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help

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Mex has \$2.65. He has only dimes and quarters.  If Mex has more quarters than dimes, how many coins does he have in total?

Jun 13, 2020

#1
+549
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maybe just do trial and error

that's what i usually do when i can't figure it out

Jun 13, 2020
edited by amazingxin777  Jun 13, 2020
#2
+307
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We set up an equation first: 0.25q + 0.10d = 2.65

Multiplying by 100 and dividing by 5, we get 5q + 2d = 53

At this point, we can just use trial and error, noting that q can only be an odd number because when it's even, 3-0 = 3, which is not divisible by 2.

After testing out a few cases, the only pair that satisfies q > d is q=9 and d=4.

Jun 13, 2020
edited by thelizzybeth  Jun 13, 2020
#3
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Let x be the number of dimes. y be the number of quarters.

y > x.

\(0.10 x + 0.25 y = 2.65\\ 2x + 5y = 53\)

You can find the total number of coins by a trial-and-error approach to find possible pairs of (x, y).

Jun 13, 2020
#4
+131
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We get the equations:

25q + 10d = 265

q > d.

Divide by 5 for the first equation yields us 5q + 2d = 53.

After trial and error the value of q = 9, d = 4

To check:

25 x 9 = 15 x 15 = 225.

4 x 10 = 40.

225+40=265...

Jun 14, 2020