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Subject :IMO    Class : Class 4

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Class : Class 9
The answer is 246,913,578.

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Subject :IMO    Class : Class 2

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Subject :IMO    Class : Class 7

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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................

Ans 2:

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

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Subject :IMO    Class : Class 7

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Class : Class 8
@ Subarno Maji, that's the correct answer

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Class : Class 7
80 Rs 5 note and 10 Rs 10 note I have solved by this equation 5x + 10(90-x) = 500

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Subject :IMO    Class : Class 8

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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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Subject :IMO    Class : Class 8

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Class : Class 8
115929

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Class : Class 8
99 hehe

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Class : Class 8
21436587 is exactly divisible by

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Subject :IMO    Class : Class 2

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Class : Class 3
90

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Class : Class 4
90

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Subject :IMO    Class : Class 6

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Subject :IMO    Class : Class 8

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Class : Class 8
The point where all 3 altitudes of a triangle intersect is called a Orthocenter.

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Subject :IMO    Class : Class 8

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