In Number Theory Tutorial ,an expression or string of symbols is intended to represent a numerical value; a properly-formed expression may be evaluated in an unambiguous way. But in practice, an expression with multiple terms and operators may be written without parentheses, in which case the intended value of the expression is determined by convention. When we evaluate the expression is both preceded and followed by an operator such as minus or times, a convention is needed to clarify which operator should be applied first; this rule is known as a precedence rule, or more informally order of operation. From the earliest use of mathematical notation[citation needed], multiplication took precedence over addition, whichever side of a number it appeared on. Thus 3 + 4 × 5 = 5 × 4 + 3 = 23. When exponents were first introduced, in the 16th and 17th centuries, exponents took precedence over both addition and multiplication, and could be placed only as a superscript to the right of their base. Thus 3 + 5 2 = 28 and 3 × 5 2 = 75. To change the order of operations, a vinculum (an overline or underline) was originally used. Today one uses parentheses (). Thus, if one wants to force addition to precede multiplication, one writes (3 + 4) × 5 = 35.We can use order of operations calculator as well.
Mnemonics are often used to help students remember the rules. One common strategy is to build an acronym from the names of the operations. For instance, in the United States, the acronym PEMDAS (for Parentheses, Exponentiation, Multiplication/Division, Addition/Subtraction) is used, sometimes expressed as the sentence "Please Excuse My Dear Aunt Sally" or one of many other variations. In other English speaking countries, Parentheses may be called Brackets, and Exponentiation may be called Indices, Powers or Orders. Also, as Multiplication and Division are of equal precedence, M and D may be interchanged in the mnemonic. (This does not mean that multiplication and division can be performed in any arbitrary order, just that it is wrong to say "M always comes before D", or "D comes before M". See below.) The same comments apply to Addition and Subtraction.
Question:-
Solve 2 - [ (7-6) + (9-19)]
Answer:-
2 - [ (7-6) + (9-19)]
Here we have many signs.
When we have more than one sign ,we have to apply PEMDAS rule
= 2 - [ (1) + (-10)]
We have to solve inside the parenthesis first
Here (+) (-) = -
= 2 - [ 1 - 10]
= 2 - [-9]
Here (-) (-) = +
= 2+9
= 11 is the Answer
Monday, August 31
Wednesday, August 19
A problem on simple interest
Interest is a fee paid on borrowed assets. It is the price paid for the use of borrowed money, or, money earned by deposited funds.Assets that are sometimes lent with interest include money, shares, consumer goods through hire purchase, major assets such as aircraft, and even entire factories in finance lease arrangements. The interest is calculated upon the value of the assets in the same manner as upon money. Interest can be thought of as "rent of money". For example, if you want to borrow money from the bank, there is a certain rate you have to pay according to how much you want loaned to you.
Let's see a problem on this from 8th grade math worksheets
Question:-
What is the interest earned on # 97500 when rate of interest is 8% and for 5 months?
Answer:-
Here is the solution from answers to math problems
This problem based on simple interest.
The formula for simple interest is
R=rate of interest =8%
T=Time of year=5 months
=5/12 in years
I=interest amount
We have to put these values in formula
This is a part of elementary algebra practice .
Simple interest is calculated only on the principal amount, or on that portion of the principal amount which remains unpaid.
The amount of simple interest is calculated according to the following simple interest formula:
Let's see a problem on this from 8th grade math worksheets
Question:-
What is the interest earned on # 97500 when rate of interest is 8% and for 5 months?
Answer:-
Here is the solution from answers to math problems
This problem based on simple interest.
The formula for simple interest is
p x T x R
I= ------------
100
p= principle=$97500R=rate of interest =8%
T=Time of year=5 months
=5/12 in years
I=interest amount
We have to put these values in formula
97500 x 8 x 5
I= --------------
100 x 12
3900000
I = ----------
1200
I=3250
So the total interest earned on $97500 is $3250.This is a part of elementary algebra practice .
Thursday, August 13
Problem on mean median mode range
This math help is for mean median mode range worksheets
Mean, median, and mode are three kinds of "averages".
The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers.
The "median" is the "middle" value in the list of numbers. To find the median, your numbers have to be listed in numerical order, so you may have to rewrite your list first.
The "mode" is the value that occurs most often. If no number is repeated, then there is no mode for the list.
The "range" is just the difference between the largest and smallest values.
EXAMPLE:-
Find the mean, median, mode, and range for the following list of values:
8, 9, 10, 10, 10, 11, 11, 11, 12, 13
The mean is the usual average:
(8 + 9 + 10 + 10 + 10 + 11 + 11 + 11 + 12 + 13) ÷ 10 = 105 ÷ 10 = 10.5
The median is the middle value. In a list of ten values, that will be the (10 + 1) ÷ 2 = 5.5th value; that is, average of the fifth and sixth numbers to find the median:
(10 + 11) ÷ 2 = 21 ÷ 2 = 10.5
The mode is the number repeated most often. This list has two values that are repeated three times.10 and 11.
The largest value is 13 and the smallest is 8,
so the range is 13 – 8 = 5.
So
mean: 10.5
median: 10.5
modes: 10 and 11
range: 5
For more help on this ,you can reply me.
Mean, median, and mode are three kinds of "averages".
The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers.
The "median" is the "middle" value in the list of numbers. To find the median, your numbers have to be listed in numerical order, so you may have to rewrite your list first.
The "mode" is the value that occurs most often. If no number is repeated, then there is no mode for the list.
The "range" is just the difference between the largest and smallest values.
EXAMPLE:-
Find the mean, median, mode, and range for the following list of values:
8, 9, 10, 10, 10, 11, 11, 11, 12, 13
The mean is the usual average:
(8 + 9 + 10 + 10 + 10 + 11 + 11 + 11 + 12 + 13) ÷ 10 = 105 ÷ 10 = 10.5
The median is the middle value. In a list of ten values, that will be the (10 + 1) ÷ 2 = 5.5th value; that is, average of the fifth and sixth numbers to find the median:
(10 + 11) ÷ 2 = 21 ÷ 2 = 10.5
The mode is the number repeated most often. This list has two values that are repeated three times.10 and 11.
The largest value is 13 and the smallest is 8,
so the range is 13 – 8 = 5.
So
mean: 10.5
median: 10.5
modes: 10 and 11
range: 5
For more help on this ,you can reply me.
Monday, July 20
A word problem on percentages
Percentages are used to express how large one quantity is, relative to another quantity. The first quantity usually represents a part of, or a change in, the second quantity, which should be greater than zero. For example, an increase of $ 0.15 on a price of $ 2.50 is an increase by a fraction of 0.15 / 2.50 = 0.06. Expressed as a percentage, this is therefore a 6% increase.
Let's see a word problem on it.
Question:-
A waitress earn an average tip of 12% of the cost of the food she serves.If she serves $ 375 worth of food in one evening ,how much money in tips will she earn on average.
Answer:-
Let's see a word problem on it.
Question:-
A waitress earn an average tip of 12% of the cost of the food she serves.If she serves $ 375 worth of food in one evening ,how much money in tips will she earn on average.
Answer:-
Waitress serves food worth=$375For more help on this,you can reply me.
She earns a tip=12%
So.
12% of $375
12
----- * 375
100
12*375
---------
375
4500
= -------
100
=$45
So the waitress earn a tip of $45.
Tuesday, July 7
problem on solving linear equations
Topic:-linear equation
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.
Linear equations can have one or more variables. Linear equations occur with great regularity in applied mathematics. While they arise quite naturally when modeling many phenomena, they are particularly useful since many non-linear equations may be reduced to linear equations by assuming that quantities of interest vary to only a small extent from some "background" state.
Question:-
solve |5b-3| = |3b+7|
So b= 5 or -1/2 is the answer
For more help on this ,please reply me.
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.
Linear equations can have one or more variables. Linear equations occur with great regularity in applied mathematics. While they arise quite naturally when modeling many phenomena, they are particularly useful since many non-linear equations may be reduced to linear equations by assuming that quantities of interest vary to only a small extent from some "background" state.
Question:-
solve |5b-3| = |3b+7|
Answer:-
This math help is from linear equations.
This is of the form |a|=|b|
which implies a = +b
or a = -b
Using this we get
5b-3=+(3b+7)
5b-3=-(3b+7)
5b-3 =+(3b+7)| 5b-3 =-(3b+7)
|
5b-3 = 3b+7 | 5b-3 = -3b-7
|
bring 3b |
left | bring 3b left
hand side | hand side
|
5b-3b=7+3 | 5b+3b= -7+3
|
2b = 10 | 8b= -4
|
b = 5 | b = -4/8
|
| b = -1/2
So b= 5 or -1/2 is the answer
For more help on this ,please reply me.
Monday, June 22
How to solve time and work problems
Topic :- Time and Work
Time and work problems are very important for test preparation when you are out of school for long period .
The time will be given and the amount of work done in that time period needs to be calculated and vice versa .
Here is a simple Time word problem Given ,Which will give an idea of solving workwordproblems
as well.
Question :-
Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?
Answer:-
Jim can fill the pool in 30 min.
So in 1 min,jim can fill 1/30 th part of pool.
Similarly in 1 min,Sue can fill 1/45th part of pool.
Now 1 1/2 hours = 60+30=90 min.
So in 1 min,Tony can fill 1/90 th part of pool.
Now,if all of them work together,then in 1 min,the part of the tak filled is
Lcm=90
So by equalizing the deniminators
Hence,If they all work together ,The tank will be filled in 15 minutes.
For more math help,You can reply me .
Time and work problems are very important for test preparation when you are out of school for long period .
The time will be given and the amount of work done in that time period needs to be calculated and vice versa .
Here is a simple Time word problem Given ,Which will give an idea of solving workwordproblems
as well.
Question :-
Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?
Answer:-
Jim can fill the pool in 30 min.
So in 1 min,jim can fill 1/30 th part of pool.
Similarly in 1 min,Sue can fill 1/45th part of pool.
Now 1 1/2 hours = 60+30=90 min.
So in 1 min,Tony can fill 1/90 th part of pool.
Now,if all of them work together,then in 1 min,the part of the tak filled is
1 1 1
= --- + ---- + ----
30 45 90
Lcm=90
So by equalizing the deniminators
1*3 1*2 1*1
------ + ----- + -------
30*3 45*2 90*1
3 2 1
= ------ + ------ + ------
90 90 90
3+2+1
= ----------
90
6
= ------
90
1
= ----
15
Hence,If they all work together ,The tank will be filled in 15 minutes.
For more math help,You can reply me .
Tuesday, May 26
Problem on parallelogram, Finding Area of parallelogram
Geometry construction section in math consists of geometrical figures like parallelogram rectangle, trapezium where we find few terms like linear pair, supplementry angles.
If a pair of angles are adjacent as well as supplementary then they are said to form a linear pair
Topic : Area of parallelogram
Below example also prove how the necessary statements are considered.
Question : In the following figure, NDBA is a rectangle, AN = NG, BF = DB, GD = 11, and DF = 8√2. What is the area of parallelogram GDFA?
If a pair of angles are adjacent as well as supplementary then they are said to form a linear pair
Topic : Area of parallelogram
Below example also prove how the necessary statements are considered.
Question : In the following figure, NDBA is a rectangle, AN = NG, BF = DB, GD = 11, and DF = 8√2. What is the area of parallelogram GDFA?
State all the statements that are necessary in your proof along with their reasons in a two-column table.
Solution :
Friday, May 8
Question on Integrating a Given Function
Here is a Multi choice question to Evaluate function by Integration.
You can find similar set of questions at TutorVista Blogs.
Topic : Integrating a Function.
Below example will help you to identify correct choice and get reasons for, why remaining choices are incorrect.
Question : At what point in the interval [0, √3] does the function y=x2 - 1 assumes its average value.
a. -1
b. 1
c. 1.5
d. 1.7
Solution :
Choice b is correct.
Average value = 1/(√3 - 0)0∫√3 (x2 - 1)dx = 0
And x2 – 1 = 0
Therefore, y= x2 -1 assumes its average at x = 1
Choice a is incorrect as -1 is not in the interval [0, √3]
Choice c is incorrect as 1.5 does not satisfy the equation x2 -1=0
Choice d is incorrect as 1.7 does not satisfy the equation x2 -1=0
If you have any queries do write to us and you will surely get help from calculus help.
You can find similar set of questions at TutorVista Blogs.
Topic : Integrating a Function.
Below example will help you to identify correct choice and get reasons for, why remaining choices are incorrect.
Question : At what point in the interval [0, √3] does the function y=x2 - 1 assumes its average value.
a. -1
b. 1
c. 1.5
d. 1.7
Solution :
Choice b is correct.
Average value = 1/(√3 - 0)0∫√3 (x2 - 1)dx = 0
And x2 – 1 = 0
Therefore, y= x2 -1 assumes its average at x = 1
Choice a is incorrect as -1 is not in the interval [0, √3]
Choice c is incorrect as 1.5 does not satisfy the equation x2 -1=0
Choice d is incorrect as 1.7 does not satisfy the equation x2 -1=0
If you have any queries do write to us and you will surely get help from calculus help.
Friday, April 10
Question on Solving a Word Problem using Proportion Method
Topic : Proportion Method
Question : Nicole enlarged the size of rectangle to a height of 28 in. What is the new width if it was originally 6 inch wide and 7 inch tall?
a). 24in. b). 112 in. c) 2in. d) 7in.
Solution :
Forming the proportion :
6 : 7 = X :28
Using law of proportion, that product of extremes is equal to the product of means
6 x 28 = 7 x X
X = 28 x 6 /7
X = 168 / 7
X = 24
So, the width is 24 inch.
Question : Nicole enlarged the size of rectangle to a height of 28 in. What is the new width if it was originally 6 inch wide and 7 inch tall?
a). 24in. b). 112 in. c) 2in. d) 7in.
Solution :
Forming the proportion :
6 : 7 = X :28
Using law of proportion, that product of extremes is equal to the product of means
6 x 28 = 7 x X
X = 28 x 6 /7
X = 168 / 7
X = 24
So, the width is 24 inch.
Monday, April 6
Question on Percentage of Sugar Solution
Topic : Word Problem on Percentage
Question : How many gallons of 15% sugar solution must be mixed with 5 gallons of 40% sugar solution to make 30% sugar solution?
Solution :
%sugar ---gallons
15% -------- x
40% -------- 5
30 % ------- x+5
so, 15% of x+ 40% of 5 = (x+5) 30%
0.15x+ 2 = 0.3x+15
x= 0.5/0.15
x= 10/3
10/3 gallons of 15% sugar solution is mixed with 5 gallons of 40% sugar solution to
make 30% sugar solution.
Question : How many gallons of 15% sugar solution must be mixed with 5 gallons of 40% sugar solution to make 30% sugar solution?
Solution :
%sugar ---gallons
15% -------- x
40% -------- 5
30 % ------- x+5
so, 15% of x+ 40% of 5 = (x+5) 30%
0.15x+ 2 = 0.3x+15
x= 0.5/0.15
x= 10/3
10/3 gallons of 15% sugar solution is mixed with 5 gallons of 40% sugar solution to
make 30% sugar solution.
Thursday, April 2
Thursday, March 26
Monday, March 23
Wednesday, March 18
Upper Bound and Lower Bound of Polynomials
Friday, March 13
Problem to Find the Unknown values of Geometrical Figure
Topic : Square Based Triangle
Question : The Square based pyramid depicted below has averticle height of 13cm. K is the mid point of CD and angle AKL =50 degrees. To the nearest tenth of degree, what is the measure of angle ACK?
Solution :

Length of Square be x
Tan 50º = 1.19175
= 13 / (x/2)
x = 21.81659041
x = 21.82
Sin 50º = (x/2) / AK
AK = (x/2) / 0.766
AK = 14.24
Tan ∟AKC = AK /(x/2) = 14.24/ 10.91 = 1.31
∟AKC = tanˉ¹(1.31)
∟AKC = 53º
Question : The Square based pyramid depicted below has averticle height of 13cm. K is the mid point of CD and angle AKL =50 degrees. To the nearest tenth of degree, what is the measure of angle ACK?
Solution :
Length of Square be x
Tan 50º = 1.19175
= 13 / (x/2)
x = 21.81659041
x = 21.82
Sin 50º = (x/2) / AK
AK = (x/2) / 0.766
AK = 14.24
Tan ∟AKC = AK /(x/2) = 14.24/ 10.91 = 1.31
∟AKC = tanˉ¹(1.31)
∟AKC = 53º
Monday, March 9
Problem to Find Measurement of Geometrical Figures
Topic : Trapezium
Question : The area of a trapezium shaped field is 960 m^2 , the distance between two parallel sides is 10 m and one of the parallel side is 40m. Find the other parallel side.
Solution :
One of the parallel side a = 40 m, let another parallel side be x, distance between sides (h) = 10 m
The given area of trapezium = 960m^2.
Area of a trapezium = ½ h (a + b)
so 960 = ½ h (a + b) = ½ x 10 x (40 + x)
or 960 = 5( 40 + x)
or 960/5 = 40 + x
or 192 = 40 + x
or 192 – 40 = x
x = 152 m
the other parallel side of the trapezium is 152 m.
Question : The area of a trapezium shaped field is 960 m^2 , the distance between two parallel sides is 10 m and one of the parallel side is 40m. Find the other parallel side.
Solution :
One of the parallel side a = 40 m, let another parallel side be x, distance between sides (h) = 10 m
The given area of trapezium = 960m^2.
Area of a trapezium = ½ h (a + b)
so 960 = ½ h (a + b) = ½ x 10 x (40 + x)
or 960 = 5( 40 + x)
or 960/5 = 40 + x
or 192 = 40 + x
or 192 – 40 = x
x = 152 m
the other parallel side of the trapezium is 152 m.
Wednesday, March 4
Question on Linear Equation
Topic : Word Problem on Linear Equation.
Question : The average of four numbers is 85.If three of the numbers are 76,78 and 81,whats the fourth number
Solution :
76+78+81+x/4=85
235+x/4= 85
235+x = 340
X= 340-235
X= 105
The fourth number is 105.
Question : The average of four numbers is 85.If three of the numbers are 76,78 and 81,whats the fourth number
Solution :
76+78+81+x/4=85
235+x/4= 85
235+x = 340
X= 340-235
X= 105
The fourth number is 105.
Monday, February 23
Question on Personal Finance Concepts
Topic : Jobs and Growth Tax Relief Reconciliation Act
Question : Bernie and Pam Britten are a young married couple beginning careers and establishing a household. They will each make about $50,000 next year and will have accumulated about $40,000 to invest. They now rent an apartment but are considering purchasing a condominium for $100,000. If they do, a down payment of $10,000 will be required.
They have discussed their situation with Lew McCarthy, an investment advisor and personal friend, and he has recommended the following investments:
• The condominium - expected annual increase in market value = 5%.
• Municipal bonds - expected annual yield = 5%.
• High-yield corporate stocks - expected dividend yield = 8%.
• Savings account in a commercial bank-expected annual yield = 3%.
• High-growth common stocks - expected annual increase in market value = 10%; expected dividend yield = 0.1.
Calculate the after-tax yields on the foregoing investments, assuming the Brittens have a 28% marginal tax rate (based on Public Law 108-27, The Jobs and Growth Tax Relief Reconciliation Act of 2003).
Solution :
a)The yield on the Brittens' investment in condominium can be taken as zero since the investment is for personal use and as such no income will be derived out of it. The yield can be realised only on liquidation of the asset.
b) As regards other assets the after tax yields have been worked out as under:
(While arriving at the yield, it has been assumed that the entire 40000 is invested in a particular asset)
Asset Yield Tax After Tax
Municipal bonds 2000 560 1440
High-yield
corporate stocks 3200 896 2304
Savings account 1200 336 864
High-growth
common stocks 4000 1120 2880
c)The yield in respect of High growth common stocks would be realised only if the assets are liquidated.
Hope the explanation will be helpful for you.
Question : Bernie and Pam Britten are a young married couple beginning careers and establishing a household. They will each make about $50,000 next year and will have accumulated about $40,000 to invest. They now rent an apartment but are considering purchasing a condominium for $100,000. If they do, a down payment of $10,000 will be required.
They have discussed their situation with Lew McCarthy, an investment advisor and personal friend, and he has recommended the following investments:
• The condominium - expected annual increase in market value = 5%.
• Municipal bonds - expected annual yield = 5%.
• High-yield corporate stocks - expected dividend yield = 8%.
• Savings account in a commercial bank-expected annual yield = 3%.
• High-growth common stocks - expected annual increase in market value = 10%; expected dividend yield = 0.1.
Calculate the after-tax yields on the foregoing investments, assuming the Brittens have a 28% marginal tax rate (based on Public Law 108-27, The Jobs and Growth Tax Relief Reconciliation Act of 2003).
Solution :
a)The yield on the Brittens' investment in condominium can be taken as zero since the investment is for personal use and as such no income will be derived out of it. The yield can be realised only on liquidation of the asset.
b) As regards other assets the after tax yields have been worked out as under:
(While arriving at the yield, it has been assumed that the entire 40000 is invested in a particular asset)
Asset Yield Tax After Tax
Municipal bonds 2000 560 1440
High-yield
corporate stocks 3200 896 2304
Savings account 1200 336 864
High-growth
common stocks 4000 1120 2880
c)The yield in respect of High growth common stocks would be realised only if the assets are liquidated.
Hope the explanation will be helpful for you.
Tuesday, February 10
Question on how to Solve the Probability
Topic : Probability
Question : How to calculate percentile rank for individualseries and group frequency distribution?
Answer :
Computing the Pth Percentile
1. Sort the data.
2. The Pth percentile is found in position P(n + 1)/100.
3. If this position is between observations round to the nearest position (or, if you know how, interpolate between the two).
Example
Let's find the 15th percentile of the starting salary data. Here's the sorted version.
08820 10800 12000 12500 13000 14000 15000 16000 16500 16600 16700 16900 16900 17000 17000 17600 17880 18000 18000 18000 18000 18000 18000 18000 18000 18000 18000 18500 18680 19100 20000 20000 20000 20000 20000 20300 20900 22000 23000 23000 23000 23000 23400 24000 25000 25000 26000 26000 27000 30000 30000 32500 37000 48000
Our position is 15(54 + 1)/100 = 15(55)/100 = 8.25. Round to position 8; the corresponding observation is shown in bold. The 15th percentile is approximately 16000.
A better approach uses interpolation. The 8.25 position is really between the 8 and 9 positions, One-quarter of the way from 16000 to 16500 is
0.75(16000)+0.25(16500) = 16125.
So 16125 is a (better) approximation to the 15th percentile. However, it's not that different from 16000, and either would suffice.
Hope the Explanation will be helpful for you.
Question : How to calculate percentile rank for individualseries and group frequency distribution?
Answer :
Computing the Pth Percentile
1. Sort the data.
2. The Pth percentile is found in position P(n + 1)/100.
3. If this position is between observations round to the nearest position (or, if you know how, interpolate between the two).
Example
Let's find the 15th percentile of the starting salary data. Here's the sorted version.
08820 10800 12000 12500 13000 14000 15000 16000 16500 16600 16700 16900 16900 17000 17000 17600 17880 18000 18000 18000 18000 18000 18000 18000 18000 18000 18000 18500 18680 19100 20000 20000 20000 20000 20000 20300 20900 22000 23000 23000 23000 23000 23400 24000 25000 25000 26000 26000 27000 30000 30000 32500 37000 48000
Our position is 15(54 + 1)/100 = 15(55)/100 = 8.25. Round to position 8; the corresponding observation is shown in bold. The 15th percentile is approximately 16000.
A better approach uses interpolation. The 8.25 position is really between the 8 and 9 positions, One-quarter of the way from 16000 to 16500 is
0.75(16000)+0.25(16500) = 16125.
So 16125 is a (better) approximation to the 15th percentile. However, it's not that different from 16000, and either would suffice.
Hope the Explanation will be helpful for you.
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