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A box contains $8.75 in nickels dimes and quarters there are 51 coins in all and the sum of numbers of nickels and dimes is three less than the number of quarters how many coins of each kind are there?

 Mar 24, 2020
 #1
avatar+2277 
0

Hint: Here are the equations

 

Let x=nickels

      y=dimes

      z=quarters.

 

5x+10y+25z=875

x+y+z=51

x+y+3=z

 

Hope this helped!

 Mar 24, 2020
 #2
avatar+26006 
+1

q  +  q-3   = 51

q = 27                  27 quarters = 6.75        there is 24 coins (n +d) and 2.00 left

 

.05n  +  .10 (24-n) = 2.00

   -.05n = -.4      n = 8  nickels          then    24-n = 16 dimes

 Mar 24, 2020
 #3
avatar+29 
0

simplist method.

5 times 51 is 255

875-255 is 620.

do the rest. iz easy.

 Mar 24, 2020

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