5 Mark Question

Question 15 Marks

A square sheet of $2\ cm$ is cut from each corner of a rectangular piece of aluminium sheet of length $12\ cm$ and breadth $8\ cm$. Find the area of the left over aluminium sheet.

Answer

We have,
Side of each square sheet cut from each corner $=2 cm$,
Length of the rectangular piece of sheet $=12 cm$ and
Breadth of the rectangular piece of sheet $=8 cm$
As, the area of the square $=($ Side $\times$ Side $)$
$=2 \times 2$
$=4 cm^2$
So, the area of four square sheets $=$ (Side $\times$ Side $)$
$=4 \times 4$
$=16 cm^2$
Also, the area of the rectangular sheet $=$ (Length $\times$ Breadth)
$=12 \times 8$
$=96 cm^2$
Now, the area of the sheet left over = (Area of the rectangular sheet - Area of the square sheet)
$=96-16$
$=80 cm^2$
So, the area of the left over aluminium sheet is $80 cm^2$.

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Question 25 Marks

Find the total cost of levelling the shaded path of uniform width 2 metres, laid in the rectangular field shown below, if the rate per $m ^2$ is Rs. 100.

Answer

The shaded path can be formed as two rectangles as shown below:

Now, the area of the shaded path $=(38 \times 2)+(50 \times 2)$
$=76+100$
$=176 m^2$
As, the rate of levelling the shaded path = Rs. 100 per $m ^2$
So, the total cost of levelling the shaded path $=176 \times 100$
$=\text { Rs. } 17,600$
Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

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Question 35 Marks

A room is 8m long and 6m wide. Its floor is to be covered with rectangular tiles of size 25cm by 20cm. How many tiles will be required? Find the cost of these tiles at Rs. 25 per tile.

Answer

We have,
Length of the room = 8m = 800cm,
Breadth of the room = 6m = 600cm,
Length of the tile = 25cm and
Breadth of the tile = 20cm
Now,
The number of the tiles required $=\frac{\text{Area of the floor of the room}}{\text{Area of a tile}}$
$=\frac{\text{Length of the room}\times\text{Breadth of the room}}{\text{Length of the tile}\times\text{Breadth of the tile}}$
$=\frac{800\times600}{25\times20}$
$=960$
So, the number of the tiles required to cover the floor of the room is 960.
Also,
Since, the cost of 1 tile = Rs. 25
So, the cost of the 960 tiles such tiles = 25 × 960 = Rs. 24,000.

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Question 45 Marks

A square sheet of side 5cm is cut out from a rectangular piece of an aluminimum sheet of length 9cm and breadth 6cm. What is the area of the aluminium sheet left over?

Answer

We have,
Side of the square sheet cut out $=5 cm$,
Length of the rectangular sheet $=9 cm$ and
Breadth of the rectangular sheet $=6 cm$
Now, the area of the square sheet cut out $=$ (Side $\times$ Side)
$=5 \times 5$
$=25 cm^2$
Also, the area of the recatngular sheet $=$ (Length $\times$ Breadth)
$=9 \times 6$
$=54 cm^2$
So, the area of the sheet left over = (Area of the rectangular sheet - Area of the square sheet cut out)
$=54-25$
$=29 cm^2$
Hence, the area of the aluminium sheet left over is $29 cm^2$

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Question 55 Marks

The area of a rectangular ground is $120 m^2$ and its length is 12 m . Find the cost of fencing the ground at the rate of Rs. 125 per metre.

Answer

We have,
Area of the rectangular ground $=120 m^2$
Length of the ground $=12 m$
Breadth of the ground $=\frac{\text { Area of the ground }}{\text { Length of the ground }}$
$=\frac{120}{12}$
10 m
Also,
Perimeter of the ground =2 (Length+ Breadth)
$=2(12+10)$
$=2 \times 22$
$=44 m$

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Question 65 Marks

The area of a square field is $36m^2$. A path of uniform width is laid around and outside of it. If the area of the path is $28m^2$, then find the width of the path.

Answer

As, the area of the square field $=36 m^2$
So, the side of the square field $=\sqrt{36}=6 m$
Also, the area of the outer square $=$ (Area of the square + Area of the path)
$=36+28$
$=64 m^2$
So, the side of the outer square $=\sqrt{64}=8\text{m}$
Now, the width of the path $=\frac{\text{side of the outer square - side of the inner square}}{2}$
$=\frac{8-6}{2}$
$=\frac{2}{2}$
$=1\text{m}$

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Question 75 Marks

The length and breadth of a rectangular park are in the ratio $5 : 3$ and its perimeter is $128\ m$. Find the area of the park.

Answer

Let the length of the park be $5 \times m$ and its breadth be $3 \times m$.
As,
perimeter of the park $=2$ (Length of the park + Breadth of the park)
$=2(5 x+3 x)$
$=2 \times 8 x$
$=16 x m$
But the perimeter of the park $=128 m$
$\Rightarrow 16 x=128$
$\Rightarrow x=\frac{128}{16}$
$\Rightarrow x=8$
$\therefore \text { The length of park }=5 x$
$=5 \times 8$
$=40 m$
The breadth of the park $=3 x =3 \times 8=24 m$
Now, the area of the park $=$ (Length $\times$ Breadth)
$=40 \times 24$
$=960 m^2$
So, the area of the park is $960 m^2$.

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