Cindy had $1.08 using 9 coins.
She had as many pennies as dimes.
There must be 4 pennies as 9 pennies would not allow for any other coins.
Since there are supposed to be as many dimes as there are pennies, 4P + 4D = 44 cents.
That leaves 65 cents to come from 1 coin which is impossible so there cannot be as many dimes as pennies.
10D + 9P = $1.09 but 19 coins
1Q + 8D + 4P = $1.09 but 13 coins
2Q + 5D + 4P = $1.09 but 11 coins
3Q + 3 D + 4P = $1.09 but 10 coins
3Q + 2D + 2N + 4P = $1.09 but 11 coins
Halves must be involved
1H leaves 59 cents from 8 coins
Must have 4 pennies
1H + 4P leaves 55 cents from 4 coins
1H + 2Q + 1N + 4P = $1.09 but 8 coins
1H + 1 Q + 3D + 4P = $1.09 and 9 coins
Hooray!!!
Cindy had $1.08 using 9 coins.
She had as many pennies as dimes.
___
Danielle had 96 coins, using ten coins.
She had the same number of nickels as all the other coins put together.
__
The coin choices are.
Half Dollar, Quarters, Dimes, Nickels, and Pennies.
Help?
4 answers
Danielle had 96 cents, using ten coins.
She has the same number of nickels as all the other coins put together.
Based on our prior experience, it is rather easy to zero in on
1Q + 4D + 6N + 1P = 96 cents in 12 coins
where 6N = 1Q + 4D + 1P coins
She has the same number of nickels as all the other coins put together.
Based on our prior experience, it is rather easy to zero in on
1Q + 4D + 6N + 1P = 96 cents in 12 coins
where 6N = 1Q + 4D + 1P coins
Danielle had 96 cents, using ten coins.
She has the same number of nickels as all the other coins put together.
Based on our prior experience, it is rather easy to zero in on
1Q + 4D + 6N + 1P = 96 cents in 12 coins
where 6N = 1Q + 4D + 1P coins
OOPS - The hand was quicker than the eye. Lets try a more exact method.
Assuming no halves are involved.
1--25Q + 10CD + 5N + 1P = 96
2--Q + D + N + P = 10
3--N = Q + D + P
4--Substituting (3) into (1) yields 30Q + 15D + 6P = 96
5--Sustituting (3) into (2) yields Q + D + P = 5
6--Multiplying (5) by 6 yields 6Q + 6D + 6P = 30
7--Subtracting (6) from (4) yields 8Q + 3D = 22
8--Dividing through by the lowest coefficient yields 2Q + 2Q/3 = 1 + 3/7
9--(2Q - 1)/3 must be an integer as does (4Q - 2)/3
10--Dividing by 3 again yields Q + Q/3 - 2/3
11--(Q - 2)/3 is an integer k making Q = 3k + 2
12--Substituting back into (7( yields D = 2 - 8k
13--Clearly, k must be 0.
14--For k = 0, Q = 2 and D = 2.
15--From (5), P = 1 and hence, N = 4.
16--Thus we have Q = 2, D = 2, N = 5 and P = 1 and 2Q's + 2D's + 1P = 5N as coins go.
17--Also, 25(2) + 10(2) + 5(5) + 1(1) = 96 cents.
At last.!!!
She has the same number of nickels as all the other coins put together.
Based on our prior experience, it is rather easy to zero in on
1Q + 4D + 6N + 1P = 96 cents in 12 coins
where 6N = 1Q + 4D + 1P coins
OOPS - The hand was quicker than the eye. Lets try a more exact method.
Assuming no halves are involved.
1--25Q + 10CD + 5N + 1P = 96
2--Q + D + N + P = 10
3--N = Q + D + P
4--Substituting (3) into (1) yields 30Q + 15D + 6P = 96
5--Sustituting (3) into (2) yields Q + D + P = 5
6--Multiplying (5) by 6 yields 6Q + 6D + 6P = 30
7--Subtracting (6) from (4) yields 8Q + 3D = 22
8--Dividing through by the lowest coefficient yields 2Q + 2Q/3 = 1 + 3/7
9--(2Q - 1)/3 must be an integer as does (4Q - 2)/3
10--Dividing by 3 again yields Q + Q/3 - 2/3
11--(Q - 2)/3 is an integer k making Q = 3k + 2
12--Substituting back into (7( yields D = 2 - 8k
13--Clearly, k must be 0.
14--For k = 0, Q = 2 and D = 2.
15--From (5), P = 1 and hence, N = 4.
16--Thus we have Q = 2, D = 2, N = 5 and P = 1 and 2Q's + 2D's + 1P = 5N as coins go.
17--Also, 25(2) + 10(2) + 5(5) + 1(1) = 96 cents.
At last.!!!
Cindy has 1 half dollar,1quater,3 dimes,1 nickel,and 3 penny's.
1 answer