Isaac and Albert wanted to take a vacation. They were debating how they could get to their hotel in the fastest manner. Isaac said, "We should go by train." But Albert said, "No, the train reaches the end of the line half way to the hotel so we would have to walk the rest of the way. We should bike to the hotel instead." Isaac disagreed. So Albert biked the whole way to the hotel, while Isaac took the train for the first half of the journey and walked for the remainder. The speed of the train turned out to be four times that of the bike's speed. The bike's speed turned out to be two times faster than walking speed. Who got to the hotel first?
Answer:
Let s be the total distance of the journey to the hotel. Let v bewalking speed. So 2v is the bike's speed, and 8v is the train's speed. Let I be the time it took for Isaac to complete the journey, and A be the time it took for Albert to complete the journey. Since distance equals rate times time, we have two equations, one for I and one for A: I = (s/2)/8v + (s/2)/v = s/16v + s/2v A = s/2v Note that I exceeds A by s/16v. Albert will reach the hotel first.
Alternate Solution #2. The problem may be solved more easily with simple logic. If thebicycle is twice as fast as walking, the time it takes to bike the whole way is equal to the time it takes to walk half the way. So if the train's speed is anything shy of infinite, biking will still be faster.
82. HOW LONG GRASS LAST?
It is known that if 10 sheep are turned out in the field, the grass will be gone in 20 days. On the other hand, if 15 sheep are turned out in the field, the grass will be gone in 10 days. If 25 sheep are turned out in the field, when will the grass be gone?
Answer:
Let r is the unit of grass grown per day. Let g be the total units of grass in the field before the sheep are turned out. Let s be the number of units of grass each sheep eats perday. We must determine the constant values of g and r from the information given in the problem. The total number of grass units eaten equals the number of days times the amount of grass the sheep eat per day. From this, we can construct two equations: g + 20r = 20(10s). g + 10r = 10(15s)Reducing the equations, we obtain: g + 20r = 200s. g + 10r = 150s. Subtract one equation from the other: 10r = 50 s. So r equals 5s. Substituting back, we discover g equals 100s. Now we construct an equation for the case of having 25 sheep turned out inthe field, where x is the number of days it takes the sheep to eat all the grass: g + xr = x(25s). 100s + x(5s) = x(25s) 20sx =100s. x = 5. So 25 sheep would consume all the grass in the field in 5 days.
83. WHAT IS WRONG?
The following is what seems to be a mathematical proof that ten equals 9.999999....
What's wrong with it?
a = 9.999999...
10a = 99.999999...
10a - a = 90.
9a = 90
a = 10
Answer:
Actually there's nothing wrong with it. Ten does equal 9.99999....... as this proof clearly shows.
84. PARTY ACQUAINTANCE.
You and your spouse invite four other couples to a party. During the course of the conversation, it is discovered that, prior to the party, each person except you was acquainted with a different number of the people present. Assuming the acquaintance relationship is symmetric (i.e., if you are acquainted with someone, that person is also acquainted with you), then how many people did your spouse know prior to the party? How many people did you know?
Answer:
Start with the assumption that everybody knows their own spouses --which means that everybody there knew at least one person. Discounting yourself, everyone knows a different number of people, which means that (again, discounting yourself) one person knows one, one person knows two, one person knows three, etc., up to one person who knows nine people (everybody else). Number the people (besides yourself) according to how many people they know, so that person 1 is the one who knows one person, person 2 is the one who knows two people, etc. Now pair up people with their spouses. If person 9 knows everybody else, s/he must be the only person who knows person 1, because person1 only knows one person. So they must be married. Person 8 knows everybody except for person 1. Person 2 therefore knows person 8 and person 9. Person 9 is married to person 1, so person 2's spouse must be person 8. Person 7 knows everybody except for persons 1 and 2. Person 3 therefore knows persons 7, 8, and 9. Persons 8 and 9 are married to persons 2 and 1 respectively, so person 3's spouse must be person 7. Person 6 knows everybody except for persons 1, 2, and 3. Person 4 therefore knows persons 6, 7, 8, and 9. The only one of those not yet paired up is person 6, so person 4 and person 6 must be married. This leaves person 5, who knows everyone except persons 1, 2, 3, and 4. These five people, therefore, must be persons 6, 7, 8, 9, and you.Since you are the only one of these five not yet paired up, person 5 must be your spouse. So your spouse knew five people prior to the party. The above also determines that the people who know you are persons 5,6, 7, 8, and 9. So you knew five people prior to the party also.
85. HOW MANY ANTS?
At least a dozen ants are marching through my kitchen! If the ants walk in rows of 7, 11, or 13, there are 2 ants left over, while in rows of 10, there are 6 left over. What is the smallest number of ants there could be satisfying the above criterion?
Answer:
One thing to notice is that 7 * 11 * 13 = 1001, so the number can be (and must be) 2 more than a multiple of 1001. This gives 1003, 2004, 3005, 4006. The last one is the first one that's 6 more than a multiple of 10. Therefore there were 4006 ants.
86. DURATION OF THE GAME.
A basketball playoff game started between 3 pm and 4 pm, and ended between 6 pm and 7 pm. The positions of the minute hand and the hour hand were reversed at the end of the game, compared to the beginning. What was the exact time the game started and ended, and how long was the game? (Try to give exact times, not rounded to the nearest anything.)
Answer:
we know the game started a little after 3:30, since the minute hand was between 6 and 7. Let's make up a variable: let "t" be the number of minutes after 3:00 ; we figure 30 < t < 35. One minute of time is 360/60 = 6 degrees of angle for the minute hand, and 1/12 of that, 1/2 degree, for the hour hand. So at the start, while the minute hand is at 6 t degrees, the hour hand starts at 3:00 (90 degrees) and after t min is at 90 + t/2 degrees. Now at the end of the game, a little after 6:15, the minute hand must be at 90 + t/2 degrees, which is (90 n+ t/2) / 6 = 15 + t/12 min past 6, causing the hour hand to move (90 + t/2) / 12 = 15/2 + t/24 degrees past 6, which has to match up with the original 6 t degrees: 180 + 15/2 + t/24 = 6 t ; mult by 24 and solving gives 4320 + 90 + t = 144 t ; 4410 = 143 t ; Starting time = t = 4410 / 143 = 30 120/143 min past 3 , or 3:30120/143 , approx. 3:30.83916; Ending time = 15 + t/12 = 15 + 2205 / 1716 = 16 489/1716 min past6 , approx 6:16:284965. Time of game = 2 1/2 hrs + 16 489/1716 min - 120/143 min = 2 hrs45 765/1716 min. Approx values in decimal seconds are : Start = 3:30:50.35 , End = 6:16:17.10 , Duration = 2:45:26.75.
87. ACHIEVING AVERAGE.
A person walks at the rate of 3 mph to his office. How fast does he have to run back home in order to average 6 mph for the round trip? that is for both trips combined, to and fro.
Answer:
He would have to be infinitely fast. This is true no matter what speed he was travelling. Anyone travelling x mph over a road nx long, would take n hours one direction. To average 2x, over 2nx (the total road length), 2nx/n+? = 2x, ? must equal zero, so it took him zero hours, or he went infinitely fast.
88. NON-EMPTY DELIVERY.
A person having 2 branches of Bananas, each branch having 50 numbers, has to cross a river in 50 boats (because after certain time the first boat man transport the person and the bunch to the second boat, the second one doing the same to the third and so on and so forth until the 50 th boat who reaches the other bank finally) but with the condition that each boatman has to be given a banana per branch he carries. Obeying the conditions stipulated strictly, the poor man cannot make it anything to his home since all 100 bananas would have been exhausted by delivering to the unscrupulous boatmen however he strikes upon a strange idea by which he could accomplish carrying something, if not all. How did he do it?
Answer:
For the first 25 boats he gives 50 bananas from the same branch so that from the 26th boat onwards, he might have only one branch and therefore is expected to give only one banana, thereby leaving 25 bananas to take it home gleefully.
89. CLASS STRENGTH.
In a certain algebra lecture class, Chris and Pat count the students and compare notes. "Hmm, 12/17 of my classmates in here are women," notes Pat. "Funny," recounts Chris, "5/7 of my classmates are women. "They were both right. How many students, men and women, were in the class, and what were the genders of Chris and Pat? Men and women are mutually exclusive sets for the purposes of this problem. Explain your reasoning.
Solution:
Let n be the total number of students in the class, w = the number of women. Now 12 / 17 is less than 5 / 7 , so Pat has to be a woman, as Chris sees more women classmates from his perspective. Then Pat sees that (w-1) / (n-1) = 12 / 17 , and Chris sees w / (n-1) = 5 / 7. The denominator (n-1) can be the product 7*17 = 119, because 12/17 = 84/119 and 5/7 = 85/119 , so since 84 and 85 are 1 apart, we're there. Pat has 84 women classmates, and Chris has 85. Don't forget the 119 doesn't count the observer, so there are 120 students in the class.
90. INCORRECT LABELS.
You are working in a store that stocks bangles. Three boxes of bangles have been incorrectly labeled. The labels say Red Bangles, Green Bangles and Red & Green Bangles. How can you re-label the boxes correctly, by just taking only one bangle from one box?
Answer:
Keep in mind that boxes are incorrectly labeled. Take out one bangle from the box labeled "Red & Green Bangles". There are 2 possibilities: Case 1: If that bangle is Red, it means that box contains Red Bangles. The box labeled as "Green Bangles" contains Red & Green Bangles. And box labeled "Red Bangles" contains Green bangles.
Case 2: If that bangle is Green, it means that box contains Green Bangles. The box labeled as "Green Bangles" contains Red Bangles. And box labeled "Red Bangles" contains Red & Green Bangles.