1. For every slice of bread a child takes, an adult takes 2 slices of bread. If there are 66 adults and children in all and 99 slices of bread are taken,
a. how many adults are there?
b. how many children are there?
2. There are 100 adults and children together. An adult could consume 3 buns while 3 children could only finish one bun. If 100 buns were consumed, how many adults were there?
no I cant
1. Let's assume the number of children is represented by "C".
a. Since for every slice of bread a child takes, an adult takes 2 slices, the total number of slices taken can be represented by the equation:
C + 2C = 99
Combining like terms, we get:
3C = 99
To find the value of C, we divide both sides of the equation by 3:
C = 99/3
C = 33
Therefore, there are 33 children.
b. We can find the number of adults by using the total number of adults and children in all:
66 - 33 = 33
Therefore, there are 33 adults.
2. Let's assume the number of adults is represented by "A".
a. Since an adult could consume 3 buns and a child could consume 1 bun, the total number of buns consumed can be represented by the equation:
3A + 1*(100 - A) = 100
Expanding the equation, we get:
3A + 100 - A = 100
Combining like terms, we have:
2A = 0
To find the value of A, we divide both sides of the equation by 2:
A = 0/2
A = 0
Therefore, there are no adults in this scenario.
To solve these problems, let's set up equations based on the given information and then solve for the variables.
1. Let's assume there are x children and y adults.
a. For every slice of bread a child takes, an adult takes 2 slices. So the total number of slices taken by children would be x, and the total number of slices taken by adults would be 2y.
We can set up the equation: x + 2y = 99.
b. We also know that there are 66 adults and children in total. So we can set up another equation: x + y = 66.
To solve these equations, we can use either substitution or elimination method. Let's use the elimination method:
Multiply the second equation by 2 to make the coefficients of y equal: 2(x + y) = 2(66), which simplifies to 2x + 2y = 132.
Now, subtract the first equation from the second equation: (2x + 2y) - (x + 2y) = 132 - 99, which simplifies to x = 33.
Substitute x = 33 into the second equation to solve for y: 33 + y = 66, which gives us y = 33.
Therefore, there are 33 adults (a) and 33 children (b).
2. Let's assume there are a adults and c children.
a. An adult could consume 3 buns, so the total number of buns consumed by adults would be 3a.
Similarly, 3 children could consume 1 bun each, so the total number of buns consumed by children would be 3c.
We can set up the equation: 3a + 3c = 100.
b. We also know that there are 100 adults and children in total. So we can set up another equation: a + c = 100.
Again, we can use the elimination method to solve these equations:
Multiply the second equation by 3 to make the coefficients of c equal: 3(a + c) = 3(100), which simplifies to 3a + 3c = 300.
Now, subtract the first equation from the second equation: (3a + 3c) - (3a + 3c) = 300 - 100, which simplifies to 0 = 200.
Since 0 is not equal to 200, there must be an error in the problem statement or a contradiction in the information provided. Please double-check the problem or provide additional information if available.
c + 2 a = 99
c + a = 66
------------------- subtract
0 + a = 33
so a = 33 and c= 33
Now you should be able to do the second one.