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09 Feb 2016, 18:16
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
77% (01:35) correct
23%
(02:11)
wrong
based on 197
sessions
History
Date
Time
Result
Not Attempted Yet
If two integers x, y (x>y) are selected from -10 to 9 (inclusive), how many possible cases are there?
A. 150
B. 180
C. 190
D. 210
E. 240
* A solution will be posted in two days.
A. 150
B. 180
C. 190
D. 210
E. 240
* A solution will be posted in two days.
09 Feb 2016, 18:57
Kudos
Bookmarks
MathRevolution
Since x>y
let us first choose y
then , we will choose x . because x is greater than y
Our choice of numbers are
-9 -8 -7 -6 ............................... 9 = total 9-(-10)+1 = 20 numbers
Now for every y that we choose ,x will always be to the right of y in the number line because x is greater
So we start from the left ,
If y = -9 , x can be chosen from 19 number in the right = > 19C1 = 19 ways
IF y= -8 , x can be chosen from 18 number in the right = > 18C1 = 18 ways
If y = -7 , x can be chosen from 17 number in the right = > 17C1 = 17 ways
.... ..... ......... ..... ......... ..... ......... ..... ......... ..... ......... ..... ......... ..... .....
.... ..... ......... ..... ......... ..... ......... ..... ......... ..... ......... ..... ......... ..... .....
If y = 8 , x can be chosen from 1 number in the right = >only 9 Left =1 way
___________________________________________________________
Total = 19+18+17+..........................................1 = n(n+1)/2=(19.20)/2 = 190
Hence the answer is (C)
11 Feb 2016, 22:52
Kudos
Bookmarks
If two integers x, y (x>y) are selected from -10 to 9 (inclusive), how many possible cases are there?
A. 150
B. 180
C. 190
D. 210
E. 240
--> 20C2=20*19/2=190. Therefore, the answer is C.
A. 150
B. 180
C. 190
D. 210
E. 240
--> 20C2=20*19/2=190. Therefore, the answer is C.
