Tricks to solve QA Questions for CAT under 1 minute

CAT 2019 is just around the corner. The entrance exam that is the key to one's selection for the prestigious IIMs is just having a day or two before you sit for those three critical hours. Time-management is something that has always prevailed as an essential element in the exam. More specifically, the time-allotted behind questions for Quantitative Aptitude often makes the difference among the applicants.

• For the maximum bunch of students, the QA section of the CAT 2019 remains to be the toughest section to crack.
• It is either the time needed for the critical thinking part of each question or the accuracy to be maintained to get a favorable score out of this tricky section.
• Within the time allotted of 60 minutes for the QA Section, somewhere between 20-23 questions, out of the total tally of 35, have to be accurately answered for a 60+ mark in this particular section.

The approach can be divided into two main rounds of problem-solving.

• The first round should be targeted towards the questions that are from the topics of your interest, not lengthy while solving and give you a clear picture by the first reading of the problem.
• This round can go upto 35 minutes of your duration from the total allotted 60 minutes and should be covering as many questions as you can.
• The second round can be targeted at questions that you were confident of solving, even at the first reading, but are a bit lengthy and critical.
• Even if you utilize 20 minutes of the remaining time, both these rounds should fetch you 23 correct answers recorded across various topics.
• If you are still left with time, you can move onto the further questions.

If you have the complexity of time assortment sorted while taking the CAT 2019 Test, more often than not, the result will be favorable. This thing is taken care of to a considerable extent when somebody is familiar with shortcut tricks, especially for time-consuming calculations. This can save time for the candidates to invest their precious minutes on the harder and more draining part of concepts and ideologies put to test.

Also Read CAT Preparation Tips

Let us dig in deeper to quench some of the highly beneficial time-saving tricks that are clinical towards solving questions for Quantitative Aptitude within a blink of an eye but with optimum accuracy.

1. Time and Work Problems

Time and Work are inter-related in terms of their equation. The steps to minimize time along with simple calculations are:

• Take the LCM of the given set of numbers
• Add or subtract as per the question and requirement
• The obtained LCM will be the total work done
• Divide the total work with the result of step 2 to get your desired answer

Ques. X, Y and Z can complete a piece of work in 5, 9 and 45 days respectively, working together, they will complete the same work in?

Sol. Take the LCM of 5, 9, and 45: 45

So individually, work of X, Y and Z is: 45/5, 45/9 and 45/45= 9, 5, 1

So, total work = 9+5+1=15

Thus, together, they can complete the work in: 45/15=3 days

2. Time, Speed and Distance Problems

Some essential formulae that make these types of questions just as the breeze:

• Multiply by 5/18 when converting km/hr to m/s
• Multiply by 18/5 when converting m/s to km/hr
• Distance = [a/(x-y)]*y, where a is a certain distance amount and x & y are two different speed quantities
• Distance = [(x*y)/(x+y)]*a, where a is a certain time duration and x & y are two different speed quantities
• Time Taken = 2*[x-(y/2)], where x is a time duration in one mode of walking/transportation and y is another time duration for a different mode/pace of walking/transportation

Ques. If a rider covers a certain distance in 11 hours on a scooter, he covers half of the distance at a speed of 35 km/h and another half distance at a speed of 20 km/h. What will be the distance covered by him?

Sol. Time duration (a) = 11 hrs, Speed I (x) = 35 km/hr, Speed II (y) = 20 km/hr

As per the formula, Distance = [(x*y)/(x+y)]*a

Thus, Distance = [(35*20)/(35+20)]*11 = 140.03 km

3. Boat and Stream Problems

Few of the critical formulae related to boats and streams:

• Net Downstream Speed = b+s, where b is the speed of the boat in still water and s is the speed of the stream. (Downstream is referred to when the boat is moving in the direction of the stream).
• Net Upstream Speed = b-s, where b is the speed of the boat in still water and s is the speed of the stream. (Upstream is referred to when the boat is moving against the direction of the stream).
• Speed of boat in still water = ½ * (Downstream speed + Upstream Speed)
• Speed of stream = ½ * (Downstream speed - Upstream Speed)

Ques. If a man rows 12 km downstream in 6 hours and 4 km upstream in 4 hours, then, how long will he take to cover 9 km in still water?

Sol. Speed of boat in still water = ½ * (Downstream speed + Upstream Speed)

Thus, Speed = ½ * [(12/6)+(4/4)] = 1.5km/hr

So, time taken for 9 km = Distance/Speed = 9/1.5 = 6 hours

4. Trains Based Problems

Some of the critical formulae related to trains that make complex problems look easy are:

Ques. A train passes a platform in 35 seconds and a man standing on the platform in 18 seconds. If the speed of the train is 54 km/h, find the length of the platform.

Sol. As the time is provided in seconds, we convert the speed of the train in m/s.

Thus, Speed = 54 * (5/18) = 15 m/s

Now, as per the formula for the platform, time (t) = [(Length of train) + (Length of Platform)] / (Speed of train)

Thus, 35 = [L(t) + L(p)]/18

=> [L(t) + L(p)] = 35*18

=> [L(t) + L(p)] = 630 ….... eq(i)

Now, for the man standing, time (t) = (Length of train) / (Speed of train)

Thus, 20 = L(t) / 15

=> L(t) = 20*15

=> L(t) = 300

Now, putting the value of L(t) in eq(i),

300 + L(p) = 630

=> L(p) = 630-300 = 330 meters

5. Percentage Based Problems

For percentage problems, mental calculation is extremely important. Some additional tips that can aid you in the cause are:

• If out of the total quantity 70% are bad, i.e., 100% - 60% = 30% are good.
• While calculating maybe, 50% of 600, the ‘00s in 600 and the ‘00s in 100 get cancelled, so simply calculate 50*6 = 300.
• 70% of a quantity can be written as simply, 0.7. So, multiply 0.7 with the given figure and put the decimal in the right spot. Quick, right?
• If the Price of Oranges, maybe, increases by 5%, then the reduction required in the consumption so as to maintain the original expenditure will be = 5/(100+5) x 100 = 4.76%
• The total number of students is given as 1850, what is the difference between the number of students studying computer and commerce and chemistry and arts?
  1. Add up the %figures for computer and commerce = 43% (as given)
  2. Add up the %figures for chemistry and arts = 31% (as given)
  3. Get the difference between the 2 sets = 43-31 = 12%
  4. Now, calculate the difference percentage = 12% * 1850 = 222 students
• If the population of a town is, say 20,000 people as of today, and if the population increases at 6% p.a., then what will be the population after 3 years?
  1. Formula being – Population x (1 +R/100)n, where, n is the number of years
  2. Population after 3 years = 20,000 x (1 + 6/100)3 = 23820 approx.

6. Ratio and Proportion Problems

When it comes to ratio and proportions, the pivotal element(s) that matter are the relations and interrelations, likewise:

• The equality of two ratios is known as proportion i.e. a/b = c/d
• So, If a/b = c/d , then it is also equal to a+c/b+d
  1. Then, a^2:b^2 is a duplicate ratio of a:b
  2. a^3:b^3 is a triplicate ratio of a:b
  3. a^1/2:b^1/2 is a sub-duplicate ratio of a:b
  4. a^1/3:b^1/3 is a sub-triplicate ratio of a:b
• If a/b = b/c = c/d =...... so on, then a,b,c,d... are in G.P.
• If a>b and the same positive number is added to each term, the ratio decreases.
• If a<b and the same positive number is added to each term, the ratio increases.
• If we multiply or divide any number, there will be no effect on the ratio.
• If a:b is a ratio,
  1. If a/b = c/d , then a+b/b = c+d/d
  2. If a/b = c/d , then a-b/b = c-d/d
  3. If a/b = c/d , then a+b/a-b = c+d/c-d

7. Simple and Compound Interest Problems

Even if problems related to these topics may churn out to be difficult just by reading it, simply get these formulae within your fingertips and it can be a different story altogether.

• Simple Interest = (P*R*T)/100, where P=Principal, R=Rate of Interest, T=Time
• Amount = P*(1+R/100)^T
• Compound Interest = P*(1+R/100)^T - P
• Amount = Principal + Compound Interest

Ques. The MRP of a bicycle is Rs.3000. The shopkeeper first gave a discount of 15% and then a discount of 20%. Find the price at which one would purchase that bicycle.

Sol. Effective discount at which the bicycle was sold: a + b - ab/100

Where a = first discount, b = 2nd discount.

Thus, 15 + 20 + 300/100 = 38%

Thus, the final selling price is: 3000 * [(100-38)/100] = 1860

These are some helpful and beneficial tricks that you may apply at the time of solving problems for the Quantitative Aptitude section of CAT. This will heavily impact on the time period for a particular problem and at the same time, make your calculations much easier. We have also compiled a detailed list of last-minute preparation tips for other sections of the CAT Exam as well. Do give it a read. Get some problems from different books or study material related to these topics. Jot down the questions, solve them with the conventional methods that you know from your basics. Alternatively, also solve the same problems with these tricks being applied. You will get to note the difference for yourself! All the best for Sunday!