Chiragjordan wrote:
Bunuel wrote:
Bunuel wrote:
If x is a positive integer, and two sides of a certain triangle have lengths 3x+2 and 4x+5 respectively, which of the following could be the length of the third side of the triangle?
I. 6x + 7
II. 6x + 9
III. 8x + 1
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II and III
Notice that the question asks which of the following
could be the length of the third side of the triangle.
What if the Question Had Asked Must be instead of could be => Then 1 and 3 will be true i am guessing ??/
Note : solving for If questions had asked for
must be true instead of
could be (then we have to plug values of
x) Now, as x is a positive integer, it can have values = {0, 1 , 2 , 3.... 10..}
(4x+5) - (3x+2) < (third side) < (3x+2) + (4x+5)
x + 3 < (third side) < 7x + 7.
Let's check options:
I. 6x + 7. For any positive value of x, x + 3 < 6x + 7 < 7x + 7. OK.
\(x + 3 < (third Side = 6x + 7) < 7x + 7\)
should hold for any positive value of x II. 6x + 9. For x = 1 or x = 2, x + 3 < 6x + 9 < 7x + 7
does not hold true but if x = 2, No.
\(x + 3 < (third Side = 6x + 9) < 7x + 7\)
\(2 + 3 < (third Side = 6.2 + 9) < 7.2 + 7\)
\(5 < 21 < 21\)
\(Nope\)III. 8x + 1. Right away we can find that if x = 1, then x + 3 < 8x + 1 < 7x + 7 is true. OK, But what about bigger value of x say x=10
does not hold true but if x = 10, No.
\(x + 3 < (third Side = 8x + 1) < 7x + 7\)
\(10 + 3 < (third Side = 8.10 + 1) < 7.10 + 7\)
\(13 < 81 < 77\)
\(Nope\)so in case of
must be true only (i) is true
in case of
could be true all 3 true