International Maths Olympiad Forum By SOF Olympiad Trainer - Page 639

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Subject :IMO    Class : Class 8

Ans 1:

Class : Class 9
ANS:DIAGONALS ARE EQUAL

Ans 2:

Class : Class 7
both the options c and d are correct

Ans 3:

Class : Class 10
Answer : Diagonals are equal. Corrected option B to say "Diagonals are equal"

Ans 4:

Class : Class 8
(D) one angle is 90 degree

Ans 5:

Class : Class 10
Answer : Diagonals are equal. Corrected option B to say "Diagonals are equal"

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Subject :IMO    Class : Class 8

Ans 1:

Class : Class 10
Answer : -3/35 We have corrected the answer. Thanks for pointing out.

Ans 2:

Class : Class 9
you are correct. the answer is wrong as given here. it should be-3/35.

Ans 3:

Class : Class 10
Answer : -3/35 We have corrected the option C as -3/35. Thanks for pointing out.

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Subject :IMO    Class : Class 8

Ans 1:

Class : Class 10
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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 8

Ans 2:

Class : Class 6
13

Ans 3:

Class : Class 3
C 13

Ans 4:

Class : Class 5
Answer is C that is 13

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Subject :IMO    Class : Class 4

Ans 1: (Master Answer)

Class : Class 1
The correct answer is B.

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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

Ans 2:

Class : Class 8
b is the answer

Ans 3:

Class : Class 4

Ans 4:

Class : Class 7
it's b because 4 circles are shaded and 1 is shaded half(5/10).

Ans 5:

Class : Class 5
Answer is B.Count the shaded part

Ans 6:

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.

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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 7

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Subject :IMO    Class : Class 4

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Subject :IMO    Class : Class 3

Ans 1:

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

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Subject :IMO    Class : Class 3

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.

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.

Ans 3:

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 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.

Ans 5:

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.

Ans 6:

Class : Class 8
there is a mistake in the answer

Ans 7:

Class : Class 6
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 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 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.

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