SET THEORY

Chapter 1: SET THEORY

Introduction

Set: A well defined collection of objects is called a set.
Finite set: a set of finite elements.
Null set: set with no elements.
Singleton set: set with one element only.
Equal set: each element of one set is in the other and vice versa.
Subset: Every element of A is element of B. Denote it as A⊆B A={1,2} and B={1,2,3} or B={1,2}. Proper subset is if there is at least one element in B but not in A but all elements in A are in B. A⊂B. E.g: A = {1,2,3} and B={1,2,3,4}.
Union of sets: A∪B = {all elements of A and B but no duplicates};
Intersection of sets: A∩B = {common elements of A and B but no duplicates}
Difference of sets: A-B = {set of all elements of A not in B}
Complement of set: If A is the subset of X then the a set of all elements in X but not in A is A^c

Note: A set of 'n' elements can have total 2ⁿ subsets.

Set Theory is an important concept of mathematics which is often asked in aptitude exams. There are two types of questions in this chapter:

Numerical questions on set theory based on venn diagrams

Logical questions based on set theory

Let us first take a look at some standard theoretical inputs related to set theory

Figure 1: Refers to the situation where there are two attributes A and B. (Let's say A refers to people who passed in Physics and B refers to people who passed in Chemistry.) Then the shaded area shows the people who passed both in Physics and Chemistry.

In mathematical terms, the situation is represented as: Total number of people who passed at least 1 subject = A + B - (A ∩ B)

Figure 2: Refers to the situation where there are three attributes being measured. In the figure below, we are talking about people who passed Physics, Chemistry and/or Mathematics.

In the above figure, the following explain the respective areas:

Area 1: People who passed in Physics only

Area 2: People who passed in Physics and Chemistry only (in other words-people who passed Physics and Chemistry but not Mathematics)

Area 3: People who passed Chemistry only

Area 4: People who passed Chemistry and Mathematics only (also, can be described as people who passed Chemistry and Mathematics but not Physics)

Area 5: People who passed Physics and Mathematics only (also, can be described as people who passed Physics and Mathematics but not Chemistry)

Area 6: People who passed Physics, Chemistry and Mathematics

Area 7: People who passed Mathematics only

Area 8: People who passed in no subjects

Also take note of the following language which there is normally confusion about:

People passing Physics and Chemistry-Represented by the sum of areas 2 and 6

People passing Physics and Maths-Represented by the sum of areas 5 and 6

People passing Chemistry and Maths-Represented by the sum of areas 4 and 6

People passing Physics-Represented by the sum of the areas 1, 2, 5 and 6

In mathematical terms, this means: Total number of people who passed at least 1 subject = P + C + M - (P ∩ C) - (P ∩ M) - (C ∩ M) + (P ∩ C ∩ M)

At the birthday party of Sherry, a baby boy, 40 persons chose to kiss him and 25 chose to shake hands with him. 10 persons chose to both kiss him and shake hands with him. How many persons turned out at the party?

Explanation :


From the figure, it is clear that the number of people at the party were 30 + 10 + 15 = 55.
We can of course solve this mathematically as below:
Let n(A) = No. of persons who kissed Sherry = 40
n(B) = No. of persons who shake hands with Sherry = 25
and n(A ∩ B) = No. of persons who shook hands with Sherry and kissed him both = 10
Then using the formula, n(A ∩ B) = n(A) + n(B) - n(A ∩ B)
n(A ∩ B) = 40 + 25 - 10 = 55

In an examination 43% passed in Math, 52% passed in Physics and 52% passed in Chemistry. Only 8% students passed in all the three. 14% passed in Math and Physics and 21% passed in Math and Chemistry and 20% passed in Physics and Chemistry. Number of students who took the exam is 200. Let Set P, Set C and Set M denotes the students who passed in Physics, Chemistry and Math respectively. Then How many students passed in Math only?

Explanation :

. Let P denote Physics, C denote Chemistry and M denote Maths.
% of students who passed in P and C only is given by
% of students who passed in P and C - % of students who passed all three = 20% - 8% = 12%
% of students who passed in P and M only is given by
% of students who passed in P and M - % of students who passed all three = 14% - 8% = 6%
% of students who passed in M and C only is:
% of students who passed in C and M - % of students who passed all three = 21% - 8% = 13%
So, % of students who passed in P only is given by:
Total no. passing in P - No. Passing in P & C only - No. Passing P & M only - No. Passing in all
threeÆ
52% - 12% - 6% - 8%- = 26%
% of students who passed in M only is:
Total no. passing in M - No. Passing in M & C only - No. Passing P & M only - No. Passing in
all threeÆ
43% - 13% - 6% - 8%- = 16%

% of students who passed in Chemistry only is
Total no. passing in C - No. Passing in P & C only - No. Passing C & M only - No. Passing in all
three 
52% - 12% - 13% - 8%- = 19%
Only Math =16% = 32 people.

16:19

6:1

16:9

1:19

Explanation :

. Let P denote Physics, C denote Chemistry and M denote Maths.
% of students who passed in P and C only is given by
% of students who passed in P and C - % of students who passed all three = 20% - 8% = 12%
% of students who passed in P and M only is given by
% of students who passed in P and M - % of students who passed all three = 14% - 8% = 6%
% of students who passed in M and C only is:
% of students who passed in C and M - % of students who passed all three = 21% - 8% = 13%
So, % of students who passed in P only is given by:
Total no. passing in P - No. Passing in P & C only - No. Passing P & M only - No. Passing in all
threeÆ
52% - 12% - 6% - 8%- = 26%
% of students who passed in M only is:
Total no. passing in M - No. Passing in M & C only - No. Passing P & M only - No. Passing in
all three
43% - 13% - 6% - 8%- = 16%

% of students who passed in Chemistry only is
Total no. passing in C - No. Passing in P & C only - No. Passing C & M only - No. Passing in all
three 
52% - 12% - 13% - 8%- = 19%
Ratio of Only Math to Only Chemistry = 16:19.

26:84

26:8

2:84

52:84

Explanation :

. Let P denote Physics, C denote Chemistry and M denote Maths.
% of students who passed in P and C only is given by
% of students who passed in P and C - % of students who passed all three = 20% - 8% = 12%
% of students who passed in P and M only is given by
% of students who passed in P and M - % of students who passed all three = 14% - 8% = 6%
% of students who passed in M and C only is:
% of students who passed in C and M - % of students who passed all three = 21% - 8% = 13%
So, % of students who passed in P only is given by:
Total no. passing in P - No. Passing in P & C only - No. Passing P & M only - No. Passing in all
threeÆ
52% - 12% - 6% - 8%- = 26%
% of students who passed in M only is:
Total no. passing in M - No. Passing in M & C only - No. Passing P & M only - No. Passing in
all three
43% - 13% - 6% - 8%- = 16%

% of students who passed in Chemistry only is
Total no. passing in C - No. Passing in P & C only - No. Passing C & M only - No. Passing in all
three 
52% - 12% - 13% - 8%- = 19%
26:84 is the required ratio

Explanation :

. Let P denote Physics, C denote Chemistry and M denote Maths.
% of students who passed in P and C only is given by
% of students who passed in P and C - % of students who passed all three = 20% - 8% = 12%
% of students who passed in P and M only is given by
% of students who passed in P and M - % of students who passed all three = 14% - 8% = 6%
% of students who passed in M and C only is:
% of students who passed in C and M - % of students who passed all three = 21% - 8% = 13%
So, % of students who passed in P only is given by:
Total no. passing in P - No. Passing in P & C only - No. Passing P & M only - No. Passing in all
threeÆ
52% - 12% - 6% - 8%- = 26%
% of students who passed in M only is:
Total no. passing in M - No. Passing in M & C only - No. Passing P & M only - No. Passing in
all three
43% - 13% - 6% - 8%- = 16%

% of students who passed in Chemistry only is
Total no. passing in C - No. Passing in P & C only - No. Passing C & M only - No. Passing in all
three 
52% - 12% - 13% - 8%- = 19%
39 % or 78 people

In the Mindworkzz club all the members participate either in the Tambola or the Fete. 320 participate in the Fete, 350 participate in the Tambola and 220 participate in both. How many members does the club have?

410

450

440

380

Explanation :


The total number of people = 100 + 220 + 130 = 450

There are 20000 people living in Defence Colony, Gurgaon. Out of them 9000 subscribe to Star TV Network and 12000 to Zee TV Network. If 4000 subscribe to both, how many do not subscribe to any of the two?

5000

2000

3000

1000

Explanation :


The required answer would be 20000 - 5000 - 4000 - 8000 = 3000.

Last year, there were 3 sections in the Catalyst, a mock CAT paper. Out of them 33 students cleared the cut-off in Section 1, 34 students cleared the cut-off in Section 2 and 32 cleared the cut-off in Section 3. 10 students cleared the cut-off in Section 1 and Section 2, 9 cleared the cut-off in Section 2 and Section 3, 8 cleared the cut-off in Section 1 and Section 3. The number of people who cleared each section alone was equal and was 21 for each section. How many cleared all the three sections?

Explanation :

Explanation :

78/21

73/21

none

Explanation :

In a locality having 1500 households, 1000 watch Zee TV, 300 watch NDTV and 750 watch Star Plus. Based on this information answer the questions that follow : The minimum number of households watching Zee TV and Star Plus is:

If we try to consider each of the households watching Zee TV and Star Plus as independent of each other, we would get a total of 1000 + 750 = 1750 households.

However, we have a total of only 1500 households in the locality and hence, there has to be a minimum interference of at least 250 households who would be watching both Zee TV and Star Plus.

Hence, the answer to this question is 250.

Q. In above case The minimum number of households watching both Zee TV and NDTV is:

In this case, the number of households watching Zee TV and NDTV can be separate from each other

Since there is no interference required between the households watching Zee TV and the households watching NDTV as their individual sum (1000 + 300) is smaller than the 1500 available households in the locality.

Hence, the answer in this question is 0.

Q. The maximum number of households who watch neither of the the three channels is

For this to occur the following situation would give us the required solution

As you can clearly see from the figure, all the requirements of each category of viewers is fulfilled by the given allocation of 1000 households.

In this situation, the maximum number of households who do not watch any of the three channels is visible as 1500 -1000 = 500.

Q. In a school, 90% of the students faced problems in Mathematics, 80% of the students faced problems in Computers, 75% of the students faced problems in Sciences, and 70% of the students faced problems in Social Sciences. Find the minimum percent of the students who faced problems in all four subjects.

In order to think about the minimum number of students who faced problems in all four subjects you would need to think of keeping the students who did not face a problem in any of the subjects separate from each other.

We know that 30% of the students did not face problems in Social Sciences, 25% of the students did not face problems in Sciences, 20% students did not face problems in computers and 10% students did not face problems in Mathematics.

If each of these were separate from each other, we would have 30+25+20+10=85% people who did not face a problem in one of the four subjects.

Naturally, the remaining 15% would be students who faced problems in all four subjects.

This represents the minimum percentage of students who faced problems in all the four subjects.

Q. For the above question, find the maximum possible percentage of students who could have problems in all 4 subjects.

In order to solve this, you need to consider the fact that 100 (%) people are counted 315 (%) times, which means that there is an extra count of 215 (%).

When you put a student into the 'has problems in each of the four subjects' he is one student counted four times — an extra count of 3.

Since, 215/3= 71 (quotient) we realise that if we have 71 students who have problems in all four subjects — we will have an extra count of 213 students.

The remaining extra count of 2 more can be matched by putting 1 student in 'has problems in 3 subjects' or by putting 2 students in 'has problems in 2 subjects'.

Thus, from the extra count angle, we have a limit of 71% students in the 'have problems in all four categories.'

However, in this problem there is a constraint from another angle — i.e. there are only 70% students who have a problem in Social Sciences — and hence it is not possible for 71% students to have problems in all the four subjects.

Hence, the maximum possible percentage of people who have a problem in all four subjects would be 70%.

Q. In the above question if it is known that 10% of the students faced none of the above mentioned four problems, what would have been the minimum number of students who would have a problem in all four subjects?

If there are 10% students who face none of the four problems, we realise that these 10% would be common to students who face no problems in Mathematics, students who face no problems in Sciences, students who face no problems in Computers and students who face no problems in Social Sciences.

Now, we also know that overall there are 10% students who did not face a problem in Mathematics;

20% students who did not face a problem in computers;

25% students who did not face a problem in Sciences and 30% students who did not face a problem in Social Sciences.

The 10% students who did not face a problem in any of the subjects would be common to each of these 4 counts.

Out of the remaining 90% students, if we want to identify the minimum number of students who had a problem in all four subjects we will take the same approach as we took in the first question of this set — i.e. we try to keep the students having problems in the individual subjects separate from each other.

This would result in: 0% additional students having no problem in Mathematics; 10% additional students having no problem in Computers;

15% additional students having no problem in Sciences and 20% additional students having no problem in Social Sciences.

Thus, we would get a total of 45% (0+10+15+20=45) students who would have no problem in one of the four subjects.

Thus, the minimum percentage of students who had a problem in all four subjects would be 90 - 45 = 45%.

Q. In a class of 80 students, each of them studies at least one language — English, Hindi and Sanskrit. It was found that 65 studied English, 60 studied Hindi and 55 studied Sanskrit. Find the maximum number of people who study all three languages

This question again has to be dealt with from the perspective of extra counting.

In this question, 80 students in the class are counted 65 + 60 + 55 = 180 times

An extra count of 100.

If we put 50 people in the all three categories as shown below, we would get the maximum number of students who study all three languages.

Q. In a class of 80 students, each of them studies at least one language — English, Hindi and Sanskrit. It was found that 65 studied English, 60 studied Hindi and 55 studied Sanskrit. Find the minimum number of people who study all three languages

In order to think about how many students are necessarily in the 'study all three languages' area of the figure (this thinking would lead us to the answer to the minimum number of people who study all three languages) we need to think about how many people we can shift out of the 'study all three category' for the previous question. When we try to do that, the following thought process emerges:

Step 1: Let's take a random value for the all three categories (less than 50 of course) and see whether the numbers can be achieved. For this purpose we try to start with the value as 40 and see what happens. Before we move on, realise the basic situation in the question remains the same — 80 students have been counted 180 times — which means that there is an extra count of 100 students & also realise that when you put an individual student in the all three categories, you get an extra count of 2, while at the same time when you put an individual student into the 'exactly two languages category', he/she is counted twice — hence an extra count of 1. The starting figure we get looks something like this:

At this point, since we have placed 40 people in the all three categories, we have taken care of an extra count of 40 × 2 = 80. This leaves us with an extra count of 20 more to manage and as we can see in the above figure we have a lot of what can be described as 'slack' to achieve the required numbers. For instance, one solution we can think of from this point is as below:

One look at this figure should tell you that the solution can be further optimised by reducing the middle value in the figure since there is still a lot of 'slack' in the figure — in the form of the number of students in the 'exactly one language category'.

Also, you can easily see that there are many ways in which this solution could have been achieved with 40 in the middle. Hence, we go in search of a lower value in the middle.

So, we try to take an arbitrary value of 30 to see whether this is still achievable.

In this case we see the following as one of the possible ways to achieve this (again there is a lot of slack in this solution as the 'only Hindi' or the 'only Sanskrit' areas can be reallocated):

Trying the same solution for 20 in the middle we get the optimum solution:

We realise that this is the optimum solution since there is no 'slack' in this solution and hence, there is no scope for re-allocating numbers from one area to another

Q. In a group of 120 athletes, the number of athletes who can play Tennis, Badminton, Squash and Table Tennis is 70, 50, 60 and 30 respectively. What is the maximum number of athletes who can play none of the games?

In order to think of the maximum number of athletes who can play none of the games, we can think of the fact that since there are 70 athletes who play tennis

Fundamentally there are a maximum of 50 athletes who would be possibly in the 'can play none of the games'.

No other constraint in the problem necessitates a reduction of this number and hence the answer to this question is 50.

In the Indian athletic squad sent to the Olympics, 21 athletes were in the triathlon team; 26 were in the pentathlon team; and 29 were in the marathon team. 14 athletes can take part in triathlon and pentathlon; 12 can take part in marathon and triathlon; 15 can take part in pentathlon and marathon; and 8 can take part in all the three games. 1. How many players are there in all?

Explanation :

Since there are 14 players who are in triathlon and pentathlon, and
there are 8 who take part in all three games, there will be 6 who take part in only triathlon and pentathlon.
Similarly,
Only triathlon and marathon = 12 – 8=4 & Only Pentathlon and Marathon = 15 – 8 = 7 
The figure above can be completed with values for each sport (only) plugged in:
The answers would be:
3 + 6 + 8 + 4 + 5 + 7 + 10 = 43

Explanation :

In a test in which 120 students appeared, 90 passed in History, 65 passed in Sociology and 75 passed in Political Science. 30 students passed in only one subject and 55 students in only two. 5 students passed no subjects. How many students passed in all the three subjects?

Explanation :

The given situation can be read as follows:
115 students are being counted 75+65+90= 230 times.
This means that there is an extra count of 115. This extra count of 115 can be created in 2 ways.
A. By putting people in the ‘passed exactly two subjects’ category. In such a case each person would
get counted 2 times (double counted), i.e., an extra count of 1.
B. By putting people in the ‘all three’ category, each person put there would be triple counted. 1
person counted 3 times – meaning an extra count of 2 per person.
The problem tells us that there are 55 students who passed exactly two subjects. This means an extra
count of 55 would be accounted for. This would leave an extra count of 115 – 55 = 60 more to be
accounted for by ‘passed all three’ category. This can be done by using 30 people in the ‘all 3’ category.

Explanation :

5% of the passengers who boarded Guwahati- New Delhi Rajdhani Express on 20 th February, 2002 do not like coffee, tea and ice cream and 10% like all the three. 20% like coffee and tea, 25% like ice cream and coffee and 25% like ice cream and tea. 55% like coffee, 50% like tea and 50 % like ice cream. The number of passengers who like only coffee is greater than the passengers who like only ice cream by

100

Explanation :
```
[(20 – 10)/10] * 100 = 100%.
```

5% of the passengers who boarded Guwahati- New Delhi Rajdhani Express on 20 th February, 2002 do not like coffee, tea and ice cream and 10% like all the three. 20% like coffee and tea, 25% like ice cream and coffee and 25% like ice cream and tea. 55% like coffee, 50% like tea and 50 % like ice cream. The percentage of passengers who like both tea and ice cream but not coffee is

100

Explanation :

5% of the passengers who boarded Guwahati- New Delhi Rajdhani Express on 20 th February, 2002 do not like coffee, tea and ice cream and 10% like all the three. 20% like coffee and tea, 25% like ice cream and coffee and 25% like ice cream and tea. 55% like coffee, 50% like tea and 50 % like ice cream. The percentage of passengers who like at least 2 of the 3 products is

100

Explanation :
```
10+10+15+15=50%
```

5% of the passengers who boarded Guwahati- New Delhi Rajdhani Express on 20 th February, 2002 do not like coffee, tea and ice cream and 10% like all the three. 20% like coffee and tea, 25% like ice cream and coffee and 25% like ice cream and tea. 55% like coffee, 50% like tea and 50 % like ice cream. If the number of passengers is 180, then the number of passengers who like ice cream only is

100

Explanation :

Only ice cream is 10% of the total. Hence, 10% of 180 =18.

In a survey among students at all the IIMs, it was found that 48% preferred coffee, 54% liked tea and 64% smoked. Of the total, 28% liked coffee and tea, 32% smoked and drank tea and 30% smoked and drank coffee. Only 6% did none of these. If the total number of students is 2000 then find The ratio of the number of students who like only coffee to the number who like only tea is

2:3

4:7

1:3

2:6

Explanation :

 If you try to draw a figure for this question, the figure would be
something like

We can then solve this as:
x - 10 + 28 - x + x + 30 - x + x + 2 + 32 - x + x - 6 = 94 so x + 76 = 94 so x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find the
missing value of students who liked all three as follows:
94 = 48 + 54 + 64 - 28 - 32 - 30 + All three so All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value - it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:

8:12 = 2:3

In a survey among students at all the IIMs, it was found that 48% preferred coffee, 54% liked tea and 64% smoked. Of the total, 28% liked coffee and tea, 32% smoked and drank tea and 30% smoked and drank coffee. Only 6% did none of these. If the total number of students is 2000 then find Number of students who like coffee and smoking but not tea is

600

240

130

150

Explanation :

 If you try to draw a figure for this question, the figure would be
something like

We can then solve this as:
x - 10 + 28 - x + x + 30 - x + x + 2 + 32 - x + x - 6 = 94 so x + 76 = 94 so x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find the
missing value of students who liked all three as follows:
94 = 48 + 54 + 64 - 28 - 32 - 30 + All three so All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value - it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:

12 % of 2000 = 240

In a survey among students at all the IIMs, it was found that 48% preferred coffee, 54% liked tea and 64% smoked. Of the total, 28% liked coffee and tea, 32% smoked and drank tea and 30% smoked and drank coffee. Only 6% did none of these. If the total number of students is 2000 then find The percentage of those who like coffee or tea but not smoking among those who like at least one of these is

more than 30

less than 30

less than 25

None of these

Explanation :

 If you try to draw a figure for this question, the figure would be
something like

We can then solve this as:
x - 10 + 28 - x + x + 30 - x + x + 2 + 32 - x + x - 6 = 94 so x + 76 = 94 so x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find the
missing value of students who liked all three as follows:
94 = 48 + 54 + 64 - 28 - 32 - 30 + All three so All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value - it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:

30/94 so more than 30%

In a survey among students at all the IIMs, it was found that 48% preferred coffee, 54% liked tea and 64% smoked. Of the total, 28% liked coffee and tea, 32% smoked and drank tea and 30% smoked and drank coffee. Only 6% did none of these. If the total number of students is 2000 then find The percentage of those who like at least one of these is

100

nil

Explanation :

 If you try to draw a figure for this question, the figure would be
something like

We can then solve this as:
x - 10 + 28 - x + x + 30 - x + x + 2 + 32 - x + x - 6 = 94 so x + 76 = 94 so x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find the
missing value of students who liked all three as follows:
94 = 48 + 54 + 64 - 28 - 32 - 30 + All three so All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value - it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:

94%

Tea only and tea and smoking only

Coffee and smoking only and tea only

Coffee and tea but not smoking and smoking but not coffee and tea.

none

Explanation :

 If you try to draw a figure for this question, the figure would be
something like

We can then solve this as:
x - 10 + 28 - x + x + 30 - x + x + 2 + 32 - x + x - 6 = 94 so x + 76 = 94 so x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find the
missing value of students who liked all three as follows:
94 = 48 + 54 + 64 - 28 - 32 - 30 + All three so All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value - it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:

the ratio turns out to be 10:20 in that case.

In a survey among students at all the IIMs, it was found that 48% preferred coffee, 54% liked tea and 64% smoked. Of the total, 28% liked coffee and tea, 32% smoked and drank tea and 30% smoked and drank coffee. Only 6% did none of these. If the total number of students is 2000 then find The number of persons who like coffee and smoking only and the number who like tea only bear a ratio

2:3

4:7

1:1

2:6

Explanation :

 If you try to draw a figure for this question, the figure would be
something like

We can then solve this as:
x - 10 + 28 - x + x + 30 - x + x + 2 + 32 - x + x - 6 = 94 so x + 76 = 94 so x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find the
missing value of students who liked all three as follows:
94 = 48 + 54 + 64 - 28 - 32 - 30 + All three so All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value - it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:

12:12 = 1:1

In a survey among students at all the IIMs, it was found that 48% preferred coffee, 54% liked tea and 64% smoked. Of the total, 28% liked coffee and tea, 32% smoked and drank tea and 30% smoked and drank coffee. Only 6% did none of these. If the total number of students is 2000 then find Percentage of those who like tea and smoking but not coffee is

14.9

less than 14

more than 15

Explanation :

 If you try to draw a figure for this question, the figure would be
something like

We can then solve this as:
x - 10 + 28 - x + x + 30 - x + x + 2 + 32 - x + x - 6 = 94 so x + 76 = 94 so x = 18.
Note: In this question, since all the values for the use of the set theory formula are given, we can find the
missing value of students who liked all three as follows:
94 = 48 + 54 + 64 - 28 - 32 - 30 + All three so All three = 18
As you can see this is a much more convenient way of solving this question, and the learning you take
away for the 3 circle situation is that whenever you have all the values known and the only unknown
value is the center value - it is wiser and more efficient to solve for the unknown using the formula
rather than trying to solve through a venn diagram.
Based on this value of x we get the diagram completed as:

14%

30 monkeys went to a picnic. 25 monkeys chose to irritate cows while 20 chose to irritate buffaloes. How many chose to irritate both buffaloes and cows?

Explanation :
```
30 = 25 + 20 - x so x = 15
```

Explanation :

Let people who passed all three be x. Then:
53 + 61 + 60 - 24 - 35 - 27 + x = 95
so x = 7.
The venn diagram in this case would become:

In the CBSE Board Exams last year, 53% passed in Biology, 61% passed in English, 60% in Social Studies, 24% in Biology & English, 35% in English & Social Studies, 27% in Biology and Social Studies and 5% in none ; If the number of students in the class is 200, how many passed in only one subject

50+

less than 40

Explanation :

Let people who passed all three be x. Then:
53 + 61 + 60 - 24 - 35 - 27 + x = 95
so x = 7.
The venn diagram in this case would become:

33% of 200 = more than 50

more than 50%

less than 50%

None of these

Explanation :

Let people who passed all three be x. Then:
53 + 61 + 60 - 24 - 35 - 27 + x = 95
so x = 7.
The venn diagram in this case would become:

If the number of students is increased by 50%, the number of students in each category would also
be increased by 50%

In the CBSE Board Exams last year, 53% passed in Biology, 61% passed in English, 60% in Social Studies, 24% in Biology & English, 35% in English & Social Studies, 27% in Biology and Social Studies and 5% in none ; What is the ratio of percentage of passes in Biology and Social Studies but not English in relation to the percentage of passes in Social Studies and English but not Biology?

5:7

7:5

4:5

None of these

Explanation :

Let people who passed all three be x. Then:
53 + 61 + 60 - 24 - 35 - 27 + x = 95
so x = 7.
The venn diagram in this case would become:

20:28 = 5:7

In the McGraw-Hill Mindworkzz Quiz held last year, participants were free to choose their respective areas from which they were asked questions. Out of 880 participants, 224 chose Mythology, 240 chose Science and 336 chose Sports, 64 chose both Sports and Science, 80 chose Mythology and Sports, 40 chose Mythology and Science and 24 chose all the three areas;The percentage of participants who did not choose any area is

23.59%

30.25%

37.46%

27.27%

Explanation :

The following figure would emerge on using all the information in the
question:

240/880 = 27.27%

In the McGraw-Hill Mindworkzz Quiz held last year, participants were free to choose their respective areas from which they were asked questions. Out of 880 participants, 224 chose Mythology, 240 chose Science and 336 chose Sports, 64 chose both Sports and Science, 80 chose Mythology and Sports, 40 chose Mythology and Science and 24 chose all the three areas; Of those participating, the percentage who choose only one area is

60%

more than 60%

less than 60%

more than 75%

Explanation :

The following figure would emerge on using all the information in the
question:

504/880 = 57.27%. Hence, less than 60

In the McGraw-Hill Mindworkzz Quiz held last year, participants were free to choose their respective areas from which they were asked questions. Out of 880 participants, 224 chose Mythology, 240 chose Science and 336 chose Sports, 64 chose both Sports and Science, 80 chose Mythology and Sports, 40 chose Mythology and Science and 24 chose all the three areas; Number of participants who chose at least two areas is

112

136

None of these

Explanation :

The following figure would emerge on using all the information in the
question:

40 + 16 + 56 + 24 = 136

Mythology & Science but not Sports: Mythology only

Mythology & Sports but not Science: Science only

Science: Sports

None of these

Explanation :

The following figure would emerge on using all the information in the
question:

Option a gives us 16:128 = 1:8.

In the McGraw-Hill Mindworkzz Quiz held last year, participants were free to choose their respective areas from which they were asked questions. Out of 880 participants, 224 chose Mythology, 240 chose Science and 336 chose Sports, 64 chose both Sports and Science, 80 chose Mythology and Sports, 40 chose Mythology and Science and 24 chose all the three areas; The ratio of students choosing Sports & Science but not Mythology to Science but not Mythology & Sports is

2:5

1:4

1:5

1:2

Explanation :

The following figure would emerge on using all the information in the
question:

40:160 so 1:4.

The table here gives the distribution of students according to professional courses. Set Theory What is the percentage of Math students over English students

50.4

61.4

49.4

None of these

Explanation :

Math Students = 130. English Students =370
130/370 = 35.13%

The table here gives the distribution of students according to professional courses. Set Theory The average number of females in all the courses is (count people doing full-time MBA and CA as a separate course)

less than 12

greater than 12

None of these

Explanation :

Number of Female Students = 10 + 8 + 10 + 2 + 10 + 6 + 3 + 1 = 50. Average number of females
per course = 50/3 = 16.66

The table here gives the distribution of students according to professional courses. Set Theory The ratio of the number of girls to the number of boys is

5:36

1:9

1:7.2

None of these

Explanation :
```
50:450 = 1:9
```

The table here gives the distribution of students according to professional courses. Set Theory The percentage increase in students of full-time MBA only over CA only is

less than 20

less than 25

less than 30

more than 30

Explanation :
```
40/140 so 28.57%
```

The table here gives the distribution of students according to professional courses. Set Theory The number of students doing full-time MBA or CA is

320

160

None of these

Explanation :

From the figures, this value would be 150+8+ 90 + 10 + 16+6+37+3= 320

A newspaper agent sells The TOI, HT and IN in equal numbers to 302 persons. Seven get HT & IN, twelve get The TOI & IN, nine get The TOI & HT and three get all the three newspapers. The details are given in the Venn diagram: Set Theory How many get only one paper?

250

260

270

280

Explanation :


Based on this figure we have:
x + x – 13 + 4 + x – 16 = 302 Æ 3x – 25 = 302 Æ x = 327. Hence, x =109.
Consequently the figure becomes:

91 + 93 + 96 = 280

more than 65%

less than 60%

64%

None of these

Explanation :


Based on this figure we have:
x + x – 13 + 4 + x – 16 = 302 Æ 3x – 25 = 302 Æ x = 327. Hence, x =109.
Consequently the figure becomes:

193/302 @ 64%

6:4:9

6:9:4

4:9:6

None of these

Explanation :


Based on this figure we have:
x + x – 13 + 4 + x – 16 = 302 so 3x – 25 = 302 so x = 327. Hence, x =109.
Consequently the figure becomes:

6:9:4 is the required ratio

none

Explanation :


Based on this figure we have:
x + x – 13 + 4 + x – 16 = 302 Æ 3x – 25 = 302 Æ x = 327. Hence, x =109.
Consequently the figure becomes:

96 – 4 = 92

A group of 78 people watch Zee TV, Star Plus or Sony. Of these, 36 watch Zee TV, 48 watch Star Plus and 32 watch Sony. If 14 people watch both Zee TV and Star Plus, 20 people watch both Star Plus and Sony, and 12 people watch both Sony and Zee TV find the ratio of the number of people who watch only Zee TV to the number of people who watch only Son

9:4

3:2

5:3

7:4

Explanation :

78 = 36 + 48 + 32 – 14 – 20 – 12 + x so x = 8.

Required ratio is 18:8 Æ 9:4

The following data was observed from a study of car complaints received from 180 respondents at Colonel Verma’s car care workshop, viz., engine problem, transmission problem or mileage problem. Of those surveyed, there was no one who faced exactly two of these problems. There were 90 respondents who faced engine problems, 120 who faced transmission problems and 150 who faced mileage problems. How many of them faced all the three problems

Explanation :

Explanation :
```
There are 30 such people.
```

Seventy percent of the employees in a multinational corporation have VCD players, 75 percent have microwave ovens, 80 percent have ACs and 85 percent have washing machines. At least what percentage of employees has all four gadgets?

cant say

Explanation :

The least percentage of people with all 4 gadgets would happen if all the employees who are not
having any one of the four objects is mutually exclusive.
Thus, 100 - 30 - 25 - 20 - 15 = 10

given below are five diagrams one of which describes the relationship among the three classes given in each of the five questions that follow. Set Theory You have to decide diagram 'A' is the most suitable which option.

Elephants, tigers, animals

Administrators, Doctors, Authors

Platinum, Copper, Gold

Gold, Platinum, Ornaments

Television, Radio, Mediums of Entertainment

Explanation :

Elephants, tigers, animals

Administrators, Doctors, Authors

Platinum, Copper, Gold

Gold, Platinum, Ornaments

Television, Radio, Mediums of Entertainment

Explanation :

Elephants, tigers, animals

Administrators, Doctors, Authors

Platinum, Copper, Gold

Gold, Platinum, Ornaments

none

Explanation :

Elephants, tigers, animals

Administrators, Doctors, Authors

Platinum, Copper, Gold

Gold, Platinum, Ornaments

Television, Radio, Mediums of Entertainment

Explanation :

Elephants, tigers, animals

Administrators, Doctors, Authors

Platinum, Copper, Gold

Gold, Platinum, Ornaments

Television, Radio, Mediums of Entertainment

Explanation :

Laws of Indices:

(i) a^m * aⁿ = a^m + ⁿ

(ii) a^m/aⁿ = a^m - ⁿ

(iii) (a^m)ⁿ = a^m * ⁿ

(v) a^-n = 1/aⁿ

(vi) ⁿ $\sqrt{a}$ ^m = a^m/ⁿ

(vii) (ab)^m= a^m ∙ b^m.

(viii) (a/b)^m = a^m/b^m

Equations

A∪B = n(A) + n(B) - n(A ∩B) and

if A ∩B = not empty then n(A-B) + n(B-A) + n(A ∩B)

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A ∩B) - n(B ∩C) -n(A ∩C) + n(A ∩B ∩C)

For the below Venn diagram if you want to find the formula for any region you can use simple addition and subtraction.
Venn diagram of multiple groups

Sample questions that can be asked are:
A = People who love football, B = people who love basketball, C = people who love carrom, U = all people. Then find out

people who love only football
people who love only carrom
people who love only basketball
people who love carrom and basketball
people who love carrom and football
people who love football and basketball.
people who love all three.
people who love none.
Region that represents any of the above. In such cases the regions shall have numbers and not equations.

Number System

Divisibility rules:

Divisible by 2: If its digit in units place is divisible by 2. E.g: 578 is divisible by 2 as units digit = 8 is divisible.
Divisible by 3: If sum of digits is divisible by 3. E.g: 372 = 3+7+2 = 12 which is divisible by 3 so 372 is also divisible.
Divisible by 4: If the number formed by the last two digits is divisible by 4. E.g: 340 is divisible as 40 is divisible.
Divisible by 5: If units digits is 5 or 0. E.g. 55 / 70.
Divisible by 6: If the number is divisible by both 2 and 3.
Divisible by 8: Number formed by last three digits is divisible by 8.
Divisible by 9: Sum of the digits is divisible by 9.
Divisible by 10: Last digit should be 0.
Divisible by 11: Difference between sum of digits at odd place and even place is 0 or divisible by 11. E.g: 121 i.e. 1+1 - 2 = 0 so divisible.
Divisible by 12: A number is divisible by 3 and 4.
Divisible by 14: A number is divisible by 2 and 7.
Divisible by 15: A number is divisible by both 3 and 5.
Divisible by 16: Last four digits form a number divisible by 16.
Divisibility By 24 : A given number is divisible by 24, if it is divisible by both 3 and 8.

Divisibility By 40 : A given number is divisible by 40, if it is divisible by both 5 and 8.

Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.

Important Notes :

If a number is divisible by "p" as well as "q", where "p" and "q" are co-primes, then the given number is divisible by "pq".

If p and q are not co-primes, then the given number need not be divisible by pq, even when it is divisible by both p and q.

Example :36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since 4 and 6 are not co-primes.

Is the following number divisible by 3 ? 541326

yes

cant say

insufficient data

Ans .

yes

Explanation :
Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3. Hence, 541326 is divisible by 3

Is the following number divisible by 3 ? 5967013

yes

cant say

insufficient data

Ans .

Explanation :
Sum of digits in 5967013 = (5 + 9 + 6 + 7 + 0 + 1 + 3) = 31, which is not divisible by 3. Hence, 5967013 is not divisible by 3.

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

(a + b)² - (a - b)² = 4ab

(a + b)² + (a + b)² = 2 (a² + b²)

(a² - b²) = (a + b) (a - b)

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a³ + b³) = (a + b)(a² - ab + b²)

(a³ - b³) = (a - b)(a² + ab + b²)

(a³ + b³ + c³ - 3abc) = (a + b + c)(a² + b² + c² - ab - bc - ac)

If a + b + c = 0, then a³ + b³ + c³ = 3abc

If we divide a given number by another number, then : Dividend = (Divisor x Quotient) + Remainder

(xⁿ - aⁿ) is divisible by (x - a) for all values of "n"

(xⁿ - aⁿ) is divisible by (x + a) for even values of "n"

(xⁿ + aⁿ) is divisible by (x + a) for odd values of "n"

Some Properties of Prime Numbers

The lowest prime number is 2. 2 is also the only even prime number. The lowest odd prime number is 3.

The remainder when a prime number p ≥ 5 is divided by 6 is 1 or 5. However, if a number on being divided by 6 gives a remainder of 1 or 5 the number need not be prime.

The remainder of the division of the square of a prime number p ≥ 5 divided by 24 is 1.

For prime numbers p > 3, p² - 1 is divisible by 24.

If a and b are any two odd primes then a² - b² is composite. Also, a² + b² is composite.

The remainder of the division of the square of a prime number p ≥ 5 divided by 12 is 1

Shortcut to Test for Prime numbers

Step 1 : A number below 49 is prime if it is not divisible by 3 (Except numbers is even or ending with 0 / 5).

Step 2 : A number between 49 - 121 is prime if it is not divisible by 3 (Except even numbers or ending with 0 / 5 or the numbers 77, 91, 119).

Step 3 : A number between 121 - 169 is prime if it is not divisible by 3 (Except even numbers or numbers ending with 0 / 5 and 133, 143, 161).

Co prime or relatively prime: Two numbers are said to be relatively prime if they have only 1 as the common factor. E.g: (4,5); (8,9). If a number is divisible by p,q where both are co primes then it is also divisible by p*q;

Test for Prime numbers: To find out if a number 'N' greater than 100 is prime; we find out the number 'x' whose square is as close but greater than that number. If the prime numbers below 'x' are not dividing 'N' then it is prime.
E.g: 191 is prime since 14^2 = 196 and prime numbers below 14 are 2,3,5,7,11,13 don't divide 191.

Multiplication short cuts:
A) a * ( b + c ) = a*b + a*c
B) a * ( b - c) = a*b - a*c;

For shortcut to solving tough looking problems use above tricks:
345345 * 9999 = 345345 * (1000 - 1) which is simpler.

Multiplication of a number by 5^n: Add 'n' zeros to the right of the number and divide it by 2^n.
E.g: 975436 * 625 = 975436 * 5^4 = 9754360000/2^4

What value will replace the question mark in each of the following equations: ? - 1936248 = 1635773

3572022

3572026

3572028

3572091

3572021

Ans .

3572021

Explanation :
Let x - 1936248=1635773. Then, x = 1635773 + 1936248 = 3572021.

What value will replace the question mark in each of the following equations: 8597 - ? = 7429 - 4358

5596

5586

5516

5526

5546

Ans .

5526

Explanation :
Let 8597 - x = 7429 - 4358. Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.

What could be the maximum value of Q in the following equation? 5P9 + 3R7 + 2Q8 = 1114

Ans .

Explanation :
When we add the numbers together, at the units place we have 9 + 7 + 8 = 24. We carried the 2 to the tens place and got P + Q + R + 2, As we need 11 at the next digit we need to carry forward 1. So P + Q + R + 2 = 11 and we get Q as 9. P = R = 0

Simplify : 5793405 x 9999

57928256595

57928256590

57928256597

57928256585

57928256565

Ans .

57928256595

Explanation :
5793405 x 9999 = 5793405(10000-1) = 57934050000-5793405 = 57928256595

Simplify : 839478 x 625

524673754

524673750

524673751

524673740

524673755

Ans .

524673750

Explanation :
839478 x 625 = 839478 x 5⁴ = 8394780000 / 16 = 524673750.

Evaluate : 986 x 237 + 986 x 863

986000

981000

989000

987000

984000

Ans .

986000

Explanation :
986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000

Evaluate : 983 x 207 - 983 x 107

98305

98300

98100

98200

98400

Ans .

98300

Explanation :
983 x 207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300

Simplify : (i) 1605 x 1605

2576025

2576035

2576015

2576005

Ans .

2576025

Explanation :
i) 1605 x 1605 = (1605)² = (1600 + 5)² = (1600)² + (5)² + 2 x 1600 x 5 = 2560000 + 25 + 16000 = 2576025.

Simplify : 1398 x 1398

1954414

1954404

1954424

1954464

Ans .

1954404

Explanation :
1398 x 1398 - (1398)² = (1400 - 2)² = (1400)² + (2)² - 2 x 1400 x 2 = 1960000 + 4 - 5600 = 1954404.

Evaluate : (313 x 313 + 287 x 287)

180368

180328

180348

180338

Ans .

180338

Explanation :
(a² + b²) = 1/2 [(a + b)² + (a - b)²] (313)² + (287)² = 1/2 [(313 + 287)² + (313 - 287)²] = 1/2 [(600)² + (26)²] = 1/2 (360000 + 676) = 180338.

Simplify : (i) 896 x 896 - 204 x 204

761200

761400

7612200

761900

Ans .

761200

Explanation :
= (896)² - (204)² = (896 + 204) (896 - 204) = 1100 x 692 = 761200.

(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114

254001

253001

252001

251001

Ans .

251001

Explanation :
(ii) Given exp = (387)² + (114)² + (2 x 387 x 114) = a² + b² + 2ab, where a = 387, b = 114 = (a+b)² = (387 + 114)² = (501)² = 251001.

81 X 81 + 68 X 68-2 x 81 X 68

179

159

169

149

Ans .

169

Explanation :
Given exp = (81)² + (68)² - 2 x 81 x 68 = a² + b² - 2ab, Where a = 81, b = 68 = (a-b)² = (81 - 68)² = (13)² = 169

541326 is divisible by 3 ?

yes

cant say

insufficient data

Ans .

yes

Explanation :
Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3. Hence, 541326 is divisible by 3

5967013 is divisible by 3 ?

yes

cant say

insufficient data

Ans .

Explanation :
Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3. Hence, 5967013 is not divisible by 3.

What least value must be assigned to * so that the number 197*5462 is divisible by 9 ?

Ans .

Explanation :
Let the missing digit be x. Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 + 2) = (34 + x). For (34 + x) to be divisible by 9, x must be replaced by 2 . Hence, the digit in place of * must be 2.

67920594 is divisible by 4 ?

yes

cant say

insufficient data

Ans .

Explanation :
The number formed by the last two digits in the given number is 94, which is not divisible by 4. Hence, 67920594 is not divisible by 4.

618703572 is divisible by 4 ?

yes

cant say

insufficient data

Ans .

yes

Explanation :
The number formed by the last two digits in the given number is 72, which is divisible by 4. Hence, 618703572 is divisible by 4.

Which digits should come in place of * and $ if the number 62684*$ is divisible by both 8 and 5 ?

4 and 0

4 and 5

2 and 0

6 and 0

Ans .

4 and 0

Explanation :
Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $. Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4. Hence, digits in place of * and $ are 4 and 0 respectively.

4832718 is divisible by 11 ?

yes

cant say

insufficient data

Ans .

yes

Explanation :
(Sum of digits at odd places) - (Sum of digits at even places) = (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11. Hence, 4832718 is divisible by 11.

52563744 divisible by 24 ?

yes

cant say

insufficient data

Ans .

yes

Explanation :
24 = 3 x 8, where 3 and 8 are co-primes. The sum of the digits in the given number is 36, which is divisible by 3. So, the given number is divisible by 3. The number formed by the last 3 digits of the given number is 744, which is divisible by 8. So, the given number is divisible by 8. Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes. So, it is divisible by 3 x 8, i.e., 24.

What least number must be added to 3000 to obtain a number exactly divisible by 19 ?

Ans .

Explanation :
On dividing 3000 by 19, we get 17 as remainder. Number to be added = (19 - 17) = 2.

Find the number which is nearest to 3105 and is exactly divisible by 21.

Ans .

Explanation :
On dividing 3105 by 21, we get 18 as remainder. So Number to be added to 3105 = (21 - 18) = 3. Hence, required number = 3105 + 3 = 3108

A number when divided by 342 gives a remainder 47. When the same number if divided by 19, what would be the remainder ?

Ans .

Explanation :
On dividing the given number by 342, let k be the quotient and 47 as remainder. Then, number - 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9. The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Find the smallest number of 6 digits which is exactly divisible by 111.

1000110

100011

300011

200011

Ans .

100011

Explanation :
Smallest number of 6 digits is 100000. On dividing 100000 by 111, we get 100 as remainder. So Number to be added = (111 - 100) = 11. Hence, required number = 100011

On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor.

177

198

188

179

Ans .

179

Explanation :
Divisor = (Dividend - Remainder ) / Quotient = ( 15968-37 ) / 89 = 179

A number being successively divided by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively. Find the respective remainders if the order of divisors be reversed

6, 4, 4

6, 4, 2

8, 4, 2

6, 5, 2

Ans .

6, 4, 2

Explanation :

Theorem of Divisibility

Calculate the number of factors of a number

Step 1 : Calculate the factors of the number "X" less than or equal to its square root or if its not a perfect square than less than or equal to a number whose square is less but closest to it. Example: For 80 it will be 1 - 8 as 9 is 81, for 25 its 1 - 5. For 99 its 1 - 9.

Step 2 : for factors above the square root limit, we just find the numbers with whom we can multiply the factors obtained from Step 1 and get the number "X". Example : For factors of 80, we find factors from 1 - 8, i.e. 1,2,4,5,8 then for factors above square root limit we get 80,40,20,16,10. As 1 * 80, 40 * 2, 20 * 4, 16 * 5 and 10 * 8.

Calculate the number of factors of a number

Calculate the Sum of factors of a number:

Step 1: Get prime factors of a number say 240

240 = 2⁴ * 3¹ * 5¹

Step 2: Sum of factors formula is

240 = (2⁰ + 2¹ + 2² + 2³ + 2⁴) * (3⁰ + 3¹) * (5⁰ + 5¹)

Step 3: 31*4*6 = 744

Calculate the Number of factors of a number:

Step 1: Get the prime factors of a number

240 = 2⁴ * 3¹ * 5¹

Step 2: Number of factors of a number.

Number of factors = ( 4 + 1 ) * ( 1 + 1) * ( 1 + 1) = 5 * 2 * 2 = 20

Thus the powers of the numbers are increased by one and multiplied.

Calculate the sum and number of even factors of a number :

Numbers which do not have any power of 2 in their prime factor will have even factors equal to 0.

Step 1: Get the prime factors of a number

240 = 2⁴ * 3¹ * 5¹

Step 2: Sum of even factors
Sum = (2¹ + 2² + 2³ + 2⁴) * (3⁰ + 3¹) * (5⁰ + 5¹)
Number of even factors = 4 * 2 * 2 = 16
Thus the powers of the numbers are increased by one and multiplied except 2.

Calculate the sum and number of odd factors of a number:

Step 1: Get the prime factors of a number

240 = 2⁴ * 3¹ * 5¹

Step 2: Sum of odd factors
Sum = (2⁰) * (3⁰ + 3¹) * (5⁰ + 5¹)
Number of odd factors = 1 * 2 * 2 = 4
Thus the powers of the numbers are increased by one and multiplied except 2.

Calculate the sum and number of factors of a number satisfying a condition:

Step 1: Get prime factors of a number say 240

30 = 2¹ * 3¹ * 5¹

Step 2: Sum of factors formula is

30 = (2⁰ + 2¹ ) * (3⁰ + 3¹) * (5⁰ + 5¹)

which is expanded as
= 2⁰*3⁰*5^{0 +}2⁰*3⁰*5^0
+2⁰*3¹*5^{1 +}2⁰*3¹*5^1
+2¹*3⁰*5⁰ + 2¹*3⁰*5⁰ + 2¹*3¹*5¹ + 2¹*3^1*5¹
These are all the factors of the number, We can apply any condition we want and remove unnecessary of them viz. remove factors that are perfect squares, not perfect squares, between higher and lower limit etc

The sum and number of all factors of 2450

Ans .

Explanation :

2450 = 50 * 49 = 2¹ * 5² * 7²
Sum and number of all factors:
Sum of factors = (2⁰ + 2¹) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7²) Number of factors = 2 * 3 * 3 = 18

The sum and number of even factors of 2450

Ans .

Explanation :

Sum of all even factors:
(2¹) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7²)
Number of even factors = 1 * 3 * 3 = 9

The sum and number of odd factors of 2450

Ans .

Explanation :

Sum of all odd factors:
(2⁰) (5⁰ + 5¹ + 5²) (7⁰ + 7¹ + 7²)
Number of odd factors = 1 * 3 * 3 = 9

The sum and number of factors divisible by 5 of 2450

Ans .

Explanation :

Sum of factors divisible by 5:
(2⁰ + 2¹) (5¹ + 5²) (7⁰ + 7¹ + 7²)
Number of factors divisible by 5 = 2 * 2 * 3 = 12

Sum and number of factors divisible by 35 of 2450

Ans .

Explanation :

Sum of factors divisible by 35:
(2⁰ + 2¹) (5¹ + 5²) (7¹ + 7²)
Number of factors divisible by 5 = 2 * 2 * 2= 8

Sum and number of all factors divisible by 245 of 2450

Ans .

Explanation :

Sum of factors divisible by 245:
(2⁰ + 2¹) (5¹ + 5²) (7²)
Number of factors divisible by 5 = 2 * 2 * 1= 4

For the number 7200 find. The sum and number of all factors.

Ans .

Explanation :

Sum and number of all factors:
Sum of factors = (2⁰ + 2¹+ 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²)
Number of factors = 6 * 3 * 3 = 54

Sum and number of even factors of 7200

Ans .

Explanation :

Sum and number of all even factors:
Sum of factors = (2¹+ 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²)
Number of factors = 5 * 3 * 3 = 30

Sum and number of odd factors of 7200

Ans .

Explanation :

Sum and number of all odd factors:
Sum of factors = (2⁰) (3⁰ + 3¹ + 3²) (5⁰ + 5¹ + 5²)
Number of factors = 1 * 3 * 3 = 9

Sum and number of factors divisible by 25: of 7200

Ans .

Explanation :

Sum and number of factors divisible by 25:
Sum of factors = (2⁰ + 2¹+ 2² + 2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5²)
Number of factors = 6 * 3 * 1 = 18

Sum and number of factors divisible by 40: of 7200

Ans .

Explanation :

Sum and number of all factors divisible by 40:
Sum of factors = (2³ + 2⁴ + 2⁵) (3⁰ + 3¹ + 3²) (5¹ + 5²)
Number of factors = 3 * 3 * 2 = 18

Sum and number of factors divisible by 150: of 7200

Ans .

Explanation :

Sum and number of factors divisible by 150:
Sum of factors = (2¹+ 2² + 2³ + 2⁴ + 2⁵) ( 3¹ + 3²) (5²)
Number of factors = 5 * 2 * 1 = 10

Sum and number of factors NOT divisible by 150: of 7200

Ans .

Explanation :

Sum and number of factors NOT divisible by 150 = Total factors - Factors divisible by 150 = 54 - 10 = 44

For the number 7200 find. The number of all factors who are perfect squares

Ans .

Explanation :

Sum and number of all factors who are perfect squares:
Sum of factors = (2⁰ + 2² + 2⁴ ) (3⁰ + 3²) (5⁰ + 5²)
Number of factors = 3 * 2 * 2 = 12

Read and Answer

New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved.

Q. What is the number of projects in which Gyani alone is involved?

Uniquely equal to zero.
Uniquely equal to 1.
Uniquely equal to 4
Cannot be determined uniquely.

Ans . D

Putting the value of M in either equation, we get G + B = 17
Hence neither of two can be uniquely determined.

Q. What is the number of projects in which Medha alone is involved?

Uniquely equal to zero.
Uniquely equal to 1.
Uniquely equal to 4
Cannot be determined uniquely.

Ans . B

G + B = M + 16 Also, M + B + G + 19 = (2 * 19) - 1
i.e. (G + B) = 18 - M Thus, M + 16 = 18 - M
i.e. M = 1

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