Question

if per cent error to determine a,b,c is 1%,2% and 3% then what will be the percent error to determine y?It is given that,y=7a^3b/c^2​

Answer 1

$y = 7((a {}^{3} b) \div c {}^{2} )$

y'= error in y

a'= error in a

b'= error in b

c'= error in c

therefore % error in y=

(y'/y)×100 = (3×(a'/a)×100) + (b'/b)×100 + (2×(c'/c)×100)

= 3×1% + 2% + 2× 3%

= 11%