Maths

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, - ∞ ).

17. Given, 3x – 2 < 2x + 1

3x – 2x< 2 + 1

x< 3.

So, x (- ∞, 3).

18. $\underset{}{}$$\underset{x?0}{lim}\frac{ax+xcosx}{bsinx}=\underset{x?0}{lim}\frac{x(a+cosx)}{bsinx}$

$=\frac{1}{b}*\frac{(a+cos0)}{1}$

$=\frac{a+1}{b}$