Samacheer Kalvi 10th Maths Unit Exercise 3 Solutions

Equate the parts two at a time.

$\dfrac{1}{3}(x+y-5)=y-z \Rightarrow x+y-5=3y-3z$

$x-2y+3z=5 \quad \ldots (1)$

$y-z=2x-11 \Rightarrow 2x-y+z=11 \quad \ldots (2)$

$2x-11=9-(x+2z) \Rightarrow 3x+2z=20 \quad \ldots (3)$

From (2), $y=2x+z-11$. Put in (1):

$x-2(2x+z-11)+3z=5 \Rightarrow -3x+z=-17$

So $z=3x-17$. Put in (3):

$3x+2(3x-17)=20 \Rightarrow 9x=54 \Rightarrow x=6$

$z=3(6)-17=1$ and $y=2(6)+1-11=2$

Check: $\dfrac{1}{3}(6+2-5)=1,\ 2-1=1,\ 12-11=1,\ 9-(6+2)=1$

Let the sections $A$, $B$, $C$ have $x$, $y$, $z$ students.

$x+y+z=150 \quad \ldots (1)$

Shifting 6 from A to C makes them equal:

$x-6=z+6 \Rightarrow x-z=12 \quad \ldots (2)$

$4$ times C exceeds A by B:

$4z-x=y \Rightarrow x+y-4z=0 \quad \ldots (3)$

$(1)-(3):\ 5z=150 \Rightarrow z=30$

From (2), $x=30+12=42$

From (1), $y=150-42-30=78$

Let the digits be $x$ (hundreds), $y$ (tens), $z$ (units).

Number $=100x+10y+z$.

Interchanging tens and hundreds digits:

$100y+10x+z=3(100x+10y+z)+54$

$\Rightarrow 70y-290x-2z=54 \Rightarrow 35y-145x-z=27 \quad \ldots (1)$

Adding 198 reverses the digits:

$100x+10y+z+198=100z+10y+x$

$\Rightarrow 99z-99x=198 \Rightarrow z-x=2 \quad \ldots (2)$

Third condition: $y-x=2(y-z) \Rightarrow y=2z-x \quad \ldots (3)$

From (2), $z=x+2$; then from (3), $y=2(x+2)-x=x+4$.

Put in (1): $35(x+4)-145x-(x+2)=27$

$-111x+138=27 \Rightarrow x=1$

So $x=1,\ y=5,\ z=3$.

Check: $513=3(153)+54$ and $153+198=351$

First expression: $k^2xy+xy+kx^2+ky^2$

$=kx(ky+x)+y(x+ky)=(x+ky)(kx+y)$

Second expression: $k^2xy-xy+kx^2-ky^2$

$=kx(ky+x)-y(x+ky)=(x+ky)(kx-y)$

Common factor $=(x+ky)$.

LCM $=(x+ky)(kx+y)(kx-y)$

Take the last two polynomials first.

Divide $x^3+3x^2+3x+1$ by $x^2+2x+1$:

quotient $x+1$, remainder $0$.

So GCD of these two is $x^2+2x+1$.

Now divide $2x^4+13x^3+27x^2+23x+7$ by $x^2+2x+1$:

$2x^2$ gives $2x^4+4x^3+2x^2$; remainder $9x^3+25x^2+23x+7$

$9x$ gives $9x^3+18x^2+9x$; remainder $7x^2+14x+7$

$7$ gives $7x^2+14x+7$; remainder $0$

Since the remainder is $0$, $x^2+2x+1$ divides all three.

Simplify the first factor.

$\dfrac{\dfrac{q+r+p}{p(q+r)}}{\dfrac{q+r-p}{p(q+r)}}=\dfrac{p+q+r}{q+r-p}$

Now the second factor:

$1+\dfrac{q^2+r^2-p^2}{2qr}=\dfrac{2qr+q^2+r^2-p^2}{2qr}$

$=\dfrac{(q+r)^2-p^2}{2qr}=\dfrac{(q+r+p)(q+r-p)}{2qr}$

Multiply:

$\dfrac{p+q+r}{q+r-p}\times\dfrac{(p+q+r)(q+r-p)}{2qr}$

Let Arul alone take $x$ hours.

Madan takes $2x$ hours, Ram takes $3x$ hours.

Work done in 1 hour together $=\dfrac{1}{6}$.

$\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{3x}=\dfrac{1}{6}$

$\dfrac{6+3+2}{6x}=\dfrac{1}{6}$

$\dfrac{11}{6x}=\dfrac{1}{6} \Rightarrow x=11$

Use the square root by long division idea: the answer is of the form $ax^2+bx+c$.

$a^2=289 \Rightarrow a=17$

$2ab=-612 \Rightarrow 34b=-612 \Rightarrow b=-18$

$b^2+2ac=970 \Rightarrow 324+34c=970 \Rightarrow c=19$

Check last term: $c^2=361$ ✓ and $2bc=-684$ ✓

So the expression $=(17x^2-18x+19)^2$

$\sqrt{2y-5}=3-\sqrt{y+1}$

Squaring: $2y-5=9-6\sqrt{y+1}+y+1$

$y-15=-6\sqrt{y+1}$, i.e. $15-y=6\sqrt{y+1}$

Squaring again: $225-30y+y^2=36(y+1)$

$y^2-66y+189=0$

$(y-3)(y-63)=0 \Rightarrow y=3$ or $y=63$

Check $y=3$: $\sqrt{4}+\sqrt{1}=2+1=3$ ✓

Check $y=63$: $\sqrt{64}+\sqrt{121}=19 \neq 3$ ✗

Let the speed of the boat in still water be $x$ km/hr.

Upstream speed $=x-4$, downstream speed $=x+4$.

$\dfrac{36}{x-4}-\dfrac{36}{x+4}=1.6$

$36\left(\dfrac{(x+4)-(x-4)}{x^2-16}\right)=1.6$

$\dfrac{36\times 8}{x^2-16}=1.6$

$288=1.6x^2-25.6$

$1.6x^2=313.6 \Rightarrow x^2=196$

$x=14$ (speed cannot be negative)

Let length $=l$, breadth $=b$.

Perimeter: $2(l+b)=320 \Rightarrow l+b=160$

Area: $lb=4800$

They are roots of $x^2-160x+4800=0$

$\Delta=160^2-4(4800)=25600-19200=6400>0$

So such a park is possible.

$x=\dfrac{160\pm 80}{2}=120$ or $40$

At $t$ minutes past 2 pm, time left to 3 pm $=60-t$ minutes.

Given $60-t=\dfrac{t^2}{4}-3$

Multiply by 4: $240-4t=t^2-12$

$t^2+4t-252=0$

$(t+18)(t-14)=0$

$t=-18$ or $t=14$; time cannot be negative.

Let the number of rows $=x$.

Then seats in each row $=x$, total seats $=x^2$.

New arrangement: rows $=2x$, seats per row $=x-5$.

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

$2x^2-10x=x^2+375$

$x^2-10x-375=0$

$(x-25)(x+15)=0 \Rightarrow x=25$

$\alpha+\beta=2$, $\alpha\beta=3$.

Sum $=\dfrac{\alpha-1}{\alpha+1}+\dfrac{\beta-1}{\beta+1}=\dfrac{2\alpha\beta-2}{\alpha\beta+\alpha+\beta+1}$

$=\dfrac{6-2}{3+2+1}=\dfrac{4}{6}=\dfrac{2}{3}$

Product $=\dfrac{\alpha\beta-(\alpha+\beta)+1}{\alpha\beta+\alpha+\beta+1}=\dfrac{3-2+1}{6}=\dfrac{1}{3}$

Polynomial $=x^2-\dfrac{2}{3}x+\dfrac{1}{3}$

Multiplying by 3:

$-4$ is a root of $x^2+px-4=0$:

$(-4)^2+p(-4)-4=0$

$16-4p-4=0 \Rightarrow 4p=12 \Rightarrow p=3$

$x^2+3x+q=0$ has equal roots, so $\Delta=0$:

$3^2-4(1)(q)=0$

$9=4q \Rightarrow q=\dfrac{9}{4}$

$CD-AB=0 \Rightarrow CD=AB$

$AB=\begin{pmatrix} 3 & 0 \\ 4 & 5 \end{pmatrix}\begin{pmatrix} 6 & 3 \\ 8 & 5 \end{pmatrix}=\begin{pmatrix} 18 & 9 \\ 64 & 37 \end{pmatrix}$

Let $D=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

$CD=\begin{pmatrix} 3a+6c & 3b+6d \\ a+c & b+d \end{pmatrix}=\begin{pmatrix} 18 & 9 \\ 64 & 37 \end{pmatrix}$

$a+c=64$ and $3a+6c=18 \Rightarrow 3(64-c)+6c=18$

$192+3c=18 \Rightarrow c=-58,\ a=122$

$b+d=37$ and $3b+6d=9 \Rightarrow 3(37-d)+6d=9$

$111+3d=9 \Rightarrow d=-34,\ b=71$