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SURAJ from NPMHSS, CHENNAI
Subject :IMO    Class : Class 3

Ans 1:

Shiven Dinkar from DPS NOIDA Sector 30, Delhi
Class : Class 4
It is 70118

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SURAJ from NPMHSS, CHENNAI
Subject :IMO    Class : Class 3

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

85+75689+4356-10012 = 70118


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SURAJ from NPMHSS, CHENNAI
Subject :IMO    Class : Class 3

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

85+75689+4356-10012 = 70118


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SURAJ from NPMHSS, CHENNAI
Subject :IMO    Class : Class 3

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

85+75689+4356-10012 = 70118


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SURAJ from NPMHSS, CHENNAI
Subject :IMO    Class : Class 4

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kushal from indian school sohar, sohar
Subject :IMO    Class : Class 2

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nimay gupta from genesis global school, noida
Subject :IMO    Class : Class 6

Ans 1:

Anil sinha from DPS, Chandigarh
Class : Class 4
Answer is B

Ans 2:

Pisal Dhruv from Veer Bhagat Singh International School, Mumbai
Class : Class 7
Breadth=w, Length is 5 times more, Therefore, length = 5w Therefore, perimeter=2(5w+w) =2*6w =12w Hence, b is the correct answer

Ans 3:

Shreyas K from Kennedy High The Global School, Hyderabad
Class : Class 7
6w

Ans 4:

R K Madhav from Kendriya Vidhyalaya Kottayam, Kottayam
Class : Class 7
P=l+b×2 L=5w B=w =5w+w×2 =6w×2 =12w cm

Ans 5:

MANAS RANJAN from Tender Heart Senior Secondary School, RANCHI
Class : Class 6
answer is 12W cm as the formula is 2*(length+breadth)=perimeter ,so w ( breadth) +5w (length) =6w (length+breadth) and then 6w *2 =12w(answer)

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Ritin Jhamb from bjbj, Jalandhar
Subject :IMO    Class : Class 6

Ans 1:

Aadvik saini from Mvn sector 17, Faridabad
Class : Class 6
yeah

Ans 2:

DEBOSMITA GARAI from G.D.BIRLA CENTRE FOR EDUCATION, kolkata
Class : Class 6
yeah....crrct

Ans 3: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

The correct answer is D

 


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Anshuman Mishra from kv no 3, Bhubaneswar
Subject :IMO    Class : Class 6

Ans 1:

Shrutisri S G from Vyasa internationa school, bangalore
Class : Class 5

Ans 2:

Sahil Tejam from NPS Indiranagar, Bengaluru
Class : Class 6
There is problem in the Q as it mentions division sign represents both addition and substraction

Ans 3:

Kavitharini Joseph from HUS, Chennai
Class : Class 6
yes pls correct the problem...

Ans 4:

Pisal Dhruv from Veer Bhagat Singh International School, Mumbai
Class : Class 7
Then the calculation will turn to 20 +8/4x3x2, =20+2x3x2, =20+6x2, =20+12 =32, so there is a problem in the question

Ans 5:

Hprakash09 from Billabong high Bhopal, Bhopal
Class : Class 7
ANSWER IS 16!!!

Ans 6:

Aditya Saha from The Lexicon International School Kalyaninagar Pune, Pune
Class : Class 7

Ans 7:

SOUVAGYA SAHA from De Nobili School, DHANBAD
Class : Class 6

Ans 8:

Aniket kumar from kendriya vidyalaya, Delhi
Class : Class 8
actually! it is 32...........plz correct it

Ans 9:

anveshi srimal from cathedral and john connon school, mumbai
Class : Class 7
yes please correct. There are 2 division signs which represent addition n subtraction which leads to various answers

Ans 10:

Arav Agarwal from Pathways School Gurgaon, GURGAON
Class : Class 6
Ya there is a problem...............answer is 32 you seee

Ans 11:

Joe Bipin from National Public School Agra, Bangalore
Class : Class 5
the answer can be 31 and 13

Ans 12:

Divisha Agarwal from DPS Noida, Noida
Class : Class 8
yes, and should be 32, which is not an option. Pl. correct this

Ans 13:

student from school, city
Class : Class 8
Yes there is a problem in the question

Ans 14:

aditya arora from Heritage Experiential Learning School, gurugram
Class : Class 8
the division sign represents both addition and subtraction and we can get to different answers

Ans 15:

DHEER PANDEY from Sentia School, Hyderabad
Class : Class 7
yes there is a problem in the question

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SHEETAL SHREYA MANJUSHA SANTHOSH from INDIAN SCHOOL MULADHA, MUSCAT
Subject :IMO    Class : Class 10

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

The correct answer is C.

In ΔAED, 
∠AE D + ∠EAD + 90° = 180° (Angle sum property) 
⇒ ∠AED + ∠EAD = 90° 
⇒ ∠EAD = 90° – ∠AED ...(i) 
Also, ∠EAD = (50° + ∠AFB) ...(ii) (Exterior angle property) 
From (i) and (ii), we have 
50° + ∠AFB = 90° – ∠AED 
⇒ ∠AFB + ∠AED = 90° – 50° = 40° 
⇒ ∠AED + ∠CFD = 40°


Ans 2:

Neha from Amity international, Delhi
Class : Class 10
C

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