Solved Class V question paper of Aryabhata interschool maths competition 2007.
Solution :
Solved Class V question paper of Aryabhata interschool maths competition 2007.
Solution :
Solved Class V question paper of Aryabhata interschool maths competition 2006.
Here we will learn the Vedic sutra ‘By the deficiency’ which means how less or more the number is from the base number. The base number as explained earlier is 10 or a multiple of 10.
7 is deficient by 3 from 10
8 is deficient by 2 from 10
29 is deficient by 1 from 30
97 is deficient by 3 from 100
Base and Complement are very important in Vedic Mathematics and form the basis of many calculations.
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.
The Complement of a number is the difference between that number and the next higher power of 10.
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.
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.
Let’s start from the basics. In arithmetic there are nine numbers and a 0. All the numbers are made using these. We need to understand the Place value of these numbers.
PLACE VALUE
We count numbers in groups of 10. Each place has a value of 10 times the place to its right.
Ten Units make a 10
Ten Tens make a 100
Ten Hundreds make a 1000 and so on…
In any number its value depends on its position/place.
In the number 2583: Value of 2 is 2000; 5 is 500; 8 is 80 and 3 is 3.
10 s: Group of 10 units/ones.
100s: Group of 10 tens
1000s: Group of 10 hundreds
10000: Group of 10 thousands
If you look at the circle closely you will notice that the opposite numbers add to 10 (5 will add to itself and make 10)
Our number system is based on the number 10 and proceeds in cycles of 10 e.g. 10, 20, 30 and so on. The reason behind this is that compared to the other numbers; ‘10’ is a very easy number to handle.
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
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.) |
![]() 3x + 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 |
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. |
![]() x = 4 |
![]() 3x = 9 |
Think about it! If one is a balanced situation… |
![]() x = 3 |
…so is the other (and vice versa)! We simply divided both sides by 3. |
The allowed operations are:
(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.
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!
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.
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Animals