SOLUTION: A wallet contains P600 in P20 and P50 bills. There are two more than P20 bills than P50 bills. How many bills of each kind are there?
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Question 1173067: A wallet contains P600 in P20 and P50 bills. There are two more than P20 bills than P50 bills. How many bills of each kind are there? Answer by greenestamps(13200) (Show Source):
let x = # of P50 bills
then x+2 = # of P20 bills (2 more than P50 bills)
The total value is P600:
Solve using basic algebra....
Even if a formal algebraic solution is needed, you can get good mental exercise by solving the problem mentally, using logical reasoning and some simple arithmetic.
There are 2 more P20 bills than P50 bills, and the total value is P600.
Take away the "extra" 2 P20 bills so that the numbers of P50 and P20 bills are the same; the total value is now P600 minus 2 times P20, or P560.
Since the numbers of P50 and P20 bills are now the same, pair them up into groups of one each. The value of each of those groups is P50 plus P20, or P70.
The number of groups of P70 needed to make the total of P560 is 560/70 = 8. So the P560 consists of 8 each of P50 and P20 bills.
Now add the other 2 P20 bills back in, giving P600 consisting of 10 P20 bills and 8 P50 bills.
