International Maths Olympiad Forum By SOF Olympiad Trainer - Page 741
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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:

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

Ans 3:

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

Ans 4:

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

Ans 5:

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

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

Ans 1:

Ishan Kelkar from Sri Sri Ravishankar Vidya Mandir, Mimbai
Class : Class 3

Ans 2:

908 from fransalian school of excellence, sambhaji nagar
Class : Class 3
how easy it was

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

Ans 1:

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

Ans 2:

Sunita from Dps, Greater Noida
Class : Class 9
a

Ans 3:

Mayank Raj from Litera velly, Patna
Class : Class 4

Ans 4:

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 5:

DIYA NOBY from VISWAJYOTHI CMI PUBLIC SCHOOL, ANGAMALY
Class : Class 8

Ans 6:

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

Ans 7:

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 8:

DIYA NOBY from VISWAJYOTHI CMI PUBLIC SCHOOL, ANGAMALY
Class : Class 8
250 GRAMS . FIRST DIVIDE 3200 BY 10 . ANS =320 THEN SUBTRACT 70 FROM 320. ANS =250 WEIGHT OF THE CYLINDER =250

Ans 9:

Kaushik from pebble, Bangalore
Class : Class 5
A

Ans 10:

MANAN AGARWAL from ,
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:

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

Ans 2:

SOUVAGYA SAHA from De Nobili School, DHANBAD
Class : Class 10

Ans 3:

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

Ans 4:

Ayaansh Srivastav from GEMS modern, new delhi
Class : Class 4

Ans 5:

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

Ans 6:

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

Ans 7:

aaaaa from aaaaa, Chennai
Class : Class 3

Ans 8:

venkatachalam ramasamy from ,
Class : Class 3
58

Ans 9:

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

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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:

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 3:

Asmi Raniga from Indian School Ghubra, Muscat
Class : Class 8
answer is told to be D not B;

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

Ans 1:

Ira Jamsandekar from Shantiniketan, KOLHAPUR
Class : Class 10
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 2:

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 3:

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.

Ans 4:

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 5:

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 6:

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 7:

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 8:

Aryan Goyal from DPS , Dwarka, New Delhi
Class : Class 3
aaaaaaabbbbbbbcccccccdddddddeeeeeeffffgggghhhhhiiiiiijjjjjjkkkkkkkllllllmmmmmnnnnnnoooooopppppqqqqrrrrssssttttuuuuvvvvwwwwwxxxxxxxxxyyyyyyyyyyyyyyyzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Ans 9:

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

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

Ans 1:

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

Ans 2:

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

Ans 3:

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

Ans 4:

Riddhiraj Saha from St. Xeviers Institution , Panihati, Kolkata
Class : Class 8
how will i solve this problem?

Ans 5:

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

Ans 6:

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 7:

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

Ans 8:

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

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

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

Ans 1:

Sarang Pande from Podar International School, Rajkot
Class : Class 4
The level of the question is certainly not for Class 3. It involves use of simultaneous equations. Answer is 35 soldiers.

Ans 2:

Shreyas Singh from Podar International School, Pune
Class : Class 6
B

Ans 3:

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

Ans 4:

Vikram from AVM, Mumbai
Class : Class 5
The way I taught my child was to try with each option, that is, start with option A, assume 40 is the number of soldiers, then calculate no of legs of soldiers, then minus that from total no. of legs, and see if the resulting number gives u the required number of horses you need in order to get a total of 45 horses and soldiers. Repeat the same for option B, C and D. Answer is B. This is how we did it: Assume 35 as no. of soldiers 35x2 = 70 legs of soldiersM Minus 70 from 110 (total legs)= 40 legs 40 horses legs = 40 divided by 4 =10 horses Therefore 35 soldiers and 10 horses = 45 soldiers and horses totally

Ans 5:

Ravi Kumar Tammineni from ,
Class : Class 3
Hey you don't understand

Ans 6:

Mahesh M G from Surabhi vidyaniketan, Hosapete
Class : Class 3
Explain the answer properly

Ans 7:

Aryan Goyal from DPS , Dwarka, New Delhi
Class : Class 3
it is not a class 3 question understand that

Ans 8:

Shreyas Singh from Podar International School, Pune
Class : Class 6
The answers in options needs to be considered while answering the question. The total number of horses and soldier in the troop is 45. Let us consider option A with 40 soldiers = 80 human legs so the balance legs are 30 which is not possible for horses as they have 4 legs. Going by option B= 35 soldiers = 70 human legs. So the balance legs remains 40, which is possible for 10 horses. Now the total number of horses and soldiers are 45 so 35 soldiers+10 horses = 45 and hence is a right answer. The balance options can be checked in similar way and will be found incorrect.

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