In a certain shop, two security cameras are positioned in different lo

If we imagine the two cameras and the cash register are the three vertices of a triangle, then from the stem, that triangle is isosceles. In any triangle, the sum of the two shorter sides must exceed the third side. From Statement 1, the triangle could have sides of length 9, 9 and 1, say, and the cameras could be 1 meter apart. But the sides could also be 9, 9, 17, say, and the cameras could be 17 meters apart. So Statement 1 is not sufficient.

From Statement 2, if the cameras are each d meters from the register, Statement 2 tells us they are d+7 meters apart. So the sides of our triangle are d, d, and d+7. Since the two short sides are d in this triangle, their sum, 2d, must be greater than the third side d+7, so 2d > d + 7, and subtracting d on both sides, d > 7. But if d > 7, then the distance between the cameras, which is d+7, must be greater than 14. So the two cameras are certainly more than 13 meters apart, and Statement 2 is sufficient. Technically, the cameras and register could make a straight line instead of a triangle, and could be exactly 14 meters apart, but that doesn't affect the answer, which is B.

What if we consider the camera positions from cash counter as radii of circle, cash counter being center. In that scenario, shaortest distance btw two camera will be the chord of circle. Now the distance between cash counter and camera being 9mts, then chord connecting the camera can be 18mts, considering the extreme location. What if the two cameras are close by on same side of arc?...the chord connecting them will be less than 9 mts....
Hence 1st staement is insufficient..agree
for 2nd statement alone is also not sufficient as we dont know what is the distance of cameras from cash counter...it may be 1 mt or 2 mts or .....
hence not sufficient alone...
however combing both we can get solution.....radii being 9mts..from 1
chord being (distance btw camera) 7mts greater than camera distance from cash counter (here radii - 9mts)
Please suggest where im logically wrong

When you looked at Statement 2 alone, you didn't use the information from Statement 2. You're looking at the problem in a good way -- you can either imagine the three points make a triangle, or you can imagine the two cameras lie on a circle with its center at the cash register. But when you use Statement 2 alone, if it were possible that each camera was 2 meters from the register, then the radius of that circle would be 2 meters. Statement 2 then tells us the cameras would be 9 meters apart. But the two cameras make a chord of the circle, and the longest chord of a circle is the diameter. The diameter of this circle is only 4 meters; it's not 9 meters or greater. So this situation is impossible, and the cameras cannot each be 2 meters from the register.

The question basically is whether the base of the isosceles triangle is greater than 13. We know that the sum of any two sides of a triangle must be greater than the third side. The first statement isn’t sufficient because the base of the isosceles triangle with sides equal to 9 can be anywhere between 0 and 18. The second statement alone is sufficient because even if the base were 14, the sum of the sides would be 14 and hence we wouldn’t have a triangle. So the base must be greater than 14.

Dear IanStewart , was the question published without your permission?