# User Forum

Subject :IMO    Class : Class 5

Class : Class 9

Class : Class 10

Class : Class 9

Class : Class 1

Class : Class 7

Class : Class 8

Class : Class 6

## Ans 8:

Class : Class 7

Subject :IMO    Class : Class 2

Class : Class 6

Class : Class 1

## Ans 3:

Class : Class 6

Subject :IMO    Class : Class 2

Class : Class 2
225

## Ans 2:

Class : Class 7

Class : Class 1

Subject :IMO    Class : Class 2

Class : Class 1
Here, 462 + 199 = 661 So, the correct match is 462 + 199 - 661, which is option B.

Subject :IMO    Class : Class 8

## Ans 1:

Class :
Find the divisor given that the dividend is2200 remainder is 13 and the divisor is one third of the quotient

## Ans 2:

Class :
Find the divisor given that the dividend is2200 remainder is 13 and the divisor is one third of the quotient

Subject :IMO    Class : Class 10

## Ans 1:

Class : Class 6
help

Subject :IMO    Class : Class 2

Class : Class 1

## Ans 2:

Class : Class 2
B

Subject :IMO    Class : Class 5

Class : Class 1
Here, the answer is A. 7 tenths is 7 over 10, if you divide 7 by 10 you get 7 tenths as a decimal which is 0.7 Similarly, 45 tenths is 45 over 10,if you divide 45 by 10 you get 45 tenths as a decimal which is 4.50 21 hundredths is 21 over 100, if you divide 21 by 100 you get 21 hundredths as a decimal which is 0.21 Hence, the sum of 7 tenths, 21 hundredths and 45 tenths is (0.7 + 0.21 + 4.50) = 5.41

Subject :IMO    Class : Class 8

## Ans 1:

Class : Class 9
a2 - 13a + 30 = a2 + 4a + 4 a - 10 a + 2 => (a2 - 13a + 30)(a + 2) = (a2 + 4a + 4)(a - 10) => a3 + 2a2 - 13a2 - 26a + 30a + 60 = a3 - 10a2 +4a2 + 4a - 40 => - 11a2 + 4a + 60 = - 6a2+ 4a - 40 => - 11a2 + 6a2 + 60 + 40 = 0 => - 5a2 + 100 = 0 => 5a2 = 100 => a2 = 20 => a = âˆš20 Insert the value of'a'and you will get the answer.

## Ans 2:

Class : Class 9
but that is not what we want

## Ans 3:

Class : Class 8
By factorization process, we get a-3 = a 2, which is in fact a linear equation without a solution. On the other hand, by solving through cross multiplication method (as shown in the solution of this question), we get aÂ²-8a-20 = 0, or (a 2)(a-10) = 0. By solving the quadratic equation, we get a as (-2) or (10), which, upon substituting in the formula given in the question, doesn't satisfy the equation. This is because the equation in the above question isn't even actually correct, as (a-3) is not equal to (a 2), as that would lead to the statement: 5 = 0, which is obviously not true.

Class : Class 9
i got 5

Class : Class 8

## Ans 6:

Class : Class 10
The question is wrong.If A is the answer, solving this quadratic eqn, we get a=10 or -2since the answer is a simplified form of the question, these values of a should also hold for question also. But if we put these vales in the question either LHS or RHS denominator will become 0 making it undefined.the actual procedure is to factorize.=> [(a-3)(a-10)]/(a-10)=[(a 2)(a 2)]/(a 2)=> a-3 = a 2=> -3 = 2 which is absurd. so the question is wrong.

Class : Class 9

## Ans 8:

Class : Class 9
This question is completely wrong as when we factorise (a^2-13a 30) and (a^2 4a 4), we get (a-3)(a-10) and (a 2)^2 respectively. After calculating more, we get a-3=a 2 which has no solution as -3Ã¢Â‰Â 2

Subject :IMO    Class : Class 4

Class : Class 6

Class : Class 4

Class : Class 4

Class : Class 5
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Class : Class 6
A

Class : Class 8

Class : Class 4

Class : Class 4

Class : Class 3

Class : Class 8
A

Class : Class 5
a