Dylan's dimes total $1.30 less than his quarters of which he has one-fourth as many as he has nickels, which total $0 more than his dimes. How many of each coin does he have?

Sturasion12345 Jan 10, 2022

#2**+1 **

@ guest I think you are wrong

reason:

'Dylan's dimes total $1.30 less than his quarters "

so the quarters have to be at least 6 of them

My attempt:

So we need 2x as many nickels as dimes

so

quarters | dimes | nickels | |||

6 | 2 | 4 | n | o | |

8 | 7 | 14 | n | o | |

10 | 12 | 24 | n | o | |

12 | 17 | 34 | n | o | |

14 | 22 | 44 | n | o |

16 | 27 | 54 | n | o |

18 | 32 | 64 | n | o | |

20 | 37 | 74 | n | o | |

22 | 42 | 84 | n | o |

24 | 47 | 94 | n | o | |

26 | 52 | 104 | y | e | s |

Thhis is acually a pretty bad method of doing it but if you noticed, there is a pattern

I'll let you find it

XxmathguyxX Jan 10, 2022

#4**+1 **

Number of nickels = x ---> cost = $0.05x

Number of quarters = x/4 ---> cost = ($0.25 * x)/4

Cost of dimes = same as of nickels = $0,05x

No. of dimes * 0.1 = 0.05x

No. of dimes = (0.05x)/0.1 = 0.5x

Also cost of dime = Cost of quarters - $1.3

0.05x = (0.25x)/4 - 1.3

=> 1.3 = 0.0125x - 0.05x

=> 1.3 = 0.0125x

=> x = 1.3/0.0125 = 104

He have 104 nickels, 104/4 = 26 quarters and 1/2 * 104 = 52 dimes

Slimesewer Jan 12, 2022