Question 320128
Let 'x' be the number of $10 coins. 

he has a total of 41 coins and since the rest are $20 coins, the number $20 coins will be 41-x.

The face value of 'x' $10 coins = 10x
Face value of (41-x) $20 coins = 20(41-x)

So the face value of all the coins taken together will be

10x + 20(41-x) = 540
10x + 20X41 - 20x = 540
10x - 20x = 540 -820
 -10x = -280
    x = -280/-10
    x = 28

so the number of $10 coins = 28
and the number of $20 coins = (41 - 28) = 13

[cross checking : 28 + 13 = 41
                 and 28X10 + 13X20 = 540]