Math, asked by gaurav286, 1 year ago

diameter of a road roller 1.2 M long is 42 CM if the cost of levelling the playground at rate 1.5 per metre square is rate 1425 determine number of revolution tax by roller for leveling the ground

Answers

Answered by vaibhavi17
17
csa of leveller=2*22/7*r*h
csa=2*22/7*1.2*21/100
=1.5840 M square
area of field
cost=rate*area
1425=1.5*x
x=1425/1.5
x=950 m square
No of revolution =area of field/csa of leveller
=950/1.584
=599.74 revolutions
approx 600

gaurav286: thanks
vaibhavi17: wlcm
Answered by divyanshukumarchanya
0

Answer:

answer is 599.74 is c. s. a cost is 599 multiple 1.5 =899

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