Imo d

St1: 2 is the least possible side of a triangle ( 0 is even but side with 0 units is not possible)

2,4,6-- possible pair
But as we know sum of 2 sides should be greater than 3rd side. Difference of 2 sides should be less than 3rd side.

6-2< 4 < 6+4
4< 4 < 10.. (not possible)

Next pair will always have perimeter greater than 15

Sufficient

St2: let sides be x, y, z

X+z= 12
We can infer that y should be less than 12

1+11---> 3rd side is between 10 to 12
It cannot be greater than 11 [ greatest side]. In all cases, perimeter is >15


7+5--> 3rd side is between 2 to 12
It can be but we need > or equal to 5 [ greater than smallest side]
Permimeter will always be >15



Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

Is the perimeter of triangle ABC less than 15 centimetres ?

(1) The lengths of the sides of triangle ABC are consecutive even integers in centimetres
so the sides are 2n, 2n+2, 2n+4
Minimum value of n is 1, so min sides are 2,4,6... But the largest side is EQUAL to sum of other two sides, so it will not be a triangle..
Next values - 4, 6, 8.....sum=4+6+8=18>15
Ans NO
Sufficient

(2) The sum of the lengths of the shortest and longest sides of triangle ABC is 12 centimetres
So if perimeter is LESS than 15, the third side can have a max value as just less than 3
The smallest side has to be less than this too, so sides are <3, <3, ~9
But such a triangle is not possible as sum of the smallest sides is LESS than the largest side.
So Ans NO
Sufficient

D

You are missing out on the rule that the sum of the smaller two sides has to be greater than the largest side, so statement II is also sufficient

Great explanation sir


Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

_________________

Statement 1) The lengths are consecutive even integers. Let's start by taking the shortest lengths, say, 2,4,6 perimeter=12, but these values don't form a triangle since triangle has a property that for any value of a, b and c, a+b>c. Here, 2+4= or 2+4 isn't greater than 6. Let's move forward. Take lengths as 4,6,8. Perimeter =18 and 4+6>8. For each next value, perimeter increases by 6 than the last value. for example, 6,8,10, perimeter =24. So, the perimeter is always greater than 15. Sufficient.

Statement 2) Let the shortest length be a and longest be b. So, a+b =12. Let a= 5, b=7. Assume second longest be x. So, in this case, 5<x<7. The perimeter will be x+12. Say, perimeter=y, Then, 12+5 <y<12+7. 17<y<19, for this case. Let's investigate further, Assume a to the least and b to be the longest, So, a=1, b=11, 1+x>11, x>10, y>1+11+10, y>22. So, the perimeter will be always greater than 15 cm. Hence, Sufficient.

The correct answer is Option D.

hi chetan2u,

if the triangle inequality theorem refers only to the longest side, then below would not be valid? please help.

(2) The sum of the lengths of the shortest and longest sides of triangle ABC is 12 centimetres

a=shortest, c=longest

a+c=12
a=5, c=7

7-5<b<7+5
2<b<12
b=2.5

perimeter = 7+5+2.5 = 12+2.5 = 14.5 < 15…