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by colakumarfanta » Wed Nov 09, 2011 5:26 am
Q. A to B is 9, B to C is 4 . Three points are not on the straight line . What is the value of A to C ? A. 3 B. 5 C. 9 D. 13 E. 15
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by user123321 » Wed Nov 09, 2011 5:38 am
colakumarfanta wrote:Q. A to B is 9, B to C is 4 . Three points are not on the straight line . What is the value of A to C ? A. 3 B. 5 C. 9 D. 13 E. 15
AB = 9, BC = 4
if three points are not on straight line, then it is possible to form a triangle. So the points will obey the properties of triangle

1) sum of two sides is always greater than third side.
we can eliminate D,E.
2) difference of two sides is always less than third side. so A & B can be eliminated.
hence C

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by Anurag@Gurome » Wed Nov 09, 2011 5:44 am
colakumarfanta wrote:Q. A to B is 9, B to C is 4 . Three points are not on the straight line . What is the value of A to C ?
Don't be tricked by the heading of the question. This is not at all a number line problem.

As the three points are not on a straight line, they must form a triangle. Now, in the triangle ABC, sum of the lengths of any two sides must be greater than the length of the third side.
  • A. If AC = 3, then (AC + BC) = (3 + 4) = 7 < 9 = AB --> NO
    B. If AC = 5, then (AC + BC) = (5 + 4) = 9 = AB --> NO
    C. 9, then
    • (AC + BC) = (9 + 4) = 13 > 9 = AB
      (AB + BC) = (9 + 4) = 13 > 9 = AC
      (AC + AB) = (9 + 9) = 18 > 4 = AB ----> YES
    D. If AC = 13, then (AB + BC) = (9 + 4) = 13 = AC --> NO
    E. If AC = 15, , then (AB + BC) = (9 + 4) = 13 < 15 = AC --> NO
The correct answer is C.
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by ArunangsuSahu » Thu Nov 10, 2011 4:37 pm
Formula for Triangle:

(a+b)>c

Now try by options

(C) is the answer
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