If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre

VERITAS PREP OFFICIAL SOLUTION

Solution: C.

Start by simplifying the question itself. The only three perfect cubes which could be expressed as a single digit are 0, 1 and 8, so the question is really asking us whether x = 0, 1 or 8.

Now consider the statements. Statement (1) is a difference of squares equation that gives us 25(a^2) - b^2 = x. Trying a few values for a and b shows that x could have many different hundreds digits: if a = 3 and b = 4, for example, then x = 225 - 16 = 209, but if a = 3 and b = 9, then x = 225 - 81 = 144; INSUFFICIENT. Statement (2) tells us nothing about x at all; INSUFFICIENT.

Statement (2) does give us two possible values for a, however: if we subtract 20a from both sides we have a quadratic equation -- a^2 - 20a + 84 = 0 -- with solutions of a = 6 and a = 14, respectively. Taking the two statements together, then, we have either:

a = 6, so x = (30+b)(30-b) = 900 - b^2
a = 14, so x = (70+b)(70-b) = 4900 - b^2

Since b is a single digit integer, b^2 must be between 0 and 100. This means that 900 - b^2 must be between 800 and 900, so its hundreds digit is 8; much the same is true of 4900 - b^2, which must be between 4800 and 4900, so its hundreds digit is also 8. SUFFICIENT; C.

use statement 2 to get value of a and then put the value of a in statement 1 and check hence C.

Hi all,
Answer is Together sufficient.
Statement 1 is not sufficient:
Given x= 25a2-b2
a=8 b=1 hundred place of x is a perfect square.
a=10 b=2 hundred place of x is not a perfect square.
So insufficient.
Statement 2 is not sufficient:
Its quadratic equation where the roots are 6 and 14
This doesn’t talks about x.
Together it is sufficient.
If we take a=6, statement 1 becomes 900-b^2 where hundreds digit of x value is always has 8 which is a perfect cube irrespective of b value chosen between 0 and 10.
If we take a=14, statement 1 becomes 4900-b^2 where hundreds digit of x value is always has 1 which is a perfect cube irrespective of b value chosen between 0 and 10.
So together it is sufficient.
So answer is C

The only perfect cubes in 0-9 are 0,1,8 , so question is asking whether hundredth digit is one of them.

Lets first examine option B which seems easy one.
B) We do not know what is X in terms of 'a' and 'b' ; NS
a^2 + 84 = 20a ; A=14,6;
even if we know what X equals from option A, as we do not have information about 'b' this stmt is NS.

A) x = (5a+b)(5a-b) ; x=25a^2 - b^2;
a=3, b=5 --> x=200; is hundred th digit 0/1/8 ? NO
a=6, b=4---> x=884; is hundred th digit 0/1/8 ? YES
so option A is also NS

Lets combine A&B
a=14,6 in both
x= 4900-b^2 when a=14
x= 900-b^2 when a=6
in both cases we know the hundred th digit is 8;

C is the answer

Hi Bunuel,

for the explanation provided for STMT 2 , we do not what is 'x' in terms of 'a' and 'b' so this option is straight forward out.
Please let me know if i am missing something.

thanks
Lucky.

Yes, from (2) we know nothing about x, hence it's not sufficient.

solution: C

1) x= 25a^2 -b^2
a=3 => x= 225-b^2
if b=1 H(hundreds digit = 2)
if b=9 H=1 perfect cube of 1
A,D eliminated

2) a^2+84=20a
=> a = 6,14
but we dont know what is x => insufficient B is eliminated

Combine

a=6 => x= 900-b^2 , but 10<b<0 so H =8 for any b => perfect cube of 2
a=14 => x= 4900-b^2 , but 10<b<0 so H =8 for any b => perfect cube of 2

Definite YES

C is the answer

It took me more a lot of time to solve this question. IS there any other way or shortcut to solve this question?

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