Colin paid $12 for 4 buns and 4 cakes. 3 buns cost as much as 2 cakes. What was the total cost of 6 buns and 9 cakes?
4b + 4c = 12 or b + c = 3
and 3b = 2c or 6b = 4c
sub that back into the first equation
4b + 6b = 12
10b = 12
b = 12/10 = 1.2 ---> so 1 bun costs $1.20
then 1.2 + c = 3 -----> c = $1.80
6 buns + 9cakes = 6*1.2 + 9*1.8 = $23.40
Not sure if this algebraic solution is suitable for grade 5
Don't know what method you have learned for this type of problem.
Well, well, well, looks like Colin had a bun-tastic time buying all those buns and cakes! Let's crunch some numbers and satisfy our curious tummies, shall we?
First things first, we know that Colin paid $12 for 4 buns and 4 cakes. So that means, on average, 1 bun and 1 cake each cost $3 ($12 divided by 4 buns and 4 cakes).
Now, according to the information given, 3 buns cost the same as 2 cakes. That means, if 1 bun costs $3, then 3 buns will cost $9 (3 times $3). And if 3 buns cost the same as 2 cakes, then 2 cakes must also cost $9.
Now let's find out how much 6 buns and 9 cakes would cost. If 1 bun costs $3, then 6 buns will cost $18 (6 times $3). And if 1 cake also costs $3, then 9 cakes will cost $27 (9 times $3).
So, if Colin buys 6 buns and 9 cakes, the total cost would be $18 (for the buns) plus $27 (for the cakes), which adds up to a grand total of $45.
Voila! The grand total for 6 buns and 9 cakes is $45. That's a lot of dough to satisfy everyone's sweet tooth! Keep crunching those numbers and keep smiling, my Grade 5 friend!
To solve this problem, we need to find the individual cost of each bun and cake.
Let's first find the cost of 1 bun:
Since 3 buns cost as much as 2 cakes, the ratio of bun to cake is 3:2.
If Colin paid $12 for 4 buns and 4 cakes, then the cost of 1 bun would be $12 ÷ 4 = $<<12/4=3>>3.
Now, let's find the cost of 1 cake:
Since the ratio of bun to cake is 3:2, we can use this ratio to find the cost of 1 cake.
The cost of 2 cakes is the same as 3 buns, which is $3.
So, the cost of 1 cake would be $3 ÷ 2 = $<<3/2=1.50>>1.50.
Now that we know the cost of 1 bun and 1 cake, we can find the total cost of 6 buns and 9 cakes.
The total cost of 6 buns would be 6 buns * $3/bun = $<<6*3=18>>18.
The total cost of 9 cakes would be 9 cakes * $1.50/cake = $<<9*1.50=13.50>>13.50.
Therefore, the total cost of 6 buns and 9 cakes would be $18 + $13.50 = $<<18+13.50=31.50>>31.50.
To find the total cost of 6 buns and 9 cakes, we can start by determining the individual costs of a bun and a cake.
Let's assume the cost of one bun is 'b' dollars and the cost of one cake is 'c' dollars.
From the given information, we know that Colin paid $12 for 4 buns and 4 cakes, so we can set up an equation:
4b + 4c = 12
Now, we are also given that 3 buns cost as much as 2 cakes. We can write this in equation form as well:
3b = 2c
To find the values of 'b' and 'c', we can solve these two equations simultaneously.
Multiply the second equation by 4 to make the coefficients of 'b' in both equations the same:
12b = 8c
Now we have two equations:
4b + 4c = 12 ...(i)
12b = 8c ...(ii)
We can solve these two equations simultaneously.
Divide equation (ii) by 8 to get:
b = c/3
Substitute this value for 'b' in equation (i):
4(c/3) + 4c = 12
Simplify the equation:
4c/3 + 4c = 12
Multiply the whole equation by 3 to get rid of the fraction:
4c + 12c = 36
Combine like terms:
16c = 36
Divide both sides of the equation by 16 to get the value of 'c':
c = 36/16
Simplify the fraction:
c = 9/4
Therefore, the cost of one cake is $9/4 or $2.25.
Now substitute the value of 'c' in the second equation:
3b = 2(9/4)
3b = 18/4
Simplify the fraction:
3b = 9/2
Divide both sides of the equation by 3 to get the value of 'b':
b = 9/6
Simplify the fraction:
b = 3/2
Therefore, the cost of one bun is $3/2 or $1.50.
Now, let's find the total cost of 6 buns and 9 cakes:
Cost of 6 buns = 6 x $1.50 = $9
Cost of 9 cakes = 9 x $2.25 = $20.25
Therefore, the total cost of 6 buns and 9 cakes is $9 + $20.25 = $29.25.