A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
588 apples
600 apples
672 apples
700 apples

Let the total number of apples the seller originally had be \( x \).

He sold 40% of the apples, so he has 60% of the apples left. This means:

\[
60\% \text{ of } x = 420
\]

We can write this as:

\[
\frac{60}{100} \times x = 420
\]

Simplifying the equation:

\[
0.6x = 420
\]

Now, solve for \( x \):

\[
x = \frac{420}{0.6} = 700
\]

Thus, the fruit seller originally had **700 apples**.

Would you like more details or have any other questions?

Here are 5 related questions for further practice:
1. If the seller had sold 50% of the apples, how many would he have left?
2. If he had 900 apples and sold 30%, how many apples would be left?
3. What fraction of the apples does the seller sell if he sells 40%?
4. If the seller had 700 apples and sold 25%, how many apples does he have left?
5. How many apples would the seller have originally if 60% equals 300 apples?

**Tip:** When working with percentages, converting them to fractions or decimals can simplify calculations.

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had: 588 apples, 600 apples, 672 apples, 700 apples.

Learn how to solve a real-world problem involving percentages and basic algebra. A fruit seller sold 40% of his apples and still had 420 left. This detailed explanation uses the percentage formula and basic algebra to find that the seller originally had 700 apples. Suitable for Grades 6-8.

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