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11 Oct 2021, 01:30
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D
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Difficulty:
75%
(hard)
Question Stats:
47% (02:00) correct
53%
(01:58)
wrong
based on 81
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Two ants leave simultaneously from a spot on the floor. One of them crawls due north at 2 inches per second, and the other crawls due west at a rate one-third faster than the first ant's rate. Approximately how many feet apart will they be after 6 seconds?
A. 83
B. 1
C. 1.33
D. 1.67
E. 20
A. 83
B. 1
C. 1.33
D. 1.67
E. 20
11 Oct 2021, 02:05
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1st ant after 6 seconds: \(12 inches\)
2nd ant after 6 seconds: \(2 \frac{2}{3} * 6\)
\(= \frac{8}{3} * 6 \)
\(= 16 inches \)
Their directions create a special 90-60-30 right angle triangle (3, 4, 5) with the hypotenuse being the distance between them.
\(3*4 = 12\)
\(4*4 = 16\)
\(5*4 = 20\)
Therefore the distance between the two ants is 20 inches, which is around 1.67 feet.
Answer D
2nd ant after 6 seconds: \(2 \frac{2}{3} * 6\)
\(= \frac{8}{3} * 6 \)
\(= 16 inches \)
Their directions create a special 90-60-30 right angle triangle (3, 4, 5) with the hypotenuse being the distance between them.
\(3*4 = 12\)
\(4*4 = 16\)
\(5*4 = 20\)
Therefore the distance between the two ants is 20 inches, which is around 1.67 feet.
Answer D
11 Oct 2021, 09:32
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Correct Option D
Ant 1 : North 2 inch / sec = 6 sec = 12 inch
Ant 2 : West :
= 1/3 times faster than Ant 1
= 33% faster than Ant 1
= 1.33 times faster than Ant 1
= Ant 1 *1.33
= 12 * 1.33
= ~16 inch
Pythagoras theorem = \(\sqrt{12^2 + 16^2}\) = 20 inch
Convert 20 inch to feet = 0.08333 *20 = 1.67
Option D
Ant 1 : North 2 inch / sec = 6 sec = 12 inch
Ant 2 : West :
= 1/3 times faster than Ant 1
= 33% faster than Ant 1
= 1.33 times faster than Ant 1
= Ant 1 *1.33
= 12 * 1.33
= ~16 inch
Pythagoras theorem = \(\sqrt{12^2 + 16^2}\) = 20 inch
Convert 20 inch to feet = 0.08333 *20 = 1.67
Option D
