[5 marks sum]

Question 15 Marks

The length (in cm) of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of another side by 1 cm. Find the length of each side. Also, find the perimeter and the area of the triangle.

Answer

Let the length of one side $=x cm$ and other side $=y cm$.
then hypotenues $=x+2$ and $2 y+1$
$
\begin{aligned}
& \therefore x+2=2 y+1 \\
& \Rightarrow x-2 y-2 \\
& \Rightarrow x-2 y=-1 \\
& \Rightarrow x=2 y-1
\end{aligned}
$
and using Pythagorous theorem,
$
\begin{aligned}
& x^2+y^2=(2 y+1)^2 \\
& \Rightarrow x^2+y^2=4 y^2+4 y+1 \\
& \Rightarrow(2 y-1)^2+y^2=4 y^2+4 y+1 \ldots[\text { From (i)] } \\
& \Rightarrow 4 y^2-4 y+1+y^2=4 y^2+4 y+1 \\
& \Rightarrow 4 y^2-4 y+1+y^2-4 y^2-4 y-1=0 \\
& \Rightarrow y^2-8 y=0 \\
& \Rightarrow y(y-8)=0
\end{aligned}
$
Either $y=0$,
but it is not possible
or
$y-8=0$,
then $y=8$
Substituting the value of $y$ in (i)
$
\begin{aligned}
& x=2(8)-1 \\
& =16-1 \\
& =15
\end{aligned}
$
$\therefore$ Length of one side $=15 cm$
and length of other side $=8 cm$
and hypotenuse
$
\begin{aligned}
& =x+2 \\
& =15+2 \\
& =17
\end{aligned}
$
$\therefore$ Perimeter
$
=15+8+17
$
$
=40 cm
$
and Area
$
\begin{aligned}
& =\frac{1}{2} \times \text { one side } x \text { other side } \\
& =\frac{1}{2} \times 15 \times 8 \\
& =60 cm ^2
\end{aligned}
$

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Question 25 Marks

A piece of cloth costs Rs. 300. If the piece was 5 metres longer and each metre of cloth costs Rs. 2 less, the cost of the piece would have remained unchanged. How long is the original piece of cloth and what is the rate per metre?

Answer

The total cost of cloth piece $=$ Rs. 300
Let the length of the piece of cloth in the beginning $= x m$
Then cost of 1 metre $=$ Rs. $\frac{300}{x}$
In second case, length of cloth $=(x+5)$
Cost of 1 metre $=$ Rs, $\frac{300}{x+5}$
According to the condition,
$\frac{300}{x}-\frac{300}{x+5}=2$
$\Rightarrow 300\left(\frac{1}{x}-\frac{1}{x+5}\right)=2$
$\Rightarrow 300\left(\frac{x+5-x}{x(x+5)}\right)=2$
$\Rightarrow \frac{300 \times 5}{x(x+5)}=2$
$\Rightarrow \frac{150 \times 5}{x(x+5)}=1 \ldots$ (Dividing by 2 )
$750= x ^2+5 x$
$\Rightarrow x ^2+5 x -750=0$
$\Rightarrow x ^2+30 x -25 x -750=0$
$\Rightarrow( x +30)-25( x +30)=0$
$\Rightarrow( x +30)( x -25)=0$
Either $x+30=0$,
then $x=-30$
which is not possible being negative
or
$x-25=0$,
then $x=25$
$\therefore$ Length of cloth piece in the begining $=25$ metres
and rate per metre $=$ Rs. $\frac{300}{25}=$ Rs. 12 .

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Question 35 Marks

A shopkeeper buys a certain number of books for Rs 960. If the cost per book was Rs 8 less, the number of books that could be bought for Rs 960 would be 4 more. Taking the original cost of each book to be Rs x, write an equation in x and solve it to find the original cost of each book.

Answer

Let original cost $=R s x$
No. of books bought $=\frac{960}{x}$
New cost of books $=$ Rs $(x-8)$
$\therefore$ No. of books bought $=\frac{960}{x-8}$
If no. of books bought is 4 more then cost $=\frac{960}{x}+4$
$\therefore$ According to conditions,
$
\begin{aligned}
& \frac{960}{x-8}-\frac{960}{x}=4 \\
& \Rightarrow 960\left(\frac{1}{x-8}-\frac{1}{x}\right)=4 \\
& \Rightarrow \frac{x-(x-8)}{x(x-8)}=\frac{4}{960} \\
& \Rightarrow \frac{x-x+8}{x^2-8 x}=\frac{4}{960} \\
& \Rightarrow \frac{8}{x^2-8 x}=\frac{1}{960} \\
& \Rightarrow x^2-8 x=8 \times 240 \\
& \Rightarrow x^2-8 x-1920=0 \\
& x=\frac{-(-8) \pm \sqrt{(-8)^2-4(1)(-1920)}}{2} \\
& =\frac{8 \pm \sqrt{64+7680}}{2} \\
& =\frac{8 \pm \sqrt{7744}}{2} \\
& =\frac{8 \pm 88}{2} \\
& =\frac{8+88}{2}, \frac{8-88}{2} \\
& =\frac{96}{2}, \frac{-80}{2} \\
& =48,-40 \ldots(\text { rejecting) } \\
& \therefore \text { cost of book = ₹ } 48 .
\end{aligned}
$

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Question 45 Marks

Solve the following equation by factorisation :
$
\sqrt{3 x^2-2 x-1}=2 x-2
$

Answer

$
\sqrt{3 x^2-2 x-1}=2 x-2
$
Squaring both sides
$
\begin{aligned}
& 3 x^2-2 x-1=(2 x-2)^2 \\
& \Rightarrow 3 x^2-2 x-1=4 x^2-8 x+4 \\
& \Rightarrow 4 x^2-8 x+4-3 x^2+2 x+1=0 \\
& \Rightarrow x^2-6 x+5=0 \\
& \Rightarrow x^2-5 x-x+5=0 \\
& \Rightarrow x(x-5)-1(x-5)=0 \\
& \Rightarrow(x-5)(x-1)=0
\end{aligned}
$
Either $x-5=0$,
$
\text { then } x=5
$
or
$
x-1=0
$
then $x=1$
Check:
$
\begin{aligned}
& \text { (i) If } x=5 \text {, then } \\
& \text { L.H.S. }=\sqrt{3 x^2-2 x-1} \\
& =\sqrt{3 \times(5)^2-2 \times 5-1} \\
& =\sqrt{3 \times 25-10-1} \\
& =\sqrt{75-10-1} \\
& =\sqrt{64} \\
& =8 \\
& \text { R.H.S. }=2 x-2 \\
& =2 \times 5-2 \\
& =10-2 \\
& =8 \\
& \because \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
$\therefore x=5$ is a root
$
\begin{aligned}
& \text { (ii) If } x =1 \text {, then } \\
& \text { L.H.S. }=\sqrt{3 x^2-2 x-1} \\
& =\sqrt{3(1)^2-2(1)-1} \\
& =\sqrt{3 \times 1-2-1} \\
& =\sqrt{3-2-1} \\
& =0 \\
& \text { R.H.S. }=2 x-2 \\
& =2 \times 1-2 \\
& =2-2 \\
& =0 \\
& \because \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
$\therefore x =1$ is also its root
Hence $x=5,1$.

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Question 55 Marks

Solve the following equation by factorisation :
$
\sqrt{x+15}=x+3
$

Answer

$
\begin{aligned}
& \sqrt{x+15}=x+3 \\
& \text { Squaring on both sides } \\
& x+15=(x+3)^2 \\
& \Rightarrow x^2+6 x+9-x-15=0 \\
& \Rightarrow x^2+5 x-6=0 \\
& \Rightarrow x^2+6 x-x-6=0 \\
& \Rightarrow x(x+6)-1(x+6)=0 \\
& \Rightarrow(x+6)(x-1)=0 \\
& \text { Either } x+6=0 \\
& \text { then } x=-6 \\
& \text { or } \\
& x-1=0, \\
& \text { then } x=1 \\
& \therefore x=-6,1
\end{aligned}
$
Check:
(i) If $x=-6$ then
$
\begin{aligned}
& \text { L.H.S. }=\sqrt{x+15} \\
& =\sqrt{-6+15} \\
& =\sqrt{9} \\
& =3 \\
& \text { R.H.S. }=x+3 \\
& =-6+3 \\
& =-3 \\
& \because \text { L.H.S. } \neq \text { R.H.S. }
\end{aligned}
$
$
\therefore x=-6 \text { is not a root }
$
(ii) If $x =1$, then
$
\begin{aligned}
& \text { L.H.S. }-\sqrt{x+15} \\
& =\sqrt{1+15} \\
& =\sqrt{16} \\
& =4
\end{aligned}
$
$
\begin{aligned}
& \text { R.H.S. }=x+3 \\
& =1+3 \\
& =4
\end{aligned}
$
$\because$ L.H.S. $=$ R.H.S.
$\therefore x=1$ is a root of this equation
Hence $x=1$.

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Question 65 Marks

A farmer wishes to grow a $100 m^2$ rectangular vegetable garden. Since he has with him only $30 \ m$ barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.

Answer

Area of rectangular garden $=100 cm ^2$
Length of barbed wire $=30 m$
Let the length of the side opposite to wall = $x$
and length of other each side $=\frac{30-x}{2}$
According to the condition,
$
\begin{aligned}
& \frac{x(30-x)}{2}=100 \\
& \Rightarrow x (30- x )=200 \\
& \Rightarrow 30 x - x ^2=200 \\
& \Rightarrow x ^2-30 x +200=0 \\
& \Rightarrow x ^2-20 x -10 x +200=0 \\
& \Rightarrow x ( x -20)-10( x -20)=0 \\
& \Rightarrow( x -20)( x -10)=0
\end{aligned}
$
Either $x-20=0$,
then $x=20$
or
$
x-10=0 \text {, }
$
then $x=10$
(i) If $x=20$,
then side opposite to the wal $=20 m$
and other side
$
\begin{aligned}
& =\frac{30-20}{2} \\
& =\frac{10^2}{2} \\
& =5 m
\end{aligned}
$
(ii) If $x=10$,
then side opposite to wall $=10 m$
and other side
$
\begin{aligned}
& =\frac{30-10}{2} \\
& =\frac{20}{2} \\
& =10 m
\end{aligned}
$
$\therefore$ Sides are $20 m , 5 m$ or $10 m$.

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Mathematics [5 marks sum]

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