Too many questions, buddy. you may want to split these into separate posts, so that each thread is dedicated to a single question - otherwise, this thread can get messy.

Q1 (the table question):
https://www.beatthegmat.com/table-question-t69806.html

Q2: h(100)+1

Let's look at h(100) = 2*4*6*8*10...96*98*100, then move to h(100)+1.
What are the prime factors of h(100)?
Is h(100) div by smallest prime factor 2? indeed it is.
So will h(100)+1 be divisible by 2? If we take a multiple of 2 and add 1, the result will Not be a multiple of 2.
Same with 3:
Is h(100) divisible 3? yes, because it has a '6'.
So will h(100) +1 be divisible by 3? no, because if we take a multiple of 3 and add 1, we won't reach the 'next' multiple of 3.
The same will also be true of prime factors 5, 7, 11, 13, etc. h(100) will be divisible by all prime factors up to 50 (for example, h(100) will be divisible by prime factor 43, since h(100 includes 86)), so consequently, h(100)+1 will Not be divisible by any of them. Whatever the smallest prime factor of h(100)+1, it is greater than 40 - therefore the answer is E.

Hi earnest.I could solve all questions but the second.

Here are my answers and explanations

1) question asks what is m+n=?

we know from chart : 4+z=m ; e+y=n; hence, 4+z+e+y=m+n.

we can find y by using : 4+y=-5 -> y=-9.

we can find e by using : 4+x=1-> x=-3 and e+x=7-. e=10

we can find z using : e+z=10 -> z=0

Therefore,m+n -> (4+0)+(10-9)=5 or B

2) couldn't solve

3) Let the number of machines be =n

Time taken by Machine R= 36 hrs
Rate of 1 machine (R) = 1/36
Rate of n machine (R) = n/36

Similarly,for machine (S) = n/18

Working together the machines complete the work in 2 hrs

Combined rate = n/18+n/36 = n/12

So, Rate*time =1
(n/12)*2 =1
n=6.

Therefore,C

4)

x/100+ 2y/100 + 3y/100 = 1.5/100 (x+y+z)
x/100 - 1.5x/100 = (1.5y/100 -2y/100) + (1.5z/100 - 3z/100)
-0.5x/100 = - 0.5 [y/100 +3z/100]
x= y+3z

Hence A

5)

x,y -> +ve Integers. xy=?

1) GCF=10. Let x=10 y =30 then xy=300; Let x=10 y=20 then xy=200
Therefore,insufficient

2) LCM=180;Let x=45 y=4 LCM=180 xy =180. Let x=180 y=2 LCM=180 xy=360
Therefore,insufficient.

Combining 1 and 2

From 1) we know x and y have 10 as a common factor then x= 10*_ and y=10*_

and LCM of x and y =180.We know 10 is the common factor x and y.

LCM=180=10*18.Because LCM is the Union of factors of x and y.
Now 18=2*3*3.

Now keep in mind you can't have 3 in both x and y because GCF will become 30 (10=2*5).

So the two terms are 90 and 20.So xy=1800.

Hence C

6)

Range= Highest value in set - Smallest value
= 112-45 =67 [initial range]

New X=56-20=36
New Y=94+24=118

New Range= Highest value in set- Smallest value
= 118-36=82

Question is What is the difference between New range and old range is = 82-67=15

7)

x+y=?

where x and y integers.

1) 690<x<y<696.

x could be 691 and y could be 692 sum =691+692

x could be 691 and y could be 693 sum=691+693

Therefore, x+y is not definitive.

2)

692<x<y<695

x could not be 694 because y is greater than x and y is an integer.
x can only be 693 and y will be 694.

therefore,sufficient.

hence B

Thank you guys....