5 Marks Questions

Question 15 Marks

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Find the mode and mean of the data given above.

Answer

Age (in years)	No. of patients (fi)	Mid point (x₁)	xifi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	320
45-55	14	50	700
55-65	5	60	300
Total	80		2830

$
\begin{aligned}
\Rightarrow \text { Mean } & =\frac{2830}{80} \\
& =35.375
\end{aligned}
$
Modal class $=(35-45)$
$
\begin{aligned}
\Rightarrow \text { Mode } & =35+\left(\frac{23-21}{2 \times 23-21-14}\right) \times h \\
& =36.81
\end{aligned}
$
Therefore, mode and mean of given data are 36.81 years and 35.375 years respectively.

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Question 25 Marks

Find the value of ' $c$ ' for which the quadratic equation
$(c+1) x^2-6(c+1) x+3(c+9)=0 ; c \neq-1$
has real and equal roots.

Answer

For real and equal roots,
$\{-6(c+1)\}^2-4(c+1) \times 3(c+9)=0\}$
$\Rightarrow 12(c+1)(2 c-6)=0$
$c \neq-1 \text { So, } c=3$

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Question 35 Marks

A train travels a distance of $90 \ km$ at a constant speed. Had the speed been $15 \ km / h$ more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.

Answer

Let the original speed be $x \ km / h$
New speed $=(x+15) \ km / h$
$\text{A.T.Q.}$
$\frac{90}{x}-\frac{90}{x+15}=\frac{1}{2}$
$\Rightarrow x^2+15 x-2700=0$
$\Rightarrow(x+60)(x-45)=0$
$x \neq-60, x=45$
The original speed of the train $=45 \ km / h$

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Question 45 Marks

Sides $AB , BC$ and the median $AD$ of $\triangle ABC$ are respectively proportional to sides $PQ , QR$ and the median PM of another $\Delta PQR$. Prove that $\triangle ABC \sim \Delta PQR$.

Answer

$\because \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}$
$\therefore \frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{A D}{P M}$
$\Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}----\text(1)$
$\Rightarrow \triangle ABD \sim \triangle PQM$
$\Rightarrow \angle B=\angle Q----\text(2)$
In $\triangle A B C$ and $\triangle P Q R$
$\frac{A B}{P Q}=\frac{B C}{Q R}$
$\angle B=\angle Q$
$\therefore \triangle ABC \sim \triangle PQR$

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Question 55 Marks

$E$ is a point on the side $AD$ produced of a parallelogram $\text{ABCD}$ and $BE$ intersects $CD$ at $F$ . Show that $\triangle ABE \sim \Delta CFB$.

Answer

In $\triangle ABE$ and $\triangle CFB$
$\angle EAB=\angle BCF$
$\angle AEB=\angle CBF$
$\Rightarrow \triangle ABE \sim \triangle CFB$

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Question 65 Marks

Two pillars of equal lengths stand on either side of a road which is $100\ m$ wide, exactly opposite to each other. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are $60^{\circ}$ and $30^{\circ}$. Find the length of each pillar and distance of the point on the road from the pillars. $($Use $\sqrt{3}=1.732 )$

Answer

Let $A B$ and $C D$ are two pillars of equal length $h\ m$ and let $P$ be the point on road $x\ m$ away from pillar $C D$.
In $\triangle CDP$
$\tan 60^{\circ}=\sqrt{3}=\frac{h}{x}$
$\Rightarrow h=\sqrt{3} x .....(i)$
In $\triangle ABP$,
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{h}{100-x}$
$\Rightarrow h=\frac{100-x}{\sqrt{3}}........(ii)$
Solving eq.$(i)$ and eq.$(ii)$
$x=25$
and $h =25 \sqrt{3}=25 \times 1.732=43.3$
The length of each pillar is $43.3 \ m$ and the distance of the point on the road from pillars is $75\ m$ and $25\ m$ respectively.

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