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Arithmetic for CAT 2018

The below article is brought to you by SDA Bocconi Asia Center and written by Dr Shashank Prabhu to help aspirants in their test prep. He is a CAT 100 percentiler, CET Rank 1, IIFT 100 percentile. He had a scaled score of 249/360 in NMAT BY GMAC 2016 and 99 percentile in each of the sections and overall score. SDA Bocconi Asia Center accepts CAT, NMAT, GMAT, GRE for its International Master in Business.

A common trend ever since the CAT became computer based is the reduction in the number of concept-based questions in favor of application based ones. Over the last decade or so, the Quantitative Ability section in CAT has become perceptibly ‘easier’. This has got a lot to do with the absence of theoretical concepts like division, factorials, base systems, complex probability based questions. These questions have been compensated by slightly deeper arithmetic questions that have a few traps. These traps are easy to miss and even the best of the candidates can get them wrong under the pressure of time and their performance in the preceding two sections. Arithmetic primarily involves questions based on the concepts of averages, percentages, ratios, mixtures, growth/interest rates, time-speed-distance, time-work, profit-loss, and simple word problems. These topics should ideally be your priority especially if you are not that great at QA and end up with underwhelming scores.
In this article, I will solve a few good questions from past year CAT papers to understand these traps and techniques.

Q1. A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead, he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train’s departure. The distance (in km) from his home to the railway station is
A: This is a classic time, speed, distance question. The most technical way to solve this is by using a variable x to denote distance and take the original time as t hours. So, we get
x = 12(t + 1/6) = 15(t – 1/6)
x = 12t + 2 = 15t – 2.5
t = 1.5 hours and x = 12 * 1.5 + 2 = 20 km, which would be the answer.
It is very easy to miss the fact that we are looking at 12t + 2 or 15t – 2.5 and substitute it directly in 12t. With no options involved, misplaced eagerness would have led to a loss of 3 marks.
Another thing to note in this context is that the problem could have been solved mentally if one were good at the basic time, speed, distance relationships.
The distance remains the same, which means that the speed is inversely proportional to the time taken. The speed increases from 12 km/hr to 15 km/hr or in other words, it becomes 5/4 times the original speed (12 * 5/4 = 15). So, the new time will become 4/5 times the original time. The reduction in time taken is by 1/5 times the original time taken. But, the question says that the second case was covered in 20 minutes fewer than the first case. So, 1/5 of original time taken corresponds to 20 minutes. So, the original time taken was 100 minutes. The original speed was 12 km/hr and so, the distance would have been 12 + 2/3rd of 12 or 20 km.

Q2. Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight members joint family is nearest to:
a. 23 years
b. 22 years
c. 21 years
d. 25 years
e. 24 years
A: Now it is probably one’s first instinct to solve this question using elaborate flowcharts and by writing down as much data as is possible. However, if you are good at looking at the bigger picture (duh!), it is much easier than that. The data talks about 10 years ago for 8 members. So, the present age should have been more than the combined age back then by 80 years. We are practically removing two people of 60 years each and replacing them by two fresh people. So, the sum should be short by 120 years. So, effectively there have been 40 years subtracted from the original sum or in other words the average is 191/8 which will be approximately 24 years.

Q3. On a 20km tunnel connecting two cities A and B there are three gutters. The distance between gutter 1 and 2 is half the distance between gutter 2 and 3. The distance from city A to its nearest gutter, gutter 1 is equal to the distance of city B from gutter 3. On a particular day the hospital in city A receives information that an accident has happened at the third gutter. The victim can be saved only if an operation is started within 40 minutes. An ambulance started from city A at 30 km/hr and crossed the first gutter after 5 minutes. If the driver had doubled the speed after that, what is the maximum amount of time the doctor would get to attend the patient at the hospital? Assume 1 minute is elapsed for taking the patient into and out of the ambulance.
PS: I am skipping the answer options in this case.
A: If you are one of those who says the answer is 15.5 minutes, you have overlooked a major data point in the question (and have solved way too many time, speed, distance questions if I may add). The key thing to understand here is that the hospital is located in city A and the patient has to be brought back to city A. City B is simply a distraction in this entire thing and the question could have practically stopped at G3.
Once you figure it out though, things become way too easy and you should be able to solve it. Let the distance between A and G1 be x and that between G1 and G2 be y. So, the distance between G2 and G3 will be 2y and that between G3 and B will be x. The total distance has been given to be 20 km which can be represented as:
2x + 3y = 20
Also, we know that the ambulance covered A to G1 in 5 minutes at the speed of 30 km/hr (or ½ km/minute). So, the value of x will come out to be 2.5 km and that of y will be 5 km. As the speed is double post the first milestone, the total time taken will be:
5 (initial A to G1) + 5 (G1 to G2) + 10 (G2 to G3) + 1 (loading the patient into the ambulance) + 10 (G3 to G2) + 5 (G2 to G1) + 2.5 (G1 to A as the speed has been doubled) + 1 (unloading the patient from the ambulance) = 38.5 minutes. So, there is still a minute and a half for the doctor to attend to the patient.

Q4. A woman sells to the first customer half her stock of apples and half an apple, to the second customer half her remaining stock of apples and half an apple and so also to a third and to a fourth customer. She finds that she has now 15 apples left. How many apples did she initially have?
A: This is again a very common question type and involves logical cascading of items. The traditional way of solving these questions involves taking the initial amount as x and then proceeding forwards by taking fractions. The easier way to solve it is to work backwards though as we will see.
Just before she had 15 apples at the end, she gave half an apple to the fourth customer. So, before giving away half an apple, she would have had 15.5 apples. But those were the apples that she had after she had given away half of what she had to the fourth customer. So, before she gave anything to the fourth customer, she would have had 15.5 * 2 = 31 apples.
Try to work backwards throughout the problem and check for yourself how it goes. The numbers that you should be getting would be 31 -> 31.5 -> 63 -> 63.5 -> 127 -> 127.5 -> 255.

Q5. Ghosh Babu took a loan for three years at 20% per annum compound interest with annual compounding. He repaid the entire loan amount in three equal annual installments of Rs. 43,200 each. What was the amount of the loan (in Rs.) taken by Ghosh Babu? (This has been simulated on existing CAT questions just to improve your understanding of the concept)
A: The best way to understand installments is by understanding the concept of the time value of money. In simple terms the money that you have is bound to appreciate with time and so, an earlier installment is more beneficial to the lender compared to an equal later installment.
So, if we consider that Ghosh Babu paid Rs. 43,200 at the end of year 1, it would mean that the company would be able to lend it to someone else for 2 more years at a compound interest of 20% per annum. So, although Ghosh Babu paid an installment of Rs. 43,200, it was worth 43200 * 1.2 * 1.2 = Rs. 62,208 at the end.
Similarly, Ghosh Babu would have paid Rs. 43,200 at the end of year 2, which would mean the lender would be able to re-lend it for 1 more year at a compound interest of 20% per annum. So, the Rs. 43,200 would be worth 43200 * 1.2 = Rs. 51,840 at the end.
The final installment does not add any additional value to the lender and so, it remains as it is.
So, the final value that was expected by the lender would have been 62,208 + 51,840 + 43,200 = Rs. 157,248 which would have been equivalent to a 20 percent compounded interest on the initial amount. So, the initial amount would have been 157248/(1.728) = Rs. 91000. Slightly calculation intensive but definitely manageable with the on-screen calculator.

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