5 Marks Questions

Question 15 Marks

If the median of the distribution given below is $28.5,$ then find the values of $x$ and $y.$

Class Interval	frequency
$0-10$	$5$
$10-20$	$X$
$20-30$	$20$
$30-40$	$15$
$40-50$	$Y$
$50-60$	$5$
$\text{Total}$	$60$

Answer

Monthly Consumption	Number of consumers $(f_i)$	Cumulative Frequency
$0-10$	$5$	$5$
$10-20$	$X$	$5 + x$
$20-30$	$20$	$25 + x$
$30-40$	$15$	$40 + x$
$40-50$	$y$	$40 + x + y$
$50-60$	$5$	$45 + x + y$
$Total$	$\sum f_i = n = 60$

Here$, \sum f_i=n=60$, then $\frac{n}{2}=\frac{60}{2}=30,$ also$,$ median of the distribution is $28.5 ,$ which lies in interval $20-30$.
$\therefore$Median class $=20-30$
So, $l =20, n =60, f =20, cf =5+ x$ and $h =10$
$\because 45+x+y=60$
$\Rightarrow x+y=15 \ldots \ldots .$
Now, Median $=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h$
$\Rightarrow 28.5=20+\left[\frac{30-(5+x)}{20}\right] \times 10$
$\Rightarrow 28.5=20+\frac{30-5-x}{2}$
$\Rightarrow 28.5=\frac{40+25-x}{2}$
$\Rightarrow 57.0=65-x$
$\Rightarrow x=65-57=8$
$\Rightarrow x=8$
Putting the value of $x$ in $eq. (i),$ we get,
$8+y=15$
$\Rightarrow y=7$
Hence the value of $x$ and $y$ are $8$ and $7$ respectively.

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Question 25 Marks

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is $14 \ cm$ and the total height of the vessel is $13 \ cm .$ Find the inner surface area of the vessel. Also, find the volume of the vessel.

Answer

Radius of hemispherical bowl $=$ radius of cylinder $=7 \ cm$
Height of cylinder $=13-7=6 \ cm$
Inner surface area of the vessel $=2 \pi rh +2 \pi r ^2$
$=2 \pi r ( h + r )=2 \times \frac{22}{7} \times 7(6+7)$
$=44 \times 13=572 \ cm^2$
Volume of the vessel $=\pi r ^2 h+\frac{2}{3} \pi r ^3$
$=\pi r ^2\left(h+\frac{2}{3} r \right)$
$=\frac{22}{7} \times 7 \times 7\left(6+\frac{14}{3}\right)$
$=\frac{4928}{3} \ cm^3$ or $1642.67 \ cm^3$

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Question 35 Marks

A student was asked to make a model shaped like a cylinder with two cones attached to its ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its total length is 12 cm . If each cone has a height of 2 cm , find the volume of air contained in the model.

Answer

For upper conical portion
Radius of the base $( r )=1.5 cm$
Height $\left(h_1\right)=2 cm$
$\therefore$ Volume $=\frac{1}{3} \pi r ^2 h_1=\frac{1}{3} \pi(1.5)^2(2)=1.5 \pi cm^3$

For lower conical portion
Volume $=1.5 \pi cm^3$
For central cylindrical portion
Radius of the base $( r )=1.5 cm$
Height $\left( h _2\right)=12-(2+2)=12-4=8 cm$
$\therefore$ Volume $=\pi r ^2 h_2=\frac{1}{3} \pi(1.5)^2(8)=18 \pi cm^3$
Therefore, volume of the model $=1.5 \pi+1.5 \pi+18 \pi=21 \pi=21 \times \frac{22}{7}=66 cm^3$
Hence, the volume of the air contained in the model that Rechel made is $66 cm^3$.

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Question 45 Marks

AD and PM are medians of triangles ABC and PQR respectively where $\triangle ABC \sim \triangle PQR$. Prove that $\frac{A B}{P Q}=$ $\frac{A D}{P M}$.

Answer

Given: In $\triangle A B C$ and $\triangle P Q R, A D$ is the median of $\triangle A B C$, $P M$ is the median of $\triangle P Q R$ and $\triangle A B C \sim \triangle P Q R$.
To Prove: $\frac{A B}{P Q}=\frac{A D}{P M}$
Proof:
Since $A D$ is the median
$
BD=CD=\frac{1}{2} BC
$
Similarly, PM is the median
$
QM=RM=\frac{1}{2} QR
$
Now,
$\triangle ABC \sim \triangle PQR .(\because$ given $)$
$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}(\because$ Corresponding sides of similar triangle are proportional $)$
So,
$
\frac{A B}{P Q}=\frac{B C}{Q R}
$ $\frac{A B}{P Q}=\frac{2 B D}{2 Q M}$ (Since AD \& PM are medians)
$
\frac{A B}{P Q}=\frac{B D}{Q M}
$
Also, since $\triangle ABC \sim \triangle PQR$.
$\angle B =\angle Q (\because$ Corresponding angles of similar triangles are equal).............(2)
Now,
In $\triangle ABD \& \triangle PQM$
$\angle B =\angle Q [\because$ from (2) $]$
$\frac{A B}{P Q}=\frac{B D}{Q M}[\because$ from (1) $]$
Hence by SAS similarly,
$\triangle ABD \sim \triangle PQM$
Since corresponding sides of similar triangles are proportional,
$
\frac{A B}{P Q}=\frac{A D}{P M}
$
Hence proved.

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Question 55 Marks

$A$ and $B$ jointly finish a piece of work in $15$ days. When they work separately, A takes $16$ days less than the number of days taken by $B$ to finish the same piece of work. Find the number of days taken by $B$ to finish the work.

Answer

Let the number of days taken by $B$ be $x$ days.
$\therefore$ number of days taken by $A =( x -16)$ days
$\frac{1}{x}+\frac{1}{x-16}=\frac{1}{15}$
$\therefore x^2-46 x+240=0$
$(x-40)(x-6)=0$
$x=40,6$ Rejected $(\because 6-16$ is $-ve)$
$\therefore$ Number of days taken by $B =40$ days

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Question 65 Marks

If $x =-2$ is a root of the equation $3 x^2+7 x+p=0$, find the value of k so that the roots of the equation$x^2+k(4 x+k-1)+p=0$ are equal.

Answer

Here $x =-2$ is the root of the equation $3 x^2+7 x+p=0$
then, $3(-2)^2+7(-2)+p=0$
or, $p =2$
Roots of the equation $x^2+4 k x+k^2-k+2=0$ are equal,then,
$16 k^2-4\left(k^2-k+2\right)=0$
or, $16 k^2-4 k^2+4 k-8=0$
or, $12 k^2+4 k-8=0$
or, $3 k^2+k-2=0$
or, $(3 k -2)( k +1)=0$
or, $k=\frac{2}{3},-1$
Hence, roots $=\frac{2}{3},-1$

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