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

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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 22 Oct 2017, 07:14
I solved this question using a very simple technique.
you have an equation given which is x+y=1. Now your second equation would be 100x+200y= ? ( put in one of the answer choice and solve for either x or y)
for option A you will get -ve value which can not be true as the question clearly states that x and y are positive. try out for other options...
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If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 22 Oct 2017, 07:23
It's a simple one.

Given that x+y=1

100x+200y=?

Ans: 100(x+2y)
=100(x+y+y)
=100(1+y)
Now if x and y are positive numbers: 0<y<1

Hence II and III
-E

+1 for Kudos
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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 26 Nov 2017, 03:10
VeritasPrepKarishma wrote:
BrainLab wrote:
If x and y are positive numbers such that x + y = 1, which of the following could be the value of 100x + 200y?

I. 80
II. 140
III. 199

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III


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)

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!
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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 26 Nov 2017, 03:15
dave13 wrote:
VeritasPrepKarishma wrote:
BrainLab wrote:
If x and y are positive numbers such that x + y = 1, which of the following could be the value of 100x + 200y?

I. 80
II. 140
III. 199

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III


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)

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.
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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 14 Dec 2018, 04:45
BrainLab wrote:
If x and y are positive numbers such that x + y = 1, which of the following could be the value of 100x + 200y?

I. 80
II. 140
III. 199

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III


Bunuel and VeritasKarishma

Can u guys post the links of few similar questions like this.. Please.
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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 26 Mar 2019, 02:02
BrainLab wrote:
If x and y are positive numbers such that x + y = 1, which of the following could be the value of 100x + 200y?

I. 80
II. 140
III. 199

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III


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.
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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 26 Mar 2019, 13:15
Hi aaronhew,

YES - thinking about the information in this prompt in this way IS a fast way to deal with the given information. In the broader sense, you're thinking in terms of the 'limits' of what's possible.

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 lower and upper "boundaries" to the possible sum.

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

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New post 26 Mar 2019, 17:56
BrainLab wrote:
If x and y are positive numbers such that x + y = 1, which of the following could be the value of 100x + 200y?

I. 80
II. 140
III. 199

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III


if x+y=1,
then 100x+100y=100
thus, 100x+200y>100
ll and lll
E
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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 12 May 2019, 01:17
BrainLab wrote:
If x and y are positive numbers such that x + y = 1, which of the following could be the value of 100x + 200y?

I. 80
II. 140
III. 199

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III


x+y=1
=>x=1-y

Now,
100(1-y)+200y
=>100-100y+200y
=>100+100y
=>100(1+y)

Both x & y are positive
Now we test the value:
If y=.2
100*1.2=120 it means that the value must be greater than 100 but less than 200.

140 & 199 are the possible values

Answer is E

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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 02 Jun 2019, 03:04
If x and y > 0 and x+y=1 (or x=1-y), then

100x+200y => 100(1-y)+200y => 100y+100, which means we won't end up with a result less than 100 -- so 80 is out

II and III are the answers => E
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Re: If x and y are positive numbers such that x + y = 1, which of the  [#permalink]

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New post 18 Nov 2019, 14:26
VeritasKarishma wrote:
BrainLab wrote:
If x and y are positive numbers such that x + y = 1, which of the following could be the value of 100x + 200y?

I. 80
II. 140
III. 199

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III


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)


karishma

-- I understand how you eliminated I

But i had to use up another 2 minutes thinking how can i actually achieve option II and option III respectively (actually calculated if x and y are 0.6 and .4 & 0.01 and 0.99 respectively)

Per your logic -- it seems like, you didnt even bother calculating what fractions are needed to achieve option II and option III

Could you clarify if that is what you would do in the exam -- i.e. -- once you eliminated I -- you immediately select option E and don't even waste another 2 mins actually thinking about WHAT fractions for X and Y do you need to get 140 and 199 respectively...you know, it can be possible (but dont care what the actual numbers are..) and thus select E

If true, i presume that is because you know 100x + 200y is between 100 < 100x + 200y < 200

Thank you !
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Re: If x and y are positive numbers such that x + y = 1, which of the   [#permalink] 18 Nov 2019, 14:26

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