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

There are many ways in which you can deal with this question. A very efficient method is using weighted average concept.

Note that \(100x + 200y = \frac{(100x + 200y)}{1} = \frac{(100x + 200y)}{(x + y)}\)

So the expression is the weighted average of 100 and 200 where x and y are the weights. The weighted average of 100 and 200 will lie between 100 and 200. So it could be 140 and 199 but it cannot be 80.

Answer (E)

Check out the post on weighted averages here:
https://anaprep.com/arithmetic-weighted-averages/
and this video: https://www.youtube.com/watch?v=_GOAU7moZ2Q
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Given that x + y = 1
or x= 1-y

we need to find possible values of 100x + 200y
plugging in x from above,
100(1-y) + 200y or 100 + 100 y
Now check for each value
1) 100+100y=80
Not possible because y would become negative in this case

2) 100+100y=140
y=40/100 = 2/5 so x =3/5
Possible

3) 100+100y = 199
y=99/100 so x=1/100
Possible

Answer:- E

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)
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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

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.

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.

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
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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)

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.
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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.

Hi suganyam,

Based on the way that the overall prompt is written, your deduction (above) is fine. However, it's not completely accurate - and on other types of questions, that type of generality could lead you to choose a wrong answer. There is an 'upper-limit' to how high 100+100Y can be in this question, meaning that it's more appropriate to say that that value is greater than 100 AND less than 200.

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

Given Condition: x + y = 1
Therefore, the values of x and y lie between 0 and 1.
i.e., 0 < (x, y) < 1

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

100x + 200y = 100(1 + y)
Because y lies between 0 and 1, the value of 1 + y is greater than 1 and lesser than 2.
From the above expression, it is evident that the value of 100(1 + y) is greater than 100 and lesser than 200. i.e., 100 < 100(1 + y) < 200

From the given answer options, II and III satisfy the above condition.

Choice E is the correct answer.

VeritasPrepKarishma

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

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.
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x+y= 1 -->x=1-y --->100(x+2y) 100(1-y+2y)--> 100+100y so any value greater than 100 for sure

The question is really identical to this one: x% of employees are men and the rest, y%, are women. If men make $100 per hour and women make $200 per hour, what could be the average hourly wage for all employees? Or at least it's identical if you think of your percentages as decimals (so you think of "10%" as "0.1"). So it's just a weighted average, and the answer can be anything between 100 and 200. That's the fastest way to look at the problem (but I'm really just saying exactly what Karishma already wrote at the top of this thread).

You can also use substitution (y = 1 - x, then plug that into 100x + 200y to get 200 - 100x) to quickly see what values are possible -- you just have to remember that the value of x is between 0 and 1.
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