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A water jug with a capacity of 20 gallons is 20 percent full. At the e
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30 May 2023, 00:36
30 May 2023, 00:36
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Question Stats:
60% (02:18) correct
40% (02:11) wrong based on 30 sessions
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A water jug with a capacity of 20 gallons is 20 percent full. At the end of every third day, water is added to the jug. If the amount of water added is equal to 50 percent of the water in the jug at the beginning of that day, how many days does it take for the jug to be at least 85% full?
A. 4
B. 6
C. 12
D. 15
E. 20
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A. 4
B. 6
C. 12
D. 15
E. 20
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Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
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Re: A water jug with a capacity of 20 gallons is 20 percent full. At the e
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02 Jun 2023, 04:00
02 Jun 2023, 04:00
1
Expert Reply
OE
This question asks for a specific number and the answer choices are integers in order from least to greatest, which means that Plug In the Answers might be a good approach. However, that would mean constructing a rather involved equation. A better approach might be to construct a short 3 column table (see below), with headings for Days, (Amount to) Add, and Gallons (in the jug).
GRE table mixture.jpg [ 55.75 KiB | Viewed 1299 times ]
The question states that the 20-gallon jug is 20% full to start. So enter “0” in the Days column and 20% × 20 = 4 in the Gallons column.
The question states that every 3 Days, an amount of water equal to 50% of the amount already in the jug is added to the jug. So in the next line enter “3” in the Days column, 50% × 4 = 2 in the Add column, and 4 + 2 = 6 in the Gallons column. In the next line, enter “6” in the Days column, 50% × 6 = 3 in the Add column, and 6 + 3 = 9 in the Gallons column. The question asks how many Days until the jug is at least 85% full. 85% × 20 = 17, so continue filling in the table until the number of gallons in the jug is greater than or equal to 17. The correct answer is (C). Note that (A) is a trap answer that represents the number of times (4) water needs to be added to the jug so the total gallons in the jug is greater than or equal to 17.
This question asks for a specific number and the answer choices are integers in order from least to greatest, which means that Plug In the Answers might be a good approach. However, that would mean constructing a rather involved equation. A better approach might be to construct a short 3 column table (see below), with headings for Days, (Amount to) Add, and Gallons (in the jug).
Attachment:
GRE table mixture.jpg [ 55.75 KiB | Viewed 1299 times ]
The question states that the 20-gallon jug is 20% full to start. So enter “0” in the Days column and 20% × 20 = 4 in the Gallons column.
The question states that every 3 Days, an amount of water equal to 50% of the amount already in the jug is added to the jug. So in the next line enter “3” in the Days column, 50% × 4 = 2 in the Add column, and 4 + 2 = 6 in the Gallons column. In the next line, enter “6” in the Days column, 50% × 6 = 3 in the Add column, and 6 + 3 = 9 in the Gallons column. The question asks how many Days until the jug is at least 85% full. 85% × 20 = 17, so continue filling in the table until the number of gallons in the jug is greater than or equal to 17. The correct answer is (C). Note that (A) is a trap answer that represents the number of times (4) water needs to be added to the jug so the total gallons in the jug is greater than or equal to 17.
Re: A water jug with a capacity of 20 gallons is 20 percent full. At the e
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21 Aug 2024, 13:15
21 Aug 2024, 13:15
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