400 Puzzles And Answers For Interview Pdf Converter
How Do You Print an N400 Citizenship Form in PDF. The N-400 form is. Must be filed with the N-400 or at the time of the applicant's interview.
400 puzzles-and-answers-for-interview. 1.
400 puzzles -collected by Bipul Kumar Bal 01. Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided. Answer for more papers please visit www.downloadmela.com www.downloadm ela.com A-PDF Watermark DEMO: Purchase from www.A-PDF.com to remove the watermark. 18 Assume that initial there were 3.X bullets. So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X - 4) bullets each. But it is given that, after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X Therefore, the equation is 3. (X - 4) = X 3.
X - 12 = X 2. X = 12 X = 6 Therefore the total bullets before division is = 3. X = 18 2. Find sum of digits of D. Let A= 19991999 for more papers please visit www.downloadmela.com www.downloadm ela.com. B = sum of digits of A C = sum of digits of B D = sum of digits of C (HINT: A = B = C = D (mod 9)) Answer the sum of the digits of D is 1. Let E = sum of digits of D.
It follows from the hint that A = E (mod 9) for more papers please visit www.downloadmela.com www.downloadm ela.com. Consider, A = 19991999 1). On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
On the nth and last day, the remaining n medals were awarded. How many days did the contest last, and how many medals were awarded altogether? Answer Total 36 medals were awarded and the contest was for 6 days. On day 1: Medals awarded = (1 + 35/7) = 6: Remaining 30 medals On day 2: Medals awarded = (2 + 28/7) = 6: Remaining 24 medals On day 3: Medals awarded = (3 + 21/7) = 6: Remaining 18 medals On day 4: Medals awarded = (4 + 14/7) = 6: Remaining 12 medals On day 5: Medals awarded = (5 +7/7) = 6: Remaining 6 medals On day 6: Medals awarded 6 for more papers please visit www.downloadmela.com www.downloadm ela.com.
I got this answer by writing small program. If anyone know any other simpler method, do submit it. A number of 9 digits has the following properties:.
The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. The number formed from the first n digits is divisible by n, 2B, B-C, C-A 2. A-B, B-C, C-B 3. A-B, B-A, C-A 4. A-B, B-A, C-B 5. A-C, C-B, B-A 6.
A-C, C-B, B-C 7. A-C, C-A, B-A 8. A-C, C-A, B-C Out of which, there are only two cases under which the ants won't collide:. A-B, B-C, C-A. A-C, C-B, B-A for more papers please visit www.downloadmela.com www.downloadm ela.com.
40. No question here. Answer There are total 8 such series: 1. Sum of 2000 numbers starting from -999 i.e. Summation of numbers from -999 to 1000. (-999) + (-998) + (-997) +.
+ (-1) + 0 + 1 + 2 +. + 997 + 998 + 999 + 1000 = 1000 2. Sum of 400 numbers starting from -197 i.e.
Summation of numbers from -197 to 202. (-197) + (-196) + (-195) +. + (-1) + 0 + 1 + 2 +. + 199 + 200 + 201 + 202 = 1000 3. Sum of 125 numbers starting from -54 i.e. Summation of numbers from -54 to 70. (-54) + (-53) + (-52) +.
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+ (-1) + 0 + 1 + 2 +. + 68 + 69 + 70 = 1000 4. Sum of 80 numbers starting from -27 i.e. Summation of numbers from - 27 to 52. (-27) + (-26) + (-25) +.
+ (-1) + 0 + 1 + 2 +. + 50 + 51 + 52 = 1000 5. Sum of 25 numbers starting from 28 i.e. Summation of numbers from 28 to 52. 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000 6. Sum of 16 numbers starting from 55 i.e.
Summation of numbers from 55 to 70. 55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000 7. Sum of 5 numbers starting from 198 i.e. Summation of numbers from 198 to 202. 198 + 199 + 200 +201 + 202 = 1000 8. Sum of 1 number starting from 1000. 1000 = 1000 for more papers please visit www.downloadmela.com www.downloadm ela.com.
41. There is a 4-character code, with 2 of them being letters and the other 2 being numbers.
How many maximum attempts would be necessary to find the correct code? Note that the code is case-sensitive. Answer The maximum number of attempts required are 16,22,400 There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52.52.10.10 ways.
These 4 characters can be arranged in 4 C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. They must be letters) and the other 2 are distinguished another way (i.e.
They must be numbers). Consider an example: Let's assume that @ represents letter and # represents number. The 6 possible ways of arranging them are: @@##, @#@#, @##@, #@@#, #@#@, ##@@ Hence, the required answer is = 52.52.10.10.6 = 16,22,400 attempts = 1.6 million approx. Thanks to Tim Sanders for opening BrainVista's brain!!! For more papers please visit www.downloadmela.com www.downloadm ela.com.
42. How many possible combinations are there in a 3x3x3 rubics cube? In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)? How many for a 4x4x4 cube? Submitted Answer There are 4.3252. 10^19 possible combinations for 3x3x3 Rubics and 7.4012.
10^45 possible combinations for 4x4x4 Rubics. Let's consider 3x3x3 Rubics first. There are 8 corner cubes, which can be arranged in 8!
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible.
Thus, total ways corner cubes can be placed = (8!). (3^8)/8 = (8!).
(3^7) Similarly, 12 edge cubes can be arranged in 12! Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!). (2^12)/2 = (12!). (2^11) Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes.
It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable for more papers please visit www.downloadmela.com www.downloadm ela.com. choice for them (not 2!). Hence, we must divide by 2.
Total different possible combinations are = (8!). (3^7). (12!). (2^11) / 2 = (8!).
(3^7). (12!). (2^10) = 4.3252.
10^19 Similarly, for 4x4x4 Rubics total different possible combinations are = (8!). (3^7). (24!). (24!) / (4!^6) / 24 = 7.4011968. 10^45 Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6).
You can switch 2 cubes without affecting the rest of the combination as 4.4.4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.
Substitute digits for the letters to make the following relation true. N E V E R L E A V E + M E - A L O N E Note that the leftmost letter can't be zero in any word. Also, there must be a one- to-one mapping between digits and letters. If you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.
For more papers please visit www.downloadmela.com www.downloadm ela.com. Answer A tough one!!! Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to solve it. Now use trial-n-error method. N E V E R 2 1 4 1 9 L E A V E 3 1 5 4 1 + M E + 6 1 - - A L O N E 5 3 0 2 1 44. One of the four people - Mr.
Nursing Questions And Answers For Interview
Clinton, his wife Monika, their son Mandy and their daughter Cindy - is a singer and another is a dancer. Clinton is older than his wife and Mandy is older than his sister. If the singer and the dancer are the same sex, then the dancer is older than the singer.
If neither the singer nor the dancer is the parent of the other, then the singer is older than the dancer. If the singer is a man, then the singer and the dancer are the same age.
If the singer and the dancer are of opposite sex then the man is older than the woman. If the dancer is a woman, then the dancer is older than the singer. Whose occupation do you know? And what is his/her occupation? Answer for more papers please visit www.downloadmela.com www.downloadm ela.com. Cindy is the Singer. Clinton or Monika is the Dancer.
From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman. CASE I: Singer is a woman and Dancer is also a woman Then, the dancer is Monika and the singer is Cindy. CASE II: Singer is a woman and Dancer is also a man Then, the dancer is Mr.
Aptitude Questions And Answers For Interview
Clinton and the singer is Cindy. In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.
There are 20 people in your applicant pool, including 5 pairs of identical twins. If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins? (Needless to say, this could cause trouble;)) Submitted Answer The probability to hire 5 people with at least 1 pair of identical twins is 25.28% 5 people from the 20 people can be hired in 20C5 = 15504 ways.
Now, divide 20 people into two groups of 10 people each: for more papers please visit www.downloadmela.com www.downloadm ela.com. G1 - with all twins G2 - with all people other than twins Let's find out all possible ways to hire 5 people without a single pair of indentical twins.