Turing

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question:Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.

answer:1. **Calculate individual painting rates:** - Abe's rate: Abe can paint the room in 15 hours, so his rate is frac{1}{15} of the room per hour. - Bea's rate: Bea is 50% faster than Abe, so her rate is frac{1}{15} times 1.5 = frac{3}{30} = frac{1}{10} of the room per hour. - Coe's rate: Coe is twice as fast as Abe, so his rate is frac{1}{15} times 2 = frac{2}{15} of the room per hour. 2. **Calculate the work done by Abe in the first 1.5 hours:** - Abe works alone for 1.5 hours, so he paints frac{1}{15} times 1.5 = frac{1.5}{15} = frac{1}{10} of the room. 3. **Calculate the combined rate of Abe and Bea:** - Together, Abe and Bea paint frac{1}{15} + frac{1}{10} = frac{2}{30} + frac{3}{30} = frac{5}{30} = frac{1}{6} of the room per hour. 4. **Determine the time Abe and Bea take to paint half the room:** - Since Abe has already painted frac{1}{10} of the room, the remaining to reach half the room is frac{1}{2} - frac{1}{10} = frac{5}{10} - frac{1}{10} = frac{4}{10} = frac{2}{5} of the room. - Time taken by Abe and Bea to paint frac{2}{5} of the room is frac{2}{5} div frac{1}{6} = frac{2}{5} times frac{6}{1} = frac{12}{5} hours. 5. **Calculate the combined rate of Abe, Bea, and Coe:** - Together, all three paint frac{1}{15} + frac{1}{10} + frac{2}{15} = frac{2}{30} + frac{3}{30} + frac{4}{30} = frac{9}{30} = frac{3}{10} of the room per hour. 6. **Determine the time all three take to paint the remaining half of the room:** - Time taken by all three to paint the remaining frac{1}{2} of the room is frac{1}{2} div frac{3}{10} = frac{1}{2} times frac{10}{3} = frac{5}{3} hours. 7. **Calculate the total time taken in minutes:** - Total time = Time Abe worked alone + Time Abe and Bea worked + Time all three worked - Convert hours to minutes: left(1.5 times 60right) + left(frac{12}{5} times 60right) + left(frac{5}{3} times 60right) = 90 + 144 + 100 = 334 minutes. Thus, the total time taken for all three to finish painting the room is boxed{334} minutes.

question:A merchant imported an item and paid a 7 percent import tax on the portion of the total value of the item in excess of a certain amount. The amount of the import tax that the merchant paid was 111.30, and the total value of the item was 2,590. What was the amount in excess of which the import tax was applied?

answer:Let's denote the amount in excess of which the import tax was applied as X. The merchant paid a 7 percent import tax on the portion of the total value of the item in excess of X. This means that the taxable amount is (2590 - X). The import tax paid on this taxable amount is 7 percent, which is equal to 111.30. So we can set up the equation as follows: 0.07 * (2590 - X) = 111.30 Now, let's solve for X: (2590 - X) = 111.30 / 0.07 (2590 - X) = 1590 Now, subtract 2590 from both sides to solve for X: -X = 1590 - 2590 -X = -1000 Multiply both sides by -1 to get the positive value of X: X = 1000 Therefore, the amount in excess of which the import tax was applied is boxed{1000} .

question:Given x, y, z are different lines or different planes in space, which of the following conditions can ensure that If x perp z and y perp z, then x parallel y is a true statement? A: x, y, z are lines B: x, y, z are planes C: x, y are lines, z is a plane D: x is a line, y, z are plane

answer:A: When x, y, z are lines, if x perp z and y perp z, then x is parallel to y or x and y are skew lines. Therefore, option A is incorrect. B: When x, y, z are planes, if x perp z and y perp z, then x is parallel to y or x intersects y. Therefore, option B is incorrect. C: According to the theorem of the properties of line-to-plane perpendicularity, option C is correct. D: When x is a line, y, z are planes, if x perp z and y perp z, then x is parallel to y or the line x is in the plane y. Therefore, option D is incorrect. Hence, the correct answer is boxed{C}.

question:was walking with my motherinlaw. We were walking slowly at a speed of 3 km/h along Saint Catherine Street in Bordeaux, which, as everyone (or maybe not everyone) knows, is straight. Suddenly, I remembered that I needed to mail a letter as soon as possible in a mailbox located a bit further down the street. Leaving my motherinlaw, who continued walking peacefully, I hurried forward, mailed the letter, and then walked back towards my motherinlaw at a speed of 5 km/h. I clearly remember that I left my motherinlaw alone for only 3 minutes. At what distance from the mailbox were we when we parted ways?

answer:1. **Convert velocities and time into consistent units:** - My speed after leaving my mother-in-law is 5 text{ km/h} which we convert to meters per minute (m/min). [ text{My speed} = frac{5000 text{ m}}{60 text{ min}} = frac{250 text{ m}}{3 text{ min}} ] - My mother-in-law's and our common walking speed is (3 text{ km/h}): [ text{Mother-in-law's speed} = frac{3000 text{ m}}{60 text{ min}} = 50 text{ m/min} ] 2. **Define the points and variables:** - Let (S) be the place where we parted, (R) be the place where we met again, and (B) be the location of the mailbox. 3. **Calculate the distance mother-in-law walked in 3 minutes:** [ SR = 50 text{ m/min} times 3 text{ min} = 150 text{ m} ] 4. **Calculate the distance I covered for the full journey (from (S) to (B) and back to (R)):** - Total time for 3 minutes at frac{250 text{ m}}{3 text{ min}}: [ SB + BR = left(frac{250 text{ m}}{3 text{ min}}right) times 3 = 250 text{ m} ] 5. **Relate the distance covered on the way back:** - Substituting into our total distance: [ SB + BR = SR + 2BR quad text{From observation that the extra distance beyond SR is returning the same BR distance twice} ] [ 250 = 150 + 2BR ] Thus, [ 2BR = 100 implies BR = 50 text{ m} ] 6. **Find the distance (SB):** [ SB = SR + RB = 150 text{ m} + 50 text{ m} = 200 text{ m} ] # Conclusion: Thus, at the moment we parted, we were at a distance of 200 meters away from the mailbox. [ boxed{200 text{ meters}} ]