Maths

27. For inequality, x+y < 5, the equation of line is x+y = 5.

We consider the tableto plot x+y=5.

Graph of x+y = 5 is given as dotted line in fig 1. This lines divides by plane in two half planes I and II. We select a point not on the line, say (0,0) which lies in region I.

Since, 0+0 < 5

0 < 5 is true.

The solution region is I. (where origin (0,0) is included)

The dotted line indicates that any point on the line does not satisfy the given inequality.

26. Let 'x' cm be the length of the shortest piece.

Then, the two other remaining length is.

(x + 3) cm and (2x). cm.

So, given that,

2x - (x + 3) ≥ 5 cm.

2x- x - 3 ≥ 5.

x ≥ 5 + 3

x ≥ 8.

And. total length ≤ 91 cm

x + (x + 3) + 2x ≤ 91

4x + 3 ≤ 91

4x + 3 ≤ 91 - 3

$\Rightarrow x\le \frac{88}{4}$

x ≤ 22.

? 8 ≤ x ≤ 22.

Hence the length of the shortest board is greater than or equal to 8 cm and less than or equal to 22 cm.

25. Let x cm be the shortest side of the triangle.

Thus, the other 2 sides are “3x” cm and “3x - 2” cm

Given, perimeter of A triangle ≥ 61

x + 3x - 3x- 2 ≥ 61

7x - 21 ≥ 61

7x ≥ 61 + 21

$\Rightarrow x\ge \frac{82}{7}$

x ≥ 9.

The minimum length of the shortest side is 9 cm.

24. Let x be the smaller no. of the two consecutive even positive integers, hence the other number is x + 2. (Two consecutive even no. differs by 2).

Then, x > 5, x + 2 > 5  => a> 3.

And x + (x + 2) < 23

2x + 2 < 23

2x < 23 - 2

2x < 21

$\Rightarrow x<\frac{21}{2}=10.5.$

So, 5 < x$\frac{21}{2},\; x\in z\; and\; x\; is\; positive.$

Hence, x can take the value 6, 8 and 10.

So, the required possible pair is given by (x, x + 2) are (6, 8), (8, 10) and (10, 12).

23. Let x be the smaller of the consecutive odd positive integers, then the. other is x + 2 {Since the consecutive odd number are differ by 2.}

then the. other is x + 2

so, x < 10 and x + 2 < 10 x< 8

and x + (x + 2) > 11.

2x > 11 - 2

2x > 9

$\Rightarrow x>\frac{9}{2}$

and hence,  $\frac{9}{2}<x<8$

4.5

As x ∈ z, x = 5 and x = 7 and x is odd

The reqd. possible pairs by (x, x + 2) are (5, 7), (7, 9).

22. Let 'x' be the marks obtained by Sunita in the 5th exam.

Then, average marks ≥ 90.

$\frac{87+92+94+95+x}{5}\ge 90$ .

368 + x ≥ 90 * 5.

368 + x ≥ 450.

x ≥ 450 - 368.

x ≥ 82.

Minimum mark required= 82.

21. Let x be the marks obtained by Ravi in third test.

Then, average mark ≥ 60

$\Rightarrow \frac{70+75+x}{3}\ge 60$

145 + x ≥ 180.

x ≥180 - 145

x ≥ 35.

So minimum mark required = 35.

20. Given,

$\frac{x}{2}\ge \frac{(5x-2)}{3}-\frac{(7x-3)}{5.}$

$\Rightarrow \frac{x}{2}\ge \frac{5(5x-2)-3(7x-3)}{3*15}$

15x ≥ 2 [25x –10 –21x + 9].

15x ≥ 2 [4x –1]

15x ≥ 8x –2

15x – 8x ≥ –2

7x ≥ –2

x ≥$-\frac{2}{7}.$

So, x $\left[-\frac{2}{7},\infty \right)$ $\left\{\therefore -\frac{2}{7}=-0.28\right\}$

19. Given, 3 (1–x) < 2 (x + 4)

3 –3x < 2x + 8

3–8 < 2x+ 3x

–5 < 5x

$\frac{-5}{5}<x$

-1< x

So, x ∈ (–1, ∞ )

18. Given, 5x – 3 > 3x 5

5x –3x≥ –5 + 3.

2x ≥ –2.

x $\frac{-2}{2}$

x ≥ –1.

So, x [1, - ∞ ).