A coin jar contains nickels, dimes and quarters. There are 46 coins in all. There are 11 more nickels than quarters. The value of the dimes is $1.00 less than the value of the quarters. How many coins of each type are in the jar?
Solve the system:
n+d+q=46
n=q+11
10d=25q-100
From the 3rd -->d=2.5q-10
The 1st: (q+11)+(2.5q-10)+q=46
q=10
n=21
d=15
solving using the elimination method and identify the system as consistent, inconsistent or dependent.
3x-4y=8
6x-2y=10
Suppose that Maria has 120 coins consisting of pennies, nickels, and dimes. The number of nickels she has is 14 less than twice the number of pennies; the number of dimes she has is 22 less than three times the number of pennies. How many coins of each kind does she have?
p+n+d=120
n=2p-14
d=3p-22
p+(2p-14)+(3p-22)=120
6p-36=120
6p=156
p=26
n=2(26)-14
n=52-14
n=38
d=3(26)-22
d=78-22
d=56
P(1, 2); Q(8, 26)
5+6
Let's solve this problem step by step.
Step 1: Assign variables.
Let's say the number of quarters is Q.
Since there are 11 more nickels than quarters, the number of nickels will be Q + 11.
The number of dimes will be the remaining coins after subtracting the number of quarters and nickels from the total number of coins, which is 46 - (Q + Q + 11) = 46 - (2Q + 11) = 35 - 2Q.
Step 2: Create equations.
We know that the total number of coins is 46, so we can write the equation:
Q + (Q + 11) + (35 - 2Q) = 46
Step 3: Simplify and solve the equation.
Combining like terms:
Q + Q + 11 + 35 - 2Q = 46
4Q + 46 = 46
Subtracting 46 from both sides:
4Q = 0
Dividing by 4:
Q = 0
Step 4: Substitute the value of Q into the other variables to find their values.
Since Q = 0, there are 0 quarters in the jar.
The number of nickels is Q + 11 = 0 + 11 = 11 nickels.
The number of dimes is 35 - 2Q = 35 - 2(0) = 35 dimes.
Therefore, there are 0 quarters, 11 nickels, and 35 dimes in the jar.