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).

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).

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

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.

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.

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.

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.

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

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

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).

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.

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.

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

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.

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

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.

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

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

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.

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

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

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

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.

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.

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

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°.

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).

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

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°.