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Prachi Shah from Holy child public school, Faridabad
Subject :IMO    Class : Class 4

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

Ans 1:

AREEN from BOLTON, HYDERABAD
Class : Class 5
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 5
Option B)

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

Ans 1:

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

Ans 2:

svayumbhu sharma from jodhamal public school, jammu
Class : Class 8

Ans 3:

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 4:

Mudit garg from La blossoms school, Jalandhar
Class : Class 8

Ans 5:

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 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.

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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:

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

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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 7
its more than easy D

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rachana from globalschool, delhi
Subject :IMO    Class : Class 3

Ans 1:

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

Ans 2:

Rajdeep Chatterjee from Ryan International School, Bolpur
Class : Class 4
IT IS OUT OF SYLLABUS.

Ans 3:

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

Ans 4:

Harshith from Cs academy, Erode
Class : Class 3

Ans 5:

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

Ans 6:

AARAV GUPTA from Khaitan Public School, Ghaziabad
Class : Class 3
Out of Syllabus

Ans 7:

AARAV GUPTA from Khaitan Public School, Ghaziabad
Class : Class 3

Ans 8:

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

Ans 9:

Krishna Nanda from Kendriyala Vidyalaya, Shaktinagar
Class : Class 4
the answer is 108

Ans 10:

Anirudh Bhattacharjee from THE LEXICON INTERNATIONAL SCHOOL KALYANI NAGAR PUNR, Pune
Class : Class 4
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 11:

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

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

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