vishal.pathak wrote: If t is a positive integer and r is the remainder when t^2+5t+6 is divided by7, what is the value of r?
1) when t is divided by 7 the remainder is 6.
2) when t^2 is divided by 7 the remainder is 1.

I posted a solution here:

https://www.beatthegmat.com/gmat-prep-2-t90357.html

GMATGuruNY wrote:

vishal.pathak wrote: If t is a positive integer and r is the remainder when t^2+5t+6 is divided by7, what is the value of r?
1) when t is divided by 7 the remainder is 6.
2) when t^2 is divided by 7 the remainder is 1.

I posted a solution here:

https://www.beatthegmat.com/gmat-prep-2-t90357.html

Thanks GMATGuru,

Your explanations always help. Any thoughts on the other 2 questions

Regards,
Vishal

+

vishal.pathak wrote: If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + y/z odd?
a. wx + yz = odd
b. wz + xy = odd

Before we evaluate the two statements, we should examine how the question stem can be rephrased.

w/x + y/z = (wz + xy)/xz.
Since w/x and y/z are integers, their sum (w/x + y/z) is an integer.
Thus, (wz + xy)/xz must also be an integer.

The question becomes: Is integer w/x + y/z -- which can be rephrased as (wz + xy)/xz -- odd?

Statement 1: wx + yz = odd.
Let w=1, x=1, y=2 and z=2, so that wx + yz = 1*1 + 2*2 = 5.
Is w/x + y/z odd?
NO, since 1/1 + 2/2 = 2.

Let w=1, x=1, y=6, and z=3, so that wx + yz = 1*1 + 6*3 = 19.
Is w/x + y/z odd?
YES, since 1/1 + 6/3 = 3.
INSUFFICIENT.

Statement 2: wz + xy = odd.
Please note the values highlighted in red:
Just as 10/2=5 is a factor of 10, and 12/3=4 is a factor of 12, so too is (wz + xy)/xz a FACTOR of wz + xy.

Since wz + xy is odd, all of its factors must be odd.
Thus, (wz + xy)/xz must be odd.
SUFFICIENT.

The correct answer is B.

GMATGuruNY wrote:+

vishal.pathak wrote: If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + y/z odd?
a. wx + yz = odd
b. wz + xy = odd

Before we evaluate the two statements, we should examine how the question stem can be rephrased.

w/x + y/z = (wz + xy)/xz.
Since w/x and y/z are integers, their sum (w/x + y/z) is an integer.
Thus, (wz + xy)/xz must also be an integer.

The question becomes: Is integer w/x + y/z -- which can be rephrased as (wz + xy)/xz -- odd?

Statement 1: wx + yz = odd.
Let w=1, x=1, y=2 and z=2, so that wx + yz = 1*1 + 2*2 = 5.
Is w/x + y/z odd?
NO, since 1/1 + 2/2 = 2.

Let w=1, x=1, y=6, and z=3, so that wx + yz = 1*1 + 6*3 = 19.
Is w/x + y/z odd?
YES, since 1/1 + 6/3 = 3.
INSUFFICIENT.

Statement 2: wz + xy = odd.
Please note the values highlighted in red:
Just as 10/2=5 is a factor of 10, and 12/3=4 is a factor of 12, so too is (wz + xy)/xz a FACTOR of wz + xy.

Since wz + xy is odd, all of its factors must be odd.
Thus, (wz + xy)/xz must be odd.
SUFFICIENT.

The correct answer is B.

Amazing answer guru, I guess the key was to mould the question in a more useable form

Thanks a lot

Regards,
Vishal

vishal.pathak wrote:For a certain set of n numbers, where n>1, is the average (arithmetic mean) equal to the median?
(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.
(2) The range of the n numbers in the set is 2(n-1).
IMO D, OA A

Statement 1: If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.
In other words, the numbers are EVENLY SPACED.
When numbers are evenly spaced, the average = the median.
SUFFICIENT.

Statement 2: The range of the n numbers in the set is 2(n-1).
If n=2, then the range = 2(2-1) = 2.
Thus, the 2 numbers could be 1 and 3.
Average = (1+3)/2 = 2, median = 2.
Average = median.

If n=3, then the range = 2(3-1) = 4.
Thus, the 3 numbers could be 1,1,5.
Average = (1+1+5)/3 = 7/3, median = 1.
Average ≠ median.
INSUFFICIENT.

The correct answer is A.