A circular park 120 m in diameter has a path of 10m wide all around outside, which is to be gravelled. The MCD received tenders from three companies as follows: | COMPANY | A | B | C | |---|---|---|---| | COST OF GRAVELLING (in Rs) | 120 per $m^2$ | 140 per $m^2$ | 10000 per 100 $m^2$ | a) Find the area of path. b) Which company will get the tender? (Hint: MCD will be selecting the tender of the company which has quoted the least) c) Calculate the total amount paid by MCD to the selected company.

> [!TLDR] A 120 m-diameter circular park has a 10 m-wide gravel path around it (area ≈4085.71 m²); the cheapest of three tenders is Company C, at a total cost ≈₹4,08,571.43. > [!FORMULA] Area of a circular ring (annulus) $=\pi(R^2-r^2)$, where $R$ = outer radius, $r$ = inner radius. Cost $=$ Area $\times$ rate per m². 1. Diameter of park $=120$ m $\Rightarrow$ inner radius $r=60$ m. 2. Path is 10 m wide all around $\Rightarrow$ outer radius $R=60+10=70$ m. 3. **(a)** Area of path $=\pi(R^2-r^2)=\dfrac{22}{7}(70^2-60^2)=\dfrac{22}{7}(4900-3600)=\dfrac{22}{7}\times1300=\dfrac{28600}{7}\approx4085.71\text{ m}^2$. 4. **(b)** Compare costs per m²: Company A = ₹120/m², Company B = ₹140/m², Company C = $\dfrac{10000}{100}=$ ₹100/m². Company C has the lowest rate. …

A circular park 120 m in diameter has a path of 10m wide all around outside, which is to be gravelled. The MCD

Exercise 2.4 · Q12

A circular park 120 m in diameter has a path of 10m wide all around outside, which is to be gravelled. The MCD received tenders from three companies as follows:

COMPANY	A	B	C
COST OF GRAVELLING (in Rs)	120 per $m^2$	140 per $m^2$	10000 per 100 $m^2$

a) Find the area of path.

b) Which company will get the tender? (Hint: MCD will be selecting the tender of the company which has quoted the least)

c) Calculate the total amount paid by MCD to the selected company.

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Concept understanding — Time, Work and Distance

Time, Work, and Distance — The Core Idea

Imagine you're walking to school. The faster you walk, the sooner you get there. That's the most basic link between time, speed, and distance. But in exam problems, "work" enters the picture — and it's really just a different kind of distance.

The Intuition

Distance is what you cover. Speed is how fast you cover it. Time is how long it takes.

If you walk at 4 km/h for 2 hours, you cover 8 km. That's it — the relationship is:

\text{Distance} = \text{Speed} \times \text{Time}

Now, work is like distance, but instead of metres or kilometres, you measure it in "jobs done" — a wall painted, a tank filled, a field ploughed. Rate of work is like speed: how much of the job you finish per unit time (per hour, per day). So:

\text{Work Done} = \text{Rate of Work} \times \text{Time}

The two ideas are structurally identical. Once you see that, the whole topic becomes one idea with two faces.

\text{Distance} = \text{Speed} \times \text{Time}

The Precise Statement

For distance problems: If an object moves at a constant speed $s$ , then in time $t$ it covers a distance $d = s \times t$ . If any two of these are known, the third is fixed.

For work problems: If a person or machine works at a constant rate $r$ (e.g., "1/3 of a wall per hour"), then in time $t$ the amount of work done is $W = r \times t$ . Again, any two determine the third.

Watch out

A common mistake is to treat "work" as a fixed number like 100. It isn't. Work is always 1 complete job unless stated otherwise. If a pipe fills a tank in 5 hours, its rate is $\frac{1}{5}$ tank per hour — not 5.

Why This Matters for Exams

Problems rarely give you the rate directly. They say: "A can do a job in 10 days." That means A's rate is $\frac{1}{10}$ job per day. Then they ask: "How long will A and B together take, if B alone takes 15 days?" You add rates:

\frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6} \text{ job per day}

So together they take 6 days. …

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A 120 m-diameter circular park has a 10 m-wide gravel path around it (area ≈4085.71 m²); the cheapest of three tenders is Company C, at a total cost ≈₹4,08,571.43.

=\pi(R^2-r^2)

Diameter of park $=120$ m $\Rightarrow$ inner radius $r=60$ m.
Path is 10 m wide all around $\Rightarrow$ outer radius $R=60+10=70$ m.
(a) Area of path $=\pi(R^2-r^2)=\dfrac{22}{7}(70^2-60^2)=\dfrac{22}{7}(4900-3600)=\dfrac{22}{7}\times1300=\dfrac{28600}{7}\approx4085.71\text{ m}^2$ .
(b) Compare costs per m²: Company A = ₹120/m², Company B = ₹140/m², Company C = $\dfrac{10000}{100}=$ ₹100/m². Company C has the lowest rate. …

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