Exercise 3.1
ANS: 1
Let the present age of Aftab be x.
And, present age of his daughter = y
Seven years ago,
Age of Aftab = x − 7
Age of his daughter = y − 7
According to the question,
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the question,
Therefore, the algebraic representation is
For
,
The solution table is
For 
,
The solution table is
The graphical representation is as follows.
ANS:2
Let the cost of a bat be Rs x.
And, cost of a ball = Rs y
According to the question, the algebraic representation is
For
,
The solution table is
For 
,
The solution table is
The graphical representation is as follows.
ANS: 3
Let the cost of 1 kg of apples be Rs x.
And, cost of 1 kg of grapes = Rs y
According to the question, the algebraic representation is
For
,
The solution table is
For 4x+ 2y = 300,
The solution table is
The graphical representation is as follows.
Exercise 3.2
ANS: 1
(i) Let the number of girls be x and the number of boys be y.
According to the question, the algebraic representation is
x + y = 10
x − y = 4
For x + y = 10,
x = 10 − y
x = 4 + y
From the figure, it can be observed that these lines intersect each other at point (7, 3).
Therefore, the number of girls and boys in the class are 7 and 3 respectively.
(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.
According to the question, the algebraic representation is
5x + 7y = 50
7x + 5y = 46
For 5x + 7y = 50,
From the figure, it can be observed that these lines intersect each other at point (3, 5).
Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.
ANS:2
(i) 5x− 4y + 8 = 0
7x + 6y − 9 = 0
Comparing these equations with
and
, we obtain
Since
,
Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.
(ii) 9x+ 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with
and, we obtain
Since
,
Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.
(iii)6x− 3y + 10 = 0
2x − y + 9 = 0
Comparing these equations with
and, we obtain
Since
,
Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.
ANS:3
(i) 3x+ 2y = 5
2x − 3y = 7
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
(ii)2x− 3y = 8
4x − 6y = 9
Since
,
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
(iii)
Since
,
Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
(iv)5x− 3 y = 11
− 10x + 6y = − 22
Since
,
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
(v)
Since
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
ANS: 4
(i)x+ y = 5
2x + 2y = 10
Since
,
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
x + y = 5
x = 5 − y
From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.
(ii)x− y = 8
3x − 3y = 16
Since
,
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
(iii)2x+ y − 6 = 0
4x − 2y − 4 = 0
Since
,
Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
2x + y − 6 = 0
y = 6 − 2x
From the figure, it can be observed that these lines are intersecting each other at the only point i.e., (2, 2) and it is the solution for the given pair of equations.
(iv)2x− 2y − 2 = 0
4x − 4y − 5 = 0
Since,
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
ANS: 5
Let the width of the garden be x and length be y.
According to the question,
y −x = 4 (1)
y +x = 36 (2)
y −x = 4
y =x + 4
y +x = 36
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.
ANS: 6
(i)Intersecting lines:
For this condition,
The second line such that it is intersecting the given line is
.
(ii) Parallel lines:
For this condition,
Hence, the second line can be
4x + 6y − 8 = 0
(iii)Coincident lines:
For coincident lines,
Hence, the second line can be
6x + 9y − 24 = 0
ANS: 7
x −y + 1 = 0
x =y − 1
3x+ 2y − 12 = 0
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).
Exercise 3.3
ANS: 1
(i) x+ y = 14 (1)
x − y = 4 (2)
From (1), we obtain
x = 14 − y (3)
Substituting this value in equation (2), we obtain
Substituting this in equation (3), we obtain
(ii)
From (1), we obtain
Substituting this value in equation (2), we obtain
Substituting in equation (3), we obtain
s = 9
s= 9, t = 6
(iii)3x− y = 3 (1)
9x − 3y = 9 (2)
From (1), we obtain
y = 3x − 3 (3)
Substituting this value in equation (2), we obtain
9 = 9
This is always true.
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x − 3
Therefore, one of its possible solutions is x = 1, y = 0.
(iv)
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
(v)
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
x = 0
x= 0, y = 0
(vi)
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
Hence, x = 2, y = 3
ANS: 2
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Putting this value in equation (3), we obtain
Hence, x= −2, y = 5
Also,
ANS: 3
(i) Let the first number be x and the other number be y such that y > x.
According to the given information,
On substituting the value of y from equation (1) into equation (2), we obtain
Substituting this in equation (1), we obtain
y = 39
Hence, the numbers are 13 and 39.
(ii) Let the larger angle be x and smaller angle be y.
We know that the sum of the measures of angles of a supplementary pair is always 180º.
According to the given information,
From (1), we obtain
x = 180º − y (3)
Substituting this in equation (2), we obtain
Putting this in equation (3), we obtain
x = 180º − 81º
= 99º
Hence, the angles are 99º and 81º.
(iii) Let the cost of a bat and a ball be x and y respectively.
According to the given information,
From (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this in equation (3), we obtain
Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.
(iv)Let the fixed charge be Rs x and per km charge be Rs y.
According to the given information,
From (3), we obtain
Substituting this in equation (2), we obtain
Putting this in equation (3), we obtain
Hence, fixed charge = Rs 5
And per km charge = Rs 10
Charge for 25 km = x + 25y
= 5 + 250 = Rs 255
(v) Let the fraction be
.
According to the given information,
From equation (1), we obtain
Substituting this in equation (2), we obtain
Substituting this in equation (3), we obtain
Hence, the fraction is
.
(vi) Let the age of Jacob be x and the age of his son be y.
According to the given information,
From (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.
Exercise 3.4
ANS: 1
(i) By elimination method
Multiplying equation (1) by 2, we obtain
Subtracting equation (2) from equation (3), we obtain
Substituting the value in equation (1), we obtain
By substitution method
From equation (1), we obtain
(5)
Putting this value in equation (2), we obtain
−5y = −6
Substituting the value in equation (5), we obtain
(ii) By elimination method
Multiplying equation (2) by 2, we obtain
Adding equation (1) and (3), we obtain
Substituting in equation (1), we obtain
Hence, x = 2, y = 1
By substitution method
From equation (2), we obtain
(5)
Putting this value in equation (1), we obtain
7y = 7
Substituting the value in equation (5), we obtain
(iii) By elimination method
Multiplying equation (1) by 3, we obtain
Subtracting equation (3) from equation (2), we obtain
Substituting in equation (1), we obtain
∴
By substitution method
From equation (1), we obtain
(5)
Putting this value in equation (2), we obtain
Substituting the value in equation (5), we obtain
∴
(iv)By elimination method
Subtracting equation (2) from equation (1), we obtain
Substituting this value in equation (1), we obtain
Hence, x = 2, y = −3
By substitution method
From equation (2), we obtain
(5)
Putting this value in equation (1), we obtain
5y = −15
Substituting the value in equation (5), we obtain
∴ x = 2, y = −3
ANS: 2
(i)Let the fraction be
.
According to the given information,
Subtracting equation (1) from equation (2), we obtain
x = 3 (3)
Substituting this value in equation (1), we obtain
Hence, the fraction is
.
(ii)Let present age of Nuri = x
and present age of Sonu = y
According to the given information,
Subtracting equation (1) from equation (2), we obtain
y = 20 (3)
Substituting it in equation (1), we obtain
Hence, age of Nuri = 50 years
And, age of Sonu = 20 years
(iii)Let the unit digit and tens digits of the number be xand y respectively. Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the given information,
x + y = 9 (1)
9(10y + x) = 2(10x + y)
88y − 11x = 0
− x + 8y =0 (2)
Adding equation (1) and (2), we obtain
9y = 9
y = 1 (3)
Substituting the value in equation (1), we obtain
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18
(iv)Let the number of Rs 50 notes and Rs 100 notes be x and yrespectively.
According to the given information,
Multiplying equation (1) by 50, we obtain
Subtracting equation (3) from equation (2), we obtain
Substituting in equation (1), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.
(v)Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the given information,
Subtracting equation (2) from equation (1), we obtain
Substituting in equation (1), we obtain
Hence, fixed charge = Rs 15
And Charge per day = Rs 3
Exercise 3.5
ANS: 1
Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication method,
∴ x = 2, y = 1
Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication,
∴
ANS: 2
For infinitely many solutions,
Subtracting (1) from (2), we obtain
Substituting this in equation (2), we obtain
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.
For no solution,
Hence, for k = 2, the given equation has no solution.
ANS:3
From equation (ii), we obtain
Substituting this value in equation (i), we obtain
Substituting this value in equation (ii), we obtain
Hence,
Again, by cross-multiplication method, we obtain
ANS: 4
(i)Letx be the fixed charge of the food and y be the charge for food per day.
According to the given information,
Subtracting equation (1) from equation (2), we obtain
Substituting this value in equation (1), we obtain
Hence, fixed charge = Rs 400
And charge per day = Rs 30
(ii)Let the fraction be
.
According to the given information,
Subtracting equation (1) from equation (2), we obtain
Putting this value in equation (1), we obtain
Hence, the fraction is
.
(iii)Let the number of right answers and wrong answers be x and y
respectively.
According to the given information,
Subtracting equation (2) from equation (1), we obtain
x = 15 (3)
Substituting this in equation (2), we obtain
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20
(iv)Let the speed of 1st car and 2nd car be ukm/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (
) km/h
Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (
) km/h
According to the given information,
Adding both the equations, we obtain
Substituting this value in equation (2), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h
(v) Let length and breadth of rectangle be x unit and y unit respectively.
Area = xy
According to the question,
By cross-multiplication method, we obtain
Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.
Exercise 3.6
ANS: 1
Let
and
, then the equations change as follows.
Using cross-multiplication method, we obtain
Putting
and
in the given equations, we obtain
Multiplying equation (1) by 3, we obtain
6p + 9q = 6 (3)
Adding equation (2) and (3), we obtain
Putting in equation (1), we obtain
Hence,
Substitutingin the given equations, we obtain
By cross-multiplication, we obtain
Putting
and 
in the given equation, we obtain
Multiplying equation (1) by 3, we obtain
Adding (2) an (3), we obtain
Putting this value in equation (1), we obtain
Puttingandin the given equation, we obtain
By cross-multiplication method, we obtain
Puttingandin these equations, we obtain
By cross-multiplication method, we obtain
Hence, x = 1, y = 2
Putting
and
in the given equations, we obtain
Using cross-multiplication method, we obtain
Adding equation (3) and (4), we obtain
Substituting in equation (3), we obtain
y = 2
Hence, x = 3, y = 2
Putting
in these equations, we obtain
Adding (1) and (2), we obtain
Substituting in (2), we obtain
Adding equations (3) and (4), we obtain
Substituting in (3), we obtain
Hence, x = 1, y = 1
ANS: 2
(i)Let the speed of Ritu in still water and the speed of stream be xkm/h
and y km/h respectively.
Speed of Ritu while rowing
Upstream =
km/h
Downstream =
km/h
According to question,
Adding equation (1) and (2), we obtain
Putting this in equation (1), we obtain
y = 4
Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.
(ii)Let the number of days taken by a woman and a man be x and yrespectively.
Therefore, work done by a woman in 1 day =
Work done by a man in 1 day =
According to the question,
Putting
in these equations, we obtain
By cross-multiplication, we obtain
Hence, number of days taken by a woman = 18
Number of days taken by a man = 36
(iii) Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,
Putting
and 
in these equations, we obtain
Multiplying equation (3) by 10, we obtain
Subtracting equation (4) from (5), we obtain
Substituting in equation (3), we obtain
Hence, speed of train = 60 km/h
Speed of bus = 80 km/h
Exercise 3.7
ANS: 1
The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.
Let the age of Ani and Biju be x and y years respectively.
Therefore, age of Ani’s father, Dharam = 2 × x = 2xyears
And age of Biju’s sister Cathy
years
By using the information given in the question,
Case (I) When Ani is older than Biju by 3 years,
x −y = 3 (i)
4x −y = 60 (ii)
Subtracting (i) from (ii), we obtain
3x= 60 − 3 = 57
Therefore, age of Ani = 19 years
And age of Biju = 19 − 3 = 16 years
Case (II) When Biju is older than Ani,
y −x = 3 (i)
4x −y = 60 (ii)
Adding (i) and (ii), we obtain
3x= 63
x = 21
Therefore, age of Ani = 21 years
And age of Biju = 21 + 3 = 24 years
ANS: 2
Let those friends were having Rs x and y with them.
Using the information given in the question, we obtain
x + 100 = 2(y − 100)
x + 100 = 2y − 200
x −2y = −300 (i)
And, 6(x− 10) = (y + 10)
6x− 60 = y + 10
6x− y = 70 (ii)
Multiplying equation (ii) by 2, we obtain
12x− 2y = 140 (iii)
Subtracting equation (i) from equation (iii), we obtain
11x= 140 + 300
11x= 440
x = 40
Using this in equation (i), we obtain
40 −2y = −300
40 + 300 = 2y
2y= 340
y = 170
Therefore, those friends had Rs 40 and Rs 170 with them respectively.
ANS: 3
Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,
Or, d = xt (i)
Using the information given in the question, we obtain
By using equation (i), we obtain
−2x + 10t = 20 (ii)
By using equation (i), we obtain
3x− 10t = 30 (iii)
Adding equations (ii) and (iii), we obtain
x = 50
Using equation (ii), we obtain
(−2) × (50) + 10t = 20
−100 + 10t = 20
10t= 120
t = 12 hours
From equation (i), we obtain
Distance to travel = d = xt
= 50 × 12
= 600 km
Hence, the distance covered by the train is 600 km.
ANS: 4
Let the number of rows be x and number of students in a row be y.
Total students of the class
= Number of rows × Number of students in a row
= xy
Using the information given in the question,
Condition 1
Total number of students = (x − 1) (y + 3)
xy = (x − 1) (y + 3) = xy − y + 3x− 3
3x− y − 3 = 0
3x− y = 3 (i)
Condition 2
Total number of students = (x + 2) (y − 3)
xy= xy + 2y − 3x − 6
3x− 2y = −6 (ii)
Subtracting equation (ii) from (i),
(3x −y) − (3x − 2y) = 3 − (−6)
− y+ 2y = 3 + 6
y = 9
By using equation (i), we obtain
3x− 9 = 3
3x= 9 + 3 = 12
x = 4
Number of rows = x = 4
Number of students in a row = y = 9
Number of total students in a class = xy = 4 × 9 = 36
ANS: 5
Given that,
∠C = 3∠B = 2(∠A + ∠B)
3∠B = 2(∠A + ∠B)
3∠B = 2∠A + 2∠B
∠B = 2∠A
2 ∠A− ∠B = 0 …(i)
We know that the sum of the measures of all angles of a triangle is 180°. Therefore,
∠A + ∠B + ∠C = 180°
∠A + ∠B + 3 ∠B = 180°
∠A + 4 ∠B = 180° …(ii)
Multiplying equation (i) by 4, we obtain
8 ∠A− 4 ∠B = 0 …(iii)
Adding equations (ii) and (iii), we obtain
9 ∠A = 180°
∠A = 20°
From equation (ii), we obtain
20° + 4 ∠B = 180°
4 ∠B = 160°
∠B = 40°
∠C = 3 ∠B
= 3 × 40° = 120°
Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.
ANS: 6
5x− y = 5
Or, y= 5x − 5
The solution table will be as follows.
3x− y = 3
Or, y= 3x − 3
The solution table will be as follows.
The graphical representation of these lines will be as follows.
It can be observed that the required triangle is ΔABC formed by these lines and y-axis.
The coordinates of vertices are A (1, 0), B (0, − 3), C (0, −5).
ANS: 7
(i)px+ qy = p − q … (1)
qx − py = p + q … (2)
Multiplying equation (1) by p and equation (2) by q, we obtain
p2x + pqy = p2 −pq… (3)
q2x − pqy = pq + q2…(4)
Adding equations (3) and (4), we obtain
p2x + q2 x = p2+ q2
(p2 + q2) x = p2+ q2
From equation (1), we obtain
p (1) + qy = p − q
qy = − q
y = − 1
(ii)ax+ by = c… (1)
bx + ay = 1 + c… (2)
Multiplying equation (1) by a and equation (2) by b, we obtain
a2x + aby = ac… (3)
b2x + aby = b + bc…(4)
Subtracting equation (4) from equation (3),
(a2 − b2) x = ac− bc − b
From equation (1), we obtain
ax + by = c
(iii)
Or, bx − ay = 0 … (1)
ax + by = a2 + b2…(2)
Multiplying equation (1) and (2) by b and a respectively, we obtain
b2x − aby = 0 … (3)
a2x + aby = a3 +ab2… (4)
Adding equations (3) and (4), we obtain
b2x + a2x = a3+ ab2
x (b2 + a2) = a (a2+ b2)
x = a
By using (1), we obtain
b (a) − ay = 0
ab − ay = 0
ay = ab
y = b
(iv) (a− b) x + (a + b) y = a2− 2ab − b2… (1)
(a + b) (x + y) = a2 +b2
(a + b) x + (a + b) y = a2+ b2… (2)
Subtracting equation (2) from (1), we obtain
(a − b) x − (a + b) x= (a2 − 2ab − b2)− (a2 + b2)
(a − b − a − b) x= − 2ab − 2b2
− 2bx = − 2b (a + b)
x = a + b
Using equation (1), we obtain
(a − b) (a + b) + (a + b)y = a2 − 2ab − b2
a2 − b2 + (a + b)y = a2− 2ab − b2
(a + b) y = − 2ab
(v) 152x− 378y = − 74
76x − 189y = − 37
…(1)
− 378x + 152y = − 604
− 189x + 76y = − 302 … (2)
Substituting the value of x in equation (2), we obtain
− (189)2 y + 189 × 37 + (76)2y = − 302 × 76
189 × 37 + 302 × 76 = (189)2 y −(76)2 y
6993 + 22952 = (189 − 76) (189 + 76) y
29945 = (113) (265) y
y = 1
From equation (1), we obtain
ANS: 8
We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
Therefore, ∠A + ∠C = 180
4y+ 20 − 4x = 180
− 4x+ 4y = 160
x −y = − 40 (i)
Also, ∠B + ∠D = 180
3y− 5 − 7x + 5 = 180
− 7x+ 3y = 180 (ii)
Multiplying equation (i) by 3, we obtain
3x− 3y = − 120 (iii)
Adding equations (ii) and (iii), we obtain
− 7x+ 3x = 180 − 120
− 4x= 60
x =−15
By using equation (i), we obtain
x −y = − 40
−15− y = − 40
y =−15 + 40 = 25
∠A = 4y + 20 = 4(25) + 20 = 120°
∠B = 3y − 5 = 3(25) − 5 = 70°
∠C = − 4x = − 4(− 15) = 60°
∠D = − 7x + 5 = − 7(−15) + 5 = 110°
ANS: 1
Let the present age of Aftab be x.
And, present age of his daughter = y
Seven years ago,
Age of Aftab = x − 7
Age of his daughter = y − 7
According to the question,
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the question,
Therefore, the algebraic representation is
For
,The solution table is
x
|
− 7
|
0
|
7
|
y
|
5
|
6
|
7
|
,The solution table is
x
|
6
|
3
|
0
|
y
|
0
|
− 1
|
− 2
|
ANS:2
Let the cost of a bat be Rs x.
And, cost of a ball = Rs y
According to the question, the algebraic representation is
For
,The solution table is
x
|
300
|
100
|
− 100
|
y
|
500
|
600
|
700
|
,The solution table is
x
|
300
|
100
|
− 100
|
y
|
500
|
600
|
700
|
ANS: 3
Let the cost of 1 kg of apples be Rs x.
And, cost of 1 kg of grapes = Rs y
According to the question, the algebraic representation is
For
,The solution table is
x
|
50
|
60
|
70
|
y
|
60
|
40
|
20
|
The solution table is
x
|
70
|
80
|
75
|
y
|
10
|
−10
|
0
|
Exercise 3.2
ANS: 1
(i) Let the number of girls be x and the number of boys be y.
According to the question, the algebraic representation is
x + y = 10
x − y = 4
For x + y = 10,
x = 10 − y
- x546y564
x = 4 + y
- x543y10−1
From the figure, it can be observed that these lines intersect each other at point (7, 3).
Therefore, the number of girls and boys in the class are 7 and 3 respectively.
(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.
According to the question, the algebraic representation is
5x + 7y = 50
7x + 5y = 46
For 5x + 7y = 50,
- x310− 4y5010
- x83− 2y− 2512
From the figure, it can be observed that these lines intersect each other at point (3, 5).
Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.
ANS:2
(i) 5x− 4y + 8 = 0
7x + 6y − 9 = 0
Comparing these equations with
and
, we obtainSince
,Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.
(ii) 9x+ 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with
and
, we obtainSince
,Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.
(iii)6x− 3y + 10 = 0
2x − y + 9 = 0
Comparing these equations with
and
, we obtainSince
,Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.
ANS:3
(i) 3x+ 2y = 5
2x − 3y = 7
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
(ii)2x− 3y = 8
4x − 6y = 9
Since
,Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
(iii)
Since
,Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
(iv)5x− 3 y = 11
− 10x + 6y = − 22
Since
,Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
(v)
Since
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
ANS: 4
(i)x+ y = 5
2x + 2y = 10
Since
,Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
x + y = 5
x = 5 − y
- x432y123
- x432y123
From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.
(ii)x− y = 8
3x − 3y = 16
Since
,Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
(iii)2x+ y − 6 = 0
4x − 2y − 4 = 0
Since
,Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
2x + y − 6 = 0
y = 6 − 2x
- x012y642
- x123y024
From the figure, it can be observed that these lines are intersecting each other at the only point i.e., (2, 2) and it is the solution for the given pair of equations.
(iv)2x− 2y − 2 = 0
4x − 4y − 5 = 0
Since
,Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
ANS: 5
Let the width of the garden be x and length be y.
According to the question,
y −x = 4 (1)
y +x = 36 (2)
y −x = 4
y =x + 4
x
|
0
|
8
|
12
|
y
|
4
|
12
|
16
|
x
|
0
|
36
|
16
|
y
|
36
|
0
|
20
|
From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.
ANS: 6
(i)Intersecting lines:
For this condition,
The second line such that it is intersecting the given line is
.(ii) Parallel lines:
For this condition,
Hence, the second line can be
4x + 6y − 8 = 0
(iii)Coincident lines:
For coincident lines,
Hence, the second line can be
6x + 9y − 24 = 0
ANS: 7
x −y + 1 = 0
x =y − 1
x
|
0
|
1
|
2
|
y
|
1
|
2
|
3
|
x
|
4
|
2
|
0
|
y
|
0
|
3
|
6
|
From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).
Exercise 3.3
ANS: 1
(i) x+ y = 14 (1)
x − y = 4 (2)
From (1), we obtain
x = 14 − y (3)
Substituting this value in equation (2), we obtain
Substituting this in equation (3), we obtain
(ii)
From (1), we obtain
Substituting this value in equation (2), we obtain
Substituting in equation (3), we obtain
s = 9
s= 9, t = 6(iii)3x− y = 3 (1)
9x − 3y = 9 (2)
From (1), we obtain
y = 3x − 3 (3)
Substituting this value in equation (2), we obtain
9 = 9
This is always true.
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x − 3
Therefore, one of its possible solutions is x = 1, y = 0.
(iv)
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
(v)
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
x = 0
x= 0, y = 0(vi)
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
Hence, x = 2, y = 3
ANS: 2
From equation (1), we obtain
Substituting this value in equation (2), we obtain
Putting this value in equation (3), we obtain
Hence, x= −2, y = 5
Also,
ANS: 3
(i) Let the first number be x and the other number be y such that y > x.
According to the given information,
On substituting the value of y from equation (1) into equation (2), we obtain
Substituting this in equation (1), we obtain
y = 39
Hence, the numbers are 13 and 39.
(ii) Let the larger angle be x and smaller angle be y.
We know that the sum of the measures of angles of a supplementary pair is always 180º.
According to the given information,
From (1), we obtain
x = 180º − y (3)
Substituting this in equation (2), we obtain
Putting this in equation (3), we obtain
x = 180º − 81º
= 99º
Hence, the angles are 99º and 81º.
(iii) Let the cost of a bat and a ball be x and y respectively.
According to the given information,
From (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this in equation (3), we obtain
Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.
(iv)Let the fixed charge be Rs x and per km charge be Rs y.
According to the given information,
From (3), we obtain
Substituting this in equation (2), we obtain
Putting this in equation (3), we obtain
Hence, fixed charge = Rs 5
And per km charge = Rs 10
Charge for 25 km = x + 25y
= 5 + 250 = Rs 255
(v) Let the fraction be
.According to the given information,
From equation (1), we obtain
Substituting this in equation (2), we obtain
Substituting this in equation (3), we obtain
Hence, the fraction is
.(vi) Let the age of Jacob be x and the age of his son be y.
According to the given information,
From (1), we obtain
Substituting this value in equation (2), we obtain
Substituting this value in equation (3), we obtain
Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.
Exercise 3.4
ANS: 1
(i) By elimination method
Multiplying equation (1) by 2, we obtain
Subtracting equation (2) from equation (3), we obtain
Substituting the value in equation (1), we obtain
By substitution method
From equation (1), we obtain
(5)Putting this value in equation (2), we obtain
−5y = −6
Substituting the value in equation (5), we obtain
(ii) By elimination method
Multiplying equation (2) by 2, we obtain
Adding equation (1) and (3), we obtain
Substituting in equation (1), we obtain
Hence, x = 2, y = 1
By substitution method
From equation (2), we obtain
(5)Putting this value in equation (1), we obtain
7y = 7
Substituting the value in equation (5), we obtain
(iii) By elimination method
Multiplying equation (1) by 3, we obtain
Subtracting equation (3) from equation (2), we obtain
Substituting in equation (1), we obtain
∴
By substitution method
From equation (1), we obtain
(5)Putting this value in equation (2), we obtain
Substituting the value in equation (5), we obtain
∴
(iv)By elimination method
Subtracting equation (2) from equation (1), we obtain
Substituting this value in equation (1), we obtain
Hence, x = 2, y = −3
By substitution method
From equation (2), we obtain
(5)Putting this value in equation (1), we obtain
5y = −15
Substituting the value in equation (5), we obtain
∴ x = 2, y = −3
ANS: 2
(i)Let the fraction be
.According to the given information,
Subtracting equation (1) from equation (2), we obtain
x = 3 (3)
Substituting this value in equation (1), we obtain
Hence, the fraction is
.(ii)Let present age of Nuri = x
and present age of Sonu = y
According to the given information,
Subtracting equation (1) from equation (2), we obtain
y = 20 (3)
Substituting it in equation (1), we obtain
Hence, age of Nuri = 50 years
And, age of Sonu = 20 years
(iii)Let the unit digit and tens digits of the number be xand y respectively. Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the given information,
x + y = 9 (1)
9(10y + x) = 2(10x + y)
88y − 11x = 0
− x + 8y =0 (2)
Adding equation (1) and (2), we obtain
9y = 9
y = 1 (3)
Substituting the value in equation (1), we obtain
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18
(iv)Let the number of Rs 50 notes and Rs 100 notes be x and yrespectively.
According to the given information,
Multiplying equation (1) by 50, we obtain
Subtracting equation (3) from equation (2), we obtain
Substituting in equation (1), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.
(v)Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the given information,
Subtracting equation (2) from equation (1), we obtain
Substituting in equation (1), we obtain
Hence, fixed charge = Rs 15
And Charge per day = Rs 3
Exercise 3.5
ANS: 1
Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication method,
∴ x = 2, y = 1
Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication,
∴
ANS: 2
For infinitely many solutions,
Subtracting (1) from (2), we obtain
Substituting this in equation (2), we obtain
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.
For no solution,
Hence, for k = 2, the given equation has no solution.
ANS:3
From equation (ii), we obtain
Substituting this value in equation (i), we obtain
Substituting this value in equation (ii), we obtain
Hence,
Again, by cross-multiplication method, we obtain
ANS: 4
(i)Letx be the fixed charge of the food and y be the charge for food per day.
According to the given information,
Subtracting equation (1) from equation (2), we obtain
Substituting this value in equation (1), we obtain
Hence, fixed charge = Rs 400
And charge per day = Rs 30
(ii)Let the fraction be
.According to the given information,
Subtracting equation (1) from equation (2), we obtain
Putting this value in equation (1), we obtain
Hence, the fraction is
.(iii)Let the number of right answers and wrong answers be x and y
respectively.
According to the given information,
Subtracting equation (2) from equation (1), we obtain
x = 15 (3)
Substituting this in equation (2), we obtain
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20
(iv)Let the speed of 1st car and 2nd car be ukm/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (
) km/hRespective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (
) km/hAccording to the given information,
Adding both the equations, we obtain
Substituting this value in equation (2), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h
(v) Let length and breadth of rectangle be x unit and y unit respectively.
Area = xy
According to the question,
By cross-multiplication method, we obtain
Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.
Exercise 3.6
ANS: 1
Let
and, then the equations change as follows.Using cross-multiplication method, we obtain
Putting
andin the given equations, we obtainMultiplying equation (1) by 3, we obtain
6p + 9q = 6 (3)
Adding equation (2) and (3), we obtain
Putting in equation (1), we obtain
Hence,
Substituting
in the given equations, we obtainBy cross-multiplication, we obtain
Putting
and in the given equation, we obtainMultiplying equation (1) by 3, we obtain
Adding (2) an (3), we obtain
Putting this value in equation (1), we obtain
Putting
andin the given equation, we obtainBy cross-multiplication method, we obtain
Putting
andin these equations, we obtainBy cross-multiplication method, we obtain
Hence, x = 1, y = 2
Putting
andin the given equations, we obtainUsing cross-multiplication method, we obtain
Adding equation (3) and (4), we obtain
Substituting in equation (3), we obtain
y = 2
Hence, x = 3, y = 2
Putting
in these equations, we obtainAdding (1) and (2), we obtain
Substituting in (2), we obtain
Adding equations (3) and (4), we obtain
Substituting in (3), we obtain
Hence, x = 1, y = 1
ANS: 2
(i)Let the speed of Ritu in still water and the speed of stream be xkm/h
and y km/h respectively.
Speed of Ritu while rowing
Upstream =
km/hDownstream =
km/hAccording to question,
Adding equation (1) and (2), we obtain
Putting this in equation (1), we obtain
y = 4
Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.
(ii)Let the number of days taken by a woman and a man be x and yrespectively.
Therefore, work done by a woman in 1 day =
Work done by a man in 1 day =
According to the question,
Putting
in these equations, we obtainBy cross-multiplication, we obtain
Hence, number of days taken by a woman = 18
Number of days taken by a man = 36
(iii) Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,
Putting
and in these equations, we obtainMultiplying equation (3) by 10, we obtain
Subtracting equation (4) from (5), we obtain
Substituting in equation (3), we obtain
Hence, speed of train = 60 km/h
Speed of bus = 80 km/h
Exercise 3.7
ANS: 1
The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.
Let the age of Ani and Biju be x and y years respectively.
Therefore, age of Ani’s father, Dharam = 2 × x = 2xyears
And age of Biju’s sister Cathy
yearsBy using the information given in the question,
Case (I) When Ani is older than Biju by 3 years,
x −y = 3 (i)
4x −y = 60 (ii)
Subtracting (i) from (ii), we obtain
3x= 60 − 3 = 57
Therefore, age of Ani = 19 years
And age of Biju = 19 − 3 = 16 years
Case (II) When Biju is older than Ani,
y −x = 3 (i)
4x −y = 60 (ii)
Adding (i) and (ii), we obtain
3x= 63
x = 21
Therefore, age of Ani = 21 years
And age of Biju = 21 + 3 = 24 years
ANS: 2
Let those friends were having Rs x and y with them.
Using the information given in the question, we obtain
x + 100 = 2(y − 100)
x + 100 = 2y − 200
x −2y = −300 (i)
And, 6(x− 10) = (y + 10)
6x− 60 = y + 10
6x− y = 70 (ii)
Multiplying equation (ii) by 2, we obtain
12x− 2y = 140 (iii)
Subtracting equation (i) from equation (iii), we obtain
11x= 140 + 300
11x= 440
x = 40
Using this in equation (i), we obtain
40 −2y = −300
40 + 300 = 2y
2y= 340
y = 170
Therefore, those friends had Rs 40 and Rs 170 with them respectively.
ANS: 3
Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,
Or, d = xt (i)
Using the information given in the question, we obtain
By using equation (i), we obtain
−2x + 10t = 20 (ii)
By using equation (i), we obtain
3x− 10t = 30 (iii)
Adding equations (ii) and (iii), we obtain
x = 50
Using equation (ii), we obtain
(−2) × (50) + 10t = 20
−100 + 10t = 20
10t= 120
t = 12 hours
From equation (i), we obtain
Distance to travel = d = xt
= 50 × 12
= 600 km
Hence, the distance covered by the train is 600 km.
ANS: 4
Let the number of rows be x and number of students in a row be y.
Total students of the class
= Number of rows × Number of students in a row
= xy
Using the information given in the question,
Condition 1
Total number of students = (x − 1) (y + 3)
xy = (x − 1) (y + 3) = xy − y + 3x− 3
3x− y − 3 = 0
3x− y = 3 (i)
Condition 2
Total number of students = (x + 2) (y − 3)
xy= xy + 2y − 3x − 6
3x− 2y = −6 (ii)
Subtracting equation (ii) from (i),
(3x −y) − (3x − 2y) = 3 − (−6)
− y+ 2y = 3 + 6
y = 9
By using equation (i), we obtain
3x− 9 = 3
3x= 9 + 3 = 12
x = 4
Number of rows = x = 4
Number of students in a row = y = 9
Number of total students in a class = xy = 4 × 9 = 36
ANS: 5
Given that,
∠C = 3∠B = 2(∠A + ∠B)
3∠B = 2(∠A + ∠B)
3∠B = 2∠A + 2∠B
∠B = 2∠A
2 ∠A− ∠B = 0 …(i)
We know that the sum of the measures of all angles of a triangle is 180°. Therefore,
∠A + ∠B + ∠C = 180°
∠A + ∠B + 3 ∠B = 180°
∠A + 4 ∠B = 180° …(ii)
Multiplying equation (i) by 4, we obtain
8 ∠A− 4 ∠B = 0 …(iii)
Adding equations (ii) and (iii), we obtain
9 ∠A = 180°
∠A = 20°
From equation (ii), we obtain
20° + 4 ∠B = 180°
4 ∠B = 160°
∠B = 40°
∠C = 3 ∠B
= 3 × 40° = 120°
Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.
ANS: 6
5x− y = 5
Or, y= 5x − 5
The solution table will be as follows.
x
|
0
|
1
|
2
|
y
|
−5
|
0
|
5
|
Or, y= 3x − 3
The solution table will be as follows.
x
|
0
|
1
|
2
|
y
|
− 3
|
0
|
3
|
It can be observed that the required triangle is ΔABC formed by these lines and y-axis.
The coordinates of vertices are A (1, 0), B (0, − 3), C (0, −5).
ANS: 7
(i)px+ qy = p − q … (1)
qx − py = p + q … (2)
Multiplying equation (1) by p and equation (2) by q, we obtain
p2x + pqy = p2 −pq… (3)
q2x − pqy = pq + q2…(4)
Adding equations (3) and (4), we obtain
p2x + q2 x = p2+ q2
(p2 + q2) x = p2+ q2
From equation (1), we obtain
p (1) + qy = p − q
qy = − q
y = − 1
(ii)ax+ by = c… (1)
bx + ay = 1 + c… (2)
Multiplying equation (1) by a and equation (2) by b, we obtain
a2x + aby = ac… (3)
b2x + aby = b + bc…(4)
Subtracting equation (4) from equation (3),
(a2 − b2) x = ac− bc − b
From equation (1), we obtain
ax + by = c
(iii)
Or, bx − ay = 0 … (1)
ax + by = a2 + b2…(2)
Multiplying equation (1) and (2) by b and a respectively, we obtain
b2x − aby = 0 … (3)
a2x + aby = a3 +ab2… (4)
Adding equations (3) and (4), we obtain
b2x + a2x = a3+ ab2
x (b2 + a2) = a (a2+ b2)
x = a
By using (1), we obtain
b (a) − ay = 0
ab − ay = 0
ay = ab
y = b
(iv) (a− b) x + (a + b) y = a2− 2ab − b2… (1)
(a + b) (x + y) = a2 +b2
(a + b) x + (a + b) y = a2+ b2… (2)
Subtracting equation (2) from (1), we obtain
(a − b) x − (a + b) x= (a2 − 2ab − b2)− (a2 + b2)
(a − b − a − b) x= − 2ab − 2b2
− 2bx = − 2b (a + b)
x = a + b
Using equation (1), we obtain
(a − b) (a + b) + (a + b)y = a2 − 2ab − b2
a2 − b2 + (a + b)y = a2− 2ab − b2
(a + b) y = − 2ab
(v) 152x− 378y = − 74
76x − 189y = − 37
…(1)− 378x + 152y = − 604
− 189x + 76y = − 302 … (2)
Substituting the value of x in equation (2), we obtain
− (189)2 y + 189 × 37 + (76)2y = − 302 × 76
189 × 37 + 302 × 76 = (189)2 y −(76)2 y
6993 + 22952 = (189 − 76) (189 + 76) y
29945 = (113) (265) y
y = 1
From equation (1), we obtain
ANS: 8
We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
Therefore, ∠A + ∠C = 180
4y+ 20 − 4x = 180
− 4x+ 4y = 160
x −y = − 40 (i)
Also, ∠B + ∠D = 180
3y− 5 − 7x + 5 = 180
− 7x+ 3y = 180 (ii)
Multiplying equation (i) by 3, we obtain
3x− 3y = − 120 (iii)
Adding equations (ii) and (iii), we obtain
− 7x+ 3x = 180 − 120
− 4x= 60
x =−15
By using equation (i), we obtain
x −y = − 40
−15− y = − 40
y =−15 + 40 = 25
∠A = 4y + 20 = 4(25) + 20 = 120°
∠B = 3y − 5 = 3(25) − 5 = 70°
∠C = − 4x = − 4(− 15) = 60°
∠D = − 7x + 5 = − 7(−15) + 5 = 110°
No comments:
Post a Comment