Class 5 Aryabhatta : Maths paper 2008 with answers

Solved Class V question paper of Aryabhata interschool maths competition 2008.

Aryabhatta_QP_2008

Solution 

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Arithmetic answers                      

aryabhattaanswers

Geometry answers

Vedic Maths: Base and Complement

Base and Complement are very important in Vedic Mathematics and form the basis of many calculations.

Base:

As explained above we work in a base 10 number system. In order to ease our calculations we can take any number ending with ‘0’ i.e. any multiple of 10 as our base.

COMPLEMENT

The Complement of a number is the difference between that number and the next higher power of 10.

10’s complement:

This includes all 1 digit numbers. It is the number that should be added to make it 10.

The complement of 6 is 4.
The complement of 7 is 3.

100’s complement:

This includes all 2 digit numbers. It is the number that should be added to make it 100.

The complement of 55 is 45.
The complement of 89 is 11.

Finding a complement of a number:

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Grade 4 Maths : Simple equations

Balance as a model of an equation

This article discusses how to use a balance to model simple linear equations in pre-algebra or algebra 1. On this page, we only deal with positive integers;

An equation is basically saying that two things (expressions, to be exact) are EQUAL. Since in a balanced situation the two sides of the balance hold equal weight, we can model simple equations with a balance.

In the pictures below, each circle represents one and the block represents the unknown x. To find out what the block weighs, you can

  • add the same amount (circles or blocks) to BOTH sides
  • take away the same amount from BOTH sides

That way both sides will maintain the balance or “the equality”.

x + 3 = 5

If this is a balanced situation…

x = 2

…so is this!
(We took away three circles from BOTH sides.)

3+ 2 = 2x + 6

Take away two blocks (two x‘s) from both sides. The balance will stay balanced.

x + 2 = 6

Take away 2 circles from both sides.The balance will stay balanced.

x = 4

Here is the solution!

Without the scale model, the solving process looks like this:

 3x + 2
-2x
=  2x + 6
-2x
(take away 2x from both sides)
x + 2
-2
= 6
-2
(take away 2 from both sides)
x = 4

Dividing

In some situations you have to divide both sides of the equation by the same number.  When is that? It’s in the fortunate situation where there are ONLY x’s (blocks) on one side but there are more than one.

2x = 8

If you take away half of the things on the left side, and similarly half of the things on the right side, the balance will stay balanced.

= 4

3x = 9

Think about it! If one is a balanced situation…

= 3

…so is the other (and vice versa)! We simply divided both sides by 3.

Combining the operations

The allowed operations are:

  • Add the same amount to both sides (either x‘s or ones)
  • Subtract the same amount from both sides (either x‘s or ones)
  • Multiply both sides by the same number (but not by zero)
  • Divide both sides by the same number (but not by zero)

(There are others, too, but they are not needed in simple equations.)

The goal is to FIRST add and subtract until we have ONLY x‘s (blocks) on one side and ONLY ones (circles) on the other. Then, if you have more than one block, you need to divide so as to arrive to the situation with only one block on the one side, which is the solved equation!

Multiplying both sides can occur if you have a fractional block (less than one block) on one side. For example, the equation  1/4x = 13  is solved by multiplying both sides by 4.  Try let your students model the equation 1/2x + 14 = 20 using a balance; they can solve it with it. More advanced students can ponder what to do about the equation 2/3x = 12.

Example of both subtracting and dividing

In this example we use all the abovementioned operations: taking away from both sides of the equation and dividing the equation by the same number.

4x + 2 = 2x + 5First we get rid of the blocks on the right side by taking away two blocks from both sides.

2x + 2 = 5Then we eliminate the circles on the left side by removing 2 circles from both sides.

2x = 3Now there are only blocks on one side and only circles on the other. To find out what 1 block weighs, we take half of both sides.

x = 1 1/2The solution is that 1 block weighs 1 1/2 circles.

Try substituting this value x = 1 1/2 into the original equation 4x + 2 = 2x + 5 and check if the equation becomes true!

Example exercises

These equations are simple enough that you can solve them using a balance model. ALWAYS check your solution by substituting it into the original equation.

  1. 2x + 3 = 5
  2. 2+ 5 = x + 9
  3. 3x + 2 = 2x + 4
  4. 3x + 3 = 5 + x
  5. 5x + 4 = 3x + 6
  6. 6x + 2 = 3x + 6
  7. 6x + 3 = 2x + 5

Grade 3 Maths (imo): Fractions

fractions_description

Some Basic Terms and Rules of Fractions

  • The numbers in a fraction are called the numerator, on the top, and the denominator, on the bottom. numerator/denominator
  • Proper fractions have a numerator smaller than the denominator.
    Examples include 1/23/4 and 7/8.
  • Improper fractions have a numerator larger than the denominator.
    Examples include 5/43/2 and 101/7.

fraction types

Comparing Fractions

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IMO for Class 3 : Syllabus

Syllabus for Grade 3 

Number sense

  • Comparing numbers
  • Abacus and place value
  • Word problems

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Class 3 IMO : Papers

Grade 3 IMO papers

imo-ieo-nso-nco

imo-ieo-nso-ncoimo-ieo-nco-nso

imo-ieo-nso-nco

grade3IMO_1

grade3IMO_2

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Class 3 NSO : papers

Grade 3 NSO papers

nso-ieo-imo-nco

nso-ieo-nco-imo

nso-ieo-imo-nco

nso-imo-ieo-nco

NSO_1

Grade3NSO_2

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Grade 3 : IEO Papers

Class 3 IEO papers

ieo-nso-imo-nco

ieo-imo-nco-nso

ieo-nco-nso-imo

ieo-nso-imo-nco

class3ieo_1

class3ieo_2a

class3ieo_2

class3ieo_3

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