A man has 21 coins in his pocket, all of which are dimes and quarters. If the total value of his change is 405 cents, how many dimes and how many quarters does he hav
number of dimes ---- x
number of quarters --- 21-x
by value:
10x + 25(21-x) = 405
10x + 525 - 25x = 405
-15x = -120
x = 8
so 8 dimes and 21-8 or 13 quarters
or
dimes ---- d
quarters ---q
based on numbers:
d+q = 21
based on value:
10d + 25q = 405 --- 2d + 5q = 81
double the first:
2d + 2q = 42
subtract from the 2nd to give
3q = 39
q = 13
back into the first:
d + q = 13
d = 8
I find the first method easier.
Well, it seems like this guy has quite the coin conundrum on his hands! Let's break it down.
Let's call the number of dimes "D" and the number of quarters "Q". Since we know he has a total of 21 coins, we can write an equation: D + Q = 21.
Now, let's talk value. Each dime is worth 10 cents and each quarter is worth 25 cents. If we multiply the number of dimes by 10 and the number of quarters by 25, we should get the total value of his change, which is 405 cents.
So, the second equation would look like this: 10D + 25Q = 405.
Now, let's unleash the clown in me to solve this!
I'm going to use a circus trick called substitution to make the math a little easier. Solving the first equation for D, we get D = 21 - Q.
Plugging that into the second equation, we get 10(21-Q) + 25Q = 405.
Expanding that, we have 210 - 10Q + 25Q = 405.
Combining like terms, we have 15Q = 195.
Dividing both sides by 15, we find that Q = 13.
So, this guy has 13 quarters. Substituting that back into the first equation, we find that D = 21 - 13 = 8.
Therefore, he has 8 dimes and 13 quarters. Who knew a pocket full of change could require so much mathematical acrobatics?
To solve this problem, we can use a system of equations.
Let's assume the number of dimes the man has is "d" and the number of quarters is "q".
We have two pieces of information given:
1. The total number of coins is 21: d + q = 21
2. The total value of the coins is 405 cents: 10d + 25q = 405
Now we can solve these equations using one of the methods, such as substitution or elimination. Let's use the elimination method:
Multiply the first equation by 10 to make the coefficients on the "d" terms the same:
10d + 10q = 210
Now, subtract the second equation from the first equation:
(10d + 10q) - (10d + 25q) = 210 - 405
This simplifies to:
-15q = -195
Divide both sides of the equation by -15 to solve for "q":
q = (-195) / (-15)
q = 13
Now substitute the value of "q" back into the first equation to solve for "d":
d + 13 = 21
d = 21 - 13
d = 8
Therefore, the man has 8 dimes and 13 quarters.
To solve this problem, we can use a system of equations. Let's assume that the number of dimes is represented by 'D' and the number of quarters is represented by 'Q'.
We know from the problem that the man has a total of 21 coins, so we can write the equation:
D + Q = 21
We also know that the total value of his change is 405 cents. Since a dime is worth 10 cents and a quarter is worth 25 cents, the equation can be written as:
10D + 25Q = 405
Now we have a system of two equations that we can solve simultaneously to find the values for D and Q.
One way to solve this system is to use the substitution method. We can rearrange the first equation to solve for D:
D = 21 - Q
Substituting this value of D into the second equation, we get:
10(21 - Q) + 25Q = 405
Now we can solve for Q:
210 - 10Q + 25Q = 405
15Q = 195
Q = 13
Substituting the value of Q back into the first equation, we can find D:
D + 13 = 21
D = 21 - 13
D = 8
Therefore, the man has 8 dimes and 13 quarters.