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Question

Pls help. Thanks in advance.

Wei Ming and Muthu played a game for 10 rounds. In each round, the winner scored 2 points while the loser was deducted 2 points. At the end of the game, Muthu’s total score was 4 points. How many rounds did Wei Ming lose?

Answer

2 X 10 = 20

20 – 4 = 16

2+2 = 4

16/4 = 4 

Wei Ming lost–> 10 – 4 = 6

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Try to use supposition/assumption here.

Assume all wins for Muthu, he would score 20 points.

Delta of assumed and actual score= 20 – 4 = 16

Difference in points between a win and a lose, 4 points. (+2 vs -2)

therefore the number of time he lost = 16/4 = 4 times

Also mean he win 6 times, also the same number of time of Wen Ming’s lost.

 

 

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A lost negates the points gained from a win.

Since Muthu has +4 points, he won 2 rounds more than he lost.

Therefore, Muthu won 6 and lost 4 out of the 10 rounds.

Answer: Wei Ming lost 6 rounds.

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Wei Ming must have lost 6 rounds.

Muthu had 4 points after 10 rounds. This means he won at least 2 rounds.

Therefore, the result of 8 rounds for Muthu must be 0. The only way this can occur is if Muthu won 4 rounds and lost 4 rounds. 

Hence Muthu won a total of 6 rounds. 

Therefore, Wei Ming lost a total of 6 rounds.

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