1. A binary number with four digits has a maximum value of 15.



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MCQ-> Following is given a set of digits and the corresponding letter code of each digit followed by certain conditions for coding. In each question one number consisting six digits followed by four combinations of letter codes. You have to find out which of the combination of letter codes represents the set of digits based on the above codes and the conditions given below and mark your answer accordingly. Otherwise give answer e:, i.e., None of these. Conditions : (I) If both the first and the last digits of the number are odd digits then both are to be coded as I. (II) If both the first and last digits of the number are even digits then both are to be coded as Y.726395
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MCQ-> The English alphabet is divided into five groups. Each group starts with the vowel and the consonants immediately following that vowel and the consonants immediately following that vowel are included in that group. Thus, the letters A, B, C, D will be in the first group, the letters E, F, G, H will be in the second group and so on. The value of the first group is fixed as 10, the second group as 20 and so on. The value of the last group is fixed as 50. In a group, the value of each letter will be the value of that group. To calculate the value of a word, you should give the same value of each of the letters as the value of the group to which a particular letter belongs and then add all the letters of the word: If all the letters in the word belong to one group only, then the value of that word will be equal to the product of the number of letters in the word and the value of the group to which the letters belong. However, if the letters of the words belong to different groups, then first write the value of all the letters. The value of the word would be equal to the sum of the value of the first letter and double the sum of the values of the remaining letters.For Example : The value of word ‘CAB’ will be equal to 10 + 10 + 10 = 30, because all the three letters (the first letter and the remaining two) belong to the first group and so the value of each letter is 10. The value of letter BUT = $$10 + 2 \times 40 + 2 \times 50 = 190$$ because the value of first letter B is 10, the value of T = 2 $$\times$$ 40 (T belongs to the fourth group) and the value of U = 2 $$\times$$ 50 (U belongs to the fifth group). Now calculate the value of each word given in questions 161 to 165 :AGE
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MCQ-> In each questions below is given a group of digits followed by four combinations of letters or symbols numbered a, b, c and d. You have to find out which of the combinations correctly represents the group of digits based on the coding system and the conditions given below and mark the number of that combination as your answer. If none of the combinations correctly represents the group of digits, mark e:, i.e. None of these’ as your answer. Conditions : (i) If the first digit is odd and the last digit is even, the codes for the first and last digits are to be interchanged. (ii) If both the first and the last digits are even, both are to be coded as *. (iii) If both the first and the last digits are odd, both are to be coded as $.215349
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MCQ->In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number?....
MCQ-> Mathematicians are assigned a number called Erdos number (named after the famous mathematician, Paul Erdos). Only Paul Erdos himself has an Erdos number of zero. Any mathematician who has written a research paper with Erdos has an Erdos number of 1.For other mathematicians, the calculation of his/her Erdos number is illustrated below:Suppose that a mathematician X has co-authored papers with several other mathematicians. 'From among them, mathematician Y has the smallest Erdos number. Let the Erdos number of Y be y. Then X has an Erdos number of y+1. Hence any mathematician with no co-authorship chain connected to Erdos has an Erdos number of infinity. :In a seven day long mini-conference organized in memory of Paul Erdos, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdos number. Nobody had an Erdos number less than that of F.On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdos number of the group of eight mathematicians to 3. The Erdos numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdos number of the group of eight to as low as 3.• At the end of the third day, five members of this group had identical Erdos numbers while the other three had Erdos numbers distinct from each other.• On the fifth day, E co-authored a paper with F which reduced the group's average Erdos number by 0.5. The Erdos numbers of the remaining six were unchanged with the writing of this paper.• No other paper was written during the conference.The person having the largest Erdos number at the end of the conference must have had Erdos number (at that time):
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