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09 Sep 2004, 12:11
Kudos
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15. A 48 gallon solution of salt and water is 10% salt. How many gallons of water must be added to the solution in order to decrease the salt to 8% of the volume?
(a) 8
(b) 12
(c) 13
(d) 14
(e) 16
. The “Racing magicâ€
(a) 8
(b) 12
(c) 13
(d) 14
(e) 16
. The “Racing magicâ€
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09 Sep 2004, 12:24
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B & D respectively......!!
1. Salt remains same, hence 4.8 = 8/100 (48+x), where x is the gallons of water added. Solve for x!!
2. The First time they meet will be LCM of 120 and 90, which is 360 seconds or 6 minutes. The second time they will meet after another 6 minutes, hence they meet in 12 minutes. BTW, since CB makes 40 rounds in 1 hr, it takes 90 sec to make one round!!
1. Salt remains same, hence 4.8 = 8/100 (48+x), where x is the gallons of water added. Solve for x!!
2. The First time they meet will be LCM of 120 and 90, which is 360 seconds or 6 minutes. The second time they will meet after another 6 minutes, hence they meet in 12 minutes. BTW, since CB makes 40 rounds in 1 hr, it takes 90 sec to make one round!!
09 Sep 2004, 13:37
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1. B.
4.8/(48+X) = 8/100
=> X= 12.
2. In circular races, A and B will meet at the starting point for the first time equal to the LCM of the time it takes A and B individually to complete one round along the circular track.
Racing magic takes 120 Sec's = 2min's.
Charging bull takes 60/40 = 3/2 = 1.5 mins.
LCM of 2 and 1.5 is 6. So, in 6 minutes they meet for the first time. They will meet at 12 mins the second time.
4.8/(48+X) = 8/100
=> X= 12.
2. In circular races, A and B will meet at the starting point for the first time equal to the LCM of the time it takes A and B individually to complete one round along the circular track.
Racing magic takes 120 Sec's = 2min's.
Charging bull takes 60/40 = 3/2 = 1.5 mins.
LCM of 2 and 1.5 is 6. So, in 6 minutes they meet for the first time. They will meet at 12 mins the second time.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
