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Pragun goel from Venkateshwar International School, Delhi
Subject :IMO    Class : Class 5

Ans 1:

AREEN from BOLTON, HYDERABAD
Class : Class 6
Ans is B because 3645279+2153467=5798746. 5798746 will be rounded to its nearest ten thousand, which is 5800000, since 98746 will be rounded to nearest ten thousand, which is 100000, 1 will be carried to the lakhs place, or, you can say that 3645279 and 2153467 will be rounded to the nearest ten thousand, which will be 3650000 and 2150000, respectively. 3650000+2150000=5800000.

Ans 2:

Arunava hury from KV1 saltlake , Kolkata
Class : Class 6
Option B)

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Shruti Hegde from Mount Litera Zee school, Bengaluru
Subject :IMO    Class : Class 8

Ans 1:

Mudit garg from La blossoms school, Jalandhar
Class : Class 7

Ans 2:

Ashwin Shridhar from Vidyashilp Academy, Bangalore
Class : Class 9
4

Ans 3:

jayashree easwaran from vibgyor High, Bengaluru
Class : Class 4
isn't the answer "A"

Ans 4:

svayumbhu sharma from jodhamal public school, jammu
Class : Class 9

Ans 5:

Hari Bhaskar from bgs, JP Nagar 7th Phase
Class : Class 6
the solution of the equation is 3*3*5*5*5*2*7 which has seven prime factors

Ans 6:

DEVRAJ DOLEY from NPS INTERNATIONAL SCHOOL, Guwahati
Class : Class 8
the answer is 7.

Ans 7:

Pranav from Ganges valley school, Hyderabad
Class : Class 8
The answer is 3. When you solve the equation you get 3*3*1*1*2*14/1*1*1. Now we see that the prime numbers left are 3,3 and 2. Therefore there are 3 prime numbers.

Ans 8:

Pranav from Ganges valley school, Hyderabad
Class : Class 8
The answer is 3. When you solve the equation you get 3*3*1*1*2*14/1*1*1. Now we see that the prime numbers left are 3,3 and 2. Therefore there are 3 prime numbers.

Ans 9:

Aman Sumod from Chinmaya Vidyalaya, KANNUR
Class : Class 8
The answer is 4. There are 4 prime factors: 2,3,5 and 7

Ans 10:

ARNAV BARANWAL from KPS SCHOOL, bhilai
Class : Class 8
ans is 7

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Advait nagale from Universal High, Thane
Subject :IMO    Class : Class 3

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Mayank Paul from Saint Edmunds School, Shillong
Subject :IMO    Class : Class 3

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Ankit Kumar Mistry from Kalyani Public School, Kolkata
Subject :IMO    Class : Class 4

Ans 1:

Midhuna from ABCD, Banglore
Class : Class 3

Ans 2:

Issac Newton from IIT Indian Institutes of Technology , Jalandhar
Class : Class 5
BIR you have written ten thousand instead of one thousand !!!

Ans 3:

Nehal from New Horizon Gurukul, Bangalore
Class : Class 4

Ans 4:

Satvik Agrawal from Delhi Public School, bhopal
Class : Class 5
dfd

Ans 5:

MANDAR PALKAR from KV DRDO, Bengaluru
Class : Class 2
100 tens = 1 thousand

Ans 6:

Issac Newton from IIT Indian Institutes of Technology , Jalandhar
Class : Class 5
The answer is D .

Ans 7:

Shraddha from Delhi public school, Hyderabad
Class : Class 4
The answer is D

Ans 8:

BIR from LOTUS VALLEYINTERNATIONAL SCHOOL, NOIDA
Class : Class 4
10000

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Astha Aswani from Fatima convent school, Kanpur
Subject :IMO    Class : Class 8

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DEVANSHI R SHAH from DIVYAPATH CAMPUS, AHMEDABAD
Subject :IMO    Class : Class 7

Ans 1:

Arshia Mittal from Ryan International School, rohini
Class : Class 8
its more than easy D

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rachana from delhi public school, NACHARAM
Subject :IMO    Class : Class 3

Ans 1:

Arpit Gupta from Treamis World School, Bengaluru
Class : Class 3
options given seems to be wrong

Ans 2:

AARAV GUPTA from Khaitan Public School, Ghaziabad
Class : Class 4
Out of Syllabus

Ans 3:

from Ryan International School, Bolpur
Class : Class 5
IT IS OUT OF SYLLABUS.

Ans 4:

AARAV GUPTA from Khaitan Public School, Ghaziabad
Class : Class 4

Ans 5:

princy gigo from Global Indian International School, Kuala Lumpur
Class : Class 4
out of syallabus

Ans 6:

Anirudh Bhattacharjee from THE LEXICON INTERNATIONAL SCHOOL KALYANI NAGAR PUNR, Pune
Class : Class 5
It is NOT out of syllabus . The sulution is that one moon is 18.So,18x18=324.To get the star devide 18 by 3=6.18-6=12

Ans 7:

Krishna Nanda from Kendriyala Vidyalaya, Shaktinagar
Class : Class 5
the answer is 108

Ans 8:

Harshith from Cs academy, Erode
Class : Class 3

Ans 9:

Prisha Jora from Adani Public School, Mundra
Class : Class 3
I don't know the answer because IT IS OUT OF SYLLABUS

Ans 10:

Raghav from Gems new millennium school, dubai
Class : Class 3
it is a little tricky but i solved it.

Ans 11:

Aamogh from Kv, Chennai
Class : Class 3
correct answer should be 33

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Sanskrit Samant from ABC, Pune
Subject :IMO    Class : Class 4

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Chiranjib kar from MOUNT LITERA ZEE SCHOOL, JAJPUR
Subject :IMO    Class : Class 5

Study the given line graph which shows the capacity of petrol Mohit used while travelling from City X to City Y from April to August and answer the following questions.

(A) How much more capacity of petrol was used in August than in May?
(B) What fraction of the total capacity of petrol used in July?

(A) (B)
A 20000 mL 7/26
B 22000 mL 8/29
C 10000 mL 7/27
D 25000 mL 9/36


Ans not understood... I doubt if the ans is correct.

Ans 1:

Arunava hury from KV1 saltlake , Kolkata
Class : Class 6
Option A)

Ans 2:

AREEN from BOLTON, HYDERABAD
Class : Class 6
Ans is A because the answer of question (A) will be 20000 ml because 40 l-20 l=20 l. Now convert 20 l to ml, which is 20000 ml, or, you can say that 20 l and 40 l will be 20000 ml and 40000 ml, respectively. Now do 20000 ml-40000 ml=20000ml. answer of question (B) will be 7/26 by adding the capacity of petrol of April, May, June, July and August, which will be 10 l+20 l+ 25 l+35 l+40 l=130 l. In July, the capacity of petrol used was 35 l, so in fraction, it will be 35 l/130 l. Now simplify the fraction, which will be 7 l/26 l.

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