Wednesday, September 6, 2023

Problem based on Direct and Inverse Variation

35 cattle can graze on a field for 18 days. After 10 days, 15 cattle are move to a different field. For how long can the remaining cattle graze on the field? b) 14.5 days c) 14 days • d) 12 days a) 15 days ​ 


Answer:

If one increases the number of cattle, then the number of days will decrease to graze the same field.

If we increase the number of days, then less cattle will be needed to graze the same field.

So the variables, cattle and number of days are inversely proportional.

Now, if we consider the variables cattle and field size ( keeping the number of days fixed), it is easy to see that they are directly proportional. ( as if you increase the field size, then the number of cattle must be increased to graze in the same day and vice versa)

Let the corresponding quantities for cattle, days and field size be C, D and F.

Then C ∝  ......(i)

⇒C = k , where k is some non-zero constant.

For the initial data, C= 35 and F = 1 ( taking field size as 1 unit ) and D = 18

Now in 1 day, 35 cattle will graze

unit of the field.

Thus in 10 days, they will graze,

unit of the field.

Thus remaining unit of field to be grazed

As 15 cattle were removed, remaining cattle

So the question, boils down to finding the number of days (D) in which 20 cattle (C) will graze

unit (F) of the field.

From (i), C = k

⇒ 

⇒   days

(c) is the correct option

Sunday, September 3, 2023

Bayes' Theorem Problem from ISC 2023 Maths Paper

In a company, 15% of the employees are graduates and 85% of the employees are non-graduates. as per the annual report of the company, 80% of the graduate employees and 10% of the non-graduate employees are in the administrative positions. find the probability that an employee selected at random from those working in administrative positions will be a graduate. 



Answer:

Step-by-step explanation:

Let G be the event that the selected employee is a graduate, and NG be the event that the selected employee is non-graduate.

Clearly, G and NG forms a mutually exclusive and exhaustive set of events.

Further, let A be the event that the selected employee works in administrative office.

According to the problem, we have to find, the selected employee is a graduate given he/she works in the administrative position P(G/A).

By Bayes' theorem,

Now, Probability of an employee to be graduate =

Probability of an employee to be non-graduate =

Probability of an employee working in administrative office given he is a graduate =

( as 80% of the graduate employee works in the administrative positions)

Probability of an employee working in administrative office given he is a non-graduate =

( as 10% of the non-graduate employee works in the administrative positions)

Substituting the values in

, we get,

Wednesday, May 24, 2023

Prove that cos pi/11 + cos 3pi/11 + cos 5pi/11 + cos 7pi/11 + cos 9pi/11...

A elegant solution for a difficult trigonometric problem using complex numbers. Many problem of trigonometry and even algebra can be beautifully solved using complex numbers. The use of different formulas from trigonometry can be avoided. #CBSE #ISC #IIT #ISI #wbjee #wbchse #jeeadvanced Prove that cos pi/11 + cos 3pi/11 + cos 5pi/11 + cos 7pi/11 + cos 9pi/11 =1/2

Integrate max { x+|x|,x-[x] } from -n to n JEE Main and Advance and Indian Statistical Institute B.Math & B.Stat

Integrate max { x+|x|,x-[x] } from -n to n Difficult Integration for JEE Main and Advance and Indian Statistical Institute Involving Limits Integration for Indian Statistical Institute B.Math & B.Stat : Integration, JEE Main and Advance, WBJEE #CBSE #ISC #wbchse #HS #jeeadvanced

Tuesday, February 14, 2023

Combinations SN Dey Solved

1. An executive committee of 6 is to be formed from 4 ladies and 7 gentlemen. In how many ways can this be formed when the committee contains (i) only 2 lady members, (ii) at least 2 lady members?

2. Find the number of committees of 5 members that can be formed from 6 gentlemen and 4 ladies if each committee has at least one lady and two gentlemen.

 

3. A committee of 5 is to be formed from six ladies and four gentlemen. In how many ways this can be done so that the committee contains (i) exactly two ladies, (ii) at least two ladies, (iii) at most two ladies?

 

4. In a cricket team of 14 players 6 are bowlers. How many different teams of 11 players can be selected keeping at least 4 bowlers in the team?

 

5. A box contains 12 lamps of which 5 are defective. In how many ways can a sample of 6 be selected at random from the box so as to include at most 3 defective lamps?

 

6. An examinee is required to answer 6 questions out of 12 questions which are divided into two groups each containing 6 questions, and he is not permitted to answer more than 4 questions from any group. In how many ways can he answer 6 questions?

 

7. A question paper contains 10 questions, which are divided into two groups each containing 5 questions. A candidate is asked to answer 6 questions only, and to choose at least 2 questions from each group. In how many different ways can the candidate make up his choice?

 

8. In how many ways can a team of 11 cricketers be chosen from 9 batsmen and 6 bowlers to give a majority of batsman if at least 3 bowlers are to be included?

 

9. The Indian Cricket Eleven is to be selected out of fifteen players, five of them are bowlers. In how many ways the team can be selected so that the team contains at least three bowlers?

10. How many combinations can be formed of eight counters marked 1, 2, 3, 4, 5, 6, 7, 8 taking them 4 at a time, there being at least one odd and one even counter in each combination?

 

11. Find the number of permutations of the letters of the words FORECAST and MILKY taking 5 at a time of which 3 letters from the first word and 2 from the second.

 

12. In how many ways can the crew of an eight-oared boat be arranged if 2 of the crew can row only on the stroke side and 1 can row only on the bow side?

 

13. Of the 17 articles, 12 are alike and the remaining 5 are different. Find the number of combinations, if 13 articles are taken at a time.

 

14. Out of 3n given things 2n are alike and the rest are different. Show that a selection of 2n things can be made from these 3n things in 2" different ways.

 

15. Show that there are 136 ways of selecting 4 letters from the word EXAMINATION.

 

16. Find the total number of ways of selecting 5 letters from the letters of the word INDEPENDENT.

 

17. (i) Find the number of combinations in the letters of the word STATISTICS taken 4 at a time.

      (ii) Find the number of permutations in the letters of the word PROPORTION taken 4 at a time.

 

18. How many different numbers of 4 digits can be formed with the digits 1, 1, 2, 2, 3, 4?

 

19. (i) From 4 apples, 5 oranges mangoes, how many selections of fruits can be made, taking at least one of each kind if the fruits of the same kind are of different shapes?

(ii) In how many ways can one or more fruits be selected from 4 apples, 5 oranges and 3 mangoes, if the fruits of the same kind be of the same shape?

20. Find the total number of combinations taking at least one green ball and one blue ball, from 5 different green balls, 4 different blue balls and 3 different red balls.

 

21. How many different algebraic quantities can be formed by combining a, b, c, d, e with + and - signs, all the letters taken together?

 

22. There are n points in space, no four of which are in the in the same place with the exception of m points, all of which are in the same plane. How many planes can be formed by joining them?

 23. n1, n2 and n3, points are given on the sides BC, CA and AB respectively of the triangle ABC. Find the number of triangles formed by taking these given points as vertices of a triangle.

 

24. A man has 7 relatives, 4 of them are ladies and 3 gentlemen; his wife has also 7 relatives, 3 of them are ladies and 4 are gentlemen. In how many ways can they invite dinner party of 3 ladies and 3 gentlemen so that there are 3 of the man's relatives and 3 of the wife's relatives?

 

25. Eighteen guests have to be seated, half on each side of long table. Four particular guests desire to sit on one particular side and three others on the other side Determine the number of ways in which the arrangements can be made. 

Sunday, February 12, 2023

Permutations SN Dey Solved XI Maths

Solved problems from the book of SN Dey Class XI WBCHSE, Permutations [ 4 and 5 marks only]

Sunday, November 20, 2022

Problem from Geometry - Circles

In the figure given below, ABD is a right-angled triangle at B. Taking AB as diameter, a circle has been drawn intersecting AD at F. Prove that the tangent drawn at point F bisects BD.



Solution:


Saturday, August 20, 2022

Trigonometry SN Dey Solved Problems - HS - Class XI

1. Trigonometric functions of Standard Angles [ 4 and 5 marks ]

2. Trigonometric functions of Associate Angles [ 4 and 5 marks ]

3. Trigonometric transformations of sums and products [ 4 and 5 marks]

4. Trigonometric functions of Compound Angles [ 4 and 5 marks]

5. Trigonometric functions of Multiple Angles [4 and 5 marks]

6. Trigonometric functions of Sub-multiple Angles [ 4 and 5 marks]

7. Trigonometric Equations  [ 4 and 5 marks]


Wednesday, August 17, 2022

Vectors SN Dey Solved Problems

 Introduction of Vectors Solved Problems SN Dey - Long Answer Type Questions 5 Marks



Product of Vectors Solved Problems SN Dey - Long Answer Type Questions 5 Marks


Thursday, August 4, 2022

Set Theory Solved Problems : Class XI

The concept of set serves as a fundamental part of the present day mathematics. Today this concept is being used in almost every branch of mathematics. Sets are used to define the concepts of relations and functions. The study of geometry, sequences, probability, etc. requires the knowledge of sets. The theory of sets was developed by German mathematician Georg Cantor (1845-1918). He first encountered sets while working on “problems on trigonometric series”. In this Chapter, we discuss some basic definitions and operations involving sets.


Empty Sets

The set with no elements or null elements is called an empty set. This is also called a Null set or Void set. It is denoted by {}.

For example: Let, Set X = {x:x is the number of students studying in Class 6th and Class 7th}

Since we know a student cannot learn simultaneously on two classes, therefore set X is an empty set.

Singleton Set

The set which has only one element is called a singleton set.

For example, Set X = { 2 } is a singleton set.

Finite and Infinite Sets

Finite sets are the one which has a finite number of elements, and infinite sets are those whose number of elements cannot be estimated, but it has some figure or number, which is very large to express in a set.

For example, Set X = {1, 2, 3, 4, 5} is a finite set, as it has a finite number of elements in it.

Set Y = {Number of Animals in India} is an infinite set, as there is an approximate number of Animals in India, but the actual value cannot be expressed, as the numbers could be very large.

Equal Sets

Two sets X and Y are said to be equal if every element of set X is also the elements of set Y and if every element of set Y is also the elements of set X. It means set X and set Y have the same elements, and we can denote it as;

X = Y

For example, Let X = { 1, 2, 3, 4} and Y = {4, 3, 2, 1}, then X = Y

And if X = {set of even numbers} and Y = { set of natural numbers} the X ≠ Y, because natural numbers consist of all the positive integers starting from 1, 2, 3, 4, 5 to infinity, but even numbers starts with 2, 4, 6, 8, and so on.

Subsets

A set X is said to be a subset of set Y if the elements of set X belongs to set Y, or you can say each element of set X is present in set Y. It is denoted with the symbol as X ⊂ Y.

We can also write the subset notation as;

X ⊂ Y if a ∊ X

a ∊ Y

Thus, from the above equation, “X is a subset of Y if a is an element of X implies that a is also an element of Y”.

Each set is a subset of its own set, and a null set or empty set is a subset of all sets.

Power Sets

The power set is nothing but the set of all subsets. Let us explain how.

We know the empty set is a subset of all sets and every set is a subset of itself. Taking an example of set X = {2, 3}. From the above given statements we can write,

{} is a subset of {2, 3}

{2} is a subset of {2, 3}

{3} is a subset of {2, 3}

{2, 3} is also a subset of {2, 3}

Therefore, power set of X = {2, 3},

P(X) = {{},{2},{3},{2,3}}

Universal Sets

A universal set is a set which contains all the elements of other sets. Generally, it is represented as ‘U’.

For example; set X = {1, 2, 3}, set Y = {3, 4, 5, 6} and Z = {5, 6, 7, 8, 9}

Then, we can write universal set as, U = {1, 2, 3, 4, 5, 6, 7, 8, 9,}

Note: From the definition of the universal set, we can say, all the sets are subsets of the universal set. Therefore,

X ⊂ U

Y ⊂ U

And Z ⊂ U

Union of sets

A union of two sets has all their elements. It is denoted by ⋃.

For example, set X = {2, 3, 7} and set Y = { 4, 5, 8}

Then union of set X and set Y will be;

X ⋃ Y = {2, 3, 7, 4, 5, 8}

Properties of Union of Sets:

X ⋃ Y = Y ⋃ X ; Commutative law

(X ⋃ Y) ⋃ Z = X ⋃ (Y ⋃ Z)

X ⋃ {} = X ; {} is the identity of ⋃

X ⋃ X = X

U ⋃ X = U

Intersection of Sets

Set of all elements, which are common to all the given sets, gives intersection of sets. It is denoted by the symbol ⋂.

For example, set X = {2, 3, 7} and set Y = {2, 4, 9}

So, X ⋂ Y = {2}

Difference of Sets

The difference of set X and set Y is such that, it has only those elements which are in the set X and not in the set Y.

i.e. X – Y = {a: a ∊ X and a ∉ Y}

In the same manner, Y – X = {a: a ∊ Y and a ∉ X}

For example, if set X = {a, b, c, d} and Y = {b, c, e, f} then,

X – Y = {a, d} and Y – X = {e, f}

Disjoint Sets

If two sets X and Y have no common elements, and their intersection results in zero(0), then set X and Y are called disjoint sets.

It can be represented as; X ∩ Y = 0



Sunday, July 17, 2022

HS Mathematics Question Paper Solved 2022 - CLASS XI

 HS Mathematics Question Paper Solved 2022 - CLASS XI

 Download the full question paper and solution of mathematics for the year 2022 under WBCHSE ( West Bengal Council of Higher Secondary Education ) HS.

HS (Class XI) 2022 Mathematics English Version

Wednesday, June 15, 2022

ICSE Class IX - Practice Set

Practice Set 32 Marks - Real Numbers and Compound Interest - Click Here
Practice Set 25 Marks -7 Questions- Real Numbers, Exponents, Compound Interest and Expansions - Set I
Practice Set 28 Marks-8 Questions- Real Numbers, Exponents, Compound Interest and Expansions - Set II
Practice Set 15 Questions - Real Numbers - Click Here
Harder Problems 22 Questions - Algebraic Identities - Click Here
Practice Set 39 Questions - Factorisations - Click Here
Practice Set 23 Questions - Expansion and Indices - Click Here
Practice Set 23 Questions - Real Numbers and Compound Interest - Click Here

Tuesday, May 3, 2022

HS MATHEMATICS PAPER English Version 2022 - XII-Question Paper and Solution - WBCHSE

 Download the full question paper and solution of mathematics for the year 2022 under WBCHSE ( West Bengal Council of Higher Secondary Education ) HS.

HS 2022 Mathematics English Version



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