Latest Exam Aptitude Question SOLUTION: Richa, Shruti and Sayali went to a restaurant to celebrate the birthday of Shruti. After having the lunch and paying the bill, they were presented by a bowl of mints. Richa went on

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(#M40038241) EXAM Category question Keep an EYE

Richa, Shruti and Sayali went to a restaurant to celebrate the birthday of Shruti. After having the lunch and paying the bill, they were presented by a bowl of mints. Richa went on to grab first and took one-third of the total mints but on being taunted by Shruti, she returned four back. Then Shruti took one-fourth of the left mints and accidentally dropped three back. Sayali took half of the remaining mints but flipped two back into the bowl. Now the bowl was left with seventeen mints.

How many mints were present originally?
a) 41
b) 38
c) 31
d) None of the above

Asked In Exam athira (9 years ago)

Unsolved Read Solution (4)

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shashank  (9 years ago)

48 is right ans.

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farahan  (9 years ago)

ans-40
let total mints=x
ruchi take 1/3 n return 4=(x/3-4)
after ruchi take left mint=x-(x/3-4)=2x+12/3
shruti take 1/4 of remng n return 3=(2x+12/12-3)
so shruti hve=(x-12/6)
after shruti rmng mint=(2x+12/3)-(x-12/6)
so rmng=x+4/2
shyli take hlf n return aftr calcu shyli hve=(x-4/4)
after add mints of 3 frnds=3x-28/4
so accrdng ques
equ is 3x-28/4+17=x(total mint)
x=40 ans

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aparna potipireddi  (9 years ago)

48 is the ri8 answer...since ,let total no. of mints=x ,richa ll have x/3-4
so remaining ll be (x-x/3+4)
sruthi ll take as follos:1/4(2x/3+4)-3
x/6-2
sayali ll take as follows:1/2(x/2+6)-2
x/4+1
x-(total no of mints all three contains)=17
x-(x/3-4+x/6-2+x/4+1)=17
x=48

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sandeep   (9 years ago)

48 is the writr answer

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