Math, asked by parasofficial04, 6 months ago

you have quarters and dimes that total $2.80. is it possible that no of quarters is 8 more than no of dimes? explain and most importantly find all the possible combinations of coins (quarters and dimes) that satisfy the condition. explain everything in detail

Answers

Answered by amitnrw
0

Given : you have quarters and dimes that total $2.80.

To Find : is it possible that no of quarters is 8 more than no of dimes?

all the possible combinations of coins (quarters and dimes) that satisfy the condition.

Solution:

A penny is worth 1 cent.

A nickel is worth 5 cents.

A dime is worth 10 cents.

A quarter is worth 25 cents.

1 $ = 100 cents

total $2.80. = 280 cents

Let say Quarters = Q  & Dimes = D

=>  25 * Q  + 10D  = 280

=> 5Q  + 2D  = 56

Q            D          

0            28             Quarters are not more than Dimes

2            23             Quarters are not more than Dimes

4             18             Quarters are not more than Dimes

6             13             Quarters are not more than Dimes

8             8               Quarters = Dimes

10            3               Quarters are 7  more  than Dimes

Hence number of quarters is 8 more than number of dimes is not possible

25(D + 8) + 10D = 280

=> 7D = 16

=> D = 16/7  not an integer  

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