2011-06-08, 07:12 AM
9/10
I didn't really look at this until now, since I had my math final on Monday and my chem final on Tuesday. I missed number 5, the spring bouncing question. Sigh. I used to be good at these types of problems.
I didn't really look at this until now, since I had my math final on Monday and my chem final on Tuesday. I missed number 5, the spring bouncing question. Sigh. I used to be good at these types of problems.
My reasoning process
Question 1: Five boxes, A through E, are stacked in descending alphabetical order with box A on top. The bottom three boxes are simultaneously removed and placed on top with their vertical order maintained. If this procedure were repeated two more times which box would end up in the middle of the stack?
Top A B C D E Bottom
Since two were removed three times, the total shift is 6, so you'd end up with box D.
Question 2: Half of all Looms are Spades, half of all Rims are Nobs, and half of all Rims are Spades. Considering the limited information above, which of the following could not be true?
Since in each category, "half" is the main divisor, we're going to imagine there are two in each category to simplify each statement.
There are 2 Looms, 1 is a Spade.
There are 2 Rims, 1 is a Nob, 1 is a Spade.
From the last statement, it's uncertain whether a Spade is a Nob, since there may be overlap. Now let's evaluate each statement:
1. Half of all Spades are Nobs. This can be possible, since it's unspecified how many spades are nobs, and there may be overlap.
2. All Looms are Rims. If all Looms are Rims, then all Rims must also be Looms, since in each category, exactly half are Spades. Since there is no specification for whether or not Looms are Rims and Rims are Looms, this is also possible.
3. All Rims are Looms, with no Looms being Nobs. From the previous conjecture, we already established the condition for all Rims being Looms. Since this equates Rims and Looms, half of all Looms must be Nobs, so this cannot be true.
Question 3: A certain metallic cube has a red top, a green bottom, two yellow sides, and a blue front and back. A woman with magnetic boots standing on the red top face walks forward onto the front blue face, turns right and walks three faces, turns left and walks three faces, turns right and walks two faces, and turns to her left and walks one face. What color is the face on which she now stands?
The first time I read this, I thought it said "a woman with magnetic boobs". Lol.
From this, we can draw a "net" diagram:
------ Blue ------
Yellow Red Yellow
------ Blue ------
----- Green -----
1. standing on the red top face walks forward onto the front blue face, turns right and walks three faces, turns left and walks three faces, turns right
At this point, we're back at our original location and orientation.
2. walks two faces, and turns to her left and walks one face
Her final face is yellow.
Question 4: One-fourth of X is one half of a number that, if quadrupled and added to X, would result in a number that is three times X. Which of the following numbers could not be X?
X/4 = 1/2 N
4N + X = 3X
Derp this is easy. From the first equation, we have:
X = 2N
From the second equation, we have:
4N = 2X
2N = X
Therefore, N must be even, but that specifies nothing about X. So all of them can satisfy X.
Question 5: Three different springs bounce at different frequencies. Spring A bounces off the ground every 2 seconds; Spring B bounces every 5 seconds; and Spring C bounces every 9 seconds. The three springs leave the ground at the same time and continue bouncing until eventually the three springs, one by one, bounce during a three consecutive second interval. Upon the second and third such intervals, respectively, which spring makes the third bounce?
Because of the nature of this problem, we note that these interrvals can only happen when C bounces. Since C bounces every 9 seconds, these "bouncing one by one" intervals can only happen at the 9, 18, 27, etc. second mark. At the 9 second mark, we have 8 being dividable by 2. 10 is dividable by 5. However, 10 is also dividable by 2, so the springs are not bouncing one by one. Actually, at any even interval, this is impossible, so let's check the odd multiples of 9. At 27, we have 25 dividable by 5 and not 2, 26 dividable by 2, but not 5. Therefore, this is the first of our intervals. Since 30 is dividable by both 2 and 5, it's not a point of interest.
Since 45 is dividable by 5, we can skip it. At 63, we have 60 dividable by both 2 and 5. However, 64 is dividable by 2 but not 5, and 65 dividable by 5 but not 2. Therefore, the bounce at 65 seconds is the last bounce, and Spring B makes this bounce.
At 81, we have 80 being dividable by both 2 and 5, and 85 is too far from 81 to fit the three second time constraint. Therefore, this is not a point of interest.
At 99, 100 is dividable by both 2 and 5, and 95 is too far out of reach.
At 117, 115 is dividable by 5 but not 2. 116 is dividable by 5 but not 9. Therefore, this is our third interval, so Spring C bounces last.
B and C is not an answer choice. The pineapple...
Question 6: There are five people of different heights. Allen is taller than Dale, who is taller than Earl. Carla is shorter than Bill, but taller than Allen. Who is the third tallest person?
Allen > Dale > Earl
Bill > Carla > Allen
Allen is the third tallest. Duh.
Question 7: The small hand and the big hand of a clock are each pointing to twelve. The small hand moves clockwise two numbers every hour while the big hand moves counterclockwise five numbers every hour. How long will it take until both hands point to the same number?
Small hand: 2, 4, 6, 8, 10, 12, 2, 4, 6, 8, 10, 12
Large hand: 7, 2, 9, 4, 11, 6, 1, 8, 3, 10, 5, 12
12 hours. As expected, since this is mod 12 and 2 and 5 don't have a reasonable least common multiple.
Question 8: A satellite moving at constant speed can orbit the moon one time in eight hours. After each complete cycle, the satellite instantly reverses direction a quarter of the way back around the moon before immediately continuing forward again for another complete cycle. Beginning in the forward direction above a particular spot of the moon, how many hours does it take the satellite to orbit that spot three times?
This is a pretty pineappleed up orbit. We're going to assume that the satellite does its reverse orbit at the same speed. Therefore, it completes 3/4th orbits in 10 hours time. So on the 10th hour, we have 3/4th orbit complete. On the 30th hour, we have 90/4, or 2.25 orbits complete. Therefore, we need 3/4ths of a remaining orbit to complete three orbits, so 30 + 6 hours is 36 hours total.
Question 9: A blanket with a width of ten feet and a length of fifteen feet is reduced in its perimeter by six feet. Assuming only whole number deductions can occur from any side, and the angle of each corner of the blanket must remain the same, what is the second greatest surface area that could result？Only whole number deductions can occur from any side.
The square is the greatest area for a given perimeter, so the first deduction we want to make is from its length. A total deduction of 3 reduces its perimeter by 6. Thus, the greatest surface area possible is 120. The next best deduction is 2 from length and 1 from width. 9*13 = 117.
Question 10: If the time between now and 8:00 p.m. is three times longer than the time between now and the time that is an hour and a half before the hour that is exactly midway between now and 8:00 p.m., what time is it now?
If 8 pm occurs some time later:
8-n = 3*(t-n)
t = (8+n)/2-1.5
8-n = 3*((8+n)/2-1.5-n)
8-n = 3*(4+0.5n-1.5-n)
8-n = 3*(2.5-0.5n)
8-n = 7.5-1.5n
n = -1
-1 pm is 11 am
If 8 pm was some time ago:
n-8 = 3*(n-t)
t = (8+n)/2-1.5
n-8 = 3*(n-(8+n)/2+1.5)
n-8 = 3*(n-4-0.5n+1.5)
n-8 = 3*(0.5n-2.5)
n-8 = 1.5n-7.5
-0.5 = 0.5n
n = -1
-1 pm is 11 am
Interestingly, the two answers agree, suggesting that this relation holds whether 11 am happened before or after 8 pm. Not surprising though, since the equation for t didn't change, and all we did for the second one was make both sides negative (which is the same as doing nothing at all).
Top A B C D E Bottom
Since two were removed three times, the total shift is 6, so you'd end up with box D.
Question 2: Half of all Looms are Spades, half of all Rims are Nobs, and half of all Rims are Spades. Considering the limited information above, which of the following could not be true?
Since in each category, "half" is the main divisor, we're going to imagine there are two in each category to simplify each statement.
There are 2 Looms, 1 is a Spade.
There are 2 Rims, 1 is a Nob, 1 is a Spade.
From the last statement, it's uncertain whether a Spade is a Nob, since there may be overlap. Now let's evaluate each statement:
1. Half of all Spades are Nobs. This can be possible, since it's unspecified how many spades are nobs, and there may be overlap.
2. All Looms are Rims. If all Looms are Rims, then all Rims must also be Looms, since in each category, exactly half are Spades. Since there is no specification for whether or not Looms are Rims and Rims are Looms, this is also possible.
3. All Rims are Looms, with no Looms being Nobs. From the previous conjecture, we already established the condition for all Rims being Looms. Since this equates Rims and Looms, half of all Looms must be Nobs, so this cannot be true.
Question 3: A certain metallic cube has a red top, a green bottom, two yellow sides, and a blue front and back. A woman with magnetic boots standing on the red top face walks forward onto the front blue face, turns right and walks three faces, turns left and walks three faces, turns right and walks two faces, and turns to her left and walks one face. What color is the face on which she now stands?
The first time I read this, I thought it said "a woman with magnetic boobs". Lol.
From this, we can draw a "net" diagram:
------ Blue ------
Yellow Red Yellow
------ Blue ------
----- Green -----
1. standing on the red top face walks forward onto the front blue face, turns right and walks three faces, turns left and walks three faces, turns right
At this point, we're back at our original location and orientation.
2. walks two faces, and turns to her left and walks one face
Her final face is yellow.
Question 4: One-fourth of X is one half of a number that, if quadrupled and added to X, would result in a number that is three times X. Which of the following numbers could not be X?
X/4 = 1/2 N
4N + X = 3X
Derp this is easy. From the first equation, we have:
X = 2N
From the second equation, we have:
4N = 2X
2N = X
Therefore, N must be even, but that specifies nothing about X. So all of them can satisfy X.
Question 5: Three different springs bounce at different frequencies. Spring A bounces off the ground every 2 seconds; Spring B bounces every 5 seconds; and Spring C bounces every 9 seconds. The three springs leave the ground at the same time and continue bouncing until eventually the three springs, one by one, bounce during a three consecutive second interval. Upon the second and third such intervals, respectively, which spring makes the third bounce?
Because of the nature of this problem, we note that these interrvals can only happen when C bounces. Since C bounces every 9 seconds, these "bouncing one by one" intervals can only happen at the 9, 18, 27, etc. second mark. At the 9 second mark, we have 8 being dividable by 2. 10 is dividable by 5. However, 10 is also dividable by 2, so the springs are not bouncing one by one. Actually, at any even interval, this is impossible, so let's check the odd multiples of 9. At 27, we have 25 dividable by 5 and not 2, 26 dividable by 2, but not 5. Therefore, this is the first of our intervals. Since 30 is dividable by both 2 and 5, it's not a point of interest.
Since 45 is dividable by 5, we can skip it. At 63, we have 60 dividable by both 2 and 5. However, 64 is dividable by 2 but not 5, and 65 dividable by 5 but not 2. Therefore, the bounce at 65 seconds is the last bounce, and Spring B makes this bounce.
At 81, we have 80 being dividable by both 2 and 5, and 85 is too far from 81 to fit the three second time constraint. Therefore, this is not a point of interest.
At 99, 100 is dividable by both 2 and 5, and 95 is too far out of reach.
At 117, 115 is dividable by 5 but not 2. 116 is dividable by 5 but not 9. Therefore, this is our third interval, so Spring C bounces last.
B and C is not an answer choice. The pineapple...
Question 6: There are five people of different heights. Allen is taller than Dale, who is taller than Earl. Carla is shorter than Bill, but taller than Allen. Who is the third tallest person?
Allen > Dale > Earl
Bill > Carla > Allen
Allen is the third tallest. Duh.
Question 7: The small hand and the big hand of a clock are each pointing to twelve. The small hand moves clockwise two numbers every hour while the big hand moves counterclockwise five numbers every hour. How long will it take until both hands point to the same number?
Small hand: 2, 4, 6, 8, 10, 12, 2, 4, 6, 8, 10, 12
Large hand: 7, 2, 9, 4, 11, 6, 1, 8, 3, 10, 5, 12
12 hours. As expected, since this is mod 12 and 2 and 5 don't have a reasonable least common multiple.
Question 8: A satellite moving at constant speed can orbit the moon one time in eight hours. After each complete cycle, the satellite instantly reverses direction a quarter of the way back around the moon before immediately continuing forward again for another complete cycle. Beginning in the forward direction above a particular spot of the moon, how many hours does it take the satellite to orbit that spot three times?
This is a pretty pineappleed up orbit. We're going to assume that the satellite does its reverse orbit at the same speed. Therefore, it completes 3/4th orbits in 10 hours time. So on the 10th hour, we have 3/4th orbit complete. On the 30th hour, we have 90/4, or 2.25 orbits complete. Therefore, we need 3/4ths of a remaining orbit to complete three orbits, so 30 + 6 hours is 36 hours total.
Question 9: A blanket with a width of ten feet and a length of fifteen feet is reduced in its perimeter by six feet. Assuming only whole number deductions can occur from any side, and the angle of each corner of the blanket must remain the same, what is the second greatest surface area that could result？Only whole number deductions can occur from any side.
The square is the greatest area for a given perimeter, so the first deduction we want to make is from its length. A total deduction of 3 reduces its perimeter by 6. Thus, the greatest surface area possible is 120. The next best deduction is 2 from length and 1 from width. 9*13 = 117.
Question 10: If the time between now and 8:00 p.m. is three times longer than the time between now and the time that is an hour and a half before the hour that is exactly midway between now and 8:00 p.m., what time is it now?
If 8 pm occurs some time later:
8-n = 3*(t-n)
t = (8+n)/2-1.5
8-n = 3*((8+n)/2-1.5-n)
8-n = 3*(4+0.5n-1.5-n)
8-n = 3*(2.5-0.5n)
8-n = 7.5-1.5n
n = -1
-1 pm is 11 am
If 8 pm was some time ago:
n-8 = 3*(n-t)
t = (8+n)/2-1.5
n-8 = 3*(n-(8+n)/2+1.5)
n-8 = 3*(n-4-0.5n+1.5)
n-8 = 3*(0.5n-2.5)
n-8 = 1.5n-7.5
-0.5 = 0.5n
n = -1
-1 pm is 11 am
Interestingly, the two answers agree, suggesting that this relation holds whether 11 am happened before or after 8 pm. Not surprising though, since the equation for t didn't change, and all we did for the second one was make both sides negative (which is the same as doing nothing at all).