Oranges

Mother divided her three children's oranges in a ratio of 6:5:4. Two children gave 45 oranges. How many oranges were there?

Correct result:

n =  75

Solution:

a:b:c=6+5+4 a+b=45∣∣b+c=45∣∣c+a=45  k1=66+5+4=25=0.4 k2=56+5+4=13≐0.3333 k3=46+5+4=415≐0.2667  q1=k1+k2=0.4+0.3333≐1115≐0.7333 q2=k1+k3=0.4+0.2667≐23≐0.6667 q3=k3+k2=0.2667+0.3333=35=0.6  n1=45/q1=45/0.7333≐67511≐61.3636 n2=45/q2=45/0.6667=1352=67.5 n3=45/q3=45/0.6=75  n=n3=75  a=k1⋅ n=0.4⋅ 75=30 b=k2⋅ n=0.3333⋅ 75=25 c=k3⋅ n=0.2667⋅ 75=20  n=75a:b:c=6+5+4 \ \\ a+b=45 || b+c=45 || c+a=45 \ \\ \ \\ k_{1}=\dfrac{ 6 }{ 6+5+4 }=\dfrac{ 2 }{ 5 }=0.4 \ \\ k_{2}=\dfrac{ 5 }{ 6+5+4 }=\dfrac{ 1 }{ 3 } \doteq 0.3333 \ \\ k_{3}=\dfrac{ 4 }{ 6+5+4 }=\dfrac{ 4 }{ 15 } \doteq 0.2667 \ \\ \ \\ q_{1}=k_{1}+k_{2}=0.4+0.3333 \doteq \dfrac{ 11 }{ 15 } \doteq 0.7333 \ \\ q_{2}=k_{1}+k_{3}=0.4+0.2667 \doteq \dfrac{ 2 }{ 3 } \doteq 0.6667 \ \\ q_{3}=k_{3}+k_{2}=0.2667+0.3333=\dfrac{ 3 }{ 5 }=0.6 \ \\ \ \\ n_{1}=45/q_{1}=45/0.7333 \doteq \dfrac{ 675 }{ 11 } \doteq 61.3636 \ \\ n_{2}=45/q_{2}=45/0.6667=\dfrac{ 135 }{ 2 }=67.5 \ \\ n_{3}=45/q_{3}=45/0.6=75 \ \\ \ \\ n=n_{3}=75 \ \\ \ \\ a=k_{1} \cdot \ n=0.4 \cdot \ 75=30 \ \\ b=k_{2} \cdot \ n=0.3333 \cdot \ 75=25 \ \\ c=k_{3} \cdot \ n=0.2667 \cdot \ 75=20 \ \\ \ \\ n=75

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