Mia is approaching her goal of 119 points for the video game. She wants to put her initials on the computerized list of Top Ten Point Scorers forr the game, She moves Freddie the frog all over the terminal screen and at last she does it. Freddie the Frog eats a total of 22 critters: worm ( worth 6 points each), bugs ( worth 7 points each). and butterfies ( worth 4 points each). if Mia earned the total of 119 points, how many critters of each type did her frog eat?
w=worms, g=bugs, f=butterflies
count critters: w+g+f = 22
count points: 6w+7g+4f = 119
6w+7g+4f = 4(w+g+f) + 2w + 5f
4(22) + 2w + 5g = 119
2w + 5g = 119-88 = 31
Now, we don't know how many w,g we have, but since 2w is even, that means 5g must be odd, which means that g must be odd.
Take a look at possibilities:
g=1: 2w+5 = 31 ==> w=13 f=8
g=3: 2w+15 = 31 ==> w=8 f=11
g=5: 2w+25 = 31 ==> w=3 f=14
Those are the only possible solutions
Good solution, Steve, but w+b+f=6+7+4=17.
worms (w) - 6
bugs (b) - 7
butterflies (f)- 4
critters: w+b+f=22
points: 6w+7b+4f=119 (the only way to get an odd number is if b is odd)
points from having one of each of the critters: w+b+f -> 6+7+4=17
so, using the points now:
6w+7b+4f=119
Break apart each set of critters into 4+something:
4w+2w+4b+3b+4f=119
Which means 4w+4b+4f+2w+3b=119
But 4w+4b+4f=4(w+b+f)=4(22) – (this is when she has equal amounts of each critter)
So 4(22)+2w+3b=119
2w+3b=119-88=31
2w+3b=31 (which can only be odd if b is odd, so try values for b)
b=1, 2w+3=31, w=14, f=7 (f=22-1-14) -> w=14, b=1, f=7
b=3, 2w+9=31, w=11, f=8 -> w=11, b=3, f=8
b=5, 2w+15=31, w=8, f=9 -> w=8, b=5, f=9
b=7, 2w+21=31, w=5, f=10 -> w=5, b=7, f=10
b=9, 2w+27=31, w=2, f=11 -> w=2, b=9, f=11
If b is larger than 9, 2w+3b will be larger than 31, so no more solutions.
Nic, your solution is a bit confusing as it contains 6+4+7. instead, try this:
worms-w-9
bugs-b-10
butterflies-f-3
11b+1w
or...
11b+3f
11b=110
1w=9
3f=9
110+9=119
SOLVED!
To solve this problem, we need to find the number of each type of critter that Mia's frog ate. Let's use algebraic notation to represent the unknowns.
Let's assume Mia's frog ate x worms, y bugs, and z butterflies. Now we can write three equations based on the given information:
1. The total number of critters Freddie the frog ate is 22:
x + y + z = 22
2. Worms are worth 6 points each, so the total points earned from worms is 6x:
6x
3. Bugs are worth 7 points each, so the total points earned from bugs is 7y:
7y
4. Butterflies are worth 4 points each, so the total points earned from butterflies is 4z:
4z
The total points earned is 119:
6x + 7y + 4z = 119
So we have a system of equations:
x + y + z = 22 (Equation 1)
6x + 7y + 4z = 119 (Equation 2)
To solve this system of equations, we can use a method such as substitution or elimination. Let's use the substitution method.
From Equation 2, we can isolate x:
6x = 119 - 7y - 4z
x = (119 - 7y - 4z)/6
Now we substitute this value of x into Equation 1:
(119 - 7y - 4z)/6 + y + z = 22
Next, let's simplify the equation:
119 - 7y - 4z + 6y + 6z = 132
-7y + 6y - 4z + 6z = 132 - 119
-y + 2z = 13 (Equation 3)
Now we have two equations:
-y + 2z = 13 (Equation 3)
x + y + z = 22 (Equation 1)
To eliminate y, let's multiply Equation 3 by 2:
-2y + 4z = 26 (Equation 4)
Now, add Equation 1 and Equation 4 to eliminate y:
(x + y + z) + (-2y + 4z) = 22 + 26
x - y + 5z = 48 (Equation 5)
Now we have two equations again:
x - y + 5z = 48 (Equation 5)
-y + 2z = 13 (Equation 3)
Let's multiply Equation 3 by -1:
y - 2z = -13 (Equation 6)
Next, add Equation 5 and Equation 6 to eliminate y:
(x - y + 5z) + (y - 2z) = 48 - 13
x + 3z = 35 (Equation 7)
Now we have two equations:
x + 3z = 35 (Equation 7)
y - 2z = -13 (Equation 6)
Let's isolate x in Equation 7:
x = 35 - 3z
Now we can substitute this value of x into Equation 6:
(35 - 3z) + 3z = 48
35 - 3z + 3z = 48
35 = 48
It seems like there is no solution that satisfies both equations. There might be an error in the initial problem setup or some missing information. Please double-check the given information and let me know if anything is missing or incorrect.