If 0<2x+3y<10 and -10<3x+2y<0, then which of the following must be tru

Hi MathRevolution,

agree with chetan2u. the OA is wrong, kindly correct it.

We see that x and y can’t be both positive or both negative. If x and y are both positive, then the second inequality will not hold. Similarly, if x and y are both negative, then the first inequality will not hold. Therefore, we must consider two separate cases: (1) x is negative and y is positive and (2) x is positive and y is negative.

Case 1. Let’s assume that x is negative and y is positive.

For example, If x = -5 and y = 5, we see that we do have 0 < 2x + 3y < 10 and -10 < 3x + 2y < 0.

Case 2. Now let’s assume that x is positive and y is negative. We see that the absolute value of y must be greater than the absolute value of x in order for the second inequality to hold.

For example, if x = 2, y has to be less than -3 in order to have -10 < 3x + 2y < 0. However, in that case, the first inequality will never hold since 2x + 3y will be negative. Thus we can’t have x as positive and y as negative.

Thus it must be true that x is negative and y is positive and in that case we also have x < y. Thus, Statements I and III must be true.

Answer: D

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Label the inequalities as follows:
0<2x+3y<10 --- (1)
-10<3x+2y<0 --- (2)
We consider each statement individually.

Statement I:
Multiplying (1) by -2 yields -20 < -4x – 6y < 0, and multiplying (2) by 3 yields -30 < 9x + 6y < 0. Adding these inequalities gives -50 < 5x < 0 or -10 < x < 0.
This statement is true.

Statement II:
Multiplying (1) by 3 yields 0 < 6x + 9y < 30, and multiplying (2) by -3 yields 0 < -6x – 4y < 20. Adding these inequalities gives 0 < 5y < 50 or 0 < y < 10.
This statement is false.

Statement III:
Multiplying (1) by – 1 yields -10 < -2x – 3y < 0. Adding this to inequality (2) yields -20 < x – y < 0.
This implies that x < y, and statement III is true.

Therefore, the answer is D.
Answer: D

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