Ans 1:
Class : Class 10
Answer: A
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.
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
Answer: A
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.
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
Answer: A
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.
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 9
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 5:
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 6:
Class : Class 10
Answer: A
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.
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 8:
Class : Class 6
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 9:
Class : Class 10
Answer: A
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.
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.
