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)
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
. 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.
. 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.
. 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
. 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
The total number of people = 100 + 220 + 130 = 450
The required answer would be 20000 - 5000 - 4000 - 8000 = 3000.
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.
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
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.
[(20 – 10)/10] * 100 = 100%.
10+10+15+15=50%
Only ice cream is 10% of the total. Hence, 10% of 180 =18.
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
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
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%
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%
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.
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
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 = 25 + 20 - x so x = 15
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:
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
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%
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
The following figure would emerge on using all the information in the question: 240/880 = 27.27%
The following figure would emerge on using all the information in the question: 504/880 = 57.27%. Hence, less than 60
The following figure would emerge on using all the information in the question: 40 + 16 + 56 + 24 = 136
The following figure would emerge on using all the information in the question: Option a gives us 16:128 = 1:8.
The following figure would emerge on using all the information in the question: 40:160 so 1:4.
Math Students = 130. English Students =370 130/370 = 35.13%
Number of Female Students = 10 + 8 + 10 + 2 + 10 + 6 + 3 + 1 = 50. Average number of females per course = 50/3 = 16.66
50:450 = 1:9
40/140 so 28.57%
From the figures, this value would be 150+8+ 90 + 10 + 16+6+37+3= 320
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
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%
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
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
78 = 36 + 48 + 32 – 14 – 20 – 12 + x so x = 8. Required ratio is 18:8 Æ 9:4
There are 30 such people.
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
(i) a^{m} * a^{n} = a^{m} + ^{n}
(ii) a^{m}/a^{n} = a^{m} - ^{n}
(iii) (a^{m})^{n} = a^{m} * ^{n}
(v) a^{-n} = 1/a^{n}
(vi) ^{n}$\sqrt{a}$^{m} = a^{m}/^{n}
(vii) (ab)^{m}= a^{m} ∙ b^{m}.
(viii) (a/b)^{m} = a^{m}/b^{m}
Divisibility rules:
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.
Ans .
yes
Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3. Hence, 541326 is divisible by 3
Ans .
no
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} = a^{2} + b^{2} + 2ab
(a - b)^{2} = a^{2} + b^{2} - 2ab
(a + b)^{2} - (a - b)^{2} = 4ab
(a + b)^{2} + (a + b)^{2} = 2 (a^{2} + b^{2})
(a^{2} - b^{2}) = (a + b) (a - b)
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)
(a^{3} + b^{3}) = (a + b)(a^{2} - ab + b^{2})
(a^{3} - b^{3}) = (a - b)(a^{2} + ab + b^{2})
(a^{3} + b^{3} + c^{3} - 3abc) = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ac)
If a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc
If we divide a given number by another number, then : Dividend = (Divisor x Quotient) + Remainder
(x^{n} - a^{n}) is divisible by (x - a) for all values of "n"
(x^{n} - a^{n}) is divisible by (x + a) for even values of "n"
(x^{n} + a^{n}) 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, p^{2} - 1 is divisible by 24.
If a and b are any two odd primes then a^{2} - b^{2} is composite. Also, a^{2} + b^{2} 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;
Ans .
3572021
Let x - 1936248=1635773. Then, x = 1635773 + 1936248 = 3572021.
Ans .
5526
Let 8597 - x = 7429 - 4358. Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.
Ans .
9
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
Ans .
57928256595
5793405 x 9999 = 5793405(10000-1) = 57934050000-5793405 = 57928256595
Ans .
524673750
839478 x 625 = 839478 x 5^{4} = 8394780000 / 16 = 524673750.
Ans .
986000
986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000
Ans .
98300
983 x 207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300
Ans .
2576025
i) 1605 x 1605 = (1605)^{ 2} = (1600 + 5)^{ 2} = (1600)^{ 2} + (5)^{ 2} + 2 x 1600 x 5 = 2560000 + 25 + 16000 = 2576025.
Ans .
1954404
1398 x 1398 - (1398)^{ 2} = (1400 - 2)^{ 2} = (1400)^{ 2} + (2)^{ 2} - 2 x 1400 x 2 = 1960000 + 4 - 5600 = 1954404.
Ans .
180338
(a^{2} + b^{2}) = 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.
Ans .
761200
= (896)^{2} - (204)^{2} = (896 + 204) (896 - 204) = 1100 x 692 = 761200.
Ans .
251001
(ii) Given exp = (387)^{2} + (114)^{2} + (2 x 387 x 114) = a^{2} + b^{2} + 2ab, where a = 387, b = 114 = (a+b)^{2} = (387 + 114)^{2} = (501)^{2} = 251001.
Ans .
169
Given exp = (81)^{2} + (68)^{2} - 2 x 81 x 68 = a^{2} + b^{2} - 2ab, Where a = 81, b = 68 = (a-b)^{2} = (81 - 68)^{2} = (13)^{2} = 169
Ans .
yes
Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3. Hence, 541326 is divisible by 3
Ans .
no
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.
Ans .
2
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.
Ans .
no
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.
Ans .
yes
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.
Ans .
4 and 0
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.
Ans .
yes
(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.
Ans .
yes
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.
Ans .
2
On dividing 3000 by 19, we get 17 as remainder. Number to be added = (19 - 17) = 2.
Ans .
3
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
Ans .
9
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.
Ans .
100011
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
Ans .
179
Divisor = (Dividend - Remainder ) / Quotient = ( 15968-37 ) / 89 = 179
Ans .
6, 4, 2
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
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 Sum of factors of a number:
Step 1: Get prime factors of a number say 240
240 = 2^{4} * 3^{1} * 5^{1}
Step 2: Sum of factors formula is
240 = (2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4}) * (3^{0} + 3^{1}) * (5^{0} + 5^{1})
Step 3: 31*4*6 = 744
Calculate the Number of factors of a number:
Step 1: Get the prime factors of a number
240 = 2^{4} * 3^{1} * 5^{1}
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 :
Step 1: Get the prime factors of a number
240 = 2^{4} * 3^{1} * 5^{1}
^{ }
Step 2: Sum of even factors
240 = 2^{4} * 3^{1} * 5^{1}
^{ }
Step 2: Sum of odd factors
30 = 2^{1} * 3^{1} * 5^{1}
^{ }
Step 2: Sum of factors formula is
30 = (2^{0} + 2^{1} ) * (3^{0} + 3^{1}) * (5^{0} + 5^{1})
Ans .
18
2450 = 50 * 49 = 2^{1} * 5^{2} * 7^{2} Sum and number of all factors: Sum of factors = (2^{0} + 2^{1}) (5^{0} + 5^{1} + 5^{2}) (7^{0} + 7^{1} + 7^{2}) Number of factors = 2 * 3 * 3 = 18
Ans .
9
Sum of all even factors: (2^{1}) (5^{0} + 5^{1} + 5^{2}) (7^{0} + 7^{1} + 7^{2}) Number of even factors = 1 * 3 * 3 = 9
Ans .
9
Sum of all odd factors: (2^{0}) (5^{0} + 5^{1} + 5^{2}) (7^{0} + 7^{1} + 7^{2}) Number of odd factors = 1 * 3 * 3 = 9
Ans .
12
Sum of factors divisible by 5: (2^{0} + 2^{1}) (5^{1} + 5^{2}) (7^{0} + 7^{1} + 7^{2}) Number of factors divisible by 5 = 2 * 2 * 3 = 12
Ans .
8
Sum of factors divisible by 35: (2^{0} + 2^{1}) (5^{1} + 5^{2}) (7^{1} + 7^{2}) Number of factors divisible by 5 = 2 * 2 * 2= 8
Ans .
4
Sum of factors divisible by 245: (2^{0} + 2^{1}) (5^{1} + 5^{2}) (7^{2}) Number of factors divisible by 5 = 2 * 2 * 1= 4
Ans .
54
Sum and number of all factors: Sum of factors = (2^{0} + 2^{1}+ 2^{2} + 2^{3} + 2^{4} + 2^{5}) (3^{0} + 3^{1} + 3^{2}) (5^{0} + 5^{1} + 5^{2}) Number of factors = 6 * 3 * 3 = 54
Ans .
30
Sum and number of all even factors: Sum of factors = (2^{1}+ 2^{2} + 2^{3} + 2^{4} + 2^{5}) (3^{0} + 3^{1} + 3^{2}) (5^{0} + 5^{1} + 5^{2}) Number of factors = 5 * 3 * 3 = 30
Ans .
9
Sum and number of all odd factors: Sum of factors = (2^{0}) (3^{0} + 3^{1} + 3^{2}) (5^{0} + 5^{1} + 5^{2}) Number of factors = 1 * 3 * 3 = 9
Ans .
18
Sum and number of factors divisible by 25: Sum of factors = (2^{0} + 2^{1}+ 2^{2} + 2^{3} + 2^{4} + 2^{5}) (3^{0} + 3^{1} + 3^{2}) (5^{2}) Number of factors = 6 * 3 * 1 = 18
Ans .
18
Sum and number of all factors divisible by 40: Sum of factors = (2^{3} + 2^{4} + 2^{5}) (3^{0} + 3^{1} + 3^{2}) (5^{1} + 5^{2}) Number of factors = 3 * 3 * 2 = 18
Ans .
10
Sum and number of factors divisible by 150: Sum of factors = (2^{1}+ 2^{2} + 2^{3} + 2^{4} + 2^{5}) ( 3^{1} + 3^{2}) (5^{2}) Number of factors = 5 * 2 * 1 = 10
Ans .
44
Sum and number of factors NOT divisible by 150 = Total factors - Factors divisible by 150 = 54 - 10 = 44
Ans .
12
Sum and number of all factors who are perfect squares: Sum of factors = (2^{0} + 2^{2} + 2^{4} ) (3^{0} + 3^{2}) (5^{0} + 5^{2}) Number of factors = 3 * 2 * 2 = 12
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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