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dracarys007
Joined: 04 Jun 2020
Last visit: 24 Feb 2022
Posts: 68
Given Kudos: 16
Location: India
Concentration: Strategy, General Management
Schools: Harvard '23 INSEAD Aug 22 intake Sloan '23
GPA: 3.4
WE:Engineering (Consulting)
Updated on: 09 Dec 2024, 05:43
Originally posted by dracarys007 on 19 Jun 2020, 09:21.
Last edited by Bunuel on 09 Dec 2024, 05:43, edited 2 times in total.
Last edited by Bunuel on 09 Dec 2024, 05:43, edited 2 times in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
72% (02:01) correct
28%
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wrong
based on 348
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A man buys 33 icecreams for a group of 7 children. Does each one eat at least one ice cream?
(1) The ratio of the number of icecreams eaten by the number one child to the number of icecreams eaten by the number two child is 2 to 3.
(2) The ratio of the number of icecreams eaten by the number two child to the number of icecreams eaten by the number three child is 5 to 2.
(1) The ratio of the number of icecreams eaten by the number one child to the number of icecreams eaten by the number two child is 2 to 3.
(2) The ratio of the number of icecreams eaten by the number two child to the number of icecreams eaten by the number three child is 5 to 2.
19 Jun 2020, 09:45
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A man buys 33 icecreams for a group of 7 children. Does each one eat at least one ice cream?
(1) The ratio of the number of icecreams eaten by the number one child to the number of icecreams eaten by the number two child is 2 to 3.
We could have a case when the number one child ate 2 ice creams, the number two child ate 3 ice creams, and one of the remaining children ate all the remaining 28 ice creams. In this case we'd have a NO answer.
But we could have a case when the number one child ate 2 ice creams, the number two child ate 3 ice creams, and the remaining 28 ice creams were distributed among the remaining 5 children so that each got at least one ice cream.
Not sufficient.
(2) The ratio of the number of icecreams eaten by the number two child to the number of icecreams eaten by the number three child is 5 to 2.
We could have a case when the number two child ate 5 ice creams, the number three child ate 2 ice creams, and one of the remaining children ate all the remaining 26 ice creams. In this case we'd have a NO answer.
But we could have a case when the number two child ate 5 ice creams, the number three child ate 2 ice creams, and the remaining 26 ice creams were distributed among the remaining 5 children so that each got at least one ice cream.
Not sufficient.
(1)+(2) The ratio of the number of ice creams eaten by the number one child to the number of ice creams eaten by the number two child to the number of ice creams eaten by the number three child is 10:15:6. The least number of the ice creams those three could have eaten is 10 + 15 + 6 = 31. So, there are only 2 ice creams left for the remaining three children. So, not all of the children ate at least one ice cream. Sufficient.
Answer: C.
(1) The ratio of the number of icecreams eaten by the number one child to the number of icecreams eaten by the number two child is 2 to 3.
We could have a case when the number one child ate 2 ice creams, the number two child ate 3 ice creams, and one of the remaining children ate all the remaining 28 ice creams. In this case we'd have a NO answer.
But we could have a case when the number one child ate 2 ice creams, the number two child ate 3 ice creams, and the remaining 28 ice creams were distributed among the remaining 5 children so that each got at least one ice cream.
Not sufficient.
(2) The ratio of the number of icecreams eaten by the number two child to the number of icecreams eaten by the number three child is 5 to 2.
We could have a case when the number two child ate 5 ice creams, the number three child ate 2 ice creams, and one of the remaining children ate all the remaining 26 ice creams. In this case we'd have a NO answer.
But we could have a case when the number two child ate 5 ice creams, the number three child ate 2 ice creams, and the remaining 26 ice creams were distributed among the remaining 5 children so that each got at least one ice cream.
Not sufficient.
(1)+(2) The ratio of the number of ice creams eaten by the number one child to the number of ice creams eaten by the number two child to the number of ice creams eaten by the number three child is 10:15:6. The least number of the ice creams those three could have eaten is 10 + 15 + 6 = 31. So, there are only 2 ice creams left for the remaining three children. So, not all of the children ate at least one ice cream. Sufficient.
Answer: C.
Updated on: 15 Jul 2020, 10:28
Originally posted by BrushMyQuant on 19 Jun 2020, 09:47.
Last edited by BrushMyQuant on 15 Jul 2020, 10:28, edited 1 time in total.
Last edited by BrushMyQuant on 15 Jul 2020, 10:28, edited 1 time in total.
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Total Ice creams 33, Total Children 7
STAT1:
The ratio of the number of icecreams eaten by the number one child to the number of icecreams eaten by the number two child is 2 to 3.
\(\frac{Icecreams eaten by no. 1 student }{ icecreams eaten by no. 2 student}\) = \(\frac{2}{3}\)
=> Icecreams eaten by no. 1 student = 2k and Icecreams eaten by no. 2 student is 3k (where k is a positive integer)
Lets assume k=1 => Icecreams eaten by no. 1 student = 2 and Icecreams eaten by no. 2 student = 3
Now, there are 33-5 = 28 icecreams left and 7-2 = 5 children
Now, any case is possible. 4 of them have remaining 28 and one doesn't get anything, or everyone gets at least one, or many other possibilities.
So, STAT 1 is NOT SUFFICIENT as we can answer "Does each one eat at least one ice cream" as both yes and no
STAT2:
The ratio of the number of icecreams eaten by the number two child to the number of icecreams eaten by the number three child is 5 to 2
\(\frac{Icecreams eaten by no. 2 student }{ icecreams eaten by no. 3 student}\) = \(\frac{5}{2}\)
=> Icecreams eaten by no. 2 student = 5t and Icecreams eaten by no. 3 student is 2t (where t is a positive integer)
Lets assume t=1 => Icecreams eaten by no. 2 student = 5 and Icecreams eaten by no. 3 student = 2
Now, there are 33-7 = 26 icecreams left and 7-2 = 5 children
Now, any case is possible. 4 of them have remaining 26 and one doesn't get anything, or everyone gets at least one, or many other possibilities.
So, STAT 2 is NOT SUFFICIENT as we can answer "Does each one eat at least one ice cream" as both yes and no
STAT 1 and STAT 2 Together
\(\frac{Icecreams eaten by no. 1 student }{ icecreams eaten by no. 2 student}\) = \(\frac{2}{3}\) and \(\frac{Icecreams eaten by no. 2 student }{ icecreams eaten by no. 3 student}\) = \(\frac{5}{2}\)
icecreams eaten by no. 2 student is a multiple of both 3 and 5 so it will be a multiple of LCM(3,5) = 15
\(\frac{Icecreams eaten by no. 1 student }{ icecreams eaten by no. 2 student}\) = \(\frac{2}{3}\) = \(\frac{10}{15}\)
\(\frac{Icecreams eaten by no. 2 student }{ icecreams eaten by no. 3 student}\) = \(\frac{5}{2}\) = \(\frac{5}{2}\) = \(\frac{15}{6}\)
=> Icecreams eaten by no. 1 student : Icecreams eaten by no. 2 student : Icecreams eaten by no. 3 student = 10:15:6
=> Icecreams eaten by no. 1 student = 10k, Icecreams eaten by no. 2 student = 15k and Icecreams eaten by no. 3 student = 6k (where k is a positive integer)
Icecreams eaten by no. 1 student + Icecreams eaten by no. 2 student + Icecreams eaten by no. 3 student = 10k + 15k + 6k = 31k
we know k is a positive integer so minimum(k) = 1
Icecreams eaten by no. 1 student = 10, Icecreams eaten by no. 2 student = 15, Icecreams eaten by no. 3 student = 6
Remaining icecreams = 33 - (31) = 2 to be divided into 7-3 = 4 children
so, everyone will NOT have at least one ice cream
So, Answer will be C
Hope it helps!
STAT1:
The ratio of the number of icecreams eaten by the number one child to the number of icecreams eaten by the number two child is 2 to 3.
\(\frac{Icecreams eaten by no. 1 student }{ icecreams eaten by no. 2 student}\) = \(\frac{2}{3}\)
=> Icecreams eaten by no. 1 student = 2k and Icecreams eaten by no. 2 student is 3k (where k is a positive integer)
Lets assume k=1 => Icecreams eaten by no. 1 student = 2 and Icecreams eaten by no. 2 student = 3
Now, there are 33-5 = 28 icecreams left and 7-2 = 5 children
Now, any case is possible. 4 of them have remaining 28 and one doesn't get anything, or everyone gets at least one, or many other possibilities.
So, STAT 1 is NOT SUFFICIENT as we can answer "Does each one eat at least one ice cream" as both yes and no
STAT2:
The ratio of the number of icecreams eaten by the number two child to the number of icecreams eaten by the number three child is 5 to 2
\(\frac{Icecreams eaten by no. 2 student }{ icecreams eaten by no. 3 student}\) = \(\frac{5}{2}\)
=> Icecreams eaten by no. 2 student = 5t and Icecreams eaten by no. 3 student is 2t (where t is a positive integer)
Lets assume t=1 => Icecreams eaten by no. 2 student = 5 and Icecreams eaten by no. 3 student = 2
Now, there are 33-7 = 26 icecreams left and 7-2 = 5 children
Now, any case is possible. 4 of them have remaining 26 and one doesn't get anything, or everyone gets at least one, or many other possibilities.
So, STAT 2 is NOT SUFFICIENT as we can answer "Does each one eat at least one ice cream" as both yes and no
STAT 1 and STAT 2 Together
\(\frac{Icecreams eaten by no. 1 student }{ icecreams eaten by no. 2 student}\) = \(\frac{2}{3}\) and \(\frac{Icecreams eaten by no. 2 student }{ icecreams eaten by no. 3 student}\) = \(\frac{5}{2}\)
icecreams eaten by no. 2 student is a multiple of both 3 and 5 so it will be a multiple of LCM(3,5) = 15
\(\frac{Icecreams eaten by no. 1 student }{ icecreams eaten by no. 2 student}\) = \(\frac{2}{3}\) = \(\frac{10}{15}\)
\(\frac{Icecreams eaten by no. 2 student }{ icecreams eaten by no. 3 student}\) = \(\frac{5}{2}\) = \(\frac{5}{2}\) = \(\frac{15}{6}\)
=> Icecreams eaten by no. 1 student : Icecreams eaten by no. 2 student : Icecreams eaten by no. 3 student = 10:15:6
=> Icecreams eaten by no. 1 student = 10k, Icecreams eaten by no. 2 student = 15k and Icecreams eaten by no. 3 student = 6k (where k is a positive integer)
Icecreams eaten by no. 1 student + Icecreams eaten by no. 2 student + Icecreams eaten by no. 3 student = 10k + 15k + 6k = 31k
we know k is a positive integer so minimum(k) = 1
Icecreams eaten by no. 1 student = 10, Icecreams eaten by no. 2 student = 15, Icecreams eaten by no. 3 student = 6
Remaining icecreams = 33 - (31) = 2 to be divided into 7-3 = 4 children
so, everyone will NOT have at least one ice cream
So, Answer will be C
Hope it helps!
