5 Marks Questions

Question 15 Marks

The following table gives the marks obtained by $50$ students in a class test:

Marks	$11-15$	$16-20$	$21-25$	$26-30$	$31-35$	$36-40$	$41-45$	$46-50$
Number of students	$2$	$3$	$6$	$7$	$14$	$12$	$4$	$2$

Calculate the mean and median for the above data.

Answer

Class Interval	Frequency $f_i$	Mid value $x_i$	$f_ix_i$	Cumulative frequency
$10.5-15.5$	$2$	$13$	$26$	$2$
$15.5-20.5$	$3$	$18$	$54$	$5$
$20.5-25.5$	$6$	$23$	$138$	$11$
$25.5-30.5$	$7$	$28$	$196$	$18$
$30.5-35.5$	$14$	$33$	$462$	$32$
$35.5-40.5$	$12$	$38$	$456$	$44$
$40.5-45.5$	$4$	$43$	$172$	$48$
$45.5-50.5$	$2$	$48$	$96$	$50$
	$\sum f_i = 50$		$\sum f_ix_i = 1600$

Mean:
$\text { Mean }=\frac{\sum f_i x_i}{\sum f_i}=\frac{1600}{50}=32$
Median:
$N=50$
$\Rightarrow \frac{N}{2}=25$
The cumulative frequency just greater than $25$ is $32 $.
Hence$,$ median class is $30.5 - 35.5$
$l=30.5, h=5, f=14, c f=18$
we know that, Median $= l+\left\{h \times \frac{\left(\frac{N}{2}-c f\right)}{f}\right\}$
$=30.5+\left\{5 \times \frac{25-18}{14}\right\}$
$=30.5+2.5=33$

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Question 25 Marks

An ice$-$cream filled cone having radius $5 \ cm$ and height $10 \ cm$ is as shown in the figure. Find the volume of the ice$-$cream in $7$ such cones.

Answer

Given,
Radius of cone $=$ Radius of hemisphere $=r=5 \ cm$
Height of cone $(h) =10 \ cm$
No. of cones $=7$
Volume of ice cream in one cone $=$ Volume of cone $+$ Volume of hemisphere
$=\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3$
$=\frac{\pi}{3} r^2(h+2 r)$
$=\frac{22}{7} \times \frac{1}{3} \times 5 \times 5(10+2 \times 7)$
$=\frac{22}{7} \times \frac{1}{3} \times 5 \times 5(10+10)$
$=\frac{22 \times 25 \times 20}{21}$
$=523.8 \ cm^3$
Volume of ice cream in $7$ cones
$=523.8 \times 7 \ cm^3$
$=3666.63 \ cm^3$
$=3.67$ litre

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Question 35 Marks

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in the figure. If the height of the cylinder is $10 \ cm$ and its base is of radius $3.5 \ cm ,$ find the total surface area of the article.

Answer

$\text { TSA of the article }=2 \pi rh +2\left(2 \pi r ^2\right)$
$=2 \pi(3.5)(10)+2\left[2 \pi(3.5)^2\right]$
$=70 \pi+49 \pi$
$=119 \pi$
$=119 \times \frac{22}{7}$
$=374 \ cm^2$

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Question 45 Marks

From a point $P$ on the ground, the angle of elevation of the top of a 10 m tall building is $30^{\circ}$. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is $45^{\circ}$. Find the length of the flagstaff and the distance of the building from the point $P$. (use $\sqrt{3}=1.73$ )

Answer

As per shown in the figure let BC is the building and AB is flag-staff.
Let $AB = x$
Then in $\triangle P B C$
$\tan 30=\frac{B C}{C P}$
$\frac{1}{\sqrt{3}}=\frac{B C}{C P}$
$C P=B C \sqrt{3}=10 \sqrt{3}$
Now in $\triangle P A C$
$\tan 45=\frac{A C}{C P}$
$1=\frac{A C}{C P}$
$AC = CP$
$A B+B C=C P$
$x +10=10 \sqrt{3}$
$x=10 \sqrt{3}-10=10 \times 1.732-10=17.32-10=7.32$
So height of flag staff $=7.32 m$
and from eqn (1),
$
C P=10 \sqrt{3}=10 \times 1.732=17.32 m
$
So the distance of point $P$ from the building $=17.32 m$

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Question 55 Marks

Two pipes together can fill a tank in $\frac{15}{8}$ hours. The pipe with larger diameter takes $2$ hours less than the pipe with smaller diameter to fill the tank separately. Find the time in which each pipe can fill the tank separately.

Answer

Let the time taken by smaller diameter tap be $x.$
Time taken by larger diameter tap is $( x -2 ).$
$\text { Therefore } \frac{1}{x-2}+\frac{1}{x}=\frac{8}{15}$
$\Rightarrow 15(2 x-2)=8 x(x-2)$
$\Rightarrow 8 x^2-46 x+30=0$
$\Rightarrow 4 x^2-23 x+15=0$
$\Rightarrow(4 x-3)(x-5)=0$
$\Rightarrow x=\frac{3}{4}, x=5$

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Question 65 Marks

A piece of cloth costs 200 Rupees. If the piece was 5 m longer and each metre of cloth costs 2 Rupees less, the cost of the piece would have remain unchanged. How long is the piece and what is the original rate per metre?

Answer

Let the length of piece be $x m$
Then , rate $=\frac{200}{x}$ per m
Now , new length $=(x+5) m$
Since, the cost remains same .
$\therefore$ New rate $=\frac{200}{x+5}$ per m .
Then, $\frac{200}{x+5}=\frac{200}{x}-2$
$\frac{200}{x+5}=\frac{200-2 x}{x}$
$\Rightarrow 200 x =( x +5)(200-2 x )$
$\Rightarrow 200 x =200 x -2 x ^2+1000-10 x$
$\Rightarrow 2 x ^2+10 x -1000=0$
$\Rightarrow x ^2+5 x -500=0$
$\Rightarrow x ^2+25 x -20 x -500=0$
$\Rightarrow x ( x +25)-20( x +25)=0$
$\Rightarrow( x -20)( x +25)=0$
Therefore, $x=20$ or $x=-25$
But length cannot be negative, therefore $x=20 m$
Therefore , length of the piece $=20 m$

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Maths 5 Marks Questions

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