Typical Relationship Problems

The following are the typical relationship problems. These problems are solved by using logic.

Problem: - Pointing out to a photograph, a man tells his friend, “She is the daughter of the only son of my father’s wife.” How is the girl in the photograph related to the man?

Solution: - The relations may be analysed as follows:

Father’s wife – Mother; Mother’s only son – Himself.

So, the girl is man’s daughter.

Problem: - Pointing to a man on the stage, Rita said, “He is the brother of the daughter of the wife of my husband.” How is the man on the stage related to Rita?

Solution: - The relations may be analysed as follows:

Wife of husband – Herself; Brother of daughter- Son.

So, the man is Rita’s son.

Problem: - A man pointing to a photograph says, “The lady in the photograph is my nephew’s maternal grandmother.” How is the lady in the photograph related to the man’s sister who has no other sister?

Solution: - Clearly, the lady is the grandmother of man’s sister’s son i.e., the mother of the mother of man’s sister’s son i.e., the mother of man’s sister.

So, the lady is man’s mother.

Problem: - Pointing out to a lady, a girl said, “She is the daughter-in-law of the grandmother of my father’s only son.” How is the lady related to the girl?

Solution: - Father’s only son- My brother; Grandmother of my brother – My grandmother;

Daughter-in-law of my grandmother – My mother.

So, the lady is girl’s mother.
Some Unsolved Typical Relationship Problems

Problem: - Pointing to a lady, a man said, “The son of her only brother is the brother of my wife.” How is the lady related to the man?           (Answer: Sister of father-in-law)

Problem: - Introducing a man to her husband, a woman said, “His brother’s father is the only son of my grandfather.” How is the woman related to this man? (Answer: Sister)

Problem: - A, B, C, D, E, F and G are members of a family consisting of four adults and three children, two of whom, F and G are girls. A and D are brothers and A is a doctor. E is an engineer married to one of the brothers and has two children. B is married to D and G is their child. How is C related to A?

(Answer: C is A’s son.)

Boolean Algebra Complement

Boolean algebra operators are AND, OR, NOT and equal. The process of NOT operator is found the complement or opposite. Through the truth table we can learn Boolean algebra operation easily. Truth table is connected with the Boolean algebra. In the truth table the expression ‘true is represented as ‘T and the expression ‘false is represented as ‘F. Let us see Boolean algebra complement and their symbols in this article.
Boolean Algebra Complement:

Only one operand is present in the NOT operator. For example ~X here the symbol ~ is used to denote expression opposite or complement. The most important operation of operator NOT is complement or opposite.

Some rules in Boolean algebra complement:

Complement of PQ:

(PQ) = P + Q

Complement of P Q

(P Q) =P + Q

Complement of P + Q + R

(P + Q + R) = P * Q * R

Complement of PQR:

(PQR) = P + Q + R

Easy way to remember this Complement rule is put * for + and put + for *

For example:

Complement of PQ = > (PQ) = P + Q
Proof for Boolean Algebra Complement:

Prove:

Complement of P Q `!=` PQ

Proof:

Given

(P Q) `!=` PQ

LHS:

P Q

Let us consider P=1 and Q =0

F = P Q

F = 1 0

F= 0 1

F = 0

Take the complement of F

We get

F =1

RHS:

PQ

P=1 and Q =0

F = 1 0

F =0

LHS is not equal to RHS therefore (P Q) `!=` PQ

Prove:

Complement of P Q = P + Q

Proof:

Given

(P Q) = P + Q

LHS:

P Q

Let us consider P=1 and Q =0

F = P Q

F = 1 0

F= 0 1

F =0

Take the complement of F

We get

F =1

RHS:

P+Q

P=1 and Q =0

F = 1 + 0

F =1

LHS is equal to RHS therefore (P Q) = P + Q.

Prove:

(PQ) = P + Q

Proof:

Given:

(PQ) = P + Q

LHS:

(PQ)

Let us consider P=1 and Q =0

F = (1 0)

F = (0)

F =1

RHS:

P + Q

Let us consider P=1 and Q =0

F =1 + 0

F =0 + 1

F = 1

LHS is equal to RHS therefore (PQ) = P + Q

Applications of Integration Work

Integration is applied on a given function within certain limits to get various output.

Here let us see three areas of integration application work. In the normal sense, the integrations are applied to get the area under the given curve between certain limits.

1. Area under the curve f (x) between the limits ‘a’ and ‘b’ is given by`int_a^bf(x)dx`

2. The arc length of a curve f (x) between the limits ‘a’ and ‘b’ is given by `int_a^bsqrt[ 1 + [dy/dx] ^2] dx` .

3. The volume of a curve f(x) between the limits ‘a’ and ‘b’ is given by `pi int_ a^b y ^2 dx`

Now let us see problems on this topic.


Example Problems on Application of Integration Work.

Ex 1: Find the area under the curve f (x) = 2

4x 2 + 3x + 2 between 2 and 4.

Soln: Given: f (x) = 4x 2 + 3x + 2

The limits are ‘2’ and ‘4’.

Therefore Area =` int_2^4(4x ^2 + 3x + 2) dx` = `[4x ^3 / 3 + 3x ^2 / 2 + 2x] ` , the limit is between 2 to 4.

= `[(4 (4 ^3) / 3 + 3 (4) ^2 / 2 + 2 (4) ) ** (4 (2) ^3 / 3 + 3 (2) ^2 / 2 + 2 (2) )]`

Therefore Area = `96 2 / 3 ` sq units.

Ex 2: Find the arc length of a curve f (x) = 4x + 3 between 0 and 1.

Soln: Given: f (x) = 4x + 3

Therefore f ‘ (x) = 4.

Therefore arc length =` int_0^1 sqrt [(4) ^2 + 1] dx ` = `sqrt 17` `int_0^11 dx`

= sqrt 17 (x) between 0 to 1 = sqrt 17 (1 – 0) = sqrt 17 units.


More Example Problem on Application of Integration Work.

Ex 3: Find the volume generated when the finite region enclosed by the curve y = x ^2 and the lines x = 2, x = 4 and the x axis.

Soln: Volume = `pi int_2^4 y ^2 dx`

=` pi int_2^4 (x ^2) ^2 dx`

= `pi` [`x ^5 / 5` ] 2 to 4

= `pi / 5 [ 4 ^5** 2 ^5] = 992 / 5` cubic units.

Direct Inverse Relationship

Chemistry is study of physical and chemical properties of elements and compounds. While studying the same, various observations are made and inference derived from it. As such it involves discovery of correlations. Correlations means dependency. Dependency may be direct or inverse.

Establishing the dependencies scientifically is  very important as this can be converted to standard laws. These standard laws are useful in study of chemistry and chemical calculations
Direct Dependencies

Direct dependency means a parameter changes as the another changes. If one increases the another too increases.

e.g. An observation is made that the volume of a gas increases as the temperature increases. How can this direct dependendeny be presented mathematically?

V `alpha ` T

This equation is further rationalised as

V   = k T

where k is the constant for the corelation

OR

V / T =Constant

This is the established equation for Charles' Law

The direct dependencies may be proportional to 1st or 2nd or 3rd or even  1/2 or 1/4 etc. power too.

As in case of famous Einstein equation

E= mc2

lt denotes that the energy released is directly proportional not to the speed of light but, to the square of  the speed of light
Inverse Dependencies

Inverse dependency means a parameter changes as the another changes.If one increases the another decreases.

e.g. An observation is made that the volume of a gas increases as the temperature decreases. How can this inverse dependendeny be presented mathematically?

V `alpha ` 1/P

This equation is further rationalised as

V   =k / P

where k is the constant

for the corelation between volume and pressure..

or PV=Constant

This is the established equation for Boyles' Law

The inverse dependencies may be proportional to 1st or 2nd or 3rd or even  1/2 or 1/4 etc. power too

Rhombus Perimeter Formula

In geometry, a rhombus or rhomb is a quadrilateral whose four sides all have the same length. The rhombus is often called a diamond, after the diamonds suit in playing cards, or a lozenge, though the latter sometimes refers specifically to a rhombus with a 45° angle. (Source: Wikipedia)

Formula for perimeter:

Perimeter of a rhombus = 4 * Side length


Example problem 1:

A Rhombus has the side length is 17 m. Find the perimeter of the rhombus.

Solution:

Given side length of the rhombus is 17 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 17 m

= 68 m

Answer:

The final answer is 68 m

Example problem 2:

Find the perimeter of the given rhombus in centimeter.

Rhombus perimeter formula - Example problem 1

Solution:

Given side length of the rhombus is 12 cm

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 12 cm

= 48 cm

Answer:

The final answer is 48 cm

Example problem 3:

Find the perimeter of the given rhombus in meter.


Rhombus perimeter formula - Example problem 2

Solution:

Given side length of the rhombus is 9.1 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 9.1 m

= 36.4 m

Answer:

The final answer is 36.4 m

Example problem 4:

Find the perimeter of the given rhombus in meter.

Rhombus perimeter formula - Example problem 4

Solution:

Given side length of the rhombus is 7.7 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 7.7 m

= 30.8 m

Answer:

The final answer is 30.8 m


Practice Problems for Rhombus Perimeter Formula

Practice problem 1:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 1

Answer:

The final answer is 128 m

Practice problem 2:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 2

Answer:

The perimeter of the rhombus is 14.4m

Practice problem 3:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 3

Answer:

The final answer is 34 m

Revenue Formula Accounting

Revenue formula is defined as one of the important formula in accounting. Revenue is defined as the income of a company.  The income of revenue is mainly obtained from the sales.  The multiplication of selling price and the income is called as the revenue formula. The revenue is directly proportional to the quantity and it always from the linear manner. In this article, we are going to study to study about the revenue formula in accounting.


Explanation to Revenue Formula Accounting

The explanation to revenue formula accounting is given as follows,

Formula:

Revenue Formula = P `xx` Q

where,

P = Price per unit
Q = Quantity

Example Problems to Revenue Formula Accounting

Problem 1: Find the revenue by using the formula, where, P = 12500, Q = 200.

Solution:

Step 1: Given: The given values for finding the revenue is as follows,

P = 12500

Q = 200

Step 2: To find:

Revenue

Step 3: Formula: The formula given for finding the revenue is as follows,

Revenue Formula = P `xx` Q

Step 4: Solve:

Revenue Formula = P `xx` Q

= 12500   `xx`   200

= 2500000

Result: Revenue = 2500000

Therefore, this is the answer for solving the revenue formula in accounting.

Problem 2: Find the revenue by using the formula, where, P = 35000, Q = 250.

Solution:

Step 1: Given: The given values for finding the revenue is as follows,

P = 35000

Q = 250

Step 2: To find:

Revenue

Step 3: Formula: The formula given for finding the revenue is as follows,

Revenue Formula = P `xx` Q

Step 4: Solve:

Revenue Formula = P `xx` Q

= 35000   `xx`   250

= 875000

Result: Revenue = 875000

Therefore, this is the answer for solving the revenue formula in accounting.


Practice Problems to Revenue Formula Accounting

Problem 1: Find the revenue by using the formula, where, P = 45000, Q = 650.

Answer: 29250000

Problem 2: Find the revenue by using the formula, where, P = 55000, Q = 65.

Answer: 3575000

Problem 3: Find the revenue by using the formula, where, P = 43000, Q = 35.

Answer: 1505000

Algebra Geometry Practice

Algebra is a branch of math that deals with properties of lines, curves and surfaces generally divided into two types, Algebraic and geometry in agree with the mathematical techniques. The relationship between Algebra and geometry was introduced by Descartes.

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The ‘Geometry’ means study of properties. A point is used to denote a position in space. such that all points lying on the line joining any two points on the surface. The algebra geometry example problems and practice problems are given below.
Example Problems for Algebra Geometry

Example problem 1:

Find the values of a and b if 3x4 + x3 + ax2 + 5x + b is exactly divisible by x + 2 and x – 1.

Let P(x) = 3x4 + x3 + ax2 + 5x + b

(x+2) and (x–1) are factors of P(x) and hence both P(–2) and P(1) = 0

P(–2) = 3(–2)4 + (–2)3 + a(–2)2 + 5(–2) + b = 30 + 4a + b

Since P(–2) = 0 we get

4a + b = –30 (1)

P(1) = 3(1)4 + (1)3 + a(1)2 + 5(1) + b = 9 + a + b

Since P(1) = 0 we get

9 + a + b = 0 or a + b = –9 (2)

(1) – (2) gives

4a + b = –30

`(a+b=-9)/(3a = 21)`

` After solving this, We get,`

` a = 7`

Substituting a = –7 in equation (2) we get

–7 + b = –9 (or) ∴ b = –2

a = –7, b = –2.

Example problem 2:

Find the remainder when x3 – 5x2 + 7x – 4 is divided by (x – 1).

Solution:

P(x) = x3 – 5x2 + 7x – 4. By using the remainder theorem when P(x) is divided by (x – 1), the remainder is P(1).

The remainder P(1) = 13 – 5(1)2 + 7(1) – 4 = –1.

Example problem 3:

If all the sides of a parallelogram touch a circle prove that the parallelogram is a rhombus.

Solution:


Given: All the sides of a parallelogram ABCD touch a circle with centre O.

To prove: ABCD is a rhombus.

Proof: Since the lengths of the tangents from an external point to a given circle are equal (Fig)

AP = AS; BP = BQ; CR = CQ; DR = DS

Adding (AP+BP)+ (CR+DR) = (AS+DS) + (BQ+CQ)

AB + CD = AD + BC (Fig)

AB + AB = AD+AD (ABCD is a parallelogram CD= AB, BC=AD)

2AB = 2AD AB= AD

But AB= CD and AD = BC ( opposite sides are equal)

AB = BC = CD = AD. Hence ABCD is a rhombus.
Practice Problems for Algebra Geometry:

Practice problem 1:

Find the quotient and the remainder using synthetic division

1.  x3 + x2 – 3x + 5 ÷ x – 1

2. 4x3 – 3x2 + 2x – 4 ÷ x – 3

Answer: 1. Remainder = 4, Quotient = x2 + 2x + 1

2. Remainder = 83, Quotient = 4x2 + 9x + 29

Practice problem 2:

In the Fig AB is the diameter of the circle, ∠BAC = 42°. Find ∠ACD

Answer: m∠ACD = 48°