54. A dog eats 7 cans of food in 3 days. At this rate, how many cans of food does the dog eat in 3 + *d* days?

F. 7/3 + *d*

G. 7/3 + *d*/3

H. 7/3 + 7/(3*d*)

J. 7 + *d*/3

K. 7 + 7*d*/3

This problem is from the 2016-17 Preparing for the ACT practice test booklet. We *could* do it (the slow way) with a proportion.

Warning! Warning! A fifth grader would not do this the slow way. |

*cans, days*) to help us set up the proportion correctly.

*Cans*are in the numerators, and

*days*are in the denominators. We used a question mark in the proportion to make it really clear that we need to solve for the number of cans, not the value of

*d*. It's really easy to mess up and solve for the wrong variable!

(If you don't want to use a question mark, you can place an

*x*there instead and draw a really big, thick, dark circle around the*x*.)We'll replace the question mark with an

*x*and then solve for

*x*:

That answer is the same as 7 + 7

*d*/3, so (K) is the correct answer.
Are you tired yet?

If you're like the high schoolers I know, you might have made a mistake somewhere. F, G, H, and J are traps designed to catch people who don't multiply or divide correctly in the steps above. This is problem 54 out of 60, and ACT problems tend to look harder ask you approach the end of the test, so we should definitely look out for trap answers.

Fortunately, every ACT question has a quick solution that takes less than 30 seconds. You can avoid unnecessary torture by looking for shortcuts a fifth grader would use.

Would you have PWNed the ACT in fifth grade? |

A fifth grader thinks

*way*less abstractly than high schoolers do. Being less abstract can help us avoid getting trapped. If we don't have to multiply an abstract unknown*x*by 7 or divide it by 3, it's a lot harder to make a mistake!
In this case, we

*know*that the dog eats 7 cans of food in three days. We can already cross of three of the answer choices because they don't give us 7 when*d*= 0!*d*days?

*d*

*d*/3

*d*)

J. 7 +

*d*/3

K. 7 + 7

*d*/3

To decide between J and K, we need a second scenario. Let's pick something fifth-grade easy: if the dog eats 7 cans of food in 3 days, it'll eat 14 cans of food in 6 days. Since we have 3 +

*d*days, the dog will eat 14 cans of food when*d*= 3.
(J) gives us 7 + 3/3 = 8 when

*d*= 3, so (J) is wrong.
(K) gives us 7 + 7*3/3 = 14 when

*d*= 3, so (K) is right.
(K) is correct because we eliminated the other four choices

*and*because it gave us the right answer when we plugged in*d*= 0 and*d*= 3. We can feel really confident about our choice now and move on to the next question.
Math teachers discourage students from thinking this way because it makes middle school math tests

*way*too easy. Fortunately, there's no rule on the ACT saying you have to show your work or solve problems algebraically, so you should always look for the easiest solution.
Generally speaking, if an ACT problem confuses you, try spending 10 seconds to look for a fifth-grade solution. If the problem

*still*confuses you, skip it and come back later after you've made your first pass through the test.
P.S. Your teacher wants you to show your work at school, but you can still double-check your answers the fifth-grade way. On a logarithm test, plug in the original problem and your simplified expression into your calculator and make sure they give you the same answer. If you're doing polynomial long division, plug

*x*= 2 into the original expression and into your answer to make sure that they spit out the same number. For an indefinite integral, evaluate your answer as a definite integral on paper and on your calculator to make sure the numbers match. Doing every problem this way will get you an A on almost any math test.Which one of these contestants is you? |

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