# User Forum

Subject :IMO    Class : Class 4

## Ans 1:

Class : Class 5
Ball A rotates in 6 seconds, ball B in 9 seconds and ball C in 12 seconds. So we have to take out the LCM of the three numbers. LCM = 36. So the answer is 36 seconds.

Class : Class 7
c)36 seconds

Class : Class 6
36 SECONDS

Class : Class 4
36

Class : Class 6
36 sec

## Ans 6:

Class : Class 5
Ball A rotates in 6 seconds, ball B in 9 seconds and ball C in 12 seconds. So we have to take out the LCM of the three numbers. LCM = 36. So the answer is 36 seconds.

Subject :IMO    Class : Class 6

Class : Class 1

Subject :IMO    Class : Class 6

Class : Class 1

Subject :IMO    Class : Class 6

Class : Class 4

## Ans 2:

Class : Class 8
ans. is D. when all the given numbers [16,34,75,21,13] are multiplied,we get 1,11,38,400.this number is ending with two zeros.if this number must end with six zeros,there must be four more zeros.to achieve this,we multiply 1,11,38,400 by 10,000.

Class : Class 4
to chaitanya LK

## Ans 4:

Class : Class 6
whats the simple solution? do we have to multiply all the numbers?

## Ans 5:

Class : Class 6
plz give solution

Class : Class 7

## Ans 7:

Class : Class 6

Subject :IMO    Class : Class 6

## Ans 1:

Class : Class 8
its 36 As the ant which walks around the square covers 12 cm in one round and the ant which walks around the rectangle covers 18 cm in one round. To find out when they will meet you can take the LCM of the distance they cover in one round. it is 36 so they will meet after both the ants cover 36 cm. Pravir

## Ans 2:

Class : Class 8
its 36 As the ant which walks around the square covers 12 cm in one round and the ant which walks around the rectangle covers 18 cm in one round. To find out when they will meet you can take the LCM of the distance they cover in one round. it is 36 so they will meet after both the ants cover 36 cm. Pravir

## Ans 3:

Class : Class 10
We have corrected the answer. Thanks for pointing out.

Solution
Perimeter of small square = 12 cm
Perimeter of big square = 18 cm
LCM of 12 and 18 = 36

Subject :IMO    Class : Class 6

## Ans 1:

Class : Class 7
but we had to find total area not area of lawn.....81 is only area of lawn.....the answer given is wrong

## Ans 2:

Class : Class 8
area of the lawn is 81 sq mtr but your question says including the path which will be 121 sq mtr. this option is not there. You need to remove including path from the question.

Class : Class 7

## Ans 4:

Class : Class 10

Solution
Let the length of one side of the path (outer square) be x.
Area of lawn + path = x2
Length of one side of the lawn = (x -2)
Area of path = x2 - (x-2)2 = 40
Solving the above equation x = 11
Side of lawn = 11 - 2 = 9
Area of lawn = 81 m2

## Ans 5:

Class : Class 7
wrong answer because the question says to include area of pathway as well as are of lawn

## Ans 6:

Class : Class 8
Correct answer is 40 (area of path) 81(area of lawn)=121 sq mtrs. This option is not given

## Ans 7:

Class : Class 8
area of the shaded region Pravir

## Ans 8:

Class : Class 8
solution let the side of the lawn(without the path) is xso the side of the lawn (with the path ) =x 2area =(x 2)^2 =x^2 4 4x40 =x^2 4 4x-x^2=40 =4 4x=40-4=4x=36=4x=x=36/4=x=9area of lawn=9*9=81area of lawn(including the path)=(9 2)*(9 2)=11*11=121

## Ans 9:

Class : Class 7
c

Subject :IMO    Class : Class 6

## Ans 1:

Class : Class 6
As mean of 5 number is 12. Therefore sum of the 5 number is 12x5=60 Mean of 4 number is 14 Sum=14x4=56. Therefore the removed no.is 60-56=4

Subject :IMO    Class : Class 6

Class : Class 8

Class : Class 5

## Ans 3:

Class : Class 8

Subject :IMO    Class : Class 4

## Ans 1:

Class : Class 5
( 25 * 110 ) + ( 32 * 120 ) =6590. So option A is correct.

Subject :IMO    Class : Class 4

Class : Class 4

Class : Class 4

Class : Class 4