For the question at 46:41- Taking approximately 0 won't be accurate right? If P is any integer greater than 0 (which it is) - Quantity B will always be greater because pq will always be negative (since q < -2). The only reason you picked D is because you approximated P to 0 which doesn't hold true. Please explain...?
You're totally right that if p and q had to be integers with p > 0 and q < -2, then the answer would be B, but the question does not say that p and q have to be integers. They could be integers, but they don't have to be. All we're told about p is that it's greater than zero. This means that p could be 0.00000000000000000..........0000000000000001 which is so close to zero it makes very little difference to the outcome of the calculations if we assume it really is zero. I hope that helps!
@@GRENinjaTutoring Though your answer is right , but you explination picking equal to zero is making a lot of confustion . you could explain it in better ways .
@@GRENinjaTutoring wow i never thought like that. In the 7 number question according to my calculation i thought you are wrong and the ans must be B but after i read your explanation i just realized i am wrong.
@@GRENinjaTutoring While I was watching your video, I was thinking to myself, "This may be confusing to people who haven't taken calculus" But you did a great job with bringing in the idea of approximating close to zero without delving into limits. @_saanchishah I am going to add this here. The GRE quant section works with all real numbers unless stated otherwise in the problem.
We could test this out by setting p = -2 as an example. If we do that, then x > -28 and y < -18. We could follow the same process Harry used with the blue text and set x = -28 and y = -18 to test the extreme value. If we do that, (x - y)/5 = (-28 - (-18))/5 = (-28 + 18)/5 = (-10)/5 = -2. However, we know x > -28 and y < -18, so we could have something like x = -27.9 and y = -18.1 and in this case, (x - y)/5 = -9.8/5 = -1.96 > -2 = p. We could test this out for any other negative value of p and get the same results, so we can say that Quantity A is greater than Quantity B even if we use negative values for p. I hope that helps!
That all depends on whether t is positive or negative. If t is positive, then it does have to be greater than 1 to satisfy that statement. However, it's also possible for us to satisfy the problem statement if -1 < t < 0. For example, if t = -0.5, then t/|t| = -0.5/|-0.5| = -0.5/0.5 = -1 which is less than t = -0.5. This means the initial problem: t/|t| < t is satisfied if -1 < t < 0 or if 1 < t. Since we can't tell whether t is greater than or less than 1, the answer to this question is (D). I hope that helps!
Thank you so much for this video. I totally understand the explanation you gave in question 8, but I am still not clear on why for question 7, you didn't apply the same rule of close estimation as you did to question 8. For question 7, both solutions gave different answers, why isn't it possible to choose option D since it wasn't stated that x , y were integers?
Thank you for the kind words! Harry did use close estimation in Q7. In this question, we know that x > 14p and y < 9p, but Harry used x = 14p and y = 9p in the first part of the solution so he could simplify quantity A. At that stage, we can say that we know that x is greater than 14p and we know that y is less than 9p, so the numerator of the fraction in quantity A will be greater than 5p and the overall value of quantity A is greater than p. This means the answer to this question is (A). The second solution Harry completed in the red ink was just an illustration of what would happen if you chose a value greater than 14p for x and a value less than 9p for y. This solution also gives a value greater than p for quantity A, so both solutions to this question give us (A) as the answer. I hope that's answered your question, but please let me know if I missed anything.
We could test this out by setting p = -2 as an example. If we do that, then x > -28 and y < -18. We could follow the same process Harry used with the blue text and set x = -28 and y = -18 to test the extreme value. If we do that, (x - y)/5 = (-28 - (-18))/5 = (-28 + 18)/5 = (-10)/5 = -2. However, we know x > -28 and y < -18, so we could have something like x = -27.9 and y = -18.1 and in this case, (x - y)/5 = -9.8/5 = -1.96 > -2 = p. We could test this out for any other negative value of p and get the same results, so we can say that Quantity A is greater than Quantity B even if we use negative values for p. I hope that helps!
on question 6, why do you have two different answers? wouldn't t=-1 so that it doesn't violate the original t/I t I is less than t ? why do you flip the sign at all?
t = -1 does violate the original expression because then t / |t| = -1 and t = -1, so t/|t| = t instead of t/|t| < t. The reason Harry needed to follow two different solution paths is because the absolute value sign does different things depending on whether t < 0 or t >= 0. From those solutions paths, we find the initial inequality is satisfied if -1 < t < 0 or 1 < t. This means we cannot tell whether t is greater than, less than, or equal to 1, so the answer to this question is (D). I hope that helps!
Hello, whats stopping us considering the value of y to be less than 0 in the 7th question and approaching a value such that the value of x-y/5 becomes negative Ex: x=15p and y=-16p
We could test this out by setting p = -2 as an example. If we do that, then x > -28 and y < -18. We could follow the same process Harry used with the blue text and set x = -28 and y = -18 to test the extreme value. If we do that, (x - y)/5 = (-28 - (-18))/5 = (-28 + 18)/5 = (-10)/5 = -2. However, we know x > -28 and y < -18, so we could have something like x = -27.9 and y = -18.1 and in this case, (x - y)/5 = -9.8/5 = -1.96 > -2 = p. We could test this out for any other negative value of p and get the same results, so we can say that Quantity A is greater than Quantity B even if we use negative values for p. I hope that helps!
In question 9 why you didn't do 3m= 24+5n and put it in quantity A then |24 + 10n | that is 24 more than to quantity B if substitute|3m| to |5n| then quantity B will be | 10n| hence Quantity A is 24 more than quantity B
We can say |24 + 10n| is 24 more than |10n| for certain values of n. However, consider what happens if n = -2. In that case |24 + 10n| = |24 + 10(-2)| = |24 - 20| = 4, and |10n| = |10(-2)| = |-20| = 20, so quantity B is greater than quantity A. This is true for all values of n less than -1.2. For n > -1.2, |24 + 10n| is greater than |10n|. From Harry's work in the video, we know quantity A equals zero. If we rearrange quantity A the way you suggest, we know |24 + 10n| = 0, so n = -2.4. As shown in the previous paragraph, quantity B will be greater than quantity A for n = -2.4. I hope that helps!
I understand by your method of explanation but why my answer is Wrong that D) don't know the answer in gre as if we open |10n +24| it makes |10n| +_24 as 24 can be greater than quantity B i.e |10n| or can be less than 24 so it should answer D) don't know the Answer in GRE
We can't say that |10n + 24| = |10n| + 24 for all values of n. This is only true for non-negative values of n. For example, if n = -1, |10n + 24| = |-10 + 24| = 14 but |10n| + 24 = 10 + 24 = 34. From what I said in the previous answer, using your work and the work Harry did in the video, we can show that n = -2.4 and m = 4. This tells us that quantity A is zero |3m + 5n| = 0 (or |10n + 24| = 0) and quantity B is 24 from |3m| + |5m| = 12 + 12 = 24 (or |10n| = 24). This is how we can show quantity B is greater than quantity A, so the answer to this question is (B). I hope that helps!
In a situation like this, we need to know whether both sides of the equation have the same sign or different signs. If we're unsure whether the signs are the same or different on the two sides of the equation, we can express this by saying 3m = 5n or 3m =-5n. This can be expressed in a single equation as 3m = +/- 5n. This is why we only need to put the +/- on one side of the equation. I hope that helps!
In the 14p question, how you know p is positive? If p is negative then 3p for instance would be smaller that the p. So, if p is -5, 3p would be -15 making B the answer.
The red text at the end of this solution is more of an illustration of something that could happen if you chose a value for p. You're right that in this case, p would have to be positive but we certainly haven't covered every possible scenario in that quick illustration. The bit to pay attention to in this solution is the blue text that shows algebraically that (x - y)/5 will always be greater than p. To see what happens when p is negative, we could choose a negative number for p. Let's say p = -5. This means x > 14p = -70 and y < 9p = -45. If x > -70, then we could say that x = - 65, and if y < -45, then we could say y = -60. This means (x - y)/5 = (-65 - (-60))/5 = -5 / 5 = -1. This means (x - y)/5 is still greater than p, so quantity A is still greater than quantity B. It doesn't matter which values you choose for x and y, as long as x > -70 and y < -45, you'll still show that (x - y)/5 is greater than p. I hope that helps!
Thanks for the tips.
For the questions such as the one in 34:50, I think it is more time efficient to do the algebra manipulation..
To simlify absolute values, see the two cases x > 0 aand x
For the question at 46:41- Taking approximately 0 won't be accurate right? If P is any integer greater than 0 (which it is) - Quantity B will always be greater because pq will always be negative (since q < -2). The only reason you picked D is because you approximated P to 0 which doesn't hold true. Please explain...?
You're totally right that if p and q had to be integers with p > 0 and q < -2, then the answer would be B, but the question does not say that p and q have to be integers. They could be integers, but they don't have to be.
All we're told about p is that it's greater than zero. This means that p could be 0.00000000000000000..........0000000000000001 which is so close to zero it makes very little difference to the outcome of the calculations if we assume it really is zero.
I hope that helps!
@@GRENinjaTutoring Though your answer is right , but you explination picking equal to zero is making a lot of confustion . you could explain it in better ways .
@@GRENinjaTutoring wow i never thought like that. In the 7 number question according to my calculation i thought you are wrong and the ans must be B but after i read your explanation i just realized i am wrong.
@@GRENinjaTutoring While I was watching your video, I was thinking to myself, "This may be confusing to people who haven't taken calculus" But you did a great job with bringing in the idea of approximating close to zero without delving into limits.
@_saanchishah I am going to add this here. The GRE quant section works with all real numbers unless stated otherwise in the problem.
love your work guys, but for question 7 how about if p was a negative won't B be larger. therefor D is our final answer
We could test this out by setting p = -2 as an example. If we do that, then x > -28 and y < -18.
We could follow the same process Harry used with the blue text and set x = -28 and y = -18 to test the extreme value. If we do that, (x - y)/5 = (-28 - (-18))/5 = (-28 + 18)/5 = (-10)/5 = -2. However, we know x > -28 and y < -18, so we could have something like x = -27.9 and y = -18.1 and in this case, (x - y)/5 = -9.8/5 = -1.96 > -2 = p.
We could test this out for any other negative value of p and get the same results, so we can say that Quantity A is greater than Quantity B even if we use negative values for p.
I hope that helps!
Great explanation. For Question 6, wouldn't t have to be greater than 1 to satisfy the given in the problem statement? "t" must be greater than t/|t|
That all depends on whether t is positive or negative. If t is positive, then it does have to be greater than 1 to satisfy that statement. However, it's also possible for us to satisfy the problem statement if -1 < t < 0. For example, if t = -0.5, then t/|t| = -0.5/|-0.5| = -0.5/0.5 = -1 which is less than t = -0.5.
This means the initial problem: t/|t| < t is satisfied if -1 < t < 0 or if 1 < t. Since we can't tell whether t is greater than or less than 1, the answer to this question is (D).
I hope that helps!
For question 3, can we take bc to the other side so we have ac-bc
You absolutely can do everything you suggest until you get to c(a-b)
Thank you so much for this video. I totally understand the explanation you gave in question 8, but I am still not clear on why for question 7, you didn't apply the same rule of close estimation as you did to question 8. For question 7, both solutions gave different answers, why isn't it possible to choose option D since it wasn't stated that x , y were integers?
Thank you for the kind words!
Harry did use close estimation in Q7. In this question, we know that x > 14p and y < 9p, but Harry used x = 14p and y = 9p in the first part of the solution so he could simplify quantity A. At that stage, we can say that we know that x is greater than 14p and we know that y is less than 9p, so the numerator of the fraction in quantity A will be greater than 5p and the overall value of quantity A is greater than p. This means the answer to this question is (A).
The second solution Harry completed in the red ink was just an illustration of what would happen if you chose a value greater than 14p for x and a value less than 9p for y. This solution also gives a value greater than p for quantity A, so both solutions to this question give us (A) as the answer.
I hope that's answered your question, but please let me know if I missed anything.
@@GRENinjaTutoring Wow. I get it now. Thanks for the explanation. You're really doing a great job at this.
Great explanation. Thanks
Thank you!
For question 7, answer choice A would assume that p is positive, right? Because if p was negative, then B would be greater?
We could test this out by setting p = -2 as an example. If we do that, then x > -28 and y < -18.
We could follow the same process Harry used with the blue text and set x = -28 and y = -18 to test the extreme value. If we do that, (x - y)/5 = (-28 - (-18))/5 = (-28 + 18)/5 = (-10)/5 = -2. However, we know x > -28 and y < -18, so we could have something like x = -27.9 and y = -18.1 and in this case, (x - y)/5 = -9.8/5 = -1.96 > -2 = p.
We could test this out for any other negative value of p and get the same results, so we can say that Quantity A is greater than Quantity B even if we use negative values for p.
I hope that helps!
on question 6, why do you have two different answers? wouldn't t=-1 so that it doesn't violate the original t/I t I is less than t ? why do you flip the sign at all?
t = -1 does violate the original expression because then t / |t| = -1 and t = -1, so t/|t| = t instead of t/|t| < t.
The reason Harry needed to follow two different solution paths is because the absolute value sign does different things depending on whether t < 0 or t >= 0. From those solutions paths, we find the initial inequality is satisfied if -1 < t < 0 or 1 < t. This means we cannot tell whether t is greater than, less than, or equal to 1, so the answer to this question is (D).
I hope that helps!
Hello, whats stopping us considering the value of y to be less than 0 in the 7th question and approaching a value such that the value of x-y/5 becomes negative
Ex: x=15p and y=-16p
We could test this out by setting p = -2 as an example. If we do that, then x > -28 and y < -18.
We could follow the same process Harry used with the blue text and set x = -28 and y = -18 to test the extreme value. If we do that, (x - y)/5 = (-28 - (-18))/5 = (-28 + 18)/5 = (-10)/5 = -2. However, we know x > -28 and y < -18, so we could have something like x = -27.9 and y = -18.1 and in this case, (x - y)/5 = -9.8/5 = -1.96 > -2 = p.
We could test this out for any other negative value of p and get the same results, so we can say that Quantity A is greater than Quantity B even if we use negative values for p.
I hope that helps!
NINJA HATTORI IS ON THE HOME :D
🥷
In question 9 why you didn't do 3m= 24+5n and put it in quantity A then |24 + 10n | that is 24 more than to quantity B if substitute|3m| to |5n| then quantity B will be | 10n| hence Quantity A is 24 more than quantity B
We can say |24 + 10n| is 24 more than |10n| for certain values of n. However, consider what happens if n = -2. In that case |24 + 10n| = |24 + 10(-2)| = |24 - 20| = 4, and |10n| = |10(-2)| = |-20| = 20, so quantity B is greater than quantity A. This is true for all values of n less than -1.2. For n > -1.2, |24 + 10n| is greater than |10n|.
From Harry's work in the video, we know quantity A equals zero. If we rearrange quantity A the way you suggest, we know |24 + 10n| = 0, so n = -2.4. As shown in the previous paragraph, quantity B will be greater than quantity A for n = -2.4.
I hope that helps!
I understand by your method of explanation but why my answer is Wrong that D) don't know the answer in gre as if we open |10n +24| it makes |10n| +_24 as 24 can be greater than quantity B i.e |10n| or can be less than 24 so it should answer D) don't know the Answer in GRE
We can't say that |10n + 24| = |10n| + 24 for all values of n. This is only true for non-negative values of n. For example, if n = -1, |10n + 24| = |-10 + 24| = 14 but |10n| + 24 = 10 + 24 = 34.
From what I said in the previous answer, using your work and the work Harry did in the video, we can show that n = -2.4 and m = 4. This tells us that quantity A is zero |3m + 5n| = 0 (or |10n + 24| = 0) and quantity B is 24 from |3m| + |5m| = 12 + 12 = 24 (or |10n| = 24).
This is how we can show quantity B is greater than quantity A, so the answer to this question is (B).
I hope that helps!
50:01 isn't it, + - 3m = + - 5n? why you only said 3m = + - 5n?
In a situation like this, we need to know whether both sides of the equation have the same sign or different signs.
If we're unsure whether the signs are the same or different on the two sides of the equation, we can express this by saying 3m = 5n or 3m =-5n. This can be expressed in a single equation as 3m = +/- 5n.
This is why we only need to put the +/- on one side of the equation.
I hope that helps!
How come the last one was the easiest one lol
Was expecting some twist..
Thanks , it was helpful
In the 14p question, how you know p is positive? If p is negative then 3p for instance would be smaller that the p. So, if p is -5, 3p would be -15 making B the answer.
The red text at the end of this solution is more of an illustration of something that could happen if you chose a value for p. You're right that in this case, p would have to be positive but we certainly haven't covered every possible scenario in that quick illustration. The bit to pay attention to in this solution is the blue text that shows algebraically that (x - y)/5 will always be greater than p.
To see what happens when p is negative, we could choose a negative number for p. Let's say p = -5. This means x > 14p = -70 and y < 9p = -45. If x > -70, then we could say that x = - 65, and if y < -45, then we could say y = -60. This means (x - y)/5 = (-65 - (-60))/5 = -5 / 5 = -1. This means (x - y)/5 is still greater than p, so quantity A is still greater than quantity B. It doesn't matter which values you choose for x and y, as long as x > -70 and y < -45, you'll still show that (x - y)/5 is greater than p.
I hope that helps!