Question 1184603
<br>
A formal algebraic solution like the one provided by the other tutor is probably what was expected from the student.<br>
But the same calculations can be done informally to reach the answer faster and with less work.<br>
The two given temperatures are 68 and 80 degrees, a difference of 12 degrees.
The two rates of chirping are 124 and 172, a difference of 48.<br>
Assuming a linear relationship (which should be stated in the problem, but which we have to assume in order to solve the problem), the rate of chirping increases by 48/12=4 for each increase of 1 degree of temperature.<br>
150 chirps per minute is 26 more than the 124 per minute at 68 degrees.  Since the rate of increase is 4 chirps for each degree increase of temperature, the increase in temperature from 68 degrees to get 150 chirps per minute is 26/4=6.5 degrees.  So<br>
ANSWER: The temperature at which the crickets will chirp at a rate of 150 per minute is 68+6.5=74.5 degrees.<br>
All the words of explanation make this sound like a long and complicated problem.  But the calculations alone are few and easy:<br>
80-68=12
172-124=48
48/12=4
150-124=26
26/4=6.5
68+6.5=74.5<br>