Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce $4$ beats per second. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies (in Hz) are:

80

and

40

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100

and

50

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44

and

22

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72

and

36

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Solution

The correct option is D $72$ and $36$
Each fork produces 4 beats per second with the previous means each fork has frequency 4 Hz more than the previous.
So the 10th fork will have frequency $9 \times 4 =$ 36Hz more than the 1st
So this is an algebraic equation as $2 x - x = 36$
Hence the frequencies are 72 and 36

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Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce 4 beats per second. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies (in Hz) are:

The correct option is D 72 and 36Each fork produces 4 beats per second with the previous means each fork has frequency 4 Hz more than the previous.So the 10th fork will have frequency 9×4=36Hz more than the 1stSo this is an algebraic equation as 2x−x=36Hence the frequencies are 72 and 36

Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce $4$ beats per second. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies (in Hz) are: