₁= -4, c₁= 8, a₂= 7, b₂= 6, c₂= -9

So, the lines are intersecting, i.e., they intersect at a point.
(ii) For, 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
a₁= 9, b₁= 3, c₁= 12, a₂= 18, b₂= 6, c₂= 24
and

So, the lines are coincident.
(iii) For, 6x - 3y + 10 = 0, 2x - y + 9 = 0
a₁= 6, b₁= - 3, c₁= 10, a₂= 2, b₂= - 1, c₂= 9


So, the lines are parallel.

Q 6.

On comparing the ratios find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x - 3y = 7
(ii) 2 x - 3y = 8; 4x - 6y = 9
(iii)
(iv) 5x - 3y = 11; -10x + 6y = -22
(v)

SOLUTION:

Soln. : (i) For 3x + 2y = 5, 2x - 3y = 7
a₁= 3, b₁= 2, c₁= - 5, a₂= 2, b₂= - 3, c₂= - 7

So, lines are intersecting i.e., they intersect at a point.
∴ It is consistent pair of equations.
(ii) For 2x - 3y = 8, 4x - 6y = 9
a₁= 2, b₁= - 3, c₁= - 8, a₂= 4, b₂= - 6, c₂= - 9


So, lines are parallel i.e., the given pair of linear equations has no solution.
∴ It is in consistent pair of equations.
(iii) , 9x - 10y = 14
, a₂= 9, b₂= -10, c₂= -14
, =
and .
So lines are intersecting.
⇒ The given pair of linear equations has one solution.
∴ It is a consistent pair of equations.
(iv) For, 5x - 3y = 11, - 10x + 6y = - 22
a₁= 5, b₁= - 3, c₁= - 11, a₂= - 10, b₂= 6, c₂= 22


So, lines are consistent.
⇒ The given pair of linear equations has infinitely many solutions.
Thus, they are consistent.
(v) , 2x + 3y = 12
, a₂= 2, b₂= 3, c₂= - 12


So, the lines are coincident i.e., they have infinitely many solutions.
⇒ The given pair of linear equations are consistent.

Q 7.

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x - y = 8; 3x - 3y = 16
(iii) 2x + y - 6 = 0, 4x - 2y - 4 = 0
(iv) 2x - 2y - 2 = 0, 4x - 4y - 5 = 0

SOLUTION:

Soln. : (i) For x + y = 5, 2x + 2y = 10
a₁= 1, b₁= 1, c₁= - 5, a₂= 2, b₂= 2, c₂= -10


∴ Their graph lines are coincident as shown.
l₁ : x + y = 5 ⇒y = 5 - x

l₂ : 2x + 2y = 10 ⇒ x + y = 5 ⇒ y = 5 - x


So, lines l₁ and l₂are coinciding. i.e., They have infinitely many solutions and are consistent.
(ii) For x - y = 8, 3x - 3y = 16
∵ a₁= 1, b₁= - 1, c₁= - 8, a₂= 3, b₂= - 3, c₂= - 16

∴ The pair of linear equations
is inconsistent and lines are parallel.
(iii) For 2x + y - 6 = 0, 4x - 2y - 4 = 0
⇒ a₁= 2, b₁= 1, c₁= - 6 and a₂= 4, b₂= -2, c₂= - 4

So, it is a consistent pair of linear
equations.


and


∵ l₁and l₂ intersects at (2, 2)
∴ x = 2 and y = 2
(iv) For 2x - 2y - 2 = 0, 4x - 4y - 5 = 0
⇒ a₁= 2, b₁= -2, c₁= - 2, a₂= 4, b₂= - 4, c₂= - 5

so, the given pair of linear
equations is inconsistent and lines are parallel.

Q 8.

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

SOLUTION:

Soln. : Let the width of the garden = x m and the length of the garden = y m
According to question, 4 + width = length
⇒ 4 + x = y
Also perimeter = 36 ⇒y + x = 36
l₁ : y = x + 4

and l₂ : x + y = 36 ⇒ y = 36 - x


The lines l₁ and l₂ intersect at (16, 20).
∴ x = 16 and y = 20
So, Length = 20 m and Breadth = 16 m

Q 9.

Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

SOLUTION:

Soln. : (i) Let the pair of linear equations be
2x + 3y - 8 = 0 ⇒ a1 = 2, b₁ = 3 and c₁ = - 8
and a₂x + b₂y + c₂ = 0.
For intersecting lines, we have:
∴ We can have a₂ = 3, b₂ = 2 and c₂ = - 7
∴ The required equation can be 3x + 2y - 7 = 0
(ii) for parallel lines
∴ Any line parallel to 2x + 3y - 8 = 0, can be taken as 2x + 3y - 12 = 0
(iii) For coincident lines
∴ Any line parallel to 2x + 3y - 8 = 0 can be 2(2x + 3y - 8 = 0) ⇒ 4x + 6y - 16 = 0

Q 10.

Draw the graphs of the equations x - y + 1 = 0 and 3x + 2y - 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

SOLUTION:

Soln. : x - y + 1 = 0 and 3x + 2y - 12 = 0
l₁ : x - y + 1 = 0 ⇒ x = -1 + y

Plotting, the points (2, 3), (-1, 0) and (3, 4), we get a straight line l₁.
Also l₂ : 3x + 2y - 12 = 0


The lines l₁ and l₂ intersect at (2, 3). Thus, co-ordinates of the vertices of the shaded triangular region are (4, 0), (-1, 0) and (2, 3).

Q 11.

Solve the following pair of linear equations by the substitution method.
(i) x + y = 14, x - y = 4
(ii) s - t = 3,
(iii) 3x - y = 3, 9x - 3y = 9
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
(v)
(vi)

SOLUTION:

Soln. : (i) x + y = 14 ..... (1) x - y = 4 ..... (2)
From the equation (1), we get x = (14 - y)
Substituting this value of x in (2), we
(14 - y) - y = 4 ⇒ 14 - 2y = 4
⇒ –2y=-10 ⇒ y = 5
Now, substituting y = 5 in (1), we have
x + 5 = 14 ⇒ x = 9
Hence, x = 9, y = 5
(ii) s - t = 3 ..... (1) ..... (2)
From (1), we have s = (3 + t) ..... (3)
Substituting this value of s in (2) we get
⇒ 2(3 + t) + 3(t) = 6 × 6
⇒ 6 + 2t + 3t = 36

Substituting t = 6 in (3) we get, s = 3 + 6 = 9
Thus, s = 9, t = 6
(iii) 3x - y = 3 ..... (1) 9x - 3y = 9 ..... (2)
From equation (1), y = (3x - 3)
Substituting this value of y in (2), 9x - 3(3x - 3) = 9
⇒ 9x - 9x + 9 = 9 ⇒ 9 = 9 which is true,
∴ The equations (1) and (2) have infinitely many solutions.
(iv)0.2x + 0.3y = 1.3 ...... (1)
0.4x + 0.5y = 2.3 ...... (2)
From the equation (1), ...... (3)
substituting the value of y in (2), we have


⇒ 0.3 × 0.4x + 0.65 - 0.1x = 0.3 × 2.3
⇒ 0.12x + 0.65 - 0.1x = 0.69
⇒ 0.02x = 0.69 - 0.65 = 0.04
From (3),
Thus, x = 2 and y = 3
(v) ...... (1) ...... (2)
From (2), we have ...... (3)
From (3) and (1), we have



Substituting y = 0 in (3), we have

Thus, x = 0 and y = 0
(vi) ......(1) ......(2)
From (2), we have
..... (3)
Substituting the value of x in (1), we get


⇒ 117 - 47y = - 24 ⇒- 47y = - 24 - 117 = - 141

Now, substituting y = 3 in (3), we have

Thus, x = 2 and y = 3.

Q 12.

Solve 2x + 3y = 11 and 2x - 4y = - 24 and hence find the value of 'm' for which y = mx + 3.

SOLUTION:

Soln. : We have 2x + 3y = 11 ....... (1)
and 2x - 4y = -24 ....... (2)
From (1), we have 2x = 11 - 3y
....... (3)
Substituting this value of x in (2), we have
⇒ 11 - 3y - 4y = -24
⇒ -7y = -24 - 11 = -35
Substituting y = 5 in (3), we get

Thus x = - 2 and y = 5
Now, y = mx + 3 ⇒ 5 = m(-2) + 3
⇒ - 2m = 5 - 3 .
Thus, m = -1.

Q 13.

Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

SOLUTION:

Soln. : (i) Let the two numbers be x and y such that
x > y
∵ Difference between two numbers = 26
∴ x - y = 26 .... (1)
Again one number = 3[the other number]
⇒ x = 3y .... (2)
Substituting x = 3y in (1), we get
3y - y = 26
Now, substituting y = 13 in (2), we have
x = 3(13) ⇒ x = 39
Thus, Two numbers are 39 and 13.
(ii) Let the two angles be x and y such that x > y
∵ The larger angle exceeds the smaller by 18°
∴ x = y + 18° .... (1)
Also, sum of two supplementary angles = 180°
∴ x + y = 180° ..... (2)
Substituting the value of (1) in (2), we get
(18° + y) + y = 180°
⇒ 2y = 180° - 18° = 162°
Substituting y = 81° in (1), we get
x = 18° + 81° = 99°
Thus, x = 99° and y = 81°
(iii) Let the cost of a bat = ₹x and the cost of a ball = ₹y
∵ [cost of 7 bats] + [cost of 6 balls] = ₹3800
⇒ 7x + 6y = 3800 ..... (1)
Again [cost of 3 bats] + [cost of 5 balls] = ₹1750
⇒ 3x + 5y = 1750 ..... (2)
From (2), we have ..... (3)
Substituting this value of y in (1), we have

⇒ 35x + 10500 - 18x = 19000 ⇒ 17x
= 19000 - 10500

Substituting x = 500 in (3), we have

Thus, x = 500 and y = 50
⇒ Cost of a bat = ₹500 and cost of a ball = ₹50
(iv) Let fixed charges be ₹x and charges per km be ₹y
Charges for the journey of 10 km = ₹ 105
∴ x + 10y = 105 ......(1)
and charges for the journey of 15 km = ₹155
∴ x + 15y = 155 ......(2)
From, (1) we have x = 105 - 10y ......(3)
Putting the value of x in (2), we get
(105 - 10y) + 15y = 155 ⇒ 5y = 155 - 105 = 50 ⇒ y = 10
Substituting y = 10 in (3), we get x = 105 - 10(10) ⇒ x = 105 - 100 = 5
Thus, x = 5 and y = 10
⇒ Fixed charges = ₹5 and
Charges per km = ₹10
Now, charges for 25 km = x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255
∴ The charges for 25 km journey = ₹ 255
(v) Let the numerator = x and the denominator = y ∴ Fraction =
Case I :
⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18 ⇒ 11x - 9y + 4 = 0 ......(1)
Case II :
⇒ 6 (x + 3) = 5 (y + 3)
⇒ 6x + 18 = 5y + 15 ⇒ 6x - 5y + 3 = 0 ......(2)
Now, from (2), ......(3)
Substituting this value of x in (1), we get

⇒ 55y - 33 - 54y + 24 = 0 ⇒y - 9 = 0 ⇒ y = 9
Now, subsituting y = 9 in (3), we get

∴ x = 7 and y = 9
(vi) Let the present age of Jacob = x years and the present age of his son = y years
∴ 5 years hence, Age of Jacob = (x + 5)years
Age of his son = (y + 5)years
Now [Age of Jacob] = 3[Age of his son]
∴ x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15
⇒ x - 3y - 10 = 0 ..... (1)
5 years ago : Age of Jacob = (x - 5)years,
Age of his son = (y - 5)years
Also [Age of Jacob] = 7[Age of his son]
∴ (x - 5) = 7(y - 5) ⇒ x - 5 = 7y - 35
⇒ x - 7y + 30 = 0 ..... (2)
From (1), x = [10 + 3y]
Substituting this value of x in (2), we get (10 + 3y) - 7y + 30 = 0
⇒ - 4y = - 40 ⇒y = 10
Now, substituting y = 10 in (3), we get
x = 10 + 3(10) ⇒x = 10 + 30 = 40
Thus, x = 40 and y = 10
⇒ Present age of Jacob = 40 years and Present age his son = 10 years

Q 14.

Solve the following pair of linear equations by the elimination method and the substitution method.
(i) x + y = 5 and 2x - 3y = 4
(ii) 3x + 4y = 10 and 2x - 2y = 2
(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7
(iv)

SOLUTION:

Soln. : (i) Elimination method : x + y = 5 ..... (1)
2x - 3y = 4 .... (2)
Multiplying (1) by 3, we get
3x + 3y = 15 .... (3)
Adding (2) & (3), we get
Now, putting in (1), we get


Substitution Method :
x + y = 5 ⇒y = 5 - x ..... (1) 2x - 3y = 4 ..... (2)
Put y = 5 - x in (2), we get
2x - 3(5 - x) = 4 ⇒ 2x - 15 + 3x = 4 ⇒ 5x = 19
From. (1),

(ii) Elimination method : 3x + 4y = 10 ..... (1)
2x - 2y = 2 ..... (2)
Multiplying equation (2) by 2, we have
⇒ 4x - 4y = 4 ..... (3)
Adding (1) and (3), we get

Putting x = 2 in (1), we get 3(2) + 4y = 10
⇒ 4y = 10 - 6 = 4
Thus, x = 2 and y = 1
Substitution Method :
3x + 4y = 10 ..... (1)
2x - 2y = 2 ⇒ x - y = 1 ..... (2)
Putting
⇒ 4x - 10 + 3x = 4 ⇒ 7x = 14

Putting x = 2 in (1), we get

Hence, x = 2 and y = 1
(iii) Elimination method : 3x - 5y - 4 = 0 ..... (1)
9x = 2y + 7 or 9x - 2y - 7 = 0 ..... (2)
Multiplying equation (1) by (3), we get
⇒9x - 15y - 12 = 0 ...... (3)
Subtracting (2) from (3),
∴ –13y–y 5 = 0 ⇒
Substituting the value of y in (1), we get



Substitution Method :
3x - 5y - 4 = 0 ..... (1)
9x - 2y - 7 = 0 ..... (2)
Putting
⇒ 45x - 6x + 8 - 35 = 0
⇒ 39x = 27
Putting in (1), we get

Hence,
(iv) Elimination method : ...... (1)
...... (2)
Multiplying equation (2) by 2, we get
...... (3)
Adding (1) and (3), we have

Putting = 2 in (2), we get y = 3(2 - 3) = 3(- 1) = - 3
Hence, x = 2 and y = - 3.

Q 15.

Form the pair of linear equations in the following problems and find their solutions if they exist by the elimination method.
(i) If we add 1 to the numerator and subtract 1 from the denominator a fraction reduces to 1. It becomes if we only add 1 to the denominator.
What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

SOLUTION:

Soln. : (i) Let the numerator = x and the denominator = y

Thus x = 3 and y = 5.
Hence, the required fraction =
(ii) Let the present age of Nuri = x years and the present age of Sonu = y years
5 years ago : Age of Nuri = (x - 5) years,
Age of Sonu = (y - 5) years
According to question :
Age of Nuri = 3[Age of Sonu]
⇒ x - 5 = 3[y - 5] ⇒ x - 5 = 3y - 15
⇒ x - 3y + 10 = 0 ....... (1)
10 years later :
Age of Nuri = (x + 10) years,
Age of Sonu = (y + 10) years
According to question :
Age of Nuri = 2[Age of Sonu]
⇒ x + 10 = 2(y + 10) ⇒ x + 10 = 2y + 20
⇒ x - 2y - 10 = 0 ....... (2)
Subtracting (1) from (2),
y - 20 = 0 ⇒ = 20
Putting y = 20 in (1), we get
x - 3(20) + 10 = 0 ⇒ x - 50 = 0 ⇒ x = 50
Thus, x = 50 and y = 20
∴ Age of Nuri = 50 years and
Age of Sonu = 20 years
(iii) Let the digit at unit's place = x and the digit at ten's place = y
∴ The number = 10y + x
The number obtained by reversing the digits
= 10x + y
∵ 9[The number] = 2 [Number obtained by
reversing the digits]
∴ 9[10y + x] = 2[10x + y]
⇒ 90y + 9x = 20x + 2y ⇒ x - 8y = 0 ........ (1)
Also sum of the digits = 9 ∴ x + y = 9 ........ (2)
Subtracting (1) from (2), we have

Putting y = 1 in (2), we get ⇒ x + 1 = 9 ⇒ x = 8
Thus, x = 8 and y = 1
∴ The required number = (10 × 1) + 8 = 10 + 8 = 18
(iv) Let the number of 50 rupees notes = x and the number of 100 rupees notes = y
According to the condition
Total number of notes = 25 ∴ x + y = 25 ....... (1)
∵ The value of all the notes = ₹ 2000
∴ 50x + 100y = 2000 ⇒ x + 2y = 40 ........ (2)
Subtracting equation (1) from (2), we get
y = 15
Putting y = 15 in (1), x + 15 = 25 ⇒ x = 25 - 15 = 10
Thus, x = 10 and y = 15
∴ Number of 50 rupees notes = 10 and
Number of 100 rupees notes = 15
(v) Let the fixed charge (for the three days) = ₹ x
and the additional charge for each extra day = ₹ y
First condition, Charge for 7 days = ₹ 27

∴ Fixed charge = ₹ 15 and
Additional charge per day = ₹ 3

Q 16.

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x - 3y - 3 = 0, 3x - 9y - 2 = 0
(ii) 2x + y = 5, 3x + 2y = 8
(iii) 3x - 5y = 20, 6x - 10y = 40
(iv) x - 3y - 7 = 0, 3x - 3y - 15 = 0

SOLUTION:

Soln. : (i) For x - 3y - 3 = 0, 3x - 9y - 2 = 0
a₁ = 1, b₁ = - 3, c₁ = - 3, a₂ = 3, b₂ = - 9, c₂ = - 2

∴ The given system has no solution.
(ii)
a₁ = 2, b₁ = 1, c₁ = - 5, a₂ = 3, b₂ = 2, c₂ = - 8
We find that
∴ The given system has a unique solution.
To solve the equation, we have

∴ The given system of linear equations has infinitely many solutions.

Now, using cross multiplication method, we have

Thus, x = 4 and y = -1.

Q 17.

(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
2x + 3y = 7, (a - b)x + (a + b)y = 3a + b - 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1, (2k - 1)x + (k - 1)y = 2k + 1

SOLUTION:

Soln. : (i) For 2x + 3y = 7 and
(a - b)x + (a + b)y = (3a + b - 2)
a₁ = 2, b₁ = 3, c₁ = - 7, a₂ = (a - b), b₂ = (a + b),
c₂ = - (3a + b - 2)
For an infinite number of solutions,

From the first two equations, we get
⇒ 2a + 2b = 3a - 3b ⇒ 2a - 3a + 2b + 3b = 0
⇒ -a + 5b = 0 ⇒ a - 5b = 0 ....... (1)
From the last two equations,
⇒ 9a + 3b - 6 = 7a + 7b
⇒ 2a - 4b = 6 ⇒ a - 2b - 3 = 0 ....... (2)
Now, to solve by cross multiplication method, we have

Thus, a = 5 and b = 1.
(ii) For 3x + y - 1 = 0, (2k - 1)x + (k - 1)y - (2k + 1) = 0
a₁ = 3, b₁ = 1, c₁ = -1, a₂ = 2k - 1, b₂ = k - 1,
c₂ = -(2k + 1)
For no solution

Taking first two, we get
3(k - 1) = 2k - 1 ⇒ 3k - 3 = 2k - 1
⇒ 3k - 2k = -1 + 3 ⇒ k = 2

Q 18.

Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9, 3x + 2y = 4

SOLUTION:

Soln. : Method-1 [Substitution method]

Thus, x = -2 and y = 5
Method -2 [Cross Multiplication Method]
For 8x + 5y - 9 = 0, 3x + 2y - 4 = 0
By cross multiplication, we have


⇒ x = 1 × - 2 = - 2 and y = 1 × 5 = 5
Thus x = - 2 and y = 5

Q 19.

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units the area increases by 67 square units. Find the dimensions of the rectangle.

SOLUTION:

Soln. :(i) Let the fixed charges = ₹x and charges of food per day = ₹y
For student A : Number of days = 20
∴ Cost of food for 20 days = ₹20y
According to the problem, x + 20y = 1000
⇒ x + 20y - 1000 = 0 ....... (1)
For student B : Number of days = 26
∴ Cost of food for 26 days = ₹26y
According to the problem : x + 26y = 1180
⇒ x + 26y - 1180 = 0 ....... (2)
Solving these by cross multiplication, we get




Thus x = 400 and y = 30
∴ Fixed charges = ₹ 400 and cost of food per day = ₹ 30
(ii) Let the numerator = x and the denominator = y
Case-I :
⇒ 3x - 3 = y ⇒ 3x - y - 3 = 0 ...... (1)
Case-II :
⇒ 4x = y + 8 ⇒ 4x - y - 8 = 0 ...... (2)
From equations (1) and (2), we have
a₁= 3, b₁= -1, c₁= -3, a₂= 4, b₂= -1, c₂= -8
Solving them by cross multiplication, we get



Thus x = 5 and y = 12

(iii) Let the number of correct answers = x and the number of wrong answers = y
Case-I : Marks for all correct answers (3 × x) = 3x
Mark for all wrong answers = (1 × y) = y
∴ According to the condition : 3x - y = 40 ..... (1)
Case-II : Mark for all correct answers = (4 × x) = 4x
Marks for all wrong answers = (2 × y) = 2y
∴ According to the condition : 4x - 2y = 50
⇒ 2x - y = 25 ..... (2)
From (1) and (2), we have a₁= 3, b₁= -1, c₁= -40,
a₂= 2, b₂= -1, c₂= -25
By cross-multiplication, we get



∴ x = 15 and y = 5
Now, total number of questions = [Number of correct
answers] + [Number of wrong answers]
= 15 + 5 = 20
Thus, required number of questions = 20.
(iv) Let the speed of car-I be x km/hr. and the speed of car-II be y km/hr.
Case-I :
Distance travelled by car-I = AC
⇒ speed × time = 5 × x km, AC = 5x
Distance travelled by car-II, BC = 5y
Since AB = AC - BC, 100 = 5x - 5y
⇒ 5x - 5y - 100 = 0 ⇒ x - y - 20 = 0 ...... (1)
Case-II :
Distance travelled by car-I = AD
∴ AD = 1 × x = x
Distance travelled by car-II in 1 hour = BD
∴ BD = 1 × y = y
Now AB = AD + DB ⇒ 100 = x + y
⇒ x + y = 100 ...... (2)
Using cross-multiplication, x - y - 20 = 0, x + y - 100 = 0, where
a₁= 1, b₁= -1, c₁= -20, a₂= 1, b₂= 1, c₂= -100, we have



Thus, speed of car-I = 60 km/hr,
Speed of car-II = 40 km/hr
(v) Let the length of the rectangle = x units and the breadth of the rectangle = y units
∴ Area of the rectangle = x × y = xy
Condition-I : (Length - 5) × (Breadth + 3) = Area - 9
⇒ (x - 5)(y + 3) = xy - 9 ⇒ 3x - 5y - 15 = - 9
⇒ 3x - 5y - 6 = 0 .... (1)
Condition-II : (Length + 3) × (Breadth + 2) = Area + 67
⇒ (x + 3)(y + 2) = xy + 67 ⇒ 2x + 3y + 6 = 67
⇒ 2x + 3y - 61 = 0 .... (2)
Now, using cross multiplication method in (1) and (2), where
a₁= 3, b₁= - 5, c₁= - 6, a₂= 2, b₂= 3, c₂= - 61, we have



Thus, length of the rectangle = 17 units and breadth of the rectangle = 9 units.

Q 20.

Solve the following pairs of equations by reducing them to a pair of linear equations.
(i) ,
(ii) ,
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

SOLUTION:

Soln. : (i)
........ (1)
........ (2)
Multiplying (1) by
....... (3)
....... (4)
Subtracting (3) from (4),


Substituting v = 3 in (3), we get

Thus, v = 3 and u = 2

(ii) We have ....... (1)
....... (2)

∴ (1) and (2) can be written as
2u + 3v = 2 ........ (3) 4u - 9v = -1 ........ (4)
Here, a₁ = 2, b₁ = 3, c₁ = -2, a₂ = 4, b₂ = -9, c₂ = 1
By cross multiplication, we get




The required solution is x = 4, y = 9
(iii) We have ......... (1)
......... (2)

The equations (1) and (2) becomes,
4p + 3y = 14 ........ (3) 3p - 4y = 23 ........ (4)
Here a₁ = 4, b₁ = 3, c₁ = -14, a₂ = 3, b₂ = - 4, c₂ = -23
Solving (3) and (4) by cross multiplication method, we get





Thus, the required solution is

∴ Equations (1) and (2) can be expressed as
5u + v = 2 ........ (3) 6u - 3v = 1 ........ (4)
Here, a₁ = 5, b₁ = 1, c₁ = -2 and a₂ = 6, b₂ = -3, c₂ = -1
Solving (3) and (4) by cross multiplication method



⇒ 3 = x - 1 ⇒ x = 4
⇒ 3 = y - 2 ⇒ y = 5
Thus, the required solution is x = 4, y = 5
(v) We have ....... (1)
....... (2)
From equation (1), .... (3)
From equation (2), ... (4)

∴ Equation (3) and (4) can be expressed as
7q - 2p - 5 = 0 ........ (5) 8q + 7p -15 = 0 ........ (6)
Here, a₁ = 7, b₁ = -2, c₁ = -5, a₂ = 8, b₂ = 7, c₂ = -15
Using cross multiplication method to solve (5) and (6), we get





Thus, the required solution is x = 1, y =1
(vi) We have 6x + 3y = 6xy ........ (1)
2x + 4y = 5xy ........ (2)
From (1), we get ...... (3)
From (2), we get ..... (4)

∴ (3) and (4) can be expressed as
6q + 3p = 6 ....... (5) 2q + 4p = 5 ........ (6)
Multiply (6) by 3, we get
6q + 12p = 15 ... (7)
Subtracting (5) from (7), we get 9p = q ⇒ p = 1
Substituting p = 1 in (5), we get 6q + 3(1) = 6



Thus, the required solution is x = 1, y =2
(vii) We have ....... (1)
....... (2)

∴ Equations (1) and (2) can be expressed as
10p + 2q = 4 ........ (3) 15p - 5q = -2 ........ (4)
Here a₁ = 10, b₁ = 2, c₁ = - 4, a₂ = 15, b₂ = - 5, c₂ = 2
By cross-multiplication, we get



........ (5)
........ (6)
Adding (5) and (6), we have

From (5), 3 + y = 5 ⇒ y = 2
Thus, the required solution is x = 3, y = 2
(viii) We have ........ (1)
........ (2)

∴ (1) and (2) can be expressed as
........ (3) ........ (4)
Multiplying equation (3) by 1/2, we get
.... (5)
Adding (4) and (5), we get


......(6)
......(7)
Adding (6) and (7),

Subtracting (7) from (6),

Thus, the required solution is x = 1, y = 1

Q 21.

Formulate the following problems as a pair of equations, and hence find their solutions.
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 mintues longer. Find the speed of the train and the bus separately.

SOLUTION:

Soln. (i) Let the speed of Ritu in still water = x km/hr and the speed of the water current = y km/hr
∴ Downstream speed = (x + y) km/hr
Upstream speed = (x - y) km/hr

.........(1)

⇒ x - y = 2 ........ (2)
Adding (1) and (2), we get

From (1), 6 + y = 10 ⇒ y = 10 - 6 = 4
Thus, speed of rowing in still water = 6 km/hr, speed of water current = 4 km/hr
(ii) Let the time taken to finish the task by one woman alone = x days
by one man alone = y days
∴ 1 woman's 1 day work
1 man's 1 day work
Since, [2 women + 5men] finish the task in 4 days

Again [3 women + 6 men], finish the task in 3 days


∴ Equation (1) and (2) are expressed as


Using cross multiplication method, we get
a₁ = 8, b₁ = 20, c₁ = -1, a₂ = 9, b₂ = 18, c₂ = -1






∴ 1 man can finish the work in 36 days and 1 woman can finish the work in 18 days.
(iii) Let the speed of the train = x km/hr and the speed of the bus = y km/hr

Case I : Total journey = 300 km
∴ Journey travelled by train = 60 km
∴ Journey travelled by bus = (300 - 60) km = 240 km
∵ Total time taken = 4 hours

......(1)
Case II : Distance travelled by train = 100 km
Distance travelled by bus = (300 - 100) km = 200 km
Total time = 4 hrs 10 mins

......(2)
Multiplying (1) by 4, we get

Subtracting (3) from (2), we get


From (1),

∴ x = 60
Thus, speed of the train = 60 km per hour and speed of the bus = 80 km per hour

Q 22.

The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

SOLUTION:

Soln. : Let the age of Ani = x years and the age of Biju's = y years
Case I : y > x
According to 1st condition : y - x = 3 ........ (1)
∵ [Age of Ani's father] = 2[Age of Ani] = 2x years
Also, [Age of Biju's sister] = [Age of Biju]

According to II^nd condition :
⇒ 4x - y = 60 ........ (2)
Adding (1) and (2), we get
3x = 63
From (1), y - 21 = 3 ⇒y = 3 + 21 = 24
∴ Age of Ani = 21 years
Age of Biju = 24 years
Case II : x > y
∴ x - y = 3 ........ (1)
According to the condition :
⇒ 4x - y = 60 ........ (2)
Subtracting (1) from (2), we get
3x = 57
From (1), 19 - y = 3 ⇒ y = 16
∴ Age of Ani = 19 years and Age of Biju = 16 years

Q 23.

One says, "Give me a hundred, friend! I shall then become twice as rich as you". The other replies, "If you give me ten, I shall be six times as rich as you". Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
[Hint : x + 100 = 2(y - 100), y + 10 = 6(x - 10)]

SOLUTION:

Soln. : Let the capital of 1^st friend = ₹x, and the capital of 2nd friend = ₹y
According to the condition, x + 100 = 2(y - 100)
⇒ x + 100 - 2y + 200 = 0 ⇒ x - 2y + 300 = 0 ..... (1)
And 6(x - 10) = y + 10 ⇒ 6x - y - 70 = 0 ..... (2) From (1), x = -300 + 2y
From (2), 6x - y - 70 = 0
⇒ 6[-300 + 2y] - y - 70 = 0 ⇒ -1870 + 11y = 0

Now, x = - 300 + 2y
= - 300 + 2(170) = - 300 + 340 = 40
Thus, 1^st friend has ₹40 and the 2^nd friend has ₹170.

Q 24.

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

SOLUTION:

Soln. : Let the actual speed of the train = x km/hr and the actual time taken = y hours
Distance = speed × time
1^st condition : (x + 10) × (y - 2) = xy
⇒ xy - 2x + 10y - 20 = xy
⇒ 2x - 10y + 20 = 0 ...... (1)
2^nd condition : (x - 10) × (y + 3) = xy
⇒ xy + 3x - 10y - 30 = xy ⇒ 3x - 10y- 30 = 0 ...... (2)
Using cross multiplication for solving (1) and (2)
a₁= 2, b₁= -10, c₁= 20 and a₂= 3, b₂= -10, c₂= -30
∴



Thus, the distance covered by the train
= 50 × 12 km = 600 km.

Q 25.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

SOLUTION:

Soln. : Let the number of students = x and the number of rows = y
∴ Number of students in each row

1^st condition :
[Number of students in a row × Number of rows
= Number of students]
........ (1)
2^nd condition :
........ (2)

∴ Equation (1) and (2) can be expressed as
p - 3y + 3 = 0 ...... (3) 2p - 3y - 6 = 0 ...... (4)
Subtracting (3) from (4), we get
p–9=0 ⇒ p = 9
From (3), we get 9 - 3y + 3 = 0 ⇒-3y = - 12



Thus, number of students in the class, x = 36

Q 26.

In a ∆ABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.

SOLUTION:

Sol.: ∵ Sum of angles of a triangle = 180°
∴ ∠A + ∠B + ∠C = 180° ...... (1)
∵ C = 3 ∠B = 2(∠A + ∠B) ...... (2)
From (1) and (2), we have
∠A + ∠B + 2(∠A + ∠B) = 180°
⇒ ∠A + ∠B + 2∠A + 2∠B = 180°
⇒ ∠A + ∠B = 60° ...... (3)
Also, ∠A + ∠B + 3∠B = 180°
⇒ ∠A + 4∠B = 180° ...... (4)
Subtracting (3) from (4), we get

From (4), ∠A + 4(40°) = 180°
⇒ ∠A = 180° - 160° = 20°
Again ∠C = 3∠B = 3 × 40° = 120°
Thus, ∠A = 20°, ∠B = 40° and ∠C = 120°.

Q 27.

Draw the graphs of the equations 5x - y = 5 and 3x - y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.

SOLUTION:

Soln. : To draw the graph of 5x - y = 5, we get

and for equation 3x - y = 3, we get

Plotting the points (1, 0), (2, 5) and (0, -5), we get a straight line l₁. Plotting the points (2, 3), (3, 6) and (0, -3), we get a straight line l₂.

From the figure, obviously, the vertices of the triangle formed are A(1, 0), B(0, -5) and C(0, -3).

Q 28.

Solve the following pair of linear equations
(i) px + qy = p - q, qx - py = p + q
(ii) ax + by = c, bx + ay = 1 + c
(iii) ax + by = a² + b²
(iv) (a - b)x + (a + b)y = a² - 2ab - b² , (a + b)
(x + y) = a² + b²
(v) 152x - 378y = -74, - 378x + 152y = -604

SOLUTION:

Soln. : (i) We have : px + qy = p - q ....... (1)
qx - py = p + q ....... (2)
Multiply (1) by pand (2) by q, we get
p² + qpy = p² - pq .... (3)
q² - qpy = q² + pq .... (4)
Adding (3) and (4), we get
p² + q² x = p² + q²
⇒ (p² + q²)x = p² + q²

From (1), p(1) + qy = p - q ⇒ p + qy = p - q ⇒ y = -1
Thus, the required solution is x = 1, y = -1
(ii) We have ax + by = c ........ (1)
bx + ay= (1 + c) ........ (2)
By cross multiplication, we have
∴



(iii) We have ........ (1)
ax + by = a² + b² ........ (2)
From (1), we have
........ (3)
Substituting in (2), we have


Substituting x = a in (3), we get
y =
Thus, the required solution is x = a, y = b.
(iv) We have : (a - b)x + (a + b)y = a² - 2ab - b² .... (1)
(a + b) (x + y) = a² + b² ..... (2)
From (2), (a + b)x + (a + b)y = a² + b² ..... (3)
Subtracting (3) from (1), we get
x (a-b)-(a-+b)=a² - 2ab - b² - a² - b²
⇒ x[a - b - a - b] = -2ab - 2b² ⇒ x(-2b) = -2b(a + b)
⇒x = a + b
Substituting x = (a + b) in (1), we get
(a - b)(a + b) + (a + b)y = a² - 2ab - b²
⇒ (a + b)y = a² - 2ab - b² - a 2 + b²
⇒ (a + b)y = -2 ab

(v) We have 152x - 378y = - 74 ........ (1)
- 378x + 152y = - 604 ........ (2)
Adding (1) and (2), we have
-226x - 226y = -678
⇒ x + y = 3 ......... (3)
Subtracting (1) from (2), we get
-530x+530y=-530
⇒ -x + y = -1 ⇒ x - y = 1 ........ (4)
Adding (3) and (4), we get
2x = 4 ⇒ x = 2
Subtracting (3) from (4), we get
-2y=-2
Thus, the required solution is x = 2 and y = 1

Q 29.

ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

SOLUTION:

Soln. : ∵ ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°
⇒ [4y + 20] + [- 4x] = 180° ⇒4y - 4x + 20° - 180° = 0
⇒ 4y - 4x - 160° = 0 ⇒ y - x - 40° = 0 ..... (1)
And [3y - 5] + [-7x + 5] = 180° ⇒ 3y - 5 + 5 - 7x - 180° = 0
⇒ 3y - 7x - 180° = 0 ........ (2)
Multiplying (1) by 7, we get
7y - 7x - 280° = 0 .... (3)
Subtracting (3) from (2), we get
- 4y + 100° = 0
Now, substituting y = 25° in (1), we get
- x = 40° - 25° = 15° ⇒ x = -15°
∴ ∠A = 4y + 20° = 4(25°) + 20° = 100° + 20° = 120°
∠B = 3y - 5° = 3(25°) - 5° = 75° - 5° = 70°
∠C = - 4x = -4(-15°) = 60°,
∠D = -7x + 5° = -7(-15°) + 5° = 105° + 5° = 110°
Thus, ∠A = 120°, ∠B = 70°, ∠C = 60°, ∠D = 110°.