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

Ans 1:

Samya Saha from Ramakrishna mission Vidyalaya Narendrapur, Burdwan
Class : Class 8
If the question is :- "in the figure (not drawn to scale), ABCD is a rectangle, FA = FG, AGE and BGF are straight lines. find x+y." then the answer is:- angle DAF is 20 degree, angle ADF is 90 degree, so, angle DFA = 180-(20+90) =70 degree. as DFC and AB are parallel, angle DFA = angle FAB =70 degree DA and CEB are parallel so, angle BEA = angle EAD = 74 degree so, angle FAG =74 - 20 =54 degree as FAG is a isoseles triangle, angle FAG = angle FGA = 54 degree so x = 180 -(54+54) degree =72 degree again, angle AGF= angle BGE = 54 degree so, y = 180 - ( 74 + 54 ) = 180 - 128 =52 degree as a result ( x + y ) = 72 + 52 =124 degree (option B). THANK YOU

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Raju Pal from sunflag school, maharastra
Subject :IMO    Class : Class 7

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SUBODH KUMAR from Delhi Public School, Bokaro
Subject :IMO    Class : Class 6

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

Ans 1:

Pulkit mittal from GAIL DAV PUBLIC SCHOOL, AURAIYA
Class : Class 9
13 OF DEC

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John Calvin from vaels billabong, kottiwakkam
Subject :IMO    Class : Class 4

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abhinav from fitjee world school, Hyderabad
Subject :IMO    Class : Class 10

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aditya from carmel convent high school, badlapur
Class : Class 10
nice question my friend

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DWIPDALAL from Queen of Angels convent school, Bharuch
Class : Class 10
What do you want to indicate by ' _' symbol

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ARNAV GARG from PRUDENCE SCHOOL, New Delhi
Subject :IMO    Class : Class 4

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SAMRUDDHI SRICHANDAN from prabhujee english medium school, BHUBANESWAR
Class : Class 9
The answer is 246,913,578.

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om sharma from The Khaitan School, Noida
Subject :IMO    Class : Class 2

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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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