Its clear now. I was always rounding based on all the digits from right to left and not taking just the first dropped digit.
For e.g. in your example of "5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5." I would always have picked 5.4 I now see how that might not work.
Thanks for the explanation. I overlooked this while preparing.
ga99 wrote:
Shouldn't the answer be C: both needed to answer.
Since d is 4 as per statement 1, the approximation also depends on e. If e=0 - 4, then x is 3.1. If e=5 - 9, then the x is 3.2. e would not matter only if d < 4.
Rounding rulesRounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.
BACK TO THE ORIGINAL QUESTION:
In the decimal 3.1de, d and e each represent a digit. If x represents the decimal 3.1de rounded to the nearest tenth, what is the value ofAccording to the rule above the tenth digit of x depends only on d, since it's the first dropped digit.
(1) d=4. Our decimal is 3.14e --> the first dropped digit is 4, so less than 5, hence 3.14e rounded to the nearest tenth is 3.1. Sufficient.
(2) e=8. Clearly not sufficient.
Answer: A.
Hope it's clear.