Question 1102836
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The condition seems to be SUSPICIOUS, since it imposes 3 (three, THREE) conditions on 2 unknowns.


So, let us look what will happen.



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d + q = 61     (1)    (counting coins)
q = d + 3      (2)    ("3 more quarters than dimes")


Substitute (2) into eq(1). You will get

d + (d+3) = 61  ====>  2d + 3 = 61  ====>  2d = 61-3 = 58  ====>  d = {{{58/2}}} = 29.


<U>Answer</U>.  Working with the first two parts of the condition (with TWO of the THREE), we get this answer:
                
         29 dimes and 29+3 = 32 quarters.


Now, let us check if the LAST condition is satisfied.

Total value = 29*10 + 32*25 = 1090.


Happily, the third condition is satisfied, so the solution and the problem itself are OK.


But still keep in mind that ONE OF THE THREE CONDITIONS is excessive and unnecessary.


The third condition is a potential source for the condition to be WRONG,
but happily it is not the case here.
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