Math  /  Algebra

QuestionQuestion 50 The temperature of a nuclear reactor rises at a steady rate of 100 degrees per minute, while the temperature of a thermal reactor doubles every minute. Given that the starting temperatures of the nuclear reactor and the thermal reactor are 400 degrees and 15 degrees, respectively, after how many whole minutes will the temperature of the thermal reactor exceed that of the nuclear reactor?
Marks:1.0 Negative Marks: 0.25

Studdy Solution
Solve the inequality for the number of minutes.
This inequality doesn't have a straightforward algebraic solution, so we will test integer values of t t to find when the inequality holds.
- For t=0 t = 0 : 15×20=15 15 \times 2^0 = 15 and 400+100×0=400 400 + 100 \times 0 = 400 . So, 15400 15 \not> 400 . - For t=1 t = 1 : 15×21=30 15 \times 2^1 = 30 and 400+100×1=500 400 + 100 \times 1 = 500 . So, 30500 30 \not> 500 . - For t=2 t = 2 : 15×22=60 15 \times 2^2 = 60 and 400+100×2=600 400 + 100 \times 2 = 600 . So, 60600 60 \not> 600 . - For t=3 t = 3 : 15×23=120 15 \times 2^3 = 120 and 400+100×3=700 400 + 100 \times 3 = 700 . So, 120700 120 \not> 700 . - For t=4 t = 4 : 15×24=240 15 \times 2^4 = 240 and 400+100×4=800 400 + 100 \times 4 = 800 . So, 240800 240 \not> 800 . - For t=5 t = 5 : 15×25=480 15 \times 2^5 = 480 and 400+100×5=900 400 + 100 \times 5 = 900 . So, 480900 480 \not> 900 . - For t=6 t = 6 : 15×26=960 15 \times 2^6 = 960 and 400+100×6=1000 400 + 100 \times 6 = 1000 . So, 9601000 960 \not> 1000 . - For t=7 t = 7 : 15×27=1920 15 \times 2^7 = 1920 and 400+100×7=1100 400 + 100 \times 7 = 1100 . So, 1920>1100 1920 > 1100 .
The smallest integer t t for which the thermal reactor's temperature exceeds the nuclear reactor's temperature is t=7 t = 7 .
The number of whole minutes after which the temperature of the thermal reactor exceeds that of the nuclear reactor is:
7 \boxed{7}

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