a box containing pens,nickels,and dimes has 13 coins with a total value of 83 cents.how many coins of each type are in the box?
number of pennies --- p
number of nickels ---- n
number of dimes ---- 13-p-n
p + 5n + 10(13-p-n) = 83
p + 5n + 130 - 10p - 10n = 83
-9p - 5n= -47
9p + 5n = 47 ---> n = (47-9p)/5
since both p and n have to be positive integers
0 < p < 5 and 0 < n< 10
try p=1 , n is a fraction
p = 2, n is a fraction
p=3 , n = 4
p = 4 , n is a fraction
so p = 3, n = 4 and subbing back, dimes = 6
check" 3(1) + 4(5) + 6(10) = 83
Well, well, looks like we have a math problem on our hands! Let's put on our detective hats and see if we can solve this mystery.
Let's assume we have x pens, y nickels, and z dimes in the box.
Now, we know we have a total of 13 coins, so we can write our first equation as:
x + y + z = 13
We also know that the total value of these coins is 83 cents. So, we can write another equation:
0.01x + 0.05y + 0.1z = 83
Now, do we need a calculator or can we solve this with a little brain power? Let's put on our math muscles and simplify those equations!
From the first equation, we can rewrite it as:
x = 13 - y - z
Now, let's substitute x in the second equation:
0.01(13 - y - z) + 0.05y + 0.1z = 83
And after doing some arithmetic magic, we can simplify this equation to:
0.13 - 0.01y - 0.01z + 0.05y + 0.1z = 83
Combining like terms, we get:
0.04y + 0.09z = 82.87
Woah, that's a funky-looking equation! But fear not, we can solve it.
Now, here's the catch... I'm going to need a little more information or a hint to find the exact values for y and z. Can you give me any additional details, like the number of pens or maybe the number of dimes?
Let's assume the number of pens in the box is P, the number of nickels is N, and the number of dimes is D.
From the given information, we can set up the following equations:
1. P + N + D = 13 (since there are a total of 13 coins)
2. 0.01D + 0.05N + 0.10D = 0.83 (since the total value of the coins is 83 cents)
To solve this system of equations, we can use the substitution method. Rearrange the first equation to solve for P:
P = 13 - N - D
Substitute this value of P into the second equation:
0.01D + 0.05N + 0.10D = 0.83
Simplify the equation:
0.11D + 0.05N = 0.83 - 13
0.11D + 0.05N = -12.17 (Equation 3)
Now substitute P = 13 - N - D into Equation 3:
0.11D + 0.05N = -12.17
0.11D + 0.05N = -12.17 (Equation 4)
Now we have a system of equations:
0.11D + 0.05N = -12.17 (Equation 4)
P + N + D = 13 (Equation 1)
Using these equations, we can solve for P, N, and D.
To solve this problem, we can set up a system of equations based on the given information.
Let's assume that the number of pens is represented by "p," the number of nickels is represented by "n," and the number of dimes is represented by "d."
According to the problem, we know two pieces of information:
1. There are 13 coins in total: p + n + d = 13
2. The total value of the coins is 83 cents: 1p + 5n + 10d = 83
Let's use these equations to find the solution.
First, we can solve the first equation for one variable in terms of the others. Let's solve for "p" in terms of "n" and "d":
p = 13 - n - d
Next, we'll substitute this value of "p" into the second equation:
1(13 - n - d) + 5n + 10d = 83
Simplifying this equation will give us:
13 - n - d + 5n + 10d = 83
13 + 4n + 9d = 83
4n + 9d = 70 ---> Equation 3 (after subtracting 13 from both sides)
Now, we have two equations:
p + n + d = 13 ----> Equation 1
4n + 9d = 70 ----> Equation 3
We can solve this system of equations to find the values of "n" and "d."
One way to solve this system is by substitution or elimination method, but considering the values, it's relatively easier to solve by trial and error.
Let's explore the different possibilities of values for "n" and "d" that satisfy both equations.
We know that the number of coins can't be negative, so both "n" and "d" have to be positive.
Let's try different values for "n" and "d" starting from 1 and onwards while observing equation 3, which is 4n + 9d = 70.
n = 1, d = 1: 4(1) + 9(1) = 4 + 9 = 13 (not equal to 70)
n = 1, d = 2: 4(1) + 9(2) = 4 + 18 = 22 (not equal to 70)
n = 1, d = 3: 4(1) + 9(3) = 4 + 27 = 31 (not equal to 70)
n = 1, d = 4: 4(1) + 9(4) = 4 + 36 = 40 (not equal to 70)
n = 1, d = 5: 4(1) + 9(5) = 4 + 45 = 49 (not equal to 70)
n = 1, d = 6: 4(1) + 9(6) = 4 + 54 = 58 (not equal to 70)
.
.
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Going through possible values, let's find the values for "n" and "d" that satisfy equation 3:
n = 5, d = 6: 4(5) + 9(6) = 20 + 54 = 74 (not equal to 70)
n = 5, d = 7: 4(5) + 9(7) = 20 + 63 = 83 (equal to 70)
By substituting n = 5 and d = 7 into equation 1, we can find the value of "p":
p + 5 + 7 = 13
p = 1
Therefore, the solution for this problem is:
- There is 1 pen, 5 nickels, and 7 dimes in the box.