The data of the problem lead to the following system of two equations:
x + 10y + 50z = 500
x + y + z = 100
We subtract the second equation from the first equation:
9y + 49z = 400
For z = 1, y = 39, x = 60
For z = 2 the system has no solution
For z = 3 the system has no solution
For z = 4 the system has no solution
For z = 5 the system has no solution
For z = 6 the system has no solution
For z = 7 the system has no solution
For z = 8 the system has no solution
For z = 9 and higher, y cannot be a positive integer, so we can ignore these.
So the only solution would be 1 half dollar, 39 dimes and 60 pennies.
That's where I end up after getting 770 Math and 650 Verbal on the SAT and 800 on the Math II achievement test (previous names of the SAT tests). *Sob...*
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