International Maths Olympiad Forum By SOF Olympiad Trainer - Page 41

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Subject :IMO    Class : Class 1

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Subject :IMO    Class : Class 1

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Subject :IMO    Class : Class 7

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Subject :IMO    Class : Class 8

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Subject :IMO    Class : Class 10

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Subject :IMO    Class : Class 10

Ans 1:

Class : Class 10
C. 4480 m VOLUME OF FRUSTUM= VOLUME OF WIRE (calculate radius using trigonometry) R=radius of original cone, H=height of original cone, r=radius of new cone, h=height of new cone, r₁²=radius of wire, h₁=height/length of wire vol. of frustum= vol. of original cone- vol. of smaller cone (cut off) => 1/3πR²H - 1/3πr²h => 1/3π*7000/3cm³ vol. of wire(cylinder) = πr₁²h₁ => π1/24*1/24*h equating and cancelling on both sides of equation, 7000= h/64 h=448000cm = 4480 m

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Class : Class 10
C. 4480 m VOLUME OF FRUSTUM= VOLUME OF WIRE (calculate radius using trigonometry) R=radius of original cone, H=height of original cone, r=radius of new cone, h=height of new cone, r₁²=radius of wire, h₁=height/length of wire vol. of frustum= vol. of original cone- vol. of smaller cone (cut off) => 1/3πR²H - 1/3πr²h => 1/3π*7000/3cm³ vol. of wire(cylinder) = πr₁²h₁ => π1/24*1/24*h equating and cancelling on both sides of equation, 7000= h/64 h=448000cm = 4480 m

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Subject :IMO    Class : Class 7

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Class : Class 8
B) because the pattern is. 5*3*2=30, 4*4*2=32 so, 8*3*5=120.

Ans 2:

Class : Class 8
B) 5*3*2=30,4*4*2=32 So, 8*3*5 = 120

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Subject :IMO    Class : Class 2

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Subject :IMO    Class : Class 6

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Subject :IMO    Class : Class 5

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