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Math Word Problems About Bees

Updated on February 22, 2017

Humans have practiced apiculture -- beekeeping -- for more than 5000 years, and some of the world's oldest artwork depicts scenes of humans attending hives or collecting honey. Much of bee life is organized with a mathematical precision not commonly seen in other social organisms. And as with all enterprises and sciences, there is plenty of math involved in the production of honey and the study of bee colonies. With this in mind, here are seven math word problems inspired by bees.

(1) New Bees -- Easy

The new queen bee of a certain hive lays lots of eggs every day, but only a small fraction of them develop into healthy worker bees and drones. Yet, as time goes on, a larger and larger proportion of her eggs become bees. Suppose during her first month on the job she produces 1000 new bees, and every month after that she produces 250 more bees than the previous month. How many new bees will she have produced after 3 years on the job?

Solution: Since 3 years is 36 months, on the 36th month she will produce 1000 + 35*250 = 9750 new bees. The total number of new bees produced over the 36 months is the sum 1000 + 1250 + 1500 + 1750 + ... + 9750. To simplify this calculation, we can rewrite it as

(1000 + 1000 + 1000 + ... + 1000) + (0 + 250 + 500 + 750 + ... + 8750)
= 36*1000 + (250 + 500 + 750 + ... + 8750)
= 36000 + 250*(1 + 2 + 3 + ... + 35)

The sum 1 + 2 + 3 + ... + 35 is the 35th triangular number. The nth triangular number T(n) can be computed with the formula T(n) = n(n+1)/2. Plugging in n = 35 gives us T(35) = 630. Going back to the original computation we get

36000 + 250*630 = 193500.

Therefore, she produces new 193500 bees during her first 3 years.

(2) The Dying Queen -- Easy

The queen of a hive is on her last legs. For the past several months she has not been able to lay as many eggs, in fact, her output has been decreasing at a non-constant rate. Today she produced 8 fewer eggs than yesterday, tomorrow she will produce 9 fewer eggs than today, the day after she will produce 10 fewer eggs than tomorrow, etc. If today she produced 100 eggs, how long will it be until she produces no eggs? And how many eggs will she have laid between today and her last day?

Solution: To solve this problem we start with 100 and successively subtract larger amounts. This gives us the sequence

  • 100
  • 100 - 9 = 91
  • 91 - 10 = 81
  • 81 - 11 = 70
  • 70 - 12 = 58
  • 58 - 13 = 45
  • 45 - 14 = 31
  • 31 - 15 = 16
  • 16 - 16 = 0

The dying queen has seven more days of egg laying ahead. In this time she will lay a total of 100 + 91 + 81+ 70 + 58 + 45 + 31 + 16 = 492 eggs.

(3) Honeycomb Cell Game -- Medium

Cindy and Lindy are naughty worker bees who would rather play games than collect nectar and pollen. They like to play a guessing game with 19 hexagonal honeycomb cells called "Find the Rock." One of them hides a tiny rock in one of the hexagonal cells, and the other tries to guess where it is with hints based on the previous guess.

The hint the seeker receives is always of the same form: The hider does a clockwise dance if the rock is in a cell adjacent to the one guessed, a counterclockwise dance if the rock is not adjacent to the guessed cell. The hider does a side-to-side dance when the seeker guesses correctly. Cindy's and Lindy's game board is shown below.

Cindy's and Lindy's honeycomb game board with 19 cells.
Cindy's and Lindy's honeycomb game board with 19 cells.

Suppose Lindy is the hider, Cindy is the seeker, and Cindy's first guess is the center cell. If Lindy does a clockwise dance on Cindy's first guess, what is the maximum number of total guesses Cindy has to make to find the rock, assuming Cindy guesses with optimal strategy after the first guess? (The total is counting the very first guess and the following guesses, including the last guess when she finds the rock.)

Solution: In this problem we need to analyze the worst case scenarios to determine the maximum number of guesses when Cindy's first guess at the center cell elicits a clockwise dance. Cindy makes the best choices available to her and "worst case scenarios" are when Cindy's failure guess the location of the rock is due to bad luck, not to poor strategy.

Lindy's clockwise dancing after the first guess indicates that the rock is hidden in one of the six cells surrounding the first guess. Cindy's second guess will naturally be one of these six cells. The six cells are equivalent in terms of their relative positions, so there is no further strategy for Cindy to apply other than to pick one at random. There are three cases for what happens next.

Case I: Lindy does a side-to-side dance indicating Cindy found the rock. This is not the worst case scenario, however, so we need to look at Cases II and III.

Case II: Lindy does a clockwise dance, indicating the rock is adjacent to Cindy's second guess. This leaves only two possibilities for where the rock is. These two cells are equivalent in terms of their relative positions to the previous guesses and there is no strategy to apply except to guess one of them. The worst case scenario in Case II is that it takes her two more guesses after the second, for a total of four guesses.

Case III: Lindy does a counter clockwise dance, indicating the rock is not adjacent to her second guess. This leaves only three possibilities for where the rock is. These three cells are connected in a chain as in the figure below.

In Case III, Cindy's best strategy is for her third guess to be a cell at one of the ends of this chain. This results in one of three sub cases:

  • Case IIIa: Cindy's third guess is right.
  • Case IIIb: Cindy's third guess is wrong and Lindy does a clockwise dance. This means the rock is in the middle of the three cells and Cindy's fourth guess will be the last.
  • Case IIIc: Cindy's third guess is wrong and Lindy does a counterclockwise dance. This means the rock is in the cell at the other end of the chain, in which case Cindy's fourth guess will still be her last.

All cases and sub-cases considered, if Cindy plays with optimal strategy at every turn, she will need to make at most four guesses to find the rock.

(4) Nectar and Pollen Proportions -- Medium

Beetrice is the hive's accountant bee and she likes to do annual summaries of all the nectar and pollen collected (by weight). This year, the colony collected 50% more pollen than nectar. Last year, the colony collected 40% more pollen than nectar. Also this year, the colony's nectar collection exceeded last year's nectar collection by 25%. By how much did this year's pollen collection exceed last year's pollen collection?

Solution: Let's say that last year the colony collected X amount of nectar. Then from the information in the problem last year's pollen collection was 1.4X. This is because a 40% increase is computed by multiplying by 1.4.

This year's nectar collection was 1.25X, 25% more than last year's. This year's pollen collection was 1.5(1.25X) = 1.875X.

To compute the increase in pollen from last year to this year, we compute the ratio 1.875X to 1.4X. This works out to 1.875/1.4 ≈ 1.3393. Therefore, this year's pollen collection exceeded last year's by about 33.93%.

(5) Finding a Shorter Path -- Hard

Buzz has to collect pollen from eight tulips before she can stop working for the day. In each of the four cardinal directions there is one tulip 10 meters from the hive. Also in each of the four cardinal directions there is a tulip 20 meters away from the hive. Starting and ending at the hive, can Buzz visit all eight of these tulips with a path that is less than 134 meters long? Assume the hive is level with the tops of the tulips so that there is no vertical distance to figure in.

Solution: There is more than one path that hits all eight tulips and measures less than 134 meters. Here is one possibility that works out to 60 + 50*sqrt(2) ≈ 130.7 meters.

In the diagram above, H is the hive and the Vs are the tulips. Each square represents 10 meters. You can slightly alter this path to create another with a total length of 60 + 20*sqrt(5) + 20*sqrt(2) meters, which is approximately 133 meters.

(6) Five Hives Riddle -- Hard

Finn is a beekeeper with five small hives he calls, conveniently, A, B, C, D, and E. He can't remember how many are in each hive so he asks the queen of each hive. Instead of giving Finn a straight answer, they collude to make him work out the answer with a math puzzle.

  • The queen of Hive A tells Finn there are 244 bees total in Hives A, B, and C.
  • The queen of Hive B tells Finn ther are 159 bees total in Hives B and E
  • The queen of Hive C tells Finn there are 297 bees total in Hives C, D, and E
  • The queen of Hive D tells Finn there are 226 bees total in Hives A and D.
  • The queen of Hive E tells Finn the product of the populations of Hives A and E is 10290.

How many bees are in each hive?

Solution: With a little algebra, we can get the population of Hive C from the first four clues. Notice that

(A+B+C) - (B+E) + (C+D+E) - (A+D) = 2C

which works out to

244 - 159 + 297 - 226 = 2C
156 = 2C
78 = C

Now that Hive C is worked out, the next step is to find the sizes of Hives A and E. The first four equations can be manipulated in a different way to give us

(A+B+C) - (B+E) - (C+D+E) + (A+D) = 2A - 2E

which works out to

244 - 159 - 297 + 226 = 2A - 2E
14 = 2A - 2E
7 = A - E

Now we know that A - E equals 7. We also know that A times E equals 10290. This means we need to find a factor pair for 10290 with a difference of 7. This can be done with trial and error or with the quadratic formula. Using the latter method we get

A*E = 10290
A*(A - 7) = 10290
A^2 - 7A - 10290 = 0
A = [7 ± sqrt(49 + 41160)]/2
A = 105 or -98

Since the population can't be negative, we get A = 105. This also gives us E = 98. Finally, working backward from the second and fourth clues we get B = 61 and D = 121.

(7) Honeypots -- Hard

Liesl's large honeypots store 2.5 times as much honey as her medium pots, which hold 4 quarts more than her small honeypots. Two large pots and one small pot together can hold as much honey as N medium pots and N small pots, where N is a positive whole number. What is the value of N and how many quarts can Liesl's various honeypots hold?

Solution: Let the small pot's capacity be X. The first statement tells us the medium pot's capacity is X + 4, and the large pot's is 2.5X + 10. The second statement tells us

2(2.5X + 10) + X = N(X + 4) + NX

This simplifies to

6X + 20 = 2NX + 4N
3X + 10 = NX + 2N
NX + 2N - 3X - 10 = 0
(NX + 2N - 3X - 6) - 4 = 0
(N - 3)(X + 2) - 4 = 0
(N - 3)(X + 2) = 4

Since we are given that N is an integer, we can start plugging in N = 4, N = 5, N = 6, etc. into the equation (N - 3)(X + 2) = 4 to see what X values they return.

  • N = 4 gives us X = 2
  • N = 5 gives us X = 0
  • N = 6 gives us X = -2/3

Going higher than N = 6 gives more negative values of X. The only value of N that returns a positive value of X is N = 4. Therefore, the capacities of the pots are 2 quarts, 6 quarts, and 15 quarts.

Bee photographs courtesy of Pixabay public domain stock.


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