# User Forum

Subject :IMO    Class : Class 4

Class : Class 1

Subject :IMO    Class : Class 3

## Ans 1:

Class : Class 5
As all the circles are devided in 10 parts. answer B in which 5 are sheded is 4 5/10 is correct

Class : Class 5

## Ans 3:

Class : Class 4
Ans: (B). There is mistake in placement of the option and the figures. Option A has only one row. while others have two rows of circles.

Class : Class 5

Class : Class 4

## Ans 6:

Class : Class 5
it's b because 4 circles are shaded and 1 is shaded half(5/10).

Subject :IMO    Class : Class 3

## Ans 1:

Class : Class 5

Subject :IMO    Class : Class 4

Subject :IMO    Class : Class 3

## Ans 1:

Class : Class 4
Answer is 5 x 4 = 20 option C

Subject :IMO    Class : Class 3

## Ans 1:

Class : Class 10

5 pencils and 6 pens.
There was a mistake in the options. It has been corrected now.

Solution:
HCF of 30 and 36 = 6
So there will be six packets and each packet will have 5 pencils and 6 pens.

## Ans 2:

Class : Class 10

5 pencils and 6 pens.
There was a mistake in the options. It has been corrected now.

Solution:
HCF of 30 and 36 = 6
So there will be six packets and each packet will have 5 pencils and 6 pens.

## Ans 3:

Class : Class 10

5 pencils and 6 pens.
There was a mistake in the options. It has been corrected now.

Solution:
HCF of 30 and 36 = 6
So there will be six packets and each packet will have 5 pencils and 6 pens.

## Ans 4:

Class : Class 4
2 pencils and 3 pens. Key is no Pens was left out, and it is fine to have pencils left out. With 2 pencils and 3 Pens, in 12 packets no pens and 6 pencils will be left out.

## Ans 5:

Class : Class 10

5 pencils and 6 pens.
There was a mistake in the options. It has been corrected now.

Solution:
HCF of 30 and 36 = 6
So there will be six packets and each packet will have 5 pencils and 6 pens.

## Ans 6:

Class : Class 5
All the answers are incorrect. The correct answer will be 5 pencils and 6 pens and as the highest common factor is 6 for both the nos. How can answer A be correct? if we put 2 pencils in each packet there will be 15 packets and 36 pens cannot be divided in 15 packets equally and if 3 pens are put in each packet then only 12 packets will get full and the remaining 3 packets will only have 5 pencils each.

## Ans 7:

Class : Class 5
there is a mistake in the answer

## Ans 8:

Class : Class 3
The question is wring if the answer is A. Where is the mention that there is no pencil should be left out ? Hence, smallest number of Pencils and Pens in each packet so that no pens left is out 2 pencils and 6 Pens. hence answer B is correct

## Ans 9:

Class : Class 10

5 pencils and 6 pens.
There was a mistake in the options. It has been corrected now.

Solution:
HCF of 30 and 36 = 6
So there will be six packets and each packet will have 5 pencils and 6 pens.

Subject :IMO    Class : Class 5

Class : Class 1

Subject :IMO    Class : Class 8

Class : Class 8

## Ans 2:

Class : Class 9
answer is 36 x root of 13 x root of 3 x root of 2.....thats what i got after calculation

## Ans 3:

Class : Class 8

Subject :IMO    Class : Class 8

Class : Class 1