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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?

Guest Jun 13, 2020

amazingxin777 · Answer 1 · 2020-06-13T05:29:37

maybe just do trial and error

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

thelizzybeth · Answer 2 · 2020-06-13T05:30:45

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.

MaxWong · Answer 3 · 2020-06-13T05:30:51

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).

Forgo · Answer 4 · 2020-06-14T06:43:08

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...