Maths Olympiad | A Tricky Maths Olympiad Questions | Algebra Problem |

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  • Опубликовано: 21 сен 2024
  • a wonderful math problem. olympiad question. Olympiad lectures. #maths #mathematics

Комментарии • 11

  • @Daniel-OConnell
    @Daniel-OConnell 16 часов назад +1

    ab=100, ac= 300 => c=3b. : bc=200, ab=100 => c=2a => a=1.5b. Thus a+b+c = (1.5 +1 + 3) b = 5.5b : bc = 3 x b sq. =200 => b sq. = 200/3 = (2 x 100 ) /3 . Thus b = sq rt 2 / sq rt 3 x10 & a+b+c = sqrt2 /sq rt3 x10 x5.5 . No pen or paper required.

  • @QWERTYQWERTY-iv7pi
    @QWERTYQWERTY-iv7pi День назад +1

    use another way more easier when we know a^2 , b^2 , c^2

  • @Misha-g3b
    @Misha-g3b День назад +1

    +_110/V6.

  • @KhinMaungSan-qc9uv
    @KhinMaungSan-qc9uv День назад

    (a+b+c)^2>>>>>

  • @manecobeirao
    @manecobeirao 2 дня назад

    Very good !!! ( Brazil )

  • @KipIngram
    @KipIngram День назад +1

    Oh, come on. There's nothing tricky about this at all.
    a = 100/b
    c = 200/b
    (200/b)*(100/b) = 300
    200*100/300 = b^2
    b = 8.165
    a = 100/b = 100/8.165 = 12.2474
    c = 24.495
    a+b+c = 44.907
    Now, what was tricky about that???

  • @bommarajuseshagirirao7520
    @bommarajuseshagirirao7520 День назад +1

    Square root of (eqn 1x2x3)
    abc=√(100×200×300)=
    √6×1000 ••••eqn4
    a=√6×5 eqn4/eqn2
    b=√6×10/3 eqn4/eqn3
    c=√6×10 eqn4/eqn1
    a+b+c=√6(5+10/3+10)=
    √6×55/3

  • @gyanprakashraj4062
    @gyanprakashraj4062 3 часа назад

    😂😂😂😂😂