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How do you convert decalitres into centilitres?

We received this question from a IMO customer yesterday and the email inquiry indicated the grade of the interested party to be in Grade 3. Here is the response provided by one of our math teachers. If you need further clarification, please email us at support@indianmathonline.com. [NOTE: the word "liter" is sometimes spelled "litre". If you were raised in the British system of education, it is not uncommon to use the spelling "liter"]

1 kilolitre = 1000 liters

1 hectolitre = 100 liters

1 decaliter = 10 liters

1 liter = Basic Unit of Volume

1 deciliter =  1/ 10th of a liter

1 centiliter = 1 /100th of a liter

1 milliliter = 1 / 1000th of a liter

The word Deca (the form used in deciliter) is derived from the Greek word “dekas”, which means group of ten.

Therefore 1 decaliter = 10 liters.

The word Centi (in centiliter) is derived from the Latin word “centum”,meaning "hundred". The mathematical meaning is “one hundred” or “hundredth part”.

Thus, we say 1 centiliter = 1/100^th of a liter.

Since 1 centiliter = 1/100^th of a liter, 100 centiliters make 1 liter.

Thus, we say1 liter = 100 centiliters.

We know that 1 decaliter = 10 liters

= 10 X 1 liter (10 times 1 liter is nothing but 10 liters )
= 10 X 100 centiliters ( since 1 liter = 100 centiliters )
= 1,000 centiliters

Therefore, 1 decaliter = 1,000 centiliters

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Calculating the slope of a line

The slope of a line (m) is defined as the ratio of the “rise” divided by the “run” between two points on a line. Rise is represented by “y” on an axis, and run is represented by “x” on an axis.  The equation to solve for a slope of a line with two points (x1, y1) and (x2, y2) on that line is:

 

m = change in y                or            m = y2 – y1

         change in x                                        x2 – x1   

 

So as long as you know of two points that make a line you can solve for the slope of that line, m. 

 

Using the points (5, 2) and (-1, 4), solve for m.

 

m = y2 – y1         = 4 – 2                  = 2          = -1

         x2 – x1            -1 – 5                   -6             3

 

So the slope of this line is -1/3 which can be read as a decrease of 1 in the “rise” and an increase of 3 in the “run”. Or some people say this slope is “down 1, and over 3 to the right”. This is the preferred method over counting the rise and run on a graph because sometimes lines are hard to match up with exact points.

 

If you were instead given a line m = -1/3 and only one point on that line (5, 2), you can draw the line by first drawing a point on your graph at (5, 2). Then to draw your next point, start from (5, 2) and count “down 1, and over 3 to the right”, this points you at the point (8, 1). You know (8, 1) is correct because the original point of (5, 2) added to the slope gives (5 +3, 2 – 1) which equals (8, 1).  Now simply connect the line between your points of (5, 2) and (8, 1).

How do I manually determine the square root of 2795?

Indian Math Online recently received a request for help on this math question:

 

How do I manually determine the square root of 2795?

 

There are a number of ways of solving this problem….without using a calculator! First you have to understand what exactly a “square root of a number” is. A square root of a number is a special number that when multiplied by its self (or squared as we say), the result is the original number. Take the number 9 for example. The number 3 multiplied by itself gives a result of 9 (3 x 3 = 9). Another way of saying this is 3 squared is 9 (3² = 9). Therefore, 3 is the square root of 9 (3 = √9).

 

On a multiple choice test where 4 possible answers are given, you could easily just square the 4 answers to see which one is the correct answer. However, that is not very challenging and it takes away all the fun.  So if you are not given 4 possible answers to choose, and you don’t have a calculator, it is time to learn how to manually calculate square roots.

 

GUESSING

The first real method would be to make educated guesses! Most students learn what are called “perfect squares” when they are in middle school. Perfect squares are integers that have integers as a square root. The number 9 is a good example of this because its square root is 3. The number 10 is not a perfect square because its square root is 3.1623.  Students memorize these perfect squares, usually as high as 625 which has a square root of 25. So to apply the guessing method to finding the square root of 155, knowing that the two closest perfect squares are 144 (which has square root of 12) and 169 (which has a square root of 13), it is safe to say that the square root of 155 is somewhere between 12 and 13. That is as far as guessing can take you.

 

PRIME FACTORIZATION

The prime factorization method only works when trying to find the square root of a number which happens to be a perfect square. So unless the question specifies “Find the square root of the perfect square 784”, this method may be a waste of your time trying to use it properly. To find the square root of 784 which is in fact a perfect square, you need to break it down into its smallest factors. The number 784 can be factored into 2 x 2 x 2 x 2 x 7 x 7 (because 2 x 2 x 2 x 2 x 7 x 7 = 784). The next step would be to group these smallest factors into pairs: 2 x 2 is a pair, 2 x 2 is a pair, and 7 x 7 is a pair.  If you multiply one of each pair together, you get 2 x 2 x 7 = 28.  The square root of 784 is 28.

 

LONG DIVISION

Finding square roots by long division can be the most difficult method to understand at first, but it will always work for any number, even decimals! To solve the question asked by our student, “what is the square root of 2795”, we must first group 2795 into pairs of 2 digits (27 and 95) and write it like a long division problem:

   ______

  | 27 95      NOTE: If the number were 12549, it would be grouped as 1, 25, and 49

 

STEP 1: Find the largest number whose square is equal to or less than the first pair. In this case, that number would be 5 because its square is 25 which is less than 27. The number 6 is too big because its square is 36 which is greater than 27, so 5 is the largest number. We now write the 5 in three spots, and its square of 25 in one spot:

 

 

       5      

5  |27 95

5    25

 

STEP 2: Then on the left side you add the numbers, and on the right side you subtract them:

 

          5    

  5 | 27 95

+ 5  -25

  10    2

 

STEP 3: Next you want to carry down the next pair of numbers, 95, and put an x after the number 10.

 

          5      

  5 |  27 95

+ 5   -25

10x     295         NOTE:  10x represents a 3-digit number, and the remainder here is 295

 

STEP 4: You now want to determine the largest number x such that the number 10x times x is equal to or less than the remainder. In this case, that largest number x is 2 because 102 times 2 = 204 which is equal to or less than 295. The number 3 gives 103 times 3 = 309 which is greater than 295, so 2 is correct. Now write the 2 in three spots, and the 204 from earlier in one spot:

 

           5 2   

  5 |  27 95

+ 5   -25

102     295

   2     204

 

STEP 5: Then on the left side you add the numbers, and on the right side you subtract them:

 

          5 2   

  5 |  27 95

+ 5   -25

102     295

+ 2     -204

104       91

 

STEP 6: You have run out of numbers to bring down, so bring down two zeros (00) which are after the decimal points. Also put an x after the number 104.

 

          5 2 .    

  5 |  27 95.00

+ 5   -25

102     295

+ 2     -204

104x      9100

 

At this point you would continue repeating steps 4, 5, and 6 until you have a remainder of zero (0), or you have gotten as many numbers after the decimal as required. The square root of 2795 is 52.86775955…. and keeps going, so usually when you are asked to find the square root manually of a number such as 2795, the question will say to only go as far as 2 or 3 decimal places.

 

Indian Math Online (IMO) has materials that explain these methods in further detail and with more examples. You can see IMO’s “Solution Explanation” on how to perform all of these methods by going here.

 

Or you can watch one of IMO’s learning assignments on this topic here.

Ask the Expert - A question from one of our subscribers

An IMO subscriber recently asked for help on this question:

Mrs. Margaret weighs 85 lb and her daughter weighs 42 lb less than her. What is the difference of their weights?

85 - 42 = 43 pounds  (easy right?!)

WRONG!

Look again; if you read the question carefully, you will notice that the daughter is not 42 pounds, but that she weighs 42 pounds less than her mother who weighs 85 pounds.  So the daughter actually weighs 43 pounds, and 42 pounds is their difference in weight. In fact, you don't even have to subtract here at all.

Got you confused? A similar question would be: A book used to cost one dollar but it now costs 10 cents less. What is the difference in the two prices?  10 cents!

This question was submitted via the "ASK THE EXPERT" link which is found on the left side of the Indian Math Online home page. Anyone is welcome to use submit a question there, and we do our best to respond in a timely manner. If the question came from an IMO practice or assessment, please reference the question number which is found at the lower right corner of each question.

How can we apply the Gauss Elimination method to solve m question?

Recently a user to the Indian Math Online web site submitted a question for one of our math experts. The user who we will refer to only by first name - Amresh - submitted this question and picked "Grade 1" as the grade. We do not believe this is a question from a 1st Grader. This is a question for higher grades, perhaps even taught in early years of Engineering programs.

Here we go - the question first;

"How can we apply the Gauss Elemination method to solve m equation? For example z1 = ax^n + bx^(n-1) + ... + m (mod p) z2 = ax^n + bx^(n-1) + ... + m (mod p) ..................................... ..................................... ..................................... zm = ax^n + bx^(n-1) + ... + m (mod p) I want the mod p vale of a,b,c... "

Our answer follows - if after reading the answer, you have more questions or need further clarifications, please click here. A pdf version of the answer can be downloaded here.

Gaussian elimination method:

This is an elementary elimination method & it reduces the system of equations to an equivalent upper triangular system which can then be solved by back substitution.

Although quite general, we shall describe the method by considering the below mentioned system of equations for the sake of clarity & simplicity.

Let the system of equations be

axⁿ+bx^(n-1)+………………….+m(mod p)= z1

axⁿ+bx^(n-1)+………………….+m(mod p) = z2

axⁿ+bx^(n-1)+………………….+m(mod p)= zm

We first form the augmented matrix

a b m(modp) z1

a b m(modp) z2

a b m(modp) zm

    

To eliminate xⁿ from the second equation, we multiply the first equation by –a/a & then add it to the second equation. Similarly, to eliminate xⁿ from the third equation, we multiply the first equation by –a/a & then add it to the third equation. This procedure can be shown thus:

 

   a b m(modp) z1

-a/a   a b m(modp) z2

-a/a   a b m(modp) z3

    a b m(modp) zm

where –a/a & -a/a are called multipliers for the first stage of elimination. In this stage, it is assumed that the first element in the first equation a≠ 0. The first equation is called the pivotal equation & the first element a is called first pivot. At the end of the first stage, the augmented matrix becomes:

    a b m(modp) z1

  0 b’ m(modp)’ z2’

  0 b’ m(modp)’ z3’

 
-a/a   0 b’ m(modp) ‘ zm’

where b’,(modp)’……….. are all changed elements. The first b’ of the second equation is the new pivot & the multiplier is –b’/b’.

After completing all stages, we should have the upper triangular system:

 

   a b m(modp) z1

  0 b’ m(modp)’ z2’

  0 0 m(modp)’ z3’

   0 0    0  zm’

 

from which the values of xⁿ,x^n-1 can be obtained by back substitution.

Those values will be substituted in the original system to get a system of equations

which can be solved to get the values of a,b, c etc

It is clear that this method will fail if one of the pivots vanishes. In such a case , the method can be modified by rearranging the rows so that the pivot is non-zero. This procedure is called partial pivoting & can be implemented on a computer.

If this is not possible, then the matrix is singular & the equations have no solution.

In particular, the number of non zero diagonal elements in the final augmented matrix will represent the rank of the original matrix.

Simple Interest - is there a simple explanation?

An IMO User sent us this question on Feb 25, 2008 -

I was wondering If you could tell me a simple way to find out Simple Interest. I don't get it at all. If you don't know what I mean then I will give you a sample problem: S. Find the interest earned on a $490 deposited in a savings account at an 8 1/2% per year for 6 months? Thank you for your time!

----------------------------Explanation-------------------

To understand the concept of Simple Interest, we first need to understand the different components that are involved in the calculation of Simple Interest.

There are five important components of Simple Interest. They are

• Principal
• Term or Period.
• Rate of interest
• Interest
• Amount    
      

The money deposited by a depositor in a Bank or the money loaned by a Bank to a borrower is called the ‘PRINCIPAL’

The duration for which the money is kept with the Bank or the duration for which money is taken on loan from the bank is called ‘PERIOD or TERM’

Banks charge a fee on the principal loaned and this is called the Interest. Bank’s major income comes from interest charged on loans.  Banks also pay interest to depositors for any money deposited with them.  Thus, in order for Banks to be profitable, they charge more interest on loans when compared to the interest paid on deposits

The interest one gets from the bank or the interest charged by a Bank on a loan is normally expressed as a percentage for a year. This is called the ‘RATE of INTEREST’.

Interest is the extra amount payable to the bank or by the bank for the money deposited or loaned

Amount is the sum total of Principal and Interest payable at the end of the term or period.

 

To calculate Simple Interest:

Simple Interest is calculated on the principal, (or on that portion of the principal which remains unpaid), for the term or period and at the Rate of interest.

When money is borrowed, interest is charged for the use of that money for a certain period of time. When the money is paid back, the principal (amount of money that was borrowed) and the interest is paid back together.

If

P = Principal amount (The sum borrowed or the sum loaned)

T = Period or Term of Deposit or loan in years

R = Rate of Interest (The interest amount on every $100 for one year)

I = Interest paid by borrower to the money lender or to the bank for use of money (P) borrowed

A =the amount paid or payable at the end of period, then (A = P+I)

For simple Interest calculation we use the formula:

Simple Interest I= PxTxR/100

Let us analyze following statement -

Find the interest earned on a $490 deposited in a savings account at an 8 1/2% per year for 6 months?

Here,

Principal P= $490

Time T = 6 months = 6/12 years = ½ year

Rate of interest R = 8 ½ % = 17/2 %

Interest =?

Now, we use the formula to find the Simple Interest

Simple Interest I= PxTxR/100

Substituting the values of P,T & R in the above formula, we have

I = (490 x ½ x 17/2) / 100

 

  =  (490 x 17) / (100 x 2 x 2)

By simplification, we have

I = 20.825

Therefore, the interest earned on a $490 deposited in a savings account at an 8 1/2% per year for 6 months is $20.825

 

Unitary method for finding simple interest

What does 8 ½ % interest mean? (The bank agrees to pay $8 1/2 for every $100 every year)

It is given that for $100, the interest is $ 8 1/2 per year (8 1/2% interest)

For $490, the interest is 8 ½ x (490/100) = (17/2) x (49/10) = $41.65 per year

Since the deposit is $ 490, the interest earned is $41.65 every year

For 6 months (1/2 years), the interest earned is 41.65 x1/2 = $20.825

Thus, the interest earned on a $490 deposited in a savings account at an 8 1/2% per year for 6 months is $20.825

In banking terms this method of arriving at interest is called ‘Simple Interest’

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