Question
Ryan bought coco powder, sugar, and wheat flour. The cost of sugar is $3 lesser than 2 times the cost of coco powder, and the white flour is $2 more than 1/2 of the cost of coco powder. The total cost is $22.56. Find the cost of each item (Estimate to the nearest tenths place).
Cost of the coco powder is $6.7, cost of the sugar is $10.4, and cost of the wheat flour is $5.4
Cost of the coco powder is $4.7, cost of the sugar is $11.4, and cost of the wheat flour is $6.4
Cost of the coco powder is $8.7, cost of the sugar is $9.4, and cost of the wheat flour is $4.4
Cost of the coco powder is $5.7, cost of the sugar is $9.4, and cost of the wheat flour is $7.4
Cost of the coco powder is $6.7, cost of the sugar is $10.4, and cost of the wheat flour is $5.4
Cost of the coco powder is $4.7, cost of the sugar is $11.4, and cost of the wheat flour is $6.4
Cost of the coco powder is $8.7, cost of the sugar is $9.4, and cost of the wheat flour is $4.4
Cost of the coco powder is $5.7, cost of the sugar is $9.4, and cost of the wheat flour is $7.4
Answers
GPT-4o mini
Let's denote the cost of cocoa powder as \( C \), the cost of sugar as \( S \), and the cost of wheat flour as \( W \).
From the problem statement, we have the following equations based on the relationships given:
1. \( S = 2C - 3 \) (the cost of sugar is $3 less than 2 times the cost of cocoa powder)
2. \( W = \frac{1}{2}C + 2 \) (the cost of wheat flour is $2 more than half of the cost of cocoa powder)
3. \( C + S + W = 22.56 \) (the total cost is $22.56)
Now we can substitute equations 1 and 2 into equation 3:
\[
C + (2C - 3) + \left(\frac{1}{2}C + 2\right) = 22.56
\]
Combining like terms:
\[
C + 2C - 3 + \frac{1}{2}C + 2 = 22.56
\]
\[
(3C + \frac{1}{2}C) - 1 = 22.56
\]
To combine \( 3C \) and \( \frac{1}{2}C \), we can convert \( 3C \) into a fraction:
\[
\frac{6C}{2} + \frac{1C}{2} - 1 = 22.56
\]
This simplifies to:
\[
\frac{7C}{2} - 1 = 22.56
\]
Adding 1 to both sides:
\[
\frac{7C}{2} = 23.56
\]
Multiplying both sides by 2:
\[
7C = 47.12
\]
Now, divide by 7:
\[
C = \frac{47.12}{7} \approx 6.73
\]
Now we can find \( S \) and \( W \):
For \( S \):
\[
S = 2C - 3 = 2(6.73) - 3 = 13.46 - 3 \approx 10.46
\]
For \( W \):
\[
W = \frac{1}{2}C + 2 = \frac{1}{2}(6.73) + 2 = 3.365 + 2 \approx 5.37
\]
Calculating the costs rounded to the nearest tenths, we find:
- \( C \approx 6.7 \)
- \( S \approx 10.4 \)
- \( W \approx 5.4 \)
Thus, we can see that the answer that matches closely with the calculations is:
**Cost of the cocoa powder is $6.7, cost of the sugar is $10.4, and cost of the wheat flour is $5.4.**
From the problem statement, we have the following equations based on the relationships given:
1. \( S = 2C - 3 \) (the cost of sugar is $3 less than 2 times the cost of cocoa powder)
2. \( W = \frac{1}{2}C + 2 \) (the cost of wheat flour is $2 more than half of the cost of cocoa powder)
3. \( C + S + W = 22.56 \) (the total cost is $22.56)
Now we can substitute equations 1 and 2 into equation 3:
\[
C + (2C - 3) + \left(\frac{1}{2}C + 2\right) = 22.56
\]
Combining like terms:
\[
C + 2C - 3 + \frac{1}{2}C + 2 = 22.56
\]
\[
(3C + \frac{1}{2}C) - 1 = 22.56
\]
To combine \( 3C \) and \( \frac{1}{2}C \), we can convert \( 3C \) into a fraction:
\[
\frac{6C}{2} + \frac{1C}{2} - 1 = 22.56
\]
This simplifies to:
\[
\frac{7C}{2} - 1 = 22.56
\]
Adding 1 to both sides:
\[
\frac{7C}{2} = 23.56
\]
Multiplying both sides by 2:
\[
7C = 47.12
\]
Now, divide by 7:
\[
C = \frac{47.12}{7} \approx 6.73
\]
Now we can find \( S \) and \( W \):
For \( S \):
\[
S = 2C - 3 = 2(6.73) - 3 = 13.46 - 3 \approx 10.46
\]
For \( W \):
\[
W = \frac{1}{2}C + 2 = \frac{1}{2}(6.73) + 2 = 3.365 + 2 \approx 5.37
\]
Calculating the costs rounded to the nearest tenths, we find:
- \( C \approx 6.7 \)
- \( S \approx 10.4 \)
- \( W \approx 5.4 \)
Thus, we can see that the answer that matches closely with the calculations is:
**Cost of the cocoa powder is $6.7, cost of the sugar is $10.4, and cost of the wheat flour is $5.4.**