Keep Your Streak Going!
A young boy on brilliant had a forty nine day streak going on. Unfortunately, on day fifty, there was a power cut in his neighborhood. Determined to keep his streak alive, the boy went to an internet cafe. The owner's son (who was also the boy's schoolmate) was in charge during the time he walked in. Jealous of his friend's success in brilliant, he decided to charge the boy 30 units of currency for each problem that he solved. The boy had only 240 units of currency in his pocket and his mother wanted him back home in three hours.
The boy decided that he would challenge himself only with level 4 and level 5 problems that day. He knew that it would, on average, take him about 45 minutes to solve a level 5 problem and around 15 minutes to solve a level 4 problem. On average, level 5 problems are worth about 250 points and level 4 problems are worth about 150 points. If the boy needs to solve x level 5 questions and y problems in level 4 in order to boost up his points as much as he can,
Assume that he always gets the right answer the first time (no decrease in points) and attempts only those questions that he thinks he can solve. He also doesn't waste time in between questions.
The answer is 1408.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
In order to solve this problem, I will use a formula I derived myself that gives the explicit solution to a 2x2 system of linear inequalities (perhaps I will help add this to the wiki some time soon). The objective function is the function for the number of points, which is given by P = 2 5 0 x + 1 5 0 y . The money constraint gives the inequality x + y ≤ 8 . The time constraint gives the inequality 3 x + y ≤ 1 2 . The solution set to a 2x2 system of linear inequalities (non-parallel lines) of the form a x + b y ≤ e c x + d y ≤ f is given by { ( u − b r − d t , v + a r + c t ) : s g n ( a d − b c ) r ≤ 0 , s g n ( a d − b c ) t ≥ 0 } where ( u , v ) represents the intersection of the two lines and s g n ( a d − b c ) represents the sign of a d − b c . Using the given information, I let a = 1 , b = 1 , c = 3 , d = 1 , e = 8 , f = 1 2 and then determine a d − b c < 0 . The intersection point of the lines is ( 2 , 6 ) . So I end up with x = 2 − r − t and y = 6 + r + 3 t , such that r ≥ 0 , t ≤ 0 . Substitute this into the objective function: P = 2 5 0 ( 2 − r − t ) + 1 5 0 ( 6 + r + 3 t ) = 1 4 0 0 − 1 0 0 r + 2 0 0 t . From here, one can easily tell the maximum is 1400, and it occurs when r = t = 0 , which corresponds to x = 2 , y = 6 .