If 1,500 is the multiple of 100 that is closest to X and 2,500 is the

Question

If 1,500 is the multiple of 100 that is closest to X and 2,500 is the multiple of 100 closest to Y, then which multiple of 100 is closest to X + Y?

(1) X < 1,500
(2) Y < 2,500

Kudos for a correct  solution .

Zhenek · Answer

#1
1451 <= x <= 1499
2451 <= y <= 2549
3902 <= x+y<= 4048
That covers numbers 3900 and 4000, which prevents us from asnwering our question. Insufficient

#2 is same, just different numbers - insufficient

#1 + #2 shrinks the distance but not enough
1451 <= x <= 1499
2451 <= y <= 2499
3902 <= x+y<= 3998
3900 and 4000 still remain as 2 options, thus insufficient.

E

Lucky2783 · Answer

i think 
1450 < x < 1550
2450 < y < 2550
so 3900<X+Y<4100

Lucky2783 · Answer

took sometime to understand the wording of the question stem and relate the fact that 1500 is also a multiple of 1500 . 1500 *1=1500

1450 < x < 1550
2450 < y < 2550
so 3900<X+Y<4100

Answer E .

EMPOWERgmatRichC · Answer

Hi Lucky2783,

Be careful with your 'rounding' here, while you did get to the correct answer, your final explanation did NOT factor in the information from the two Facts. After incorporating ALL of the given information, the inequality should be 3900 < (X+Y) < 4,000.

GMAT assassins aren't born, they're made,
Rich

Lucky2783 · Answer

god knows when will i stop making these callous mistakes !!
I overlooked .
thanks for pointing that out.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

The first step here is to try and understand what the question is asking. It can be a little confusing so you might have to read it more than once to correctly paraphrase it. Essentially some number X exists and some number Y exists, and the question is asking us what X + Y would be. The only information we get about X is that 1,500 is the closest multiple of 100 to it, meaning that X essentially lies somewhere between 1,450 and 1,550. Any other number would lead to a different number being the closest multiple of 100 to it. Number Y is similar, but offset by 1,000. It must lie between 2,450 and 2,550. At this point we may note that the problem would be exactly the same with 100 and 200 instead of 1,500 and 2,500, so the magnitude of the numbers is simply meant to daunt the reader.

Without even looking at the two statements, let’s see what we can determine from this problem: Essentially if we add X and Y together, the smallest amount we could get is (1,450 + 2,450 =) 3,900. The largest number we could get is (1,550 + 2,550 =) 4,100. The sum can be anywhere from 3,900 to 4,100, and therefore the closest multiple of 100 could be 3,900, 4,000 or 4,100, depending on the exact values of X and Y. This tells us that we have insufficient information through zero statements, which isn’t particularly surprising, but it also sets the limits on what we need to know. There aren’t dozens of options; we’ve already narrowed the field down to three possibilities.

(1) X < 1,500

Looking at statement 1, we can narrow down the scope of value X. Instead of 1,450 ? X ? 1,550, we can now limit it to 1,450 ? X < 1,500. This reduces the maximum value of X + Y from 4,100 to under 4,050. This statement alone has eliminated 4,100 as an option for the closest multiple of 100, but it still leaves two possibilities: 3,900 and 4,000. Statement 1 is thus insufficient.

(2) Y < 2,500

Looking at statement 2 on its own, we now have an upper bound for Y, but not for X. This will end up exactly as the first statement did, as we can now limit the value of Y as 2,450 ? Y < 2,500. This is fairly clearly the same situation as statement 1, and we shouldn’t spend much time on it because we’ll clearly have to combine these statements next to see if that’s sufficient.

(1) X < 1,500
(2) Y < 2,500

Combining the two statements, we can see that the value of X is: 1,450 ? X < 1,500 and the value of Y is 2,450 ? Y < 2,500. If we tried to solve for X + Y, the value could be anywhere between 3,900 and 4,000 (exclusively), so 3,900 ? X+Y < 4,000. This still leaves us in limbo between two possible values. To illustrate, let’s pick X to be 1,460 and Y to be 2,460. Both satisfy all the given conditions and give a sum of 3,920, which is closest to 3,900. If we then picked X to be 1,490 and Y to be 2,490, we’d get a sum of 3,980. The second situation clearly gives 4,000 as the closest multiple. If we can solve the equation using valid arguments and yield two separate answers, we have to pick  answer choice E .

These types of questions can be daunting because of the big numbers and the ambiguous wording, but the underlying material on these questions will never be something that can’t be solved in a matter of minutes. The difficulty often lies in determining how much work we really need to do to solve the question at hand. The old adage is that you get A for effort, but that’s applicable when you tried earnestly and failed. On the GMAT, you want to put in as much effort as is needed, but the only A you want to get is for Awesome GMAT Score (admittedly an AGMATS acronym).

EMPOWERgmatRichC · Answer

Hi All,

This question is based on 'rounding' rules. We're told that 1500 is the multiple of 100 that is closest to X and that 2500 is the multiple of 100 that is closest to Y. This means that X and Y have the following ranges...

1450 <= X < 1550 
2450 <= Y < 2550

We're asked for the multiple of 100 that is closets to (X+Y). We can TEST VALUES to answer this question.

(1) X < 1,500

IF.... 
X = 1499, Y = 2450, then (X+Y) = 3949 and the answer to the question is 3900 
X = 1499, Y = 2499, then (X+Y) = 3998 and the answer to the question is 4000 
Fact 1 is INSUFFICIENT

(2) Y < 2,500

The same two TESTs that I used in Fact 1 also 'fit' Fact 2...

IF.... 
X = 1499, Y = 2450, then (X+Y) = 3949 and the answer to the question is 3900 
X = 1499, Y = 2499, then (X+Y) = 3998 and the answer to the question is 4000 
Fact 2 is INSUFFICIENT

Combined, we already have two TESTs that fit both Facts and produce different answers. 
Combined, INSUFFICIENT.

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

ocelot22 · Answer

Since 1,500 is the closest multiple of 100 to X we can write 1450<=x <=1549, and since 2500 is the closest multiple of 100 to Y we can write 2450<=y<=2549

(1) x<1500 ---> then 1450<=x<1500 What about Y? NS

(2) Y<2,500. Then 2450<=Y<2500 . What about X? NS

(1) and (2) Since inequalities are facing the same direction we can sum:3,900<=x+Y<4,000. It is a trap at this point to think that x+Y must equal 3,900. Since we have an interval, try to write down some potential values for x+Y: Let x+y=3,900. Then x+Y is closest to 3,900. Let X+Y =3950. Then x+Y is closest to 4,000 NS E

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