Question 1188238
<br>
Yes, this particular problem is probably solved most easily by working backward.<br>
He finished with 160 cards, after buying 40 more, so before buying those 40 he had 120.
The 120 cards was after he used 1/3 of what he had.  This part is a bit tricky to do mentally.  Since the 120 was after he used 1/3 of what he had before, the 120 is 2/3 of what he had before; therefore the number he had before was (3/2) times 120, which is 180.<br>
That's the first couple of steps; I'll leave it to you to go the last couple.<br>
If you are going to solve it "forwards" using formal algebra, take a moment to see what is happening in the problem before deciding how to set the problem up.<br>
The algebraic solution from one of the other tutors starts immediately by letting x be the initial number of cards; this leads quickly to equations involving fractions, which are always a bit harder to work with.<br>
Instead, noticing that the first thing that happens is he give away 1/10 of his cards, start with the initial number of cards being 10x. That gives us....<br>
initial number: 10x
after giving 1/10 away: 9x
after buying 45 more: 9x+45
after using 1/3 of those (2/3 of 9x+45 are left): 6x+30
after buying 40 more: 6x+70<br>
At that point he had 160 cards:<br>
6x+70=160
6x=90
x=90/6=15<br>
ANSWER: The number he started with was 10x=150<br>