2 Marks Questions

Question 12 Marks

A room is $4.5m$ long and $4m$ wide. The floor of the room is to be covered with tiles of size $15\ cm$ by $10\ cm.$ Find the cost of covering the floor with tiles at the rate of $Rs. 4.50$ per tile.

Answer

Given, lenght of room $= 4.5m,$ of room $= 4m$ and
size of title $= 15cm \times 10cm$
$\because$ Area of room $=l \times \mathrm{b}=4.5 \times 4=18 \mathrm{~m}^2=18 \times(100)^2 \mathrm{~cm}^2=180000 \mathrm{~cm}^2[\because 1 \mathrm{~m}=100 \mathrm{~cm}]$
$\therefore$ Area of $1$ title $=15\ c\ 10=150 \mathrm{CM}^2$
So, number of titles $=\frac{\text { Area of room }}{\text { Area of } 1 \text { title }}=\frac{180000}{150}=1200$
$\therefore$ Cost of covering the floor with title $= Rs. 4.50 \times 1200= Rs. 5400$

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Question 22 Marks

Find the total cost of wooden fencing around a circular garden of diameter $28\ m,$ if 1m of fencing costs $Rs. 300.$

Answer

Give, diameter of a circular garden $= 28\ m$
Length of the fencing $=$ Circumference of circle
$=\pi\text{d}=\frac{22}{7}\times 28 = 88\text{m}$
$\therefore $ Total cost of fencing $= 88 × 300 = Rs. 26400$

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Question 32 Marks

The moon is about $384000\ km$ from earth and its path around the earth is nearly circular. Find the length of path described by moon in one complete revolution. $($Take $\pi = 3.14)$

Answer

Length of path described by moon in one complete revolution $=\pi\text{r} = 2 \times 3.14 \times 384000$
$ [\because$ radius = distance of moon from the earth$] = 2411520\ km$

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Question 42 Marks

A hedge boundary needs to be planted around a rectangular lawn of size $72m \times 18m.$ If $3$ shrubs can be planted in a metre of hedge, how many shrubs will be planted in all$?$

Answer

Here, length of rectangular lawn $= 72m$ and
breadth of rectangular lawn $= 18$
$\because$ Perimeter of rectangle $= 2\ \times ($Length $+$ Breadth$)$
$\therefore$ Perimeter of rectangular lawn $= 2(72 + 18) = 2(90) = 180m$
If $3$ shrubs can be planted in a meter of hedge.
Then, number of shrubs $= 3\ \times $ Perimeter of rectangular lawn $= 3 \times 180 = 540$

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Question 52 Marks

Three squares are attached to each other as shown in Fig. Each square is attached at the mid point of the side of the square to its right. Find the perimeter of the complete figure.

Answer

$\therefore$ Perimeter of the complete figure $= 6 + 6 + 6 + 3 + 1.5 + 1.5 + 1.5 + 3 + 3 + 1.5 = 33\ m$

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Question 62 Marks

How much distance, in metres, a wheel of $25\ cm$ radius will cover if it rotates $350$ times$?$

Answer

Given, radius of wheel $(r) = 25\text{cm}=\frac{25}{100}$
$\text{m}=\frac{1}{4}\text{m}$
$\big[\because1\text{cm} = \frac{1}{100}\text{m}\big]$
$\because$ Distance travelled in one rotation $=2\pi\text{r}=2\times\frac{22}{7}\times\frac{1}{4}=\frac{11}{7}\text{m}$
$\therefore$ Distance travelled in $350$ rotation $=\frac{11}{7}\times350=550\text{m}$
Hence, the wheel coveres $550m$ distance.

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Question 72 Marks

Calculate the area of shaded region in Fig. where all of the short line segments are at right angles to each other and $1\ cm$ long.

Answer

Length of the larger rectangle $=1 \times 9 \mathrm{~cm}=9 \mathrm{~cm}$
Breadth of the larger rectangle $=1 \times 9 \mathrm{~cm}=9 \mathrm{~cm}$
$\therefore$ Area of shaded region $=$ Area of larger square $-$ Area of $41$ small identical square
$=9 \times 9-41 \times 1 \times 1=81-41=40 \mathrm{~cm}^2$

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Question 82 Marks

The dimensions of a plot are $200m × 150m$. A builder builds $3$ roads which are $3m$ wide along the length on either side and one in the middle. On either side of the middle road he builds houses to sell. How much area did he get for building the houses$?$

Answer

Given, that, dimersions of piot $= 200m × 150m$ and width of rode $= 3m$
$\therefore \text { Totale area available for house }=\text { Area of totale plot }- \text { Area of } 3 \text { roads }$
$=200 \times 150-3 \times(3 \times 200)[\because \text { area of rectangle }=\text { length } \times \text { breadth }]$
$=30000-1800=28200 \mathrm{~m}^2$

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Question 92 Marks

In Fig. $ABCD$ is a square with $AB = 15\ cm.$ Find the area of the square $BDFE$.

Answer

Given, $ABCD$ is a square and $AB = 15\ cm$
$\therefore$ Diagonal of square $ ABCD =\sqrt{2}\text{a}=\sqrt{2}\times15=15\sqrt{2}\text{cm}$
From the figure, diagonal of square $ABCD$ is the side of square $BDEF.$
$\therefore$ Area of the square $BDFE = (\text{Side})^2 = (15\sqrt{2})^2=15\times15\times\sqrt{2}\times\sqrt{2}$
$=225\times2=450\text{cm}^2$

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Question 102 Marks

A rectangular field is $48m$ long and $12m$ wide. How many right triangular flower beds can be laid in this field, if sides including the right angle measure $2m$ and $4m,$ respectively$?$

Answer

Given, dimenstions of a rectangular field $= 48m \times 12m$ and
dimensions of a right angled triangle $= 2m \times 4m$
$\therefore$ Number of right triangle florer beds $=\frac{\text{Area of filed}}{\text{Area of right angled triangle}}$
$=\frac{48\times12}{\frac{1}{2}\times2\times4}= 144$
$\Big[\because$ area of rectangle $=$ lenght $\times $ breadth and area of a right angled $=\frac{1}{2}\times\text{height}\times\text{breadth}\Big]$

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