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100X + 200Y

by pappueshwar » Wed Mar 14, 2012 6:32 am
If x, y are positive numbers and x+y=1, which of the following can be the value of 100x+200y?
I. 80 II. 140 III. 199

A. I only
B. â…¡ only
C. â… ,â…¡ only
D. â…¡,â…¢ only
E. â… ,â…¡ and â…¢

OA IS C WHICH COMBINATION OF X AND Y SHOULD BE TAKEN ?
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by GMATGuruNY » Wed Mar 14, 2012 7:07 am
pappueshwar wrote:If x, y are positive numbers and x+y=1, which of the following can be the value of 100x+200y?
I. 80 II. 140 III. 199

A. I only
B. â…¡ only
C. â… ,â…¡ only
D. â…¡,â…¢ only
E. â… ,â…¡ and â…¢

OA IS C WHICH COMBINATION OF X AND Y SHOULD BE TAKEN ?
Put 100x+200y in terms of one of the variables.

Rephrasing x+y=1 as y=1-x and substituting for y in 100x+200y, we get:
100x + 200(1-x) = 100x + 200 - 200x = 200-100x.

Since x+y=1 and both x and y are positive, 0<x<1.
To determine the viable range here, convert 0<x<1 to 200-100x:
0 < x < 1
0 > -100x > -100
200 > 200-100x > 100.

Thus, 200-100x -- which is the equivalent of 100x+200y -- can be any value between 100 and 200.

The correct answer is D.

II: 140
200-100x = 140
-100x = -60
x=.6, implying that y=.4.
100(.6) + 200(.4) = 60+80 = 140.

III: 199
200-100x = 199.
-100x = -1
x = .01, implying that y=.99.
100x + 200y = 100(.01) + 200(.99) = 1 + 198 = 199.

I: 80
200-100x = 80
-100x = -120
x = 1.2, implying that y=-.2.
Doesn't work, since y must be positive.

Only II and III are possible.
Last edited by GMATGuruNY on Wed Mar 14, 2012 7:14 am, edited 1 time in total.
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by killer1387 » Wed Mar 14, 2012 7:09 am
pappueshwar wrote:If x, y are positive numbers and x+y=1, which of the following can be the value of 100x+200y?
I. 80 II. 140 III. 199

A. I only
B. â…¡ only
C. â… ,â…¡ only
D. â…¡,â…¢ only
E. â… ,â…¡ and â…¢

OA IS C WHICH COMBINATION OF X AND Y SHOULD BE TAKEN ?
I doubt the OA
imo D
FOR X=3/5;Y=2/5
100X+200Y=140 HENCE II IS CORRECT.

FOR X=0.01 AND Y=0.99
100X+200Y=199
HENCE III IS CORRECT.

HENCE D
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by icemanKK » Sat Mar 24, 2012 6:28 pm
Hi guys,

Do you think the following approach can be applied.

Multiplying 100 with x+y = 1
we get -> 100x+100y = 100 -----> 1

Multiplying 200 with x+y = 1
we get -> 200x+200y = 200 -----> 2

Now the question is 100x+200y = ?

Since the numbers are both positive, Using equations 1 and 2 the following can be applied.

100x+200y > 100
and
100x+200y < 200

The only 2 values in this range are 140 and 199. Hence D[/spoiler]
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by jsfinol » Sat Jul 23, 2016 5:03 pm
Does anyone know is the score difficulty of this question?
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by [email protected] » Sat Jul 23, 2016 5:57 pm
Hi jsfinol,

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: D

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by Matt@VeritasPrep » Thu Aug 04, 2016 9:57 pm
It might be easier to say that

x + y = 1

so

100x + 100y = 100

and

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

So we have everything in terms of one variable. Then we know that y > 0, so 100 + 100y > 100 + 100*0, meaning our solution must be greater than 100.

We also know that x > 0, so 1 > y. That means 100 + 100y < 100 + 100, so our answer must be less than 200.

From here, anything between 100 and 200 is fine, so we've got two possible answers of the three given.
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by Matt@VeritasPrep » Thu Aug 04, 2016 10:00 pm
jsfinol wrote:Does anyone know is the score difficulty of this question?
I'd need to see response data, but I'd guess it's around the 50th percentile of the current quant scale, maybe a little higher. (What'd be really interesting is the average time each test taker takes to complete it, though! I'd guess it's more than 2 minutes.)
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