Exercise 2.3 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

EX 2.3
Question 1.
Represent the following inequalities in the interval notation:
Solution:

$
\Rightarrow \mathrm{x} \in[-1,4)
$
[] closed interval, end points are included () $\rightarrow$ open interval end points are excluded
(ii) $x \leq 5$ and $x \geq-3[i] x \leq 5$ and $x \geq-3$

Solution:

$
x \in[-3,5)
$
(iii) $\mathrm{x}<-1$ or $\mathrm{x}<3$
Solution:

$
\begin{aligned}
& \mathrm{x} \in(-\infty,-1) \text { or } \mathrm{x} \in(-\infty, 3) \\
& \text { (iv) }-2 \mathrm{x}>0 \text { or } 3 \mathrm{x}-4<11 \\
& \text { Solution: } \\
& -2 \mathrm{x}>0 \Rightarrow 2 \mathrm{x}<0 \Rightarrow \mathrm{x}<0 \\
& \mathrm{x} \in(-\infty, 0) \\
& 3 \mathrm{x}-4<11 \\
& \Rightarrow 3 \mathrm{x}-4+4<11+4 \\
& 3 x<15 \Rightarrow \frac{3 x}{3}<\frac{15}{3} \\
& \text { (i.e.) } \quad x<5 \\
& x \in(-\infty, 5)
\end{aligned}
$
Question 2.
Solve $23 \mathrm{x}<100$ when
(i) $\mathrm{x}$ is a natural number,
(ii) $\mathrm{x}$ is an integer.
Solution:
$
23 \mathrm{x}<100
$

$
\begin{aligned}
& \Rightarrow \quad \frac{23 x}{23}<\frac{100}{23} \\
& \text { (i.e.,) } \mathrm{x}>4.3
\end{aligned}
$
(i) $\mathrm{x}=1,2,3,4(\mathrm{x} \in \mathrm{N})$
(ii) $\mathrm{x}=\ldots .-3,-2,-1,0,1,2,3,4(\mathrm{x} \in \mathrm{Z})$
Question 3.
Solve $-2 x \geq 9$ when
(i) $\mathrm{x}$ is a real number,
(ii) $\mathrm{x}$ is an integer,
(iii) $\mathrm{x}$ is a natural number.
Solution:
$
\begin{aligned}
&-2 \mathrm{x}>9 \Rightarrow 2 \mathrm{x} \leq-9 \\
& \frac{2 x}{2} \leq-\frac{9}{2} \\
& x \leq-\frac{9}{2}(=-4.5)
\end{aligned}
$
(ii) $\mathrm{x}=\ldots .-3,-2,-1,0,1,2,3,4$
(iii) $\mathrm{x}=1,2,3,4$
Question 4.
Solve (i) $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$
Solution:
$
\begin{array}{rlll} 
& \frac{3(x-2)}{5} & \leq & \frac{5(2-x)}{3} \\
\Rightarrow \quad & \leq & 25(2-x) \\
9(x-2) & \leq & 50-25 x \\
9 x-18 & \leq & 50+18 \\
9 x+25 x & \leq & 68 \\
34 x & \leq & \frac{68}{34} \\
\frac{34 x}{34} &
\end{array}
$
(ii)
$
\frac{5-x}{3}<\frac{x}{2}-4
$

Solution:
$
\begin{aligned}
\frac{5-x}{3} & <\frac{x-8}{2} \\
\Rightarrow \quad 2(5-x) & <3(x-8) \\
10-2 x & <3 x-24 \\
10+24 & <3 x+2 x \\
34 & <5 x \\
\Rightarrow \quad 5 x>34 & \Rightarrow \frac{5 x}{5}>\frac{34}{5} \\
x>\frac{34}{5} \text { (i.e.) } & x \in\left(\frac{34}{5}, \infty\right)
\end{aligned}
$
Question 5.
To secure A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If one scored $84,87,95,91$ in first four subjects, what is the minimum mark one scored in the fifth subject to get $\mathrm{A}$ grade in the course?
Solution:
Required marks $=5 \times 90=450$
Total marks obtained in 4 subjects $=84+87+95+91=357$
So required marks in the fifth subject $=450-357=93$
Question 6.
A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Solution:
$12 \%$ solution of acid in $6001 \Rightarrow 600 \times \frac{12}{100}=721$ of acid
$15 \%$ of $6001 \Rightarrow 600 \times \frac{15}{100}=901$
$18 \%$ of $6001 \Rightarrow 600 \times \frac{18}{100}=1081$
Let $\mathrm{x}$ litres of $18 \%$ acid solution be added
Given, $(600+x) \frac{15}{100} \geq 72+\frac{30 x}{100}$
$(600+\mathrm{x}) 15 \geq 7200+30 \mathrm{x}$
$9000+15 \mathrm{x} \geq 7200+30 \mathrm{x}$
$1800 \geq 15 \mathrm{x}$
$\mathrm{x} \leq 120$
Let $\mathrm{x}$ litres of $18 \%$ acid solution be added
Given, $(600+x) \frac{18}{100} \leq 72+\frac{30}{100} x$

$
\begin{aligned}
& (600+\mathrm{x}) 15 \geq 7200+30 \mathrm{x} \\
& 9000+15 \mathrm{x} \geq 7200+30 \mathrm{x} \\
& 1800 \geq 15 \mathrm{x} \\
& \mathrm{x} \leq 120
\end{aligned}
$
Let $\mathrm{x}$ litres of $18 \%$ acid solution be added
$
\begin{aligned}
& \text { Given, }(600+x) \frac{18}{100} \leq 72+\frac{30}{100} x \\
& (600+x) 18 \leq 7200+30 x \\
& 10800+18 \leq 7200+30 \mathrm{x} \\
& 3600 \leq 12 \mathrm{x}
\end{aligned}
$
$
\mathrm{x}>300
$
The solution is $120 \leq \mathrm{x}>300$
Question 7.
Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40 .
Solution:
Let the two numbers be $\mathrm{x}$ and $\mathrm{x}+2$
$
\begin{aligned}
& \mathrm{x}+\mathrm{x}+2<40 \\
& \Rightarrow 2 \mathrm{x}<38 \\
& \frac{2 x}{2}<\frac{38}{2} \\
& \Rightarrow \mathrm{x}<19 \text { and } \mathrm{x}>10 \\
& \text { so } \mathrm{x}=11 \Rightarrow \mathrm{x}+2=13 \\
& \mathrm{x}=13 \Rightarrow \mathrm{x}+2=15 \\
& \mathrm{x}=15 \Rightarrow \mathrm{x}+2=17
\end{aligned}
$
When $\mathrm{x}=17 \Rightarrow \mathrm{x}+2=19$
So the possible pairs are $(11,13),(13,15),(15,17),(17,19)$

Question 8 .
A model rocket is launched from the ground. The height $\mathrm{h}$ of the rocket after $\mathrm{t}$ seconds from lift off is given by $h(t)=-5 \mathrm{t}^2+100 \mathrm{t} ; 0 \leq \mathrm{r} \leq 20$. At what time the rocket is 495 feet above the ground?
Solution:
$
\begin{aligned}
& \mathrm{h}(\mathrm{t})=-5 \mathrm{t}^2+1 \text { oot } \\
& \text { at } \mathrm{t}=0, \mathrm{~h}(0)=0 \\
& \text { at } \mathrm{t}=1, \mathrm{~h}(1)=-5+100=95 \\
& \text { at } \mathrm{t}=2, \mathrm{~h}(2)=-20+200=180 \\
& \text { at } \mathrm{t}=3, \mathrm{~h}(3)=-45+300=255 \\
& \text { at } \mathrm{t}=4, \mathrm{~h}(4)=-80+400=320 \\
& \text { at } \mathrm{t}=5, \mathrm{~h}(5)=-125+500=375 \\
& \text { at } \mathrm{t}=6, \mathrm{~h}(6)=-180+600=420 \\
& \text { at } \mathrm{t}=7, \mathrm{~h}(7)=-245+700=455 \\
& \text { at } \mathrm{t}=8, \mathrm{~h}(8)=-320+800=480 \\
& \text { at } \mathrm{t}=9, \mathrm{~h}(9)=-405+900=495
\end{aligned}
$
So, at 9 secs, the rocket is 495 feet above the ground.
Question 9.
A Plumber can be paid according to the following schemes: In the first scheme he will be paid Rs. 500 plus Rs. 70 per hour, and in the second scheme he will be paid Rs. 120 per hour. If he works $\mathrm{x}$ hours, then for what value of $\mathrm{x}$ does the first scheme give better wages?
Solution:
I scheme with $\mathrm{x}$ hr
$
\begin{aligned}
& 500+(\mathrm{x}-1) 70=500+70 \mathrm{x}-70 \\
& =430+70 \mathrm{x}
\end{aligned}
$
II scheme with $\mathrm{x}$ hours
$
\begin{aligned}
& 120 \mathrm{x} \\
& \text { Here I }>\text { II } \\
& \Rightarrow 430+70 \mathrm{x}>120 \mathrm{x} \\
& \Rightarrow 120 \mathrm{x}-70 \mathrm{x}<430
\end{aligned}
$
$
\begin{aligned}
& 50 \mathrm{x}<430 \\
& \frac{50 x}{50}<\frac{430}{50} \\
& \mathrm{x}<8.6 \text { (i.e. }
\end{aligned}
$
$x<8.6$ (i.e.) when $x$ is less than 9 hrs the first scheme gives better wages.

Question 10.
A and $B$ are working on similar jobs but their annual salaries differ by more than Rs 6000 . If B earns Rs. 27000 per month, then what are the possibilities of A's salary per month?
Solution:
A's monthly salary $=₹ \mathrm{x}$
B's monthly salary $=₹ 27000$
Their annual salaries differ by ₹ 6000
A's salary $-27000>6000$
A's salary > ₹ 33000
B's salary - A's salary $>6000$
$27000-$ A's salary $>6000$
A's salary < 21000
A's monthly salary will be lesser than ₹ 21,000 or greater than ₹ 33,000