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Examples of coin problem with solution? - Answers

Eugenia had five times as many quarters as dimes. If the total value of her coins was $16.20 how many of each kind of coin did she have?Let x= the amount of dimes Eugenia had (dimes are worth 10)Let 5x= the amount of quarters Eugenia had (quarters are worth 25)Equation: 10(x) + 25(5x)=1620 (because that is how many cents there are)10(x)+25(5x)=162010x+125x=1620135x=16201/135(135)=1/135(1620)1x= 12x=12check10(x) + 25(5x)=162010(12)+25(5x12)=1620120+25(60)=1620120+1500=16201620=1620x=125x=1500Therefore Eugenia had 12 dimes and 60 quarters



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Examples of coin problem with solution? - Answers

https://math.answers.com/other-math/Examples_of_coin_problem_with_solution

Eugenia had five times as many quarters as dimes. If the total value of her coins was $16.20 how many of each kind of coin did she have?Let x= the amount of dimes Eugenia had (dimes are worth 10)Let 5x= the amount of quarters Eugenia had (quarters are worth 25)Equation: 10(x) + 25(5x)=1620 (because that is how many cents there are)10(x)+25(5x)=162010x+125x=1620135x=16201/135(135)=1/135(1620)1x= 12x=12check10(x) + 25(5x)=162010(12)+25(5x12)=1620120+25(60)=1620120+1500=16201620=1620x=125x=1500Therefore Eugenia had 12 dimes and 60 quarters



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https://math.answers.com/other-math/Examples_of_coin_problem_with_solution

Examples of coin problem with solution? - Answers

Eugenia had five times as many quarters as dimes. If the total value of her coins was $16.20 how many of each kind of coin did she have?Let x= the amount of dimes Eugenia had (dimes are worth 10)Let 5x= the amount of quarters Eugenia had (quarters are worth 25)Equation: 10(x) + 25(5x)=1620 (because that is how many cents there are)10(x)+25(5x)=162010x+125x=1620135x=16201/135(135)=1/135(1620)1x= 12x=12check10(x) + 25(5x)=162010(12)+25(5x12)=1620120+25(60)=1620120+1500=16201620=1620x=125x=1500Therefore Eugenia had 12 dimes and 60 quarters

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      Eugenia had five times as many quarters as dimes. If the total value of her coins was $16.20 how many of each kind of coin did she have?Let x= the amount of dimes Eugenia had (dimes are worth 10)Let 5x= the amount of quarters Eugenia had (quarters are worth 25)Equation: 10(x) + 25(5x)=1620 (because that is how many cents there are)10(x)+25(5x)=162010x+125x=1620135x=16201/135(135)=1/135(1620)1x= 12x=12check10(x) + 25(5x)=162010(12)+25(5x12)=1620120+25(60)=1620120+1500=16201620=1620x=125x=1500Therefore Eugenia had 12 dimes and 60 quarters

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