in a factory the production of motorbikes was 40000 in a particular year which goes to 48400 in two years. find the rate of growth per annum if it was uniform during 2 years
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Answered by
143
Amount = Principal (1 + R/100 ) ^n
48400. = 40000 ( 1 + R/ 100 ) ^2
48400/ 40000 = ( 1+ R/ 100 ) ^2
484 / 400 = ( 1+R/100 ) ^2
(22/20) ^2 = ( 1 + R / 100 ) ^2
22 / 20 =( 1 + R / 100 )
22/ 20 - 1 = R / 100
22- 20 / 20 = R / 100
2/ 20 = R / 100
R = 2/ 20 * 100
R = 2× 5
R = 10 %
I think u r concept is clear.
48400. = 40000 ( 1 + R/ 100 ) ^2
48400/ 40000 = ( 1+ R/ 100 ) ^2
484 / 400 = ( 1+R/100 ) ^2
(22/20) ^2 = ( 1 + R / 100 ) ^2
22 / 20 =( 1 + R / 100 )
22/ 20 - 1 = R / 100
22- 20 / 20 = R / 100
2/ 20 = R / 100
R = 2/ 20 * 100
R = 2× 5
R = 10 %
I think u r concept is clear.
Answered by
85
Hey mate!!!
The question is from Ch-6 COMPOUND INTEREST, WS- 4 of D.A.V. BOOK.
Hope my answer will help you.
And, plz mark it as Brainliest...
The question is from Ch-6 COMPOUND INTEREST, WS- 4 of D.A.V. BOOK.
Hope my answer will help you.
And, plz mark it as Brainliest...
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