6x - 2y+ 8z

Given, base of a parallelogram = 5 cm
and height of a parallelogram = 4.8 cm
\(\therefore\) Area of a parallelogram = Base x Height
                                     = 5 x 4.8 = 24 cm².

When, p = 1
then 4p - 3 = 4(1) - 3 = 4 - 3
= 1 \(\neq \) 13
So, P = 1 is not a solution of the given equation.

Interchanging the sides, we have
5(p - 2) = 4
Dividing both sides by 5, we have
\(\frac{5(p-2)}{2}=\frac{4}{5}\)
or p-2=\(\frac{4}{5}\) or p=\(\frac{4}{5}\)+2
[Transposing (-2) from L.H.S. to RH.S.]
or \(p=\frac{4+10}{5}=\frac{14}{5}\)
\(\therefore\) p=\(\frac{14}{5}\) is required solution.

Since, \(3\left( x+\frac { 1 }{ 2 } \right) =18\)
Dividing both sides by 3, we get
\(x+\frac { 1 }{ 2 } =\frac { 18 }{ 3 } \)
\(\Rightarrow \quad x+\frac { 1 }{ 2 } =6\)
\(\Rightarrow x=6-\frac { 1 }{ 2 } \)
[ On transposing to RHS]
\(\Rightarrow \quad x=\frac { 12-1 }{ 2 } =\frac { 11 }{ 2 } \)

25

Like terms are - xy² and 2xy² (have 2 same algebraic factors x, y, y); -4yx² and 20x² y ((have same algebraic factors x, x, y); 8x², -11x² and -6x² (have some algebric factors x, x) 7y and y ( have same algebraic factors y); - 100x and 3x (have same algebraic factor x); -11 yx and 2xy (have same algebric factors x, y)

Given, perimeter of a rectangular sheet = 100 cm and length of a rectangular sheet, I = 35 cm
Let b be the breadth of a rectangular sheet .
We know that, perimeter of a rectangular sheet = 2 (/ + b)
\(\Rightarrow\)  2(/ + b) = 100  \(\Rightarrow\) 2(35 + b) = 100
\(\Rightarrow\) \(35+b=\frac{100}{2}\)    [dividing by 2 on both sides]
\(\Rightarrow\) 35 + b = 50  \(\Rightarrow\) b = 50 - 35  \(\Rightarrow\) b = 15 cm
\(\therefore\) Area of a rectangular sheet = I X b = 35 x 15 = 525 cm²
Hence, the breadth and area of a rectangular sheet are 15 cm and 525 cm², respectively.

2

Let the radii of the given circles be r₁ and r₂. Let their circumference be 'C₁' and 'C₂' respectively.
\(\therefore\)C₁ = 2\(\pi\)r₁ and C₂ = 2\(\pi\)r₂
Since, C₁ : C₂ = 5 : 6
\(\therefore\)(2\(\pi\)r₁):(2\(\pi\)r₂)=5:6
\(\Rightarrow\)\(\frac { 2\pi r_{ 1 } }{ 2\pi r_{ 2 } } =\frac { 5 }{ 6 } \Rightarrow \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 5 }{ 6 } \)
Now, let the area of the given circles be A₁ and A₂
\(\therefore\)A₁ = \(\pi\)r₁² and A₂ = \(\pi\)r₂²
\(\Rightarrow\)\(\frac { { A }_{ 1 } }{ A_{ 2 } } =\frac { { { \pi r }_{ 1 } }^{ 2 } }{ { { \pi r }_{ 2 } }^{ 2 } } =\frac { { { r }_{ 1 } }^{ 2 } }{ { { r }_{ 2 } }^{ 2 } } \Rightarrow \frac { A_{ 1 } }{ A_{ 2 } } =\left( \frac { { { r }_{ 1 } } }{ { { r }_{ 2 } } } \right) ^{ 2 }\)
\(\Rightarrow\)\(\frac { { A }_{ 1 } }{ A_{ 2 } } =\left( \frac { 5 }{ 6 } \right) ^{ 2 }=\frac { 25 }{ 36 } \Rightarrow \)A₁:A₂
= 25:36
Thus, the areas of the two given circles are in the ratio 25 : 36.

Substituting a = 3 and b = 2 in
(a) a + b, we get
a + b = 3 + 2 = 5
(b) 7a - 4b,  we get
7a - 4b = 7 x 3 - 4 x 2
= 21 - 8 = 13
(c) a² + 2ab + b², we get
a² + 2ab + b² = 3² + 2 x 3 x 2 + 2²
= 9 + 2 x 6 + 4
= 9 + 12 + 4 = 25.
(d) a3 - b3, we get
a³ - b³ = 3³ - 2³ = 3 x 3 x 3 - 2 x 2 x 2
= 27 -8 = 19.

Age of Ram's father = 49 years
Let the age of Ram be x years
\(\therefore\) 3x + 4 = 49
\(\Rightarrow\) 3x = 49 - 4
\(\Rightarrow\) 3x = 45
\(\Rightarrow\) \(x=\frac { 45 }{ 3 } \)
x = 15
\(\Rightarrow\) Ram's age = 15 years