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A collection of 20 coins made up of only nickels, dimes, and quarters has a total value of $3.35. If the dimes were nickels, the

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Answer: The required number of quarters in the collection is 11.

Step-by-step explanation: Given that a collection of 20 coins made up of only nickels, dimes and quarters has a total value of $3.35.

If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $2.75.

We are to find the number of quarters in the collection.

Let x, y and z represents the number of nickels, dimes and quarters respectively in the collection.

We will be using the following values of nickels, dimes and quarters in form of dollar :

1 nickel = $ 0.05, 1 dime = $ 0.10 and 1 quarter = $0.25.

Then, according to the given information, we have

$x+y+z=20\\\\\Rightarrow x=20-y-z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05x+0.10y+0.25z=3.35\\\\\Rightarrow 5x+10y+25z=335\\\\\Rightarrow x+2y+5z=67~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\0.05y+0.25x+0.10z=2.75\\\\\Rightarrow 5y+25x+10z=275\\\\\Rightarrow y+2z+5x=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of x from equation (i) in equations (ii) and (iii), we have

$(20-y-z)+2y+5z=67\\\\\Rightarrow y+4z=47\\\\\Rightarrow y=47-4z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)$

and

$y+2z+5(20-y-z)=55\\\\\Rightarrow -4y-3z=-45\\\\\Rightarrow y=\dfrac{45-3z}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(v)$

Comparing the values of y from equations (iv) and (v), we get

$47-4z=\dfrac{45-3z}{4}\\\\\Rightarrow 188-16z=45-3z\\\\\Rightarrow 16z-3z=188-45\\\\\Rightarrow 13z=143\\\\\Rightarrow z=\dfrac{143}{13}\\\\\Rightarrow z=11.$