Conclusion after finishing quant sections

Well, this morning I was feeling pretty down… seeing the 750+ question videos etc… (which btw not sure if they really are official or not), really demotivated me sometimes as I was falling in all traps, etc… So anyway, as I have little time left I had to start really looking into official questions to understand where I am at after these 10 days of study.

So, athough I know it is not the most accurate, I decided to open the Official GMAT 2017 guide for the first time and take the diagnostic test on the QUANT section. Very interesting conclusions where drawn from it. I timed each question in the process, and after the test passed all my stats to the computer.

My approach with this test was NOT to test myself with the time fully constrained. Of course I timed everything, but the main priority is understanding… given the tools I have acquired, can I actually solve these problems? Well it turns out the answer is a decently SOLID YES. I got 22/24 right on PS and 21/24 right on DS. Does this mean I can be pleased and down my guard on Quant? NO WAY.

Actually, this test just shows that I have the tools to solve the problems, but this IS NOT enough. I need to be able to solve the problems, AND IN THE ALLOWED TIME.

Each question should take max (MIN*2.02). My PS took (MIN*2.95) which is WAY OFF. And my DS took (MIN*2.1) which is pretty close to where I need to be.

LEARNINGS

For each question I got wrong, or took too long, I analyzed and found improvements and solutions to most of them! If I am able to retain these learnigns and apply them on the next tests, I should see big improvements in terms of time, which is what I need to improve on. Also of course, a question that took me 10 min / 5 min is unacceptable on the test day. On the test day I should never go above 3min on a question.

Also careful. Absolutes in inequalities (which is a weak topic I currently have) didn’t appear for instance in the questions… so once those appear maybe I will be more prone to errors. So I MUST do many more questions ASAP to keep training and seeing where my weaknesses really are especially in terms of content. Geometry wasn’t actually tested that much either on those questions (maybe just 2 or 3) and that might also have helped not sure. I def. need to take more test to have a better understanding.

CONCLUSION

I am extremely happy I have been able to assimialte the content in 10 days. From 0 to where I am is a big improvement. This means I can start focusing on methods, improving my timings and basically practice practice … I feel my basic calculations are speeding up the more I do, which is really good.

I need a true gauge of where I am at. Must take a formal practice test ASAP. Tomorrow if I feel confident enough might take it in the evening and see where I am at. I just feel I want to try and train more questions before doing it :P, as for example doing the exercise below really taught me a lot actually.

 

 

Ok so as it happens, no, today wasn’t a full gmat day. In fact I don’t think I will be able to start proper study until 7pm. Nothing bad happened but I had to get some tests done in the morning, then microwave oven broke and the technician is here repairing it.

I will start with Probability

Probability

Probability: Likelihood that an event will occur

Probability goes from 0 to 1. The more likely the probability, the closer tha value will be to 1.

P(A) = Number of outcomes where A occures / total number of possible outcomes

Complement

The complement of selecting a green ball out of 7 total balls is the probability of selecting a ball that is not green.

P(event happens) + P (event NOT happens) = 1

The question below is an example of a question in which knowing the complement rule is essential!

As you can see, it is much easier to calculate what n of events will give an odd number in that case.

This is helpful for cases when the complement is easier to calculate, and also when “at most” “at least” can be an indication that complement might be best approach .

Mutually exclusive events

These are events that can not occur together.

“OR” probabilities

P(A or B) = P(A) + P(B) – P(A and B)

Hmm interesting. I am sure this will come handy in the future. Also important to understand that mutually exclusive events P(a and b) will be 0 as they can’t happen both. Hence the formula above for those cases becomes = P(A) + P(B) + 0.

This question is an example where the formula is required:

Hmm this course is being so helpful! Just what I was looking for. Kudos to GMATPrepNow.

Important notation + AND formula

Important to check notation. P(b|a) menas probability og b after a has occurred.

It is very important to remember that formula too.

If question involves the occurrence of 2 or more events, I might need that AND formula.

Consider question below:

I am quite happy I got it right! Took 2min 15s though :(. I am def. having issues with time. Will have to find a solution once I start taking proper tests.

Independent events

“With replacement” so means second selection is not affected by first one. Like tossing a coin.

To understand if two events are dependent or independent, video suggests to ask “Does the occurrence of one event affect the probability of the other?”

So one thing I see is that being strong in counting helps in probabiliy. I am strong in counting so this section shouldn’t be the biggest challenge.

Consider this question. It is freaking hard:

 

I am putting my guard a bit down… getting a bit tired now that reached the end.

DONE! Finished all sections!!!!!

Final conclusion of quant study

Well, I am very happy that I have a very clear idea of the topics that come into GMAT now. By no means I am ready or prepared, but I am light years ahead from where I was 10 days back. I definitely know I will need to work hard on inequalities, word problems, geometry etc… if I am able to perfect those topics quickly enough, I might have a chance to get a high schore.

What will I do now?

1. Will watch these +750 GMAT series
2. Will probably open GMAT Official book and will try some questions from the first assesment test, just to see how much it differes from what I have done here.

OMG – +750 exercises

Damn damn damn… I am starting to watch the videos and WOW, the hard exercises are HARD indeed… damn… seems that I am gonna need to keep watching videos of high level problems. Sometimes methods are so unique to a problem… consider this question.

I mean… GOOD LUCK trying to answer it. Seriously.

Even after seeing the answer I am not fully clear of why the method actually works. OH WAIT! I know why…! Damn… ok, finally it dawned on me.

First of all, the question asks about a percentage. This menas that we can assume a population of X, and the results still we can calculate because no matter what population we assume, the % still applies.

Then the next thing to realize is that if we are talking about how many % directors are there… the SALARY thing CAN be used… why? Because if I am talking about average salaries etc… the formula SUM of whatever = n*Avg of whatever still applies. This means that it must be true that D*Ad + M*Am = 100*At. I know that this formula is dealing with salaries not people, but remember that the D,M,100 part of it IS people numbers, not salaries. And the CRAZY thing that one doesn’t know is that if i know how much Ad and Am differ from A, and substitute those in the equation, and also realize that M = 100-D and substitute, then everything cancels out but D. Why it cancels out?! Not really sure to be honest.

And WTF, there seems to be a 10 s solution by realizing that at the end of the day, what we are saying is that M are 5000 away from “fulcrum” and D 15000 so in order to balance the scale, M has to be 3/1 in ratio with D. Means that out of 4, 1 D is there hence 25% which is the answer.

WTFFFF…. these questions are insane.

Try solve the one above. GOOD LUCK. You think you got it right? Think again. Damn…

What I learn from this is that… well… questions can get VERY VERY hard. I am still not sure how to handle this, but is good to be aware that this level of trickiness is embedded in the system.

Ok so I am still watching the videos in the series… in math section main learnings have been to be aware of certain common constructions such as X squared is a two digit odd integer… this doesn’t mean X is an integer etc. Also some interesting techniques such as 8! + 9 I know its not prime because both share a 3 inside (and a 6 in fact) hence that num will be divisible by 3 or 6 etc.

 

Tips on reading comprehension

Now he is talking about some tips on reading comprehension. One thing I learned is that RC seems to weight quite a lot in the exam. Also , “disinterested” means “non biased”… one needs to be careful because it does not mean “uninterested”. Also, there is a technique overall to keep in mind… OCTAVE… opinion of author, contrast, tone, avoid superlatives in answers, verify and eliminate, dont select. I like the approach and although I don’t plan to apply it religiously, I am gonna def keep it in mind.

Also critiqur is not necessarily bad. It means analyzing + and – and then giving an opinion. Similar with “Assesment”.

AND, OR, BUT must maintain strong parallelism from R to L.

If 2 part dependent on 1st then we need to use ING on the second. If something is INTENDED and something is NOTINTENDED then they can’t be put in parallel forms.

Conclusion 2

Hmm woring on very hard questions demotivatea a littles. Damn I hope I am in the right direction. I must take a sample test soon to see where I am at. Maybe tomorrow. Lets see.

 

 

Statistics, counting and probability… the last 3 topics. Had a long day, managed to get quite a lot of verbal reasoning and sentence correction done in between meetings etc using prep4gmat app… really reccomend it. I wanna get now 4 hours of quant done. Tomorrow I think I am gonna have a FULL free day!!!! I am SO looking forward to it. If it does hapen then I will have so much time to push.

Common let’s do this!

Statistics

Average, arithmetic mean, mean

This is easy. Sum of n numbers divided by n.  Influenced by outliers.

Median

This is basically the middle number on an odd list of arranged numbers from low to high, or the average of 2 middle ones if list is even.

The median is simply the point where 50% of the data is above and 50% of the data is below.  It is also the point that minimizes the distance from it to any of the other points in the dataset. It is a statistic not influenced by outliers.

Mode

The number that occur most frequently in the set.

Rule 1: In this set 4,4,5,5,7 for example, the mode is 4 and 5. This is important to know.

Properties

Rule 1: If the numbers in a set are evenly spaced, then the mean and median of that set are equal. (Evenly spaced means difference between 2 adjacent numbers is same as any 2 adjacent numbers of the set). This also works for negative numbers too. We can not say though that if mean and median same, the numbers are evanly spaced. That doesn’t work.

Weighted averages

If you have 2 groups of numbers with same number of numbers in both, the average of the combined 2 groups is the same as the average of the averages of each group. If the groups don’t have equal numbers thought, then you need to “weight” each average. Actually when only 2 groups with same… at the end of the day you are wighting the average by 1/2 each, so it’s better to just learn the below.

Weighted average: (Group average1) (proportion) + (Group average 2) (proportion) …

3 cases come from the above that one needs to keep in mind. Imagine u have 2 populations A and B with averages a and b. Well if popA = popB then avg is (a+b)/2. If A > B then we know average will be closer to a than b and viceversa.

Measures of dispersion

These are ways to measure how much a group of values is “spread out”.

RANGE: Greatest number in the set – Smallest number in the set. (Limitation is that it only tells us about the extreme values in the set)

STANDARD DEVIATION: Measures the extent to which all data values deviate from the mean. In other words “Is the average distance the data values are away from the mean”.

To calculate the standard deviation we need to know the mean and the number of values.

Below you can see how it is calculated (not required for GMAT). Basically measure how far each value is from the mean, square to make all positive, take the average of those distances, and eliminate squaring by taking the root.

VARIANCE

This is the value right before we take the last root on the Standard Deviation to “reverse” the squares we did before. So effectively Variance is the standard deviation squared.

UNITS OF STANDARD DEVIATION

How many SD’s u are from the mean.

Consider this question:

Hmm… although took right approach, I made a mistake in thinking that if 2 numbers add to 30, then one of then must be greater or equal than 25 which is notright as I can have 15 15. In the future if ever I encounter such situation for example… a+b = 35, so what are possible values for a and b? One thing I can do is to divide by 2to see values close to each other… in this case 17 and 18. thats the closest they are gonna get… So by dividing by 2 I will make sure I know all cases.

More interesting properties.

If I have a set of numbers with mean M, and add K to each number, the new mean will be M+K and the standard deviation will be the same as it was before.

Consider this question:

Now, if you remember that in a set of even spaced numbers, the MEAN and the MEDIAN is the same, immediately you know answer is 0. I forgot the rule so had to calculate which gave me right answer but took 2,5 min which is not good.

May this be a lesson for the future. Know your properties.

Consider this question:

Now hmm the thing is, I used the weighted average approach which gave me the right value but its not the quickest. The other way to solve it that I missed would have been realizing that again… if I am given n of population and average, I can know the sum of the population. In this case I am given the total n and the total average hence I know the total sum. I am also given n and sum for 2 and 3 guys and I am asked to find the value of the remaining guy. So quickly one can see that I just need to do mean n + mean2 n2 + remaining guy = totalmean + ntotal

Which is a faster approach.

Consider this question:

Now in this question the main problem was that I forgot about the “Positive Integers” part. Everything else I did correctly, but by forgetting that key bit, I got the wrong answer. Again, hmm… when writing the integers as a, b etc… on my paper, I didn’t put any note regarding them having to be positive. So next time if a condition like that is given, I must NOTE it on my paper once transcribing the question otherwise this will happen.

Once you have to solve a question in less than 2 min, there are no stupid mistakes. Every mistake is a mistake that was caused by something… and those mistakes and the speed is what I guess differenciates an OK score from a great score.

D0ne with this section! 😀

I am 70% confident with this section. Will have to surely revise, but I need to revise Geometry etc more.

Counting

So here we are! In counting! This is a very interesting and fun section. If you remember on my first day I did a bit of it and tackled some interesting problems. I am looking forward to revise this again.

Listing and counting – Hm… many cases not fastest way but sometimes might be considered. Important if listing to do it in a systematic way. Counting tree.

Counting tree is a beautiful way to list and also to understand structure.

Omg, one of the first problems is this one… later I found it is especially hard maybe even too hard for gmat.

How to list systemtically in such case is very important. I didn’t know how to do it.

Fundamental Counting Principle

This makes sense. Similar to “BOX” method I use.

Next video actually is the box method explained:

Hmm… important to be diligent when using the method.

Factorials

This we have done before.

If we have n unique objects, we can arrange them in n! ways.

The simple counting problems I am confident and can solve in 30 s or so, but yeah, complex ones take time.

Consider this question.

This one I solved in 1m 10s.

Restriction Rule: N of ways to follow restriction = N ways to ignore the restriction – N ways to break the restriction

I don’t think I will have to use the above though.

WRONG I do it seems.

Consider this question:

Now this question if you don’t use the above rule, it becomes pretty pretty hard, so I MUST learn that rule.

Now first of all if I wanted to follow normal method then how would I do it? Well I didn’t think of it but basically one needs to realize there are 3 cases… numbers with 1 seven, numbers with 2 and 1 number with 3.

Now that I think of it this could be a viable strategy too. There is only 1 num with 3 sevens. There is 9+8+9 numbers with 2 sevens as you can arrange the boxes as: (8*1*1) (1*1*9) (1*9*1) . And then there are 8*9 + 9*8 + 9*9 numbers with 1 seven. As you can arrange the boxes as (8*9*1) (8*1*9)(1*9*9). Note that in the places where 0 can be put , there are 9 digits that can go, in the places 7 is there, there are 1 choice only, and in the places where 0 can’t be there are 8 options only (1,2,3,4,5,6,8,9)

Now the other approach is to think… let me calcualte all numbers with 3 digits (9*10*10) = 900 and then lets substract all numbers that have NO 7’s… (8*9*9) = 648 hence the result = 252 = D

Hmm hard problem indeed.

TIP: In the future questions “at least” or “at most” indicate usually that the fastest way is to use the restriction rule.

Arranging letters where some are same.

To arrange letters NANNY for example… its important that we eliminate the effect of the 3 N’s by dividing by 3 factorial. Rule is below:

Combinations

Order of selected objects does not matter.

The important thing to remember again is that order does not matter.

Quick way to do combinations

Actually the method explained here, is what I actually do when I tackle a combination

Important combiantions property.

If we have 5 objects, and we have to see how many combinations are there if I chose 0,1,2,3,4,5 objects, there is a way one can calculate this easily.

 

 

You can see the method above. Basically if we do all the way from nC0 to nCn the result is 2 to the power of n.

Another interesting property

Hard to visualize but basically taking 6 people and selecting 2 on comitee and its same as taking 6 people and selecting 4 of them not to be on the committee.

Selecting 1 person to be on comitee is same as selecting 6 people NOT to be on comitee.

The generalization of the formula is:

DONE :D!

Hence the current progress is as follows:

Only Probability is left! After that, I plan to do a bit of quick exploration on some videos regarding advanced strategies for quant section, and then probably its time to do a practice test using the OfficialBook see what happens. Then after that one I can correct mistakes and then probably take a proper practice test using the software? We will see! I wanna revise geometry, word problems and also every other topic briefly too … don’t wanna waste the tests unnecessarily.

 

 

GMAT Day 8

Ok, so as it happens the last 2 days I have had too much work and was not able to do much (yesterday post is actually from 2 days back I think? I just forgot to post it and got saved as a draft. While I am between meetings I use prep4gmat app to learn verbal reasoning and so far I am really enjoying it and it is helping me quite a lot I would say. I definitely recomend that app .

Geometry

At last… we get into… Geometry! For some reason I am not looking  forward… so let’s re-wire the mind. Let’s enjoy it, let’s have fun, let’s be great at geometry, let’s go.

Angles

Ok so went through the basics of angles on a line. Important learned the notation <ABC , this means the angle is coming from B.

Properties of lines easy no need to write .

Triangles

• All angles add to 180 degree
• The longest side is opposite to the largest angle
• The shorter side is opposite the smallest angle
• Those two can be summarized as if a<b<c then A<B<C. This is important to really understand… so if one asks is angle x > than y is the same as asking is the opposite side to x larger than opposite side to y?
• The sum of the lengths of any two sides of a triangle must be greater than the third side. For example if sides are 3 and 7, what lenghts are possible for the third side? a positive number <10 and > 4 is the answer. (So any 3rd side has to be greater than abs(difference other two sides) and less than sum of other two sides.

Consider this question

This I am happy to solve correctly and quickly in < 1min. i knwo it’s easy but still… happy its working. Important to celebrate small victories XD!

Assumptions

On GMAT all lines that seems straight are straight. However this doesn’t mean u can assume angles. So if it looks perpendicular doesn’t mean its perpendicular for example.

Can’t assume 2 lines are paralel because they look like.

Angles that appear on diagram are bigger than 0 degrees.

Triangles isosceles  / equilateral

2 sides are equal and the 2 angles opposite are equal too of course.  If we know 1 angle, we can always know the other 2 angles.

Equilateral has 3 equal sides and 3 angles same, so 60 degree each.

Area of triangle

a = bh/2

Area of equilateral triangle = (Root(3)*side^2)/4 – I don’t like to memorize formulas but well, I guess I will have to.

Rule: Altitudes of equilateral and isosceles triangles always bisect the base

Right and special triangles properties

Hypotenuse is longest side (as it’s the one oposite of the 90 degree angle which will always be the largest angle in a right triangle).

Pythagoras theorem. Ina  right triangle  h^2 = a^2 + b^2

Pythagorean triple: A set of 3 integers that can be the sides of a right triangle. Here seems good to memorize these ones: 3-4-5 , 5-12-13, 8-15-17, 7-24-25. Their multiples are also pythagorean triples. This can only be used if one has 2 CORRESPONDING sides. Just one side or 2 non correspomding sides doesn’t cut it.

Consider this question

Now it is easy however, remember it asks the height. I calculated side right, and made a mistake as it asks height not width. I MUST always try as best as possible to re-write question to make sure I am answering what they are asking.

A special triangle is the 45-45-90 or 90-60-30 as below:

This is important because it is the only way to for example find one side of a special triangle… if I only have 1 angles besides the right one, and one side.

Thesespecial triangles hide between squares and also equilateral triangles. Watch out for them

Here an example of a tricky question that basically can only be answered if you know those rules:

Now, I actually got confused, first because angle ABC I confused with angle ABD. Didn’t look careful enough on the diagram. Then later once I fixed that, I again made another error concluding that an isosceles right triangle has angles 90 55 55 which obviously doesnt make sense, it is 90 45 45, which allows us to use the properties, also this shows that the bottom triangle is a 90-60-30 one, which again allows us to easily solve for X.

 

Similar triangles have 3 angles in common

So one is the same triangle magnified.

Rule: With similar triangles, the ratio of any pair of corresponding sides is the same

Quadrilateral

It is a poligon with 4 sides. Square, Parallellogram, rectangle, trapezoid rombus

Rule 1: Angles add to 360 degrees

Parallelogram has 2 pairs of parallel sides (opposite angles equal, sides opposite equal). Rectangle is a type of parallelogram. Rombus also. Opposite angles equal and opp sides equal… actually in rombus 4 equal sides too.

Trapezoid is not paralelogram, but has 2 parallel.

Other important rules. In rombus, diagonals are perpendicular bisectors. And in rectangle and square diagonas have equal length.

Area of quadrilateral is b*h

Area of trapezoid is ((base1 + base 2)/2) * height (So basically the average of two parallel sides multiplied by height).

Area of rombus can also be : (diagonal1 * diagonal2) / 2

Polygon

Closed figure formed by 3 or more lines.

All interior angles inside are less than 180 degrees. We have already seen triangle and quadrilateral.

 

Important rule to find sum of angles inside a polygon.

Angles inside polygon = 180*(N-2)

One can see why this is the case if draw line from one point to all other intersections. You have (N-2) number of triangles, and polygon must have all those angles inside hence the formula.

A “REGULAR” polygon means it has equal sides and equal angles.

Circles

• Points all at a given distance from the circle (Radius)
• A line from a point of the circle to another is called a CHORD. The LONGEST POSSIBLE CHORD on a circle is the Diameter.
• Arc is the perimeter between 2 points (minor arc).
• Circumference is 2*(PI)*radius.
• Area is PI*r^2
• Length of an arch of angle X is X/360 of a circumference
• Area of a sector of angle X is X/360 of a circumference

Circles properties

INSCRIBED ANGLE (is the angle opposite to a chord, when we have a triangle formed by a chord and one other point in the circle)

RULE 1: Inscribed angles at the same side of the chord and holding the same chord/arc are always equal

RULE2: Inscribed angles holding chords of equal length are equal

RULE 3: Inscribed angles holding de diameter is a right angle

An angle is a central angle when its vertex is at the center of the circle. Holds chord AB

RULE 4: If we have a central angle holding chord AB and we have an inscribed angled holding chord AB, then the CENTRAL ANGLES IS 2X the INSCRIBED ANGLE

RULE 5: I didn’t know this rule but basically inscribed angle creates a chord yes, but also an intercepted arc. The rule is that the intercepted arc is twice the measure of the inscribed angle. So the inscribed angle is half the intersected measure. 

A line that touches cercle side, is TANGENT to the circle.  A line from the center to the tangential line is 90 degree.

Volume and surface area

Volume of cilinder is Area of base * height .

Surface area of a rectangle with 6 faces. 12 edges and 8 vertices. Just add areas of 3 faces.

Surface area of a cylinder is the sum of the 2 circles + the side area which is perimeter of circle * the height.

Data sufficiency

To find one length we need at least one other length. Usually sufficient if forces the shape to having one length.

Three hours went by. I finished this section, now there are 14 questions. Will try finish them now.

Consider this:

Geometry questions can be damn hard. Hmm…

Big conclusions

Geometry is still a big weakness for me. I managed to answer most questions but its taking me way too long to answer them so it is totally insufficient. Need to practice this a lot after doing the general revision.

Geometry can be pretty tricky because there are so many relations one needs to take into account, formulas also… and you need to apply everything very quickly.

Gonna be tough guys to get this all right with the few time I have… but I will try!

 

 

 

 

 

 

 

 

Day 7 – Soon getting into the hardest part

Ok, so here I am, it’s been an extremely long day, reached home 8pm absolutely shattered. I have 2 taks that take some hours that I will have to do before tomorrow ends and there’s no way I can do them today. I have taken a 30min nap (otherwise it’d be impossibe to function) and although it was tough to wake up, managed and here I am ready for probably about 2 hours max 2:30 of focused study. I know the hardest part comes soon and I am not looking forward for it, so I  need to re-wire my thoughts. Bring it on. Bring the hardest part in, once I learn it I will be able to do so much more! Let’s enjoy it, let’s have fun with it.

First Im gonna finish the word problems section, and I will then go to Geometry.

Finishing word problems

DId some work on rates etc similar like speed nothing new.

Double matrix method ^ Venn diagrams

This seems a very popular question type so need to be careful.

Ok I did this question right, but got blocked beause I missed the statement “7 students take physics and sociology” :(. Stupid me. And I actually double checked to see that I wasn’t missing anything.

Need to be more careful next time.

Ok I have a huge problem with venn diagram which is a lack of process that causes the probelm to take way longer than it should. So I am gonna practice a process.

Ok, main issue is that I am writing the equations etc. I should just fill the venn diagram with the numbers directly. Doing that it took me 2 min to find the answer. Will try this approach from here onwards.

Sequences

Nice, sequences is an interesting hard topic. Hmm simple sequence I see can be given by a simple equation… then can be used to calculate other values. This should be fine. The only little issue might be me having to come up with the equations myself. That might cause some confusions if I am not sharp on the day.

Recursive Sequences

Hmm… so far seems one just needs to substitute and need the values of the starting of the sequence to calculate the ones in the future. Let’s see if in the future they give a method to calculate a far term like t80 by using a formula or something.

Ok this exercise totally blocked me…. because I am given a recursive series but no terms to play with.

Ok, so right after pasting that image above and not having watched the answer yet, I realized I missed something very important. Basically I do have the information I need. I know the formula for Xn can be solved if I know Xn-2 and Xn-1. Hence to know n = 4 I need to know 3 and 2. I know 3 already. So just need to know 2. And the thing is, because I already know 3, and I know 1 also, I can use the formula of X3 to find X2, hence find X4 which is the solution.

Hmm… ok got it and will realize much quicker next time.

Sum of Sequences

Ok I know this kind of stuff comes up so need to learn it well.

Ok so the technique I already knew, but learning the formula might be be nice to speed things up. For the moment I will just use logic. Also then it teaches a formula to calculate how many numbers from 1 to 20 for example (inclusive). This also I already know that is 20 – 1 + 1 = 20. It’s very easy to know if u check 1,2,3 then you do 3-1 = 2 which is wrong, so you know you need to add 1 to the result to get the number of integers in that series.

Alarm sounded 1:45 passed. Put 1:45 more.

Ok so consider this question:

Now, I was doing it almost perfect, understood the whole dynamics of it quickly. My mistake was with the timings. I took L (in the pic they call it P, but I chose L) as being the volume NOW, but in my mind I confused it by thinking it is the volume 6 hours ago… which doesn’t make sense.

If I had just chosen L to be the volume 6 hours ago, and then multiplied by 3/4 to get the one now , and then 2 times again to get the one in 12 hours… then It’d all have been fine.

The concepts are there, I am just making some pretty careless mistakes… that I will have to tackle in the future with diligence.

Ok consider this question:

Here I took a wrong approach by splittin the salt and water components. Hmm It works also for sure but hmm I guess it made me do more errors? Main error was that I did Xsalt/Xother = 80/100 which is wrong . it should be Xsalt/Xother+Xsalt = 80/100. Then one can solve the equation.

It’s probably better to do what the video suggest. Every addition of volume of the solution X brings 0.8 salt and o.2 of other. Hence the new solutino is (50 + 0.8X added)/(250 + xadded). And then solve. This approach is much much quicker. The key is to realize that adding X volume adds 0.8X salt. Next time will solve using this approach.

Ok 1 hour passed since last time I put clock. Im done with the section. Hmm… I should go to sleep now but there is still 12 questions related to this section that I haven’t done. Should I do them or not… If I do them it’ll certainly take me 1 hour or so I guess… hmm… ok look I’ll take them. Then tomorrow I can focus on the other sections. Btw I see the remaining sections have less videos than some of the ones I did before so hopefully by Tuesday I should have done all… let’s see, I will try.

Consider this question:

I have a big issue with this question. When evaluating statement 1 I got this wquation 11p+5b = 100. Now… although with 1 equation and 2 variables we cant know the answer… if p and b are integers… then we might. I already knew this… and when looking at the equation I thought… damn… looks non sufficient but I bet you it is… but then I decided to ignore it and just put NS. XD! As I had thought, statement 1 IS sufficient. My big problem is that one needs to do the table and check every case… and that’s very time consuming. I want to know a technique that can tell me if an equation likea bove has just 1 solution if variables are integers… not sure if it exist.

Consider this question:

Now this took me 3min 29 sec unnecessarily. The main issue was that I was not clear how many terms are there between 5 and 200 when split in 5’s. It turns out it is (200-5)/5 and then add 1. This I should have known much faster. Then it’s easy.

Another approach which is also smart is the one given in the video. It relies on realizing that 5 + 10 + 15 … is equal to 5(1+2+3…). I will totally keep this in mind because it might be helpful for some other kind of questions. By doing that then one just needs to solve sum from 1 to 40 which is easy as it is 41*20 and then multiply by 5 and you have your answer.

Ok consider this question:

 

Well it caught me totally of guard, made me waste a ton of time . For problems with 2 variables like this one… I need to try the matrix approach instead of venn… I need to practice this.

Ok, so I am done with this section