A roller is 120 cm long and has diameter 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of 30 paise per sq. M.
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Answered by
3
given
l=120cm =120/100=1.2m
d=84cm=84/100=0.84m
revolution =500 of level the ground
cost of level=?
1m^2= 30paise
answer step;
circumference of cricle =pi×d=22/7×0.84=2.95m
b=2.95m
A=l×b=1.2×2.95=3.545m^2
1rev= 3.545m^2 of area
500rev = 500×3.545= 1772.5m^2 0f area
area of 1m^2 =30paise
area of 1772.5m^2= 1772.5×30= 53175 paise
l=120cm =120/100=1.2m
d=84cm=84/100=0.84m
revolution =500 of level the ground
cost of level=?
1m^2= 30paise
answer step;
circumference of cricle =pi×d=22/7×0.84=2.95m
b=2.95m
A=l×b=1.2×2.95=3.545m^2
1rev= 3.545m^2 of area
500rev = 500×3.545= 1772.5m^2 0f area
area of 1m^2 =30paise
area of 1772.5m^2= 1772.5×30= 53175 paise
Answered by
13
= 120 cm
= 1.2m
Also,
= 84 cm
= 0.84 m
We know that :-
= 0.42 m
= 2πrh
= 2 × 22 × 0.06 × 12
= 3.168 m²
= 3.168 m²
Now,
= 500 × 3.168 m²
= 1584 m²
Cost of levelling at 0.30 per m²
= 0.30 × 1584
= 475.20
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