Question

the length of a room is 50% more than its breadth.the cost of carpeting the room at the rate of Rs. 38.50m^2is Rs. 924and the cost of painting the walls at the rate of Rs.5.50 m^2 is Rs. 1320. Find the dimension of room.

Answer 1

Let length (l)=x+1/2x=3/2x
Let breadth (b)=x
Area of floor=924×100/3850
l×b=24
3/2x×x=24
3/2x^2=24
x^2=24×2/3
x^2=4^2
x=4
Area of 4walls=1320×100/550
2 (l+b) h=240
10×h=240/2
h=12m
l=3/2×4=6m
b=x=4m

Answer 2

Let,
the breadth of the room be $\bf{b}$ m.

According to the question,
the length of the room is 50% more than its breadth.
length $\bf{(l)}$ = $\bf{b}$ + 50% of $\bf{b}$

length $\bf{(l) }$ = $\bf{b}$+ $\frac{50}{100}$ × $\bf{b}$

length $\bf{(l) }$ = $\bf{b}$ + $\frac{b}{2}$

length $\bf{(l) }$ = $\frac{3b}{2}$

Now,

Area of the floor = $\frac{924}{38.50}$

[ as the cost of carpeting the room at the rate of ₹ 38.50 m² is ₹ 924 ]

=> Area of the floor = 924 ÷ $\frac{3850}{100}$

=> length × breadth = 924 × $\frac{100}{3850}$

=> $\frac{3b}{2}$ × b = 24

=> \frac{3 {b}^{2} }{2} = 24

=> 3b² = 24 × 2

=> 3b² = 48

=> b² = $\frac{48}{3}$

=> b² = 16

=> b = √16

=> b = 4m

Therefore,
the breadth (b) is 4 m.

and

length (l) = $\frac{3b}{2}$ = $\frac{3×4}{2}$ = $\frac{12}{2}$ = $\bf{6m}$

Now,

Area of four walls = $\frac{1320}{5.50}$

[ as the cost of painting the walls at the rate of ₹ 5.50 per m² is ₹ 1320 ]

=> Area of four walls = 1320 ÷ $\frac{550}{100}$

=> 2(length+breadth) × height = 1320 × $\frac{100}{550}$

=> 2(6 + 4) × height = 240

=> 2 × 10 × height = 240

=> 20 × height = 240

=> height = $\frac{240}{20}$

=> height = 12 m

Therefore,
the height of the room is 12m.

Hence,
the dimensions of the room are
$\bf{length (l)}$ = 6m
$\bf{breadth (b)}$ = 4m
$\bf{height (h) }$ = 12m