Ant trouble

[kushal.adhia]

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

I chose C

Would appreciate it if someone could throw some light on why it is wrong.

Kushal

[GMATGuruNY]

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

I chose C

Would appreciate it if someone could throw some light on why it is wrong.

Kushal

1/3 faster means a rate that is 4/3 of the slower rate. Since the slower rate is 2 inches second, the faster rate is (4/3)*2 = 8/3 inches per second. Since the ants are traveling for the same amount of time, a rate that is 4/3 faster will result in a distance that is 4/3 of the shorter distance.

Distance north = r*t = 2*6 = 12.
Distance west = (4/3)*12 = 16.
These two distances form the legs of a right triangle. The distance between the ants is the hypotenuse. This right triangle is a multiple of a 3:4:5 triangle. 4*(3:4:5) = 12:16:20. Thus, the hypotenuse -- which is the distance between the ants -- is 20 inches.

The correct answer is E.

[GMATGuruNY]

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

I chose C

Would appreciate it if someone could throw some light on why it is wrong.

Kushal

1/3 faster means a rate that is 4/3 of the slower rate. Since the slower rate is 2 inches per second, the faster rate is (4/3)*2 = 8/3 inches per second. Since the ants are traveling for the same amount of time, a rate that is 4/3 faster will result in a distance that is 4/3 of the shorter distance.

Distance north = r*t = 2*6 = 12.
Distance west = (4/3)*12 = 16.
These two distances form the legs of a right triangle. The distance between the ants is the hypotenuse. This right triangle is a multiple of a 3:4:5 triangle. 4*(3:4:5) = 12:16:20. Thus, the hypotenuse -- which is the distance between the ants -- is 20 inches. Since 12 inches = 1 foot, 20/12 = 5/3 = 1.67 feet.

The correct answer is D.

[ikaplan]

Use Pythagorean theorem to solve this problem.

The paths of the two ants represent 2 sides of a right triangle. The distance between two ants is in fact the hypotenuse of the triangle.

The first ants (heading north) will travel 12 inches for 6 seconds (6 seconds * 2 inches/second= 12 inches)
The second ant (heading west) will travel 16 inches for the same time (rate: 2+1/3*2=8/3; 6 seconds * 8/3= 16 inches)

Therefore, the distance between two ants after 6 seconds equals:

d^2=12^2 + 16^2
d= SQRT (144+256)
d= SQRT (400)
d= 20

My answer is E (20).

[MAAJ]

The hyppotenuse is 20 inches but it asks "how many feet apart will they be after 6 seconds". Because 1 foot = 12 inches, the answer would be 20/12 = 1.67 which is Answer (D).

However I dont know if its a typo of the question stem because a real Gmat question would make reference to the relationship between a foot and an inch.