Solving Sticker Problems Through Algebra

Algebra is a powerful tool for solving various types of mathematical problems, from simple arithmetic to complex word problems. One classic type of problem involves the distribution of items, such as stickers, between two individuals. This article explores several examples of such problems and demonstrates how algebra can be used to find the solutions. We will walk through step-by-step solutions for each problem, making this content accessible and easy to follow.

Example 1

Sally and Nicholas had a total of 1405 stickers. If Sally had 4 times as many as Nicholas, how many stickers did Sally have?

Solution

Let's denote the number of stickers Nicholas has as ( x ). Therefore, Sally has ( 4x ) stickers.

The total number of stickers is given as 1405:

[ x 4x 1405 ]

This simplifies to:

[ 5x 1405 ]

Solving for ( x ):

[ x frac{1405}{5} 281 ]

Nicholas has 281 stickers. Since Sally has 4 times as many stickers as Nicholas:

[ mathrm{Sally's stickers} 4 times 281 1124 ]

Example 2

Sally has a certain number of stickers, and Nicholas has a fraction of Sally's stickers. The total number of stickers is 1405. If Sally has 4 times as many as Nicholas, how many stickers does Sally have?

Solution

Let's denote Nicholas's stickers as ( x ). Therefore, Sally has ( 4x ) stickers. The total number of stickers is:

[ x 4x 1405 ]

[ 5x 1405 ]

Solving for ( x ):

[ x frac{1405}{5} 281 ]

Nicholas has 281 stickers. Consequently, Sally has:

[ mathrm{Sally's stickers} 4 times 281 1124 ]

Example 3

Given a ratio of 4:1 between Sally and Nicholas, if Sally has 80% of the total stickers, and Nicholas has 20% of the total stickers, how many stickers does Sally have?

Solution

The total ratio is 4 1 5. Therefore, Sally's share is 4 out of 5 parts of the total 1405 stickers:

[ mathrm{Sally's stickers} frac{4}{5} times 1405 ]

[ mathrm{Sally's stickers} frac{4 times 1405}{5} frac{5620}{5} 1124 ]

Nicholas's share is:

[ mathrm{Nicholas's stickers} frac{1}{5} times 1405 frac{1405}{5} 281 ]

Example 4

Let Annie's original number of stickers be ( X ) and Sarah's original number of stickers be ( Y ). Initially, Annie and Sarah had the same number of stickers, so ( X Y ). After Annie bought 38 more stickers and Sarah lost 25 stickers, Annie had 4 times as many stickers as Sarah. We can set up the equation based on this information:

[ X 38 4(Y - 25) ]

Since ( X Y ), we substitute ( Y ) for ( X )

[ X 38 4(X - 25) ]

[ X 38 4X - 100 ]

Rearranging terms:

[ 3X 138 ]

Solving for ( X ):

[ X frac{138}{3} 46 ]

Initially, Annie and Sarah each had 46 stickers.

Conclusion

Algebra provides a systematic and logical approach to solving sticker distribution problems. By setting up equations based on given conditions, we can determine the quantities involved and solve for unknowns. This method is not only applicable to sticker problems but can be used in a wide range of real-world scenarios. Practicing these types of problems enhances analytical skills and mathematical proficiency.