Errata – Equations, Inequalities, & VICs, 4th Edition
Edition 4.2 |
Edition 4.1 |
Edition 4.0 |
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Back Cover, Bottom Right CornerThe Equations, Inequalities, & VICs Guide (192 pages) covers algebra in all its various forms (and disguises) on the GMAT. Master fundamental techniques and nuanced strategies to help you solve for unknown variables of every type.
4.1
| Page | Loc | Description | Erroneous Text | Correction |
|---|---|---|---|---|
| 65 | Bot | Problem #8 should include units. | Problem #8: … The volume of the model is 8. | Problem #8: … The volume of the model is 8 cm3. |
| 118 | Mid | First sentence in the third paragraph mistakenly uses the variable z instead of y. | Because we are told that x percent of z must equal y percent less than z… | Because we are told that x percent of y must equal y percent less than z… |
| 103 | Mid | The final paragraph of the explanation is misplaced and is not relevant to the question. | The upper extreme for x is less than 6. The upper extreme for y is also less than 6, as long as it is less than x. Therefore, x + y must be less than 12. The largest prime number less than 12 is 11. | |
| 151 | Bot | The last equation has an incorrect number inside the absolute value signs. |
Last equation:
|2(5/3) – 2| = 1/3. |
|2(5/3) – 3| = 1/3. |
| 123 | Mid | Answer choice A for question #6 should be (t/j + 2). | (A) (t + 2)/j + 1 | (A) t/j + 2 |
| 126 | Bot | Answer choice A for question #6 only works when j = 2. The correct answer should be (t/j + 2). Ignore the explanation. | ||
| 157 | Bot | There is no answer choice (E). | Problem #9: … (D) 5 | Problem #9: … (D) 5 (E) 7 |
| 158 | Mid | The variable should be x, not z. | Problem #16: … what is z? | Problem #16: … what is x? |
| 93 | Top | In the third row of the table, the Min value of b should be -7, not 7. | ||
| 167 | Top | As the problem is now written, the value of k can be either 1/2 or -1/2. Nothing says that k must take on the positive value. To determine a single value of k and thus the recursive rule, we need another constraint in the problem itself. |
If each term is equal to the previous term times a constant number, what is the recursive rule for this sequence?
… 1/4 = k2 1/2 = k |
If each term is equal to the previous term times a constant number, and if all terms in the sequence are positive, what is the recursive rule for this sequence?
… 1/4 = k2 1/2 = k or -1/2 = k However, the requirement that all terms be positive eliminates the possibility that k = -1/2, since a negative k would force every other term to be negative. Thus, we know that k = 1/2. |
| 195 | Mid | Answer choices (B) and (C) are identical. | Problem #4: … (B) 100yz/x (C) 100yz/x … | Problem #4: … (B) 100yz/x (C) 100y/(xz) … |
4.0
| Page | Loc | Description | Erroneous Text | Correction |
|---|---|---|---|---|
| 65 | Bot | Problem #8 should include units. | Problem #8: … The volume of the model is 8. | Problem #8: … The volume of the model is 8 cm3. |
| 118 | Mid | First sentence in the third paragraph mistakenly uses the variable z instead of y. | Because we are told that x percent of z must equal y percent less than z… | Because we are told that x percent of y must equal y percent less than z… |
| 105 | Top | Diagram for #12(E) should have shaded dots at -7 and -3. | ||
| 126 | Bot | Answer choice A for question #6 only works when j = 2. The correct answer should be (t/j + 2). Ignore the explanation. | ||
| 123 | Mid | Answer choice A for question #6 should be (t/j + 2). | (A) (t + 2)/j + 1 | (A) t/j + 2 |
| 93 | Top | In the third row of the table, the Min value of b should be -7, not 7. | ||
| 103 | Mid | The final paragraph of the explanation is misplaced and is not relevant to the question. | The upper extreme for x is less than 6. The upper extreme for y is also less than 6, as long as it is less than x. Therefore, x + y must be less than 12. The largest prime number less than 12 is 11. | |
| 139 | Mid | Incorrect rephrasing. | Does [the symbol] mean either of these 2 operations: addition or multiplication? | Does [the symbol] mean multiplication? |
| 146 | Bot | Erroneous inclusion of problem #77 from the Quant Review among VIC problem set. | Quantitative Review: 1, 29, 32, 42, 52, 60, 69, 77, 85, 99, … | Quantitative Review: 1, 29, 32, 42, 52, 60, 69, 85, 99, … |
| 151 | Bot | The last equation has an incorrect number inside the absolute value signs. |
Last equation:
|2(5/3) – 2| = 1/3. |
|2(5/3) – 3| = 1/3. |
| 157 | Bot | There is no answer choice (E). | Problem #9: … (D) 5 | Problem #9: … (D) 5 (E) 7 |
| 158 | Mid | The variable should be x, not z. | Problem #16: … what is z? | Problem #16: … what is x? |
| 167 | Top | As the problem is now written, the value of k can be either 1/2 or -1/2. Nothing says that k must take on the positive value. To determine a single value of k and thus the recursive rule, we need another constraint in the problem itself. |
If each term is equal to the previous term times a constant number, what is the recursive rule for this sequence?
… 1/4 = k2 1/2 = k |
If each term is equal to the previous term times a constant number, and if all terms in the sequence are positive, what is the recursive rule for this sequence?
… 1/4 = k2 1/2 = k or -1/2 = k However, the requirement that all terms be positive eliminates the possibility that k = -1/2, since a negative k would force every other term to be negative. Thus, we know that k = 1/2. |
| 168 | Top | Incorrect definitions of variables in the equation for exponential growth. | Any exponential growth can therefore be written as y(t) = y0 · kt, in which y0 represents time, k is the value of the quantity at time zero, and k represents the constant multiplier. | Any exponential growth can therefore be written as y(t) = y0 · kt, in which t represents time, y0 is the value of the quantity at time zero, and k represents the constant multiplier. |
| 177 | Top | 2 equations have confusing line breaks. | Problem #12,
(B) (6 – 32)(8 – 42) = (-3) (-8) = 24 (D) (2 – 3)2(2 – 4)2 = (-1)2 (-2) 2 = 4 |
Problem #12,
(B) (6 – 32)(8 – 42) = (-3)(-8) = 24 (D) (2 – 3)2(2 – 4)2 = (-1)2(-2)2 = 4 |
| 190 | Top | Incorrect answer printed. | Problem #6. 49: … | Problem #6. 16: … |
| 190 | Bot | Incorrect rephrasing of Statement 2 in #9. | Problem #9:
Statement (2) tells is that x2 > x, so x < -1 OR x > 1. INSUFFICIENT. |
Problem #9:
Statement (2) tells is that x2 > x, so x < 0 OR x > 1. INSUFFICIENT. |
| 195 | Mid | Answer choices (B) and (C) are identical. | Problem #4: … (B) 100yz/x (C) 100yz/x … | Problem #4: … (B) 100yz/x (C) 100y/(xz) … |