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C; Let x be the length of each unpainted section
2xn+x(n-1)=203
3xn-x=203
x=203/(3n-1)
Hence, n has to be even; ABE out; D is 203/41; Clearly not divisible;
Check C;203/29=7

D

Correction - Its C

Can u explain why u say ’n’ has to be even?

Sure. From the question we know that length "x" of each unpainted (hence painted) section is integer.
In, x=203/(3n-1) ; Because numerator is odd, for "x" to be an integer, "3n-1" must be odd, else x will become a fraction value.
3n-1 can be odd only for even integer value of n.

is this E?

as i can find individual number of gemstoned but cant find say which is 3 or 4 or 5 making total as 12

Even I believe so, I was confused since in the mentioned answer they have only considered D,R and S as 4,3 and 5 respectively. They havnt considered 5,4 and 3 as an option, hence wanted to clarify.

Two observations:
1. why does statement-1 says number of balls are distinct in place of gemstones
2. there are 3 cases with both statement combined
D,R,S - (5,4,3) (5,3,4) (4,3,5)
Answer is E; else question must be incomplete and needs few corrections

PS Question 1 - May 27

If the least common multiple of 2x and y is 400 and the greatest common divisor of 6x and 3y is 30, what is the value of (x*y)/2

A. 100
B. 500
C. 1000
D. 1500
E. 2000

Source: GMATWhiz | Difficulty: Hard

1) Consider the following cases:
D: 6
R: 4
S: 2
You can change these numbers in any order and get different results. Therefore, NS.

2) All this tells us is that p(d)>p(r). We do not know the exact split, nor do we know how many types of gemstones there are. NS

3) We know there are 3 types.
We also know that none of the types have more than 6.
We also know that the p(d)>p(r).
We can make a few combinations:
d=5
r=4
s=3
or
d=5
r=3
s=4
or
d=4
r=3
s=5

Notice in the last one, that d=4, so (4/12) chance, but in the first cases its 5/12.
Because we do not know the relationship between p(s) and p(r), it cannot be determined.

E

DS Question 1 - May 27

A set of integers, S, contains more than one element. Is the range of S greater than its mean?

(1) S does not contain positive integers.
(2) The mean of S is negative.

Source: 800 Score | Difficulty: Hard

PS Question 1 - May 27

If the least common multiple of 2x and y is 400 and the greatest common divisor of 6x and 3y is 30, what is the value of (x*y)/2

A. 100
B. 500
C. 1000
D. 1500
E. 2000

Source: GMATWhiz | Difficulty: Hard

I am out of practice, but let me try this:

LCM(2x,y) is 400.
GCF(6x, 3y)=30
recall that
xy=LCM(x,y)*GCF(x,y)
Now, we don’t have xy so this is unusable.
What we can do, is this:
LCM(6x,3y)=3*400=1200.
So we have that
18xy=1200*30=36000
xy=2000
xy/2=1000
C

1) Consider the case where S is just composed of 0. The range and the mean are exactly the same. However, if you consider the case where S is composed of negatives, or 0 with some distinct negatives, the range must be higher than the mean. INS.
2) The mean of S is negative.
Consider the following cases:
the max of S is a positive number, and the min of x is some negative number. In this case, range is greater than mean.
the max of s is 0, and the min of x is some negative number. Range>mean.
the max of s is the mean, and the min of x is the mean (both negative numbers). range>mean (0>negative numbers)
B should be suff.

Very nice question.. Enjoyed it

B
stmt1: for all 0, range=mean; for at least 1 -ve, Range > mean
Hence, Insufficient
stmt2: Minimum value of range can be 0. Hence, sufficient

C
GCD(6x,3y) = 30 => GCD(2x,y)=10; LCM(2x,y)=400
2x*y=gcd*lcm=4000; xy/2=1000

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should be B ) 36

use n+r-1 C r-1 as we are dividing identical to distinct

B ?

You can consider this as various arrangement of 7 identical objects and two identical partitions.

9! / 2! * 7!

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This is used when, each person can get at least one object right? but we can also consider 0 as mentioned. Am i wrong ?

Where do we get 9 from ?

7 (identical objects) + 2 (identical partitions) = 9