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Strategy 1: Translate from Words to an Arithmetic or Algebraic Representation

Word problems are often solved by translating textual information into an arithmetic or algebraic representation. For example, an “odd integer” can be represented by the equation 2 n plus 1. where n is an integer; and the statement “the cost of a taxi trip is $3.00, plus $1.25 for each mile” can be represented by the equation c equals 3 plus 1 point 2 5 m.. More generally, translation occurs when you understand a word problem in mathematical terms in order to model the problem mathematically.

• This strategy is used in the following two sample questions.

This is a Multiple-Choice – Select One Answer Choice question.

  1. A car got 33 miles per gallon using gasoline that cost $2.95 per gallon. Approximately what was the cost, in dollars, of the gasoline used in driving the car 350 miles?

    (A) $10
    (B) $20
    (C) $30
    (D) $40
    (E) $50

     

    Explanation

    Scanning the answer choices indicates that you can do at least some estimation and still answer confidently. The car used 350 over 33 gallons of gasoline, so the cost was Open parenthesis, 350 over 33, close parenthesis, times 2.95 dollars. You can estimate the product Open parenthesis, 350 over 33, close parenthesis, times 2.95 by estimating 350 over 33 a little low, 10, and estimating 2.95 a little high, 3, to get approximately 10 times 3 = 30 dollars. You can also use the calculator to compute a more exact answer and then round the answer to the nearest 10 dollars, as suggested by the answer choices. The calculator yields the decimal 31 point two, eight, seven, dot, dot, dot. which rounds to 30 dollars. Thus, the correct answer is Choice C, $30.

 

This is a Numeric Entry question.

  1. Working alone at its constant rate, machine A produces k liters of a chemical in 10 minutes. Working alone at its constant rate, machine B produces k liters of the chemical in 15 minutes. How many minutes does it take machines A and B, working simultaneously at their respective constant rates, to produce k liters of the chemical?

    The answer space consists of a box for the answer followed by the word, minutes. minutes

     

    Explanation

    Machine A produces k over 10 liters per minute, and machine B produces k over 15 liters per minute. So when the machines work simultaneously, the rate at which the chemical is produced is the sum of these two rates, which is The fraction k over 10, +, the fraction k over 15, which is equal to k times, open parenthesis, one tenth + one fifteenth, close parenthesis, which is equal to k times, open parenthesis, 25 over 150, close parenthesis, which is equal to k over 6 liters per minute. To compute the time required to produce k liters at this rate, divide the amount k by the rate k over 6 to get The fraction with numerator k and with denominator k sixths equals 6. Therefore, the correct answer is 6 minutes (or equivalent).

    One way to check that the answer of 6 minutes is reasonable is to observe that if the slower rate of machine B were the same as machine A's faster rate of k liters in 10 minutes, then the two machines, working simultaneously, would take half the time, or 5 minutes, to produce the k liters. So the answer has to be greater than 5 minutes. Similarly, if the faster rate of machine A were the same as machine B's slower rate of k liters in 15 minutes, then the two machines, would take half the time, or 7.5 minutes, to produce the k liters. So the answer has to be less than 7.5 minutes. Thus, the answer of 6 minutes is reasonable compared to the lower estimate of 5 minutes and the upper estimate of 7.5 minutes.