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by wilson4mba » Fri Nov 19, 2010 9:56 pm
At a particular moment, a restaurant has x biscuits and y patron(s), with x > 2 and y > 1. How many values of y are there, such that all the biscuits can be distributed among the patrons, with each patron receiving an equal number of whole biscuits and with no biscuits left over?

(1) x = a^2.b^3, where a and b are different prime numbers.

(2) b = a + 1
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Source: — Data Sufficiency |

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by Rahul@gurome » Sat Nov 20, 2010 1:22 am
wilson4mba wrote:At a particular moment, a restaurant has x biscuits and y patron(s), with x > 2 and y > 1. How many values of y are there, such that all the biscuits can be distributed among the patrons, with each patron receiving an equal number of whole biscuits and with no biscuits left over?
  • (1) x = a^2.b^3, where a and b are different prime numbers.
    (2) b = a + 1
Given: x > 2 and y > 1 and x is completely divisible by y or in other words y is a factor of x.
Thus, we have to find the number of factors of x.

Statement 1: x = a²b³, a and b are different prime numbers.
There are infinite numbers of x of the given form.

Not sufficient.

Statement 2: b = (a + 1)
a and b has no relation with question.

Not sufficient.

1 & 2 Together: x = a²b³, a and b are different prime numbers and b = (a + 1)
Only possible values of a and b are 2 and 3 respectively.
Thus, x = 2²3³ = 4*27 = 108
Now we can easily find the number of factors of x!

Sufficient.

The correct answer is C.
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by Curguar » Sun Nov 21, 2010 5:43 pm
Statement 1: x = a²b³, a and b are different prime numbers.
There are infinite numbers of x of the given form.
But x = a²b³, does it not mean that x has 12 factors, and thus y can have 11 values (y has to be different than 1)?

In which case, OA would be A?
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by diebeatsthegmat » Sun Nov 21, 2010 7:25 pm
Rahul@gurome wrote:
wilson4mba wrote:At a particular moment, a restaurant has x biscuits and y patron(s), with x > 2 and y > 1. How many values of y are there, such that all the biscuits can be distributed among the patrons, with each patron receiving an equal number of whole biscuits and with no biscuits left over?
  • (1) x = a^2.b^3, where a and b are different prime numbers.
    (2) b = a + 1
Given: x > 2 and y > 1 and x is completely divisible by y or in other words y is a factor of x.
Thus, we have to find the number of factors of x.

Statement 1: x = a²b³, a and b are different prime numbers.
There are infinite numbers of x of the given form.

Not sufficient.

Statement 2: b = (a + 1)
a and b has no relation with question.

Not sufficient.

1 & 2 Together: x = a²b³, a and b are different prime numbers and b = (a + 1)
Only possible values of a and b are 2 and 3 respectively.
Thus, x = 2²3³ = 4*27 = 108
Now we can easily find the number of factors of x!

Sufficient.

The correct answer is C.
]
first of all i also think the same, and concluded that the answer is C
later, i think if x=108 y might be anu number such as 3,5,6
thus if y=5 , 108:5=21 with the remander 3
can you explain again, please?
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by goyalsau » Sun Nov 21, 2010 10:07 pm
Curguar wrote:
Statement 1: x = a²b³, a and b are different prime numbers.
There are infinite numbers of x of the given form.
But x = a²b³, does it not mean that x has 12 factors, and thus y can have 11 values (y has to be different than 1)?

In which case, OA would be A?
Even i thought that answer should be A,
What is the OA, Wilson.
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by beat_gmat_09 » Sun Nov 21, 2010 10:29 pm
x = a^2 * b^3
a and b are different prime numbers. Number of factors of x are - (2+1)*(3+1) = 12
This is regardless of prime numbers and what x turns out, number of factors will always be same.
Therefore y can have (12-1) =11 values.
A is sufficient.
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by TOPGMAT » Mon Nov 22, 2010 7:23 am
Hi Rahul,
Do we need to know a & b ?
Waiting for your reply
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by hitesht » Thu Nov 25, 2010 11:10 pm
Can any expert please tell us what the answer might be...
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