If 3x + 5y 15, which of the following must be true?

Question

If 3x + 5y < 15 and 5x + 3y > 15, which of the following must be true? I. x > y II. x < y III. x > 3 (A) I only (B) II only (C) III only (D) I and II only (E) I and III only

SaquibHGMATWhiz · Accepted Answer

Solution: We have 3x + 5y < 15.....(i) and 5x + 3y > 15 or -5x-3y<-15......(ii) (We do this because we need to keep the inequality sign same to do the addition) When we see the options, we see that we need to compare x and y and the value of x if it is greater than 3 or not Adding inequality i and ii , we get: 3x-5x+5y-3y<15-15 ⇒-2x+2y<0 ⇒2y<2x ⇒yy MUST BE true. Thus we can eliminate options B, C and D Now we have 3x + 5y < 15.....(i) and -5x-3y<-15......(ii) Multiplying equation (i) with 3, equation (ii) with 5, and adding the equations, we get: 9x+15y-25x-15y<45-75 ⇒-16x<-30 ⇒16x>30 ⇒x>\frac{30}{16} ⇒x>1. imes imes So, statement III i.e., x>3 is NOT a MUST. It can be true but not a MUST. Hence the right answer is Option A

Braintree · Answer

I did it by logic as follows:
a. First see that x has to be greater than y, because we are replacing 2y by 2x to the first inequality, resulting in the sum increasing from less than 15 to greater than 15. Eliminate B, C, and D.
b. also see that logically, both I and II cannot be correct, another reason to eliminate D.
c. y can be any number, as long as it’s less than x. so in an extreme case in which x and y are almost equal, and 3x+5y is almost equal to 15 – therefore at that extreme case – x just needs to be greater than 15/8, and need not be greater than 3. Eliminate E.

Answer is A.

AmritaISBAspirant · Answer

can you please elaborate the solution? I didn't understand.
What about the condition 5x+3y>15

AmritaISBAspirant · Answer

Thank you so much for writing the detailed solution.
The values we got is x > 1.8
y < x
when we plug these values in the inequalities :
assume x=2
y=1
5(x)+3(y)>15
5(2)+3(1)>15 which is not true. Why is this happening...

SaquibHGMATWhiz · Answer

This is a very common yet interesting mistake that a lot of test takers make. In the case of inequalities, plugging values doesn't really work in the same way as it does for linear equations . Let's understand why. Firstly, let us try and get the range of y in the same way we got the range of x We have 3x + 5y < 15.....(i) and -5x-3y<-15......(ii) Multiplying equation (i) with 5, equation (ii) with 3, and adding the equations, we get: 15x+25y-15x-9y<75-45 ⇒16y<30 ⇒y<\frac{30}{16} ⇒y<1.8 Now does it mean we can take any values of x > 30/18 and any value of y < 30/18 and the inequalities will be satisfied? No. Let us also understand this using a graph which should give you a lot clearer idea. graphinequality.png The red part is the region of 3x + 5y < 15 while the blue is the region of 5x + 3y > 15 and the overlapping part is what we are concerned with. If you look closely, coordinates like (2, 1) do not lie in the overlapping range. The idea behind is really simple. All the values in the overlapping range will follow the trend of x > y, x > 1.8, and y < 1.8. However, all the values following the trend of x > y, x > 1.8, and y < 1.8 will not necessarily fall in the overlapping range. It is similar to "all alkalies are bases, but all bases are not alkali"

Hope this helps.

AmritaISBAspirant · Answer

I can't thank you enough for this explanation!!!!

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If 3x + 5y < 15 and 5x + 3y > 15, which of the following must be true?

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