Exercise 3.12 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.12$
Question $1 .$
If the difference between a number and its reciprocal is $\frac{24}{5}$, find the number.
Solution:
Let a number be $x$.
Its reciprocal is $\frac{1}{x}$
$\begin{aligned}
&x-\frac{1}{x}=\frac{24}{5} \\
&\frac{x^{2}-1}{x}=\frac{24}{5} \\
&5 x^{2}-5-24 x=0 \Rightarrow 5 x^{2}-24 x-5=0 \\
&5 x^{2}-25 x+x-5=0 \\
&5 x(x-5)+1(x-5)=0 \\
&(5 x+1)(x-5)=0 \\
&x=\frac{-1}{5}, 5
\end{aligned}$
$\therefore$ The number is $\frac{-1}{5}$ or 5 .
 

Question $2 .$
A garden measuring $12 \mathrm{~m}$ by $16 \mathrm{~m}$ is to have a pedestrian pathway that is ' $w$ ' meters wide installed all the way around so that it increases the total area to $285 \mathrm{~m}^{2}$. What is the width of the pathway?
Solution:
Area of $\mathrm{ABCD}=16 \times 12^{2}$
$-192 \mathrm{~m}^{2}$
Area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}(12+2 w)(16+2 w)$
$192+32 w+24 w+4 w^{2}=285$

$\begin{aligned}
&4 w^{2}+56 w-93=0 \\
&4 w^{2}+62 w-6 w-93=0 \\
&2 w(2 w+31)-3(2 w+31)=0 \\
&(2 w-3)(2 w+31)=0 \\
&w=1.5 \text { or } \frac{-31}{2}=15.5 \\
&w=-15.5 \text { cannot possible } 3 \\
&\therefore w=\frac{3}{2}=1.5 \mathrm{~m}
\end{aligned}$
(w cannot be (-ve))
The width of the pathway $=1.5 \mathrm{~m}$.
 

Question 3.
A bus covers a distance of $90 \mathrm{~km}$ at a uniform speed. Had the speed been $15 \mathrm{~km} /$ hour more it would have taken 30 minutes less for the journey. Find the original speed of the bus.
Solution:
Let $x \mathrm{~km} / \mathrm{hr}$ be the constant speed of the bus.
The time taken to cover $90 \mathrm{~km}=\frac{90}{x}$ hrs.
When the speed is increased bus $15 \mathrm{~km} / \mathrm{hr}$.
$=\frac{90}{x+15}$
It is given that the time to cover $90 \mathrm{~km}$ is reduced by $\frac{1}{2}$ hrs.
$\begin{aligned}
\Rightarrow \frac{90}{x}-\frac{90}{x+15} &=\frac{1}{2} \\
\frac{90(x+15)-90 x}{x(x+15)} &=\frac{1}{2} \Rightarrow \frac{90 x+1350-90 x}{x^{2}+15 x}=\frac{1}{2}
\end{aligned}$

$\begin{aligned}
x^{2}+15 x &=2700 \\
x^{2}+15 x-2700 &=0 \\
&=\frac{-15 \pm \sqrt{225+10800}}{2} \\
=\frac{-15 \pm \sqrt{11025}}{2} & \Rightarrow \frac{-15+105}{2}=\frac{-15-105}{2} \\
=\frac{-15 \pm 105}{2} & \begin{array}{l}
\text { as the roots are } \\
\text { real and equal }
\end{array} \\
& \Rightarrow 45,-60
\end{aligned}$
The speed of the bus cannot be -ve value
$\therefore$ The original speed of the bus is $45 \mathrm{~km} / \mathrm{hr}$.

Question $4 .$
A girl is twice as old as her sister. Five years hence, the product of their ages (in years) will be 375 .
Find their present ages.
Solution:
Let the age of the girl be $=2 y$ years
Her sister's age is $=y$ years
$\begin{aligned}
(2 y+5)(y+5)=375 \\
2 y^{2}+5 y+10 y+25-375=0 \\
2 y^{2}+15 y-350=0 \\
y &=\frac{-15 \pm \sqrt{15^{2}-4 \times 2 \times-350}}{2 \times 2} \\
&=\frac{-15 \pm \sqrt{3025}}{4} \\
&=\frac{-15 \pm 55}{4} \text { or } \frac{-15-55}{4} \\
&=\frac{+40}{4} \text { or } \frac{-70}{4}
\end{aligned}$
$y=10$, y cannot be (-ve).
$\therefore$ Girls age is $2 \mathrm{y}=20$ years.
Her sister's age $=\mathrm{v}=10$ vears.

Question $5 .$
A pole has to be erected at a point on the T boundary of a circular ground of diameter $j 20 \mathrm{~m}$ in such a way that the difference of its i distances from two diametrically opposite $j$ fixed gates $P$ and $Q$ on the boundary is $4 \mathrm{~m}$. Is $i$ it possible to do so? If answer is yes at what $j$ distance from the two gates should the pole $j$ be erected?
Solution:
$\mathrm{PQ}=20 \mathrm{~m}$
$P X-X Q=4 m$

Squaring both sides,
$\begin{aligned}
&\mathrm{PX}^{2}+\mathrm{XQ}^{2}-2 \mathrm{PX} \cdot \mathrm{QX}=16\left(\because \angle \mathrm{Q} \times \mathrm{p}=90^{\circ}\right) \\
&\mathrm{PQ}{ }^{2}-2 \mathrm{P} \times \mathrm{QX}=16 \\
&400-16=2 \mathrm{PX} \times \mathrm{QX} \\
&384=2 \mathrm{PX}-\mathrm{QX} \\
&\mathrm{PX} \cdot \mathrm{QX}=192 \\
&\therefore(\mathrm{PX}+\mathrm{QX})^{2}=\mathrm{PX}{ }^{2}+\mathrm{QX}^{2}+2 \mathrm{PX} \cdot \mathrm{QX} \\
&=400+2 \times 192 \\
&=784=28^{2} \\
&\therefore \mathrm{PX}+\mathrm{QX}=28 \\
&\text { From (1) }) \text { \& }(2) 2 \mathrm{PX}=32 \Rightarrow \mathrm{PX}=16 \mathrm{~m} \mathrm{QX}=12 \mathrm{~m} \\
&\therefore \text { Yes, the distance from the two gates to the pole } \mathrm{PX} \text { and } \mathrm{QX} \text { is } 12 \mathrm{~m}, 16 \mathrm{~m} . \\
&\text { Question 6. } \\
&\text { From a group of hlack hees } 2 \mathrm{x}^{2} \text {, square root of half of the group went to a tree Again eight- ninth } \\
&\text { of the hees went to the same tree. The remaining two got caught up in a fragrant lotis. How many } \\
&\text { bees were there in total? } \\
&\text { Solution: } \\
&\text { Total no. of bees }=2 \mathrm{x}^{2}
\end{aligned}$
No. of bees that went to a tree $=\sqrt{\frac{1}{2} \times 2 x^{2}}=$ $\sqrt{x^{2}}=x$

Second batch of bees that went to tree $=\frac{8}{9} \times 2 x^{2}$
After this, only 2 are left.
$\therefore 2 x^{2}-x-\frac{16}{9} x^{2}=2$
$\begin{aligned}
&18 x^{2}-9 x-16 x^{2}=2 \times 9 \\
&2 x^{2}-9 x-18=0 \\
&(x-6)(2 x+3)=0 \\
&x=6, x=\frac{-3}{2} \text { (it is not possible) }
\end{aligned}$
No. of bees in total $=2 \mathrm{x}^{2}$
$=2 \times 6^{2}=72$

Question 7.
Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of $70 \mathrm{~m}$. Where should a person stand for hearing the same intensity of the singers voice? (Hint: The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances).
Solution:
Let the person stand at a distance 'd' from 2 nd gallery having 9 singers.

Given that ratio of sound intensity is equal to the square of the ratio of their conesponding distance.
$\begin{aligned}
&\therefore \frac{9}{4}=\frac{d^{2}}{(70-d)^{2}} \\
&4 \mathrm{~d}^{2}=9(70-\mathrm{d})^{2} \\
&4 \mathrm{~d}^{2}=9\left(70^{2}-140 \mathrm{~d}+\mathrm{d}^{2}\right) \\
&4 \mathrm{~d}^{2}=9 \times 70^{2}-9 \times 140 \mathrm{~d}+9 \mathrm{~d}^{2} \\
&\therefore 5 \mathrm{~d}^{2}-9 \times 140 \mathrm{~d}+9 \times 70^{2}=0
\end{aligned}$
$5 \mathrm{~d}^{2}=1260 \mathrm{~d}+44100=0$
$\mathrm{d}^{2}-252 \mathrm{~d}+8820=0$
$d=\frac{252 \pm \sqrt{252^{2}-4 \times 1 \times 8820}}{2 \times 1}$
$=\frac{252 \pm \sqrt{63504-35280}}{2}$
$=\frac{252 \pm \sqrt{98784}}{2}$
$=\frac{252 \pm 168}{2}$

$\begin{aligned}
&=\frac{252+168}{2} \text { or } \frac{252-168}{2} \\
&=\frac{420}{2} \text { or } \frac{84}{2} \\
&=120 \text { or } 42
\end{aligned}$
$\therefore$ The person stand at a distance $28 \mathrm{~m}$ from the first and $42 \mathrm{~m}$ from second gallery.
 

Question $8 .$
There is a square field whose side is $10 \mathrm{~m}$. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at $\square 3$ and $\square 4$ per square metre respectively is $\square 364$. Find the width of the gravel path.
Solution:
Area of the flower bed $=\mathrm{a}^{2}$
Area of the gravel path $=100-\mathrm{a}^{2}$
Area of total garden $=100$
given cost of flower bed $+$ gravelling $=\square 364$
$\begin{aligned}
&3 a^{2}+4\left(100-a^{2}\right)=\square 364 \\
&3 a^{2}+400-4 a^{2}=364 \\
&\therefore a^{2}=400-364
\end{aligned}$
$=36 \Rightarrow a=6$

width of gravel path $=\frac{10-6}{2}=\frac{4}{2}=2 \mathrm{~cm}$
 

Question 9.
Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: "If I had your eggs, I would have earned $\square 15$ ", to which the second replied: "If I had your eggs, I would have earned $\square 6 \frac{2}{3}$ ". How many eggs did each had in the beginning? Answer:
Number of eggs for the first women be ' $x$ '
Let the selling price of each women be ' $y$ '
Selling price of one egg for the first women $=\frac{y}{100-x}$
By the given condition
$(100-x) \frac{y}{x}=15$ (for first women)
$y=\frac{15}{100-x} \cdots \ldots(1)$ $x \times \frac{y}{(100-z)}=\frac{20}{3}$ [For second women] $y=\frac{20(100-x)}{3 x}$
$y=\frac{20(100-x)}{3 x} \ldots \ldots . .(2)$ From $(1)$ and $(2)$ We get
From (1) and (2) We $\frac{15}{100-x}=\frac{20(100-x)}{3 x}$
$45 \mathrm{x}^{2}=20(100-\mathrm{x})^{2}$
$(100-\mathrm{x})^{2}=\frac{45 x^{2}}{20}=\frac{9}{4} \mathrm{x}^{2}$ $\therefore 100-\mathrm{x}=\sqrt{\frac{9}{4} x^{2}}$
$100-x=\frac{3 x}{2}$
$3 x=2(100-x)$
$3 x-200-2 x$
$3 \mathrm{x}+2 \mathrm{x}=200 \Rightarrow 5 \mathrm{x}=200$
$x=\frac{200}{5} \Rightarrow x=40$
Number of eggs with the first women $=40$
Number of eggs with the second women $=(100-40)=60$

Question $10 .$
The hypotenuse of a right-angled triangle is $25 \mathrm{~cm}$ and its perimeter $56 \mathrm{~cm}$. Find the length of the smallest side.
Solution:

$\begin{aligned}
&\mathrm{AB}+\mathrm{BC}+\mathrm{CA}=56 \mathrm{~cm}\\
&\mathrm{AC}=25 \mathrm{~cm}\\
&\mathrm{AB}+\mathrm{BC}=56-25=31\\
&A B^{2}+B C^{2}=A C^{2}\\
&(\mathrm{AB}+\mathrm{BC})^{2}-2 \mathrm{AB} \cdot \mathrm{BC}=\mathrm{AC} 2\left[\because \mathrm{a}^{2}+\mathrm{b}^{2}=(\mathrm{a}+\mathrm{b})^{2}-2 \mathrm{ab}\right]\\
&31^{2}-2 \mathrm{AB}, \mathrm{BC}=25^{2}\\
&-2 \mathrm{AB}, \mathrm{BC}=625-961\\
&\begin{array}{rr}
-2 \mathrm{AB} \cdot \mathrm{BC}=-336 & \text { Hint: } \\
\mathrm{AB} \cdot \mathrm{BC}=168 & \because \alpha+\beta=31 \\
\alpha \beta=168
\end{array}\\
&\begin{array}{r}
\therefore \text { Quadratic equation is } \\
x^{2}-31 x+168=0
\end{array}\\
&x=\frac{-(-31) \pm \sqrt{(-31)^{2}-4 \times 1 \times 168}}{2 \times 1}\\
&=\frac{31 \pm \sqrt{961-672}}{2}\\
&=\frac{31 \pm \sqrt{289}}{2}\\
&=\frac{31 \pm 17}{2} \Rightarrow \frac{31+17}{2} \text { or } \frac{31-17}{2}\\
&=\frac{48}{2}, \frac{14}{2} \Rightarrow 24,7\\
&\therefore \text { The length of the smallest side is } 7 \mathrm{~cm} \text {. }
\end{aligned}$