# Interest and Probability

Interest questions represent the easiest questions you will encounter in the IPAT. With a little practice, you should nail this section with ease.

Here is a primer on the interest formulae that would be tested in the IPAT.

 Simple Interest Total Interest = Principal x Rate x Time/100 Accumulated sum = Principal + Interest Compound Interest Accumulated sum = Principal x (Interest Rate/100 + 1)n The rule of 72 Under compound interest, the time it takes to double your principal is roughly equivalent to 72 divided by the interest rate.

Now,let’s warm up with some practice questions.

Q) If a plot of land, originally costing \$1 million, appreciated by 7% every year for 5 years and inflation is 2% annually. What is the worth of the plot in today’s dollars?

A] 1.75 million

B] 1.50 million

C] 2.40 million

D] 1.27 million

E] 1.64 million

Solution

Solution – D] 1.27 million

Appreciated value of plot of land = \$1 million x (1 + 7/100)5 = 1.402 million.

However, 1 dollar today is worth 1 x (1+2/100)5 = 1.104 dollars.

Therefore value of land in today’s dollars = 1.402 million/1.104 = 1.27 million dollars

Inflation does eat up on your returns!

Q) John has to choose between two bank investment accounts. One is offering 3% compound interest on the principal per annum and one 7% simple interest on the principal per annum. Given that his investment window is 4 years, what is the value of (final returns investment 1 – final returns investment 2) if the initial principal invested in \$P in both cases?

A] \$1.055P

B] \$ -0.155P

C] \$1.320P

D] \$ -0.320P

E] \$ -0.04P

Solution

Solution – B] \$ -0.155P

John invests \$P as principal

Investment account 1

Appreciated value of principal = P x (1 + 3/100)4 = \$1.125P

Investment account 2

Interest on investment = P x Rate x Time/100 = P x 7 x 4/100 = 0.28P

Total returns = Principal + Interest = P + 0.28P = \$1.28P

(final returns investment 1 – final returns investment 2) = \$1.125P – \$1.28P = \$ -0.155P

We get to see some very interesting problems as far as probability is concerned. Here is a quick primer on probability for the IPAT.

 Probability of an event P(E) = Number of favored outcomes for event/ Total number of outcomes Negation Rule Probability of event happening = 1 - Probability of event not happening Conditional Probability P(A given B happens) = P(A happens and B happens)/P(B)

Now let’s explore some of the topics that have been tested in the IPAT.

Q) In a group of 3 people what is the probability that at least 2 have a common birthday?

A] 0.0001

B] 0.05

C] 0.008

D] 0.04

E] 0.005

Solution

Solution – C] 0.008

In probability questions, it is always helpful to think about the negation of the problem, so let’s think about the probability of no two people’s birthday being the same

The first person can have any date in the year for their birthday which translates to a probability of 365/365 = 1 (365 is the number of days in a year, let’s neglect leap year in this computation as its significance in the final answer will be negligible)

The second person must have any date as their birthday except for the one belonging to the first person, this translates to the probability of 364/365

The third person must have any date as their birthday except for the two belonging to the first and second person, this translates to the probability of 363/365

Therefore the probability of no common birthday = 365/365 x 364/365 x 363/365

Therefore the probability of at least 2 people having a common birthday = 1 – probability of no common birthday = 1 – 365/365 x 364/365 x 363/365 =~ 0.008

Incidentally, this problem is formally called the Birthday problem and has garnered a lot of academic attention.

Q) Given that the numbers coming out of two independent dice throws are different, find the probability that that sum of the numbers is 4.

A] 1/15

B] 1/30

C] 1/5

D] 1/3

E] 1/25

Solution

Solution – A] 1/15

Referring to our formula sheet,

P( A given B happens ) = P(A happens and B happens)/P(B)

Here, A = Sum of numbers is 4 and B = Different numbers are thrown in the two dices

Let’s look at P(A happens and B happens) first – What are the two different numbers in dice throws that add up to 4. (2,2) wont work here so we are left with (1,3) and (3,1) – Notice that the ordering of the numbers matters. What is the total number of outcomes in this case? 6×6 as the first throw can yield 6 numbers and the second throw can again yield 6 numbers.

Therefore P(A and B) = Favored Outcomes/ Total Outcomes = 2/36 = 1/18.

Now let’s turn out attention to P(B). What are the outcomes that would yield different numbers? A good way to look at it is to think about the negation of this. What are the outcomes that would yield same numbers in the two dice throws? The answer is (1,1),(2,2),(3,3),(4,4),(5,5) and (6,6) – a total of 6 outcomes. This 36 – 6 = 30 outcomes would represent different numbers turning up.

Therefore P(B) = 30/36 = 5/6

P(A given B happens) = 1/18 / 5/6 = 1/15 which is our final answer.

This brings us to the end of our primer on IPAT preparation. For serious test takers wanting more practice, tips and tricks, do check out our comprehensive IPAT train plus practice material here

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