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ratio, probablity

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ratio, probablity

by ska7945 » Mon Aug 11, 2008 8:46 am
1)A company has assigned a distinct 3-digit code number to each of its 330 employees. Each code number was formed from the digits

2, 3, 4, 5, 6, 7, 8, 9

and no digit appears more than once in any one code number. How many unassigned code numbers are there?

A.6 B.58 C.174 D.182 E.399

Q17:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

A. 1/20 B. 1/6 C. 1/5 D. 4/21 E. 5/21


please explain as well. thx.
answers: [spoiler]A/B[/spoiler]
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by sudhir3127 » Mon Aug 11, 2008 9:11 am
please post the question separately so that its helpful to everyone.

for 1st question my answer is A.

the number of ways 3 digit code can be formed is 8*7*6 = 336.

total number of employes. 330.

hence the number of unassigned codes would be 336-330 =6.

for 2nd question

my answer is B

Let the sum of 20 nos be=X
hence mean=X/20
n=(x/20)*4=x/5

sum of 21 nos=x+x/5=6x/5

n/x=(x/5)/(6x/5)= 1/6
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