Answer:
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Step-by-step explanation:
Consider the provided information.
Machine A can fill an order of widgets in a hours. Machine B can fill the same order of widgets in b hours.
Thus, in 1 hr machine A's work is 1/a and machine B's work is 1/b.
We need to find whether the Machine A's rate the same as that of Machine B?
Statement 1: Machines A and B finish the order at exactly 4:48 p.m.
Total time, A and B worked = 4:48 = 4+
hrs = 4+
hrs=
hours
Thus, ![\frac{1}{a}+\frac{1}{b} = \frac{5}{24}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%20%3D%20%5Cfrac%7B5%7D%7B24%7D)
Let say Machine A's rate the same as that of Machine B
![\frac{1}{a}+\frac{1}{a} = \frac{5}{24}\\\\\frac{2}{a} = \frac{5}{24}\\\\a = \frac{48}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Ba%7D%20%3D%20%5Cfrac%7B5%7D%7B24%7D%5C%5C%5C%5C%5Cfrac%7B2%7D%7Ba%7D%20%3D%20%5Cfrac%7B5%7D%7B24%7D%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B48%7D%7B5%7D)
It is given that a and b are even integers, but
is not an even integer.
Hence, Machine A's rate is not same as that of Machine B.
Therefore, statement (1) ALONE is sufficient.
Statement 2: ![(a + b)^2 = 400](https://tex.z-dn.net/?f=%28a%20%2B%20b%29%5E2%20%3D%20400)
![(a+b)^2 = 400\\a+b=20](https://tex.z-dn.net/?f=%28a%2Bb%29%5E2%20%3D%20400%5C%5Ca%2Bb%3D20)
There are many possible case in which a and b are even integer and there sum is 20.
If a = b = 10 (both even), then Machine A's rate is same as that of Machine B.
if a = 6 and b = 14 (both even), then Machine A's rate is not same as that of Machine B.
Therefore, No unique answer with statement 2.
Hence, statement 2 alone is not sufficient.