86 Solving Word Problems with Multiple Variables
Mathematics
Algebra
Word Problems
System of Equations
Problem Solving
Synthetic
86.1 Description:
This task involves interpreting word problems, identifying relevant variables, and constructing equations to solve complex scenarios. It evaluates the model’s ability to translate real-world situations into mathematical expressions and solve systems of equations.
86.2 Modality:
Text only
86.3 Examples:
86.3.1 Example 1:
Input:
A farm has chickens and cows. If there are 50 heads and 140 legs in total, how many chickens and cows are there?Output:
Let's solve this step by step:
1. Define variables:
x = number of chickens
y = number of cows
2. Set up equations:
Equation 1 (heads): x + y = 50
Equation 2 (legs): 2x + 4y = 140
3. Solve the system of equations:
From Equation 1: y = 50 - x
Substitute into Equation 2: 2x + 4(50 - x) = 140
Simplify: 2x + 200 - 4x = 140
-2x = -60
x = 30
4. Find y:
y = 50 - 30 = 20
Therefore, there are 30 chickens and 20 cows.86.3.2 Example 2:
Input:
A rectangle's length is 5 cm more than twice its width. If the perimeter of the rectangle is 70 cm, what are its dimensions?Output:
Let's solve this step-by-step:
1. Define variables:
w = width of the rectangle
l = length of the rectangle
2. Express length in terms of width:
l = 2w + 5
3. Set up equation using perimeter formula:
2l + 2w = 70
2(2w + 5) + 2w = 70
4. Solve for w:
4w + 10 + 2w = 70
6w = 60
w = 10 cm
5. Calculate length:
l = 2(10) + 5 = 25 cm
Therefore, the rectangle's dimensions are 10 cm wide and 25 cm long.