5 Marks Questions

Question 15 Marks

A 5 m 60 cm high vertical pole casts a shadow $3\ m$ $20\ cm$ long. Find at the same time the height of a pole which casts a shadow 5m long.

Answer

Let the height of the vertical pole be $x\ m$ and the length of the shadow by $y\ m$.
As the height of the vertical pole increases, the length of the shadow also increases in the same ratio, so it is a case of direct proportion.
We make use of the relation of the type $\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$
Here,
$x_1= 5m$ $60\ cm = 560\ cm$
$y_1= 3m$ $20\ cm = 320\ cm$
$x_2= 5m$ $00\ cm = 500\ cm$
Therefore, $\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$ gives
$\frac{{560}}{{320}} = \frac{{{x_2}}}{{500}}$
$\therefore$ $320x_2= 560$ $\times$ $500$
$\therefore$ ${x_2} = \frac{{560 \times 500}}{{320}}$
$\therefore$ $x_2= 875$ $cm = 8\ m\ 75\ cm$

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Question 25 Marks

A $5\ m$ $60\ cm$ high vertical pole casts a shadow $3\ m 20\ cm$ long. Find at the same time the length of the shadow cast by another pole $10\ m$ $50\ cm$ high.

Answer

Let the height of the vertical pole be $xm$ and the length of the shadow by $ym$.
As the height of the vertical pole increases, the length of the shadow also increases in the same ratio, It is a case of direct proportion.
We make use of the relation of the type $\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$.
Here,
$x_1= 5 m$ $60\ cm = 5.60\ m$
$y_1= 3 m$ $20\ cm = 3.20\ m$
$x_2= 10 m$ $50\ cm = 10.50\ m$
Therefore, $\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$ gives
$\frac{{5.6}}{{3.2}} = \frac{{10.5}}{{{y_2}}}$
$\therefore $ $5.6y_2= 3.2$ $\times$ $10.5$
$\therefore $ ${y_2} = \frac{{3.2 \times 10.5}}{{5.6}}$
$\therefore $ $y_2= 6$
Hence, the length of the shadow is $6\ m$.

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Question 35 Marks

In a model of a ship, the mast is $9\ m$ high, while the mast of the actual ship is $12\ m$ high. If the length of the ship is $28\ m$, how long is the model ship?

Answer

Let the length of the model ship be $x\ m$ and the height of the mast be $y\ cm$.
We form a table as shown below:

Length of the ship (in metres)	$28$	$x$
Height of the mast (in metres)	$12$	$9$

More the length of the ship, more would be the length of its mast. Hence, this is a case of direct proportion. That is,
$\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$
$\therefore$ $\frac{{28}}{{12}} = \frac{x}{9}$
$\therefore$ $12x = 28$ $\times$ $9$
$x = \frac{{12 \times 9}}{{12}}$
$\therefore$ $x = 21$
Hence, the length of the model ship is $21\ m$.

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Question 45 Marks

A photograph of a bacteria enlarged $50000$ times attains a length of $5\ cm$ as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged $20000$ times only, what be its enlarged length?

Answer

Actual length of the bacteria
$\frac{5}{{50000}}cm$
$ = \frac{1}{{10000}}cm.$
$= 10^{-4} cm$
More the number of times a photograph of a bacteria is enlarged, more the length attained. So, the number of times a photograph of a bacteria is enlarged and the length attained are directly proportional to each other.
So, $\frac{{{x_1}}}{{{x_2}}} = \frac{{{x_2}}}{{{y_2}}}$
$\therefore$ $\frac{{50000}}{5} = \frac{{20000}}{{{y_2}}}$
$\therefore$ 50000 $y_2= 5 \times 20000$
$\therefore$ ${y_2} = \frac{{5 \times 20000}}{{50000}}$
$\therefore$ $y_2= 2$
Hence, its enlarged length would be $2\ cm$.

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Question 55 Marks

If $1$ part of a red pigment requires $75\ mL$ of base, how much red pigment should we mix with $1800\ mL$ of base ?

Answer

Let the number of parts of red pigment be $x$ and the amount of base be $y\ mL$.
As the number of parts of red pigment increases, amount of base also increases in the same ratio.
It is a case of direct proportion. We make use of the relation of the type.
$\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$
Here,
$x_1= 1$
$y_1= 75$ and $y_2= 1800$
Therefore, $\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$ gives
$\frac{1}{{75}} = \frac{{{x_2}}}{{1800}}$
$\therefore $$ 75x_2= 1800$
$\therefore $ ${x_2} = \frac{{1800}}{{75}}$
$\therefore $ $x_2= 24$
Hence, $24$ parts of the red pigment should be mixed.

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Question 65 Marks

A mixture of paint is prepared by mixing $1$ part of red pigments with $8$ parts of base. In the following table, find the parts of base that need to be added.

Parts of red pigment	$1$	$4$	$7$	$12$	$20$
Parts of base	$8$	...	...	...	...

Answer

Let the ratio of parts of red pigment and parts of base be $\frac{a}{b}.$
Here $a_1= 1, b_1= 8$
$\Rightarrow \frac{a_{1}}{b_{1}}=\frac{1}{8} = k$ (say)
When $a_2= 4, b_2= ?$
$k=\frac{a_{2}}{b_{2}} \Rightarrow b_{2}=\frac{a_{2}}{k}=\frac{4}{\frac{1}{8}}=4 \times 8 = 32$
When $a_3= 7_3, b = ?$
$k=\frac{a_{3}}{b_{3}}\Rightarrow b_{3}=\frac{a_{3}}{k}=\frac{7}{\frac{1}{8}}=7 \times 8 = 56$
When a$_4= 12, b_4= ?$
$k=\frac{a_{4}}{b_{4}} \Rightarrow b_{4}=\frac{a_{4}}{k}=\frac{12}{\frac{1}{8}}=12 \times 8 = 96$
When $a_5= 20, b_5= ?$
$k=\frac{a_{5}}{b_{5}} \Rightarrow b_{5}=\frac{a_{5}}{k}=\frac{20}{\frac{1}{8}}=20 \times 8 = 160$
Hence the table becomes,

Parts of red pigment	$1$	$4$	$7$	$12$	$20$
Parts of base	$8$	$32$	$56$	$96$	$160$

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Question 75 Marks

The cost of $5$ metres of a particular quality of cloth is $₹210$. Tabulate the cost of $2, 4, 10$ and $13$ metres of cloth of the same type.

Answer

Suppose the length of cloth is $x$ metres and its cost, in $₹$ is $y$

$x$	$2$	$4$	$5$	$10$	$13$
$y$	$y_2$	$y_3$	$210$	$y_4$	$y_5$

As the length of cloth increases, cost of the cloth also increases in the same ratio. It is a case of direct proportion.
We can use the relation of type $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$
$i.$ Here $x_{1} = 5, y_1= 210$ and $x_2= 2$
Therefore, $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$
we get $\frac{5}{210}=\frac{2}{y_{2}}$
or $5 y_2= 2\times 210$
or $y_2=\frac{2 \times 210}{5}$
$y_2= 84$
$ii.$ If $x_3= 4,$
then $\frac{5}{210}=\frac{4}{y_{3}}$
or $5y_3= 4\times 210$
or $y_3=\frac{4 \times 210}{5}$
$y_3= 168$
$iii.$ If $x_4= 10,$
then $\frac{5}{210}=\frac{10}{y_{4}}$
or $y_4=\frac{10 \times 210}{5}$
$y_4= 420$
$iv.$ If $x_5= 13,$
then $\frac{5}{210}=\frac{13}{y_{5}}$
or $y_5=$ $\frac{13 \times 210}{5}$
$y_5= 546$
$[$Note that here we can also use $\frac{2}{84}$ or $\frac{4}{168}$ or $\frac{10}{420}$ in the place of $\frac{5}{210}]$

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MATHS 5 Marks Questions

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