IN a fectory the production of motor. Bike was 40000in a particular year 0which rose to 48400 in 2 years find the rate of groth per annum if it was uniform during. Two year.
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Let the uniform rate of growth per annum= n%
initial production = 40,000.
after one year, production= 40,000(1 + n/100).
after two year, production= 40,000(1 + n/100)^2.
Thus 40,000(1 + n/100)^2 = 48400.
(1 + n/100)^2 = 48400/40000.
(1 + n/100)^2 = 1.21 = 1.1^2.
1 + n/100 = 1.1
n/100 = 1.1 - 1 = 0.1
n = 100 × 0.1 = 10
So rate of growth per annum is 10%.
initial production = 40,000.
after one year, production= 40,000(1 + n/100).
after two year, production= 40,000(1 + n/100)^2.
Thus 40,000(1 + n/100)^2 = 48400.
(1 + n/100)^2 = 48400/40000.
(1 + n/100)^2 = 1.21 = 1.1^2.
1 + n/100 = 1.1
n/100 = 1.1 - 1 = 0.1
n = 100 × 0.1 = 10
So rate of growth per annum is 10%.
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