Mukul has$1.60 in nickels, dimes, and quarters. He has nine more dimes than quarters and six more nickels than quarters. How many of each coin are in his pocket.
d=# dimes
n=# nickels
q= # quarters
d+n+q=1.60
d+9=q
n+6=q
0.1d+0.05n+0.25q=1.60
10 d + 5 n + 25 q = 160
d = q + 9
n = q + 6
substituting
... 10(q + 9) + 5(q + 6) + 25 q = 160
solve for q,
then substitute back to find d and n
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Let's solve the system of equations step-by-step:
1) From the given information, we have the following equations:
d + n + q = 1.60 (equation 1)
d + 9 = q (equation 2)
n + 6 = q (equation 3)
0.1d + 0.05n + 0.25q = 1.60 (equation 4)
2) Let's rearrange equations 2 and 3 to express d and n in terms of q:
d = q - 9 (equation 2 rearranged)
n = q - 6 (equation 3 rearranged)
3) Substitute equations 2 and 3 into equation 1 to eliminate d and n:
(q - 9) + (q - 6) + q = 1.60
3q - 15 = 1.60
4) Simplify equation 4:
0.3q - 0.15 + 0.05q + 0.25q = 1.60
0.6q = 1.60 + 0.15
0.6q = 1.75
5) Divide both sides of equation 4 by 0.6:
q = 1.75 / 0.6
q = 2.92 (approximately)
Since the number of coins must be whole numbers, this means that q (quarters) must be 3.
6) Substitute q = 3 into equations 2 and 3 to find the values of d and n:
d = 3 - 9
d = -6 (This negative value for d is not possible, so let's disregard this solution)
n = 3 - 6
n = -3 (This negative value for n is not possible, so let's disregard this solution)
Therefore, there are no valid solutions for the given problem.
To solve this problem, we can set up a system of equations based on the given information.
Let's assign variables to represent the number of dimes, nickels, and quarters in Mukul's pocket:
d = number of dimes
n = number of nickels
q = number of quarters
We can create three equations based on the given information:
1. The total value of the coins is $1.60:
0.1d + 0.05n + 0.25q = 1.60
2. The number of dimes is nine more than the number of quarters:
d = q + 9
3. The number of nickels is six more than the number of quarters:
n = q + 6
Now we have a system of three equations with three variables. We can solve this system by substitution or elimination.
Let's rearrange equations (2) and (3) to express q in terms of d and n:
Equation (2): d = q + 9 ⟹ q = d - 9
Equation (3): n = q + 6 ⟹ q = n - 6
Now we can substitute q in equation (1) with the expressions above:
0.1d + 0.05n + 0.25(d - 9) = 1.60
Simplifying the equation, we get:
0.1d + 0.05n + 0.25d - 2.25 = 1.60
0.35d + 0.05n = 3.85
To eliminate the decimal coefficients, we can multiply each term by 20:
7d + n = 77
Now we have a simplified system of equations:
0.35d + 0.05n = 3.85
7d + n = 77
Now we can solve this system of equations using substitution or elimination.