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.- I need little hint please. My answer is coming out wrong.
The best hint is for you to study the answers you've gotten for your other similar questions.
Since the number of quarters is the basis of comparison,
let the number of quarters be x
then the number of dimes = x+9
and the number of nickels = x+6
equation based on the VALUE of the coins
25x + 10(x+9) + 5(x+6) = 160
solve for x, plug back into my definitions are you are done.
To solve this problem, you can use a system of equations. Let's denote the number of quarters as q, the number of dimes as d, and the number of nickels as n.
From the given information, we can create three equations:
1. The total value of the coins: 0.25q + 0.10d + 0.05n = 1.60 (since each quarter is worth $0.25, each dime is worth $0.10, and each nickel is worth $0.05).
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, you have a system of equations that can be solved simultaneously. Try substituting equations 2 and 3 into equation 1 to eliminate d and n in terms of q. Then, you can solve for q and find the values of d and n.
Let me know if you need help with the math or if you have any other questions!