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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
avatar+549 
+1

maybe just do trial and error 

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

 Jun 13, 2020
edited by amazingxin777  Jun 13, 2020
 #2
avatar+307 
+1

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
avatar+8341 
-2

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
avatar+131 
+4

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

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