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Pair of Linear Equations in Two Variables - NCERT Questions
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 :
ncert
ncert
From equation (2), we have
ncert
ncert
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),
ncert
ncert
and for equation (2),
ncert
ncert
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
ncert

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
ncert
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
ncert
ncert
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 ncert
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 ncert 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
ncert
And from equation (2), we have ncert
ncert
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
ncert
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
ncert and ncert
ncert
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
ncert
ncert
So, the lines are parallel.

Q 6.

On comparing the ratios ncert 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) ncert
(iv) 5x - 3y = 11; -10x + 6y = -22
(v) ncert

SOLUTION:

Soln. : (i) For 3x + 2y = 5, 2x - 3y = 7
a1= 3, b1= 2, c1= - 5, a2= 2, b2= - 3, c2= - 7
ncert
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
ncert
ncert
So, lines are parallel i.e., the given pair of linear equations has no solution.
∴ It is in consistent pair of equations.
(iii) ncert, 9x - 10y = 14
ncert, a2= 9, b2= -10, c2= -14
ncert, = ncert
and ncert.
ncert 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
ncert
ncert
So, lines are consistent.
⇒ The given pair of linear equations has infinitely many solutions.
Thus, they are consistent.
(v) ncert, 2x + 3y = 12
ncert, a2= 2, b2= 3, c2= - 12
ncert ncert
ncert
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
ncert
ncert
∴ Their graph lines are coincident as shown.
l1 : x + y = 5 ⇒y = 5 - x
ncert
l2 : 2x + 2y = 10 ⇒ x + y = 5 ⇒ y = 5 - x
ncert
ncert
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
ncert
ncert ∴ 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
ncert
ncert So, it is a consistent pair of linear
equations.
ncert
ncert
and ncert
ncert
ncert
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
ncert
ncert 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 ncert perimeter = 36 ⇒y + x = 36
l1 : y = x + 4
ncert
and l2 : x + y = 36 ⇒ y = 36 - x
ncert
ncert
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: ncert
∴ We can have a2 = 3, b2 = 2 and c2 = - 7
∴ The required equation can be 3x + 2y - 7 = 0
(ii) for parallel lines ncert
∴ Any line parallel to 2x + 3y - 8 = 0, can be taken as 2x + 3y - 12 = 0
(iii) For coincident lines ncert
∴ 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
ncert
Plotting, the points (2, 3), (-1, 0) and (3, 4), we get a straight line l1.
Also l2 : 3x + 2y - 12 = 0 ncert
ncert
ncert
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, ncert
(iii) 3x - y = 3, 9x - 3y = 9
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
(v) ncert
(vi) ncert

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) ncert ..... (2)
From (1), we have s = (3 + t) ..... (3)
Substituting this value of s in (2) we get
ncert⇒ 2(3 + t) + 3(t) = 6 × 6
⇒ 6 + 2t + 3t = 36
ncert
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), ncert ...... (3)
substituting the value of y in (2), we have
ncert
ncert
⇒ 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 ncert
From (3), ncertncert
Thus, x = 2 and y = 3
(v) ncert...... (1) ncert...... (2)
From (2), we have ncert...... (3)
From (3) and (1), we have
ncert
ncert
ncert
Substituting y = 0 in (3), we have
ncert
Thus, x = 0 and y = 0
(vi) ncert......(1) ncert ......(2)
From (2), we have ncertncert
ncert ..... (3)
Substituting the value of x in (1), we get
ncert
ncert
⇒ 117 - 47y = - 24 ⇒- 47y = - 24 - 117 = - 141
ncert
Now, substituting y = 3 in (3), we have
ncertncert
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
ncert ....... (3)
Substituting this value of x in (2), we have
ncert ⇒ 11 - 3y - 4y = -24
⇒ -7y = -24 - 11 = -35 ncert
Substituting y = 5 in (3), we get
ncertncert
Thus x = - 2 and y = 5
Now, y = mx + 3 ⇒ 5 = m(-2) + 3
⇒ - 2m = 5 - 3 ncert.
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 ncert
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° ncert
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 ncert ..... (3)
Substituting this value of y in (1), we have
ncert
⇒ 35x + 10500 - 18x = 19000 ⇒ 17x
= 19000 - 10500
ncert
Substituting x = 500 in (3), we have
ncert
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 = ncert
Case I :
ncert⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18 ⇒ 11x - 9y + 4 = 0 ......(1)
Case II :
ncert⇒ 6 (x + 3) = 5 (y + 3)
⇒ 6x + 18 = 5y + 15 ⇒ 6x - 5y + 3 = 0 ......(2)
Now, from (2), ncert ......(3)
Substituting this value of x in (1), we get
ncert
⇒ 55y - 33 - 54y + 24 = 0 ⇒y - 9 = 0 ⇒ y = 9
Now, subsituting y = 9 in (3), we get
ncertncert
x = 7 and y = 9 ncert
(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) ncert

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 ncert
Now, putting ncert in (1), we get
ncert ncert
ncert
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 ncert
From. (1), ncert
ncert
(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
ncert
Putting x = 2 in (1), we get 3(2) + 4y = 10
⇒ 4y = 10 - 6 = 4 ncert
Thus, x = 2 and y = 1
Substitution Method :
3x + 4y = 10 ncert ..... (1)
2x - 2y = 2 ⇒ x - y = 1 ..... (2)
Putting ncert
ncert ⇒ 4x - 10 + 3x = 4 ⇒ 7x = 14
ncert
Putting x = 2 in (1), we get
ncert
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 ⇒ ncert
Substituting the value of y in (1), we get
ncertncert
ncert
ncert
Substitution Method :
3x - 5y - 4 = 0 ncert ..... (1)
9x - 2y - 7 = 0 ..... (2)
Putting ncert
ncert ⇒ 45x - 6x + 8 - 35 = 0
⇒ 39x = 27 ncert
Putting ncert in (1), we get
ncert
Hence, ncert
(iv) Elimination method : ncert...... (1)
ncert ...... (2)
Multiplying equation (2) by 2, we get
ncert ...... (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 + 2yx - 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
ncert
ncert
ncert
ncert
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 ncert
Case-I : ncert
⇒ 3x - 3 = y ⇒ 3x - y - 3 = 0 ...... (1)
Case-II : ncert
⇒ 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
ncert
ncert
ncert
Thus x = 5 and y = 12
ncert
(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
ncert
ncert
ncert
∴ 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 : ncert
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 : ncert
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
ncert
ncert
ncert
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
ncert
ncert
ncert
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) ncert, ncert
(ii) ncert, ncert
(iii) ncert
(iv) ncert
(v) ncert
(vi) ncert
(vii) ncert
(viii) ncert

SOLUTION:

Soln. : (i) ncert
ncert ........ (1)
ncert ........ (2)
Multiplying (1) by ncert
ncert ....... (3)
ncert ....... (4)
Subtracting (3) from (4),
ncert
ncert
Substituting v = 3 in (3), we get
ncert ncert
Thus, v = 3 and u = 2
ncert
(ii) We have ncert....... (1)
ncert ....... (2)
ncert
∴ (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
ncert
ncert
ncert
ncert
The required solution is x = 4, y = 9
(iii) We have ncert ......... (1)
ncert......... (2)
ncert
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
ncert
ncert
ncert
ncert
ncert
Thus, the required solution is ncert
ncert
∴ 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
ncert
ncert
ncert
ncert ⇒ 3 = x - 1 ⇒ x = 4
ncert ⇒ 3 = y - 2 ⇒ y = 5
Thus, the required solution is x = 4, y = 5
(v) We have ncert ....... (1)
ncert ....... (2)
From equation (1), ncert ncert.... (3)
From equation (2), ncert ncert... (4)
ncert
∴ 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
ncert
ncert
ncert
ncert
ncert
Thus, the required solution is x = 1, y =1
(vi) We have 6x + 3y = 6xy ........ (1)
2x + 4y = 5xy ........ (2)
From (1), we get ncert ...... (3)
From (2), we get ncert..... (4)
ncert
∴ (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 = qp = 1
Substituting p = 1 in (5), we get 6q + 3(1) = 6
ncert
ncert
ncert
Thus, the required solution is x = 1, y =2
(vii) We have ncert ....... (1)
ncert ....... (2)
ncert
∴ 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
ncert
ncert
ncert
ncert ncert ........ (5)
ncert ........ (6)
Adding (5) and (6), we have
ncert
From (5), 3 + y = 5 ⇒ y = 2
Thus, the required solution is x = 3, y = 2
(viii) We have ncert ........ (1)
ncert ........ (2)
ncert
∴ (1) and (2) can be expressed as
ncert ........ (3) ncert ........ (4)
Multiplying equation (3) by 1/2, we get
ncert .... (5)
Adding (4) and (5), we get
ncert
ncert
ncert ncert ......(6)
ncert ncert......(7)
Adding (6) and (7),
ncert
Subtracting (7) from (6),
ncert ncert
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
ncert
ncert .........(1)
ncert
x - y = 2 ........ (2)
Adding (1) and (2), we get
ncert ncert
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 ncert
1 man's 1 day work ncert
Since, [2 women + 5men] finish the task in 4 days
ncert ncert
Again [3 women + 6 men], finish the task in 3 days
ncert
ncert
∴ Equation (1) and (2) are expressed as
ncert
ncert
Using cross multiplication method, we get
a1 = 8, b1 = 20, c1 = -1, a2 = 9, b2 = 18, c2 = -1
ncert
ncert
ncert
ncert
ncert
ncert
∴ 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
ncert
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
ncert
ncert ......(1)
Case II : Distance travelled by train = 100 km
Distance travelled by bus = (300 - 100) km = 200 km
Total time = 4 hrs 10 mins
ncert
ncert ncert ......(2)
Multiplying (1) by 4, we get
ncert
Subtracting (3) from (2), we get
ncert
ncert ncert
From (1), ncert
ncert ncert
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] = ncert [Age of Biju]
ncert
According to IInd condition : ncert
⇒ 4x - y = 60 ........ (2)
Adding (1) and (2), we get
3x = 63 ncert
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 : ncert
⇒ 4x - y = 60 ........ (2)
Subtracting (1) from (2), we get
3x = 57 ncert
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 1st 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
ncert
Now, x = - 300 + 2y
= - 300 + 2(170) = - 300 + 340 = 40
Thus, 1st friend has ₹40 and the 2nd 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
1st condition : (x + 10) × (y - 2) = xy
xy - 2x + 10y - 20 = xy
⇒ 2x - 10y + 20 = 0 ...... (1)
2nd 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
ncert
ncert
ncert
ncert
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
ncert
1st condition : ncert
[Number of students in a row × Number of rows
= Number of students]
ncertncert ........ (1)
2nd condition : ncert
ncertncert ........ (2)
ncert
∴ 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
ncert
ncert
ncert
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
ncert ncert
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
ncert
and for equation 3x - y = 3, we get
ncert
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.
ncert
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) ncertax + 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
ncert
From (1), p(1) + qy = p - qp + qy = p - qy = -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
ncert
ncert
ncert
ncert
(iii) We have ncert ........ (1)
ax + by = a2 + b2 ........ (2)
From (1), we have
ncert ........ (3)
Substitutingncert in (2), we have
ncert ncert
ncert
Substituting x = a in (3), we get
y = ncert
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 - 2b2x(-2b) = -2b(a + b)
ncertx = 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 ncert
ncert
(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 ncert
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.
ncert

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 ncert
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°.