30, 42

Rs 20

x = 3.

\(\frac{92}{3}\)

We have, y + 4 = - 4
On subtracting 4 from both sides, we get
y + 4 - 4  = -4 - 4 \(\Rightarrow\) y = -8
Hence. y = -8 is the solution of the given equation.

Let the lowest marks (score) be l.
\(\therefore\) Twice the lowest marks = 2l
According to the question,
Highest marks = (Twice the lowest marks + 7) = 2l + 7
But the highest score is 87.
Hence, the required equation is 2l+ 7 = 87.

According to the question,
Number b divided by 5 =\(\frac { b }{ 5 } \) and the quotient = 6  Hence, the required equation is \(\frac { b }{ 5 } \)=6

Given equation is n+5 = 19 (n=1)
Here, LHS = n+5
Puting n = 1 in LHS we get
LHS = 1+5 = 6 ≠ 19
∵ LHS ≠ RHS
So, n = 1is not a solution of the given equation

Other forms for equation\(\frac { m }{ 5 } -2=6\) =6 are as follows:
(a) Subtract 2 from one-fifth of a number m to get 6.
(b) One-fifth of m is greater by 2 than 6.

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]
\(\Rightarrow \quad x=\frac { 12-1 }{ 2 } =\frac { 11 }{ 2 } \)

250

Let x be the number.
As per the given information, we have
         2x - 12 = 16
Adding 12 to both the sides, we get
        2x -12 + 12 = 16 + 12  \(\Rightarrow\) 2x = 28
Now, dividing both sides by 2, we get
          \(\frac{2x}{2}=\frac{28}{2}\Rightarrow x=14\)
Hence, the required number is 14.

Let ABC be an isosceles triangle, whose base angles are equal and let measure of base angle be x", Also, vertex angle is 40°.
Since the sum of three angles of a triangle is 180°
\(\therefore \quad \angle A+\angle B+\angle C=180^o\)
\(\Rightarrow\)          40^o + x + x = 180^o
\(\Rightarrow\)              40^o + 2x = 180^o
which is the required equation.
To solve this equation, transposing (+ 40) from LHS to RHS, we get
        2x = 180° - 40°  \(\Rightarrow\)  2x = 140°
On dividing both sides by 2, we get
        \(\frac{2x}{2}=\frac{140^o}{2}=70^o\quad \Rightarrow x=70^o \)
Hence, the base angles of the triangle are of measure 70°.

Let the number of mangoes in each smaller box be x.
\(\therefore\) Number of mangoes in 8 such boxes = 8x
According to the question,
Number of mangoes in each larger box = 8x + 4 and each larger box contains 100 mangoes (given).
So, the equation becomes 8x + 4 = 100
On transposing (+4) from LHS to RHS, we get
        8x = 100 - 4 ~ 8x = 96
On dividing both sides by 8, we get
        \(\frac{8x}{8}=\frac{96}{8}\quad \Rightarrow x=12\)
Hence, the required number of mangoes in each smaller box is 12.

Given equation is 5P + 2 = 17.
When P = 0, then LHS = 5\(\times\)0 + 2 = 0 + 2 = 2
and    RHS = 17
∴ LHS ≠ RHS
When p = 1, then LHS = 5\(\times\)1+ 2 = 5 + 2 = 7
and RHS = 17
∴  LHS ≠ RHS
When p = 2, then LHS = 5\(\times\)2 + 2 = 10 + 2 = 12
and  RHS =17
∴  LHS ≠ RHS
When p = 3, then LHS = 5\(\times\)3 + 2 = 15 + 2 = 17
∴  LHS = RHS
So, p = 3 is the solution of the given equation.

Given, \(\frac { 2x-3 }{ 5 } +\frac { x+3 }{ 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 4(2x-3) }{ 5\times 4 } +\frac { 5(x+3) }{ 5\times 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12 }{ 20 } +\frac { 5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12+5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 13x+3 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow\) 7(13x + 3) = 20(4x + 1)
\(\Rightarrow\) 91x + 21 = 80x + 20
\(\Rightarrow\) 91x - 80x = 20 - 21
\(\Rightarrow\) 11x = -1
Thus, \(x=-\frac { 1 }{ 11 } \)