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Three TestPrep problems

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Three TestPrep problems

by moneyman » Fri May 25, 2007 5:16 am
PLs explain these problems..


1-x^4+y^4=100 , greatest value of x is between
a)0-3 b)3-6 c)6-9 d)9-12 e)12-15


2-For which of the following functions f(a+b)=f(a)+f(b) for all positive numbers of a and b?
a)f(x)=x^2
b)f(x)=x+1
c)f(x)=sq.root x
d)f(x)=2/x
e)f(x)=3x

3-A Certain company employs 6 senior officers(SO) and 4 junior officers(JO).If a commitee is to be created that is made up of 3 SO and 1 JO,how many different commitees are possible?
a)8 b)24 c)58 d)80 e)210



Thanks
Maxx
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by Cybermusings » Fri May 25, 2007 5:35 am
1-x^4+y^4=100 , greatest value of x is between
a)0-3 b)3-6 c)6-9 d)9-12 e)12-15

Greatest value of x would be when y=0...hence x^4 = 100 or x^2 = 10 (ignore -10 since we are trying to find out the maximum value)....x^2 = 10...hence x would lie between 3-6


2-For which of the following functions f(a+b)=f(a)+f(b) for all positive numbers of a and b?
a)f(x)=x^2
b)f(x)=x+1
c)f(x)=sq.root x
d)f(x)=2/x
e)f(x)=3x

Start with Choice C - f(a+b) = sqr rt.(a+b)
f(a) + f(b) = sqr rt.(a) + sqr rt.(b)...clearly not true

Choice D - f(a+b) = 2/(a+b)...f(a) + f(b) = 2/a + 2/b = 2b + 2a /(ab)
Hence not true

Choice E - f(a+b) = 3(a+b) = 3a+3b
f(a) + f(b) = 3a + 3b
Hence true

3-A Certain company employs 6 senior officers(SO) and 4 junior officers(JO).If a commitee is to be created that is made up of 3 SO and 1 JO,how many different commitees are possible?
a)8 b)24 c)58 d)80 e)210

Committees possible = 6C3 * 4C1 = 20*4 = 80 ways
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Can u help me..

by moneyman » Fri May 25, 2007 8:54 am
Hello cybermusings,
Thanks a lot for those explanations...I am just blank in permutations and combinations..Just to be on the safer side on the test day,can u pls give me some basic perm and comb stuff to remember..will be a great help..Thanks
Maxx
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