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Aditya Patro from New Horizon Scholar School, Navi Mumbai
Subject :IMO    Class : Class 4

Ans 1:

Viraj Singh from Reliance Foundation School, Navi Mumbai
Class : Class 6
answer can be 3 also

Ans 2:

Devyani bansal from Blue bell school international, New Delhi
Class : Class 5

Ans 3:

Viraj Singh from Reliance Foundation School, Navi Mumbai
Class : Class 6

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Sriyansh from Surabhi , Hosapete
Subject :IMO    Class : Class 8

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nitin bhutada from bhavans b p vidya mandir trimurti nagar nagpur, Nagpur
Subject :IMO    Class : Class 7

Ans 1:

Abhjith S Kumar from Airport Senior Secondary school Karipur, Kozhikode Intl. Airport, Kozhikode Airport
Class : Class 9
THE ANSWER SHOULD BE 20.

Ans 2:

Bhagyesh Kamal from DPS, Mumbai
Class : Class 8

Ans 3:

Aniket Patankar from Symbiosis International School, Pune
Class : Class 8
it should be 20 because the formula for number of diagonals from a vertices is n-3, so for one vertice is 10 diagonals so two is gonna be 20

Ans 4:

Aditya Sai Vemparala from Amanora school, Pune
Class : Class 8
The answer should be 20

Ans 5:

Aniket Patankar from Symbiosis International School, Pune
Class : Class 8

Ans 6:

Ramya from brindavan vidyalaya, Trichy
Class : Class 9
ANSWER MUST BE 20 SINCE FORMULA IS N-3 SO FOR ONE VERTICE IT IS 10 SO TWO WILL BE 20

Ans 7:

Rick Banerjee from Calcutta Boys School, Kolkata
Class : Class 8

Ans 8:

AKSHIT CHHABRA from APEEJAY SCHOOL NOIDA, NOIDA
Class : Class 8
answer should be 20

Ans 9:

SATYAPRAKASH KAR from LUMBINI PUBLIC SCHOOL , BHUBANESWAR
Class : Class 7
If the answer is 22 then the polygon should be 14 sided.

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Aditya Patro from New Horizon Scholar School, Navi Mumbai
Subject :IMO    Class : Class 4

Ans 1:

Vihaan Binaykia from Riverside, AHMEDABAD
Class : Class 5
C

Ans 2:

Prisha anand from Saplings excel, New delhi
Class : Class 4

Ans 3:

Saiansh Tapuriah from The Shri Ram School, Aravali, Gurgaon
Class : Class 5

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Aditya Patro from New Horizon Scholar School, Navi Mumbai
Subject :IMO    Class : Class 4

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Armani from Jainendra public school, Panchkula
Subject :IMO    Class : Class 5

Ans 1:

Arunava hury from KV1 saltlake , Kolkata
Class : Class 6
D)

Ans 2:

Sejal Mittal from dlf, Ghaziabad
Class : Class 5
It should be 1 not 2 because half half = 1

Ans 3:

Bhavya from ABC, Delhi
Class : Class 6
IT SHOULD BE 1 NOT 2. 1/2 1/2 = 1

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Pranav from Ganges valley school, Hyderabad
Subject :IMO    Class : Class 8

Fill in the blanks.
(i) The perimeter of circle and its diameter vary __P__ with each other.
(ii) If two quantities p and q vary inversely with each other then __Q__ of their corresponding values remain constant.
(iii) When x and y are in indirect proportion and if y doubles then x becomes __R__

P Q R
A Inversely Ratio Double
B Directly Product Half
C Inversely Ratio Double
D Directly Product Half


A and c , b and d are same.

Ans 1:

from kv2, kalpakkam
Class : Class 9
answer is both B and D

Ans 2:

Anirudh from 989820Delhi Pubic school Rohini , Delhi
Class : Class 8
B and D are identical

Ans 3:

Sneh from The Shriram Millennium School, Noida
Class : Class 8
How can they give they two same options and mark one wrong

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Aditya vardhan roy from Bharatiya vidya bhavan vidyashram jaipur , Jaipur
Subject :IMO    Class : Class 7

Ans 1:

harsh shukla from DPS Bangalore East, Bengaluru
Class : Class 7
This can be solved like this: take length as 2x and breadth as x. Applying the pythagoras theorem, (x)^2 (2x)^2 = (9 * root 5)^2. Equal to x squared 4x squared = 9 squared * root 5 squared 5x squared = 81 * 5. 5 and 5 cancel on both sides. x squared = 81 x = 9 = breadth 2x = 18 = length perimeter = 2 (l b) 2 * 27 = 54

Ans 2:

Ashwin Shridhar from Vidyashilp Academy, Bangalore
Class : Class 9
give explanation please.

Ans 3:

S SAHITHYAA RAMAN from Padma Seshadri T Nagar, Chennai
Class : Class 7
w=xl=2xd=9root52(l b)=?2(x 2x)=?using pythogras theorem,xsquare 4 x square=(9root5)squared5xsquare=405x square=81x=9hence substitute values with x,perimeter=54

Ans 4:

S SAHITHYAA RAMAN from Padma Seshadri T Nagar, Chennai
Class : Class 7
please provide explanation soon

Ans 5:

Roy N from DSB Rishikesh, Dehradun
Class : Class 8
pl give solution, very confusing question

Ans 6:

Mehul Bansal from Pawar Public School, mumbai
Class : Class 7
it is very simple and can be solved with Pythagoras.Length=2y Breadth=yroot of [(2y)square ysquare]=9 x root of 5root of [4ysquare ysquare]=9 x root of 5root of 5ysquare=9 x root of 5 = y x root of 5thus y = 9 and 2y = 18p=54

Ans 7:

Mehak Narang from Bluebells School International, New Delhi
Class : Class 9
explaination of answer not provide....how to check the solution??

Ans 8:

Karthikeyan from Velammal bodhi campus, Thanjavur
Class : Class 7
Sorry diagonal of a rectangle is l^2 b^2

Ans 9:

Karthikeyan from Velammal bodhi campus, Thanjavur
Class : Class 7
First of all,diagonal of a rectanhle=square root of l^2 b^2,so 9×square root of 5 is square root of 81 × 5 that is 405 so here 2x is l and x is b.if i m taking l is 18 and b is 9 and squaring them we get 405 so perimeter of the rectangle is 2(18 9) that is 54.

Ans 10:

Rishith Sunil from BVM global, CHENNAI
Class : Class 6

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Trijal Kaushik from DPS Numaligarh, Golaghat
Subject :IMO    Class : Class 5

Ans 1:

Arunava hury from KV1 saltlake , Kolkata
Class : Class 6
C)

Ans 2:

Arunava hury from KV1 saltlake , Kolkata
Class : Class 6
C)

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Bhoomika from Cornerstone, Hyderabad
Subject :IMO    Class : Class 6

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