3 kinds of ice cream, chocolate, vanilla, and strawberry are to be selected. In a survey of 60 people about the preference of the ice cream, 3/5 will put the vanilla on the last position, 1/3 put the vanilla ahead of the chocolate, and 1/10 put the vanilla ahead of strawberry. How many people will put the strawberry on the first position?
A. 2
B. 4
C. 6
D. 12
E. 24
Dear i_m_wat_i_m, can you explain the solution.
can some one please explain the solution??
OK then, I can give you people a hint only:
If from the remaining 24 people, 20 put the vanilla ahead of the chocolate, and 6 put the vanilla ahead of strawberry, then only 2 must have put vanilla at first position, leaving at least 18 people putting strawberry at first position (needless to know that at least 4 people will put chocolate at first position). Now see choices and apply common sense.
dear Sanju, If I am correct, the answer should be 24. The following is the explanation.
The possibilities of arranging Strawberry (S), Vanilla (V) and Chocolate© are :- SVC+SCV+VSC+VCS+CSV+CVS=60.—(i) 1) Vanilla as third option are 36. That means CSV+SCV = 36. —(ii) Hence, VSC+VCS+CVS+SVC=24. —> (iii) 2) Vanilla ahead of Chocolate are 20. That means, VSC+SVC+VCS = 20. —> (iv) Hence, CVS+CSV+SCV = 40—> (v) Now, (iii) – (iv) gives us CVS = 4. —(vi)
3) Vanilla ahead of strawberry are 6. That means VSC+VCS+CVS = 6. —> (vii) Now substitute (vi) in (vii) and we get VSC + VCS = 2. Now substitue this in (iii). We get SVC = 18. Therefore VCS+VCS = 2. CVS = 4. SVC = 18. and SCV+CSV = 36. Since no other addl. info is available, and the required answer is >= 18, the answer is 24.
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