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If a and b are two consecutive

This topic has 4 expert replies and 0 member replies

If a and b are two consecutive

Post Mon Sep 25, 2017 7:20 pm
If a and b are two consecutive even integers such that a > b and 3a < 2b. then which of the following describes the range of possible values for b?

A) b > 0
B) b < 0
C) b < -6
D) b > 6
E) b < 4

The OA is C.

I need help. Experts can you explain this PS question to me? Thanks.

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Post Fri Dec 15, 2017 9:50 am
Vincen wrote:
If a and b are two consecutive even integers such that a > b and 3a < 2b. then which of the following describes the range of possible values for b?

A) b > 0
B) b < 0
C) b < -6
D) b > 6
E) b < 4
Since a and b are consecutive even integers and a > b, we can say that a = b + 2.

We can substitute b + 2 for a in the inequality 3a < 2b and we have:

3(b+2) < 2b

3b + 6 < 2b

b < -6

Answer: C

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Post Tue Sep 26, 2017 5:10 pm
We could also set a in terms of b at the outset: a = b + 2, since a > b and they're consecutive evens.

With that, replace a with b + 2 in the second inequality:

2b > 3 * (b + 2)

2b > 3b + 6

-6 > b

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Post Mon Sep 25, 2017 9:36 pm
Vincen wrote:
If a and b are two consecutive even integers such that a > b and 3a < 2b. then which of the following describes the range of possible values for b?

A) b > 0
B) b < 0
C) b < -6
D) b > 6
E) b < 4

The OA is C.

I need help. Experts can you explain this PS question to me? Thanks.
It is given that a and b are two consecutive even integers such that a > b and 3a < 2b.

3a < 2b => 1.5a < b

Given that a > b, if a were positive, 1.5a > b; however, it is given that 1.5a < b. It implies that a and b are negative even integers.

The correct answer must be either B or C.

To choose between B and C, let's choose a test value for b that lies between 0 and -6.

Say b = -4, thus a = -2 (consecutive even and a > b).

Test this in 1.5a < b.

At a = -2 and b = -4, 1.5a < b => 1.5*(-2) ? -4 => -3 > -4. This is incorrect possible value of b, thus b must not lie between 0 and -6. Thus, the correct answer is C.

Hope this helps!

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Post Tue Sep 26, 2017 5:08 pm
We want a common term in each inequality in order to chain them together, so multiply the first inequality by 3.

With that, we've got 3a > 3b and 2b > 3a. That chains into 2b > 3a > 3b, or 2b > 3b.

2b > 3b isn't possible for any positive b, so b is negative. But if b = -2, then a = 0 and 2b > 3a doesn't hold. Likewise, if b = -6, then a = -4 and 2b > 3a doesn't hold, so b < -6.

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