5 Marks Questions

Question 15 Marks

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is $14 \ mm$ and the diameter of the capsule is $4 \ mm$ , find its surface area. Also, find its volume.

Answer

$\text { Radius of hemisphere }=\text { radius of cylinder }=2 mm$
$ \text { Length of cylindrical part } =14-4=10 mm .$
$\text { Surface area of the capsule } = CSA \text { of cylinder }+2( CSA \text { of hemisphere) }$
$ =2 \times \frac{22}{7} \times 2 \times 10+2 \times 2 \times \frac{22}{7} \times 2 \times 2$
$ =176 mm^2$
$ \text { Volume of the capsule } =\text { volume of cylinder }+2(\text { volume of hemisphere) }$
$ =\frac{22}{7} \times 2 \times 2 \times 10+2 \times \frac{2}{3} \times \frac{22}{7} \times 2 \times 2 \times 2$
$ =\frac{3344}{21} mm^3 \text { or } 159.24 mm^3$

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Question 25 Marks

A solid iron pole consists of a solid cylinder of height $200 \ cm$ and base diameter $28 \ cm$ , which is surmounted by another cylinder of height $50 \ cm$ and radius $7 \ cm$ . Find the mass of the pole, given that $1 \ cm^3$ of iron has approximately $8 g$ mass.

Answer

Radius of lower cylinder $=14 \ cm$
Volume of pole $=\frac{22}{7} \times 14 \times 14 \times 200+\frac{22}{7} \times 7 \times 7 \times 50$
$=130900 \ cm^3$
$\text { Mass of the pole } =8 \times 130900$
$ =1047200 \ gm$ or $1047.2 \ kg$

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Question 35 Marks

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20 m$ high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower.

Answer

In $\triangle BPA$
$\tan 45^{\circ}=1=\frac{20}{x}$
$\Rightarrow x=20 m...... (i)$
Now, In $\triangle CPA$
$\tan 60^{\circ}=\sqrt{3}=\frac{h+20}{x}$
$\Rightarrow h+20=x \sqrt{3} \ldots . .(ii)$
Solving $(i)$ and $(ii)$
$h=20(\sqrt{3}-1) m$
$\therefore$ Height of the tower is $20(\sqrt{3}-1) m$.

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Question 45 Marks

State and prove Basic Proportionality theorem.

Answer

For correct statement
For correct given, to prove, construction and figure
For correct Proof

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Question 55 Marks

The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $2 \frac{16}{21}$, find the fraction.

Answer

Let numerator be x ,
then denominator be $(2 x +1)$
Fraction $=\frac{x}{2 x+1}$
A.T.Q.
$
\begin{aligned}
& \frac{x}{2 x+1}+\frac{2 x+1}{x}=\frac{58}{21} \\
\Rightarrow & 11 x^2-26 x-21=0 \\
\Rightarrow & (x-3)(11 x+7)=0 \\
& x \neq-\frac{7}{11} \text { So, } x=3 \\
\therefore & \text { Fraction }=\frac{3}{7}
\end{aligned}
$

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Question 65 Marks

In a flight of $2800 \ km$ , an aircraft was slowed down due to bad weather. Its average speed is reduced by $100 \ km / h$ and by doing so, the time of flight is increased by $30$ minutes. Find the original duration of the flight.

Answer

Let original speed of aircraft be $\times \ km / hr$.
$\text{A.T.Q.}$
$\frac{2800}{x-100}-\frac{2800}{x}=\frac{1}{2}$
$\Rightarrow x^2-100 x-560000=0$
$\Rightarrow(x-800)(x+700)=0$
$x \neq-700 \text { So, } x=800$
Original Duration $=\frac{2800}{800}=\frac{7}{2} hrs$ or $3$ hrs $30$ min .

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Maths 5 Marks Questions

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