etestkart

  Baba had yellow shoes and blue socks. Boba had green shoes and yellow socks. Bilbo had red shoes and green socks. Babil had blue shoes and red socks.

  Sage had a 1, Rosemary had an 11, and Tim had a 2. From Sage's statement, he had to have an odd number. He knew that the sum of all three numbers was even, so if his number was odd, one of the other two numbers would have to be odd, and the other even, leaving them unequal. Rosemary could deduce this, too, so when she saw an 11, she knew that Sage had a 1 and Tim had a 2. If she had seen a 12, she would have known that the other two boys each had a 1, which is contradicted by her saying that she already knew they all had different numbers. If she saw any number other than an 11 (say 9, for example), she would not have know which odd number Sage held. (In our example, he could have had a 1 and Tim a 4, or he could have had a 3 and Tim a 2). Tim, understanding all of this, therefore knew their numbers.

  If there are exactly N liars, then the last N men are truthtellers, leaving the first (12-N) men to be liars. Therefore, N=12-N, N=6. The first 6 men are liars, and the last 6 men are truthtellers.

  She wants to say, "6." The series of numbers she will say is 6, 17, 28, 39, 50. Since she wants to say 50, she needs Henry to say a number between 40 and 49. Therefore, she wants to say 39. Knowing she wants to say 39, she needs Henry to say a number between 29 and 38. So she wants to say 28. Following this same logic recursively, she will want to say 17, and she will want to say 6 to start the game, and be assured victory.

Mike is the victim. If Mike's sister and Jack's sister are the same person, then (a) reads "Lily argued exactly once with (Mike or Jack)" and (b) reads "Lily argued twice with (the same Mike or Jack)". These two statements contradict each other. Therefore Mike's sister must be different from Jack's sister. If this is the case then there are two possibilities: 1) Jack and Carol are brother and sister. Mike and Lily are brother and sister. Jack is married to Lily and Mike is married to Carol. If this is the case, then (a) reads "Carol argued once with Mike" and (b) reads "Lily argued twice with herself", leaving Jack dead. Since Lily did not likely argue with herself, this is not the solution. 2) Jack and Lily are brother and sister. Mike and Carol are brother and sister. Jack is married to Carol and Lily is married to Mike. If this is the case, then (a) reads "Lily argued once with Jack" a

  Cyclops. Why? Because the other names don't have any repeated letters

  Left: Tom, yellow, 5; Middle: Jim, red, 7; Right: Sam, blue, 6. From clue 4, the red suit's number must be greater than Sam's number, and must be odd. By trial and error, the possible combinations are (3,2), (5,4), (7,3), (7,6). From clue 1, the square can only be 324 or 576, where 3 or 7 is the red suit's number. From clue 3, the square must be 576, and 5 is the yellow suit's number and 6 the blue one. From clue 2, Tom must be the 5 and Jim the 7.

  Suppose that a second pair of hands turns together with the wrong pair of hands, but then in the correct way. When the wrong pair is in the same position as the correct pair, this means that the time is shown in the right way. First look at the hour-hands that are at the twelve. One turns with the correct speed, the other with a speed that is twelve times as small. These two hands are in the same position again when the 'slow' hand has progressed x minutes. The fast hand then has progressed 60+x minutes. For the time x that passed, then holds: (60+x)/12 = x. This means that x = 5 5/11 minutes. For the minutes-hands that start at six holds the same. The confused clock therefore shows the correct time again at 5 5/11 minutes past 7. 

  Back page of 24 and 41 .. those are 23 and 42 Also joining pages of 24 and 41 those are 37 and 20 also their back pages 38 and 19. Answer is : 19 20 23 37 38 42

  A and B cross first using up 2 minutes. A comes back making it 3 C and D cross making it 13 minutes then B crosses back over making it 15 minutes. And finally A and B cross together to make it 17 minutes!

  By putting 8 boxes in a box, the total number of empty boxes increases by 8 - 1 = 7. If we call x the number of times that 8 boxes have been put in a box, we know that 11 + 7x = 102. It follows that x=13. In total, 11 + 13 × 8 = 115 boxes have been used.

  18 days; Working backwords in 19 days it will be half of full i.e. 648 ft sq. In 18 days it will 324 ft sq.

  0 because in that case only Edvard is right

  B says: "Suppose I have red-red. A would have said on her second turn: 'I see that B has red-red. If I also have red-red, then all four reds would be used, and C would have realized that she had green-green. But C didn't, so I don't have red-red. Suppose I have green-green. In that case, C would have realized that if she had red-red, I would have seen four reds and I would have answered that I had green-green on my first turn. On the other hand, if she also has green-green [we assume that A can see C; this line is only for completeness], then B would have seen four greens and she would have answered that she had two reds. So C would have realized that, if I have green-green and B has red-red, and if neither of us answered on our first turn, then she must have green-red. "'But she didn't. So I can't have green-green either, and if I can't have green-green or red-red, then I must have green-red.'

  The treasurer's plan was to drink a weak poison prior to the meeting with the king, and then he would drink the pharmacist's strong poison, which would neutralize the weak poison. As his own poison he would bring water, which will have no effect on him, but the pharmacist who would drink the water, and then his poison would surely die. When the pharmacist figured out this plan, he decided to bring water as well. So the treasurer who drank poison earlier, drank the pharmacist's water, then his own water, and died of the poison he drank before. The pharmacist would drink only water, so nothing will happen to him. And because both of them brought the king water, he didn't get a strong poison like he wanted.

  The first person must take either stool 9 or 17 (because of symmetry, it doesn't matter which). Assume they pick seat 9. The next person will pick seat 25, since it is the furthest from seat 9. The next two people will take Seats one and 17. The next three will occupy 5, 13, and 21. The next six will occupy 3, 7, 11, 15, 19, and 23. This seats the maximum of 13 people, and no one is sitting next to another person. If a seat other than 9 or 17 is chosen first, the total bar patrons will be less than 13.

  White. The king would not select two white hats and one black hat. This would mean two princes would see one black hat and one white hat. You would be at a disadvantage if you were the only prince wearing a black hat. If you were wearing the black hat, it would not take long for one of the other princes to deduce he was wearing a white hat. If an intelligent prince saw a white hat and a black hat, he would eventually realize that the king would never select two black hats and one white hat. Any prince seeing two black hats would instantly know he was wearing a white hat. Therefore if a prince can see one black hat, he can work out he is wearing white. Therefore the only fair test is for all three princes to be wearing white hats. After waiting some time just to be sure, you can safely assert you are wearing a white hat.

  Harry would see seven "Shen Express" trains go past in the opposite direction. Let's say that Harry travelled on the "Shen Express" on Wednesday 10:00PM from Ronville. The first train he would have crossed is the one scheduled to arrive at Ronville at 11:30 PM the same day i.e. the one that left Gothamville at 10:00 PM on last Sunday. He will pass 7 other trains before arriving at his destination. The first, he will see approximately 45 minutes after departure. Then he will see one train approximately every 12 hours for three days. This is because the trains leave from each end of the track once every 24 hours travelling toward each other. Also, he would have crossed the last train just before reaching Gothamville on Saturday. Thus, Harry must have crossed 7 "Shen Expresses" during his journey.

   Ans : 16 Name the 10 people with letters: A, B, C, ... and so on. A through F each call any one of G, H, I, or J (it doesn't matter which one). That makes six calls so far. Then G calls H and I calls J, after which G calls I and H calls J. Now we've used 10 calls, and G, H, I, and J all know everything. Finally, each of A through F is called by someone in G through J - 6 more calls to get everyone knowing everything. 16 calls in all.

  If today is January 1st, and December 31st was Chris' birthday and she turned 8, and on December 30 Chris was 7, and this year Chris will turn 9, then next year Chris is turning 10.

  6210001000 There are 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights, and 0 nines.

  89 eggs. In order to see this, look at the smallest way possible to guarantee a win; you leave me with 101 eggs. If I take 1 egg, you take all 100; if I take 100 eggs, you take the last one. Therefore you would always win. If you try with 202, I can take from 1 to 100, so there is 201 to 102 left. Then you would take 100 or 1 egg(s) in order to leave me with 101 again. By simple analysis, you must take 101 eggs over two rounds in order to get down to 101. 19*101 = 1919, so we must take 2008-1919 = 89 eggs to start off with a multiple of 101.

   If you look at the question, you will see that the large letters each match up with the beginning of a word. The C is for Characters and the T for Two.

  If there is only one monk with red eyes, then he sees all the others are brown-eyed, so he must be the red-eyed one. He kills himself the first night. If there are two monks with red eyes, then each sees one monk with red eyes and reasons that if this other monk is the only monk with red eyes, he will kill himself the first night. Neither monk kills himself the first night, so they each reason that they must have red eyes too. Both kill themselves the second night. If there are three, each expects the other two to commit suicide the second night. This doesn't happen, so each deducts that he must be a third, and the suicides happen the third night. Extends to four, five, etc. If the suicides happened n midnights after the tourist's remark, then there are n monks with red eyes.

  16 ounces (about half of pot A).  The amount of tea that can be kept within each pot is determined by the height of the spout opening.  The tea level cannot rise above the spout opening since any extra tea would merely spill out from the spout.  A simple visual estimate would conclude that the spout of teapot B is approximately half the height of that of teapot A, therefore providing only half of the capacity, or 16 ounces. 

  1. The time is 2:12. 2. The time is 9:48. The hour hand is exactly on a line every 12 minutes, so we only need to look at times ending in 00, 12, 24, 36, and 48. The hands exactly overlap at 12:00. They are 5 lines apart at 11:00 and 1:00, and farther apart at other hours. At times ending in 12, the hands are closest at 2:12, where they are 1 line apart. At times ending in 24, the hands are closest at 4:24, where they are 2 lines apart. At times ending in 36, the hands are closest at 7:36, where they are 2 lines apart. At times ending in 48, the hands are closest at 9:48, where they are 1 line apart.

Create a subgroup of any 128 pennies. Then flip over all 128. That group of 128 and the group of all the remaining pennies will have the same number of heads facing up. This works because the "tails" that you grab while making your subset of 128 will equal the "heads" that are left in the original pile. Once you flip your 128 coins over, these "tails" will turn into "heads" and the two groups will have a matching number of heads-up coins.

    The number 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0 on a US keyboard when you hold the shift button down.

If the proper order is chosen, you can determine the breaking point with a maximum of 14 drops. Here's how to do it: First drop the first ball from the 14th floor. If it breaks you can determine the exact breaking point with the other ball in at most 13 more droppings, starting at the bottom and going up one floor at a time. If the first ball survives the 14 floor drop then drop it again from the 27th (14+13) floor. If it breaks you can determine the exact breaking point with at most 12 more droppings. If the first ball survives the 27 floor drop then drop it again from the 39th (14+13+12) floor. If it breaks you can determine the exact breaking point with at most 11 more droppings. Keep repeating this process always going up one less floor than the last dropping until the first ball breaks. If it breaks on the xth dropping you will only need at most 14-x more droppings with the second ball to find the breaking point. By the 11th dropping of the first ball

   Since only 1 option is correct, it can only be either odd or even so C and D are eliminated, and a number which is not prime and composite between 1 and 20 is 1. So mike number is 1.

The prisoner told them: "You will hang me."

If there are two pirates left (#4 & #5), #4 has no options. No matter what he proposes, pirate #5 will disagree, resulting in a 1-1 vote (no majority). #5 will kill #4 and will keep all of the gold. Now say there are 3 pirates left. #4 has to agree with whatever #3 decides, because if he doesn't #3 will be killed (because #5 won't vote for #3's proposal no matter what it is). #3 will just propose that he keep all of the gold and will get a 2-1 vote in his favor. Now if there are 4 pirates left: #3 won't vote for #2's proposal because if #2's fails, #3 will get all of the gold. #4 and #5 know that they will get nothing if the decision goes to #3, so they will vote for #2's proposal if he gives them one gold piece each. Therefore, #2 would keep 998 gold, and #4 and #5 would each get one gold. So let's wrap this up: Pirate #1 needs 2 other votes. He will not get a vote from #2 because #2 will get 998 gold if #1's plan fails. #1 offers #3 o

  You place a quarter right in the center of the table. After that, whenever the devil places a quarter on the table, mimic his placement on the opposite side of the table. If he has a place to place a quarter, so will you. The devil will run out of places to put a quarter before you do.

What is etestkart?   It’s an online test platform for students of various domains such as Engineering, Medical, MBA, etc. The website facilitates students with free tests, where they can give tests and have complete analysis of their strength and weakness. The tests have been divided into four categories - Topic-wise, Module, Schedule, Random-question tests Topic-wise : This category is not based on any particular competitive exam. It focuses on various topics of particular subjects. The topics are, further, divided into three levels - easy, moderate and difficult. Module : This test focuses on the latest pattern of various competitive exams and has many mock tests along with the previous years’ question papers. Schedule : As the name suggests, this test is scheduled by us every month and the result is analyzed with all the users in India appearing for this test. The test can be attempted by the user at a convenient time during the window provided by us . Ra