If x and y are positive numbers such that x + y = 2, then which of the

Question

Question-1

If x and y are positive numbers such that x + y = 2, then which of the following could be the value of 300x+200y?
 1. 156
2. 397
3. 533

Options

A. I only
B. II only
C. III only
D. I and II
E. II and III

EgmatQuantExpert · Accepted Answer

Solution

Given:  
 • x and y are positive integers
• x + y = 2

To find:  
 • From the given options, which one could be the value of 300x + 200y

Approach and Working:   
We can rewrite the given expression 300x + 200y in the following manner:
 • 300x + 200y = 200x + 200y + 100x = 100x + 200 (x + y) = 100x + 200*2 = 100x + 400

Now, as x is positive, 100x > 0
 • Hence, 400 + 100x > 400

Only one value is more than 400 (i.e. 533)

Hence, the correct answer is option C.

Answer: C

Hero8888 · Answer

300x+200y =200*(3x+2y)=200(2x+2y+x)=200(2+x)=400+200x - so the value will depend on  x  only

We are given that x and y are positive numbers such that x + y = 2

x will be  min  = 0.00..1 when y=1.99...9
x will be  max  = 1.99...9 when y=0.00..1

Obviously 400<400+200x<800 . Only 533 is in that range

Answer C. III only

gmatzpractice · Answer

300x+200(2-x)
400+100x
x>0
Answer C. Greater than 400

Posted from my mobile device

KSBGC · Answer

This question is designed based on weighted average method. This question can be solved easily following few steps.

Given,

300x + 200y

=  \frac{600x + 400y}{2} 
 = \frac{600x + 400y}{x+y}

Now we know that weighted average value must be between 400 to 600 where x and y are their weights.

533 falls between 400 to 600.

Only option match is C.

NDND · Answer

both x & y are positive numbers meanwhile their sum equals to 2
Therefore, x&y could be (1,1)or (0,2) or (2,0)

therefore 300x+200y= 300(1)+200(1)= 500
Or = 300(0)+200(2)= 400
Or = 300(2)+200(0)=600

Therefore, C is the correct answer, as all the answers are greater than 400

lybeaver · Answer

x + y = 2 => x = 2-y and y = 2-x

x, y are positive numbers => \(0\leq{x}\) < 2 and \(0\leq{y}\) <2

300x+200y = 300(2-y)+200y = 600-100y. Because \(0\leq{y}\) <2 => 600-100y < 600 (1)

300x+200y = 300x+200(2-x)=400+100x. Because \(0\leq{x}\) => 400 + 100x > 400 (2)

Combine (1) and (2), the answer falls between 400 and 600, hence the correct answer is 533 (Option C)

ColumbiaEMBA · Answer

IIRC, 0 is neither positive nor negative (it is however even) so you cannot set one of the variables to 0 as the stem states that x and y are  positive  numbers.

NDND · Answer

I guess you are right, which means that the only option we have is (1,1), and this would save us at least half the time, right? Thanks for correcting me

NDND · Answer

Thanks EgmatQuantExpert could you comment on the approach used on this post? https://gmatclub.com/forum/if-x-and-y-a ... l#p2088234 Is it ok? Thx

EgmatQuantExpert · Answer

Hey  NDND ,
can you specify which solution you are referring to? The link you have provided somehow reverts back to the above post only.

NDND · Answer

Here you go:

both x & y are positive numbers meanwhile their sum equals to 2
Therefore, x&y could be (1,1)or (0,2) or (2,0)

therefore 300x+200y= 300(1)+200(1)= 500
Or = 300(0)+200(2)= 400
Or = 300(2)+200(0)=600

Therefore, C is the correct answer, as all the answers are greater than 400

EgmatQuantExpert · Answer

Although the approach and final answer is absolutely correct, there is a very small error in the calculation part. In the question it was mentioned that both x and y are positive integers, hence they are always greater than 0. • Non-negative integers (Whole Numbers) start from 0. • Positive integers (Natural Numbers) start from 1. As both x and y are positive integers, they can't be equal to 0, hence, the only possible case will be (1, 1). You have identified that also in one of your previous replies.

axmanX · Answer

You don't have to solve for x&y or even suppose values.

given: that x + y = 2, 
to find: possible values for 300x+200y?

Solution:
 300x+200y = 100x + 200(x+y) = 100x + 200(2) = 100x + 400

Since x is a +ve value, so whatever it may be, it won't reduce the value of the equation from 400 in any case

So, any solution would have 2 be more than 400, which leaves us with just Option 3!

gracie · Answer

if x+y=2,
then 300x+300y=600
testing options:
1. subtracting 300x+200y=156,
100y=600-156=444
y=4.44 NO, must be less than 2
2. subtracting 300x+200y=397,
100y=600-397=203
y=2.03 NO, must be less than 2
subtracting 300x+200y=533,
100y=600-533=67
y=.67 YES, less than 2
III only
C

Rishovnits · Answer

Given,

X+Y=2.................1)
300x+200Y=?

Multiplying Eqn 1) with 2 ,
200X+200Y=200x2=400..............2)
Add another 100X to the equation derived.
300X+200Y=400+100X
It is going to be greater than 400 since X is positive.

Hence answer C.

NandishSS · Answer

HI  VeritasKarishma

Can you help me how to solve this question using weighted average concept ?

KarishmaB · Answer

First of all, I am glad you thought of weighted averages here.

Now, x and y are positive numbers so they could be the weights allotted to 300 and 200 respectively.

Avg = (300x + 200y)/(x + y)

Avg = (300x + 200y)/2

(300x + 200y) = 2 * Avg

Now, in any case, the weighted avg of 200 and 300 will lie between 200 and 300 only. This means 2*Avg will lie between 400 and 600.

Only 533 satisfies this so answer is (C).

Pallavimalik2597 · Answer

If we consider x =0, and it is given that x+y=2 then it means y=2

And, putting x=0,y=2 in the given equation we will get to know the max range is 600, now do the same thing for y=0, x=2 we see min is 400

So, x will lie between 400-600 only option is C

Posted from my mobile device

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