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 Aptitude Interview Questions And Answers 56. How many number of times will the digit ‘7' be written when listing the integers from 1 to 1000? Sol:7 does not occur in 1000. So we have to count the number of times it appears between 1and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etcThis means that 7 is one of the digits and the remaining two digits will be any of the  other 9 digits (i.e 0 to 9 with the exception of 7) You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.In each of these numbers, 7 is written once. Therefore, 243 times. 2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers. In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times. 3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.       Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300. 57. There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song? Sol:  There are 2n ways of choosing ‘n’ objects. For e.g. if n = 3, then the three objects can be chosen in the following 23 ways - 3C0 ways of choosing none of the three, 3C1 ways of choosing one out of the three, 3C2 ways of choosing two out of the three and 3C3 ways of choosing all three. In the given problem, there are 5 Rock songs. We can choose them in 25 ways. However, as the problem states that the case where you do not choose a Rock song does not exist (at least one rock song has to be selected), it can be done in 25 - 1 = 32 - 1 = 31 ways. Similarly, the 6 Carnatic songs, choosing at least one, can be selected in 26 - 1 = 64 - 1 = 63 ways. And the 3 Indi pop can be selected in 23 = 8 ways. Here the option of not selecting even one Indi Pop is allowed. Therefore, the total number of combinations = 31 * 63 * 8 = 15624. 58. A takes 3 min 45 seconds to complete a kilometre. B takes 4 minutes to complete the same 1 km track. If A and B were to participate in a race of 2 kms, how much start can A give B in terms of distance? Sol:     A can give B a start of 15 seconds in a km race.                      B takes 4 minutes to run a km. i.e = 250 m/min = m/sec Therefore, B will cover a distance of = 62.5 meters in 15 seconds. The start that A can give B in a km race therefore, is 62.5 meters, the distance that B run in 15 seconds. Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or 30 seconds. 59. P can give Q a start of 20 seconds in a kilometer race. P can give R a start of 200 meters in the same kilometer race. And Q can give R a start of 20 seconds in the same kilometer race. How long does P take to run the kilometer? Sol: P can give Q a start of 20 seconds in a kilometer race. So, if Q takes 'x' seconds to run a kilometer, then P will take x – 20 seconds to run the kilometer. Q can give R a start of 20 seconds in a kilometer race. So, if R takes 'y' seconds to run a kilometer, then Q will take y – 20 seconds to run the kilometer. We know Q takes x seconds to run a kilometer Therefore, x = y – 20 Therefore, P will take x – 20 = y – 20 – 20 = y – 40 seconds to run a kilometer. i.e. P can give R a start of 40 seconds in a kilometer race, as R takes y seconds to run a kilometer and P takes only y – 40 seconds to run the kilometer. We also know that P can give R a start 200 meters in a km race. This essentially means that R runs 200 meters in 40 seconds. Therefore, R will take 200 seconds to run a km. If R takes 200 seconds to run a km, then P will take 200 – 40 = 160 seconds to run a km. 60. How many squares can be formed using the checkered 1 * 1 squares in a normal chessboard? Solution: The number of squares that can be formed using the 1 * 1 checkered squares of a chess board are given by the relation 12 + 22 + 32 + 42 + ... + 82 = 204 .  