moneyman wrote:If 500 is the multiple of 100 that is closest to x and 400 is the multiple of 100 that is closest to y , which multiple of 100is closest to x+y ?

(1) x<500

(2)y<400

Ans E

For 500 to be the multiple of 100 closest to x, x has to be between 450 and 550. If x is less than 450 e.g 438 then the multiple of 100 closest to x will be 400 and not 500. Similarly if x is greater than 550 e.g. 578 then the multiple of 100 closest to x will be 600 and not 500.

Using the same logic, the number y will have to between 350 and 450. So the information that we have of the question stem is that x is between 450 and 550 and y is between 350 and 450

Now, the first statement says x<500, so this narrows down our range of x from 450-500 to 450-499. But this still does not tell us anything about y and it also does not give us a exact figure for x, so the range of possibilities for x+y is huge. Take an example let x = 470 and y =360 then x+y= 830 and in this case the multiple of 100 closest to 830 is 800. Take another example suppose x=470 but y= 430, x+y=900 and in this case obviously the multiple of 100 closest to x+y is 900. So we do not get a definite answer and hence statement 1 is not sufficient.

The second statement says y < 400, this narrows down our range of y from 350-450 to 350-400. Which using the same logic above is not sufficient

When we take both the statements together we have x is between 450-500 and y is between 350-400. Now within this range let us assume some values for x and y. Let x= 470 and y =370 we get x+y = 840, in this case the multiple of 100 closest to x+y is 800. Now, let x=480 and y =380 in this case x+y = 860 and the multiple of 100 closest to x+y in this case is 900. So as you can see inspite of taking both the statements together we do not get one definite answer. So the answer is E.

Note: - The important thing to learn over here is how to compute the range of x and y. Once you get the range you can assume values for x and y within that range and get the answer.

Let me know if you have any problem with the solution.