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Comment on Rephrasing the Target Question
Question at 5:23 - this is a
Statement 2 essentially tells
In order to conclude that statement 2 is sufficient, it must be the case that we can answer the rephrased target question (Is x < 0?) with absolute certainty.
Statement 2 essentially tells us that x < 2.5. This information does not allow us to answer the rephrased target question (Is x < 0?) with absolute certainty.
Consider these two cases:
case a: x = -1, in which case x IS less than 0
case b: x = 1, in which case x is NOT less than 0.
Since we can't answer the rephrased target question with certainty, statement 2 is not sufficient.
I have the same question too
In the Avoiding Common
In the Avoiding Common Mistakes - Part I video, the target question is "Does x = 6?"
Statement 2 says x + 10 = 60, which means x = 50
Since x = 50, we can be certain that x does NOT equal 6
So, we can answer the target question (Does x = 6?) with a definitive: NO, x does NOT equal 6)
Since we can answer the question with certainty, statement 2 is sufficient.
---------------------------
In the video question above, we can rephrase the target question as: Is x negative?
Statement 2 tells us that x < 2.5 (indirectly)
Does this information provide enough information to answer the target question (Is x negative?)?
No.
If x < 2.5, then x could equal 1, in which case, the answer to the target question is: NO, x is NOT negative.
Alternatively, if x < 2.5, then x could equal -1, in which case, the answer to the target question is: YES, x IS negative.
Since we cannot answer the target question with certainty, statement 2 is not sufficient.
Does that help?
Cheers,
Brent
I was thinking the same thing
Can you tell what part of the
Can you tell what part of the explanation is confusing you?
could you please explain me
How P-B = 10,000. i.e please make me understand how statement 2 is sufficient.
P = total pay and B = base
P = total pay and B = base salary
Statement 2 says the total pay (P) is 10,000 greater than the base salary (B)
So, we can write: P = B + 10,00
Another way to say this is P-B = 10,000
We phrased the question earlier to be, "What is the value of (P-B)/0.05
Since we now know that P-B = 10,000, we'll take (P-B)/0.05 and replace P-B with 10,000 to get: 10,000/0.05
This evaluates to be 200,000, which means we have successfully answered the rephrased target question. SUFFICIENT
@7:25... I'm having trouble
We might ask ourselves, "What
We might ask ourselves, "What are some numbers that CAN be expressed as the product of two integers, where each integer is greater than 1?"
6 works, since we can write 6 = 2 x 3 (2 and 3 are both greater than 1)
20 works, since we can write 20 = 4 x 5 (4 and 5 are both greater than 1)
36 works, since we can write 36 = 6 x 6
70 works, since we can write 70 = 7 x 10
Now ask, "What are some numbers that CANNOT be expressed as the product of two integers, where each integer is greater than 1?"
Well, 5 is one such number. We CANNOT write 5 as the product of two integers, where each integer is greater than 1.
11 is another such number.
So is 2, and so is 5.
Notice that the numbers that CAN expressed as the product of two integers (where each integer is greater than 1) are all COMPOSITE numbers, and the numbers that CANNOT expressed as the product of two integers (where each integer is greater than 1) are all PRIME numbers.
Hello, could you explain me
That formula/rule is in our
That formula/rule is in our Circle Properties video (https://www.gmatprepnow.com/module/gmat-geometry/video/880) starting around 2:30
Great material, I have
Sorry to hear that the red
Sorry to hear that the red (activated) flags appear the same as the flags that aren't activated. One way to get around this is to manually note the videos you'd like to revisit (e.g., Statistics video #7).
As for your second question, it's important to note that this video appears very early in the course. So, many of the questions we're looking at in this video involve concepts that are covered later in the course. So, in the video, I encourage students to return to this video once they cover certain concepts (e.g., circle properties) in the future.
@GMAT-Admin :I would like to
Hi Sophia,
Hi Sophia,
I have used the technique of rephrasing the target question TONS of times on the Beat The GMAT forum and on the GMAT Club forum.
So, if you go to those forums and search "rephrasing the target question," you'll find over 100 questions where I've used that technique.
In the last question why do
Can S ever be greater than k+ 1?
All integers greater than 1
All integers greater than 1 are either prime numbers or non-prime numbers (aka composite numbers).
All prime numbers, k, are such that the sum of their positive divisors = k + 1. For example, the divisors of 5 (a prime number) are 1 and 5. So, the sum (6) is 1 greater than 5. Likewise, the divisors of 11 (a prime number) are 1 and 11. So, the sum (12) is 1 greater than 11.
All non-prime numbers, k, are different. For them, the sum of their divisors is GREATER THAN k+1. For example, For example, the divisors of 8 (a non-prime number) are 1, 2, 4 and 8. So, the sum (15) is greater than 8+1. Likewise, the divisors of 15 (a non-prime number) are 1, 3, 5 and 15. So, the sum (24) is greater than 15+1.
So, to answer your question, S CAN be greater than k+1. In fact, if k is a non-prime number, then S is certainly greater than k+1
So, asking "Is S > k+1?" is exactly the same as asking "Is k a non-prime number?"
This lesson appears very early in the course. You'll learn more about divisors and prime number in the Integer Properties module: https://www.gmatprepnow.com/module/gmat-integer-properties
Thank you for this well
Hi Brent,
in the end of the video you tell us to keep track of original and rephrased questions. How do you suggest to do this? Excel, paper, etc.? Also, do you have a file with a list like that to be downloaded?
Best,
Guilherme
Hi Guilherme,
Hi Guilherme,
You can use pen and paper, pencil and paper, Excel, Word, hieroglyphics, etc :-)
It's your choice.
I don't have a downloadable list of such questions, however if you go to the Beat The GMAT and GMAT Club forums, and perform a search for "This is a good candidate for rephrasing the target question" (a preset phrase I often type when answering Data Sufficiency questions), you'll find all of the instances in which I've rephrased the target question.
I hope that helps.
Cheers,
Brent
Hi,
First of all, loved your videos..they help in a very subtle way but nevertheless important. i have just one 'silly' doubt. It's regarding |x| < 1. Doesnt that mean x < 1, -1 or in other words, x is less than 1 and x is less than -1...if that's true, then the range will not be -1<x<1, isn't it?
Quote: "...in other words, x
Quote: "...in other words, x is less than 1 and x is less than -1"
That conclusion isn't correct.
We can show why it isn't correct by testing some x-values that are less than -1
For example, -2 is less than -1, so let's plug this into our given inequality, |x| < 1
We get: |-2| < 1
Evaluate to get: 2 < 1
Since 2 is not less than 1, we can see that -2 is not a possible value of x.
Here's the video on solving inequalities involving absolute value: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...
Cheers,
Brent
I just want to thank you for
That's very nice of you to
That's very nice of you to say. You made my day!
Cheers,
Brent
Hello there,
Thanks for the great videos.
I am from Saudi Arabia, and I just got confused why did we rephrase 4x<3x to is x<0? and is X negative?
Good question.
Good question.
We can take the inequality: 4x < 3x
And subtract 3x from both sides to get: x < 0
So, asking "Is 4x < 3x?" is the same as asking "Is x < 0?"
For more information about working with inequalities, watch: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...
Cheers,
Brent
Hi Brent, could you please
Hi Minh,
Hi Minh,
Target question: If k is a positive integer greater than 1, and S is the sum of all positive divisors of k, is S > k + 1?
To answer your question, let's examine the sums of a few positive integers
If k = 2, then the positive divisors of k are {1, 2}
So, S = 1 + 2 = 3
In other words, S = k + 1
If k = 5, then the positive divisors of k are {1, 5}
So, S = 1 + 5 = 6
In other words, S = k + 1
If k = 11, then the positive divisors of k are {1, 11}
So, S = 1 + 11 = 12
In other words, S = k + 1
Notice that 2, 5 and 11 are all PRIME numbers.
Also notice that when k is prime, there are exactly 2 divisors: 1 and k
So, when k is PRIME, S = k + 1
------------------------
Now let's see what happens when k is NOT prime.
If k = 6, then the positive divisors of k are {1, 2, 3, 6}
So, S = 1 + 2 + 3 + 6 = 12
In this case, S > k + 1
If k = 9, then the positive divisors of k are {1, 3, 9}
So, S = 1 + 3 + 9 = 13
In this case, S > k + 1
If k = 20, then the positive divisors of k are {1, 2, 4, 5, 10, 20}
So, S = 1 + 2 + 4 + 5 + 10 + 20 = 46
In this case, S > k + 1
So, when k is NOT prime, S > k + 1
--------------------------
So, rather than ask "Is S > k + 1?", we can REPHRASE the question as "Is k not prime?"
Does that help?
Cheers,
Brent
Hi Brent,
Target question: If k is a positive integer greater than 1, and S is the sum of all positive divisors of k, is S > k + 1?
If I choose 4 as an example (k=4), then the positive divisors of k are {1, 2, 2}
So, S = 1 + 2+2=5
K+1=4+1=5
In this case, S = k + 1
5=5
please help me to find mistake in my logic.
Be careful. The positive
Be careful. The positive divisors of 4 are: 1, 2, and 4 (not 1, 2, and 2)
Cheers,
Brent
If x and y are integers and x
As you point out in the videos, x to an even power will return a positive value (or integer in this case).
But you re-worded the question as is y even? Shouldn't it be is y even and not equal to 0? Since, any number raised to the power of 0 would be 0.
Also please get rid of the captcha verification as I have to constantly update it to type the comments.
Be careful. (any non-zero
Be careful. (any non-zero value)^0 = 1 (not zero)
For example:
4^0 = 1
12^0 = 1
(-1)^0 = 1
(-8)^0 = 1
Sorry about the Captcha verification.
Even with Captcha, the site still receives lots of spam posts. I'd rather not see what happens if Captcha were removed altogether.
Having said all of that, I was under the impression that once you verify with Captcha the first time, you aren't required to keep verifying after that. Are you saying you have to verify every time?
Cheers,
Brent
Thanks Brent. My memory
At first I thought it was the Captcha field that prevented me from typing my queries on the comment section. But actually, its the mini-pop up box that keeps appearing at the bottom right corner (free question of the day email) - if I don't minimise it, I can't type my query. But it's no big deal for me.
If you click "Don't show this
If you click "Don't show this message again," you won't have that problem.
Is |x| < 1 ? => -1 < x < 1?
In video above,
Why not 0 < x < 1? -> 0 because the regardless whatever shall be the value of X will be changed to 's Absolute Value that can only be Positive or Zero. Therefore, 0 < x < 1.
Good question.
Good question.
Later in the course, you'll learn more about absolute value inequalities here: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...
In the meantime, let's examine some possible values of x that satisfy the inequality |x| < 1
For example, x = 0 works, since |0| = 0, and 0 < 1
x = 0.22 also works, since |0.22| = 0.22, and 0.22 < 1
x = 0.5 also works, since |0.5| = 0.5, and 0.5 < 1
x = -0.5 also works, since |-0.5| = 0.5, and 0.5 < 1
x = -0.713 also works, since |-0.713| = 0.713, and 0.713 < 1
As you can see, all values of x between -1 and 1 will satisfy the inequality |x| < 1
So, we can say that the inequality |x| < 1 is the same as the inequality -1 < x < 1
Cheers,
Brent
Hi Brent,
Hope you are doing good.
have a concern regarding the rephrasing of the following question.
Is p + pz = p?
(1) p = 0
(2) z = 0
I saw your solution on GMATClub.
I rephrased it as follows.
p(1+z)=p => Is z=0?
I cancelled the p out, and I see that I landed on the wrong answer as option B.
I can certainly see that statement 1 is sufficient though.
I think I am realizing my mistake as I am typing.Was canceling out one of the variables the mistake I made?
Appreciate your help.
Thanks,
Pritish
In this case, you took the
In this case, you took the equation p(1+z) = p, and divided both sides by p to get: 1 + z = 0.
However, when we divide by a variable, the resulting equation may not be accurate.
For example, if xy = xz, we can't necessarily conclude that y = z.
It could be the case that x = 0, y = 1 and z = 2, in which case, it's true that xy = xz
However, if we divide both sides by x to get y = z, we see that this is equation is not true (since 1 ≠ 2)
Here's why we can't divide both sides of an equation by a variable (unless we're certain the variable does not equal zero):
First, notice that x/x = 1 for all non-zero values of x.
That is, 2/2 = 1 and 7.3/7.3 = 1, and (-4)/(-4) = 1, etc.
However, 0/0 does NOT equal 1.
In fact, 0/0 is not a defined number.
We can use this property to simplify fractions.
For example, 15/20 = (5)(3)/(5)(4) = (5/5)(3/4) = (1)(3/4) = 3/4
Likewise, 14/63 = (7)(2)/(7)(9) = (7/7)(2/9) = (1)(2/9) = 2/9
In both cases, we were able to use the fact that x/x = 1 to simplify the fraction.
Now let's TRY to use the same strategy with: xy = xz
If we divide both sides by x we get: xy/x = xz/x
Rewrite as follows: (x/x)(y) = (x/x)(z)
At this point, we must ask "Does x/x equal 1 or is it undefined?"
Since we don't know whether x is zero, we can't answer that question.
Therefore, we can't be certain that x/x = 1.
As such, we CAN'T simplify the equation to say y = z
Does that help?
Cheers,
Brent
Thanks! You Rock!
Hi Brent,
Sorry,Have another question related to the rephrasing for the following.
Does x + c = y + c ?
(1) x = y
(2) x = c
I can see the solution that you provided for the rephrasing ,the variable c was cancelled out.
So is it like, we can only cancel out variables in order to rephrase ,when they are represented as a sum and not when they are in form of a product?
Kindly help.
Thanks,
Pritish
In the question you asked
In the question you asked above, we were DIVIDING both sides of an equation by a variable.
When we divide by a variable, we risk the chance of accidentally dividing by 0 (which can cause problems).
For the equation x + c = y + c, we can safely SUBTRACT the variable c from both sides to get x = y.
There's no concern about SUBTRACTING any number (including zero) from both sides of an equation.
The resulting equation will always be valid.
Does that help?
Cheers,
Brent
Yes, it does help.Thanks! You
Hi Brent,
For the following question, wanted to know what your thought process is when simplifying the 1st statement.
Is (x - 2)² > x²?
(1) x² > x
(2) (1/x) > 0
Here is your explanation for Statement 1.
Statement 1: x² > x
First, since x² > x, we can conclude that x ≠ 0
So, we know that x² is POSITIVE
So, let's divide both sides of the inequality by x² to get: 1 > 1/x
This means that EITHER x < 0 OR x > 1
So there are two possible cases to consider.
case a: If x < 0, then it IS the case that x < 1
case b: If x > 1, then it is NOT the case that x < 1
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT
Wanted to know why you divided the inequality by x^2 and not by x to get x>1.
Appreciate your help.
Thanks,
Pritish
KEY CONCEPT: If we divide
KEY CONCEPT: If we divide both sides of an inequality by a NEGATIVE value, we must REVERSE the direction of the inequality sign.
For more on this, watch: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...
So, if we take x² > x and divide both sides by x we get: x (> or <) 1
Since we don't know whether x is positive or negative, we don't know whether the resulting inequality should be x > 1 OR x < 1.
At this point, we're kind of stuck if we insist on dividing by x.
However, since we can be certain that x² is positive, we know that, when we divide both sides of the inequality by x², the direction of the inequality sign stays the same.
Does that help?
Cheers,
Brent
Yes, it does help.Thanks! You
Please, explain
Can the positive integer p be expressed as the product of two integers, each of which is greater than 1?
(1) 31 < p < 37
(2) p is odd.
Here's my full solution:
Here's my full solution: https://gmatclub.com/forum/can-the-positive-integer-p-be-expressed-as-th...
Cheers,
Brent
Hi Brent,
https://www.beatthegmat.com/ds-t157326.html
You have rephrased the target question as 'Is x>y?' from 5x + x^2 > 5y + y^2 step.
I see that the rephrasing is true for positive numbers and negative proper fractions, and not negative integers. I don't understand the logic behind. Would you please explain it?
Thanks
Question link: https://www
Question link: https://www.beatthegmat.com/ds-t157326.html
Good question. I should have added one more step.
Start with: Is 5x + x² > 5y + y² ?
Factor: Is x(5 + x) > y(5 + y)?
Notice that both sides are in the same form: variable(5 + variable)
Since we're told that x and y are POSITIVE, we can see that:
- if x > y, then x > y (this part is obvious)
- if x > y, then (5+x) > (5+y)
So, if x > y, then we can be sure that x(5+x) > y(5+y)
Does that help?
Cheers,
Brent
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