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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries - Mathematics

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Question

Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice?

(Hint: The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances)

Sum

Solution

Number of singers in the first group = 4

Number of singers in the second group = 9

Distance between the two galleries = 70 m

Let the distance of the person from the first group be x and the distance of the person from the second group be 70 – x

By the given condition

4 : 9 = x2 : (70 – x)2 ...(by the given hint)

`4/9 = x^2/(70 - x)^2`

`2/3 = x/(70 - x)`  ...[taking square root on both sides]

3x = 140 – 2x

5x = 140

x = `140/5` = 28

The required distance to hear same intensity of the singers voice from the first galleries is 28m

The required distance to hear same intensity of the singers voice from the second galleries is (70 – 28) = 42m

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Quadratic Equations
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Chapter 3: Algebra - Exercise 3.12 [Page 117]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 3 Algebra
Exercise 3.12 | Q 7 | Page 117
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