Question

A water tank 15m long, 10m wide and 6m depth is made. Upper part of tank is open. Find the cost of the iron sheet at rate 2.50 per metre if width of sheet is 5m ?

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

 

Given :-

Length of tank = 15 m (l)

Breadth = 10 m (b)

Depth = 6m (h)

TSA of Tank without upper part which is open

= 2(l × b + b × h + l × h) - lb

= 2(15 × 10 + 10 × 6 + 15 × 6) - 15 × 10

= 2(150 + 60 + 90) - 150

= 2 × 300 - 150

= 600 - 150

= 450 m²

${\boxed{\sf\:{Area\;of\;metal\;sheet\;of\;tank\;made = 450 m^2}}}$

Assumption

Length of metal sheet be p m

Hence

Area of sheet :-

= Length × Breadth

= (p × 5)m²

5p = 450

${\implies\dfrac{450}{5}}$        

p = 90 m

$\textbf{\underline{Cost\;of\;1m\;metal\;sheet = 2.50}}$

Now,

$\textbf{\underline{Cost\;of\;90m\;length\;sheet}}$

= 2.50 × 90

= 225

Answer 2

Answer:

${\boxed{\bold{Cost= Rs.225}}}$

Step-by-step explanation:

Given,

• Length = 15 m
• Breadth = 10 m
• Depth = 6 m
• Rate = 2.5 per meter.
• Width of sheet = 5 m
• Cost of iron sheet = ?

We know that,

${\boxed{\bold{TSA = 2(lb + bh + hl)}}}$

Now, without the area of upper part of tank ,

$\tt TSA = 2(15 \times 10 + 10\times 6 + 6\times 15) - (15\times 10)\:{m}^{2} \\\rightarrow\tt TSA = 2(150 + 60 + 90) - 150 \: {m}^{2} \\\rightarrow\tt TSA = 2\times 300 - 150 \:{m}^{2}\\\rightarrow\tt TSA = (600 - 150)\:{m}^{2} \\\rightarrow{\boxed{\tt {TSA = 450\: {m}^{2} }}}$

$\bold{Let\:length\:of\:iron\:sheet=X\:m}$

Given, width of sheet = 5 m

$\tt Now, 450 = X \times 5 \\\rightarrow\tt 450 = 5X \\\rightarrow\tt X =\frac{450}{5} \\\rightarrow{\boxed{\tt {X = 90\:m}}}$

$\therefore\bold{\underline{Length\:of\:iron\:sheet = 90\:m}}$

$\rightarrow\tt Cost = Rate \times Length \\\rightarrow\tt Cost = Rs.(2.5 \times 90)\\\rightarrow{\boxed{\tt {Cost = Rs.225}}}$