Answer :
In the word problem, if mike has 10 coins consisting of dimes and quarters and if the dimes were quarters and the quarters dimes, he would have 30 cents more, he has 2 quarters.
A word problem refers to a mathematical exercise where significant information on the problem is given in ordinary language and not mathematical notation. Let x be the number of dimes (10 cents) and y be the number of quarters (25 cents). Hence, the total number of coins is:
x + y = 10
The amount of the total coins would be 0.1x + 0.25y.
If the dimes were quarters and quarters were dimes, he would have 30 cents more than the amount he has now.
Hence,
0.25x + 0.10y = 0.1x + 0.25y + 0.9
Solving,
0.25x – 0.1x = 0.25y – 0.1y +0.9
0.15x = 0.15y + 0.9
As x + y = 10, x = 10 – y
Substituting,
0.15 (10 – y) = 0.15y + 0.9
1.5 - 0.15 y = 0.15y + 0.9
0.15y + 0.15y = 1.5 – 0.9
0.3y = 0.6
y = 0.6/0.3 = 2
Hence, the number of quarters are 2
Learn more about of Word problem here
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