CAT Coaching Onlne - Links across topics in Math

Some simple interesting patterns emerge when we look across topics. Very often, thinking about these patterns helps us wind our heads around one or the other topic. In this post, we discuss a few of those links. Nothing profound, but just an outline to help remember simple ideas

Simple Interest Compound Interest and Progressions.

The amounts at the end of each year form an AP if interest is calculated on a simple interest basis, and form a GP if interest is calculated on a compound interest basis.

For instance if amount invested is Rs. 1000 and interest rate were 10%, then amount at the end of yr 1, yr 2, yr 3,.. would be Rs. 1100, 1200, 1300,...Each year's amount is previous year's amount + Rs. 100

The same amount of Rs. 1000 invested at same 10% but on a compound interest basis would give us amounts at the end of yr 1, yr 2 of Rs. 1100, 1210, 1331...Each year's amount is previous year's amount * 1.1. This is the basis for the formula for amount on principal invested in Compound Interest basis

Combinatorics - This can be linked with almost any topic
See if you can spot some idea in some other topic that is 'sitting inside' the following combinatorics questions

1. Consider a shelf that has 4 copies of a book and 3 copies of a painting. In how many ways can we select at least one article from this shelf?

2. Consider a shelf that has 4 different books and three different paintings. In how many ways can we select at least one article from this shelf?

3. In how many ways can be place 5 different toys into 3 different boxes such that all 5 toys are allotted and no box is empty?

How else these can be interpreted is given below.

1. Consider a shelf that has 4 copies of a book and 3 copies of a painting. In how many ways can we select at least one article from this shelf?
How many factors greater than 1 does the number 2^4 & 3^3 have?

2. Consider a shelf that has 4 different books and three different paintings. In how many ways can we select at least one article from this shelf?
How many non-empty subsets does the set {1, 2, 3, 4, 5, 6 7} have?

3. In how many ways can be place 5 different toys into 3 different boxes such that all 5 toys are allotted and no box is empty?
How many onto fucntions can be defined from the set {1, 2, 3, 4, 5} to the set {a, b, c}?

Weighted averages and Mixtures.
Class A has 40 students whose average mark is 40, class B has 60 students whose average mark is 50. What is the overall average? Class A has an average mark of 40 and class B has a an average mark of 50. If the overall average is 46, what is the ratio of students in class A and class B? These two questions are two sides of the same coin. Sometimes we teach one in weighted averages and the other in mixtures.

Mathematicians are supposed to love connections. There are far more powerful connections across topics than the ones I have mentioned above. It is instructive to observe even these simple connections.

CAT Preparation Online - Permutation and Combination

Question

Product of the distinct digits of a natural number is 60. How many such numbers are possible?

Explanation

60 = 2^2 * 3 * 5

60 cannot be written as a product of two single digit numbers. SO, the number in question should either have 3 or more digits.

Three-digit numbers

The digits could be 345 or 265

Digits being 345 - there are 3! such numbers

There are six numbers for each of these outlines. So, there are 3! + 3! = 12 three-digit numbers

Four-digit numbers

The digits could be 1345, 1265 or 2235.

Digits being 1345 - there are 4! such numbers

Digits being 1265 - there are 4! such numbers

Digits being 2235 - this is not possible as digits have to be distinct.

So, there are totally 24 + 24 four-digit numbers possible. 48 four-digit numbers.

Total number of numbers = 12 + 48 = 60.

(this post has been modified to plug error)

CAT preparation - Ratio and Proportion (Tough)

Question

John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If there are more chocolates of type C than of type B, and more of type B than of type A, what is the minimum possible number of chocolates overall?

Explanation

Again, big thanks to Mukund Sukumar for excellent solution.

Let john have 3x chocolates of type A and 7x of type B

Let Mike have 5y chocolates of type B and 4y of type C

Let john have 3z chocolates of type C and 5z of type A

So in total A=3x+5z ; B=7x+5y ; C=4y+3z

Since C>B we get solving y<3z-7x --->(1)

Since B>A we get solving 5y>5z-4x ---> (2)

What gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only y=1 as choice, for which second condition doesnt satisfy.

So, when x=1,z=4, we get y<5 from first condition and when y > 3.2 from second condition. so which gives choice the only y=4.

Hence x=1,y=4 and z=4 works and is the best possible answer.

For these values, we get A=23,B=27,C=28.

Minimum possible number of chocolates overall is 76. 

CAT Preparation Online - Simple one from Quadratic Equations

Question

How many integer solutions exist for the equation x² - 8|x| - 48 =0?

Explanation

One approach is to solve when x > 0 and then solve for x < 0. However, there is a slightly simpler method.

Note that x² is the same as |x|², so we can treat the equation as a quadratic in |x|.

Or, |x|² – 8|x| - 48 = 0

(|x| - 12|) (|x| + 4) = 0

|x| could be 12. |x| cannot be -4.

If |x| could be 12, x can be -12 and 12.

Two possible solutions exist. 

CAT - Question from Pipes and Cisterns

Question

There are n pipes that fill a tank. Pipe 1 can fill the tank in 2 hours, Pipe 2 in 3 hours, Pipe 3 in 4 hours and so on. Pipe 1 is kept open for 1 hour, pipe 2 for 1 hour, then pipe 3 and so on. In how many hours will the tank get filled completely?

Explanation

Almost all questions can be approached well by asking the question "What happens in 1 hour" (or 1 day, or 1 minutes)

So, let us start with that

In 1 hour, pipe 1 fills 1/2 of the tank. So, in the first hour, the tank will not be filled
In 1 hour, pipe 2 fills 1/3 of the tank. So, in two hours we would have filled 1/2 + 1/3 of the tank, or 5/6 of the tank. So, by the end of the second hour, the tank would still not be filled.
Let us move to the third hour. In 1 hour, pipe 3 fills 1/4 of the tank. So, by the end of the third hour, we should have filled 5/6 + 1/4 = 13/12 of the tank.

Oops, one cannot fill 13/12 of a tank. What this tells us is that the tank gets filled in the 3rd hour. 

When exactly during the third hour?

At the beginning of the third hour, we still have 1/6 of the tank still to fill. Pipe 3 can fill at the rate of 1/4 of the tank per hour. Or, pipe 3 will take 2/3 hours to fill the tank.

Or, the tank will be filled in 2 hours and 40 minutes.

This is an example of a sequence that is a harmonic progression. The formulae for HP are needlessly confusing. So, simple step-by-step approach works best.