Chapter 6: Linear Inequalities Exercise 6.1

Ans: Let us assume $\text{x}$ as the marks obtained by Ravi in the third unit test.

As it is given that the student must have an average of $\text{60}$ marks,

Therefore, the expression is:

 $\frac{\text{70+75+x}}{\text{3}}\geq\text{60} $ 

$ \Rightarrow \text{145+x}\geq\text{180} $ 

$ \Rightarrow \text{x}\geq\text{180-145} $ 

$ \Rightarrow \text{x}\geq\text{35}$

Therefore, we can say that the student should obtain greater than or equal 

to $35$ marks.

Ans: Let us assume $\text{x}$ as the fifth examination marked by Sunita.

As it is given that to obtain ‘A’ one should obtain an average of $\text{90}$ marks or

more in five examinations.

So, the expression is:

$\frac{87+92+94+95+\text{x}}{5}\geq90 $ 

$ \Rightarrow \frac{368+\text{x}}{5}\geq90 $ 

$ \Rightarrow 368+\text{x}\geq450 $ 

$ \Rightarrow \text{x}\geq450-368 $ 

$ \Rightarrow \text{x}\geq82$ 

Therefore, we can say that sunita must obtain $\text{82}$ or greater than it to obtain 

‘A’ grade.

Ans: Let us assume $\text{x}$ as the smallest number in two consecutive odd integers. Then, the other odd integer will be $\text{x+2}$.

As it is given that both the integers are smaller than $\text{10}$, we get 

$\text{x+2}<10 $ 

$ \Rightarrow \text{x}<\text{10-2} $ 

$ \Rightarrow \text{x}8$

Let us assume it as inequality $\left( 1 \right)$, we get

$\text{x}<8.............\left( 1 \right)$

Also, it is given that the sum of the two integers is more than $\text{11}$.

So, 

$\text{x+}\left( \text{x}+2 \right)>11 $ 

$ \Rightarrow 2\text{x}+2>11 $ 

$ \Rightarrow 2\text{x}>11-2 $ 

$ \Rightarrow 2\text{x}>\text{9} $ 

$ \Rightarrow \text{x}>\frac{9}{2} $ 

$ \Rightarrow \text{x}>\text{4}\text{.5}$

Let us assume this as inequality $\left( 2 \right)$

$\text{x}>4.5........\left( 2 \right)$

Therefore, from inequality $\left( 1 \right)$ and $\left( 2 \right)$, we get that

$\text{x}$ can take the values of 

$5$ and $7$. 

Therefore, the required pairs are $\left( 5,7 \right)$ and $\left( 7,9 \right)$.

Ans: Let us assume $\text{x}$ as the smallest number in two consecutive even integers. 

Then, the other odd integer will be $\text{x+2}$.

As it is given that both the integers are greater than $5$, we get

$\text{x}>\text{5}..........\left( 1 \right)$

Also, it is given that the sum of the two integers is less than $23$.

So, 

$\text{x+}\left( \text{x}+2 \right)<23 $ 

$ \Rightarrow 2\text{x}+2<23 $ 

$ \Rightarrow 2\text{x}<\text{23}-2 $ 

$ \Rightarrow 2\text{x}<\text{21} $ 

$ \Rightarrow \text{x}<\frac{21}{2} $ 

$ \Rightarrow \text{x}<\text{10}\text{.5}$

Let us assume this as inequality $\left( 2 \right)$

$\text{x}<10.5........\left( 2 \right)$

From $\left( 1 \right)$ and $\left( 2 \right)$, we obtain

$5<\text{x}<\text{10}\text{.5}$.

Therefore, we can say that $\text{x}$satisfies $\text{6,8}$ and $\text{10}$ values.

Therefore, the pairs that are possible are $\left( 6,8 \right)$, $\left( 8,10 \right)$ and $\left( 10,12 \right)$.

Ans: Let us assume the shortest side of the given triangle as $\text{x}$.

As it is given that the longest side is the three times the shortest side, we

get longest side$\text{=3x}$ 

Therefore, the length of the other side will be $\left( 3x-2 \right)$

As it is given that the perimeter should be at least $61$ cm, we can write it as:

$\text{xcm+3xcm+}\left( 3\text{x}-2 \right)\text{cm=61cm} $ 

$ \Rightarrow \text{7x-2}\geq\text{61} $ 

$ \Rightarrow \text{7x}\geq\text{61+2} $ 

$ \Rightarrow \text{7x}\geq\text{63} $ 

$ \Rightarrow \text{x}\geq\text{9}$

Therefore, we can say that the shortest side of the given triangle should 

satisfy the inequality$\text{x}\geq\text{9}$.

Ans: Let us assume the length of the shortest piece be $\text{x}$ cm.

From the question we can write that the second and third piece will be $\left( \text{x+3} \right)$ cm and $\text{2x}$ respectively.

As it is given that the three lengths are single piece of board of length $\text{91}$ cm,

$\text{x cm+}\left( \text{x+3} \right)\text{cm+2xcm=91cm}$

$ \Rightarrow \text{4x+3}\leq\text{91} $ 

$ \Rightarrow \text{4x}\leq\text{91-3} $ 

$ \Rightarrow \text{4x}\leq\text{88} $ 

$ \Rightarrow \frac{4}{4}\text{x}\leq\frac{88}{4} $ 

$ \Rightarrow \text{x}\leq22........\left( 1 \right) $

Also, the third piece is at least $5\text{cm}$ longer than the second piece.

$\therefore \text{2x}\geq\left( \text{x+3} \right)+5 $ 

$ \Rightarrow 2\text{x}\geq\text{x+8} $ 

$ \Rightarrow \text{x}\geq\text{8}..........\left( 2 \right)$

Therefore, from $\left( 1 \right)$and $\left( 2 \right)$, we get

$8\leq\text{x}\leq22$

So, we can say that the shortest board should be greater than or equal to 

$8$ cm but less than or equal to $22$ cm.