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 Online - Tough one from Permutation and Combination

Question

A wonderful, but very tough question from Permutation and Combinations.

In how many ways, can we rearrange the word MONSOON such that no two adjacent positions are taken by the same letter? (Tough one. Tougher than what we will see in CAT)

Explanation

First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2 N’s. 

Let us focus on the three O’s.

We can place the three O’s in some blanks around the other letters. Or, three O’s can be placed in 3 slots out of the 5 in __M__N__S__N__ . This can be done in ⁵C₃, or 10 ways.

Or, the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1, 4, 7} or {1, 5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7}  - Whew.

Now, for each of these arrangements, there are 4!/2! = 12 arrangements for the other 4 letters. But the one thing we need to keep in mind now is the fact that 2 N’s could be adjacent in these arrangements. We will need to eliminate these.

O’s in slots {1, 3, 5} or O__O__O__ __ - Ns could be in the last two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are 2! = 2 

words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 7} or O__O__ __ __O - Ns could be in the slots {4, 5} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 4, 6} or O__ __O__O__  - Ns could be in the slots {2, 3}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 4, 7} or O__ __O__ __O - Ns could be in slots {2, 3} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 5, 7} or O__ __ __O__ O - Ns could be in the slots {2, 3} or {3, 4}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {2, 4, 6} or __O__O__O__  - Tehre are no possible slots for N. So, we count all 12 words on this list.

O’s in slots {2, 4, 7} or __O__O__ __ O - Ns could be in slots {5, 6}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {2, 5, 7} or __O__ __O__ O - Ns could be in slots {3, 4}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10.

O’s in slots {3, 5, 7} or __ __ O__O__O - Ns could be in the first two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.

Phew!.

There is a far more elegant method for accounting for the words where the 2 N’s appear together. This one came from Mukund Sukumar.

We need to account for the number of possibilities of N,N being together. So to subtract that part, consider 'NN' being together as one letter and place O's. In 3 out of the 4 slots in _M_NN_S_

The O’s can be selected in ⁴C₃ ways. The MNNS can be rearranged in 3! Ways if the N’s have to appear together.

Or, we get ⁴C₃ * 3! = 24 ways. So, we have 120-24 = 96 ways totally.

Inequalities Question and Solution

Question

How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x - 12).......(x - 100) < 0? 

Explanation

Inequalities Question and Solution

Question

|x + 2| + | x - 3| + | x - 5| < 15, find the range of x that satisfies this inequality.

Explanation

Functions Question and Solution

Question

Consider set A with 'a' elements, set B with 'b' elements, set C with 'c' elements. We can define a function that is one-one but not onto from set A to set B, a function that is onto but not one-one from set B to set C; and a function that is injective but not surjective from C to A. Arrange a, b, c in ascending order.

Explanation

Algebra Question and Solution

Question

|x| + |2y| + |3z| = 13, x * y * z is non-zero; x, y, z are all integers. How many sets of values are possible?

Answer
64 sets of values

Explanation

Inequalities - Interesting Question

Question
How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x - 12).......(x - 100) < 0?

Answer: 30

Explanation

Let us try out a few values to see if that gives us anything.

When x = 8, 10, 12, ....100 this goes to zero. So, these cannot be counted.
When x = 101, 102 or beyond, all the terms are positive, so the product will be positive. 

So, straight-away we are down to numbers 1, 2, 3, ...7 and then odd numbers from there to 99.

Let us substitute x =1,

All the individual terms are negative. There are totally 47 terms in this list (How? Figure that out). Product of 47 negative terms will be negative. So, x = 1 works. So, will x =2, 3, 4, 5, 6, and 7.

Remember, product of an odd number of negative terms is negative; product of even number of negative terms is positive. Now, this idea sets up the rest of the question.

When x = 9, there is one positive terms and 46 negative terms. So, the product will be positive. 
When x = 11, there are two positive terms and 45 negative terms. So, the product will be negative. 
When x = 13, there are three positive terms and 44 negative terms. So, the product will be positive. 

and so on.

Essentially, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7...and then 11, 15, 19, 23, 27, 31,..... and so.

What will be the last term on this list? 
99, because when x = 99, there are 46 positive terms and 1 negative term. 

So, we need to figure out how many terms are there in the list 11, 15, 19,....99. These can be written as 
4 * 2 + 3, 
4 * 3 + 3, 
4 * 4 + 3
4 * 5 + 3
4 * 6 + 3
...
4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30. 30 positive integer values of x exist satisfying the condition. 

Profit and Loss Question

Question:

A merchant can buy goods at the rate of Rs. 20 per good. The particular good is part of an overall collection and the value is linked to the number of items that are already on the market. So, the merchant sells the first good for Rs. 2, second one for Rs. 4, third for Rs. 6…and so on. If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell?

A.  24

B.  18

C.  27

D.  32

Correct Answer: (C)

Explanation:

Let us assume he buys n goods.

Total CP = 20n

Total SP = 2 + 4 + 6 + 8 ….n terms

Total SP should be at least 40% more than total CP

 2 + 4 + 6 + 8 ….n terms > 1.4 * 20 n

2 (1 + 2 + 3 + ….n terms) > 28n

n(n + 1) > 28n

n² + n  > 28n

n² -  27n  > 0

n > 27

He should sell a minimum of 27 goods.

Answer Choice (C)

CAT - Logarithms

Question:
If log₂X + log₄X = log_0.25and x > 0, then x is

A.  6^-1/6

B.  6^1/6

C.  3^-1/3

D.  6^1/3

Correct Answer: (A)

Explanation:

log₂x + log₄x = log_0.25

log₂x + = log_0.25

log₂x * = log_0.25

log₂x * 3 = 2log_0.25

log₂x³ = log_0.256

log₂x³ = -log₄6

log₂x³ =

log₂x³ =

2log₂x³ = -log₂6

2log₂x³ + log₂6 = 0

log₂6x⁶ = 0

6x⁶ = 1

x⁶ =

x =  Choice (A).

Level of difficulty 2