Pan Balance Math Problems

A balance is a beam balance or laboratory balance or pan balance used to measure the accurate value of mass of an object. A weighting pan and scale pan span is hold above the machine. Pan balance is used to distribute weight between two containers in a machine balance.  By using pan balance we have to calculate the weight of the objects and evenly distribute the objects. Let us see about pan balance math problems in this article.


Worked Examples to Pan Balance Math Problems

Example 1 to Pan Balance Math Problems:

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the two rectangles in the above problem?

Solution:

Step 1:

Given weight in left side object = 6 kg

Given weight in the right side object = 11 kg

Step 2:

In left side, the pan is not balanced if we remove the two rectangles.

So, subtract the right side object weight – left side present object weight, we will get the result.

Step 3:

Subtract the amount on left side object from the right side object, we get,

11 – 6 = 5 kg

Step 4:

Two rectangles have the total of 5 kg.

Step 5:

Two rectangles in the left side of the pan balance are of the same size.

Therefore, each rectangle is of 2.5 kg.

Example 2 to Pan Balance Math Problems:

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the diamond object in the above problem?

Solution:

Step 1:

Total weight in the right side of the pan balance = 17 kg

Triangle object = 4 kg

Square object = 8 kg

Diamond object =?

Step 2:

In left side, the pan is not balanced if we remove the diamond object.

So, subtract the right side object weight – left side present object weight, we will get the result.

Step 3:

Adding the triangle and square shaped object weight = 4 + 8 = 12

Left side = 12 kg

Step 4:

Subtracting the right side objects weight – left side objects weight = 17 – 12 = 5 kg

Step 5:

Therefore, the weight of the diamond object is 5 kg.



Practice Problems to Pan Balance Math Problems

Solve the pan balance math problem.

Pan Balance Math Problems

Find the weight of the diamond object in the above problem?

Solution:

1. Pan balance weight in the right side is 10 kg.

Proving Identities Calculator

An identity is a relative which is tautologically factual. These are typically taken to denote something that is factual by meaning, either straight by the definition, or as a result of it. For model, algebraically, this occurs if an equation is content for all values of the concerned variables. Definitions are frequently indicated by the 'triple bar' sign ≡, such as A2 ≡ x·x. A calculator is a little (often small), typically cheap electronic machine used to do the basic operations of mathematics. Present calculators are additional moveable than most computers, although most PDAs are similar in size to handheld calculators.


Proving Algebra and Trigonometric Identities Calculator:

proving identities calculator

c + 0 = c (additive identity)
c + (-c) =0 (additive inverse)
c x 1 = c (the multiplicative identity)
c  (1/c )= 1 (the multiplicative inverse)
sin2a≡ 1 – cos2a
cos2a≡ 1 – sin2a
tan2a ≡ sec2a − 1

Examples of Proving Algebra Identities:

Study example 1:

21+ 0 = 21

Solution:

(We know that the additive identity in algebra (c+0 = c). Therefore the answer is 21)

Study example 2:

23+(-23) = 0

Solution:

(We know that the additive inverse in algebra (c+(-c) = 0) this problem answer 0.)

Study example 3:

1x14= 14

Solution:

(We know that the multiplicative identity  (1x c = c). So this problem answer 14.)

study example 4:

23 (1/23) = 1

Solution:

(We know that the multiplicative inverse  (c x 1/c = 1). So this problem answer is 23.)

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Examples for Proving Trigonometric Identities:

Study example 1:

Prove the following identity: (sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = tan M cot M

Solution:

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = tan M cot M

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 =(sin M/cos M)xx(cos M/sinM)

(sin M)4 +2(sin M)2(cos M)2 + (cos M)4 = 1
((sin2M)2 + (cos2M)2)((sin2M)2 + (cos2M)2) = 1       { (a+b)2=a2+b2+2ab}

( (Sin2M)+cos2M )2 =1

(1)(1) = 1

Study example 2:

Prove the following identity: tan(p)cos(p) = sin(p)

Solution:

tan(p)cos(p) = sin(p)

(sinP/cosP) cos p= sin(p)

sin(p) = sin(p)

Logic Problems With Solutions

logic problems with solutions set 1

Problem: - If in the code, ALTERED is written as ZOGVIVW, then in the same code, RELATED would be written as?

Solution: - Clearly, each letter of the ALTERED is replaced by the letter which occupies the same position from the other end of the English alphabet, to obtain the code. Thus, A, the first letter of the alphabet, is replaced by Z, the last letter. L, the 12th letter from the beginning of the alphabet, is replaced by O, the 12th letter from the end. T, the 7th letter from the end of the alphabet is replaced by G, the 7th letter from the beginning of the alphabet and so on.

Similarly, in the word RELATED, R will be coded as I, E as V, L as O, A as Z, T as G and D as W. Thus, the code becomes IVOZGVW.                            (Answer)

Problem: - Pointing to a photograph, a lady tells X, “I am the only daughter of this lady and her son is your maternal uncle.” How the speaker is related to X’s father?

Solution: - Clearly, the speaker’s brother is X’s maternal uncle. So, the speaker is X’s mother or his father’s wife.

Problem: - In a school, there were five teachers. A and B were teaching Hindi and English. C and B were teaching English and Geography. D and A were teaching Mathematics and Hindi. E and B were teaching History and French. Answer the following questions:

1)      Who among the teachers was teaching maximum number of subjects?

2)      Who among the teachers, teaches Hindi?

3)      D, B and A were teaching which subjects?

Solution: - The given information can be analysed as under:

Prob 1


Clearly, B teaches maximum number of subjects, i.e. 5.

Three teachers were teaching Hindi – A, B and D.

D, B and A were teaching Hindi.
Logic Problems with Solutions Set 2

Problem: - A man walks 1 km towards East and then he turns to South and walks 5 km. Again he turns to East and walks 2 km, after this he turns to North and walks 9 km. Now, how far is he from his starting point?

Solution: - The movements of the man are as shown in the figure. (A to B, B to C, C to D, D to E).

Prob2

Clearly, DF = BC = 5 km.

EF = (DE – DF) = (9-5) km = 4km

BF = CD = 2 km

AF = AB + BF = AB + CD = (1 + 2) km = 3 km.

Therefore man’s distance from starting point A

= AE = `sqrt(AF^2 +EF^2)`  = `sqrt(3^2 + 4^2)`

= `sqrt25`   =   5 km                                      (Answer)

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Problem: - Which of the following diagrams correctly represents the relationship among the classes: Tennis fans, Cricket players, Students?

Prob2

Solution: - Some students can be cricket players. Some cricket players can be tennis fans.

Some students can be tennis fans. So, the given items are partly related to each other. So the correct diagram which represents the relationship between the above is option (A).

Loci and Concurrency Theorems Tutorial

In this tutorial, we will study about definition of loci and concurrency, loci and concurrency theorems.

Definition of Loci:

If a point moves in some way which satisfies some given geometrical condition at every instant during its motion, then the path traced out by the moving point is called loci of a point.

Every point satisfies the given geometrical condition is a point of loci, and

Every point of loci should satisfy the given geometrical condition.

Singular form of loci is called as locus.

Definition of concurrency:

If three or more than three lines pass through a point, then the common point is called point of concurrency.

Let us see about loci and concurrency theorems in this tutorial.
Tutorial - Loci Theorem:

Definition:

The loci of a point which is equidistant from the two given points, is the perpendicular bisector of the line segment and joining the two given points.

Given:

There are two given fixed points M and N.

A is a moving point that is A is loci of two given points M and N.

AM = AN.

To Prove:

The loci of A is the perpendicular bisector of MN line segment.

Construction:

Triangle-Loci Theorem

Join MN. Bisect MN at B. Join AM, AN and AB.

Proof:

Therefore, A moves such that AM = AN.

As we know that,

A must pass through the point B of the line segment MN, because BM = BN.

Thus A passes through the mid-point of MN.

Now, in ∆MAB and ∆NAB,

Given    AM = AN

By construction, BM = BN

AB = AB

By sss congruence rule,

∆MAB = ∆NAB

angle MBA = angle NBA       (1)

But angle MBA + angle NBA = 180°  (2)

Substitute (1) into (2),

2 angle MBA = 180°

Divide by 2 each side.

angle MBA = 90°

Therefore, AB _|_ MN

Thus, AB is proved as perpendicular bisector of MN.
Tutorial - Concurrency Theorem:

Definition:

A concurrency point of the perpendicular bisectors of sides of a triangle is called circum-center of the triangle.

A concurrency point of the median of a triangle is called centroid of the triangle.

The concurrency diagram is shown in triangle figure beliow.

Triangle-Concurrency Theorem

A,B,C are mid-points of PQ, QR and QR respectively. PB, QC and RA are medians of ∆PQR. Here O is Concurrency point.

As we know from congruence rule,

OB _|_ QR

OA _|_ PQ

OC _|_ PR

These are all perpendicular bisectors of triangle. Therefore, O is concurrency point.

Primary Series Sequence

The sequence is a set of things that are in order. The sequence goes on to infinite number and it is called an infinite sequence. Otherwise it is finite sequence. They are different sequences like arithmetic, geometric, Triangular sequences these are special sequences. Series representation of sequence depends on the difference between two digits and the difference should be same thought of sequence then represent with mathematical induction formulas and simplify the equation

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Problems on Primary Series Sequence:

The series and sequence contains arithmetic progression, geometric progression, sum of first n natural numbers and sum of square of n natural numbers etc.  The following problems based on these concepts.

Example 1 in primary series sequence:

Problem:

To find general term for given sequence, suggest possible next three terms. The sequence is 8, 16, 24…

Solution:

The given sequence is 8, 16, 24….

The first term = 8.

The common difference is = 16-8=8.

The next three terms in the given sequence =24+8=32, 32+8=40, 40+8=48,

The series are 8, 16, 24, 32, 40, and 48.

Example 2 in primary series sequence:

Problem:

To find the numbers the sum of 3 numbers in GP is 16 and their product is 64.

Solution:

Let the three numbers x/y, x, xy.

x/y + x+ xy =16

(x/y) x (xy) =64.  X3 = 64; x= 4.

x/y(1 + y +y2 )=16

4(1 + y +y2 ) =16y.

4 y2  + 4y -16y +4=0

y2  -3y+1=0

y=2.

Where y=2 then the numbers are 2, 4, 8.

Example 3 in Primary Series Sequence:

Problem:

To find general term for given sequence, suggest possible next four terms. The sequence is 3, 6, 9…

Solution:

The given sequence is 3, 6, 9….

The first term = 3.

The common difference is = 6-3=3.

The four terms in the given sequence =9+3=12, 12+3=15, 15+3=18, 18+3=21.

The series are 3, 6, 9, 12, 15, 18, and 21.

Example 3 in primary series sequence:

Problem:

To find sum to 5th in the sequence numbers in the 2,4,8..

Solution:

Here first term a=2

The ratio r = 4/2=2

Sum =  arn-1

n=5, sum = 2(25-1) =32.

Practice Doing Functions in Math

In mathematics, function is nothing but depending on the input value, the output value to be determined. We can map the variables of the function into a coordinate system. Function is denoted in the form of y = f(x). If a function contains only one variable then it is called as one variable function. For example: f(x) = 12x + 4. Here, x is a variable. Now, we are going to discuss some of the problems for doing functions in math.


Example Problems- Practice Doing Functions Math:

Example problem 1:

Find the ordered pairs of the function: f(x) = 10x + 2

Solution:

f(x)= 10x + 2

Substitute x=0

f (0) =  10(0) + 2

y = 2

Therefore the ordered pair (x, f(0)) is (0, 2).

Substitute x = 1

f(1) = 10(1) + 2

y = 12

Therefore the ordered pair (x, f(1)) is (1, 12).

Substitute x=2

f(2) = 10(2) + 2

y = 22

Therefore the ordered pair (x, f(2)) is (2, 22).

Substitute x=3

f(3) = 10(3) + 2

y = 32

Therefore the ordered pair (x, f(3)) is (3, 32).

The ordered pairs of the function f(x) = 10x + 2 is (0, 2), (1, 12), (2, 22), (3, 32).

Example problem 2:

Find whether the relation {(-2, 4), (-1, 6), (-1, 8), (0, 10)} is a function?

Solution:

The given ordered pair is {(-2, 4), (-1, 6), (-1, 8), (0, 10)}.

Functions / relations

Here, more than one ordered pair with the same x coordinate, but with different y coordinates. The ordered pairs (-1, 6) and (-1, 8) have the same x coordinate and y coordinates are different (i.e.,) two y coordinate 6 and 8 corresponds to a single x coordinate -1. Therefore, this relation cannot be a function.

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Additional Problems- Practice Doing Functions Math:

Example problem 3:

Find the zeros of the one variable function f(x) = x2 - 173x + 172.

Solution:

Set the equation equal to zero.

0 = x2 -173 x + 172

Factor the quadratic function and solve for x.

Here, a = coefficient of x2 = 1

b = coefficient of x = -173

c = constant term = 172

We find a × c = 1× 172 = 172 = -1*-172, (-1) + (-172) = -173 = b.

x2 - 173x + 172 = 0

x2 + (- 1 - 172)x + 162 = 0

x2 – 1 x – 172 x + 172 = 0

x (x - 1) – 172 (x - 1) = 0

 (x – 1) (x – 172) = 0

x = 1, 172

So, the zeros occur when x equals 1 and 172.

Example problem 4:

Is the function f(x) = 10x + x3 even function or odd function?

Solution:

f(x) = 10x + x3

Substitute the value –x in the place of x.

f(-x) = 10(-x) + (-x)3

f(-x) = -10x – x3 = -(10x + x3)

f(-x) = -f(x)

So, the given function f(x) = 10x+ x3 is an odd function.


Practice Problems for Doing Functions in Math:

1) Find the zeros of the one variable function f(x) = x2 - 172x + 171. (Answer: 1, 171).

2) Is the function f(x) = 2x2 + x4 even function or odd function? (Answer: even function).

Precalculus Trigonometry

Precalculus is an advanced form of  the secondary schools algebra, is a foundational mathematical discipline. It is also called Introduction to the Analysis. In many school, precalculus is actually two separate courses: Algebra and Trigonometry.

Trigonometry is a branch of a mathematics that studies triangles, particularly right triangles. Trigonometry deal  with the relationships between the sides and the angles of triangles, and with trigonometric functions, which describe those relationships and angles in general, and the motion of waves such as sound and light waves.                                           (Source.Wikipedia)
Precalculus with Trigonometry Identities:

sin2x+ cos2 x≡ 1

1 + tan2x≡ sec2 x

1 + cot2x ≡ cosec2 x

sin2x≡ 1 – cos2x

cos2x≡ 1 – sin2x

tan2x ≡ sec2x − 1

sec2x − tan2x ≡ 1

cot2x ≡ cosec2x − 1

cosec2x − cot2x ≡ 1
precalculus trigonometry ratios:

sin (90° − x) = cosx

cos (90° − x) = sin x

tan (90° − x) = cot x

cot (90° − x) = tan x

sec (90° − x) = cosecx

cosec (90° − x) = sec x
Examples for Precalculus with Trigonometry:

Example 1:

Find x° if cosec x° = sec 55°.

Solution:

Since cosec x° =sec(90°− x°),we have sec(90°− x°)= sec 45°. ∴ 90°− x°=45°.

∴ x° = 90°− 45°= 55°.

Example 2:

Evaluate sin 30° tan 60° sec 30°

Solution:

sec 30° = sec(90°−60°) = cosec 30°=1/sin 30°

∴sin 30° tan 60° sec 70° = sin 30° tan 60° cosec 30°

=sin 30° × 3× 1/sin 30°  = 3

Examples for trigonometry identities:

Problem 1:

Prove that sin4a + cos4 a= 1 – 2sin2a cos2a.

Solution:

LHS = sin4a + cos4a = (sin2a) 2 + (cos2a) 2

= [sin2a + cos2a] 2 – 2 (sin2a) (cos2a)                 ( a2 + b2 = (a + b)2 – 2ab)

= (1)2 – 2sin2a cos2a

= 1 – 2sin2a cos2a

= RHS

Hence proved

Problem 2:

Prove that sin4b − cos4b = sin2b − cos2b

Solution:

LHS = sin4b − cos4b = (sin2b) 2 – (cos2b) 2

= (sin2b + cos2b) (sin2 b − cos2b) = (1) (sin2b − cos2b)

= sin2b − cos2b = RHS

Problem 3:

Prove that (sec c+ cos c ) (sec c − cosc ) = tan2c + sin2c.

Solution:

LHS = (sec c+ cosc ) (sec c − cosc ) = sec2c − cos2c

= (1 + tan2c) – cos2c= tan2c + (1 – cos2c)

= tan2c + sin2c = RHS