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Here is my retake and I hope it clears the air:

we know stmt 1 doesnt suffice. So, A and D are out.
Stmt 2 gives us: (x-2) is a multiple of 5. This can we written as (x-2)= 5 x k(natural number)
or x= 5k+2. now the possible values of k could be (keeping in mind X is a prime number).
if k=0, then x=2
if k=1, then x=7
if k=3, then x=17 and so on.
Clearly stmnt 2 is not sufficient in itself.
Combining Stmt1 and Stmt2, we get x is less than 15 and therefore, x can have 2 and 7 as valid values. Since we have to find a unique value of x, both the stmts are not sufficient.

There fore, E.
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Senthil7
If x is a prime number, what is the value of x?

(1) x is less than 15
(2) (x-2) is multiple of 5.
.
A prime number is a whole number greater than 1, whose only two whole-number factors are 1 and itself. If this is the constraint then considering statement 2 it means x-2 is multiple of 5 or in other words x-2 is divisible by 5. Therefore the only values possible to satisfy the constraint of being a prime number and x-2 being divisible by 5 are:

7 and 17 => x-2 is 5 and x-2 is 15 both divisible by 5. Take 1 and 2 together x<15 , x is prime and x-2 divisible by 5 only integer value possible is 7. Hence I think correct answer is C. I disagree with above explanation by Bunuel as 0 is not prime as given in question constraint.


he took x=2,which is a prime and not x=0 :!:
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In first option we have x=2,3,5,7 ----- so not possible

In second option we have

if x=2 then x-2 = 2-2= 0 is a multiple of 5.

Same as x=7,17--- will follow. So not possible..

If we combine 2 statements then x=2,7. Not sufficient. Answer is E
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x is a prime number.
We need to get x

statement 1 => x<15 => x can be => {2,3,5,7,11,13} => insuff
statement 2 => x=5k+2 => x can be 2 or 7 => insuff
combining the two statements => x can be 2 or 7 .
No exact values=> clearly insuff


Hence E.
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If x is a prime number, what is the value of x?

(1) x is less than 15
(2) (x-2) is multiple of 5.

1-x<15 i.e. x=3,5,7,11,13
2-x-2=5n i.e. x=7,17,37,47 etc

7 is common hence C

x=0 is not the case since 0 is not a prime number ?
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