SOLUTION: If Qasim has 4 times as many quarters as nickels and they have a combined value of 735 cents, how many of each coin does he have?

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Question 1200308: If Qasim has 4 times as many quarters as nickels and they have a combined value of 735 cents, how many of each coin does he have?
Found 3 solutions by Theo, josgarithmetic, ikleyn:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
25 * x + 5 * y = 735
x is the number of quarters; each quarter is worth 25 cents.
y is the number of nickels; each nickel is worth 5 cents.
you are given that x = 4 * y (4 times as many quarters as nickels)
your two equations are:
25 * x + 5 * y = 735
x = 4 * y
in the first equation, replace x with 4 * y to get:
25 * 4 * y + 5 * y = 735
simplify to get:
100 * y + 5 * y = 735
combine like terms to get:
105 * y = 735
solve for y to get:
y = 735 / 105 = 7
since x = 4 * y, then x = 28
25 * x becomes 25 * 28 = 700
5 * y becomes 5 * 7 = 35
25 * x + 5 * y becomes 700 + 35 = 735 cents.
solution is 28 quarters and 7 nickels.



Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!
"they have a combined,..." You mean ALL of the coins?


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Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
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If Qasim has 4 times as many quarters as nickels and they have a combined value
of 735 cents, how many of each coin does he have?
~~~~~~~~~~~~~~~~~~~ 

Let x be the number of nickels. 
Then the number of quarters is 4x.


Write the total money equation

    25 *(4x) + 5x = 735 cents.

Simplify and find x

    100x + 5x = 735

        105x  = 735

           x  = 735/105 = 7.


ANSWER.  7 nickels and 7*4 = 28 quarters.

Solved.

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It can be solved mentally  (the same solution as above,  only re-told in wording form). 

Group coins in the sets in a way that each group consists of 1 nickel and 4 quarters.

According to the problem, such grouping is possible.

Each group is worth  4*25 + 5 = 105 cents.

The number of group is  735/105 = 7.


It gives the answer above.

Solved in two ways,  for your better understanding.



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