Question
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:
Answer: Option A
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Let the total number of shots be x. Then,
Shots fired by A = \(\frac{5}{8}x\)
Shots fired by B = \(\frac{3}{8}x\)
Killing shots by A =\(\frac{1}{3} of \frac{5}{8}x = \frac{5}{24}x\)
Shots missed by B = \(\frac{1}{2} of \frac{3}{8}x = \frac{3}{16}x\)
Therefore \(\frac{3x}{16} = 27 or x =\left(\frac{27\times16}{3}\right)=144. \)
Birds killed by A = \(\frac{5x}{24} = \left(\frac{5}{24}\times144\right) = 30.\)
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