81. One day Rapunzel meets Dwarf and Byte in the Forest of forgetfulness. She knows that Dwarf lies on Mondays, Tuesdays and Wednesdays, and tells the truth on the other days of the week. Byte, on the other hand, lies on Thursdays, Fridays and Saturdays, but tells the truth on the other days of the week. Now they make the following statements to Rapunzel - Dwarf: Yesterday was one of those days when I lie. Byte: Yesterday was one of those days when I lie too. What day is it?
Answer:
Explanation:Answer is Thursday
As the condition says that the dwarf lie on Monday, Tuesday & Wednesday so they will speak truth that they lied on Thursday also on the other side the bytes speak lie on Thursday and will tell that they spoke lie on Wednesday. So the day has to be Thursday as no other option satisfies the condition also.
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82. If a three digit number 'abc' has 3 factors, how many factors does the 6-digit number 'abcabc' have?
Answer:
Explanation:abc' has exactly 3 factors, so 'abc' should be square of a prime number. (This is an important inference, please remember this).
Any number of the form paqbrc will have (a + 1) (b + 1) (c + 1) factors, where p, q, r are prime. So, if a number has 3 factors, its prime factorization has to be p2.
'abcabc' = 'abc' * 1001 or abc * 7 * 11 * 13 (again, this is a critical idea to remember)
Now, 'abc' has to be square of a prime number. It can be either 121 or 169 (square of either 11 or 13) or it can be the square of some other prime number.
When abc = 121 or 169, then 'abcabc' is of the form p3q1r1 1, which should have 4 * 2 * 2 = 16 factors. When 'abc' = square of any other prime number (say 172 which is 289), then 'abcabc' is of the form
p1q1r1s2 , which should have 2 * 2 * 2 * 3 = 24 factors So, 'abcabc' will have either 16 factors or 24 factors.
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83. Alice and Bob play the following coins-on-a-stack game. 20 coins are stacked one above the other. One of them is a special (gold) coin and the rest are ordinary coins. The goal is to bring the gold coin to the top by repeatedly moving the topmost coin to another position in the stack. Alice starts and the players take turns. A turn consists of moving the coin on the top to a position i below the top coin (0 ≤ i ≤ 20). We will call this an i-move (thus a 0-move implies doing nothing). The proviso is that an i-move cannot be repeated; for example once a player makes a 2-move, on subsequent turns neither player can make a 2-move. If the gold coin happens to be on top when it's a player's turn then the player wins the game. Initially, the gold coinis the third coin from the top. Then
Answer:
Explanation:In order to win, Alice's first move should be a 1- move.Because there are two possibilities after 1- move by Alice:-
1. When bob takes 0-move, then the coin config is same. But now Alice can neither take 1-move nor 0- move, so she has to take 2 or greater move (let it be 2 moves).Now we have only one coin above the gold coin. Bob can neither take 1-move nor 0-move nor 2- moves. So she has to take 3 or greater move. At the move greater than 2-move brings gold coin on top that makes Alice win.
2. Now bob takes 2-move after Alice's first move. Now there is only one coin above the gold coin. So Alice takes 0-move.Now we have only one coin above the gold coin. Bob can neither take 1-move nor 0-move nor 2-move. So she has to take 3 or greater move.
But the move greater then 2-move brings gold coin on top that makes Alice win.
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84. When 40! Is expressed in base 8 form, what is the last non-zero digit in the base 8 expansion?
Answer:
Explanation:We need to find the largest power of 8 that divides 40!.
We need to find the largest power of 2 that divides 40! This is given by (40/2) and then successive division by 2. = 20 + 10 + 5 + 2 + 1 = 38
So, 238 divide 40! Or, (23)12 × 22 divides 40! (23)12 divides the number or the base 8
representation ends with 12 zeroes. Now, the base 8
representation of this number will be some (abcd…n)8
× (1000000000000)8. Now, (abcd…n)8 does not end in 0 and is a multiple of 22. The last digit has to be 4.
The last non-zero digit is 4.
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85. The IT giant Tirnop has recently crossed a head count of 150000 and earning of $7 billion. As one of the forerunners in the technology front, Tirnop continues to lead the way in products and services in India. At Tirnop, all programmers are equal in every respect. They receive identical salaries and also write code at the same rate Suppose 16 such programmers take 16 minutes to write 16 lines of code in total. How many lines of code can be written by 96 programmers in 96 minutes? The IT giant Tirnop has recently crossed a head count of 150000 and earning of $7 billion. As one of the forerunners in the technology front, Tirnop continues to lead the way in products and services in India. At Tirnop, all programmers are equal in every respect. They receive identical salaries and also write code at the same rate Suppose 16 such programmers take 16 minutes to write 16 lines of code in total. How many lines of code can be written by 96 programmers in 96 minutes?
Answer:
Explanation:(16 men)*(16 min)/ (16 lines) = (96 men) (96 min)/x (lines)
Hence x=576
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86. Abhishek starts to paint a fence on one day. On the second day, two more friends of Abhishek join him. On the third day 3 more friends of him join him and so on. If the fence is completely painted this way in exactly 20 days, then find the number of days in which 10 girls painting together can paint the fence completely, given that every girl can paint twice as fast as Abhishek and his friends (Boys)? (Assume that the friends of Abhishek are all boys).
Answer:
Explanation:Number of men working on first day = 1 Number of men working on second day = 3
Number of men working on second day = 6 and so on.. Total number of boys till the end of the work = [n(n + 1) (n + 2)] / 6
= [20 x 21 x 22] / 6 = 1540
Given that every girl paints twice as fast as Abhishek's friends.
Hence, 20 girls work is being done.
Thus, the number of days taken to paint the fence = 1540/20 = 77.
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87. Events A, B, C are mutually exclusive events such that: The set of possible values of x are in the interval
88. Fill dirt or fill soil is usually the sub-soil removed from an excavation site and is used to level a place or create artificial mounds. If the average density of sub- soil removed from a site is 3gm/cu cm and it weight 400 kg. How many hemispherical pits each of volume 240 cubic cm, can this sub-soil fill?
Answer:
Explanation:3 gm per cm cube and there are hemispheric pits of capacity 240 cm cube
So each hemisphere can contain 3*240=720 cm cube soil.
So total of 400 kg i.e. 400*1000=400000 gms will be held in 400000/720=555.55 so approximately 555.
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89. Kailash faces towards North. Turning to his right, he walks 25 meters. He then turns to his left and walks 30 meters. Next, he moves 25 meters to his right. He then turns to his right again and walks 55 meters. Finally he turns to the right and moves 40 meters. In which direction is he now from his starting point?
Answer:
Explanation:Facing north turns towards right and walks 25km=east Then turns towards left walks 30 m=north
Again he turns towards his right and walks 25m=north east
He moves towards his right and walks 55m=south Finally he turns right and walks 40m=south east.
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90. A group of friends Tom, Tim, Dick, Diana, Harry, and Harriet go out to a fair three hundred meters from the McDonalds which is five Km away. They see a weighing machine and decide to have some fun. However the girls refuse to step on the weighing machine. So Tom, Dick and harry, weigh themselves in a particular order. First Tom, Dick, and Harry weigh themselves individually and then tom and Dick, Dick and Harry, Tom and Harry and then Tom, Dick and harry together respectively. The recorded weight for the last measure is 158 kgs. The average of all the 7 measures is
Answer:
Explanation:Let tom,dick and harry be a, b and c.
Now there is totally 7 rounds of weight measure. First Tom, Dick, and Harry weigh themselves individually and then Tom and Dick, Dick and Harry,
Tom and Harry and then Tom, Dick and Harry together respectively.
Sum of total 7 rounds written as, a+b+c+(a+b)+(b+c)+(c+a)+(a+b+c) = 4(a+b+c) 4*158=632
Average of 7 weighing is 632/7=90.29 kg
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