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SAMRUDDHI SRICHANDAN from prabhujee english medium school, BHUBANESWAR
Subject :IMO    Class : Class 7

Ans 1:

Om Kumar from Rps ser sec school, Khatod mahendragarh
Class : Class 8
Let tom has Rs. X Let Jasmine has Rs y A.T.Q x+y = 41 x = 41-y 1/4 * x = 2 + 1/7*y 1/4*(41-y) = 2 + 1/7*y 41/4 - y/4 = 2 + y/7 41/4 - 2 = y/7 + y/4 33/4 = 11y/28 33/4*28/11 = y 21 = y x+21=41 X = 41-21 = 20

Ans 2:

Oliva Rajguru from Siliguri girls' high school, Siliguri, Darjeeling ;West bengal
Class : Class 8
Tom and jasmine have Rs 41 altogether . let, Jasmine has Rs x.. therefore tom has Rs( 41-x) {the rest of the money} according to the question, 1/4 of Tom's money is 2 more than 1/7 of jasmine's money. therefore, 1/4 x (41-X) = (1/7 x X)+2 or, ( 41-X)/4=X/7+2 or, (41-X)/4=(X+14)/7 or; (7x41)-7X=4X+(4x14) or; 287-7X=4X+56 or; 287-56=4X+7X or, 231=11X: 11X=231 X=231/11 =21 THE ANSWER IS 21 OR TOM HAS RS 21................

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G. Priyanka from Chennai Public School, Chennai
Subject :IMO    Class : Class 7

Ans 1:

Subarno Maji from The Assembly of god church school, Purulia , Purulia
Class : Class 7
80 Rs 5 note and 10 Rs 10 note I have solved by this equation 5x + 10(90-x) = 500

Ans 2:

G. Priyanka from Chennai Public School, Chennai
Class : Class 8
@ Subarno Maji, that's the correct answer

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adityak1135@gmail.com from R.N. Podar School, Mumbai
Subject :IMO    Class : Class 8

Ans 1:

RiptideMel from Modern Vidya Niketan Senior Secondary School Sec 17, Faridabad
Class : Class 10
There exist two such numbers : (1-sqrt(5))/2 and (1+sqrt(5))/2 This can be taken out by making a quadratic equation and using the discriminant formula, which is: x=(b*sqrt(b^2-4ac))/2a

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Shaikishak7558 from St.adams high school, Golden heights colony
Subject :IMO    Class : Class 8

Ans 1:

OM KANADE from Euro School, Bengaluru
Class : Class 8
115929

Ans 2:

Shaikishak7558 from St.adams high school, Golden heights colony
Class : Class 8
21436587 is exactly divisible by

Ans 3:

Anidroid ,Aniruddha from Modern DPS, Faridabad
Class : Class 8
99 hehe

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Amitava Pushilal from Global Indian International School, Singapore
Subject :IMO    Class : Class 2

Ans 1:

Sreenath from BLUEBELLS SCHOOL INTERNATIONAL, New Delhi
Class : Class 4
90

Ans 2:

anudip from Ramkrishna Mission Vidyalaya , Agartala
Class : Class 3
90

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mrktrichy@yahoo.com from vishwa bharati public school, Ghaziabad
Subject :IMO    Class : Class 6

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CHANDRASHEKAR BHAT from a e c s kakrapar, Surat
Subject :IMO    Class : Class 8

Ans 1:

OM KANADE from Euro School, Bengaluru
Class : Class 8
The point where all 3 altitudes of a triangle intersect is called a Orthocenter.

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Yogeshwaran_TP from Chrysalis high, Bengaluru
Subject :IMO    Class : Class 8

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Ravi Ranjan Anand from THE AIR FORCE SCHOOL, SUBROTO PARK
Subject :IMO    Class : Class 8

Ans 1:

adityak1135@gmail.com from R.N. Podar School, Mumbai
Class : Class 8
Yes, because of the simple reason, that other polygons have edges! Circle has the same distance from the center at each and every point, whereas the distance from the center to the edges is generally lesser than that to the vertices in other polygons Thus, more area is covered in case of circles.

Ans 2:

OM KANADE from Euro School, Bengaluru
Class : Class 8
yes, it was mentioned in the maths trivia a few days ago. but idk the reason.

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Ishita from Chrysalis high, Bengaluru
Subject :IMO    Class : Class 4

Ans 1:

mohanananda from delhipublicschool, Dhanbad
Class : Class 6
you have to practice atleast 2 to 3 hours in a day maths , follow the syllabus.

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