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An asteroid travels through space at a constant rate of 2.6 million fe
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08 Apr 2024, 22:59
08 Apr 2024, 22:59
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Question Stats:
77% (01:03) correct
22% (02:10) wrong based on 9 sessions
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An asteroid travels through space at a constant rate of 2.6 million feet per day. If the distance between the asteroid and a certain planet is 10.2 million feet, approximately how many seconds will it take the asteroid to reach the planet?
A. 80,000
B. 120,000
C. 200,000
D. 250,000
E. 350,000
A. 80,000
B. 120,000
C. 200,000
D. 250,000
E. 350,000
Re: An asteroid travels through space at a constant rate of 2.6 million fe
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19 May 2024, 08:34
19 May 2024, 08:34
Expert Reply
OE
Plug the given values into the R × T = D formula. The rate is 2.6 million feet per day and the distance is 10.2 million feet. Thus: 2.6 million × T = 10.2 million → T = 10.2 million/2.6 million = approximately 4 days.
Now, convert days to seconds. 1 day = 24 hours = 60(24) minutes = 60(60)(24) seconds = 86,400 seconds. So, the number of seconds in 4 days is 86,400 x 4 = 345,600 seconds. The closest answer is 350,000.
Plug the given values into the R × T = D formula. The rate is 2.6 million feet per day and the distance is 10.2 million feet. Thus: 2.6 million × T = 10.2 million → T = 10.2 million/2.6 million = approximately 4 days.
Now, convert days to seconds. 1 day = 24 hours = 60(24) minutes = 60(60)(24) seconds = 86,400 seconds. So, the number of seconds in 4 days is 86,400 x 4 = 345,600 seconds. The closest answer is 350,000.
gmatclubot
Re: An asteroid travels through space at a constant rate of 2.6 million fe [#permalink]
19 May 2024, 08:34
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