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Adrian John from DBX, Egmore, Chennai
Subject :IMO    Class : Class 3

Ans 1:

Priansh Nair from Bhavans Vidya mandir eroor, Ernakulam
Class : Class 4

Ans 2:

Priansh Nair from Bhavans Vidya mandir eroor, Ernakulam
Class : Class 4

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docritika from DPS International edge, Gurgaon
Subject :IMO    Class : Class 4

Ans 1:

Subhashini M from Rastrotthana Vidya Kendra, Bengaluru
Class : Class 4

Ans 2:

Pradeep Mishra from chirec, Hyderabad
Class : Class 4
your answer is wrong the correct answer is ans3

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HIMANISH BASU from APS BANGALORE , Bangalore
Subject :IMO    Class : Class 3

Ans 1:

Daivik Ghosh from Delhi Public School, Kolkata
Class : Class 8
[B]

Ans 2:

Daivik Ghosh from Delhi Public School, Kolkata
Class : Class 8
[B]

Ans 3:

Daivik Ghosh from Delhi Public School, Kolkata
Class : Class 8
[B]

Ans 4:

Vansh Thakker from The Somaiya school, Mumbai
Class : Class 5

Ans 5:

AREEN from BOLTON SCHOOL, Secunderabad
Class : Class 5
Answer is B.Add only animals

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Ankur Agrawal from Greenwood High, Bengaluru
Subject :IMO    Class : Class 3

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Anchita from Euroschool Wakad, Pune
Subject :IMO    Class : Class 4

Ans 1:

Sunita from Dps, Greater Noida
Class : Class 6
a

Ans 2:

DIYA NOBY from VISWAJYOTHI CMI PUBLIC SCHOOL, ANGAMALY
Class : Class 7

Ans 3:

DIYA NOBY from VISWAJYOTHI CMI PUBLIC SCHOOL, ANGAMALY
Class : Class 7
250 GRAMS . FIRST DIVIDE 3200 BY 10 . ANS =320 THEN SUBTRACT 70 FROM 320. ANS =250 WEIGHT OF THE CYLINDER =250

Ans 4:

Mayank Raj from Litera velly, Patna
Class : Class 4

Ans 5:

Praket Tejas from Delhi Public School Indira Nagar Lucknow, Lucknow
Class : Class 4
answer is A

Ans 6:

Kaushik from pebble, Bangalore
Class : Class 5
A

Ans 7:

Kushal from Chirec, Hyderabad
Class : Class 7
a+b=70g 10a+10b+10c=3200 10(a+b+c)=3200 a+b+c=320 c=320-70=250

Ans 8:

MANAN AGARWAL from ,
Class : Class 4
A

Ans 9:

from , Allahabad
Class : Class 6
If large square = A small square =B cylinder=C then, A+B=70gm 10A+10B=700gm 10C=3200gm-700gm =2500gm C=2500/10 =250

Ans 10:

Aleeza Khan from Ekya School ITPL, Bengaluru
Class : Class 4
A

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Ayush Banerjee from DAV Pune, Pune
Subject :IMO    Class : Class 3

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1
The correct answer is D.

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Ayush Banerjee from DAV Pune, Pune
Subject :IMO    Class : Class 3

Ans 1:

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

Ans 2:

AREEN from BOLTON SCHOOL, Secunderabad
Class : Class 5
Answer is C that is boundary of the rectangle

Ans 3:

Vansh Thakker from The Somaiya school, Mumbai
Class : Class 5
how did we arrive this answer can you help

Ans 4:

Admin from SOF, Noida
Class : Class 10
Longer side of the rectangle paper = 7 + 6 + 4 + 4 + 4 = 25 cm
Smaller side of the rectangle paper = 4 cm
Total length (perimeter) = 25 + 25 + 4 + 4 = 58 cm

Ans 5:

Utkarsh Chakravarty from Pinnacle High International School, Mumbai
Class : Class 6
58 cm

Ans 6:

aaaaa from aaaaa, Chennai
Class : Class 3

Ans 7:

vineet8026@gmail.com from dps, Doha
Class : Class 1
I thought it was C

Ans 8:

venkatachalam ramasamy from ,
Class : Class 3
58

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Anil Gupta from DAV PUBLIC SCHOOL , Ludhiana
Subject :IMO    Class : Class 8

Ans 1:

Samedh from Tree House High School, Pune
Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

Ans 2:

Soham Raniga from Indian School Ghubra, Muscat
Class : Class 9
answer is told to be D not B;

Ans 3:

Samedh from Tree House High School, Pune
Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

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debangshi das from Springdale High School, Kalyani
Subject :IMO    Class : Class 3

Ans 1:

Aditi Shiv Shankar Jaiswal from Delhi Public School Jeddah, Jeddah
Class : Class 4

Ans 2:

Sarang Pande from Podar International School, Rajkot
Class : Class 4
Answer: (B) Area of square = 12*12 = 144 sq cm Area of 1 rectangle cut = 3*2 = 6 sq cm Area of 4 rectangles cut = 4*6 = 24 sq cm Area of remaining paper = 144 - 24 = 120 sq cm

Ans 3:

Keerthi menon from mvm, Hyd
Class : Class 5
Total area of the square=12 x 12 = 144 cm2 Area of one rectangle cut = 3 x 2 = 6 cm2 Area of 4 rectangles cut = 4 x 6 = 24 cm3 Area of the square after cutting = 144 - 24 = 120 cm2

Ans 4:

Ira Jamsandekar from Shantiniketan, KOLHAPUR
Class : Class 6
Total area of uncut square is 144. 4 rectangles of (3 x 2) area 6 are cut. Which means 24 square unit area is cut. That leaves 144 - 24 = 120 square cm area. So answer is B

Ans 5:

Admin from SOF, Noida
Class : Class 10
Answer: B

Solution
Area of original square paper sheet = 12 x12 = 144cm2
Area of 1 rectangle cut = 3 x 2 = 6 cm2
Area of 4 such rectangles - 6 x 4 = 24 cm2
Area of paper left = 144 - 24 = 120 cm2

Ans 6:

Admin from SOF, Noida
Class : Class 10
Area of original square paper sheet = 12 x12 = 144cm2
Area of 1 rectangle cut = 3 x 2 = 6 cm2
Area of 4 such rectangles - 6 x 4 = 24 cm2
Area of paper left = 144 - 24 = 120 cm2

Ans 7:

kashish sah from Bal Bharati Public School, New Delhi
Class : Class 7
the area of square paper sheet without cutting corners is Breadth x lenght = 12 x 12 = 144, now the area of the cut rectangle is breadth x length = 3 x 2 = 6, and there are four corners cutted. therefore 6 x 4 =24, now minus the cut corners area from total area. so 144-24 =120. therefore the answer is B

Ans 8:

Aryan Goyal from DPS , Dwarka, New Delhi
Class : Class 3
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Ans 9:

Shreyas Singh from Podar International School, Pune
Class : Class 6
Its was a square 12X12=144sq cm. cut part area= 2x3=6 sq cm. total cut area = 6x4= 24 sq cm. Area of remaining sheet = 144-24 = 120. so, B is the answer.

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debangshi das from Springdale High School, Kalyani
Subject :IMO    Class : Class 3

Ans 1:

Sai Lakshmi Rao from Maneckji Cooper Education Trust, Mumbai
Class : Class 4
Shouldnt it be 15? The equation is (X 2) X [(9/5)X] = 59

Ans 2:

Darsh Vijay from The Orbis School, Pune
Class : Class 5

Ans 3:

sklogic from SSM School, Chennai
Class : Class 3
Let Janak points = J | Adil A = J + 2 | Rohit R = (9/5) * J | Adil + Janak + Rohit = 59 | A + J + R = 59 | Replace as above | (J + 2) + J + (9/5)*J = 59 | 2J + 2 + 9J/5 = 59 | To remove 5 in denominator, multiply entire equation by 5 | 10J + 10 + 9J = 295 | 19J = 295 - 10 | 19J = 285 | J = 15 | Hence Janak's points = 15 | => Adil = J + 2 = 17 points |

Ans 4:

Riddhiraj Saha from Don Bosco School Berhampore, Berhampore
Class : Class 5
how will i solve this problem?

Ans 5:

Aryan Goyal from DPS , Dwarka, New Delhi
Class : Class 3
how did you get the answer

Ans 6:

LALIT MANI from K R Mangalam World School, New Delhi
Class : Class 6
Got it

Ans 7:

Darsh Vijay from The Orbis School, Pune
Class : Class 5

Ans 8:

LALIT MANI from K R Mangalam World School, New Delhi
Class : Class 6
how will we solve this?

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