Subject :IMO Class : Class 7

Subject :IMO Class : Class 6

## Ans 5:

Class : Class 6

answer is D because maximum water that cant be consumed is 4:1 but in the options all the ratios have more quantity of water

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Subject :IMO Class : Class 4

## Ans 4:

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.

## Ans 5:

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.

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Subject :IMO Class : Class 6

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Subject :IMO Class : Class 6

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Subject :IMO Class : Class 6

## Ans 1:

Class : Class 4

gone mad or what !!!!!! read question carefully i,m more reading questioncarefully

## Ans 6:

Class : Class 7

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.

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Subject :IMO Class : Class 6

## Ans 1:

Class : Class 10

Answer: B

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

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

## 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 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

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Subject :IMO Class : Class 6

## Ans 1:

Class : Class 7

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.

## 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 7

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

## Ans 7:

Class : Class 10

Answer : D

Solution

Let the length of one side of the path (outer square) be x.

Area of lawn + path = x

Length of one side of the lawn = (x -2)

Area of path = x

Solving the above equation x = 11

Side of lawn = 11 - 2 = 9

Area of lawn = 81 m

Solution

Let the length of one side of the path (outer square) be x.

Area of lawn + path = x

^{2}Length of one side of the lawn = (x -2)

Area of path = x

^{2}- (x-2)^{2}= 40Solving the above equation x = 11

Side of lawn = 11 - 2 = 9

Area of lawn = 81 m

^{2}## 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

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Subject :IMO Class : Class 6

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Subject :IMO Class : Class 6