Question 166436: a piggy bank contains pennies, nickels, and dimes---a total of 222 coins. if the bank helb 3 times as many dimes, half as many pennies, and 8 times as many nickels, it would contain twice as much money. how many of each type of coin are actually in the bank?
--i need to know how you did it too
thanks
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! OK
FIRST OF ALL, WE HAVE MORE UNKNOWNS IN THIS PROBLEM THAN WE HAVE EQUATIONS. IN THIS SITUATION, WE SOMETIMES HAVE MULTIPLE ANSWERS AND WE ALMOST ALWAYS WILL HAVE TO DO SOME TRIAL AND ERROR TO GET AT THE SOLUTIONS. HAVING SAID THAT, LETS SOLVE THE PROBLEM:
Let x=number of pennies
Let y=number of nickels
Let z=number of dimes
Let A=amount of money in the initial 222 coins
Now we are told that:
x+y+z=222-----------------------------------------------------eq1
And we know that (lets deal in pennies):
x+5y+10z=A----------------------------------------------------eq2
Now we are also told that:
"If the bank held 3 times as many dimes, half as many pennies, and 8 times as many nickels, it would contain twice as much money." This translates into the following equation (deal in pennies again):
(x/2)+5*8y+10*3z=2A multiply each term by 2 and simplify
x+80y+60z=4A---------------------------------------------------eq3
Next, multiply eq2 by 4 (we get: 4x+20y+40z=4A) and subtract eq3 from it and this yields:
4x+20y+40z=4A
-x-80y-60z=-4A
which equals:
3x-60y-20z=0------------------------------------------------------eq3a
Next, we multiply eq1 by 3 (we get: 3x+3y+3z=666) and subtract it from eq3a and this gives us:
-63y-23z=-666 multiply each term by -1
63y+23z=666 solve for z
z=(666-63y)/23-----------------------------------------------------eq3b
Eq3b gives us a relationship between the number of dimes and the number of nickels:
Now we know the following information about this problem which will help greatly in solving it:
(1) We cannot have negative coins--------our answers have to be positive numbers
(2) We cannot have fractions of coins-----our answers have to be whole numbers
Armed with the above information, by inspection, we can see from eq3b that y has to be less than or equal to 10, otherwise we start getting negative dimes. So, we have established that:
y<=10
Next, we will deal with the fractional aspect of this problem. We'll do this by calculating the value of z for the values of y between 1 and 10. Any whole numbers that we get are possible solutions:
y=1, z=26.22------------------------no good
y=2, z=23.5-------------------------no good
y=3, z=20.74------------------------no good
y=4, z=18---------------------------BINGO!!!!!!! possible answer
y=5, z=15.3-------------------------no good
y=6, z=12.5-------------------------no good
y=7, z=9.8--------------------------no good
y=8, z=7.04-------------------------no good
y=9, z=4.3--------------------------no good
y=10, z=1.57------------------------no good
So, out of all the possible values for y, we have only one possible solution:
y=4 and
z=18
Substitute y=4 and z=18 into eq1 and solve for x:
x+4+18=222 solve for x
x=200
So our answer is
x=200 pennies
y=4 nickels
z=18 dimes
CK
Original amount of money:
200*1+4*5+10*18=200+20+180=400
"If the bank held 3 times as many dimes, half as many pennies, and 8 times as many nickels, it would contain twice as much money." This translates into the following amount:
(200/2)*1+5*4*8+10*3*18=800
100+160+540=800
800=800
Niche problem!!!!
Hope this helps---ptaylor
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