Now that you’ve learned some strategies for the quantitative comparisons portion of the ISEE, let’s test your readiness for the ISEE with some practice questions.
D: If you tried to do this one by Picking Numbers and all you picked were small integers like 1, 2, or 3, you’d think the answer was (A). Try taking a value for x that’s greater than 5. When x = 6, for example, Column A is 29 and Column B is 30. More than one relationship is possible, so the answer is (D).
A: Don’t calculate; compare piece by piece. The first fraction in Column A is greater than the first fraction in Column B, and the second fraction in Column A is greater than the second fraction in Column B. Therefore, the sum in Column A is also greater.
B: Note that if you factor 19 out of Column A, you end up with 19 times 100, which is 1 less than 1,901. 19(56) + 44(19) = 19(56 + 44) = 19(100) = 1,900. Column B is larger.
D: This is another Picking Numbers type of problem. Let’s begin by plugging in a positive value for x that is consistent with the centered information. Let x = 2. So Column A is 22 = 2 × 2, or 4, and Column B is 23 = 2 × 2 × 2, or 8. In this case, Column B is greater. Now pick a negative value for x that is consistent with the centered information. Let x = –2. Now Column A is (–2)2 = (–2) × (–2) = 4 and Column B is (–2)3 = (–2) × (–2) × (–2) = –8. Column A is greater. Since more than one relationship between the columns is possible, (D) is correct.
QC Practice Question 4 Answer
D: Since the variables could be positive or negative, pick different kinds of numbers for the variables to see if different relationships between the columns are possible. Remember that the values you pick must be consistent with the centered information, which is that a > b > c > d.
If a = 4, b = 3, c = 2, and d = 1, then the value of Column A is a^2 + c = 4^2 + 2 = 16 + 2 = 18, and the value of Column B is b^2 + d = 3^2 + 1 = 9 + 1 = 10. Column A is greater. If you pick only positive numbers, then it will always be true that a^2 + c > b^2 + d. You could fall for the trap here of thinking that a2 + c is always greater than b2 + d if you don’t let some or all of the variables be negative. Let a = –1, b = –2, c = –3, and d = –4. These values are consistent with the centered information. This time the value of Column A is a^2 + c = (–1)^2 + (–3) = 1 – 3 = –2, and the value of Column B is b^2 + d = (–2)^2 + (–4) = 4 – 4 = 0. So in this case, Column B is greater. More than one relationship between the columns is possible.