Question

A packet contains silver powder of mass (20.23+-0.01)g Some of the powder of mass (5.75+-0.01)g taken out from it. the mass of the powder left back is....

Answer 1

mass of silver powder in packet = 20.23+-0.01g
mass taken out = 5.75+-0.01g
mass of powder left = (20.23+-0.01)g - (5.75+-0.01)g
= (20.23-5.75)g +- (0.01+0.01)g
errors are always added because we always  find the maximum possible error
= 14.48+-0.02gm.

Answer 2

Given from the question:

Initial Mass of the silver ($M_s$) = 20.23 + -0.01g

Mass of the silver taken out from the packet ($M_o$) = 5.75 + -0.01g

Mass of the silver that is left = $M_s$ – $M_o$

                       = (20.23 + -0.01g ) – (5.75 + -0.01g)

                       = (20.23 – 5.75) – (0.01g + 0.01g)

                       = (14.48 + -0.02) g

Some errors occurring must also be included.