Page 1 of 2 WASHINGTON STATE MATHEMATICS COUNCIL 2018 MIDDLE SCHOOL MATH OLYMPIAD Session I: PROBLEM SOLVING Design, Operation and Maintenance of a Swimming Pool The City is planning to build a swimming pool for its rapidly growing community. The swimming pool is to be 25 m long and 12.5 m wide. The City decided to build a pool whose depth changes from one end to the other as shown in the figure below. The shallow end of the pool is to have water to a depth of 1.5 m; at least a 10 m length of the pool at this end needs to have water to depth of 1.5 m. The diving or deep end of the pool is to have water to a depth of 3.6 m; at least a 6 m length of the pool at the diving end needs to have water to a depth of 3.6 m. The depth of water is to increase gradually from 1.5 m to 3.6 m in central part of the pool. For the safety and comfort of the swimmers, the depth of the pool should not change by more than 1 m over a horizontal length of 3 m as shown in the figure. The water surface is 15 cm (or 0.15m) below the deck surrounding the pool. The City wanted a pool with the least volume that satisfies these requirements, because the cost of constructing and operating the pool increase with the volume of water in the pool. Your team needs to decide on some dimensions on the side view. You will also need to calculate the volume of the water in the pool.
For safety reasons, diving from the deck is allowed only at water depths of 2 m or more. There must be pool markings on the side walls and on the floor of the pool to indicate where the water is 2 m deep. How far from the shallow end does the pool have water to a depth of 2 m? For safety reasons, there is a limit on the number of people who can be in the pool at one time. Safety regulations specify that each person needs 2.5 m 2 of water surface area if they are in the part of the pool that has water to a depth of more than 1.5 m; each person needs 0.93 m 2 of water surface area if they are in the part of the pool that has water to a depth of 1.5 m or less. What is the maximum number of people who can be in the pool at one time? Lap Cumulative time (minutes:seconds to the hundredth) 1 0:35.00 2 1:10.35 3 1:46.05 4 2:22.11 5 2:58.53 6 3:35.31 Jo, a 13 year old, has been training to be a competitive swimmer. Jo s coach noted the time it takes for Jo to swim each lap of the 1500 m swim. A lap is defined as swimming from one end of the pool to the other end, lengthwise, and back for a total distance of 50 m in this case. The lap times for the first six laps, to the nearest hundredths of a second are given in table to the right. Use this information to find estimates of the lap time for the 30 th lap. Kim is responsible for creating the schedule for the pool for Saturday. Activities are to be scheduled in one hour slots, beginning on the hour, from 8:00 am until 8:00 pm. Kim needs to schedule 4 hours of lap swim, 2 hours of family swim/play in the shallow end and diving at the deep end (both scheduled at the same time), 3 hours of school swim teams, 3 hours for public swim. How many different schedules can Kim create? And now we are in the year 2020; the pool has been constructed and has been in use for more than a year. Measured values of chlorine concentration in the pool during the first year of operation are in the table to the right. Chlorine is a common disinfectant used in pools for the health and safety of the swimmers. According to regulations the water in the pool must have a minimum chlorine concentration of 3 mg/l. For every day that the chlorine concentration falls below Minimum daily chlorine concentration (mg/l) Number of days during which this minimum concentration occurred the 3 mg/l guideline, the city will have to pay a fine of $100. The 3.5 3 3.6 22 current chlorination plan costs $20.00 per day and should have 3. 8 resulted in a chlorine concentration of 5.0 mg/l. The differences 3.8 141 between the 5.0 mg/l and the measured values are due to the loss of 3.9 88 chlorine to reaction with the ammonia secreted and excreted by the 4.0 22 swimmers and to volatilization into the air. Use the data to recommend whether City should continue with the current 4.1 2 chlorination plan or whether it should decrease the amount of chlorine it adds to the water by 10%.
Page 1 of 6 Write the Responses Here. Show all of your work. Use the back of the pages if needed. Your work will be evaluated on: Your understanding of the problem The strategies that you used and your reasoning Your communication (both verbal and quantitative) of how you arrived at the solutions Your solutions and your checks for reasonableness of the solutions where appropriate. 1a: Determine the dimensions of the pool that satisfies all the requirements of the City. (4@1 point for dimensions, 5@1 point for reasoning and/or calculations.) The length of the deep end is short as possible to minimize the volume; 6 m. The sloping floor is as steep as possible to minimize the volume; 1:3. The change in depth = 3.6 1.5 = 2.1 m The horizontal length of the sloping floor = 2.1 (1/3) = 6.3 m. The length of the shallow end = 25 6 6.3 = 12. m. 1b: Determine the area of the side wall of the pool shown in 1a, that is below the water level. (1 point for area, 4@1 for reasoning and/or calculations.) Rectangle at the shallow end:- 12. x 1.5 = 19.05 m 2. Rectangle at the deep end:- 6 x 3.6 = 21.6 m 2. Central trapezoid:- 6.3 x (1.5 + 3.6)/2 = 16.065 m 2. Total area = 56.15 m 2.
Page 2 of 6 1c: Determine the volume of water in the pool. You may round the volume of the pool to three significant digits. This means that you may express the answer in a form where only the three highest place digits can be non-zero, all lower place digits are zero. (1 point for volume, 1 for reasoning and/or calculations.) Volume of water = area of side x width = 56.15 x 12.5 = 08.935 m 3 = 09 m 3. Or (if they do it this way, they can recover any corresponding reasoning credit that they may have lost in 1b.) Volume of the shallow end:- 12. x 12.5 x 1.5 = 238.125 m 3. Volume of the deep end:- 6 x 12.5 x 3.6 = 20 m 3. Volume of the central trapezoidal prism:- 6.3 x 12.5 x (1.5 + 3.6)/2 = 200.8125 m 3. Volume of water = 08.935 m 3 = 09 m 3. 2: Determine the distance from the shallow end to the line where the depth of water is 2 m. (1 point for distance, 3 for reasoning and/or calculations.)
Page 3 of 6 3: Determine the maximum number of bathers/swimmers who can be in the pool at the same time. Why are there different criteria for the parts of the pool depending on whether the depth of water is > 1.5 m or 1.5 m (1 point limit, 4 for reasoning and/or calculations. 2 points for plausible reasons for criteria) Water surface area where water depth 1.5 m:- 12. x 12.5 = 158.5 m 2. Limit to bathers who can stand in this part of the pool = 158.5 0.93 = 10.6 = 10 Rounding down to stay within limit. Water surface area where water depth > 1.5 m:- (6.3+6) x 12.5 = 153.5 m 2. Limit to bathers who can swim in this part of the pool = 153.5 2.5 = 61.5 = 61 Rounding down to stay within limit. The maximum number of bather allowed = 10 + 61 = 231. Accept 232 if they argue that the combined fractional spaces are sufficient for 1 more person. (1 point loss total for answer and reasoning if they round up prior to adding and give 233 as the limit.) A person will likely be swimming in the deeper depths and will be horizontal, need space lengthwise from toes to the extended finger tips and need space across for arms and feet to spread out depending on the stroke. Also need some separation between swimmers for safety. A person could stand in the shallower depths and will require less space.
Page 4 of 6 4a: Fill out the table below and use that information to calculate two estimates for Jo s lap time for the 30 th lap. (1 point each for the estimate, 2 points for each of the three columns of numbers, 1 point for each average, 2 points each for showing how to calculate the estimate for the 30 th lap time) Lap Cumulative time (minutes:seconds to the hundredths) mm:ss.ss Lap time (Seconds to the hundredths) ss.ss 1 0:35.00 35.00 Difference between current lap time and previous lap time (Seconds) Ratio between current lap time ad previous lap time To the thousandths 2 1:10.35 35.35 0.35 1.010 3 1:46.05 35.0 0.35 1.010 4 2:22.11 36.06 0.36 1.010 5 2:58.53 36.42 0.36 1.010 6 3:35.31 36.8 0.36 1.010 Average 0.356 1.010 Use the two averages computed above to find two estimates of the 30 th lap time Estimate 1 Estimate 2 30 Estimated lap time for the 30 th lap, to the hundredth of a second 45.32 sec 46.1 sec Assuming that the lap times increase linearly or are in an arithmetic sequence., the 30 th lap time = 35 + 0.356 x 29 = 45.62 seconds Assuming that the lap times increase at a constant ratio or are in a geometric sequence., the 30 th lap time = 35 x 1.01 29 = 46.1 seconds 4b: What should Jo s coach have done to allow you to decide which of these two methods was a better fit to Jo s lap times. (1 point) If Jo s coach had timed a few more laps, say 4 more laps, that would have been sufficient to decide whether either sequence as a good fit to the lap times.
Page 5 of 6 5a: Determine the number of different schedules Kim can create if multiple hours of the same activity must be scheduled consecutively. (1 point, 2 points for reasoning) If the multiple hours of the same activity have to be scheduled consecutively then there are four activities to be arranged; lap swim, family swim/play in the shallow end and diving at the deep end (both scheduled at the same time), school swim teams, and public swim The number of distinct schedules = 4! = 4 x 3 x 2 x 1 = 24 5b: Determine the number of different schedules Kim can create if multiple hours of the same activity do not have to be scheduled consecutively. (1 point, 3 points for reasoning) If the multiple hours of the same activity do not have to be scheduled consecutively then there are 12 activities to be arranged; 4 lap swims, 2 family swim/play in the shallow end and diving at the deep end (both scheduled at the same time), 3 school swim teams, and 3 public swim The number of distinct schedules = 12! 4! 2! 3! 3! = 11x10x9x8xx5 = 2200
Page 6 of 6 6a: What is the maximum amount of chlorine that was lost to reaction with ammonia or to volatilization on any day during the past year. (2 points) Maximum amount of chlorine lost in a day during the past year:- 5.0 3.5 = 1.5 mg/l 6b: What should the chlorine concentration be under the 10% reduction plan if there are no losses to reaction with ammonia nor to volatilization. (2 points) Chlorine concentration under the 10% reduction plan if there are no losses:- 5.0 x 90% = 4.5 mg/l 6c: Estimate the lowest minimum daily concentration under the proposed plan of reducing the chlorination rate by 10%. Assume that the losses to reaction with ammonia and to volatilization will not be affected by the chlorination plan. (2 points) Estimate of lowest minimum daily concentration of chlorine under the 10% reduction plan:- 4.5 1.5 = 3.0 mg/l 3.0 mg/l The expected distribution of the lowest minimum daily concentration of chlorine under the 10% reduction plan:= The concentrations are expected to shift down by 0.5 mg/l (= 5.0 4.5) The frequencies are expected to be similar 6d: Make a recommendation as to whether the rate of chlorination can be reduced by 10% after computing the cost implications of this reduction. (4 points for justification of decision 4 points for cost implications) (3 points) Annual Cost savings from reduced use of chlorine (or chlorinating agent):- $20.00/day x 365 day x 0.1= $30 Anticipated fines ;- $0. Recommendation:- Chlorinate at 90% of last year s rate as the (residual) chlorine concentration is not expected to go below the limit. Compare the observed distribution of the chlorine measurements under the new plan with the tabulated expected distribution and adjust the plan if necessary.