1. Andres bought 60 kilograms of fruits consisting of pomelo and seedless red grapes. He bought P4800 worth of pomelo and P8400 worth of seedless red grapes. If the cost per kilogram of seedless red grapes is P30 less than that of pomelo, how many kilograms of each fruit did he buy?

Let's define the variables to set up the problem:

- Let \( x \) be the cost per kilogram of pomelo in pesos.
- Let \( y \) be the cost per kilogram of seedless red grapes. 
- Let \( a \) be the kilograms of pomelo bought.
- Let \( b \) be the kilograms of seedless red grapes bought.

Given conditions:

1. Andres bought a total of 60 kilograms of fruit:
   \[
   a + b = 60
   \]

2. The total cost of pomelo is \( 4800 \) pesos:
   \[
   a \cdot x = 4800
   \]
   Thus,
   \[
   x = \frac{4800}{a}
   \]

3. The total cost of seedless red grapes is \( 8400 \) pesos:
   \[
   b \cdot y = 8400
   \]
   Thus,
   \[
   y = \frac{8400}{b}
   \]

4. The cost per kilogram of seedless red grapes is \( 30 \) pesos less than that of pomelo:
   \[
   y = x - 30
   \]

### Step-by-Step Solution

From equation 2:
\[
x = \frac{4800}{a}
\]

From equation 3:
\[
y = \frac{8400}{b}
\]

Substitute the cost relationship from equation 4 into the expression for \( y \):
\[
\frac{8400}{b} = \frac{4800}{a} - 30
\]

Now, we have two equations:
1. \( a + b = 60 \)
2. \( \frac{8400}{b} = \frac{4800}{a} - 30 \)

### Solving the Equations

From the first equation:
\[
b = 60 - a
\]

Substitute \( b = 60 - a \) into the second equation:
\[
\frac{8400}{60 - a} = \frac{4800}{a} - 30
\]

Multiply all terms by \( a(60 - a) \) to clear the denominators:
\[
8400a = 4800(60 - a) - 30a(60 - a)
\]

Expand each term:
\[
8400a = 288000 - 4800a - 1800a + 30a^2
\]

Combine like terms:
\[
0 = 30a^2 - 15000a + 288000
\]

### Simplifying the Quadratic Equation

Divide the entire equation by 30:
\[
a^2 - 500a + 9600 = 0
\]

Now, solve this quadratic equation using the quadratic formula:
\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{500 \pm \sqrt{500^2 - 4 \times 9600}}{2}
\]

Calculate the discriminant:
\[
500^2 - 4 \times 9600 = 250000 - 38400 = 211600
\]

Find the square root of the discriminant:
\[
\sqrt{211600} = 460
\]

Substitute back into the formula:
\[
a = \frac{500 \pm 460}{2}
\]

This gives us two possible solutions:
1. \( a = \frac{500 + 460}{2} = 480 \div 2 = 40 \)
2. \( a = \frac{500 - 460}{2} = 40 \div 2 = 20 \)

Since we want a positive amount, let's use \( a = 40 \).

### Finding \( b \)

\[
b = 60 - a = 60 - 40 = 20
\]

### Final Answer

Andres bought:
- **40 kilograms of pomelo**
- **20 kilograms of seedless red grapes**

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. What is the cost per kilogram of the pomelo?
2. What is the cost per kilogram of the seedless red grapes?
3. How much would Andres pay if he bought double the amount of each fruit?
4. If the cost per kilogram of seedless red grapes increased by 10%, how would the total cost change?
5. What would be the new proportions if he bought an equal cost of each fruit instead?

**Tip:** When dealing with word problems, defining variables and setting up equations based on given conditions is a crucial first step.

Andres bought 60 kilograms of fruits consisting of pomelo and seedless red grapes. He bought P4800 worth of pomelo and P8400 worth of seedless red grapes. If the cost per kilogram of seedless red grapes is P30 less than that of pomelo, how many kilograms of each fruit did he buy?

Learn how to solve a real-world algebra problem where Andres buys 60 kilograms of pomelo and seedless red grapes. This detailed step-by-step solution explains how to use linear and quadratic equations to find out how many kilograms of each fruit were bought. Suitable for Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation