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Sam has forgotten his friend’s seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number nine appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed?

  1. 1000
  2. 2430
  3. 3402
  4. 3006

 

asked in Quantitative Aptitude by (7.9k points) 115 216 353 | 120 views

1 Answer

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      635_ _ _ _

OR

      674_ _ _ _

First 3 digit 4th digit 5th digit 6th digit 7th digit No. of trial & error
635 1 way (can place only no. 9) 9 ways (can place 0 to 8 no.s) 9 ways (can place 0 to 8 no.s) 4 ways (can place 1/3/5/7) 1*9*9*4 = 324
635 9 ways (can place 0 to 8 no.s) 1 way (can place only no. 9) 9 ways (can place 0 to 8 no.s) 4 ways (can place 1/3/5/7) 9*1*9*4 = 324
635 9 ways (can place 0 to 8 no.s) 9 ways (can place 0 to 8 no.s) 1 way (can place only no. 9) 4 ways (can place 1/3/5/7) 9*9*1*4 = 324
635 9 ways (can place 0 to 8 no.s) 9 ways (can place 0 to 8 no.s) 9 ways (can place 0 to 8 no.s) 1 way (can place only no. 9) 9*9*9*1 = 729

The total no. of Trial and Error process with 635 as prefix is (324 + 324 + 324 + 729) = 1701.

This 1701 combinations will repeat with 674 as prefix also.

Minimum no. of trials = 1701 + 1701

                                  = 3402 (option 3)

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