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Question 157328: Solve the problem. Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9.90. How many dimes does he have?
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! let n = original number of nickels and d = original number of dimes.
a nickel is worth .05 cents so value of n nickels is .05*n.
a dime is worth .10 cents so value of d dimes is .10*n.
first equation is .05*n + .10*d = 6.00 dollars.
second equation says if you add 6 nickels and double the number of dimes you get $9.90.
second equation would then be .05*(n+6) + .10*(2*d) = 9.90
this becomes .05*n + .05*6 + .2*d = 9.90.
this becomes .05*n + .30 * .2*d = 9.90.
this becomes .05*n + .2*d = 9.90 - .30
this finally becomes .05*n + .2*d = 9.60.
now we have 2 equations that need to be solved simultaneously.
.05*n + .1*d = 6.00
.05*n + .2*d = 9.60
since if we subtract the top equation from the bottom equation, the n will cancel out, we don't have to do anything to manipulate these equations in order to eliminate one of the unknowns.
subtracting the top equation from the bottom equation yields .1*d = 3.60.
d must then = 36 because 36 * (.1) = 3.60.
going back to the first equation, we have (.05)*n + (.1)*36 = 6.00
which becomes (.05)*n + 3.60 = 6.00
which becomes (.05)*n = 2.40
which becomes n = 2.40/.05
which becomes n = 48.
so number of nickels is 48 and number of dimes is 36.
plugging these numbers in the second equation we get
(.05)*48 + (.2)*36 = 2.4 + 7.2 = 9.6 which confirms that the number of nickels is 48 and the number of dimes is 36.
in the bottom equation, (.2)d is really (.1)*2*d where d is the original number of dimes and .01 is the value of each dime and 2 is because we doubled the number of dimes.
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