If x and y are positive numbers such that x + y = 1, which of the

Another method is to take extreme values of x and y to figure out the range

x + y = 1
Say x is almost 1 (infinitesimally smaller than 1) and y is almost 0 (infinitesimally greater than 0)
Then 100x + 200y = 100*1 + 100*0 = approx. 100

Say y is almost 1 (infinitesimally smaller than 1) and x is almost 0 (infinitesimally greater than 0)
Then 100x + 200y = 100*0 + 200*1 = approx. 200

Now if x = 1/2 and y is 1/2,
Then 100x + 200y = 100*(1/2) + 200*(1/2) = 150

So we see that value of the expression will vary from 100 to 200.

Hence answer will be (E)

given, x+y=1 [0<x,y<1]

100x+200y=?
Or 100(x+2y)=?
100(x+y+y)=100(1+y)=?

100(1+y) can be of any value more than 100 and less than 200, only II and III match with this.

E

Hi All,

Certain Quant questions are built around relatively simple short-cuts; this prompt has a great built-in logic shortcut that you can take advantage of (and avoid lots of unnecessary calculations). As such, instead of thinking of the 'level' of this question, you should try thinking in terms of whether you could have gotten it correct in a reasonable amount of time or not.

We're told that X and Y are POSITIVE. That is an important 'restriction' that impacts how the math 'works' and we can use it to our advantage. Next, we're told that X+Y = 1. With this information, we know that both X and Y will end up being positive fractions.

We're asked for what COULD be the value of 100X and 200Y.

To start, it helps to think about the 'extreme' possibilities.

IF.... X=1 and Y=0, then the sum would be 100(1) + 200(0) = 100
IF... X=0 and Y=1, then the sum would be 100(0) + 200(1) = 200

Now, neither of those is a possible outcome (remember that BOTH X and Y have to be positive, and 0 doesn't fit that restriction), but they do provide the limits to the possible sum.

If we made X really small, then the bulk of the total would be in Y (eg. X=0.01 and Y= 0.99), so the sum would be REALLY close to 200. In that same way, if we made Y really small, then the bulk of the total would in X (eg. X= 0.99 and Y=0.01), so the sum would be REALLY close to 100. Moving the values in tiny increments would give us every possible value between 100 and 200, but NOT 100 or 200. Thus, Roman Numerals II and III are possible, while Roman Numeral I is not.

Final Answer:

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

Solved this one using min/max principle...

1. 80 < min 100*0,9=90 x
2. 100*0,99=99, 200*0,01=2 ok
3. 200*0,99=198 + 100*0,01=1 =>199 ok (E)

100x+200y = 100(x+y) + 100y
Given X+Y =1,
100x+200y = 100 + 100y
And Y's range is 0<Y<1,
Only options II and III are correct.

VeritasPrepKarishma

How did you know that the weighted average method applies to this problem?

It's a lot about familiarity achieved through practice. x and y are given to be positive and x + y = 1 is given. Then you are given (100x + 200y). In all, it reminds one of the weighted average formula terms.

But if it doesn't come to your mind, it is alright. That is why I have shown the other method too.

You can solve it pretty quickly using regular algebra and subtracting 2 equations to isolate a variable.

Keep in mind the given fact that both x and y are positive.

Eq 1: x + y = 1

Eq 2: 100x + 200y = ?

There are 3 possible solutions to Eq 2 given the question: I. 80; II. 140; III. 199

Plug in each scenario into Eq 2, reduce, then subtract Eq 1.

Scenario I: 100x + 200y = 80; or x + 2y = 0.8. Subtract Eq 1 (x+y=1) to isolate what y equals. This gives y = -0.2 which violates the given info.

Scenarios II and III yield solutions greater than 1, so when you subtract Eq 1 you get a positive value for y (edit: positive value that is less than 1). Therefore both II and III are valid.

x + y = 1

So, x = 1-y

Substituting in 100x + 200y:
100(1-y) + 200y
100 +100y
100(1+y)

Since y is positive, so clearly 100(1+y) cannot be less than 100. Hence, it cannot be 80. 140 and 199 are possible. So, E.

x+y=1
100x+100y=100
200x+200y=200

Since both values could be close to 0 (but not 0), we can safely assume that 100<100x+200y<200.

Given: x + y = 1
Given: x and y are positive numbers. So, if x + y = 1, then x and y are each less than 1

100x + 200y = 100x + 100y + 100y
= 100(x + y) + 100y
= 100(1) + 100y
= 100 + 100y

Since y is a POSITIVE number and since y < 1, we know that: 0 < 100y < 100
So, 100 < 100 + 100y < 200
In other words 100 + 100y (aka 100x + 200y) can have any value between 100 and 200

Answer: E

NOTE: If anyone needs more convincing, consider these two cases:
case a: If x = 0.6 and y = 0.4, then 100x + 200y = 60 + 80 = 140 (value II)
case b: If x = 0.01 and y = 0.99, then 100x + 200y = 1 + 198 = 199 (value III)

We are given that x and y are positive numbers and that x + y = 1. We must determine possible values of 100x + 200y. An easy way to determine whether 80, 140, or 199 could be values of 100x + 200y is to use the given fact that x + y = 1 to determine the possible range of 100x + 200y. Since 200 is greater than 100, the high end of our range will be when y is the largest, and the low end of our range will be when x is the largest.

High end of range:

y = 1 and x = 0

100x + 200y = 100(0) + 200(1) = 200

Low end of range:

y = 0 and x = 1

100(1) + 200(0) = 100

Finally we must remember that x and y both must be positive, which means neither x nor y can be zero. They must each be a decimal between zero and one. Thus, the low end of the range cannot actually be 100 and the high end of the range cannot actually be 200. Therefore, we can create the following inequality:

100 < 100x + 200y < 200

Only 140 and 199 are greater than 100 and less than 200.

Answer: E

Will someone confirm that my logic works for this question:

I wanted to find the range of numbers that x could be without testing as that would take a very long time. We know that since x & y are positive integers and together them sum to 1, individually they can range from 0 to 1 as long as the sum is 1. So x could be 1 and y would be 0 or x could be 0 and y would be 1. In order to determine the maximum of the range we would need to apply the largest value (1) to the 200 y. Therefore 100(0) + 200(1)=200. Again, this is the maximum of the range. Also, 100(1) and 200(0)=100, which is the low end of the range. So my answer choices will range from 100 to 200. This eliminates roman numeral 1.

Thoughts with my logic?

Your logic is perfectly correct. This is nothing but weighted average in your own words. The weight from 0 to 1 has to be allotted to 100 and 200. You could allot the entire 1 to 100 in which case you get the minimum or you could allot the entire 1 to 200 in which case you get the maximum. The overall sum will lie between 100 and 200 only.

karishma hello why are you multiplying by 0. Zero is neither negative nor positive. No ? and condition says that "x and y are positive numbers" have a great day!

Since we know that x and y are positive numbers but may not be integers, we have assumed that x is very close to 0 but not 0 e.g. x = 0.000000001 (which is a positive number).
100 multiplied by this will become 0.0000001 which is almost 0.

If 0<x, 0<y, & x+y=1, systematically breakdown possible scenarios where this is true:
.1 + .9 = 1
.2 + .8 = 1
.3 + .7 = 1
.4 + .6 = 1
.5 + .5 = 1

if you quickly apply this range of possible values to x & y, 100x + 200y cant possibly equal 80. If I am overgeneralizing, please let me know. But if my understanding is correct, this is probably the best approach to saving time and keeping things simple.