SOLUTION: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but three times as many quarters as dimes. The total value of the coins was $11.4

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but three times as many quarters as dimes. The total value of the coins was $11.4      Log On


   

Question 190271: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but three times as many quarters as dimes. The total value of the coins was $11.40. how many coins of each type did Sam find? : Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but three times as many quarters as dimes. The total value of the coins was $11.40. how many coins of each type did Sam find?
Answer by jim_thompson5910(35256) About Me  (Show Source):
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#9 Q: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but three times as many quarters as dimes. The total value of the coins was $11.40. how many coins of each type did Sam find?

A:

Let n=# of nickels, d=# of dimes, and q=# of quarters


Since "He found six more dimes than nickels but three times as many quarters as dimes", this means that d=n%2B6 and q=3d. We'll call these equations 1 and 2.


Furthermore, because "The total value of the coins was $11.40", this means that 0.05n%2B0.10d%2B0.25q=11.40


Note: the total value of the nickels alone is 0.05n (ie the value of ONE nickel multiplied by the number of nickels). The same is applied to the dimes and quarters. These expressions are then added up to get the total value. This is probably where you're stuck.


0.05n%2B0.10d%2B0.25q=11.40 Start with the last equation.


100%280.05n%29%2B100%280.10d%29%2B100%280.25q%29=100%2811.40%29 Multiply EVERY term by 100 to make every number a whole number


5n%2B10d%2B25q=1140 Multiply. Let's call this equation 3.



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So we have the equations


Equation 1: d=n%2B6

Equation 2: q=3d

Equation 3: 5n%2B10d%2B25q=1140


You may see this as a system of equations, which looks like

system%28d=n%2B6%2Cq=3d%2C5n%2B10d%2B25q=1140%29



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5n%2B10d%2B25q=1140 Start with the third equation


5n%2B10d%2B25%2Ahighlight%28%283d%29%29=1140 Plug in q=3d. Notice how the variable "q" is no longer in the equation.


5n%2B10d%2B25%283d%29=1140


5n%2B10d%2B75d=1140 Multiply


5n%2B10%2Ahighlight%28%28n%2B6%29%29%2B75%2Ahighlight%28%28n%2B6%29%29=1140 Plug in d=n%2B6. Now the variable "d" is gone.


Now we're left with a simple equation with one unknown variable.


5n%2B10%28n%2B6%29%2B75%28n%2B6%29=1140


5n%2B10n%2B60%2B75n%2B450=1140 Distribute.


90n%2B510=1140 Combine like terms on the left side.


90n=1140-510 Subtract 510 from both sides.


90n=630 Combine like terms on the right side.


n=%28630%29%2F%2890%29 Divide both sides by 90 to isolate n.


n=7 Reduce.


So this means that there are 7 nickels.


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d=n%2B6 Go back to the first equation


d=7%2B6 Plug in n=7


d=13 Add


So there are 13 dimes


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q=3d Move onto the second equation


q=3%2813%29 Plug in d=13


q=39 Multiply


So there are 39 quarters.



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Answer:

So there are 7 nickels, 13 dimes and 39 quarters.