International Maths Olympiad Forum By SOF Olympiad Trainer - Page 387
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Mohammed Najl from Credence high school Dubai, Dubai
Subject :IMO    Class : Class 5

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Vindhya sawhney from Sanskriti , Gurgaon
Subject :IMO    Class : Class 5

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

The correct answer is option D.


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Soumyajit Roy from Delhi Public School Jamnagar, Jamnagar
Subject :IMO    Class : Class 7

Ans 1:

Adrita Bej from Vidyasagar Shishu Niketan , Paschim Medinipur
Class : Class 7
Adrita Bej of Vidyasagar Shishu Niketan Class : 7A - is the correct answer

Ans 2:

SHAURYA PRATHAM from MAY FLOWER SCHOOL, Patna
Class : Class 7
A will be the correct answer.

Ans 3:

Ishan gupta from Silver Oak School, Alwer
Class : Class 7
b is the correct answer not C bcoz

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Deborah Sunil George from Airforce School , Bengaluru
Subject :IMO    Class : Class 4

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

(C):

GH = AB + CD + EF
17 cm = 5 cm + CD + 4 cm

17 cm = 9 cm + CD
 CD = 17 cm – 9 cm
CD = 8 cm
 


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Prisha Jhunjhunwala from PODAR international, Thane
Subject :IMO    Class : Class 4

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

(B):

There are a total of 24 hours in a day.
1 day = 24 hours
So, 5/12 of a day = (5/12) x 24 = 10 hours
 


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manya murali from sri sankara senior secondry school, CHENNAI
Subject :IMO    Class : Class 6

Ans 1:

Aahan Jain from ,
Class : Class 1
Answer is A-1 line because there can be only one line which can pass through a given plane as asked in question.

Ans 2:

Raashid Sarfraz from IIPS, Riyadh
Class : Class 6

Ans 3:

Aahan Jain from ,
Class : Class 1

Ans 4:

Ryan from Deens, Bangalore
Class : Class 6
A

Ans 5:

Raashid Sarfraz from IIPS, Riyadh
Class : Class 6

Ans 6:

Harshith mh from Freedom International School, bangalore
Class : Class 7

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Prisha Jhunjhunwala from PODAR international, Thane
Subject :IMO    Class : Class 4

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1
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Amitava Pushilal from International Schooling, Singapore
Subject :IMO    Class : Class 5

Ans 1:

ARADHYA AGARWAL from NEW ERA PROGRESSIVE SCHOOL, Korba
Class : Class 8

Ans 2:

Samreen shams from Ashok Hall Girls higher secondary school , Kolkata
Class : Class 7
c is the correct option

Ans 3: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

The correct answer is option C. 


Ans 4:

Risha Jakhotiya from Sanskriti School , Pune
Class : Class 3
C is the correct answer

Ans 5:

Neha Samanvita from GIIS TOKYO, Tokyo
Class : Class 5
Cccccccc

Ans 6:

divish from Delhi Public School, gurgaon
Class : Class 5
c is the correct answer

Ans 7:

Pratham Dave from GNS, Bangalore
Class : Class 7
C is the answer to this question

Ans 8:

Ashwak from Cambridge, Jalandhar
Class : Class 9
c

Ans 9:

rethish from Veltech, Avadi
Class : Class 3
C is the correct answer.

Ans 10:

Student from School, City
Class : Class 8
c

Ans 11:

divish from Delhi Public School, gurgaon
Class : Class 5
c is the correct answer

Ans 12:

Aharshi Kauri from Bishnupur High School, Bishnupur Bankura
Class : Class 7
C

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Amitava Pushilal from International Schooling, Singapore
Subject :IMO    Class : Class 5

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

(D):

A. 4th multiple of 52 = 208
B. 8th multiple of 37 = 296
C. 5th multiple of 25 = 125
D. 7th multiple of 50 = 350
So, 350 is the greatest value amongst all.


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Samiha Grewal from Sat Paul Mittal School, Ludhiana
Subject :IMO    Class : Class 3

Ans 1:

Nihira Tiwari from Delhi public school electronic city, Bengaluru
Class : Class 6
Seee

Ans 2:

ANIMESH BEHERA from TIMPANY, VISAKHAPATTNAM
Class : Class 3
(D):To tip the scale below to the right,the weight ofchocolates should be greater than 37kg.Weight of 5 chocolates =5 x 7=35kgWeight of 6 chocolates=6x7=42kgSo,minimum number of chocolates needed = 6

Ans 3:

Nihira Tiwari from Delhi public school electronic city, Bengaluru
Class : Class 6

Ans 4: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

(D): To tip the scale below to the right, the weight of chocolates should be greater than 37kg.
Weight of 5 chocolates = 5 x 7 = 35kg
Weight of 6 chocolates = 6 x 7 = 42kg
 As, 42kg > 37kg
So, minimum number of chocolates needed = 6


Ans 5:

Ankit from Delhi Public School Megacity Kolkata, KOLKATA
Class : Class 5

Ans 6:

Vivaan Nigam from New Horizon Gurukul, Bangalore
Class : Class 3
its 7

Ans 7:

Nihira Tiwari from Delhi public school electronic city, Bengaluru
Class : Class 6
This question is wrong as 7x9 is 63 and that is also bigger than 37.

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