Solve the following Question.(1 Marks)

Question 11 Mark

The longest side of a triangle is twice the shortest side and the third side is $2 \mathrm{~cm}$ longer than the shortest side. If the perimeter of the triangle is more than $166 \mathrm{~cm}$ then find the minimum integer length of the shortest side.

Answer

Let the shortest side be $x$.
Then longest side length $=2 x$
and third side length $=x+2$
Perimeter $=x+2 x+x+2=4 x+2$
Given, perimeter $>166$
$
\begin{aligned}
& \therefore 4 x+2>166 \\
& \therefore 4 x>164 \\
& \therefore x>41
\end{aligned}
$
$\therefore$ Minimum integer length of shortest side is $42 \mathrm{~cm}$.

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Question 21 Mark

Solve the following inequations: $|7 x-4|<10$

Answer

$
\begin{aligned}
& |7 x-4|<10 \\
& -10<7 x-4<10 \ldots . . .[|x|<k \text { is same as }-k<x<k]
\end{aligned}
$
Adding 4 on both sides, we get
$
-6<7 x<14
$
Dividing both sides by 7 , we get
$
\begin{aligned}
& -\frac{6}{7}<x<\frac{14}{7} \\
& \therefore-\frac{6}{7}<x<2
\end{aligned}
$
$\therefore x$ takes all real values between $-\frac{6}{7}$ and 2 .
$\therefore$ Solution set $=\left(-\frac{6}{7}, 2\right)$

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Question 31 Mark

Solve the following inequations: $0<\frac{x-5}{4}<3$

Answer

$
\begin{aligned}
& 0<\frac{x-5}{4}<3 \\
& 0<x-5<12
\end{aligned}
$
Adding 5 on both sides, we get
$
5<x<17
$
$x$ takes all real values between 5 and 17 .
$\therefore$ Solution set $=(5,17)$

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Question 41 Mark

Solve the following inequations: $7 x-25 \leq-4$

Answer

$
7 x-25 \leq-4
$
Adding 25 on both sides, we get
$
\begin{aligned}
& 7 x-25+25 \leq-4+25 \\
& \therefore 7 x \leq 21
\end{aligned}
$
Dividing both sides by 7 , we get $x \leq 3$
$\therefore \mathrm{x}$ takes all real values less or equal to 3 .
$\therefore$ Solution Set $=(-\infty, 3]$

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Question 51 Mark

Solve the following inequations: $3 \mathrm{x}-36>0$

Answer

$
3 x-36>0
$
Adding 36 both sides, we get
$
\begin{aligned}
& 3 x-36+36>0+36 \\
& \therefore 3 x>36
\end{aligned}
$
Dividing both sides by 3 , we get
$
\begin{aligned}
& \frac{3 x}{3}>\frac{36}{3} \\
& \therefore x>12 \\
& \therefore x \text { takes all real values more than } 12 \\
& \therefore \text { Solution set }=(12, \infty)
\end{aligned}
$
$\therefore x$ takes all real values more than 12
$\therefore$ Solution set $=(12, \infty)$

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Question 61 Mark

Write the inequations that represent the interval and state whether the interval is bounded or unbounded: $(-\infty, 3)$

Answer

$(-\infty, 3)$
Here, $x$ takes values between $-\infty$ and 3
$\therefore$ the required inequation is $-\infty<\mathrm{x}<3$
$\therefore$ it is an unbounded (open) interval.

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Question 71 Mark

Write the inequations that represent the interval and state whether the interval is bounded or unbounded: $(-11,-2)$

Answer

$(-11,-2)$
Here, $x$ takes values between -11 and -2
$\therefore$ the required inequation is $-11<\mathrm{x}<-2$
$\therefore$ it is a bounded (open) interval.

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Question 81 Mark

Write the inequations that represent the interval and state whether the interval is bounded or unbounded: $[5, \infty)$

Answer

$[5, \infty)$
Here, $x$ takes values between 5 and $\infty$ including 5 .
$\therefore$ the required inequation is $5 \leq \mathrm{x}<\infty$
$\therefore$ it is an unbounded (semi-left closed) interval.

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Question 91 Mark

Write the inequations that represent the interval and state whether the interval is bounded or unbounded: $(-\infty, \infty)$

Answer

$(-\infty, \infty)$
Here, $x$ takes values between $-\infty$ and $\infty$
$\therefore$ the required inequation is $-\infty<\mathrm{x}<\infty$
$\therefore$ it is an unbounded (open) interval.

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Question 101 Mark

Write the inequations that represent the interval and state whether the interval is bounded or unbounded: $(0,0.9]$

Answer

$(0,0.9]$
Here, $x$ takes values between 0 and 0.9 , including 0.9 and excluding 0 .
$\therefore$ the required inequation is $0<\mathrm{x} \leq 0.9$
$\therefore$ it is a bounded (semi-right closed) interval.

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Question 111 Mark

Write the inequations that represent the interval and state whether the interval is bounded or unbounded: $\left[-4, \frac{7}{3}\right]$

Answer

$\left[-4, \frac{7}{3}\right]$
Here, $x$ takes values between -4 and $\frac{7}{3}$ including -4 and $\frac{7}{3}$
$\therefore$ the required inequation is $-4 \leq \mathrm{x} \leq \frac{7}{3}$
$\therefore$ it is a bounded (closed) interval.

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Maths (commerce) Solve the following Question.(1 Marks)

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