A student buys a sandwich at lunchtime from the school canteen for $4.20

Question

A student buys a sandwich at lunchtime from the school canteen for $4.20 And pays the exact amount using 50 cent coins, 20 cent coins and 10 cent coins. If the number of 20 cent coins is the same as half the 10 cent coins plus 4 times the number of 50 cent coins and the student pays with 22 coins in total, how many coins of each type did the student use?

 

please tell me how to work it out! Thanks 

Answer 1

x = number of 50 cent coins

y = number of 20 cent coins

z = number of 10 cent coins

 

We can dispense with the decimals by putting all totals in number of cents:

 

50x + 20y + 10z = 420   {equation 1}

y = z/2 + 4x                   {equation 2} 

x + y + z = 22                {equation 3}

 

from equation 2:

2y = z + 8x

z = 2y - 8x

 

Substitute that into the other 2 equations

50x + 20y + 10(2y - 8x) = 420

50x + 20y + 20y - 80x = 420

-30x + 40y = 420       {equation 1a}

 

x + y + 2y - 8x = 22

-7x + 3y = 22             {equation 3a}

 

-30x + 40y = 420        1a

-7x + 3y = 22              3a

 

Multiply 1a by -3.  Multiply 3a by 40.  Add equations

   90x - 120y = -1260

-280x + 120y = 880

-----------------------------

-190x = -380

x = 2

 

There are 2 of the 50 cent coins

Substitute into equation 3a

 

-7(2) + 3y = 22

-14 + 3y = 22

3y = 36

y = 12

 

There are 12 of the 20 cent coins

 

x + y + z = 22

2 + 12 + z = 22

z + 14 = 22

z = 8

 

There are 8 of the 10 cent coins

 

Check the monetary total:

50x + 20y + 10z = 420

50(2) + 20(12) + 10(8) = 420

100 + 240 + 80 = 420

420 = 420

 

Two 50 cent coins

Twelve 20 cent coins

eight 10 cent coins