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Bunuel
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He used up 91 chocolates.Therefore 91 = chocolates per pack * number of guests
HCF(21,28,42)=7
Therefore each packet will have 3 Flavour A,4 Flavour B and 6 Flavour C chocolates
We have 13 chocolates per pack which we are giving to 7 guests
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nick1816
Statement 1-

Number of guests must be the common factor of 21, 28 and 42 or factor of GCD(21,28,42)=7; hence number of guests can be 1 or 7

If number of guests = 1; number of chocolates in a packet \(= 21+28+42 > 50\) (not possible)

If number of guests = 7; number of chocolates in a packet \(= \frac{21}{7}+\frac{28}{7}+\frac{42}{7} = 13\)

Sufficient

Statement 2-

Number of chocolates in each pack is multiple of 13 that is less than 50 [13, 26 or 39]

We have no idea about number of guests [ can be from 1 to infinity]

Insufficient




Bunuel
Peter gifted a packet of chocolates to each of the guests who came to his birthday party. Each packet was identical and contained x chocolates of flavor A, y chocolates of flavor B and z chocolates of flavor C, where x, y and z are positive integers. If the total number of chocolates in a packet was not greater than 50, then how many guests attended his birthday party?

(1) Peter used a total of 21 chocolates of flavor A, 28 chocolates of flavor B and 42 chocolates of flavor C to prepare all the packets.
(2) The numbers of chocolates of flavors A, B and C in each packet were in the ratio 3: 4: 6 respectively.


Hi Nick ,

Could you please explain me why you are taking the LCM of the number of chocolates of each flavour to find out the number of guests .

Number of guests can be anything ... then why we are talking the LCM to find the same ...

Am I missing something , kindly guide

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