[30 birds] A fires 5 shots to Bs 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has m

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A fires 5 shots to Bs 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:

30 birds

60 birds

72 birds

90 birds

Explanation:

Let the total number of shots be $ x $. Then,

Shots fired by A =$ \dfrac{5}{8} x $

Shots fired by B =$ \dfrac{3}{8} x $

Killing shots by A =$ \dfrac{1}{3} $of$ \dfrac{5}{8} x $=$ \dfrac{5}{24} x $

Shots missed by B =$ \dfrac{1}{2} $of$ \dfrac{3}{8} x $=$ \dfrac{3}{16} x $

$\therefore \dfrac{3x}{16} $= 27 or $ x $ =$ \left(\dfrac{27 \times 16}{3} \right) $= 144.

Birds killed by A =$ \dfrac{5x}{24} $=$ \left(\dfrac{5}{24} \times 144\right) $= 30.

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