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neokeynesian
10-29-2004, 06:13 AM
Please see the attachment

lmtuan
10-29-2004, 06:54 AM
we have 800 integers between 200 and 999.
The number of the integers between 200 and 999, inclusive ,begin with an 8 or a 9 and end with an odd digit is: 2 * 10 * 5 = 100.
Thus, the percent of the integers between 200 and 999, inclusive ,begin with an 8 or a 9 and end with an odd digit is (100/800) * 100% = 12.5%.
Answer: C

vn_snoopy
10-30-2004, 02:37 AM
What do they mean by "between"? Do they count the two numbers at the beginning and the end (which are 200 and 999) as "between" 200 and 999?

If they do not, we have 798 numbers between 200 and 999, and the approximation percentage is also 12.5%.

About the number required between 200 and 999: we have 2 choices for the hundreds digit (8 and 9), 5 choices for the units digit (the 5 odd numbers: 1,3,5,7,9), and 10 choices for the tens digit (0-9). Hence, we have 2*5*10 choices... as lmtuan has stated. ;)

calm_J
11-06-2004, 10:43 AM
here it's
the count of no. between 200 and 999 inclusive is 999-200+1 (this one because he said inclusive) which will equal 800 No.

no we have only 200 No. out of those 800 that have 8 or 9 as a start digit.

from those 200 No we are searching for the folowing criteria

8_1
8_3
8_5
8_7
8_9
9_1
9_3
9_5
9_7
9_9

this means two things
1- in each 10 consecutive No. 's between 800 -999 there exists 5 No. that meets our criteria
2- we have 20 tens in those 200 No.

then 20 * 5 = 100 No. as a total that meets our criteria

then the percentage is 100/800 *100% = 12.5