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Explanation:

Let number of satellites exclusively serving B be ‘10x’ and number of satellites serving O be ‘z’.

∴ number of satellites exclusively serving C = number of satellites exclusively serving S = 3a

As the number of satellites serving C = number of satellites serving S, the number of satellites serving C and B but not S = number of satellites exclusively serving S and B but not C = y (assume).

∴ 10x + 2y + 100 + 6x + 2z

= 1600 ⇒ 8x + y + z

= 750 … (i)

From (1), 10x + 2y + 100

= 6x + 2y + 2z + 200

∴ z = 2x – 50 … (ii)

Thus, the minimum value that ‘x’ can takes is 25.

From (i) and (ii),

10x + y = 800 ⇒ y = 800 – 10x

Thus, the maximum value that ‘x’ can takes is 80.

The number of satellites serving

C = 3x + y + z + 100 = 3x + (800 – 10x) + (2x – 50) + 100 = 850 – 5x

As value of ‘x’ is between 25 and 80, value of (850 – 5x) must be between 450 and 725.

Hence, option (c).

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