- Based on Figure 1, what will the output be closest to at 18:00?
Answer: 0.3 kW
Output in Figure 1 peaks at 5.0 kW at noon and then falls, reaching 4.1 kW at 14:00 and 2.0 kW at 16:00, a drop that is getting steeper. With sunset shortly after 18:00, output at 18:00 will be near zero. Options at or above 2.0 kW would require the afternoon decline to stop.
- Based on Table 1, what would the expansion be at a beam temperature of 50 °C?
Answer: 6.0 mm
Expansion in Table 1 increases by 2.4 mm for every 20 °C, or 1.2 mm for every 10 °C. A temperature of 50 °C is halfway between the 40 °C and 60 °C rows, so the expansion is halfway between 4.8 mm and 7.2 mm: 6.0 mm.
- If the pattern in Table 1 continues, what will the activity be 15 days after the tracer was prepared?
Answer: 125 kBq
The activity in Table 1 halves every 3 days: 4000, 2000, 1000, 500 kBq. Continuing that pattern gives 250 kBq at 12 days and 125 kBq at 15 days. Halving never reaches zero, so 0 kBq cannot be right, and 250 kBq is the 12-day figure.
- Based on Table 1, after about how many cycles will the battery reach the 80% replacement point?
Answer: 1,000 cycles
Capacity in Table 1 falls 4 percentage points every 200 cycles. From 84% at 800 cycles, one more step of 200 cycles brings it to 80%, so the battery reaches the replacement point at about 1,000 cycles.
- If the vehicle continues at the same speed, at about what total distance will the tank be empty?
Answer: 500 km
Figure 1 shows 12 L used every 100 km. After 400 km, 12 L remain, which is enough for exactly 100 km more, so the tank empties at about 500 km. Choosing 440 km or 480 km stops short of using the last 12 L.
- If the trend in Figure 1 continues, in which year will the percentage first reach 60%?
Answer: 2025
The percentage rises 6 points every year in Figure 1. Continuing from 42% in 2022 gives 48% in 2023, 54% in 2024, and 60% in 2025. That makes 2025 the first year at 60%.
- Based on the pattern in Table 1, what will the amplitude of the sixth swing be closest to?
Answer: 6.5 cm
Each swing in Table 1 is about 0.8 times the one before it (16.0 ÷ 20.0 = 0.80, 12.8 ÷ 16.0 = 0.80, 10.2 ÷ 12.8 ≈ 0.80). Continuing gives about 8.2 cm for the fifth swing and about 6.5 cm for the sixth. Subtracting a fixed amount each time instead would predict a much smaller amplitude.
- If the drink is left on the bench for several hours, its temperature will be closest to which of the following?
Answer: 20 °C
The drops between readings in Figure 1 get smaller and smaller (17, 12, 9, 6, 4 °C), so the curve is flattening rather than continuing downward at a steady rate. It is levelling off at the 20 °C temperature of the room, which is as cool as the drink can get. Extending the early steep slope in a straight line would wrongly predict 0 °C or below.
- Based on Figure 1, what reading is expected in May 2025?
Answer: 427 ppm
The May readings in Figure 1 rise by 3 ppm each year: 418 in 2022, 421 in 2023, and 424 in 2024. May 2025 should therefore be about 427 ppm. Comparing a May reading with a September reading instead would suggest the concentration is flat or falling, which is why 418 ppm is tempting.
- An engineer recorded the distance a test cart traveled after different numbers of identical pushes on a smooth track. Based on the pattern in the table, what distance would most likely be traveled after 6 pushes?
Answer: 24 m
The data indicate a linear increase in the distance traveled by the cart with each additional push. The distances are 4 m after 1 push, 8 m after 2 pushes, 12 m after 3, 16 m after 4, and 20 m after 5, with a consistent addition of 4 m per push. This quantifies the pattern as a linear relationship where each push adds 4 m to the total distance. To predict the distance after 6 pushes, add 4 m to the 5-push distance: 20 + 4 \= 24 m. One might mistakenly assume a multiplicative pattern, such as doubling, which would incorrectly predict 40 m, but the data show additive increases.
- If the level keeps falling at the same rate, on the first day of which month will the recorded level first be below 22.0 m?
Answer: July
The level in Table 1 drops 3.5 m between readings. Continuing from 28.0 m on 1 May gives 24.5 m on 1 June and 21.0 m on 1 July, so the July reading is the first one below 22.0 m. The June reading is still 2.5 m above the limit.
- If the pattern in Figure 1 continues, what percentage of colonies would be expected to bleach at an anomaly of 2.5 °C?
Answer: 100%
The increases in Figure 1 are getting larger: 13, 22, and 30 percentage points. Adding another step of roughly 38 points to 70% gives about 108%, which is impossible because a site cannot have more than 100% of its colonies bleached. The prediction is therefore essentially total bleaching, 100%. Choosing 130% extends the arithmetic past what the quantity allows.
- Based on Figure 1, what pressure would the sample reach at 400 K?
Answer: 160 kPa
Pressure in Figure 1 rises 20 kPa for every 50 K, a constant rate. Continuing from 140 kPa at 350 K gives 160 kPa at 400 K. The other options overshoot the steady 20 kPa step.
- Based on Figure 1, what would the sustained swimming speed most likely be at 35 °C?
Answer: 4 cm/s
Speed in Figure 1 climbs to 32 cm/s at 25 °C and then drops sharply to 18 cm/s at 30 °C, so the trend has already reversed before the last reading. Continuing that decline puts the speed at 35 °C well below 18 cm/s: repeating the 14 cm/s drop seen from 25 °C to 30 °C gives 18 − 14 = 4 cm/s. That is also the only option below the 30 °C reading, and the options above 32 cm/s would require the early rise to resume.
- A materials scientist measured the thickness of a coating after each pass of a sprayer. Based on the pattern in the table, what thickness would most likely be measured after 6 passes?
Answer: 27 µm
The data show a linear accumulation in coating thickness with each sprayer pass. The thicknesses are 4.5 µm after 1 pass, 9 µm after 2, 13.5 µm after 3, 18 µm after 4, and 22.5 µm after 5, increasing by 4.5 µm per pass. This quantifies the pattern as a constant addition of 4.5 µm each pass. To predict the thickness after 6 passes, add 4.5 µm to the 5-pass value: 22.5 + 4.5 \= 27 µm. Some might assume doubling instead, leading to 45 µm, but the data confirm additive growth.
- Based on Figure 1, what mass of ice will remain at 25 minutes?
Answer: 0 g
The block loses 50 g of ice every 5 minutes in Figure 1 and reaches 0 g at 20 minutes. Once no ice is left, none can melt, so the mass stays at 0 g at 25 minutes. Extending the straight line past 20 minutes would give a negative mass, which is not possible.
- If the pattern in Figure 1 continues, what will the cell count be at 6 hours?
Answer: 12,800 thousand per mL
The count doubles every hour in Figure 1: 200, 400, 800, 1,600, 3,200. Doubling twice more gives 6,400 at 5 hours and 12,800 at 6 hours. Adding a constant 1,600 each hour instead would give 6,400, which is why that option is tempting.
- Based on the pattern in Table 1, what height will the ball reach after the fifth bounce?
Answer: 13.0 cm
Each bounce in Table 1 reaches 0.6 times the previous height (60 ÷ 100, 36 ÷ 60, and 21.6 ÷ 36 all equal 0.60). The fifth bounce therefore reaches 21.6 × 0.6 = 13.0 cm. Applying the ratio twice by mistake gives 7.8 cm, which is the height after the sixth bounce rather than the fifth.
- Based on the pattern in Table 1, what would the intensity be at 8 m from the speaker?
Answer: 2.5 mW/m²
Doubling the distance in Table 1 divides the intensity by four: 160 mW/m² at 1 m becomes 40 at 2 m, and 40 at 2 m becomes 10 at 4 m. Doubling again from 4 m to 8 m gives 10 ÷ 4 = 2.5 mW/m². Simply halving the intensity would give 5.0 mW/m², 10 mW/m² is the reading already listed at 4 m, and 1.25 mW/m² comes from dividing the 4 m reading by eight instead of by four.
- Based on the pattern in Table 1, at about what time will the next high tide occur?
Answer: 02:50
Each high tide in Table 1 comes 12 hours 25 minutes after the one before it: 01:10 to 13:35, 13:35 to 02:00, and 02:00 to 14:25. Adding another 12 hours 25 minutes to 14:25 gives 02:50 the following morning.
