There are 76 persons. 53 can read hindu,46 can read times,39 can read deccan and 15 can read all.if 22 can read hindu and deccan and 23 can read deccan and times then what is the number of persons who read only times and hindu
I would like to share my answer and approach, even if I may not take the accenture exam.Those who speak Hindu and Deccan (HD) include those who speak all the languages. When we subtract (22 (HD) -15 (ALL)), we get the people who speak only the HD combination. We do the same thing (23 (DT) -15 (ALL)) for people who speak only Deccan and Times (DT). This leads us to a subtraction of these people from the greater whole. For the Hindu speakers, we get (53 (H) -15 (ALL) -7 (HD)=) 31 candidates who could either be only Hindu speakers or Hindu-Times speakers. For the Times speakers, we get (46 (T) -15 (ALL) -8 (DT)=) 23 candidates. We figure that there is some intersection between these figures. So we subtract those non-candidates from the running by subtracting all Deccan speakers from the total (76 (tot) -39 (D)=37(left)). The total is the number of people we have left who can speak either Hindu, Times or Hindu-Times only. We add the possible candidates up (23+21) and from the total, subtract the people we have left (54-37(left)). The answer I got is 17. When we add all of the kinds of speakers up:ALL DECCAN HINDU TIMES DEC/HIN DEC/TIM HIN/TIM TOTAL15 9 14 6 7 8 17 76You get the correct total.
I calculate the answer to be 32 !!! = 15 (given by the problem) + 17 (calcd as 76-23= 37 = (31-y) + (23 -y) + y , where y represented the missing piece between H&T; this piece would be added to 15, so as to represent the ENTIRE H&T section - which is the question..
I used part Venn diagrams to portay it, and then equations to do the final calc (its been a while since I took a discrete math class..so my answer wasnt elegant).
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