Folks, sorry for the delay (I was on a vacation.)

hemantsood has given the correct answer to the previous question.

Here is the next one.

What is the sum of all the digits of the four-digit even number?

I. The sum of the first and the last digit is 6 more than the sum of the middle two digits.

II. The sum of the first three digits is 2 less than the last digit. The four digit number comprises 4 distinct digits and no zeros.

Let abcd be num
we need a+b+c+d
1. a+d=b+c+6 (not suff) 6000 or 8020
2. a+b+c=d-2 (not suff) as 1+2+3 then d=8 or 1+2+4 then d =9

Combining both number could be

2149 or 2438

Ans E

sureshbala wrote: What is the sum of all the digits of the four-digit even number?

I. The sum of the first and the last digit is 6 more than the sum of the middle two digits.

II. The sum of the first three digits is 2 less than the last digit. The four digit number comprises 4 distinct digits and no zeros.

IMO E.

Question is asking what is the value of a+b+c+d = ?

Stmt I : a+d=b+c+6 ==> a+a+d-6+d ==> 2(a+d)-6 (Insuff)

Stmt II: a+b+c = d-2 ==> d-2+d ==> 2d-2 (Insuff)

Combining both the statements, we still cannot determine the value of a+b+c+d

hemantsood wrote:Let abcd be num
we need a+b+c+d
1. a+d=b+c+6 (not suff) 6000 or 8020
2. a+b+c=d-2 (not suff) as 1+2+3 then d=8 or 1+2+4 then d =9

Combining both number could be

2149 or 2438

Ans E

I haven't checked your math, but if you're correct about the 2 possible values, the answer is (C), since the question tells us that abcd is EVEN, which means that 2149 is NOT possible.

However, if you were just giving those as possible examples and there are others, the answer may in fact be (E).

So we have a number ABCD

I. NS (as in prev. posts)

II.
A+B+C=D-2

Now, since the number is even D can only be 0,2,4,6 or 8

0 - no zeros
2 - A+B+C=0, no zeros, not possible
4 - A+B+C=2, not possible, since # are distinct
6 - A+B+C=4, not possible, since # are distinct
8 - A+B+C=6, possible

So, SUFFICIENT

(B)

IMO B,

1) example number 6000 or 1218 and many other numbers possible

2) numbers must contain 1,2,3 and 8 as last digit
1238,1328,2318...but sum of digits is same-15
so B is the answer

Thanks

sureshbala wrote:Here is the next one...

The roots of a quadratic equation ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

1. Sum of the roots is positive.

2. Product of the roots is positive.

Alirhgt let R1 and R2 be the roots

Statement 1: R1 + R2 = +ve
Statement 2: R1* R2 = + ve. This statement is true when either both roots are negative or both are positive. Substitute both the cases in statement 1 (i.e. take either both positive or both negative), therefore, you see that both roots will have to be positive.

B

Assume the number to be ABCD,,, where D can be anything from set of 2,4,6,8,0

1. A+D = B+C+6

numerous numbers available.. as B and C need not be different as well...

2. A+B+C = D-2

A != B != C ! = D ! = 0

Only possible value D = 8

Because the minimum sum of ABC has to be 6

hence B

what is OA ?

Answer is definitely B for this
1) many numbers are possible
2) only no possible are 1238, 2138, 3128 etc
and all digits add to 14 always . Hence option B is correct