Root 27 is a little more than root 25 which equals 5. Similarly, root 50 is a little more than root 49 which equals 7. So the answer is near 5/7. Answer b) is the only answer less than 1.
Sqrt(27) is 3*sqrt(3). Sqrt(50) is 5*sqrt(2). If we rationalize the denominator by multiplying with sqrt(2) we get 3*sqrt(2*3)/ (5*2) or 3*sqrt(6)/10 or answer b)
The result is less than 1. This rules out a, c, and d. So the answer is b. To evaluate multiply by sqrt(2). This leads to sqrt(54)/10, which is the same as 3×sqrt(6)/10. The only thing that is difficult is how to make a 20 minutes clip out of a trivial problem...
I don't think I ever had a Math teacher give multiple choices with only one possible answer. The numerator is smaller than the denominator therefore the answer is less than 1 only one choice has any possibility of being less than 1.
I'm a tail-end baby boomer/gen X and was always a B+ or A- student in middle and high school and had to work hard at math. I am convinced that had I had a tutor like you (and YT) back then, I would have aced my math courses consistently. Nevertheless, because it's never too late to learn, I'm relearning basic math skills. Why, some might ask? It's because it gives me great joy to master what I know I can, reinforces what I already know, and keeps my mind sharp.
Take the squares out of each you end up with 3✓3 / 5✓2. Now multiply by ✓2/✓2 ( since that is one and you can get multiply anything by one and get away with it! ) and you end up with 3✓2✓3 / (5 *2) or 3✓6/10
I start with a quick approximation. ✓27 is a bit larger than 5 (✓25), and ✓50 slightly larger than 7 (✓49). So, the result must be less than 1. The ONLY proposed answer that is less than 1 is (3✓6)/10, or about 0.735, (B). Many years ago, I solved a LOT of similar problems in my math classes. Thanks for the refresh.
As a 74 year old former English major who had a career in IT I didn't need the coaching on test-taking, but the explanation on how to derive the answer algebraicly was worth the viewing.
but it makes it easier ? to compute an exacter number - the trick to multiply by a suitable substitution for 1 is something I would not have thought of
why do i need to calculate anything? 27 SQRT(27)0) => SQRT(27)/SQRT(50) < 1 b) is the only solution 3*SQRT(6) < 9) Just 30 seconds to solve, because this is multiple choice
Not too sure about this bur here goes: 27^0.5/50^0.5 = (9^0.5 x 3^0.5)/ (25^0.5 x 2^0.5) = (3 x 3^0.5) / (5 x 2^0.5), multiply the square root components top and bottom by 2^0.5 = (3 x 6^0.5) / (5 x 4^0.5) = 3 x 6^0.5 / 10
It's (b). sqrt(27)/sqrt(50) = sqrt(27/50)=sqrt(54/100). Not take sqrt (1/100) outside the radical and you have sqrt(54)/10. 54 = 9*6, so you can take sqrt(9)=3 outside the radical. Now you have 3 sqrt(6)/10. Q.E.D.
Took your way at first, got stuck at 11:22, then decided to take another step and expanding (hope that is the correct english word) Root27/Root50 by Root2 and got Root54/Root100 = Root54/10 = Root(9*6)/10 = 3*Root6/10, means b) is the correct solution
A great way to look at it, but I never assume that any one of the choices is necessarily correct. So, while it couldn't possibly be a), c), or d), I checked that b) was, in fact, the answer. I didn't watch the video because I don't want to know how a three step solution could take so long to explain: sqrt(27)/sqrt(50) => 3×sqrt(3)/(5×sqrt(2)) => 3×sqrt(3)×sqrt(2)/(5×sqrt(2)×sqrt(2)) => 3×sqrt(6)/10
😅I don't know how to solve this mathematically but given a multiple choice option, it's clear that only b) comes close enough to my thinking which is sqr 27 is about 5, sqr 50 is about 7, 5/7 is about 0.7. Close enough to get the answer right using logic and a bit of arithmetic
When a number is divided by a larger number, the answer is always less than one. Answer b is the only option less than one, so it is the correct answer. Simple logic.
Thats how I got it but it seemed sus. Ran it through a calculator, and sure enough. Then realized 27 is 9 x 3 (god, idiot!) and the entire method was clear to me. Embarassing a little, so I gave him a thumbs up and watched the video as punishment. :)
I don't if anybody had already said this but you could have done Sqrt(27)/sqrt(50) = sqrt(27)*sqrt(50)/sqrt(50)*sqrt(50) Sqrt(1350)/50 = sqrt(9)*sqrt(25)*sqrt(6)/50 Sqrt(9)*sqrt(25)*sqrt(6) = 3*sqrt(6)/10 Explanation: If your denominator is a root, you to get rid of the root without changing the value. For the first step, you can multiply the numerator and denominator by the sqrt(50) (Note that you can't sqaure both sides because that would give a different number eg sqrt(2)²/sqrt(3)² ≠ sqrt(2)*sqrt(3)/sqrt(3)*sqrt(3) ) After multiplying both by sqrt(50) you should get sqrt(50)*sqrt(27)/50. Our next step is to combine both root in to one root which would be sqrt(50*27)/50 = sqrt(1350)/50 After this, we should find the prime factors of 1350. Which is 1350=2*(3^3)*(5^2) Now we are only interested into getting the square numbers, now one of them is 5^2 but another one is 3^2 meaning we can rewrite this as 1350=6*(3^2)*(5^2) Which means, sqrt(1350)=sqrt(6)*sqrt(9)*sqrt(25) Because our denominator is 50, we can divide both side by the sqrt(25) giving us Sqrt(6)*sqrt(9)/±10 Next is to find the sqrt(9) which is ±3 So now our eqaution is, ±3*sqrt(6)/±10 Because there is ± both on the numerator and denominator we can cancel it out. Making our final answer, 3*sqrt(6)/10
That's impressively complicated (and you can drop the ± stuff as there's no ± here). Giving yourself the task of finding the prime factors of a number like 1350 when you can just see from inspection that 27 is 3×3² and 50 is 2×5² is a good one 😂
Nothing. You just stopped too early. 3√3 = 3 so now we have 3/5√2 We can't leave a root in the denominator (for reasons I only semi-understand, so I won't try to explain), so we multiply top and bottom by √2 3√2/5√2 × √2 = 3√2/5√4 3√2/5√4 = 3√2/(5 ×2) 3√2/(5 ×2) = 3√2/10
Numerator: 27 is just 3 * 3 * 3 Denominator: 50 is just 5 * 5 * 2 Numerator is 3(3^1/2) Denominator is 5(2^1/2) we have a root in the denominator so we multiply: Numerator: (3 * 2)^1/2 gives us 3(6^1/2) Denominator (2 * 2)^ 1/2 5(4 ^ 1/2) is 5 (2) which is 10 Final answer [3(6 ^ 1/2)] / 10 This is answer b since 1/2 as exponent means square root.
Let us do it in a very simple way. The question is √27/√50. We can write it as √ (27/50) which also equal to √ (54/100). Let us convert all possible answers in this form. (a) 5√8 = √ 25. √8 = √ 200. (b) 3√6/10 = √9.√6/√100 = √(54/100) (c) √ 1350 (d) 8√6 = √ (64.6) √ 384. Clearly, the answer is (b).
√27 is 3√3 √50 is 5√2 we can't have a √ in the denominator so multiply top and bottom by √2 numerator is 3√6 denominator is 5 (2) or 10 Can't reduce further so (3√6)/10 is the answer.
If you are going to do multiple choice questions, make them at least A LITTLE realistic. You don’t need any math skill other than knowing that the answer to this problem will be LESS THAN 1 to be able to simply guess the correct answer. All the other choices are MAGNITUDES greater than any remotely possible guess. If you’re going to “teach” this way, you might incorporate some test taking skills into the mix.
Correct answer can be seen at a glance of the choices. Square root of 27 is ~5, of 50 ~7. So the answer should be around .7, 5-7ths or ~5x.14. Answers a, c and d are too big to way-too-big. So only answer b remains. On second glance, you can see that answer b is 3/10 of square root of 6. if that was SR of 4, 2x.3 = .6. Close enough to .7. So you can get the correct answer easily.
Easy: SR27 is 3SR3, and SR50 is SR100 divides by SR2. Thé result is so 3SR3 divides by SR(100 /2) , that GI es: 10/SR2, resulting in thé second solution.
Fun. For once I didn't guesstimate first, but looked for factors in the original question. They just looked factorable. Once I had 3 root 3, and 5 root 2, the rest was just picking up the pieces.
This is his RU-vid channel. So, we can’t be rude, picking on his style of teaching. Each of our minds are unique and each one operates the way it has been programmed and maybe according to the genes too. We have an advanced human brain to realize that and let’s be appreciative of what we are offered in life. Maybe there are some others who are not that smart as you all who complain, and they must be learning and benefiting from the hard work of this teacher. It is impossible to please everyone, when it comes to anything in this world. Maybe you all can have your own channel to display your expert math skills!
You did okay - a little long but you were taking the scenic route - up to 17:00. That was all you needed to do. Getting the radical out of the denominator is fine, but then why did you leave it as a fraction rather than converting it to 0.3 sqrt 6? A matter of taste, I suppose.
Multiple answers are not bad, especially for this problem, but your explanations are very boring - too many unnecessary steps even for the lower level students
All the answers except (b) are clearly larger than one while the answer must be smaller than one. I had to watch the video to remember the "trick" of getting to (b).
You should be teaching your students to always try to estimate the correct answer as a first step in any multiple choice test instead of wasting every ones time with a lot of nonsense.
Dear Mr Tablet I like to do two of these in the morning. I CANNOT spend 40 - 50 minutes doing this. PLEASE cut the running times in half or do two versions. Thank you. Dr S.
This is the second of your videos I have seen in 2 days with a trivial answer. The answer is obviously less than 1 (since the square root of 50 is larger than the square root of 27) and only b is less than 1. Perhaps you should pose more challenging problems for your viewers.
I didn't watch the video since it's over 20 minutes long. Here's my try, though: ✓1350 is definitely wrong. I can see an easy way to get 10 as denominator, so let's try that: ✓50=✓(100•½)=10✓½ ✓27=✓(3•9)=3✓3 (or, if you'd rather prefer, ✓27=✓3³=✓(3²•3¹)=3✓3) (3✓3)/(10✓½)=3✓(3/½)/10=3✓(3•2)/10=(3✓6)/10, so B it is. Now, looking at A and D I realize that I should have had a look at those before beginning to calculate, since they are both impossible answers to this question, but calculating is one of the fun bits in maths anyway, so I'm OK with that.
Yes. Twenty seconds on Google found me a NASA article which contains an equation with multiple square-roots, having something to do with flow rates through a rocket engine.
this is exactly why in my country you NEVER have multiple choice in Math unless show your work is required. I actually had a challenge for my class once when they begged me to have a multiple choice test. I told them I would do one, but they would have to show each and every step NOTHING in there head. In exchange any student who got the questions 100% right would have their grade raised a full letter and anybody who got even one wrong answer would have their grade lowered one full letter. To be fair the test was optional and they did NOT have to take my challenge. All of them accepted though. I threw a curve ball on one of the questions and did NOT have the proper answer listed. Only 2 students of a class of 18 got their grade raised everyone else didn't even come close to getting it and I did NOT need to throw the curve ball.
the answer is obviously less than one, and three of the options are obviously more than one, but i’ll prove it anyway. it’s 3sqrt(3)/5sqrt(2), which is 3sqrt(3/2)/5. multiply both parts by 2(put the 2 in the square root) and you get 3sqrt(6)/10, which is answer choice b
Break each number into their prime factors, 27 = 3^3 and 50 = 2x5^2, so sqrt(27) = 3 sqrt(3) and sqrt(50) = 5 sqrt(2), then rationalise the denominator by multiplying by sqrt(2)/sqrt(2) and you’re done.
Excellent John sir thanks for telling us that you are teaching for s decade. Do you know that 1year experience multiplied by 10 is not the same as 10 years experience multiplied by 1.
I guessed the answer by approximating the original two square roots, and then comparing that estimate to the four choices of answers. Only one of the four could have been close, so although I forgot the details of solving this, I was confident just using approximations. But now I know how to solve it precisely!
