Data Sufficiency

Dear All,

I will post interesting complex data sufficiency problems. The level of the questions is 670-700+.

1. What is the median of a group of numbers?
(1) At least half of the numbers less than or equal to 15.
(2) At least half of the numbers greater than or equal to 15.

2. 19. The median of Set S and T are 12 and 18, respectively, then S and T are combined, is the median of new set greater than the greatest number in Set S?
(1) The range of Set S is 6.
(2) The range of Set T is 6.

3. Is m > n?
(1) 0 < 2m + n < 10.
(2) -10 < m - 2n < 0.

4. If axby = 200, where a, b, x, y are positive integers, what is the value of x + y?
(1) x > y.
(2) xy = 6.

5. In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangle region RCS?
(1) The area of triangular region AÐ’Ð¥ is 32.
(2) The length of one of the altitudes of triangle ABC is 8.

6. Is the median of the numbers in set S less than the average of the numbers in the set?
(1) The median is of the sum of all of the numbers in set S.
(2) There are 9 numbers in set S.

7. Is the sum of seven different positive integers greater than 99?
(1) The largest number is 18.
(2) The range of the seven numbers is 10.

Wish you good luck with the above The answers and explanations will be given later

Answer1:
IMO E
1)take the numbers 3 3 3 -median 3
take numbers 14 14 14 -median 14
NOT SUFF

2)same as 1
take numbers 20 20 20- median 20
take numbers 25 25 25-median is 25
NOT SUFF
combined
take 3,3,3,20,20,20-median is 11.5
take 4 4 4 20 20 20-median is 12

I don't think stmt 2) xy = 6 in problem 4 is correct bcz according to stmt 2 xy has prime factors of 2 & 3 which by definition should be present in axby since a,x,b and y are all integers but the prime factors of 200 is 2,2,2,5,5. no 3!

stmt 1: SUFF
ABX = 32
=> ABC = 64 (since BX devides ABC in half)

now, RCS = 1/2.height.SC
= 1/2. 1/4*height of ABC . 1/4*BC
= 1/16 * (1/2*height of ABC * base of ABC)
= 1/16 * (area of ABC)
= 1/16 * 64 (from above)
= 4

stmt 2: INSUFF
we don't have enough info to derive anything

___________________________________________________
Prob 5: E
stmt 1: INSUFF
18+6+5+4+3+2+1 <99
18+17+16+15+14+13+12 >99

stmt 2: INSUFF
18+13+12+11+10+9+8 <99
18+17+16+15+14+13+8 >99

together, same as stmt 2: INSUFF

My 2 cents.

6. Is the median of the numbers in set S less than the average of the numbers in the set?
(1) The median is of the sum of all of the numbers in set S.
(2) There are 9 numbers in set S.

Answer: (C)

Stmt 1:
median = sum(#s).
average = sum(#s)/L. (L = length of series)

If L>1, stmt is sufficient.

Stmt 2:
{1 1 1 1 1 1 1 1 8} --> median = 1, mean = 2 --> median<mean
{1 1 1 1 1 1 1 1 0} --> median = 1, mean = 8/9 --> median>mean

Insufficient.

But both combined, establishes that L>1. So median>mean and answers the question without ambiguity.

The standard rule is:
Always the median of a combined set will lie between the medians of the individual sets (endpoint inclusive).
So for Q.2, median of combined set lies between 12, 18 (inclusive). Largest element of Set S could be as big as 18 or as small as 12. So we cannot say unambiguously whether the largest element of S is greater than the median of the combined sets. So (E).