Subject :IMO Class : Class 3
Sohail had 30 pencils and 36 pens. The pencils are stacked equally into some packets. The pens were also stacked equally in the same packets, so that no pen was left out.
What is the smallest number of pencils and pens that he put in each packet?
A2 pencils and 3 pens
B2 pencils and 6 pens
C10 pencils and 3 pens
D15 pencils and 3 pens
Kindly explain how A is the answer
Ans 1:
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 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 8
there is a mistake in the answer
Ans 4:
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 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 7:
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 8:
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 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.