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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?

Jan 10, 2022

#1
0

Dylan has 4 dimes, 2 quarters, and 8 nickels.

Jan 10, 2022
#2
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@ 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

Jan 10, 2022
#3
0

Yes I noticed it

Guest 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

Jan 12, 2022