The manager of a snack bar buys bottled water in packs of 35 and candy bars in packs of 20. Then, she sells the items individually. Yesterday, she bought the same number of water bottles as candy bars to sell at the snack bar. If she bought the lowest number of items possible, how many packs of each item did she buy?

5 packs of bottled water and 5 packs of candy bars
4 packs of bottled water and 7 packs of candy bars
8 packs of bottled water and 14 packs of candy bars
20 packs of bottled water and 35 packs of candy bars

1 answer

Let's denote the number of packs of bottled water the manager buys as \( x \) and the number of packs of candy bars as \( y \). The number of bottled water she buys is \( 35x \) and the number of candy bars is \( 20y \).

According to the problem, the manager buys the same number of items (bottled water and candy bars), which means:

\[ 35x = 20y \]

To find the least number of items possible, we simplify the equation. Dividing both sides by 5 gives:

\[ 7x = 4y \]

From this equation, we can express \( y \) in terms of \( x \):

\[ y = \frac{7}{4}x \]

Since \( y \) must be an integer, \( x \) needs to be a multiple of 4. Let's denote \( x = 4k \) where \( k \) is a positive integer. Substituting into the equation gives:

\[ y = \frac{7}{4}(4k) = 7k \]

Now we can determine the total number of packs of each item:

  • Packs of bottled water: \( x = 4k \)
  • Packs of candy bars: \( y = 7k \)

Next, we want to find the lowest number of items possible.

Setting \( k = 1 \) gives:

  • Packs of bottled water: \( x = 4 \times 1 = 4 \)
  • Packs of candy bars: \( y = 7 \times 1 = 7 \)

This means she would buy 4 packs of bottled water and 7 packs of candy bars, which corresponds to the selling amounts:

  • Total water bottles: \( 35 \times 4 = 140 \)
  • Total candy bars: \( 20 \times 7 = 140 \)

Both quantities equate to 140, confirming the number of items bought is indeed the same.

Hence, the number of packs she bought is:

4 packs of bottled water and 7 packs of candy bars.