Exercise 2.3-Additional Questions - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Solve: $\frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$
Solution:
$
\begin{aligned}
& \frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5} \\
\Rightarrow \quad & \frac{x}{2} \geq \frac{5(5 x-2)-3(7 x-3)}{15} \\
\Rightarrow & \frac{x}{2} \geq \frac{25 x-10-21 x+9}{15} \\
\Rightarrow & \frac{x}{2} \geq \frac{4 x-1}{15}
\end{aligned}
$
Multiplying both sides by 30 , we get $15 x \geq 2(4 x-1) \Rightarrow 15 x \geq 8 x-2 \Rightarrow 15 x-8 x \geq-2$

$
\begin{aligned}
& 15 x \geq 2(4 x-1) \\
\Rightarrow & 15 x \geq 8 x-2 \Rightarrow 15 x-8 x \geq-2 \\
\Rightarrow & 7 x \geq-2 \Rightarrow x \geq \frac{-2}{7} \\
\therefore & \text { The solution set }=\left[-\frac{2}{7}, \infty\right)
\end{aligned}
$
The graphical representation of the solutions is given in the figure below.

Question 2.
Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution:
Let $x$ be the marks obtained by Ravi in the third test.
Then, $\frac{70+75+x}{3} \geq 60$
$
\Rightarrow 145+\mathrm{x} \geq 180 \Rightarrow \mathrm{x}>180-145
$
$
\Rightarrow \mathrm{x} \geq 35
$
Thus, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.
Note. A minimum of 35 marks.
$\Rightarrow$ Marks greater than or equal to 35 .
Question 3.
To receive Grade ' $\mathrm{A}$ ' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in first four examinations are 87, 92, 94 and 95 , find minimum marks that Sunita must obtain in fifth examination to get Grade ' $\mathrm{A}$ ' in the course.
Solution:
Let $\mathrm{x}$ be the marks obtained by Sunita in the fifth examination. Then,
$
\begin{aligned}
& \frac{87+92+94+95}{5} \geq 90 \\
& \Rightarrow 368+x \geq 450 \Rightarrow \mathrm{x} \geq 450-368 \\
& \Rightarrow \mathrm{x} \geq 82
\end{aligned}
$
Thus, Sunita must obtain marks greater than or equal to 82 ,
i. e., a minimum of 82 marks.
Question 4.
Find the pairs of ceonsecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11 .
Solution:
Let $x$ be the smaller of the two consecutive odd positive integers, then the other is $x+2$.
According to the given conditions.

$
\begin{aligned}
& \mathrm{x}<10 \mathrm{x}+2<10 \text { and } \mathrm{x}+(\mathrm{x}+2)>11 \\
& \Rightarrow \Rightarrow \mathrm{F}=\frac{108.6}{2.3}=47.23 \quad(\because x<8 \text { automatically }
\end{aligned}
$
$\Rightarrow$ Bo the ranges of $f$ alue $19_{\text {ere }} 42.87<$ smatlest of the lesser than)
and
$
x>\frac{9}{2}
$
From (1) and (2), we get 9
$
\frac{9}{2}<\mathrm{x}<8
$
Also, $x$ is an odd positive integer. $x$ can take values 5 and 7 .
So, the required possible pairs will be $(x, x+2)=(5,7),(7,9)$
Question 5.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23 .
Solution:
Let $x$ be the smaller of the two consecutive even positive integers, then the other is $x+2$. According to the given conditions.

Also, $\mathrm{x}$ is an even positive integer.
$\mathrm{x}$ can take the values 6,8 and 10 .
So, the required possible pairs will be $(x, x+2)=(6,8),(8,10),(10,12)$
Question 6.
Forensic Scientists use $h=61.4+2.3 \mathrm{~F}$ to predict the height $h$ in centimetres for a female whose thigh bone (femur) measures $F \mathrm{~cm}$. If the height of the female lies between 160 to $170 \mathrm{~cm}$ find the range of values for the length of the thigh bone?
Solution:
$
\begin{aligned}
& \text { Given } \mathrm{h}=61.4+2.3 \mathrm{~F} \\
& \text { Given } \mathrm{h}=160 \Rightarrow 160=61.4+2.3 \mathrm{~F} \\
& \Rightarrow 2.3 \mathrm{~F}=160-61.4=98.6 \\
& \mathrm{~F}=\frac{98.6}{2.3}=42.87
\end{aligned}
$
Given $\mathrm{h}=170 \Rightarrow 170=61.4+2.3 \mathrm{~F}$
$
\begin{aligned}
& \Rightarrow 170-61.4=2.3 \mathrm{~F} \\
& 2.3 \mathrm{~F}=108.6 \\
& \Rightarrow F=\frac{108.6}{2.3}=47.23
\end{aligned}
$
So the ranges of values are $42.87<\mathrm{x}<47.23$