GMAT Math Review Part 2

Provided by TCU Neeley Graduate programs, GMAT Math Review Part 1 and 2 cover the primary mathematical concepts included on the GMAT. Both videos are prerequisites for the free GMAT prep class offered by TCU Neeley.
For more information, go to www.neeley.tcu.edu/gmatprep.

[@tireswing]

The white board was so clean in the first video when he started.

[@thomasmr.9931]

Haha yes, I didn't notice while recording but later when I saw this part 2 I thought "whoooaaa what is that?!"

[@shafiabdullahhauque7933]

Please upload more tutorials. You're the only KZbin tutor who made me understand things.

[@Ashley_429]

These clear explanations have helped me so much!! Thank you for taking the time to make the videos.

[@mathtutoring8276]

excellent! for everyone! I particularly liked how he took out the mystery out of the 'distance formula'!

[@Chris08160]

This second part is just great! thanks very useful and easy to follow

[@Scheport]

This second part is so good, Thank you!

[@sumana301]

Best tutorial on the internet! THANK YOU THANK YOU

[@edgarchang9447]

Thank you for this video. I got through MBA and preparing for PhD. Super thanks prof

[@Marcelllu23]

Thanks for the amazing video!

[@cycle00001]

very helpful review of math, please do more videos

[@leftbrainprep]

This videos is excellent. I cannot recommend highly enough.

[@lola_hiss3628]

great video! thank you!

[@marinabaskakova2333]

You’re the best, thanks for these videos!

[@ThomasCambridgeMaths]

Welcome Marina!

[@ammarahmed9754]

what a great teacher

[@najibmaftah79]

Thank you so much, this review is amazing, very helpful

[@vincenttanudjaya7397]

Your video helps me a lot, I deeply thank you for that.

[@ThomasCambridgeMaths]

Welcome Vincent!

[@sakethgayam178]

Thank you so much.

[@chumbaamalelthedoc8816]

A great instructor

[@giusyniclaorlando6842]

JUST GREAT. THANKS A LOT

[@mendypeters8489]

Holy shite every time he says HyPOTthenuse I crack up, forgot that was a real word, key and peele man

[@javierbercedo395]

thank you a lot

[@ThomasCambridgeMaths]

Welcome Javier!

[@Onlybelieveable]

at 32:24 I think you might have made a mistake on the slope formula. isn't it y2-y1 /x2-x1 ?

[@ShinichiroSatoshi]

Yes but as long as you understand that the hypotenuse is downward sloping it doesn't matter whether it's (y2 - y1) / (x2 - x1) or the way he did it in the video.

[@hassanmahfouz4997]

Is there more ?

[@michaelbarron6073]

Super helpful, but I'm not sure the trapezoid area at 14:50 was done right. It sounds like you used non-perpendicular sides as your bases and heights. Shouldn't both areas be equal since you "cut" the trapezoid in half diagonally? Please help me understand what I'm missing.

[@ThomasCambridgeMaths]

Hi Michael, you are correct that using two sides as base and height would not work as there is no right angle; note the measurement of 8 is not a side length, but the given height, which by definition is perpendicular to the base. The "bottom half" triangle has a larger area because it has a larger base; the diagonal cuts the figure "in half" in English, but geometrically the resulting two triangles are not equal.

[@bassamal-kaaki3253]

Shouldn’t slope = y2 - y1 / x2 - x1 = 7/5, since (4- -3) / (-2 - 3) = 7/5?

[@abdullahhafeez2210]

Thank you, pink shirt man. God Bless

[@arpitamitra8363]

Hey is there any data suffiency and data lnterpretation video that you made.

[@thomasmr.9931]

Hi Arpita, we only did the presentation of the GMAT Guide math review chapter, but search youtube, lots of good content out there.

[@chukwumahenurah1989]

slope is rise over run so y2-y1/x2-x1

[@ThomasCambridgeMaths]

Agreed! Tnanks Chuk

[@shafiabdullahhauque7933]

Can we solve this using the formula you've given in the video? Please help. 300 g of salt solution has 40% salt in it. How much salt should be added to make it 50% in the solution?

[@shafiabdullahhauque7933]

@@rockindoodcm Thank you for the answer. I fail to equate this formula in some other problems. Is there anyone on KZbin who has explained this formula for multiple problems? I have tried using this in many problems, but I failed to solve using this.

[@azamtokhi6882]

For mixture you made it too hard even it can be done easy 15%L+5*8%=10% 15%L+5*0.08=0.1 multiply 100 both side 15L+40=10 15L=10-(+40) 15L=-30 now divid the both side by 15 L=-2 Means we are missing 2 liters I am learning it and your explanation is great

[@ThomasCambridgeMaths]

Hi Azam, often the content has muliple possible methods, and the best method to use is the one that works for you, so for any mixture questions do use the method you show.

[@spider4633]

He didn’t really explain the short cut with the 30-60-90 or 45-45-90 at all.

[@ThomasCambridgeMaths]

Hi spider4633, take a look at another of my videos, starting at the 5 minute mark: kzbin.info/www/bejne/m6CUhWOXfrOGgbM for this topic with examples.

[@shafiabdullahhauque7933]

@@ThomasCambridgeMaths Hello, can you help me solve this using the formula given in this video. 300 g of salt solution has 40% salt in it. How much salt should be added to make it 50% in the solution?

[@ThomasCambridgeMaths]

@@shafiabdullahhauque7933 Using the variable S to represent the required salt, formula is 300 x 0.4 + S x 1 = (300 + S) x 0.5, which in English is "300g x 40% + SALT grams x 100% (the new solution is all salt, thus 100%) equals the new amount of 300g + SALT grams, which is 50% salt. Let me know your answer using this formula and I'll let you know mine!

[@shafiabdullahhauque7933]

@@ThomasCambridgeMaths At first, I had a complete different answer using the formula you've explained in this video. Then, I've found the right way to solve that using that formula. My only confusion is sometimes we use 0 and sometimes 100. Using that formula, can you please clarify me when to use 0 and when to use 100? It has been a big confusion. I want to do all mixture problems using the formula you've given.

[@ThomasCambridgeMaths]

@@shafiabdullahhauque7933 The 1 (100%) I used is based on the percentage of the solution, salt is "all" salt, thus 100%. if the solution were 80% salt it would be 0.8. If the new solution were sugar, that would be 0.0 (0% salt). A clue is that the final percentage must be between the two contributing percentages. In this case, 40% + ___% = 50% tells you that the missing % must be more than 50%, thus if you're unsure between 0% and 100%, the only one of those two which is more than 50% is 100%.

[@randomchica23]

"they give a simple equation that will work for any polygon but let me not show you that AT ALL and show you one that will work for only this pentagon" ......................

[@ThomasCambridgeMaths]

Hi, can you let me know where in the video, thanks.