A student measured the height of a pole as 5.98 m which is less than the actual height. If the percentage error is 5%, find correct to two d.p the actual height of the pole.
-
6.29m
- 7.67m
- 7.18m
- 6.65m
The correct answer is: A
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).