# Lesson: Guessing Strategically

## Comment on Guessing Strategically

### At 2:11 you say that the

At 2:11 you say that the correct answer is C. But why? When using the table-method I can see that x has to be at least 4. And if x would be -4, you cannot solve the equitation. So I think that the answer is B. What am I doing wrong?

### For statement 2, consider

For statement 2, consider these two cases:

Case a: x = -1. Notice that this satisfies the given inequality. Here, x < 3

Case b: x = 10. Notice that this satisfies the given inequality. Here, x > 3

So, statement 1 alone is not sufficient.

Thank you!

### Solving statement 2: we get x

Solving statement 2: we get x^3-9x>0 or x*(x^2-3^2)>0
x(x-3)(x+3)>0 if we consider x=1, we will get a negative value, and if consider x=3 than we will get zero, only if we consider a value above 3 we will satisfy the second statement. The tricky part is when u combime statements for values of x=1,2 from statememt 1 you get negative values for statement 2 and end up marking option E as the answer. In order to avoid such mistakes using a number line can be one of the option. Great queation cheers!

### You're referring to the

You're referring to the question that starts at 1:30.
Just to be clear, the correct answer for this question is C (I wasn't sure if you were suggesting that the answer is actually E)

### Thanks for the response. I

Thanks for the response. I was able to understand the reasoning behind the correct choice (c). My comments were just to highlight a situation where someone could mark (E) as the answer.

### I figured so. Yes, a good

I figured so. Yes, a good point and great advice about the number line.

### What is number line? can you

What is number line? can you please elaborate?

### gary391 is referring how we

gary391 is referring how we can use a number line to help us solve higher-order inequalities.

This video (starting at 0:31) shows how this technique works: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

### Hi Brent,

Hi Brent,
What is the best way to solve the question that starts at 1.30? I plugged in values for x in statement 2 but it took me more than 2 min to solve it. Is there a way to solve it algebraically without testing values for x? What is the difficulty level of the question?
Thanks for a quick response to my previous doubts as well.

### Here's an algebraic solution

Here's an algebraic solution to the inequality in statement 2:

First, we'll solve the EQUATION: sqrt(x³ - 9x + 4) = 2
Square both sides to get: x^3 - 9x + 4 = 4
Subtract 4 from both sides to get: x^3 - 9x = 0
Factor: x(x² - 9) = 0
Factor more: x(x+3)(x-3) = 0
So, x = 0, 3 and -3 [for the EQUATION: sqrt(x³ - 9x + 4) = 2]

These values (0, 3 and -3) divide the number line into 4 REGIONS
REGION 1: x < -3
REGION 2: -3 < x < 0
REGION 3: 0 < x < 3
REGION 4: x > 3
-------------------------------------------------------------------
If x is in REGION 1, then it is NOT the case that sqrt(x³ - 9x + 4) > 2 (try replacing x with any value less than -3 and see what happens)
So, x is NOT in REGION 1
---------------------------------------------------------------------
If x is in REGION 2, then it IS the case that sqrt(x³ - 9x + 4) > 2
So, x COULD be in REGION 2
In other words, it COULD be the case that -3 < x < 0
---------------------------------------------------------------------
If x is in REGION 3, then it is NOT the case that sqrt(x³ - 9x + 4) > 2
So, x is NOT in REGION 3
---------------------------------------------------------------------
If x is in REGION 4, then it IS the case that sqrt(x³ - 9x + 4) > 2
So, x COULD be in REGION 4
In other words, it COULD be the case that x > 3
---------------------------------------------------------------------
STATEMENTS 1 & 2 COMBINED
Statement 2 tells us that EITHER -3 < x < 0 OR 3 > x
Statement 1 tells us that x > 0, which means we can RULE OUT the possibility that -3 < x < 0
So, it MUST BE the case that x > 3

Does that help?

Cheers,
Brent

### Yes, that helps a lot! Thank

Yes, that helps a lot! Thank you so much Brent :)

### Hi it helps thank you :)

Hi it helps thank you :)

### By the way, here's the video

By the way, here's the video on solving quadratic inequalities: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

### Hi Brent!

Hi Brent!

I absolutely love your material. It is super informative and truly insightful. Thank you for providing me with it.

I had a question. I was hoping you could help me clarify it:

Could you please provide the solution for statement 2 in the first question (timestamp: 00:36) i.e. by providing the solution for the mentioned inequality and elaborating why it would be insufficient?

Thank you in advance

### Thanks for the kind words

Thanks for the kind words about the course!

TARGET QUESTION: Is x > 2?

Statement 2: |x² - 3x + 2| < 5

Applying a rule for absolute inequalities, we can rewrite the inequality as follows:
-5 < x² - 3x + 2 < 5

We might even subtract 2 from all 3 sides to get: -7 < x² - 3x < 3

At this point, rather than try to solve this mess for x, it's probably fastest to find solutions by testing x-values that are above and below 2

TRY x = 1
If x = 1, we get: -7 < 1² - 3(1) < 3
Simplify: -7 < -2 < 3.
Perfect, the inequality holds true, which means x = 1 IS a solution to the given inequality.
In this case, the answer to the target question is "NO, x is NOT greater than 2"

TRY x = 3
If x = 3, we get: -7 < 3² - 3(3) < 3
Simplify: -7 < 0 < 3.
Perfect, the inequality holds true, which means x = 3 IS a solution to the given inequality.
In this case, the answer to the target question is "YES, x IS greater than 2"

So, statement 2 is not sufficient.

Cheers,
Brent

### Thank you Brent! :)

Thank you Brent! :)

### Hi Brent, could you elaborate

Hi Brent, could you elaborate how is 7 < -2 < 3 make x is NOT greater than 2 and same with -7 < 0 < 3 part please? Thanks.

If I try to solve it separately, x² - 3x - 3 = 0 make x = 3 or 1 but I'm not sure how to proceed for second part from x² - 3x + 7 = 0 ?

### If |x² - 3x + 2| < 5, then we

If |x² - 3x + 2| < 5, then we can conclude that -5 < x² - 3x + 2 < 5
So, we're looking for values of x that make this inequality true.

Let's test some possible values of x to see if they satisfy the inequality -5 < x² - 3x + 2 < 5

If x = 1, we get: -5 < (1)² - 3(1) + 2 < 5
This simplifies to: -5 < 0 < 5.
It works! So, x = 1 is one possible solution to the inequality.

The target question asks "Is x > 2?"
Since x COULD equal 1, the answer to the target question is "NO, x is not greater than 2.

Now let's test another possible value of x.

If x = 3, we get: -5 < (3)² - 3(3) + 2 < 5
This simplifies to: -5 < 2 < 5.
It works! So, x = 3 is another possible solution to the inequality.

The target question asks "Is x > 2?"
Since x COULD equal 3, the answer to the target question is "YES, x is greater than 2.

Since our two possible x-values yield different answers to the target question (NO and YES), statement 2 is not sufficient

### Thanks Brent. Make sense now.

Thanks Brent. Make sense now.
To clarify this seems a better method compared to number lines when it involve Modulus? Is my understanding correct?

Lastly this x² - 3x + 7 = 0 has no solution right?

### For this particular question,

For this particular question, I think the about strategy is better than using number lines.

That's right, the equation x² - 3x + 7 = 0 has no solution.
We know this because, for this equation a = 1, b = -3 and c = 7
So, when we plug these values into the the discriminant,
we get: b² - 4ac = (-3)² - 4(1)(7) = -19
Since the discriminant is negative, the quadratic equation doesn't have a solution.

For more on this property, check out the following video, starting at 7:40
https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

### Great explanation as always.

Great explanation as always. Thanks Brent.

### Hi! could you please explain

Hi! could you please explain your reasoning (at 1:41)
1) x>0 is not sufficient
why it is not sufficient?

### I'm happy to help!

I'm happy to help!

Target question: Is x > 3 (Is x greater than 3)?

Statement 1: x > 0 (x is greater than 0)
There are several values of x that satisfy statement 1. Here are two:
CASE A: x = 1. In this case, the answer to the target question is NO, x is NOT greater than 3.
CASE B: x = 5. In this case, the answer to the target question is YES, x IS greater than 3.

Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Does that help?

Cheers,
Brent

### Al, Pablo, and Marsha shared

Al, Pablo, and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance on the trip?

(1) Al drove 1 hour longer than Pablo but at an average rate of 5 miles per hour slower than Pablo.
(2) Marsha drove 9 hours and averaged 50 miles per hour.

Hi Brent
Combining both the statements, we get an equation
ST + (S-5)(T+1) = 1050

Though there are two variables in the equation but there should be only 1 combination, which should atisfy the equation, so I chose C. The answer, however, is option E.

Is there a method through which we can quickly solve these questions?

### Question link: https:/

An equation with two variables can almost never be solved for the two variables.
Once you get to the point where you have the equation ST + (S-5)(T+1) = 1050, you can either try to solve it for S and T (which you won't be able to do)
At that point you might just assume that the correct answer is E (since we can't solve that equation for S and T)

Alternatively, you might test some possible cases.

Statement 2 tells us that Marsha drove 450 miles of the total 1500 miles. So we automatically know that Marsha didn't drive the greatest distance on the trip. So we're down to either Al or Pablo driving the greatest distance.

Also, if Marsha drove 450 miles, then Al and Pablo shared driving duties for the other 1050 miles.

So far you have:
T = Pablo's driving time (in hours)
T + 1 = Al's driving time (in hours)

S = Pablo's driving speed (in mph)
S - 5 = Al's driving speed (in mph)

So, ST = Pablo's total distance driven
And (S-5)(T+1) = Al's total distance driven
-------------------
CASE A:
If S = 10, about then our equation becomes: (10)T + (10-5)(T+1) = 1050
Simplify: : 10T + 5(T+1) = 1050
Expand: 10T + 5T + 5 = 1050
Simplify: 15T + 5 = 1050
So: 15T = 1045
Solve: T ≈ 70 (in actuality T = 69 2/3, but we don't need to be that precise here)
So, if S = 10 then T ≈ 70
This means Pablo's total distance ≈ (10)(70) ≈ 700 miles
1050 - 700 = 450, which means Al's total distance was 450 miles
In this case, Pablo drove the greatest distance

CASE B:
If S = 100, about then our equation becomes: (100)T + (100-5)(T+1) = 1050
Simplify: : 100T + 95(T+1) = 1050
Expand: 100T + 95T + 95 = 1050
Simplify: 195T + 95 = 1050
So: 195T = 955
Solve: T ≈ 5 (in actuality T = 4.8974, but we don't need to be that precise here)
So, if S = 100 then T ≈ 5,
This means Pablo's total distance ≈ (100)(5) ≈ 500 miles
1050 - 500 = 550, which means Al's total distance was 550 miles
In this case, Al drove the greatest distance