Question 1203578
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According to the problem, the coins can be grouped in sets, containing 3 nickels and one dime in each set.


Thus each set is worth  3*5 + 10 = 25 cents, and the number of sets is  {{{525/25}}} = 21.


It gives the <U>ANSWER</U>:  there are 21 dimes and 3*21 = 63 nickels.


<U>CHECK</U>.  21*10 + 63*5 = 526 cents, in total, or $5.25.   ! correct !
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Solved (mentally).