Unit Exercise 3 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Unit Exercise 3
Question $1 .$
Solve $\frac{1}{3}(x+y-5)=y-z=2 x-11=9-(x+2 z)$
Solution:

$x=\frac{18}{3}=6$
(1) becomes, $6-2 \mathrm{y}+3(1)=5 \Rightarrow 9-2 \mathrm{y}=5$
$\Rightarrow 9-5=2 \mathrm{y} \Rightarrow 2 \mathrm{y}=4$ $\therefore \mathrm{y}=\frac{4}{2}=2$ $\therefore$ Solution set is $\{6,2,1\}$
 

Question 2.
One hundred and fifty students are admitted to a school. They are distributed over three sections A, $B$ and C. If 6 students are shifted from section A to section C, the sections will have equal number of students. If 4 times of students of section $\mathrm{C}$ exceeds the number of students of section $A$ by the number of students in section B, find the number of students in the three sections.
Solution:
Let the students in section $A, B, C$ be $a, b, c$, respectively.

Question $3 .$
In a three-digit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number.
Solution:
Let the three digits numbers be $100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}$.
$100 b+10 a+c=3(100 a+10 b+c)+54$
$100 a+106+c+198=100 c+106+a$
$(b-a)=2(b-c) \ldots \ldots . .(3)$
$(1) \Rightarrow 100 \mathrm{~b}+10 \mathrm{a}+\mathrm{c}=300 \mathrm{a}+30 \mathrm{~b}+3 \mathrm{c}+54$
$\Rightarrow 290 \mathrm{a}-70 \mathrm{~b}+2 \mathrm{c}=-54$
(2) $\Rightarrow 99 \mathrm{a}-99 \mathrm{c}=-198 \Rightarrow \mathrm{a}-\mathrm{c}=-2$
$\Rightarrow \mathrm{a}=\mathrm{c}-2$
(3) $\Rightarrow \mathrm{a}+\mathrm{b}-2 \mathrm{c}=0 \Rightarrow \mathrm{a}+\mathrm{b}=2 \mathrm{c}$
$\Rightarrow \mathrm{b}=2 \mathrm{c}-\mathrm{c}+2$
$\Rightarrow \mathrm{b}=\mathrm{c}+2$
Substituting a, b in (1)
$290(c-2)-70(c+2)+2 c=-54$
$290 \mathrm{c}-580-70 \mathrm{c}-140+2 \mathrm{c}=-54$
$222 \mathrm{c}=666 \Rightarrow \mathrm{c}=3$
$a=1,6=5$
$\therefore$ The number is 153 .

Question $4 .$
Find the least common multiple of
$x y\left(k^{2}+1\right)+k\left(x^{2}+y^{2}\right)$ and
$x y\left(k^{2}-1\right)+k\left(x^{2}-y^{2}\right)$
Answer:

$\therefore$ G.C.D of $2 x^{4}+13 x^{3}+21 x^{2}+23 x+7, x^{3}+3 x^{2}+3 x+1, x^{2}+2 x+1$ is $(x+1)^{2}$
 

Question $6 .$
Reduce the given Rational expressions to its lowest form
(i) $\frac{x^{3 a}-8}{x^{2 a}+2 x^{a}+4}$
(ii) $\frac{10 x^{3}-25 x^{2}+4 x-10}{-4-10 x^{2}}$
Solution:
(i) $\frac{x^{3 a}-8}{x^{2 a}+2 x^{a}+4}$
$=\frac{\left(x^{a}\right)^{3}-8}{\left(x^{a}\right)^{2}+2 x^{a}+4}=\frac{\left(x^{a}\right)^{3}-2^{3}}{x^{2 a}+2 x^{a}+4}$
$=\frac{\left(x^{a}-2\right)\left(x^{2 a}+2 x^{a}+4\right)}{x^{2 a}+2 x^{a}+4}=\left(x^{a}-2\right)$

$=\frac{\left(x^{a}-2\right)\left(x^{2 a}+2 x^{a}+4\right)}{x^{2 a}+2 x^{a}+4}=\left(x^{a}-2\right)$
(ii) $\frac{10 x^{3}-25 x^{2}+4 x-10}{-4-10 x^{2}}$
$=\frac{5 x^{2}(2 x-5)+2(2 x-5)}{-2\left(5 x^{2}+2\right)}=\frac{\left(5 x^{2}+2\right)(2 x-5)}{-2\left(5 x^{2}+2\right)}$
$=\frac{(2 x-5)}{-2}=-x+\frac{5}{2}$

Question 7.
$\frac{\frac{1}{p}+\frac{1}{q+r}}{\frac{1}{p}-\frac{1}{q+r}} \times\left[1+\frac{q^{2}+r^{2}-p^{2}}{2 q r}\right]$
Solution:

Question $8 .$
Arul, Ravi and Ram working together can clean a store in 6 hours. Working alone, Ravi takes twice as long to clean the store as Arul does. Ram needs three times as long as Arul does. How long would it take each if they are working alone?
Solution:
Let Aral's speed of working be $x$
Let Ravi's speed of working be y
Let Ram's speed of working be z
given that they are working together. ,
I et $V$ be the quantum of work, $x+y+z=\frac{w}{6}$
Also given that Ravi takes twice the time as Aral for finishing the work.

$\therefore \frac{w}{y}=2 \times \frac{w}{x} \quad \therefore x=2 y$
$\therefore y=\frac{x}{2}$

$\therefore y=\frac{x}{2}$
Also Ram takes 3 times the time as Aral for finishing the work.
$\therefore \frac{w}{z}=3 \times \frac{w}{x}$
$\therefore \mathrm{x}=3 \mathrm{z} \therefore \mathrm{z}=\frac{x}{3}$
Substitute (2) and (3) in (1),
$\begin{aligned}
x+\frac{x}{2}+\frac{x}{3} &=\frac{w}{6} \\
\therefore 6 x+3 x+2 x &=w \\
11 x &=w \\
x &=\frac{w}{11}, y=\frac{w}{22}, z=\frac{w}{33}
\end{aligned}$
Working alone time taken as
Arul: $\frac{w}{x}=\frac{w}{w / 11}=11 \mathrm{hrs}$.
Ravi: $\frac{w}{y}=\frac{w}{w / 22}=22 \mathrm{hrs}$.
$\operatorname{Ram}: \frac{w}{z}=\frac{w}{w / 33}=33$ hrs.

Question 9.
Find the square root of $289 x^{4}-612 x^{3}+970 x^{2}-684 x+361$

Solution:

Question $10 .$

Solution:
Squaring both sides
$\begin{aligned}
&(\sqrt{y+1}+\sqrt{2 y-5})^{2}=3^{2} \\
&y+1+2 y-5+2(\sqrt{y+1} \sqrt{2 y-5})=9 \\
&3 y-4-9=-2 \sqrt{y+1} \sqrt{2 y-5} \\
&9 y^{2}-78 y+169=4(y+1)(2 y-5) \\
&9 y^{2}-78 y+169=4\left(2 y^{2}+2 y-5 y-5\right) \\
&9 y^{2}-78 y+169=8 y^{2}+8 y-20 y-20 \\
&9 y^{2}-78 y+169-8 y^{2}+12 y+20=0 \\
&y^{2}-66 y+189=0 \\
&y^{2}-63 y-3 y+189=0 \\
&y(y-63)-3(y-63)=0 \\
&(y-63)(y-3)=0 \\
&y=63,3
\end{aligned}$

Question 11.
A boat takes $1.6$ hours longer to go $36 \mathrm{kms}$ up a river than down the river. If the speed of the water current is $4 \mathrm{~km}$ per hr, what is the speed of the boat in still water?
Solution:
Let the speed of boat in still water be ' $\mathrm{v}$ '
$\begin{aligned}
&\therefore \text { speed }=\frac{\text { distance }}{\text { time }} \Rightarrow \text { time }=\frac{\text { distance }}{\text { speed }} \\
&\therefore \frac{36}{v-4}-\frac{36}{v+4}=\frac{96}{60}=\frac{8}{5} \quad\left(\because 1.6 \mathrm{hrs}=\frac{96}{60}\right)
\end{aligned}$

$\begin{aligned}
&\Rightarrow 36(\mathrm{v}+4)-36(\mathrm{v}-4)=\frac{8}{5}(\mathrm{v}-4)(\mathrm{v}+4) \\
&\Rightarrow 36 \mathrm{v}+144-36 \mathrm{v}+144=\frac{8}{5}\left(\mathrm{v}^{2}-4 \mathrm{v}+4 \mathrm{v}-16\right) \\
&\Rightarrow 288=\frac{8}{5} \mathrm{v}^{2}-\frac{128}{5} \Rightarrow 8 \mathrm{v}^{2}-128=1440 \\
&\Rightarrow 8 \mathrm{v}^{2}=1568 \Rightarrow \mathrm{v}^{2}=196 \mathrm{v}=\pm 14 \\
&\therefore \text { Speed of the boat }=14 \mathrm{~km} / \mathrm{hr} .(\because \text { speed cannot be }-\mathrm{ve})
\end{aligned}$

Question $12 .$
Is it possible to design a rectangular park of perimeter $320 \mathrm{~m}$ and area $4800 \mathrm{~m}^{2}$ ? If so find its length and breadth.
Solution:
Let the length and breadth of the rectangle be $1 \mathrm{~m}$ and $\mathrm{bm}$
Given $2(1+b)$
$\Rightarrow 1+b=160$
Also $1 \mathrm{~b}=4800$
$\quad b=\frac{4800}{l}$
Substituting $(2)$ in (1) we get
$\begin{aligned}
l+\frac{4800}{l}=& 160 \\
\Rightarrow l^{2}+4800 &=160 l \\
\Rightarrow(l-120)(l-40) &=0
\end{aligned}$

$\Rightarrow l=120$ or 40
When $l=120, b=\frac{4800}{120}=40$
When $l=40, b=\frac{4800}{40}=120$
$\therefore$ Length and breadth of the rectangular park is $120 \mathrm{~m}$ and $40 \mathrm{~m}$
 

Question $13 .$
At $t$ minutes past $2 \mathrm{pm}$, the time needed to $3 \mathrm{pm}$ is 3 minutes less than $\frac{t^{2}}{4}$ Find $\mathrm{t}$.
Solution:
$60-\mathrm{t}=\frac{t^{2}}{4}-3$
$\Rightarrow t^{2}-12=240-4 t$
$\Rightarrow t^{2}+4 t-252=0$
$\Rightarrow t^{2}+18 t-14 t-252=0$
$\Rightarrow t(t+18)-14(t+18)=0$
$\Rightarrow(t+18)(t-14)=0$
$\therefore t=14$ or $t=-18$ is not possible.

Question $14 .$
The number of seats in a row is equal to the total number of rows in a hall. The total number of seats in the hall will increase by 375 if the number of rows is doubled and the number of seats in each row is reduced by 5 . Find the number of rows in the hall at the beginning.
Solution:
Let the no of seats in each row be $\mathrm{x}$
$(x-5)(2 x)=x^{2}+375$

$\begin{aligned}
&\Rightarrow 2 x^{2}-10 x=x^{2}+375 \\
&\Rightarrow x^{2}-10 x-375=0 \\
&\Rightarrow x^{2}-25 x+15 x-375=0 \\
&\Rightarrow x(x-25)+15(x-25)=0 \\
&\Rightarrow(x-25)(x+15)=0 \\
&\Rightarrow x=25, x=-15, x>0
\end{aligned}$
$\therefore 25$ rows are in the hall.
 

Question $15 .$
If $\mathrm{a}$ and $\mathrm{b}$ are the roots of the polynomial $f(x)=x^{2}-2 x+3$, find the polynomial whose roots are
(i) $\alpha+2, \beta+2$
(ii) $\frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}$
Solution:
$f(x)=\frac{1 x^{2}}{a}-\frac{2 x}{b}+\frac{3}{c}$
Sum of the roots $(\alpha+\beta)=\frac{-b}{a}=-\frac{(-2)}{1}$
Product of the roots $(\alpha \beta)=\frac{c}{a}=\frac{3}{1}=3$
(i) $\alpha+2, \beta+2$ are the roots (given)
Sum of the roots $=\alpha+2+\beta+2$
$\begin{aligned}
&=\alpha+\beta+4 \\
&=2+4=6
\end{aligned}$
Product of the roots $=(\alpha+2)(\beta+2)$
$\begin{aligned}
&=\alpha \beta+2 \alpha+2 \beta+4 \\
&=\alpha \beta+2(\alpha+\beta)+4 \\
&=3+2 \times 2+4 \\
&=3+4+4=11
\end{aligned}$
$\therefore$ The required equation $=\mathrm{x}^{2}-6 \mathrm{x}+11=0$.

$\begin{aligned}
&\text { (ii) } \frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}\\
&\Rightarrow \text { sum of the roots }=\frac{\alpha-1}{\alpha+1}+\frac{\beta-1}{\beta+1}\\
&\text { Sum }=\frac{(\alpha-1)(\beta+1)+(\beta-1)(\alpha+1)}{(\alpha+1)(\beta+1)}\\
&=\frac{\alpha \beta-\beta+\alpha-1+\alpha \beta-a+\beta-1}{\alpha \beta+\beta+\alpha+1}\\
&=\frac{\alpha \beta+\alpha \beta-2}{\alpha \beta+(\alpha+\beta)+1}=\frac{2 \alpha \beta-2}{\alpha \beta+(\alpha+\beta)+1}\\
&=\frac{2(3)-2}{3+2+1}=\frac{2}{6}=\frac{2}{3}\\
&\text { Product }=\frac{\alpha-1}{\alpha+1} \times \frac{\beta-1}{\beta+1}=\frac{(\alpha-1)(\beta-1)}{(\alpha+1)(\beta+1)}\\
&=\frac{\alpha \beta-\beta-\alpha+1}{\alpha \beta+\beta+\alpha+1}=\frac{\alpha \beta-(\alpha+\beta)+1}{\alpha \beta+(\alpha+\beta)+1}\\
&=\frac{3-2+1}{3+2+1}\\
&=\frac{2}{6}=\frac{1}{3}\\
&\therefore \text { Required equation }=x^{2}-\frac{2}{3} x+\frac{1}{3}=0\\
&\Rightarrow 3 \mathrm{x}^{2}-2 \mathrm{x}+1=0
\end{aligned}$

Question $16 .$
If $=4$ is a root of the equation
$x^{2}+p x-4=0$ and if the equation
$x^{2}+p x+q=0$ has equal roots, find the values of $p$ and $q$.
Answer:
Let $p(x)=x^{2}+p x-4$
$-4$ is the root of the equation
$\begin{aligned}
&P(-4)=0 \\
&16-4 p-4=0 \\
&-4 p+12=0 \\
&-4 p=-12 \\
&p=\frac{12}{4}=3
\end{aligned}$
The equation $x^{2}+p x+q=0$ has equal roots
$x^{2}+3 x+q=0$
Here $a=1, b=3, c=q$
since the roots are real and equal
$\begin{aligned}
&b^{2}-4 a c=0 \\
&3^{2}-4(1)(q)=0 \\
&9-4 q=0 \\
&9=4 q \\
&q=\frac{9}{4}
\end{aligned}$
The value of $p=3$ and $q=\frac{9}{4}$

Question $17 .$
Two farmers Senthil and Ravi cultivates three varieties of grains namely rice, wheat and ragi. If the sale (in $\square$ ) of three varieties of grains by both the farmers in the month of April is given by the matrix.

May month sale (in $\sqcup$ ) is exactly twice as that of the April month sale for each variety.
(i) What is the average sales of the months April and May.
(ii) If the sales continues to increase in the same way in the successive months, what will be sales in the month of August?
Solution:
$\begin{aligned}
A &=\left[\begin{array}{lll}
500 & 1000 & 500 \\
2500 & 1500 & 500
\end{array}\right] \\
\text { May } &=2 \times A=\left[\begin{array}{lll}
1000 & 2000 & 3000 \\
5000 & 3000 & 1000
\end{array}\right]=M
\end{aligned}$
(i) Average $=\frac{\mathrm{A}+\mathrm{M}}{2}=\left[\begin{array}{lll}750 & 1500 & 2250 \\ 3750 & 2250 & 750\end{array}\right]$

(ii)
$\begin{aligned}
\text { May } &=2 \mathrm{~A} \\
\text { June } &=2 \times \text { May }=4 \mathrm{~A} \\
\text { July } &=2 \times \text { June }=8 \mathrm{~A} \\
\text { August } &=2 \times \text { July }=16 \mathrm{~A} \\
\Rightarrow \quad \text { August } &=\left[\begin{array}{lll}
8000 & 16000 & 24000 \\
40000 & 24000 & 8000
\end{array}\right]
\end{aligned}$

Question $18 .$
$\begin{aligned}
&\text { If } \cos \theta\left[\begin{array}{ll}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
x & -\cos \theta \\
\cos \theta & x
\end{array}\right] \\
&=\text { I }_{2} \text {. Find } x .
\end{aligned}$
Solution:

$\begin{aligned}
&\text { L.H.S }=\cos \theta\left[\begin{array}{ll}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{ll}
x & -\cos \theta \\
\cos \theta & x
\end{array}\right] \\
&=\left[\begin{array}{rr}
\cos ^{2} \theta & \cos \sin \theta \\
-\sin \theta \cos \theta & \cos ^{2} \theta
\end{array}\right]+\left[\begin{array}{rr}
x \sin \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & x \sin \theta
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{rr}
\cos ^{2} \theta+x \sin \theta & 0 \\
0 & \cos ^{2} \theta+x \sin \theta
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \text { given } \\
&\therefore \cos ^{2} \theta+x \sin \theta=1 \\
&x \sin \theta=1-\cos ^{2} \theta \\
&x=\frac{\sin 2}{\sin \theta}=\sin \theta
\end{aligned}$

Question $19 .$
Given $A=\left[\begin{array}{ll}p & 0 \\ 0 & 2\end{array}\right], B=\left[\begin{array}{ll}0 & q \\ 1 & 0\end{array}\right], C=\left[\begin{array}{rr}2 & -2 \\ 2 & 2\end{array}\right]$ and if $B A=C^{2}$, find $p$ and $q$.

Solution:
$\mathrm{A}=\left[\begin{array}{ll}p & 0 \\ 0 & 2\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}0 & -q \\ 1 & 0\end{array}\right], \mathrm{C}=\left[\begin{array}{ll}2 & -2 \\ 2 & 2\end{array}\right]$
$\mathrm{BA}=\mathrm{C}^{2} \Rightarrow\left[\begin{array}{ll}0 & -q \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}p & 0 \\ 0 & 2\end{array}\right]$
$=\left[\begin{array}{ll}2 & -2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}2 & -2 \\ 2 & 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}0 & -2 q \\ p & 0\end{array}\right]=\left[\begin{array}{ll}(4-4) & (-4-4) \\ (4+4) & (-4+4)\end{array}\right]$
$\begin{tabular}{r|r}
$-2 q=-8$ & $p=8$ \\
$q=4$ & $q=4$
\end{tabular}$

Question $20 .$
$A=\left[\begin{array}{ll}3 & 0 \\ 4 & 5\end{array}\right], B=\left[\begin{array}{ll}6 & 3 \\ 8 & 5\end{array}\right], C=\left[\begin{array}{ll}3 & 6 \\ 1 & 1\end{array}\right]$ find the matrix D, such that $C D-A B=0$.
Solution:
$\begin{aligned}
&\mathrm{A}=\left[\begin{array}{ll}
3 & 0 \\
4 & 5
\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}
6 & 3 \\
8 & 5
\end{array}\right], \mathrm{C}=\left[\begin{array}{ll}
3 & 6 \\
1 & 1
\end{array}\right] \\
&\mathrm{CD}-\mathrm{AB}=0 \Rightarrow \mathrm{CD}=\mathrm{AB} \\
&\mathrm{AB}=\left[\begin{array}{ll}
3 & 0 \\
4 & 5
\end{array}\right]\left[\begin{array}{ll}
6 & 3 \\
8 & 5
\end{array}\right]=\left[\begin{array}{ll}
(18+0) & (9+0) \\
(24+40) & (12+25)
\end{array}\right]
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
&=\left[\begin{array}{ll}
18 & 9 \\
64 & 37
\end{array}\right] \\
\Rightarrow \mathrm{AB}=\mathrm{CD} &=\left[\begin{array}{ll}
18 & 9 \\
64 & 37
\end{array}\right]
\end{aligned}\\
&\begin{aligned}
\text { Let } \mathrm{D} &=\left[\begin{array}{ll}
x & y \\
z & w
\end{array}\right] \\
{\left[\begin{array}{ll}
3 & 6 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
x & y \\
z & w
\end{array}\right] } &=\left[\begin{array}{ll}
18 & 9 \\
64 & 37
\end{array}\right] \\
{\left[\begin{array}{ll}
3 x+6 z & 3 y+6 w \\
x+z & y+w
\end{array}\right] } &=\left[\begin{array}{ll}
18 & 9 \\
64 & 37
\end{array}\right] \\
3 x+6 z &=18 \\
x+z &=64
\end{aligned}
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
&\text { Sub. } x=122 \text { in }(2) \\
&\begin{aligned}
122+z &=64 \\
z &=64-122=-58 \\
3 y+6 w &=9 \\
y+w &=37
\end{aligned}
\end{aligned}\\
&x=122
\end{aligned}$

$\begin{aligned}
&\text { (3) }-3(4) \Rightarrow 3 y+6 w=9\\
&\frac{\begin{array}{c}
3 y+3 w \\
(-)
\end{array}=\begin{array}{c}
111 \\
(-)
\end{array}}{3 w=-102}\\
&w=-34\\
&\text { Sub. } w=-34 \text { in (4) }\\
&y-34=37\\
&y=37+34=71\\
&\therefore \text { Solutions: } x=122\\
&\begin{aligned}
&y=71 \\
&z=-58
\end{aligned} \quad \therefore \mathrm{D}=\left[\begin{array}{cc}
122 & 71 \\
-58 & -34
\end{array}\right]\\
&w=-34
\end{aligned}$