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Okay, so I need to figure out how to solve this coin problem. Let me read it again carefully. The problem says: "You have 10 coins, all of which are either quarters or dimes. The total value of the coins is $2.50. How many quarters and dimes are there?" Alright, let's break this down.
First, I know that there are two types of coins: quarters and dimes. Quarters are worth 25 cents each, and dimes are 10 cents each. The total number of coins is 10, and the total value is $2.50, which is 250 cents. Maybe working in cents will make the calculations easier without dealing with decimals.
Let me define variables. Let's say the number of quarters is Q and the number of dimes is D. Since there are 10 coins in total, I can write the equation:
Q + D = 10
That's straightforward. Now, for the total value. Each quarter is 25 cents, so the total value from quarters is 25Q. Each dime is 10 cents, so the total value from dimes is 10D. Together, these should add up to 250 cents. So the second equation is:
25Q + 10D = 250
Now I have a system of two equations:
Q + D = 10
25Q + 10D = 250
I need to solve this system to find Q and D. Let me think about the best way to do this. Since it's a system of linear equations, substitution or elimination would work. Let's try substitution since the first equation can easily express one variable in terms of the other.
From equation 1, I can solve for D:
D = 10 - Q
Then substitute this into equation 2:
25Q + 10(10 - Q) = 250
Let me compute that step by step. First, expand the 10 into the parentheses:
25Q + 100 - 10Q = 250
Combine like terms. 25Q - 10Q is 15Q:
15Q + 100 = 250
Now subtract 100 from both sides:
15Q = 250 - 100
15Q = 150
Divide both sides by 15:
Q = 150 / 15
Q = 10
Wait, that can't be right. If Q is 10, then D would be 10 - 10 = 0. But the total value would be 2510 + 100 = 250 cents, which is $2.50. Hmm, actually, that works. But does that mean there are 10 quarters and 0 dimes? Let me check again.
But intuitively, 10 quarters would be $2.50, which matches the total. So maybe that's the answer. But wait, the problem says "all of which are either quarters or dimes," so having all quarters is allowed. But maybe I made a mistake in the calculation. Let me verify.
Starting with Q = 10, D = 0. Total coins: 10 + 0 = 10. Total value: 2510 + 100 = 250. Yes, that's correct. So the answer is 10 quarters and 0 dimes. But wait, that seems too straightforward. Let me try elimination method to confirm.
Using elimination on the two equations:
Equation 1: Q + D = 10
Equation 2: 25Q + 10D = 250
If I multiply equation 1 by 10, I get:
10Q + 10D = 100
Then subtract this from equation 2:
25Q + 10D - (10Q + 10D) = 250 - 100
Which simplifies to:
15Q = 150
Again, Q = 10. So same result. So it's correct. But maybe the problem expects some dimes? Wait, but according to the math, it's possible to have all quarters. Let me check if there's another possible solution.
Suppose there are 9 quarters. Then 9 quarters = 225 cents. The remaining 1 coin would be a dime, which is 10 cents. Total value would be 225 + 10 = 235 cents, which is 2.35, less than 2.50. So that's not enough.
If there are 8 quarters: 8*25 = 200. Then 2 dimes = 20. Total 220, still less.
7 quarters: 175 + 30 = 205
6 quarters: 150 + 40 = 190
5 quarters: 125 + 50 = 175
4 quarters: 100 + 60 = 160
3 quarters: 75 + 70 = 145
2 quarters: 50 + 80 = 130
1 quarter: 25 + 90 = 115
0 quarters: 0 + 100 = 100
None of these add up to 250. So the only solution is 10 quarters and 0 dimes. Therefore, the answer is 10 quarters and 0 dimes.
But wait, the problem says "all of which are either quarters or dimes," which doesn't exclude the possibility of all quarters. So maybe that's the answer. But sometimes these problems expect at least one of each, but the problem doesn't specify that. So according to the math, the answer is 10 quarters and 0 dimes.
Alternatively, maybe I made a mistake in setting up the equations. Let me check again.
Total coins: Q + D = 10
Total value: 25Q + 10D = 250
Yes, that's correct. So solving gives Q=10, D=0. So that's the answer. Maybe the trick is that you can have all quarters. So the answer is 10 quarters and 0 dimes.
But let me think again. If I have 10 quarters, that's $2.50 exactly. So that's correct. So the answer is 10 quarters and 0 dimes. Therefore, the solution is 10 quarters and 0 dimes.
To solve the problem of determining how many quarters and dimes there are among 10 coins totaling $2.50, we can set up and solve a system of equations.
Step 1: Define Variables
Let ( Q ) be the number of quarters.
Let ( D ) be the number of dimes.
Step 2: Set Up Equations
Total number of coins:
( Q + D = 10 )
Total value in cents:
( 25Q + 10D = 250 ) (since 2.50 = 250 cents)
Step 3: Solve the System
From the first equation, express ( D ) in terms of ( Q ):
( D = 10 - Q )
Substitute ( D ) into the second equation:
( 25Q + 10(10 - Q) = 250 )
Simplify and solve for ( Q ):
( 25Q + 100 - 10Q = 250 )
( 15Q = 150 )
( Q = 10 )
Substitute ( Q = 10 ) back into ( D = 10 - Q ):
( D = 10 - 10 = 0 )
Conclusion
There are 10 quarters and 0 dimes.
Verification
Total coins: ( 10 + 0 = 10 )
Total value: ( 10 \times 25 + 0 \times 10 = 250 ) cents = $2.50
Thus, the solution is correct.
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发表于 2025-11-4 00:40:47