Question

the length and breadth of a playground are 18 and 35 metre respectively
first question find the cost of levelling it at rupees 4.50 per square metre
second question. how long will a boy take to go three times round the field if he walks at rate of 1.5 metre per second ​

Answer 1

Answer :-

$\: \\ \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Areas of Rectangle and Distance - Time Relationship has been used. We see that for the first question, we need to calculate the area of the Playground. When we calculate the area, we can multiply with the rate to get the total cost of levelling it. Then, we need to calculate its perimeter to find the length of its border. The perimeter multiplied by 3 will be the distance needed to be covered by the boy. After that we can apply the formula, and get our answer.

Let's do it  !!

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★ Formula Used :-

$\: \\ \: \large{\boxed{\boxed{\sf{Area \: of \: Rectangle \: = \: \bf{Length(L) \: \times \: Breadth(B)}}}}}$

$\: \\ \: \large{\boxed{\boxed{\sf{Total \: cost \: of \: levelling \: = \: \bf{Area \: \times \: Rate_{(in \: per \: m^{2})}}}}}}$

$\: \\ \: \large{\boxed{\boxed{\sf{Perimeter \: of \: Rectangle \: = \: \bf{2\: \times \: (Length(L) \: + \: Breadth(B)}}}}}$

$\: \\ \: \large{\boxed{\boxed{\sf{Time \: = \: \bf{\dfrac{Distance}{Speed}}}}}}$

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★ Question :-

The length and breadth of a playground are 18 and 35 metre respectively .

i) Find the cost of levelling it at rupees 4.50 per square metre .

ii) How long will a boy take to go three times round the field if he walks at rate of 1.5 metre per second .

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★ Solution :-

Given,

» Length of the playground = L = 18 m

» Breadth of the playground = B = 32 m

» Rate of cementing the ground per m² = Rs. 4.5

» Rate of walking around the ground = 1.5 m/s

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~ For the Area to be levelled :-

$\: \\ \qquad \: \large{\sf{:\Rightarrow \: \: \: Area \: of \: Rectangular \: Part \: = \: \bf{Length(L) \: \times \: Breadth(B)}}}$

$\: \\ \qquad \: \large{\sf{:\Rightarrow \: \: \: Area \: of \: Rectangular \: Part \: = \: \bf{18 \: m \: \times \: 35 \: m \: \: = \: \: \underline{\underline{630 \: m^{2}}}}}}$

$\: \\ \: \large{\boxed{\boxed{\tt{Area \;\: of \;\: the \;\: ground \;\: = \; \bf{630 \: m^{2}}}}}}$

~ For the Total Cost of Levelling :-

$\: \\ \qquad \: \large{\sf{:\longrightarrow \: \: \: Total \: cost \: of \: levelling \: = \: \bf{Area \: \times \: Rate_{(in \: per \: m^{2})}}}}$

$\: \\ \qquad \: \large{\sf{:\longrightarrow \: \: \: Total \: cost \: of \: levelling \: = \: \bf{630 \: m^{2} \: \times \: 4.5 \: per \: m^{2} \: \: = \: \: \underline{\underline{Rs. \: 2835}}}}}$

$\: \\ \: \large\underline{\underline{\boxed{\boxed{\tt{Total \;\: cost \;\: of \;\: cementing \;\: the \:\; playground \:\; = \; \bf{Rs. \: 2835}}}}}}$

~ For the Perimeter of the Playground :-

$\: \\ \qquad \: \large{\sf{:\Rightarrow \: \: \: Perimeter \: of \: Rectangle \: = \: \bf{2\: \times \: (Length(L) \: + \: Breadth(B)}}}$

$\: \\ \qquad \: \large{\sf{:\Rightarrow \: \: \: Perimeter\:of\:Rectangle\:=\: \bf{2\: \times \: (18 \: m \: + \: 35 \: m)}}}$

$\: \\ \qquad \: \large{\sf{:\Rightarrow \: \: \: Perimeter \: of \: Rectangle \: = \: \bf{2\: \times \: (53 \: m) \: \: = \: \: \underline{\underline{106 \: m}}}}}$

$\: \\ \: \large{\boxed{\boxed{\tt{Perimeter \;\: of \;\: the \;\: ground \;\: = \; \bf{106 \: m}}}}}$

~ For the Time Taken by the Boy to go Thrice the length of Ground  :-

$\: \\ \qquad \: \large{\sf{\Longrightarrow \: \: \: Time \: = \: \bf{\dfrac{Distance}{Speed}}}}$

$\: \\ \qquad \: \large{\sf{\Longrightarrow \: \: \: Time \: = \: \bf{\dfrac{3 \: \times \: Perimeter}{Speed}}}}$

Here we are multiplying by 3 , because its given in question that the boy goes three times round the field.

$\: \\ \qquad \: \large{\sf{\Longrightarrow \: \: \: Time \: = \: \bf{\dfrac{\cancel{3} \: \times \: 318 \: \cancel{m}}{\cancel{1.5} \: \cancel{m}s^{-1}} \: \: = \: \: \underline{\underline{ 212 \: second}}}}}$

$\: \\ \: \large{\underline{\underline{\boxed{\boxed{\tt{Time \:\; taken \;\: by \;\: the \;\: boy \;\: = \; \bf{212 \: seconds}}}}}}}$

$\large{\underline{\underline{\rm{\mapsto\:\:\:Thus,\:total\:cost\:of\:levelling\:the\:playground\:is\:\:Rs.\:2835\:and\:time\:taken\:by\:boy\:is\:\:212\:seconds}}}}$

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$\: \: \large{\underbrace{\underbrace{\sf{Let's \: know \: more \: :-}}}}$

• Area of Square = (Side)²

• Area of Triangle = ½ × Base × Height

• Area of Parallelogram = Base × Height

• Perimeter of Square = 4 × Side

• Area of Circle =  πr²

• Perimeter of Square =  2πr

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Answer 2

Given :

The length and breadth of a playground are 18 and 35 metre respectively.

To find :

1. The cost of levelling it at rupees 4.50 per square metre

2. how long will a boy take to go three times round the field if he walks at rate of 1.5 metre per second ​

Solution for question 1 :

Given , length and breadth = 18 m and 35 m

⇒ Area of playground = 18 * 35

⇒ Area of playground = 630 m²

Now given rate of levelling = Rs. 4.5/m²

⇒ Cost of levelling = Area * Rate

⇒ Cost of levelling = 630 * 4.5

⇒ Cost of levelling = Rs. 2835

Therefore,

Cost of levelling the playground = Rs. 2835

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Solution for question 2 :

Now, length and breadth are 18m and 35m

⇒ Perimeter of playground = 2(18 + 35)

⇒ Perimeter of playground = 2 * 53

⇒ Perimeter of playground = 106 m

As the boy goes 3 times,

⇒ Distance covered = 3 * Perimeter

⇒ Distance covered = 3 * 106

⇒ Distance covered = 318 m

Given, speed = 1.5 m/s

Now, we know:

⇒ Speed = Distance/Time

⇒ 1.5 = 318/Time

⇒ Time = 318/1.5

⇒ Time = 212 s

⇒ Time = 3 min 32 s

Therefore,

It will take 3 min 32 s for the boy to go 3 times round the field.