Samacheer Kalvi 10th Maths Exercise 3.12 Solutions

Let the number be $x$. Its reciprocal is $\dfrac{1}{x}$.

Given $x-\dfrac{1}{x}=\dfrac{24}{5}$.

$\dfrac{x^2-1}{x}=\dfrac{24}{5}$

$5x^2-5=24x$

$5x^2-24x-5=0$

$5x^2-25x+x-5=0$

$5x(x-5)+1(x-5)=0$

$(5x+1)(x-5)=0$

$x=5$ or $x=-\dfrac{1}{5}$.

Check: $5-\dfrac{1}{5}=\dfrac{24}{5}$ and $-\dfrac{1}{5}+5=\dfrac{24}{5}$. Both are valid.

Garden is $12$ m by $16$ m and the path of width $w$ runs all around it.

New length $=16+2w$, new breadth $=12+2w$.

Total area $=285$ m$^2$:

$(12+2w)(16+2w)=285$

$192+24w+32w+4w^2=285$

$4w^2+56w-93=0$

$a=4,\ b=56,\ c=-93$

$w=\dfrac{-56\pm\sqrt{3136+1488}}{8}=\dfrac{-56\pm\sqrt{4624}}{8}=\dfrac{-56\pm 68}{8}$

$w=\dfrac{12}{8}=\dfrac{3}{2}$ or $w=-\dfrac{124}{8}$.

Width cannot be negative, so $w=\dfrac{3}{2}=1.5$ m.

Let the original speed be $x$ km/hr.

Time at original speed $=\dfrac{90}{x}$ hr; time at increased speed $=\dfrac{90}{x+15}$ hr.

$30$ minutes $=\dfrac{1}{2}$ hr, so

$\dfrac{90}{x}-\dfrac{90}{x+15}=\dfrac{1}{2}$

$90\left(\dfrac{x+15-x}{x(x+15)}\right)=\dfrac{1}{2}$

$\dfrac{1350}{x^2+15x}=\dfrac{1}{2}$

$x^2+15x=2700$

$x^2+15x-2700=0$

$(x+60)(x-45)=0$

$x=-60$ or $x=45$. Speed cannot be negative, so $x=45$.

Let the sister's present age be $x$ years. Then the girl's age is $2x$ years.

After $5$ years: sister $=x+5$, girl $=2x+5$.

$(x+5)(2x+5)=375$

$2x^2+5x+10x+25=375$

$2x^2+15x-350=0$

$2x^2+35x-20x-350=0$

$x(2x+35)-10(2x+35)=0$

$(2x+35)(x-10)=0$

$x=-\dfrac{35}{2}$ or $x=10$. Age cannot be negative, so $x=10$.

Let $R$ be the position of the pole on the circle and $PQ$ the diameter $=20$ m.

Angle in a semicircle is $90^\circ$, so $\angle PRQ=90^\circ$.

Let $QR=x$ m. Then $PR=x+4$ m (difference is $4$ m).

By Pythagoras, $PR^2+QR^2=PQ^2$:

$(x+4)^2+x^2=20^2$

$x^2+8x+16+x^2=400$

$2x^2+8x-384=0$

$x^2+4x-192=0$

$(x+16)(x-12)=0$

$x=-16$ or $x=12$. Distance cannot be negative, so $x=12$.

Then $PR=12+4=16$ m.

Since a real positive solution exists, the pole can be erected.

Total number of bees $=2x^2$.

Half the group $=x^2$; its square root $=x$ bees went to the tree.

Eight-ninth of the group $=\dfrac{8}{9}(2x^2)=\dfrac{16x^2}{9}$ bees went to the same tree.

Remaining bees $=2$.

$x+\dfrac{16x^2}{9}+2=2x^2$

Multiply by $9$: $9x+16x^2+18=18x^2$

$2x^2-9x-18=0$

$2x^2-12x+3x-18=0$

$2x(x-6)+3(x-6)=0$

$(2x+3)(x-6)=0$

$x=-\dfrac{3}{2}$ or $x=6$; take $x=6$.

Total bees $=2x^2=2(36)=72$.

Let the person stand at a distance $x$ m from the first gallery ($4$ singers).

Then the distance from the second gallery ($9$ singers) is $(70-x)$ m.

Ratio of intensities $=$ square of ratio of distances:

$\dfrac{4}{9}=\dfrac{x^2}{(70-x)^2}$

$9x^2=4(70-x)^2$

$9x^2=4(4900-140x+x^2)$

$9x^2=19600-560x+4x^2$

$5x^2+560x-19600=0$

$x^2+112x-3920=0$

$(x+140)(x-28)=0$

$x=-140$ or $x=28$. Distance cannot be negative, so $x=28$.

Distance from the second gallery $=70-28=42$ m.

Side of the square field $=10$ m, so its area $=100$ m$^2$.

Let the width of the gravel path be $x$ m.

Side of the flower bed $=10-2x$, so its area $=(10-2x)^2$.

Area of the gravel path $=100-(10-2x)^2$.

Total cost: $3(10-2x)^2+4\left[100-(10-2x)^2\right]=364$

$400-(10-2x)^2=364$

$(10-2x)^2=36$

$10-2x=6$ (taking the positive value)

$2x=4\Rightarrow x=2$.

Let the two legs be $a$ and $b$, with hypotenuse $25$ cm.

Perimeter: $a+b+25=56\Rightarrow a+b=31$.

Pythagoras: $a^2+b^2=25^2=625$.

$(a+b)^2=a^2+b^2+2ab$

$961=625+2ab\Rightarrow 2ab=336\Rightarrow ab=168$.

So $a,b$ are the roots of $t^2-31t+168=0$.

$t=\dfrac{31\pm\sqrt{961-672}}{2}=\dfrac{31\pm\sqrt{289}}{2}=\dfrac{31\pm 17}{2}$

$t=24$ or $t=7$.

The sides are $7$ cm, $24$ cm and $25$ cm.