SOLUTION: an ant can walk 1 cm/s faster than a beetle. from points that are 50cm apart the ant and the beettle start walking directly towards each other and meet in 10 secs. how fast does ea
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Question 262214: an ant can walk 1 cm/s faster than a beetle. from points that are 50cm apart the ant and the beettle start walking directly towards each other and meet in 10 secs. how fast does each of them walk. asume they both started walking at the same time Found 2 solutions by drk, palanisamy:Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! This is an RTD question. Here is a table based on the above information:
animal . . . . . . ..rate . . . . . . . . .time . . . . . .. . .distance
ant . .. . . . . . . . b+1. . . . . . . . .10. . . . . . . . . . . . . .10b + 10
beetle. . . . . . . .b. . . . . . . . . . . .10. . . . . . . . . . . . .10b
total. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . . . . 50
looking at the third column, we have
combine like terms to get
subtract to get
divide to get
ant walks 3 cm/s
beetle walks 2 cm/s
You can put this solution on YOUR website! Let the speed of a beetle = x cm/s
Then,the speed of the ant = x+1 cm/s
Given, from points that are 50cm apart the ant and the beettle start walking directly towards each other and meet in 10 secs.
speed*time = distance
(x+x+1)*10 = 50
Dividing by 10,
2x+1 = 5
2x = 5-1
2x = 4
x = 2
Therefore, the speed of a beetle = 2 cm/s
And,the speed of the ant = 2+1 = 3 cm/s
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