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How is the answer to this C when the solution set comes out as -1/3=<j<= 5? Doesn't solution C of j<=5 also extends for all values that are lesser than 5 ?
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How is the answer to this C when the solution set comes out as -1/3=<j<= 5? Doesn't solution C of j<=5 also extends for all values that are lesser than 5 ?

\(|\frac{7 - 3j}{2}| \leq 4\) is equivalent to \(-\frac{1}{3} \leq j \leq 5\).

So, if \(-\frac{1}{3} \leq j \leq 5\) then it must be true that j is less than or equal to 5. Or in another way, any possible j from the given range (\(-\frac{1}{3} \leq j \leq 5\)) will for sure be less then or equal to 5. That's why the answer is C.

Check similar questions here: Trickiest Inequality Questions Type: Confusing Ranges.

Hope it helps.
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Kinshook
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If \(|\frac{7 - 3j}{2}| \leq 4\), then which of the following must be true?


A. \(j \leq -\frac{1}{3}\)

B. \(j \leq \frac{3}{7}\)

C. \(j \leq 5\)

D. \(j \geq \frac{3}{7}\)

E. \(j \geq 5\)

Asked: If \(|\frac{7 - 3j}{2}| \leq 4\), then which of the following must be true?

\(|\frac{7 - 3j}{2}| \leq 4\)
\(|7 - 3j| \leq 8\)
|3j - 7| <=8

- 8 <= 3j - 7 <=8
-1 <= 3j <= 15
-1/3 <= j <= 5

IMO C

Hi Kinshook,
Could you please explain the logic behind flipping the values inside the modulus in the highlighted part?
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Kinshook
Bunuel
If \(|\frac{7 - 3j}{2}| \leq 4\), then which of the following must be true?


A. \(j \leq -\frac{1}{3}\)

B. \(j \leq \frac{3}{7}\)

C. \(j \leq 5\)

D. \(j \geq \frac{3}{7}\)

E. \(j \geq 5\)

Asked: If \(|\frac{7 - 3j}{2}| \leq 4\), then which of the following must be true?

\(|\frac{7 - 3j}{2}| \leq 4\)
\(|7 - 3j| \leq 8\)
|3j - 7| <=8

- 8 <= 3j - 7 <=8
-1 <= 3j <= 15
-1/3 <= j <= 5

IMO C

Hi Kinshook,
Could you please explain the logic behind flipping the values inside the modulus in the highlighted part?


You don't have to do it, I think he just did it so it looks cleaner. You'll end up flipping the signs when you divide by -3 anyway.
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Bunuel
If \(|\frac{7 - 3j}{2}| \leq 4\), then which of the following must be true?


A. \(j \leq -\frac{1}{3}\)

B. \(j \leq \frac{3}{7}\)

C. \(j \leq 5\)

D. \(j \geq \frac{3}{7}\)

E. \(j \geq 5\)

Multiplying both sides of the inequality by 2, we have:

|7 - 3j| ≤ 8

When we have a “less than or equal to” inequality, we can re-express it as a “between” statement, as shown below, and this allows us to drop the absolute value sign.

-8 ≤ 7 - 3j ≤ 8

-15 ≤ -3j ≤ 1

5 ≥ j ≥ -1/3

Answer: C
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For me, Visualization helps in solving mod questions.
as I know |a| ≤ 4 means the solutions would lie between -4 to +4 (inclusive) on the number line as modulus means the distance from the origin to the solution on number line.

hence 7−3j => -8 to +8(inclusive)
−3j => -15 to 1 (inclusive)
-j => -5 to 1/3 (inclusive)
j => reverse the solutions on number line => -1/3 to 5 (inclusive)
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Bunuel
If \(|\frac{7 - 3j}{2}| \leq 4\), then which of the following must be true?


A. \(j \leq -\frac{1}{3}\)

B. \(j \leq \frac{3}{7}\)

C. \(j \leq 5\)

D. \(j \geq \frac{3}{7}\)

E. \(j \geq 5\)

\(j=0\) satisfies \(|\frac{7 - 3j}{2}| \leq 4\).
Since the correct range must include 0, eliminate A, D and E.

\(j=1\) satisfies \(|\frac{7 - 3j}{2}| \leq 4\).
Since the correct range must include 1, eliminate B.

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neo92
How is the answer to this C when the solution set comes out as -1/3=<j<= 5? Doesn't solution C of j<=5 also extends for all values that are lesser than 5 ?

\(|\frac{7 - 3j}{2}| \leq 4\) is equivalent to \(-\frac{1}{3} \leq j \leq 5\).

So, if \(-\frac{1}{3} \leq j \leq 5\) then it must be true that j is less than or equal to 5. Or in another way, any possible j from the given range (\(-\frac{1}{3} \leq j \leq 5\)) will for sure be less then or equal to 5. That's why the answer is C.

Check similar questions here: Trickiest Inequality Questions Type: Confusing Ranges.

Hope it helps.


I have a doubt with this. If it is <= 5 then it can also be a value that is - infinity. And if we suppose that then the equation j>= -1/3 will be invalid. Hence what is the logic behind this?

Posted from my mobile device
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Varane
I have a doubt with this. If it is <= 5 then it can also be a value that is - infinity. And if we suppose that then the equation j>= -1/3 will be invalid. Hence what is the logic behind this?

Posted from my mobile device

You have correctly determined that the following statement is not valid:
If \(j ≤ 5\), then \(|\frac{7 - 3j}{2}| \leq 4\).
Since values less than -1/3 do not satisfy the given equation, the statement above is not justified.

However, option C implies the following statement:
If \(|\frac{7 - 3j}{2}| \leq 4\), then \(j ≤ 5\).
The statement above is true, since every value that satisfies the given equation is less than or equal to 5.
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