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Updated on: 13 Apr 2023, 03:26
Originally posted by gmatophobia on 13 Apr 2023, 03:21.
Last edited by gmatophobia on 13 Apr 2023, 03:26, edited 3 times in total.
Last edited by gmatophobia on 13 Apr 2023, 03:26, edited 3 times in total.
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gmatophobia
D
gmatophobia
C
DS Question 1 - Apr 13
If y=|x|+x, where y is an integer. Is y = 0 ?
(1) x!+7 is prime
(2) y < 2
Source: Others | Difficulty: Hard
PS Question 1 - Apr 13
Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?
A. 1/216
B. 1/36
C. 5/108
D. 11/216
E. 1/18
Source: Veritas Prep | Difficulty: Hard
Updated on: 13 Apr 2023, 03:52
Originally posted by AbhinavKumar on 13 Apr 2023, 03:50.
Last edited by AbhinavKumar on 13 Apr 2023, 03:52, edited 1 time in total.
Last edited by AbhinavKumar on 13 Apr 2023, 03:52, edited 1 time in total.
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gmatophobia
if y=|x|+x
is y = 0 ?
it is possible only if x is negative
Statement 1:
x!+7 = prime
and that prime can be 2,3,5,7,11,13, so on
BUT
x! = -5,-4,-4,0,4 for primes 2 to 11
so min value of prime for which x!+7 = prime = 13
or x! = 6
or x = 3
therefore x >=3 (always positive) hence y != 0
Statement 2:
y<2 then y can be 0 or (say) 1, so not a sufficient statement
hence going for option A
(edit )***
BUT
x! = -5,-4,-2,0,4 for primes 2 to 11 s
(such values of x! which is not possible)
so min value of prime for which x!+7 = prime = 13
or x! = 6
or x = 3
wickedvikram
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Updated on: 13 Apr 2023, 05:48
Originally posted by wickedvikram on 13 Apr 2023, 05:40.
Last edited by wickedvikram on 13 Apr 2023, 05:48, edited 1 time in total.
Last edited by wickedvikram on 13 Apr 2023, 05:48, edited 1 time in total.
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D - 11/216
wickedvikram
Solution: I attempted this with an imaginary 2 face dice and extrapolated all the scenarios for 6 faced dice
Case 1: All numbers are same 11-11,22-22,33-33,.....66-66 = 6 cases
Case 2: Here, there is a small catch, 12 for pam can be 12 &21 for robin. Why? because it’s a selection not arrangement between the two.
So, for one number 5*2 cases (we already counted same numbers)
for 6 numbers = 6*5*2=60 cases
total = 66
we already know for four dices number of cases are 6*6*6*6
Probability 66/6*6*6*6 =11/216
