An algebra problem by Digi Verse
Another relatively classic math problem: Two men have arranged to meet in a cafe between 2 and 4 pm. However, they did not arrange an exact time to meet. Therefore, they will both show up at a random time. These men are impatient, so they will only wait 15 minutes each before leaving. Find the probability that they will meet, and then add the denominator and numerator to plug in to the box. Assume that if one arrives at 3:59 pm, and it strikes 4, they will leave, even though they have not waited the 15 minutes yet. Make sure the fraction is in simplest forms before adding.
The answer is 79.
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1 solution
You will need graphing paper for this particular problem. Lets call 2 pm 0 on the graph, and each number is 1 minute. Plan approximately 120 numbers, on both x and y axis, since there are 120 minutes from 2pm to 4pm. Lets call the first guy x and the other one y. X<120 and Y<120, since they both will leave at 4pm. The two men will wait 15 minutes, so absolute(x-y)<15. You will get a common area where all these inequality meet, which is the area in which the two men see each other. You could calculate the area of the shaded area, and divide it by the total area, which would be 3375/14400, which simplifies to 15/64. 15+64 is 79.