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Algebra / Linear inequalities in one or two variables Difficulty: Hard

$I equals, V over R$

The formula above is Ohm’s law for an electric circuit with current I, in amperes, potential difference V, in volts, and resistance R, in ohms. A circuit has a resistance of 500 ohms, and its potential difference will be generated by n six-volt batteries that produce a total potential difference of $6 n$ volts. If the circuit is to have a current of no more than 0.25 ampere, what is the greatest number, n, of six-volt batteries that can be used?

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Explanation

The correct answer is 20. For the given circuit, the resistance R is 500 ohms, and the total potential difference V generated by n batteries is $6 n$ volts. It’s also given that the circuit is to have a current of no more than 0.25 ampere, which can be expressed as $I is less than 0 point 2 5$ . Since Ohm’s law says that $I equals, V over R$ , the given values for V and R can be substituted for I in this inequality, which yields $6 n over 500 is less than 0 point 2 5$ . Multiplying both sides of this inequality by 500 yields $6 n is less than 125$ , and dividing both sides of this inequality by 6 yields $n is less than 20 point 8 3 3$ . Since the number of batteries must be a whole number less than 20.833, the greatest number of batteries that can be used in this circuit is 20.