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mr Goodwill [35]
3 years ago
13

Suppose there is a pile of​ quarters, dimes, and pennies with a total value of ​​$1.08

Mathematics
1 answer:
WINSTONCH [101]3 years ago
6 0

Answer:

a)

if 1 quarter = $ 0.25

  1 dime = $ 0.10

  1 penny = $ 0.01

so to make the total of $1.08 and its is also required to calculate the number of each coins present without being able to make change for a dollar

therefore we say;

1 Quarter + 8 dimes + 3 penny

: ( 1 × 0.25 ) + ( 8 × 0.10 ) + ( 3 × 0.01 )

: 0.25 + 0.80 + 0.03 = $ 1.08

b)

Now if you have a 4 Quarters, you have change for $1.

If you have 5 dimes, you have change for 2 Quarters.

If you have nickel; one of those can combine with 2 dimes to have a change for a Quarter.

If you have 5 pennies, you have enough change for 1 nickel

Therefore

(4-1)×0.25 + (5-1)×0.1 + (0×0.05) + (5-1)×0.01 = x

(3 × 0.25) + ( 4 × 0.1) + 0 + ( 4 × 0.01) = x

x = 0.75 + 0.4 + 0.04

x = $ 1.19

PROVED

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