Problem Solving

ketkoag wrote: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

sanju09 wrote:

ketkoag wrote: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

Among the 60 people surveyed, if 36 put vanilla in the last then 24 must put vanilla on first or second position; and out of those 24, 20 put the vanilla ahead of the chocolate and 6 put the vanilla ahead of strawberry, then [spoiler]2[/spoiler] of them must put vanilla ahead of both chocolate and strawberry, or on the first position. Mine [spoiler]A[/spoiler].

sir here s the sol. but its too long first, we will observe that the ppl putting vanilla ahead of strawberry are automatically eliminated in our search, since u cant put something ahead of the first position

so, 1/10 or 6 ppl are out, leaving us with only 54

out of these 54 ppl, now consider those (1/3 of 60, i.e. 20 ppl) who put vanilla ahead of chocolate…they can do that in 2 ways…say, ‘x’ ppl put chocolate in the 2nd position, (thereby putting vanilla 1st and strawberry 3rd) and (20 – x) ppl who put chocolate 3rd (thereby putting vanilla 2nd and automatically strawberry 1st)....so we have (20 – x) “favourable” ppl…the earlier party of ‘x’ ppl are also to be eliminated since they are putting vanilla first, contrary to the problem.

finally, 3/5 or 36 ppl put vanilla in the last posn, and so among these ppl are : 1. who put strawberry first (say ‘y’)

2. who put chocolate first [ the rest (36 - y) ppl]now we observe further tht….the 20 and 36 just considered are “mutually exclusive” since u cant put vanilla ahead of chocolate and in the last position at the same time!

now if we add (20 + 36) u get 56 > 54, which means, 2 ppl are common between the 6 ppl eliminated at the outset and the remaining 54 ppl.

i.e., these two ppl can’t be from the last grp, since the last grp (36 ppl) put vanilla last, but the first group has put vanilla ahead of strawberry!

clearly, these 2 ppl are from the grp of 20 ppl putting vanilla ahead of chocolate…ie these 2 ppl put vanilla first as also ahead of chocolate..thus they have 2 options…(choc 2nd, straw 3rd) or (straw 2nd, choc 3rd)

only one of these two options matches the 2nd grp

now if we study the conditions carefully, it will be clear that x = 2

ie, 2 ppl have ranked vanilla 1st, chocolate 2nd and strawberry 3rd…and are eliminated.

from the group of 20, only 18 ppl rank straw first (refer to the upper paragraph) and are included in our reqd group

now we come to the 3rd group…these 36 ppl have two options : either (straw 1st, choc 2nd) ie ‘y’ ppl, or (choc 1st, straw 2nd) ie (36 – y) ppl. clearly the 3rd grp is independant of the 1st 2 grps

out of the 3rd grp, the ‘y’ ppl are of our interest.

now observe that since these 36 ppl are independant of the 1st 2 grps, it is impossible to know the value of ‘y’

what we do know is that y>0 for certain

total no of ppl favourable to our reqd result are (18 + y) > 18….so the only possible answer is 24

cudnt think up a shorter method. plss…............tl me any shorter method …............i can’t write asong today my hand ache wrt nw…......... 2nd query noooooo idea….... bye



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PAB2706 wrote:hey,

I am not all that great when it comes to quant

I not sure but i got the answer as 24...

not sure of the method i used..if i am right i will post my method bcos i dnt want to confuse the other members who are equally weak in quant as i am..

sanju09 wrote:Oopsy! My faux pas!! Sorry for late reply as I was not well in the gone days, and thanks too!! May be because, I like vanilla more than strawberry. Okay, my answer in that case is 24, and that is more due to deductions than due to the real calculations. My process includes "Box Method", it will be very confusing if I post it right now, as I am still not in a very perfect frame of mind (thanks to medicines), I have posted same question to one of my friend, "Namita Lohani", on a forum whose link is given below, during my casual conversation with her, Namita is a gifted girl with logics, she sent to me the following solution for this problem, which I am posting as it is for other people to read and let us know if that is alright. I am not able to read and comprehend through such a big text at the moment, and one more thing Mr Ketkoag, why do I feel that this cannot be answered in 2 minutes, this feeling is so rare with me. Is it because I am not well, or it is really so? Anyways, here's Namita's take:

Namita Lohani wrote:

sir here s the sol. but its too long first, we will observe that the ppl putting vanilla ahead of strawberry are automatically eliminated in our search, since u cant put something ahead of the first position

so, 1/10 or 6 ppl are out, leaving us with only 54

out of these 54 ppl, now consider those (1/3 of 60, i.e. 20 ppl) who put vanilla ahead of chocolate…they can do that in 2 ways…say, ‘x’ ppl put chocolate in the 2nd position, (thereby putting vanilla 1st and strawberry 3rd) and (20 – x) ppl who put chocolate 3rd (thereby putting vanilla 2nd and automatically strawberry 1st)....so we have (20 – x) “favourable” ppl…the earlier party of ‘x’ ppl are also to be eliminated since they are putting vanilla first, contrary to the problem.

finally, 3/5 or 36 ppl put vanilla in the last posn, and so among these ppl are : 1. who put strawberry first (say ‘y’)

2. who put chocolate first [ the rest (36 - y) ppl]now we observe further tht….the 20 and 36 just considered are “mutually exclusive” since u cant put vanilla ahead of chocolate and in the last position at the same time!

now if we add (20 + 36) u get 56 > 54, which means, 2 ppl are common between the 6 ppl eliminated at the outset and the remaining 54 ppl.

i.e., these two ppl can’t be from the last grp, since the last grp (36 ppl) put vanilla last, but the first group has put vanilla ahead of strawberry!

clearly, these 2 ppl are from the grp of 20 ppl putting vanilla ahead of chocolate…ie these 2 ppl put vanilla first as also ahead of chocolate..thus they have 2 options…(choc 2nd, straw 3rd) or (straw 2nd, choc 3rd)

only one of these two options matches the 2nd grp

now if we study the conditions carefully, it will be clear that x = 2

ie, 2 ppl have ranked vanilla 1st, chocolate 2nd and strawberry 3rd…and are eliminated.

from the group of 20, only 18 ppl rank straw first (refer to the upper paragraph) and are included in our reqd group

now we come to the 3rd group…these 36 ppl have two options : either (straw 1st, choc 2nd) ie ‘y’ ppl, or (choc 1st, straw 2nd) ie (36 – y) ppl. clearly the 3rd grp is independant of the 1st 2 grps

out of the 3rd grp, the ‘y’ ppl are of our interest.

now observe that since these 36 ppl are independant of the 1st 2 grps, it is impossible to know the value of ‘y’

what we do know is that y>0 for certain

total no of ppl favourable to our reqd result are (18 + y) > 18….so the only possible answer is 24

cudnt think up a shorter method. plss…............tl me any shorter method …............i can’t write asong today my hand ache wrt nw…......... 2nd query noooooo idea….... bye



Read more: https://learnhub.com/conversation/