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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.
The approach can be divided into two main rounds of problem-solving.
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.
Time and Work are inter-related in terms of their equation. The steps to minimize time along with simple calculations are:
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
Some essential formulae that make these types of questions just as the breeze:
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
Few of the critical formulae related to boats and streams:
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
Some of the critical formulae related to trains that make complex problems look easy are:
|Scenario||Formulae of time to cross|
|An object is standing still & of negligible length||t = (length of train)/(speed of train)|
|An object is standing still & of some length||t = [length of (train+object)]/(speed of train)|
|An object is moving and is of negligible length||t = (length of train)/[speed of (train-object)]|
|An Object is moving and has some length||t = [length of (train+object)]/[length of (train-object)]|
|If the object is moving in the opposite direction||t = [length of (train+object)]/[length of (train+object)]|
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
For percentage problems, mental calculation is extremely important. Some additional tips that can aid you in the cause are:
When it comes to ratio and proportions, the pivotal element(s) that matter are the relations and interrelations, likewise:
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.
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!
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