Ncert Solutions Maths class 11th

62. It is given that

$68<\; F<77.$

Putting $f=\frac{9}{5}c+32$ we get,

$68<\frac{9}{5}c+32<77.$

$\Rightarrow 68-32<\frac{9}{5}c+32-32<77-32.$  (Subtracting 32 throughout)

$\Rightarrow 36<\frac{9}{5}c<45$

Multiplying by $\frac{5}{9}$ throughout,

$\Rightarrow 36*\frac{5}{9}<\frac{9}{5}*c*\frac{5}{9}<45*\frac{5}{9}$

$\begin{array}{l}\\ \Rightarrow 20<c<25\end{array}$

Hence, the required range of temperature is between $20^{°}25^{°}.$

51. The given system of inequality is

x+2y≤ 10- (1)

x+y≥ 1 - (2)

x – y ≤ 0 - (3)

x≥ 0 and y≥ 0 - (4)

The corresponding equation of (1), (2) and (3) are

x + 2y = 10

and x + y =1

and x – y = 0

Putting (2,0)= (x, y) in inequality (1), (2) and (3),

2+2 * 0 ≤ 10 =>  2≤ 10 is true.

and 2+0 ≥ 1 =>  2 ≥ 1 is true.

and 2 – 0 ≤ 0  => 2 ≤ 0 is false.

So, the solution of inequality (1) and (2) is the plane that includes point (2,0) whereas the solution of inequality (3) is the plane which includes point (2, 0)

∴ The shaded region represents the solution of the given system of inequality.

50.The given system of inequality is

3x+2y≤ 150- (1)

x+4y≤ 80- (2)

x≤ 15 - (3)

y≥ 0 and x≥ 0 - (4)

The corresponding equation of (1) and (2) are

3x + 2y = 150

and x + 4y =80

Putting (0,0)= (x, y) in inequality (1) and (2) we get,

3 * 0+2 * 0 ≤ 150  => 0 ≤ 150 is true.

and 0+4 * 0 ≤ 80  => 0 ≤ 80 is true.

So, the solution plane of both inequality (1) and (2) includes the origin (0,0).

∴ The shaded region is the solution of the given system of inequality.

49. The given system of inequality is

4x+3y≤ 60- (1)

y≥ 2x- (2)

x≥ 3- (3)

andx, y ≥ 0- (4)

The corresponding equation of inequality (1) and (2) are

4x+3y= 60

and y = 2x

Putting (1,0) in inequality (1) and (2) we get,

4 * 1+3 * 0 ≤ 60

4 ≤ 60 which is true.

and 0 ≥ 2 * 1

0 ≥ 2 which is false.

So, solution of inequality (1) includes the plane with point (1,0) whereas the solution of inequality (2) excludes the plane with point (1,0).

? The shaded region is the solution of the given system of inequality.

48. The given system of inequality is

x – 2y≤ 3 - (1)

3x – 4y≥12- (2)

x ≥ 0 - (3)

y≥ 1 - (4)

The corresponding equation of (1) and (2) are

x – 2y= 3

and 3x – 4y=12

Putting (x, y)= (0,0) in inequality (1) and (2),

0 – 2 * 0 ≤ 3   => 0 ≤ 3 is true.

and 3 * 0+4 * 0 ≥ 12   => 0 ≥ 12 is false.

So, solution of inequality (1) includes plane wilt origin (0,0) while solution plane of inequality (2) includes the origin.

∴ The shaded portion determines the solution region of the given system of inequality.