Solution:

Adding the given equations,

Dividing both sides of (iii) by 8,
  x + y = 2 ... (iv)

∴ The correct option is: (A) 2

Solution:

Let, the length of the rectangle be x and breadth be y.

When 5 is subtracted from length and breadth, the new length is x − 5 and new breadth is y − 5.
The perimeter of the new rectangle is 26.
∴ 2[(x − 5) + (y − 5)] = 26.
∴ 2x + 2y − 20 = 26
∴ 2x + 2y = 46
∴ x + y = 23 ... (i)

∴ The correct option is: (C) x + y = 23.

Solution:

Let, Ajay's age be A years and Vijay's age be V years.

From the first condition:
 A = V − 5 ... (i)

From the second condition:
 A + V = 25 ... (ii)

Substituting the value of A in equation (ii), we get,
 A + V = 25 ... (ii)
∴ V − 5 + V = 25
∴ 2V − 5 = 25
∴ 2V = 25 + 5
∴ 2V = 30

∴ \(\displaystyle \text V = \frac {30}{2}\)

∴ V = 15 years ... (iii)

Substituting the value of V in equation (i), we get,
 A = V − 5 ... (i)
∴ A = 15 − 5
∴ A = 10 years ... (iv)
∴ Ajay’s age is 10 years

∴ The correct option is: (C) 10

Solution:

Let, 2x + y = 5 ... (i)
and 3x − y = 5 ... (ii)

Adding (i) and (ii),

∴ \(\displaystyle x = \frac {10}{5}\)

∴ x = 2 ... (iii)

Substituting the value of x in (i), we get,
 2x + y = 5 ... (i)
∴ 2 × 2 + y = 5
∴ 4 + y = 5
∴ y = 5 − 4
∴ y = 1 ... (iv)

∴ The solution of the given simultaneous equations is x = 2 and y = 1.

Solution:

Let, 2x − y = 5 ... (i)
and 3x + 2y = 11 ... (ii)

Multiplying equation (i) by 2, we get,
 4x − 2y = 10 ... (iii)

Adding (ii) and (iii),

∴ \(\displaystyle x = \frac {21}{7}\)

∴ x = 3 ... (iii)

Substituting the value of x in (i), we get,
 2x − y = 5 ... (i)
∴ 2 × 3 − y = 5
∴ 6 − y = 5
∴ − y = 5 − 6
∴ − y = −1
i.e. y = 1 ... (iv)

∴ The solution of the given simultaneous equations is x = 3 and y = 1.

Solution:

Let, x + y = 11 ... (i)
and 2x − 3y = 7 ... (ii)

Multiplying equation (i) by 3, we get,
 3x + 3y = 33 ... (iii)

Adding (ii) and (iii),

∴ \(\displaystyle x = \frac {40}{5}\)

∴ x = 8 ... (iv)

Substituting the value of x in (i), we get,
 x + y = 11 ... (i)
∴ 8 + y = 11
∴ y = 11 − 8
∴ y = 3 ... (v)

∴ The solution of the given simultaneous equations is x = 8 and y = 3.

Solution:

Let, 2x + y = − 2 ... (i)
and 3x − y = 7 ... (ii)

Adding (i) and (ii),

∴ \(\displaystyle x = \frac {5}{5}\)

∴ x = 1 ... (iii)

Substituting the value of x in (i), we get,
 2x + y = − 2 ... (i)
∴ 2 × 1 + y = − 2
∴ 2 + y = − 2
∴ y = − 2 − 2
∴ y = − 4 ... (iv)

∴ The solution of the given simultaneous equations is x = 1 and y = − 4.

Solution:

Let, 2x − y = 5 ... (i)
and 3x + 2y = 11 ... (ii)

Multiplying equation (i) by 2, we get,
4x − 2y = 10 ... (iii)

Adding (ii) and (iii),

∴ \(\displaystyle x = \frac {21}{7}\)

∴ x = 3 ... (iv)

Substituting the value of x in (i), we get,
 2x − y = 5 ... (i)
∴ 2 × 3 − y = 5
∴ 6 − y = 5
∴ − y = 5 − 6
∴ − y = − 1
i.e. y = 1 ... (iv)

∴ The solution of the given simultaneous equations is x = 3 and y = 1.

Solution:

Let, x − 2y = − 2 ... (i)
and x + 2y = 10 ... (ii)

Adding (i) and (ii),

∴ \(\displaystyle x = \frac {8}{2}\)

∴ x = 4 ... (iii)

Substituting the value of x in (ii), we get,
 x + 2y = 10 ... (ii)
∴ 4 + 2y = 10
∴ 2y = 10 − 4
∴ 2y = 6
∴ \(\displaystyle y = \frac {6}{2}\)
∴ y = 3 ... (iv)

∴ The solution of the given simultaneous equations is x = 4 and y = − 3.

Solution:

Let, 3x − 4y = 7 ... (i)
and 5x + 2y = 3 ... (ii)

Multiplying equation (ii) by 2, we get,
10x + 4y = 6 ... (iii)

Adding (i) and (iii),

∴ \(\displaystyle x = \frac {13}{13}\)

∴ x = 1 ... (iv)

Substituting the value of x in (ii), we get,
 5x + 2y = 3 ... (ii)
∴ 5 × 1 + 2y = 3
∴ 5 + 2y = 3
∴ 2y = 3 − 5
∴ 2y = − 2

∴ \(\displaystyle y = \frac {- 2}{2}\)

∴ y = − 1 ... (v)

∴ The solution of the given simultaneous equations is x = 1 and y = − 1.

Solution:

Let, 5x + 7y = 17 ... (i)
and 3x − 2y = 4 ... (ii)

Multiplying equation (i) by 2, we get,
10x + 14y = 34 ... (iii)

Multiplying equation (ii) by 7, we get,
21x − 14y = 28 ... (iv)

Adding (iii) and (iv),

∴ \(\displaystyle x = \frac {62}{31}\)

∴ x = 2 ... (v)

Substituting the value of x in (i), we get,
5x + 7y = 17 ... (i)
∴ 5 × 2 + 7y = 17
∴ 10 + 7y = 17
∴ 7y = 17 − 10

∴ 7y = 7

∴ \(\displaystyle y = \frac {7}{7}\)

∴ y = 1 ... (vi)

∴ The solution of the given simultaneous equations is x = 2 and y = 1.

Solution:

Let, x − 2y = − 10 ... (i)
and 3x − 5y = − 12 ... (ii)

Multiplying equation (i) by 3, we get,
3x − 6y = − 30 ... (iii)

Subtracting (iii) from (ii),

Substituting the value of y in (i), we get,
x − 2y = − 10 ... (i)
∴ x − 2 × 18 = − 10
∴ x − 36 = − 10
∴ x = − 10 + 36
∴ x = 26 ... (v)

∴ The solution of the given simultaneous equations is x = 26 and y = 18.

Solution:

Let, 4x + y = 34 ... (i)
and x + 4y = 16 ... (ii)

Multiplying equation (i) by 4, we get,
16x + 4y = 136 ... (iii)

Subtracting (ii) from (iii),

∴ \(\displaystyle x = \frac {120}{15}\)

∴ x = 8 ... (iv)

Substituting the value of x in equation (ii), we get,
x + 4y = 16 ... (ii)
∴ 8 + 4y = 16
∴ 4y = 16 − 8
∴ 4y = 8
∴ \(\displaystyle y = \frac {8}{4}\)
∴ y = 2 ... (v)

∴ The solution of the given simultaneous equations is x = 8 and y = 2.

Solution:

Let, \(\displaystyle \frac {x}{3} + \frac {y}{4} = 4\) ... (i)

and \(\displaystyle \frac {x}{2} - \frac {y}{4} = 1\) ... (ii)

Multiplying equation (i) by 12, we get,

 \(\displaystyle \frac {x}{3} \times 12 + \frac {y}{4} \times 12 = 4 \times 12\)

∴ 4x + 3y = 48 ... (iii)

Multiplying equation (ii) by 12, we get,

\(\displaystyle \frac {x}{2} \times 12 - \frac {y}{4} \times 12 = 1 \times 12\)

∴ 6x − 3y = 12 ... (iv)

Adding (iii) from (iv),

∴ \(\displaystyle x = \frac {60}{10}\)

∴ x = 6 ... (v)

Substituting the value of x in equation (iii), we get,
4x + 3y = 48 ... (iii)
∴ 4 × 6 + 3y = 48
∴ 24 + 3y = 48
∴ 3y = 48 − 24
∴ 3y = 24

∴ \(\displaystyle y = \frac {24}{3}\)

∴ y = 8 ... (vi)

∴ The solution of the given simultaneous equations is x = 6 and y = 8.

Solution:

Let, \(\displaystyle \frac {x}{3} + 5y = 13\) ... (i)

and \(\displaystyle 2x + \frac {y}{2} = 19\) ... (ii)

Multiplying equation (i) by 12, we get,

 \(\displaystyle \frac {x}{3} \times 12 + 5y \times 12 = 13 \times 12\)

∴ 4x + 60y = 156 ... (iii)

Multiplying equation (ii) by 2, we get,

\(\displaystyle 2x \times 2 + \frac {y}{2} \times 2 = 19 \times 2\)

∴ 4x + y = 38 ... (iv)

Subtracting (iv) from (iii),

∴ \(\displaystyle y = \frac {118}{59}\)

∴ y = 2 ... (v)

Substituting the value of x in equation (iv), we get,
4x + y = 38 ... (iv)
∴ 4x + 2 = 38
∴ 4x = 38 − 2
∴ 4x = 36

∴ \(\displaystyle x = \frac {36}{4}\)

∴ x = 9 ... (vi)

∴ The solution of the given simultaneous equations is x = 9 and y = 2.

Solution:

Let, \(\displaystyle \frac {2}{x} + \frac {3}{y} = 13\) ... (i)

and \(\displaystyle \frac {5}{x} - \frac {4}{y} = - 2\) ... (ii)

Now, let \(\displaystyle \frac {1}{x} = a\) and \(\displaystyle \frac {1}{y} = b\)

The given equations,
  2a + 3b = 13 ... (iii)
and 5a − 4b = − 2 ... (iv)

Multiplying equation (iii) by 4, we get,
 8a + 12b = 52 ... (v)

Multiplying equation (iv) by 3, we get,
15a − 12b = − 6 ... (vi)

Adding (v) and (vi),

∴ \(\displaystyle a = \frac {46}{23}\)

∴ a = 2 ... (vii)

Substituting the value of a in equation (iii), we get,
 2a + 3b = 13 ... (iii)
∴ 2 × 2 + 3b = 13
∴ 4 + 3b = 13
∴ 3b = 13 − 4

∴ 3b = 9

∴ \(\displaystyle b = \frac {9}{3}\)

∴ b = 3 ... (viii)

Resubstituting the value of a,

 \(\displaystyle \frac {1}{x} = a\)

∴ \(\displaystyle \frac {1}{x} = 2\)

∴ \(\displaystyle \frac {1}{2} = x\)

i.e. \(\displaystyle x = \frac {1}{2}\) ... (ix)

Resubstituting the value of b,

 \(\displaystyle \frac {1}{y} = b\)

∴ \(\displaystyle \frac {1}{y} = 3\)

∴ \(\displaystyle \frac {1}{3} = y\)

i.e. \(\displaystyle y = \frac {1}{3}\) ... (x)

∴ The solution of the given simultaneous equations is \(\displaystyle x = \frac {1}{2}\) and \(\displaystyle y = \frac {1}{3}\)

Solution:

∴ Let, the digit in the ten’s place be x and the number digit in the unit’s place be y.
∴ That number = 10x + y

From the first condition,
 10x + y = 4(x + y) + 3
∴ 10x + y = 4x + 4y + 3
∴ 10x + y − 4x − 4y = 3
∴ 6x − 3y = 3

Dividing both sides by 3,
  2x − y = 1 ... (i)

And the number obtained by interchanging the digits
= 10y + x

From the second condition,
 10x + y + 18 = 10y + x
∴ 10x + y − 10y − x = − 18
∴ 9x − 9y = − 18

Dividing both sides by 9,
 x − y = − 2 ... (ii)

Subtracting equation (ii) from equation (i),

Substituting the value of x in equation (ii),
 x − y = − 2 ... (ii)
∴ 3 − y = − 2
∴ − y = − 2 − 3
∴ − y = − 5
i.e. y = 5 ... (iv)
∴ That number is 35.

Solution:

Let, the cost of 1 book be ₹ x and the cost of 1 pen be ₹ y.

From the first condition,
 5x + 7y = 79 ... (i)

From the second condition,
 7x + 5y = 77 ... (ii)

Adding equation (i) and equation (ii),

Dividing both sides by 12,
x + y = 13 ... (iii)

Subtracting equation (i) from equation (ii),

Dividing both sides by 2,
x − y = − 1 ... (iv)

Adding equation (iii) and equation (iv),

\(\displaystyle \therefore x = \frac {12}{2}\)

∴ x = ₹ 6 ... (v)

Substituting the value of x in (iii), we get,
 x + y = 13 ... (iii)
∴ 6 + y = 13
∴ y = 13 − 6
∴ y = 7 ... (vi)

Hence, the cost of one book is ₹ 6 and the cost of one pen is ₹ 7.

∴ The cost of 1 book and 2 pens
= x + 2y
= 6 + 2 × 7
= 6 + 14
= ₹20
∴ The cost of 1 book and 2 pens is ₹ 20.

Solution:

Let the income of first person be ₹ x and the income of second person be ₹ y respectively.

From the first condition,

 \(\displaystyle \frac {x}{y} = \frac {9}{7}\)

By cross multiplication,
 7x = 9y
∴ 7x − 9y = 0 ... (i)

Now, every person saves ₹ 200.
Therefore, expenses of first person
= x − 200
And expenses of second person
= y − 200
But, their expenses are in the ratio 4 : 3. ... (Given)
∴ From the second condition,

 \(\displaystyle \frac {x - 200}{y - 200} = \frac {4}{3}\)

By cross multiplication,
 3(x − 200) = 4(y − 200)
∴ 3x − 600 = 4y − 800
∴ 3x − 4y = − 800 + 600
∴ 3x − 4y = − 200 ... (ii)

Multiplying equation (i) by 4,
 7x − 9y = 0 ... (i)
∴ 7x × 4 − 9y × 4 = 0 × 4
∴ 28x − 36y = 0 ... (iii)

Multiplying equation (ii) by 9,
∴ 3x − 4y = − 200 ... (ii)
∴ 3x × 9 − 4y × 9 = − 200 × 9
∴ 27x − 36y = − 1800 ... (iv)

Subtracting (iv) from (iii),

Substituting the value of x in equation (ii),
 3x − 4y = − 200 ... (ii)
∴ 3 × 1800 − 4y = − 200
∴ 5400 − 4y = − 200
∴ − 4y = − 200 − 5400
∴ − 4y = − 5600

\(\displaystyle \therefore y = \frac {- 5600}{-4}\)

∴ y = 1400
∴ The income of the first person is ₹ 1800 and the income of the second person is ₹ 1400.

Solution:

Let, the length of that rectangle be l units and its breadth be b unit.
∴ It’s area = l × b sq. unit

From the first condition,
 (l − 5)(b + 3) = lb − 9
∴ lb + 3l − 5b − 15 = lb − 9
∴ 3l − 5b − 15 = − 9
∴ 3l − 5b = − 9 + 15
∴ 3l − 5b = 6 ... (i)

From the second condition,
 (l − 3)(b + 2) = lb + 67
∴ lb + 2l − 3b − 6 = lb + 67
∴ 2l − 3b − 6 = 67
∴ 2l − 3b = 67 + 6
∴ 2l − 3b = 73 ... (ii)

Multiplying equation (i) by 2,
 3l − 5b = 6 ... (i)
∴ 6l − 10b = 12 ... (iii)

Multiplying equation (ii) by 3,
 2l − 3b = 73 ... (ii)
∴ 6l − 9b = 219 ... (iv)

Subtracting (iii) from (iv),

Substituting the value of b in (ii),
 2l − 3b = 73 ... (ii)
∴ 2l − 3 × 207 = 73
∴ 2l − 621 = 73
∴ 2l = 73 + 621
∴ 2l = 694

∴ \(\displaystyle l = \frac {694}{2}\)

∴ l = 347 ... (vi)
∴ The length of that rectangle is 347 unit and it’s breadth is 207 unit.

Solution:

Let, the speed of the car starting from point ‘A’ be x km/hr and the speed of the car starting from point ‘B’ be y km/hr.
Here, x > y

From the first condition,
  7x − 7y = 70
Dividing both sides by 7,
 x − y = 10 ... (i)

From the secons condition,
 x + y = 70 ... (ii)

Adding equation (i) and equation (ii),

\(\displaystyle \therefore x = \frac {80}{2}\)

∴ x = 40 km/hr ... (iii)

Substituting the value of x in (ii), we get,
 x + y = 70 ... (ii)
∴ 40 + y = 70
∴ y = 70 − 40
∴ y = 30 km/hr ... (iv)

Hence, the speed of the car starting from point ‘A’ is 40 km/hr and the speed of the car starting from point ‘B’ is 30 km/hr.

Solution:

Let, the digit in the ten’s place be x and the digit in the unit’s place be y.
∴ That number = 10x + y

And the number obtained by interchanging the digits
= 10y + x

From the given information,
 10x + y + 10y + x = 99
∴ 11x + 11y = 99

Dividing both sides by 11,
 x + y = 9 ... (i)

When x = 1, y = 8
That number is 18.

When x = 2, y = 7
That number is 27.

When x = 3, y = 6
That number is 36. ... ...

∴ Those numbers are 18, 27, 36 45, ... ...

(The information given in the problem is insufficient.)