Math: a04050d8

ID: a04050d8 

(SAT Suite Question Bank > Find Questions > Assessment: SAT + Test: Math + Domain: Algebra)

Comment: The answer choices may be presented in a less direct way than you might be expecting, but that does not change the facts.

Method 1: The only number given in the passage is 180 food calories, which we can think of as total calories. The table provides information on macronutrients in terms of Energy per Gram, and we see 4-9-4 for protein, fat, and carbohydrate, respectively, in the Food Calories column. If the problem does not deal in joules, then we can ignore the other column. Think about modeling the food:

4 calories * p grams of protein + 9 calories * f grams of fat + 4 calories * c grams of carbohydrate = 180 total calories

To isolate fat, we would just need to treat the above as an algebra problem.

4p + 9f + 4c = 180

9f = 180 - 4p - 4c

f = 180/9 - (4/9)p - (4/9)c

The correct answer will have to look like the above. Get rid of (A) and (D) for adding p and c to the total calories. Finally, get rid of (C) because, if you were to distribute -4/9 into the group (p - c), c would be positive. The correct answer must be (B). Note the similarity to the above equation:

f = 20 - (4/9)(p + c)

Method 2: With three unknowns and no equations given except for the answer choices, it might be best to skip Desmos, unless you wanted to use it for arithmetic.

Method 3: Building off of the previous advice, there is nothing wrong with substituting numbers into an algebraic answer and checking to see what makes sense. For instance, it must be true that a 180-calorie food can contain no more than 20 grams of fat, since 20 * 9 = 180. Thus, if f = 20, then p and c must each equal 0. Such an understanding would not help you eliminate any answer choices right away, but it does open the door to other possibilities. What if there were 10 grams of fat instead of 20? That means there would be 90 calories that would have to come from some combination of protein and carbohydrate. Work out the problem:

4p + 9f + 4c = 180

4p + (90) + 4c = 180

4p + 4c = 90

p + c = 90/4

You can now substitute these values—10 for f and 90/4 for the sum p + c—into the answer choices to test for equivalence.

A. (10) = 20 + (4/9)(90/4) --> 10 = 20 + 10 X

B. (10) = 20 - (4/9)(90/4) --> 10 = 20 - 10 √

C. (10) = 20 - (4/9)(p - c) --> We do not know what the quantity p - c may be; it cannot be the same as the sum of p and c unless c = 0.

D. (10) = 20 + (9/4)(90/4) --> 10 = 20 + 810/16 X Regardless of the actual sum, 20 plus anything nonnegative cannot be 10.

Between (B) and (C), we know that (B) must be true, whereas (C) works only if a specific condition is introduced (c = 0). That should be sufficient cause to steer away from (C) to the safer option. To be clear, I would not recommend this kind of tinkering over method 1 above, but sometimes working with numbers is easier than working with variables.