A student measured the height of a pole as 5.98 m which is less than

Explanation

%error=5%, measured height = 5.98m

let the actual height = y

error=x - 5.98 (since 'y' is more than 5.98)

%error = \(\frac{error}{actual height}\times 100%\)

5% = \(\frac{y - 5.98}{y}\times 100%\)

\(\frac{5}{100} = \frac{y - 5.98}{y}\)

5y = 100(y - 5.98)

5y = 100y - 598

5y - 100y = - 598

-95y = - 598

y = \(\frac{-598}{-95}\)

y = 6.29m( to 2 d.p).