For future reference.

The first thing you need to know is the circuit sign convention.

A battery is drawn with the longer of the two lines denoting the positive and the shorter the negative.

Conventional current flows from ~~negative to positive. ~~ Oops Edit positive to negative

(There was recently an ill advised movement to change this convention which still causes much confusion to many people). (Edit serious confused me anyway)

You circuit is drawn (thankfully) using conventional current so the readings on the ammeters are positive when convention current flows (from positive to negative through them).

The next thing to note is that all the battery current flows through ammeter 5 and since it has no other path from the negative to the positive the current in A_{5} must equal the sum of the currents in all the available paths from negative to positive.

It is quite simple in this case to see that this is equal to A_{1} + A_{2} + A_{3}

It is also fairly obvious that the only way current can flow through A_{2} and A_{3 }is what is left after A_{1} has left the main flow. ie the current through A_{4} = A_{2} + A_{3}

**Edited by studiot, 2 January 2017 - 10:59 PM.**