QA is the section that gets the maximum attention of test-takers of all stripes and there is always a litany of frustrations and queries that plague aspirants —
- I am good at Math and like Math but my score just does not seem to go up!
- Should one attempt the long Arithmetic questions?
- I feel every problem is do-able!
- I get stuck for long with one problem without realising it
- I realise there were many problems I could have solved when I analyse the test
The answer to all of these questions lies in the way you select questions and the way you navigate between them.
What should be your single biggest agenda on the CAT — Return on Time Invested (RoTI).
This is toughest to achieve in the QA section because there are no sets for you to choose questions in bulk. So how does go about question selection?
Does it make sense to first skim through all the questions?
This is a reasonably good strategy but I feel that there some drawbacks to this.
- Firstly, since you have to go through 34 questions you will really try to read the question fast, say in under 30 seconds and take a call whether to do it or not.
- Secondly, once you have selected a question, there is a high chance that you get stuck on those questions because you feel since you have selected this and rejected others you have to do this.
So test-takers using this approach tend to
- spend very little time selecting
- and get stuck on questions they have chosen
What you ideally be doing is
- spending enough time in the beginning to evaluate the question properly
- never get stuck on a question for more than 3 minutes
A better approach is the A-B-C or Now-Later-Never approach
After you finish reading a question the first task is to decide whether to do the question — NOW (A), LATER (B) or NEVER (C).
How do you classify these questions into these categories?
Your approach should be the one that you would follow if you were required to pluck fruits from a tree — you will pick up all the lowest hanging fruit first and then go on to picking the ones that are higher up.
While it is easy to judge the height of fruit hanging on trees, it is quite another matter to judge the difficulty level of questions that appear on the QA section of the CAT.
How do test-takers normally select, or rather, reject questions:
- I am good at Numbers (or I just did Numbers last week) so I will do this question.
- Geometry is better left alone, it is always time-taking
- P & C and Probability, no way.
- These Arithmetic questions are too long and will consume a lot of reading time
- I haven’t learned and have never liked Logarithms and Functions, so better leave.
So what happens when you follow this approach?
No Arithmetic, No Geometry, No Logs, No P&C, No Probability, Only basic Algebra, No Polynomials!
You are left with very few questions! You will get at least 5-7 questions from each of the 5 Areas — Numbers, Arithmetic, Algebra, Geometry and Modern Math.
If you refuse to do questions from more than 2 areas you will be left with around 15-20 questions to take a shot at and will end up attempting around 12-15 and getting 10-12 right and have a score in the 30s to 40s.
So the first rule is that you shall not discriminate against a question based on area, topic, or length!
You will use your powers of discrimination to identify whether the question is Easy (NOW), Medium (LATER) or Difficult (NEVER).
Some of you might have a different problem
- I know all topics and hence try to solve all questions!
Discriminate on the wrong grounds and not discriminating at all are equally punishable with low scores!
A question should be categorised as
NOW: If you can see the clear steps to the answer, the number of steps are few and the question can be solved in 2 minutes
LATER: If you feel you can do it but you will know the way only after you start doing it — you cannot see the four steps to the answer right now — or if you feel that there are too many steps and it will take more than 3 minutes
NEVER: You do not have any idea how to go about the question
So in effect, your QA score will depend on your ability to
- make 34 good selection decisions and
- exiting questions without getting stuck
How to classify a question as Now, Later or Never
The process listed above is easier said than done (management entrance tests are in that way very similar to management itself — it is not very easy to execute everything that is told in the book)
So let’s take the questions in the QA section of an old SimCAT and try out this process.
SET 1: For the set of questions below, classify the questions into NOW, NEVER or LATER based on your current approach
A small caveat before I give my classification:
Each of us can make different choices based on our abilities, so my choices will differ from yours but the underlying agenda is to choose the right questions and maximise RoTI.
I have classified each question as NOW, LATER, or NEVER and the time eventually spent on the question.
QUESTION 1: LATER (< 1 min.)
As I read this question, I know they have given me a clear equation to write ak and that once I write that equation and try to find a pattern for a1, a2 etc. I will get a pattern. But what that pattern will be I will know only after I get. Hence, I would mark this for later.
QUESTION 2: NOW or NEVER (or Reverse Sweep!) (< 1 min. or 3 min.)
This question is directly based on a concept in Number Systems. We know that 72 is 8*9 or 2³*3².
If any of the options does not completely divide 200! then it means that 200! does not have as many 2s and 3s as there are in that option.
So one needs to find how many powers of 2 and 3 are there in 200!.
So at this point, it depends on whether you know the method to find how many powers of 2 and 3 are there in 200! or not. If you know you do it immediately if you do not you leave to not come back to this question again.
I could not recollect the formula (as most of my students know I do not know advanced formulas that were not covered in school)
So what did I do?
I just looked at the answer options and thought through the logic.
If (72)^41 is the answer — it does not completely divide 200! — that means it has more 2s and 3s than 200!
Now if (72)^41 has more 2s and 3s than 200! then (72)^44 will have even more and the other two options even more since all of them are higher powers.
So if option 1 cannot divide 200! then options 2, 3 and 4 also cannot divide so all four have to be right but you can have only 1 answer. So by the same logic, it cannot be option 2 or 3 since then 2, 3 and 4 should also be right. So the only logically possible answer is option 4, the highest power of 72, (72)^50.
QUESTION 3: LATER (< 1 min.)
This is a problem involving counting and drawing up a list — two perfect squares whose sum will end in zero. Such problems usually tend to take time and also have a scope for error. So at this point, I will mark this question for later and move ahead.
In the first 10 questions you should always be decisive in your choices and move quickly with the focus being on solving only those questions you are absolutely sure about. Like in a T20 or in a ODI match you should have a sense of urgency in this phase.
QUESTION 4: NOW (2 min)
This question has to be done immediately since you can clearly see the steps to the answer:
- You need to take the number of 10 rupee notes as x, it is given that number of 5 rupee notes is 2x + 1 and 1 rupee note is 4x + 3.
- The total value will be 4x + 3 + 5(2x+1) + 10x or 24x + 8, option 2.
Do not reject Arithmetic questions based on length — if you do not have time at the beginning you will not have time at the end! Read and reject only if the logic is complicated.
Arithmetic questions are going to be the staple of every test you take. So you have to step up your game as far as this area is concerned.
If Arithmetic is a problem then it follows that your English and/or your Logical Reasoning are a problem, making the odds of you clearing the test really tough. So if you are a Number Systems person who does not like Arithmetic then it is time you remove your bias, practice Arithmetic and start scoring!
QUESTION 5: NOW (3 min)
They have given a simple case of the roots being equal, the coefficients being positive and in AP and they have asked for the sum of roots. So only a few steps to the answer.
If roots are unequal then a quadratic is written as (x-a)(x-b) = 0. Since the roots are equal it should be (x-a)(x-a) = 0 or x² – 2ax + a² = 0.
Since the coefficients — 1, -2a, a² — are in A.P:
– 4a = a² + 1 or a²+ 4a + 1 = 0 or (a + 2)² – 3 = 0 or a + 2 = +√3 or a = √3 – 2 or – √3 – 2.
Since all the coefficients of x² – 2ax + a² are positive, -2a should be positive hence a has to be – √3 – 2 and the sum of roots is 2a or -2(√3 + 2).
If some of you marked this as LATER it is fine, but the key thing is that it is not a question to reject.
QUESTION 6: LATER (< 1 min.)
Since this question involves a could-not-be type of situation, I know that I have to find a pattern and then substitute values. The logic is not complex so it can be done later.
QUESTION 7: NOW (4 min.)
This is the kind of Arithmetic question that you need to crack to move up to the next level in terms of scores.
The key to cracking a long Arithmetic question is to wait till the end. They will make the question seem heavy in the beginning in order to either scare test-takers away without reading or make them lose interest mid-way. If you reach the end keeping your logical bearings intact you will see that in the end they really simplify it for you.
They have given the price of the flight and first AC tickets in terms of second AC, tickets — 2x, 1.3x, x.
Flight tickets have discounts based on the number of tickets bought (this information takes up a lot of lines but can basically be skimmed through)
This question right at the end mentions that the total amount spent and the number of tickets purchased is the same for flight tickets and first AC tickets and just makes the whole thing really simple.
If both Total Amount Spent and Number of Tickets Bought is the same, then the price has to be the same for both flight tickets and first-class tickets — first AC is 1.3x and flight is 2x — since there is a discount on flight fares based on a number of tickets bought and first AC fares cannot be increased, the common price has to be 1.3x, with the discount on flight fare.
So you know that all you need to do is find the number of flight tickets bought based on the discounts.
If the price of flight tickets fell from 2x to 1.3x then the discount was 35%.
The discounts are
- First 10 – 0
- Second 10 – 10
- Third 10 – 20
- Fourth 10 – 30
- Fifth 10 – 40
So, if you buy 50 tickets the average discount will be 20% since as you can see there are 10 tickets in each slab and the discounts percentages are in A.P so you can take the middle value ( if they were not in AP you can take the simple average of the discount percentages since the number of tickets in each slab is the same).
The actual discount is 35%, so more than 50 tickets were bought, the ones above 50 at 50%
So by buying
- 50 tickets at 20%
- X tickets at 50%
- the total discount is 35%
If you see 35% is a simple average of 20% and 50% this can happen only if the number of tickets bought at 20% and 50% is the same! So there are 50 tickets at 20% and 50 tickets at 50% so a total of 100 flight tickets.
(What if the resultant was not a simple average? Always use the see-saw concept and nothing else for alligation and weighted averages)
100 flight tickets and 100 first AC tickets at 1.3x, making the total amount 260x, which is the amount spent on second AC tickets.
At x per ticket, 260 second AC tickets could have been bought.
QUESTION 8: NOW (3 min)
This is again a problem where you can see clear steps to the answer. When you read the question you know that you need to maximize the value of y³*x². This can be possible only giving y a higher value since it has the higher power. The values of x and y have to be from 1 to 9 since x + y = 10. So all you need to do is to try out values.
When listing values do not just take any numbers randomly. Start from one end of the list and work your way to the other. This is will ensure that you do not miss out on some values.
So you should start from one end (9,1) and then move on to (8,2) (7,3), (6,4) and (5, 5). There is no point checking for values of y lower than x since that will not maximize the function.
You will get (6,4) and 3456 as the answer.
QUESTION 9: LATER, < 1 min
While it is easy to figure that using these 4 numbers you can form 4! or 24 numbers it will be a bit time-consuming to find the sum of these 24 numbers. Also since there is a chance of making a silly mistake by not considering a case it is best to return to this later.
SET 1: Questions 1 to 9
Ideally, you should have spent about 15-20 minutes and got 5 questions right, 15 marks. Even if you spent 12 minutes and got 3 questions right, 9 marks, it is a decent showing.
SET 2: On this set try to do the classification not as you usually do but as per the strategy outlined so far.
QUESTION 10: NOW (3 min.)
Do not leave probability, P & C or counting questions just because they are thought of to be tough. This paper had a few of both, we left a few for later and we choose to do a few. Let the question tell you and not the topic!
In this one you now that you can choose three numbers in 8C3 or 56 ways. Now the number of ways can you not form GPs is huge.
One of the standard ways of doing probability questions if you cannot calculate the probability of x happening is to find the probability of x not happening and subtract it from 1 to get the required probability.
This rule has to be used as a strategy as it is the case in this problem. We can find the number of ways of forming GPs using 3 numbers.
Even in this question do not randomly look at the set and start forming combinations, start from one end and work your way to the other.
Start with (1, 2, 4) then the next one with 1 can be (1,3,9), (1,4,16), then move on to starting with (2,4,8) then (3,9,27) and (4,8,16).
There are thus 6 ways out of 56 of forming a GP, the remaining 50 are thus the number of ways a GP cannot be formed, the probability of which is hence 50/56 or 25/28.
QUESTION 11: LATER (< 1 min.)
While the steps in this problem are definitely clear there are too many of them to take up at this point in time, so it is better to leave it for later.
QUESTION 12: LATER (< 1 min.)
If 1729 and 4104 evoke some cubes in your mind then you can try this problem; 1729 rang a bell since 1728 is 12³ but 4104 did not, so I left it for a later.
QUESTION 13: NOW (2 min.)
This is an absolute sitter that has to be done now. Just remember the golden rule with such problems that give you chances to form equations. Do not start solving and taking variables before you reach the end, you will end up taking up the wrong or inconvenient variables; the answer is option 1.
QUESTION 14: NOW (2 min.)
This is again an absolute sitter that has to be done now since you can see the steps to the answer — you just need to form one equation and equate it to $9200 to determine the profit and then the profit percentage.
A small trick with problems involving ratios is to take convenient variables for the unknown.
So do not rush to take profit as x:
- B gets 3/7 of one-third, if this should not be a fraction then the total should be a multiple of 21,
- B gets 1/3 of the remaining two-third, so if this also should not be a fraction then the total has to be a multiple of 9
- LCM of 21 and 9 is 63, so take the total profit as 63x
- 21 x and 42 x, b gets 9x and 14 x
- 23x = 9200 or x = 400
- Profit percentage = (63*400/84000)*100 or 30%
Even if you did not take the total as 63x but took one-third as 7x and calculated it will still be easier than taking the profit as x.
The ket to solving this in 2 min is not your conceptual or your application skills but pure brute force execution skills. I solved it without putting pen on paper visualizing all the calculations.
QUESTION 15: NOW (3 min.)
The big Geometry leave! Most students will leave this question without batting an eyelid
- Oh rhombus, I don’t know what a rhombus is!
- rhombus and rectangle together, better leave
We are a few months away from the test and if you still do not know what a rhombus is then I am afraid you are really wasting your chances. Also, you cannot wait for the Geometry class to happen so that then you can learn what a rhombus is! You have the books and videos with you and as I have mentioned in previous posts, just brush up the basics quickly.
Those who know what a rhombus is will know that the only difference between rhombus and square is that in a square all angles are 90 but in a rhombus, they are not; the two sets of opposite angles are equal.
If the sides of the rectangle are, a and 3a, then the diagonal is √10a, this is equal to the diagonal of the rhombus √3x, so a = (√3/√10)x!
Not doing this problem can only be termed as a willful donation of marks. I know social work is good to have on your resume but don’t donate marks!
QUESTION 16: LATER (< 1 min.)
This is not a straightforward P & C question and might take time, so better done later.
QUESTION 17: LATER (< 1 min.)
While it seems like a mixtures and alligations problem there is actually no resultant here. No values are given so everything has to be a different variable, using which we have to form two equations and try to solve it from there. So better left for later.
QUESTION 18: NOW (4 min.)
Yet another long Arithmetic problem that test-takers would have left just looking at the size. But invariably long questions have a huge give away at the end, all you need to do is follow the story. The reason you need to solve this problem is that the entire information is presented to you and you just need to execute the steps. You do not need to take anything as X!
Let the A and D be the initial positions of the rescue team and the body. You need to find the distance AD
Let them meet at B, from B to C the boat floats at 1 m/s for 14 seconds while they resuscitate the body so BC = 1*14 = 14m
From C to A they come back at (25 + 1) m/s in 41 seconds, AC = 26*41 = 1066. So, AB = AC + BC = 1080m.
AD = AB + BD, so all you need is BD, which is the distance the body floats for.
While the rescue team covers the distance AB, the body floats from D to B. If AB = 1080, the rescue team would have covered that distance at (25-1)m/s in 1080/24 = 45 seconds. In these 45 seconds, the body would have floated at 1m/s, 45m. So BD = 1125m
Again these are 3 marks that you will get if you invest that 3-4 minutes.
SET 2: 10 to 18
Again another 15-18 minutes and 5 questions, 15 marks. At the least 12 minutes, 9 marks.
SET 3: See if you can put all your learnings so far into selecting the right question.
QUESTION 19: NOW ( 1 min.)
Again a big leave since it seems to be Geometry plus Co-ordinate Geometry! This is the easiest question in the whole paper you need not even do any calculation!
Equations of three sides are given, DE is the line joining the midpoints of two sides, BC and AC, and hence it is always parallel to the third side, AB, and half of it.
And what do we know about equations of parallel lines?
They are mirror equations, one is just the multiple of the other only the constant value changes.
So all you need to look for is a multiple of the equation of AB, 5x – 3y, only option 3, 10x – 6y!
I repeat, the easiest question you can ever get.
QUESTION 20: LATER or NEVER
If you know the rules governing polynomials, you should do it later since if you cannot see the steps at this point, you will only know once you get it, else leave the problem.
QUESTION 21: NOW (2 min.)
You can see that all you need to do is to formulate a simple equation taking the number as xy, so you should do it right away.
You will get the equation 19x – 8y = 80, the difference has to be 80, the units digit 0, then the last digits of 19x and 8y should be the same. This is possible only if x = 9 and y = 8 (because 9*8 = 8*9) . The answer is 145.
QUESTION 22: LATER (< 1 min.) or NEVER
If you have good visualization skills you should do it later since it will take some time and there is a scope for error, else leave.
QUESTION 23: NOW ( 2 min.) or NEVER
If you know the rules governing Logarithms then you should do it now since it is a very simple construction (take one term as x and the other one as y, for each one apply log on both sides and you will get x = 2 and y = 5). If you do not know logs then leave; the answer is 5.
QUESTION 24: NOW ( 3 min.) or NEVER
Either you know the rules governing factors, in which case you should do it else better left alone; the answer is 76.
QUESTION 25: NOW (< 2 min.)
This is the simplest P & C problem possible. You just need to group all Chemistry books together and take them as one unit, making the total 4 books. If all Physics books were different then the answer would have been 4! But the 3 Physics books are identical, making the situation similar to a 4-letter word with three letters repeating and hence it has to be 4!/3!
QUESTION 26: NOW (2 min.)
This is a very, very standard problem. When squares and circles are infinitely inscribed
- The area of each subsequent circle will be half the previous one since the radius becomes r/√2
- The side of the first inscribed square is always radius of circle multiplied by √2.
- The subsequent squares will have an area half the previous one.
So, if the area of the first circle is πr² then the subsequent ones will be half of that, so its an infinite GP will common ratio 1/2, the sum of which is a/1-r or πr²/(1-1/2) or 2πr².
For the square area of the first one will be 2r², the sum of all squares again follows the same pattern, 2r²/(1-1/2), 4r².
The required ratio will be π/2.
QUESTION 27: NOW (2 min.)
The steps cannot be clearer than in this problem. All you need to do is calculate the interest paid by Chirag to Bilal and the interest paid by Bilal to Afzal, both are simple interests. You can get the difference and hence the profit. The only trick is to convert those interest rates into fractions 21/2 and 15/2 so that calculation becomes easy; the answer is Rs.2000.
QUESTION 28: NEVER
The problem involves too many calculations of taking different sets of points and seeing where slopes become equal. So it will end up being too time-taking and hence better left alone.
SET 3: 19 to 28
So, of these 9 questions, you can do 5 questions and get 15 marks in 15 minutes. Even if you leave questions, you can still get 9 marks in under 12 minutes.
The QA cut-off for this SimCAT was 27.
If you did at least 3 questions( and left the rest quickly) in each set you should have cleared the cut-off at this point by spending about 35 minutes.
If you did 5 at this point you will be at 45 marks with 15 minutes left.
SET 4: Since you are reaching the end you should be absolutely spot on with your selection and do only sure-shot questions.
QUESTION 29: NEVER (< 1 min.)
Since it involves counting the distinct points along the road it would involve drawing and also the possibility of making a mistake, hence better left alone
QUESTION 30: NOW (3 min.)
Again a big giveaway at the end about how much A is in container M, if you read through. You do not need to calculate how much A and B are there in each container since they will always be there in the ratio 3:7.
Your calculation will become easier if you take the amount in J as 130 since it is divided in the ratio 5:8 into K and L making the portions 50 and 80 and then M and N will have 15 and 35. M will have 3/10 of 45 or 4.5 ml of A, which is given as 45, so the original solution should also be 10 times or 1300 out of which 70% or 910 is B.
QUESTION 31: NEVER (< 1 min.)
Too many knots to try at this stage of the test, no clear path to answer better left.
QUESTION 32: NEVER (< 1 min.)
Involves knowledge of pyramidal figures and visualization, so better left alone at this point.
So on these 4 questions — around 5 minutes and 3 marks.
SET 5: The last two questions
QUESTION 33: NEVER ( or Reverse Sweep)
At this stage, unless you have no questions marked you should not get into drawing and attempting this question before you have done the marked ones.
I always have a fondness for symmetric figures since one can trust one’s eyesight and approximate. Circles are as symmetric as they come and they can be fully circumscribed only in symmetric triangles — equilateral.
But to do that I first need to know how much leeway I have to approximate – how far apart the answer options are.
A quick calculation will give you the answer options as — 2.73, 3, 7.29, 5.76 — far enough to take a swing.
So I took the radius if the inner circles as r, making the side of the inner triangle, PQR, 2r.
Now I need the side of the outer triangle. I decide to estimate it, from its altitude, the line that passes through, A and P and touches BC.
I take it as 2r + 2r + r.
- 2r + 2r – the heights or diameters of the two circles
- r – the small part that is left over at the top, between the circle that contains P and the vertex A.
I know I am taking only a marginally higher value since to the naked eye it is almost the same.
So if the altitude of the outer equilateral triangle is 5r then the side will be 2/√3 times it, or 10r/√3.
The ratio of the sides of outer to inner triangles will be (10r/√3)/2r or 5/√3, so ratio of areas will be a square of that or 25/3, around 8, and you know the highest option is 7.29 or option 3!
QUESTION 34: NEVER
One will not only know after one gets into the problem, hence better left alone.
A score of 45 was definitely possible on this test, some of the questions that I marked for later are relatively simple and some of the questions that require knowledge of specific concepts are also very do-able — 1, 6, 17, 23 and 24.
It might be that some you might disagree with the choices I have suggested but as long as you did not get stuck on any question and managed a score above 45, you are fine. If that has not happened then it means that you have to rework your strategy.
Geometry & Arithmetic can be big scoring areas
In this test, there were 16 questions from Arithmetic and Geometry of which I suggested that 10 should have been done. If even that is a bit unrealistic at least 8 of them were there for the taking — 24 marks from these two areas alone.
In another test maybe there might be more marks to be gained from Numbers and Algebra. But one thing is for certain, questions will get divided across all areas and unless you are conversant across all areas you will be always leaving questions that should have been taken.
What prevents you from scoring is not the question but your perception of it, your biases
Through this exercise, you would have realized that in most cases the questions that you left were, in fact, do-able and the ones that you entered, a few of them would have sunk your time.
This is because we are never really reading the question for what it is. We are looking at all the outer trappings of the question and making a decision about the same.
Do not judge a question by its wrapper, look beyond it and into the logic of the question and that will tell you the truth about the question.
This might seem like a tall order but it isn’t. It just means that your head should be absolutely still while reading the question and the processing will automatically happen from that center of stillness.
What you need is to do though to achieve that stillness is to first remove all the notions, biases, strategies that you have force-fitted into your head.
It don’t matter if it’s TITA or not as long as it is easy
Of the questions I chose 11 were MCQ and 8 were TITA. So, in my evaluation of a question the nature of the question does not matter at all. Most of the time this is also an artificial barrier that might exist in your sub-conscious. Check if in your evaluation you ended up not choosing a bulk of these TITA questions because you automatically deemed them best left for later.
The area you love the most can hurt you the most
All of us have our favourite areas and sometimes it can be our favourite areas that can hurt us the most in terms of time.
Since we are more attached to some areas we get sucked into tough problems since we feel it is a variation of something we have done or we feel that it can be solved using a particular technique; this is unavoidable. But what you should avoid at all costs is getting too attached at that point and waste time trying to solving the question again from the beginning.
You have to be aware at all points of the most precious commodity in an aptitude test — time.
The Elegant Solution
There was a milk and water problem, problem number that I left for later since there were no values. I returned to it later to find that I had overlooked something very basic, with respect to milk water problems —
- profit is always the result of number of liters of water
- the more the water the more the profit, milk only adds to the cost
- assuming water is free, if there is 1 litre of water, it means there is a profit equal to the cost of 1 litre of milk
Milk and water were mixed in a certain ratio. When the mixture was sold at the price of milk, the profit obtained was Rs. 70. When the mixture was sold at
the price of water, the loss incurred was Rs. 56. What is the ratio in which milk and water was mixed?
This problem makes things even more easier because in both cases the selling prices are not the a resultant third price but the price of milk (X) and water itself.
I straightway assumed the price of water to be zero since the profit or loss stems of the difference between the two prices.
If the price of water is zero, then by selling it at the price of water the loss — Rs.56 — is nothing but the entire cost of m litres milk, which is nothing but m(X) = 56 — (I)
The profit — Rs. 70 — is nothing but the money saved by putting in w litres of water at zero instead of milk at X, w(X) = 70 — (II)
Dividing (I) by (II), m : w = 4 : 5, which is what the problem is asking for.
|(1) 2||(2) 4||(3) ||(4) 2||(5) 2||(6) 2||(7) |
|(8) 2||(9) ||(10) 3||(11) 3||(12) ||(13) 1||(14) [30%]|
|(15) 2||(16) ||(17) 2||(18) ||(19) 3||(20) ||(21) |
|(22) 4||(23) ||(24) ||(25) 2||(26) 2||(27) ||(28) 4|
|(29) ||(30) ||(31) 2||(32) 3||(33) 3||(34) 2|
Use the comments tool to ask any doubts that you might have in the questions we took up in this post or point out any typos; this has turned out to be a super-long post!
In the next post, I will take up the ways and means by which you can increase your QA speed.