# User Forum

Subject :IMO    Class : Class 3

## Ans 1:

Class : Class 4
It is 70118

Subject :IMO    Class : Class 3

Class : Class 1

85+75689+4356-10012 = 70118

Subject :IMO    Class : Class 3

Class : Class 1

85+75689+4356-10012 = 70118

Subject :IMO    Class : Class 3

Class : Class 1

85+75689+4356-10012 = 70118

Subject :IMO    Class : Class 4

Subject :IMO    Class : Class 2

Subject :IMO    Class : Class 6

Class : Class 4

## Ans 2:

Class : Class 7
Breadth=w, Length is 5 times more, Therefore, length = 5w Therefore, perimeter=2(5w+w) =2*6w =12w Hence, b is the correct answer

Class : Class 7
6w

## Ans 4:

Class : Class 7
P=l+b×2 L=5w B=w =5w+w×2 =6w×2 =12w cm

## Ans 5:

Class : Class 6

Subject :IMO    Class : Class 6

Class : Class 6
yeah

## Ans 2:

Class : Class 6
yeah....crrct

Class : Class 1

Subject :IMO    Class : Class 6

Class : Class 5

## Ans 2:

Class : Class 6
There is problem in the Q as it mentions division sign represents both addition and substraction

## Ans 3:

Class : Class 6
yes pls correct the problem...

## Ans 4:

Class : Class 7
Then the calculation will turn to 20 +8/4x3x2, =20+2x3x2, =20+6x2, =20+12 =32, so there is a problem in the question

Class : Class 7

Class : Class 7

Class : Class 6

## Ans 8:

Class : Class 8
actually! it is 32...........plz correct it

Class : Class 7

## Ans 10:

Class : Class 6
Ya there is a problem...............answer is 32 you seee

## Ans 11:

Class : Class 5
the answer can be 31 and 13

## Ans 12:

Class : Class 8
yes, and should be 32, which is not an option. Pl. correct this

## Ans 13:

Class : Class 8
Yes there is a problem in the question

## Ans 14:

Class : Class 8
the division sign represents both addition and subtraction and we can get to different answers

## Ans 15:

Class : Class 7
yes there is a problem in the question

Subject :IMO    Class : Class 10

Class : Class 1

In ΔAED
AE D + ∠EAD + 90° = 180° (Angle sum property)
⇒ ∠AED + ∠EAD = 90°
⇒ ∠EAD = 90° – ∠AED ...(i)
Also, ∠EAD = (50° + ∠AFB) ...(ii) (Exterior angle property)
From (i) and (ii), we have
50° + ∠AFB = 90° – ∠AED
⇒ ∠AFB + ∠AED = 90° – 50° = 40°
⇒ ∠AED + ∠CFD = 40°

Class : Class 10
C