Micheal o rata go palama baesekele. O ne a e palama bobotlana makgetlho a le 5 ka beke mme a tsamaya dikilometara di le 25 nako le nako. O dirile jalo ka dibeke di le nne, mme o bo o swetsa ka gore a e palame makgetlho a le 2 fela ka beke, mme a tsamaye dikilometara di le 60 nako le nako, mme a dira jalo ka dibeke di le 3. Ke dikilometara di le kae tse Micheal a di tsamaileng ka kakaretso? 

Brinley is in Mr. Bert's math class. Mr. Bert gives six tests each semester. He allows the students to remove the lowest score from the average each semester. If Brinley has scores of 89, 71, 92, 100 and 86 on the first five tests, what score does she need on the sixth test to get an average of 93?
Mo karolong ya ntlha, Micheal o tsamaile 5 * 25 = <<5*25=125>>125 dikilometara ka beke.
O dirile jalo ka dibeke di le nne, se se rayang gore o tsamaile 125 * 4 = <<125*4=500>>500 dikilometara mo nakong eo.
Morago ga moo, o ne a tsamaya 2 * 60 = <<2*60=120>>120 dikilometara ka beke.
Mme Micheal o dirile jalo ka dibeke di le tharo, ka jalo o tsamaile 120 * 3 = <<120*3=360>>360 dikilometara mo nakong eo.
Ka kakaretso Micheal o tsamaile 500 + 360 = <<500+360=860>>860 dikilometara.
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First, identify the lowest score, which is 71, and remove it.
Since Brinley wants a score of 93, we know her total score needs to be 93 (average) * 5 (tests) = <<93*5=465>>465 points
So far, Brinley has earned 89 + 92 +100 + 86 = <<89+92+100+86=367>>367 points
The final test must be 465 - 367 = <<465-367=98>>98 points.
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