An egg seller had some chicken eggs and duck eggs. He 156 more chicken eggs than duck eggs. After he sold 3/4 of chicken eggs and 2/3 of duck eggs, he found that he had 284 eggs left. Find the total number of eggs he had first.

Let's denote the number of duck eggs as \( x \) and the number of chicken eggs as \( y \).

According to the problem, the relationships are:

1. \( y = x + 156 \) (since there are 156 more chicken eggs than duck eggs)
2. After selling, the remaining eggs total 284. He sold \(\frac{3}{4}\) of the chicken eggs, so \(\frac{1}{4}\) of the chicken eggs remain, and he sold \(\frac{2}{3}\) of the duck eggs, so \(\frac{1}{3}\) of the duck eggs remain. Therefore:

\[
\frac{1}{4}y + \frac{1}{3}x = 284
\]

Now, let's substitute the first equation \( y = x + 156 \) into the second equation:

\[
\frac{1}{4}(x + 156) + \frac{1}{3}x = 284
\]

To eliminate the fractions, let's multiply the entire equation by 12 (the least common multiple of 4 and 3):

\[
12 \times \frac{1}{4}(x + 156) + 12 \times \frac{1}{3}x = 12 \times 284
\]

This simplifies to:

\[
3(x + 156) + 4x = 3408
\]

Expand and combine like terms:

\[
3x + 468 + 4x = 3408
\]

\[
7x + 468 = 3408
\]

Subtract 468 from both sides:

\[
7x = 2940
\]

Now, solve for \( x \):

\[
x = \frac{2940}{7} = 420
\]

So, there are 420 duck eggs. Since \( y = x + 156 \), the number of chicken eggs is:

\[
y = 420 + 156 = 576
\]

The total number of eggs he had initially is:

\[
x + y = 420 + 576 = 996
\]

Therefore, the egg seller originally had **996** eggs.

Would you like more details, or have any other questions?

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Here are 5 related questions for further exploration:

1. What if the egg seller had sold half of the chicken eggs instead of 3/4?
2. How would the problem change if the seller had 200 more chicken eggs than duck eggs?
3. What would happen if the remaining eggs after the sale were 300 instead of 284?
4. Can you solve a similar problem with different fractions, such as selling \( \frac{2}{5} \) of chicken eggs and \( \frac{

Explore how an egg seller manages his stock of chicken and duck eggs through detailed problem-solving. Learn about equations and fractional operations in this math problem. Suitable for students in Grades 7-9.

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