Qn: If two sidesof a triangle have lengths 2 and 5, which of the following could be the perimeter of the triangle?
I. 9
II. 15
III. 19

I thought that the third side would be greater than 3 but less 7 and therefore the perimeter could be 2+5+4=11 OR 2+5+6=13. Did i just solve my problem?

BLISSFUL: Yes..you did right!
Perimeter= 2+5+x = 7+x where x <7 which means Perimeter
is less than 14 --> it can only be I.
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Posted: Thu Nov 23, 2006 2:17 am Post subject: PS & DS

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Hi,

Can anyone explain how the following qn was solved, because from what I recall about the third side rule of a triangle, its length should be greater than the difference of the given two sides, and less than the sum of the given two sides...(correct me if I'm wrong)

Qn: If two sidesof a triangle have lengths 2 and 5, which of the following could be the perimeter of the triangle?
I. 9
II. 15
III. 19

I thought that the third side would be greater than 3 but less 7 and therefore the perimeter could be 2+5+4=11 OR 2+5+6=13. Did i just solve my problem?

How would you solve the following:

Qn: If the operation * is defined for all integers a and b by a*b=a+b-ab, which of the following must be true for all integers a, b, and c?


I. a*b = b*a
II. a*0 = a
III. (a*b)*c= a*(b*c)

BLISSFUL:
substitite 0 in place of b --> I and II are right
III. LHS= (a+b-ab) +c-(a+b-ab)*c = a+b+c-ab-bc-ca+abc

Looking at the symmetry, you could conclude RHS will be same
( you could check by doing the donkey work..
RHS= a+ (b+c-bc)-a(b+c-bc)= a+b+c-ab-bc-ca+abc=LHS

Therefore, I, II, II are all right

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Qn: If d is the standard deviation of x, y and z,what is the sandard deviatioon of x+5, y+5, and z+5?

a. d
b. 3d
c. 15d
d. d+5
e. d+15


BLISSFUL:
ave of x+5, y+5,z+5 = (x+y+z+15)/3 = original ave +5

Now while calculating new SD , we do..x-ave x , then sqr bla bla..

here, x+5-(old ave+5)..they cancel out..

THerefore with a glance , you could conculde SD is same..that is, answer a)