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
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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