Bunuel
Is x = y ?
(1) The results when x and y are each rounded to the nearest tenth are equal to each other.
(2) The results when x and y are each rounded to the nearest hundredth are equal to each other.
Disproving the individual statements is quite simple, but combined is what requires some might
S1: The results when x and y are each rounded to the nearest tenth are equal to each other [Insufficient]1. Let's assume that \(x = 1.35\) and \(y = 1.42\)
2. Rounding them to the nearest tenth yields \(x = 1.4\) and \(y = 1.4\)
3. Is \(x = y\)? No
4. Let's assume that \(x = 1.42\) and \(y = 1.42\)
5. Rounding them to the nearest tenth yields \(x = 1.4\) and \(y = 1.4\)
6. Is \(x = y\)? Yes
S2: The results when x and y are each rounded to the nearest hundredth are equal to each other [Insufficient]1. Let's assume that \(x = 1.356\) and \(y = 1.362\)
2. Rounding them to the nearest hundredth yields \(x = 1.36\) and \(y = 1.36\)
3. Is \(x = y\)? No
4. Let's assume that \(x = 1.362\) and \(y = 1.362\)
5. Rounding them to the nearest hundredth yields \(x = 1.36\) and \(y = 1.36\)
6. Is \(x = y\)? Yes
S1 + S2 [Insufficient]1. Let's assume that \(x = 1.356\) and \(y = 1.362\)
2. Rounding them to the nearest tenth yields \(x = 1.4\) and \(y = 1.4\)
3. Rounding them to the nearest hundredth tenth yields \(x = 1.36\) and \(y = 1.36\)
4. Is \(x = y\)? No
5. Let's assume that \(x = 1.412\) and \(y = 1.412\)
6. Rounding them to the nearest tenth yields \(x = 1.4\) and \(y = 1.4\)
7. Rounding them to the nearest hundredth tenth yields \(x = 1.41\) and \(y = 1.41\)
8. Is \(x = y\)? Yes
Ans. E