Is the permeter of square S greater than the perimeter of equilateral triangle T?
(1) The ratio of the length of a side of S to the length of a side of T is 4:5
(2) The sum of the lengths of a side of S and a side of T is 18
< Standard DS Sufficiency Choices >
My Solution:
Let side of the square = a
Let side of the triangle = l
Q Rephrase: Is 4a > 3l or a/l > 3/4 ?
(1) a/l=4/5 = 0.8 > 3/4. That is, perimeter of square is greater than triangle. Therefore, (1) is sufficient.
(2) a+l = 18
l = 18 - a
4a / 3l = 4a / 3(18-a) = 4a / 54-3a
Suppose 4a /3l > = 1 or 4a / 54-3a > = 1
4a >= 54 - 3a
7a >= 54
a >= 22 ... (a)
Since l = 18 - a and cannot be -ve, a < 18. Intersection with (a) gives No solution. Hence, the above is UNTRUE.
Suppose 4a/3l <= 1 or 4a / 54-3a <= 1
4a <= 54 - 3a
7a <= 54
a <= 22 ... (b)
Since l = 18 - a and cannot be -ve, a < 18. Intersection with (b) gives solution a < 18. Or , according to statement (2) , as long as a < 18, the condition of 4a < 3l is met. Hence, the above supposition is TRUE.
That is, 4a / 3l < 1 or 4a < 3l or Perimeter of square is greater than perimeter of triangle.
Therefore, (2) is sufficient.
MY ANSWER = (C) , Either of the two is sufficient to answer the question.
The online lab says the answer is (A) - Statement 1 only is sufficient.