Linear Equations — Practice Set
Bank code: ALG-LIN · Section: Algebra · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
Clear fractions, distribute carefully, collect, isolate, check - and read 'less than' backwards, because it reverses the order.
Every question below is live in the question bank under the ID shown — the sheet and the portal are the same questions. Attempt a level with the clock running, then check Part C.
Part A — Questions
Quantitative Comparison — the four choices are always the same, so they are not reprinted: (A) Column A is greater · (B) Column B is greater · (C) The two quantities are equal · (D) The relationship cannot be determined from the information given
Level 1 · Easy — 5 questions · ~4 min
warm-up — these must be automatic
Q1 · ALG-LIN-004 · QC · 40s
x + 4 = 10
Column A: x Column B: 7
Q2 · ALG-LIN-005 · MCQ · 45s
If 8 - 3x = 2, what is the value of x?
(A) -2
(B) 2
(C) -2/3
(D) 6
(E) -6
Q3 · ALG-LIN-011 · Numeric Entry · 40s
If (x + 3)/4 = 5, what is the value of x? Enter your answer as a number.
Numeric entry — write the number.
Q4 · ALG-LIN-012 · Select all that apply · 55s
The equation 3x - 5 = 7 has the solution x = 4. Which of the following statements are true? Select all that apply.
(A) 6x - 5 = 14 has the solution x = 4
(B) 3x - 5 + x = 7 + x has the solution x = 4
(C) (3x - 5)/2 = 7 has the solution x = 4
(D) x = 4 is the only solution of 3x - 5 = 7
(E) 5 - 3x = 7 has the solution x = 4
Q5 · ALG-LIN-013 · MCQ · 45s
If 3(x + 4) = 21, what is the value of x?
(A) 3
(B) 7
(C) 9
(D) 11
(E) 17
Level 2 · Medium — 8 questions · ~10 min
two or three steps, one planted trap each
Q6 · ALG-LIN-017 · Numeric Entry · 70s
If (2x - 1)/3 = (x + 4)/2, what is the value of x? Enter your answer as a number.
Numeric entry — write the number.
Q7 · ALG-LIN-018 · MCQ · 75s
If 3(2x - 4) - 2(x + 1) = 14, what is the value of x?
(A) 4
(B) 5
(C) 6
(D) 25/4
(E) 7
Q8 · ALG-LIN-021 · MCQ · 70s
A number is increased by 12 and the result is then multiplied by 3. The final result is 51. What is the original number?
(A) 5
(B) 13
(C) 15
(D) 17
(E) 29
Q9 · ALG-LIN-024 · QC · 60s
5x - 3 = 3x + 9
Column A: x Column B: x + 1
Q10 · ALG-LIN-027 · Numeric Entry · 75s
The perimeter of a rectangle is given by P = 2l + 2w, where l is the length and w is the width. A certain rectangle has perimeter 56 and its length is 3 times its width. What is the width? Enter your answer as a number.
Numeric entry — write the number.
Q11 · ALG-LIN-030 · MCQ · 70s
In the equation ax - b = c, the constant a is not 0. Which of the following correctly expresses x in terms of a, b and c?
(A) x = (c - b)/a
(B) x = (b + c)/a
(C) x = (b - c)/a
(D) x = c/a - b
(E) x = c - b/a
Q12 · ALG-LIN-033 · Select all that apply · 85s
Consider the statement: three more than a number n is twice as much as the number decreased by 9. Which of the following must be true? Select all that apply.
(A) n + 3 = 2(n - 9)
(B) n + 3 = 2n - 9
(C) n - 3 = 2(n - 9)
(D) n is a multiple of 7
(E) n is less than 20
Q13 · ALG-LIN-038 · Select all that apply · 85s
Which of the following equations have x = 2 as a solution? Select all that apply.
(A) (x^2 - 4)/(x - 2) = 4
(B) 3(x - 2) + 5 = 5
(C) (x + 6)/(x + 2) = 2
(D) 3x - 2(x + 1) = 4
(E) (x + 4)/3 + (x - 4)/2 = 1
Level 3 · Hard — 12 questions · ~21 min
where 162+ is won or lost
Q14 · ALG-LIN-034 · MCQ · 110s
In the formula S = a/(1 - r), the constants satisfy S is not 0, a is not 0, r is not 1 and r is not -1. A second quantity T is defined by T = a/(1 + r). Which of the following expresses T in terms of S and a?
(A) T = aS/(2S - a)
(B) T = S
(C) T = aS/(S + a)
(D) T = (2S - a)/(aS)
(E) T = a(2S - a)/S
Q15 · ALG-LIN-040 · Numeric Entry · 105s
Solve for x: (2x + 5)^2 - (2x - 3)^2 = 176. Enter your answer as a number.
Numeric entry — write the number.
Q16 · ALG-LIN-041 · QC · 95s
p and q are nonzero constants, and the equation px + 5 = qx + 3 has no solution.
Column A: p Column B: q
Q17 · ALG-LIN-042 · MCQ · 110s
Jian's age now is 4 years more than twice Ben's age now. Six years ago, Jian was three times as old as Ben was then. In how many years from now will Jian be exactly twice as old as Ben?
(A) 2
(B) 4
(C) 16
(D) 20
(E) 40
Q18 · ALG-LIN-043 · Numeric Entry · 105s
The equations 3(x - n) = 2x + 5 and x + 2n = 20 have the same solution for x, where n is a constant. What is the value of n? Enter your answer as a number.
Numeric entry — write the number.
Q19 · ALG-LIN-044 · MCQ · 105s
Solve for x in terms of a, b and c: (x - a)/b = (x + a)/c, where b and c are nonzero and b is not equal to c.
(A) x = a(b + c)/(c - b)
(B) x = a(c - b)/(b + c)
(C) x = -a
(D) x = a(c + b)/(b - c)
(E) x = 2a/(c - b)
Q20 · ALG-LIN-045 · QC · 105s
In the equation 4 - c(x + 2) = k, c is a nonzero constant, k is a given constant and x is the resulting solution.
Column A: the value of x when k = 1 Column B: the value of x when k = 7
Q21 · ALG-LIN-049 · QC · 105s
n is a positive integer, and x satisfies (n + 1)x = n^2 + 2n.
Column A: x Column B: n + 1
Q22 · ALG-LIN-050 · MCQ · 105s
In the equation 2(3x - a) + 4 = 6x + b, a and b are constants and the equation has infinitely many solutions. What is the value of b + 2a?
(A) -8
(B) -4
(C) 0
(D) 4
(E) 8
Q23 · ALG-LIN-051 · Numeric Entry · 105s
In the equation (3x + c)/(x - 4) = 5, c is a constant and x is not equal to 4. For one value of c the equation has no solution. What is that value of c? Enter your answer as a number.
Numeric entry — write the number.
Q24 · ALG-LIN-052 · MCQ · 105s
In the equation 4x + k = 2x + 18, the constant k is an integer with 1 <= k <= 30, and the solution x is a positive integer. What is the sum of all possible values of k?
(A) 36
(B) 72
(C) 90
(D) 153
(E) 240
Q25 · ALG-LIN-053 · Select all that apply · 110s
A linear equation in x has the form px + q = rx + s, where p, q, r and s are real constants. Which of the following conditions guarantee that the equation has exactly one solution? Select all that apply.
(A) The number p - r has a multiplicative inverse
(B) p = r and q is not equal to s
(C) q is not equal to s
(D) p = 2r and r is not equal to 0
(E) p + r = 0
Level 4 · Extreme — 5 questions · ~11 min
165+ — expect to need the insight, not the grind
Q26 · ALG-LIN-039 · MCQ · 125s
For how many integer values of k does the equation (2x + 4)/(x - 1) = k have a solution x that is an integer? (x is not equal to 1.)
(A) 4
(B) 6
(C) 8
(D) 9
(E) Infinitely many
Q27 · ALG-LIN-054 · MCQ · 130s
At one firm every employee's net pay N is computed as N = G - r*G - F, where G is that employee's gross pay, r is the tax rate applied to gross pay and F is a fixed deduction of $170. The same r and the same F apply to every employee. Employee X has gross pay $1,200 and net pay $850. Employee Z's net pay is exactly 80 percent of Z's gross pay. What is employee Z's gross pay?
(A) $340
(B) $680
(C) $850
(D) $2,720
(E) $3,400
Q28 · ALG-LIN-055 · QC · 125s
a and b are real numbers, a + b is not 0, and 3a + 5b = 7a + 3b + 12.
Column A: (a + 2b)/(a + b) Column B: 5/3
Q29 · ALG-LIN-056 · Numeric Entry · 130s
A collection of tokens consists only of green tokens and white tokens, and exactly 30 of the tokens are white. When 12 green tokens are removed from the collection, the green tokens that remain make up 1/4 of the tokens that remain. Green tokens are then added to that reduced collection until green tokens make up 2/5 of it. How many tokens are in the collection at that point? Enter your answer as a number.
Numeric entry — write the number.
Q30 · ALG-LIN-057 · Select all that apply · 130s
In the equation (2x + a)/(x - b) = 3, a and b are integers and x is not equal to b. If x = 5 is a solution, which of the following must be true? Select all that apply.
(A) a + 3b = 5
(B) 2b - a = 10
(C) a - 2 is divisible by 3
(D) b can be any integer
(E) a + b = 5
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | ALG-LIN-004 |
easy | QC | B |
| 2 | ALG-LIN-005 |
easy | MCQ | B |
| 3 | ALG-LIN-011 |
easy | Numeric Entry | 17 |
| 4 | ALG-LIN-012 |
easy | Select all that apply | B, D |
| 5 | ALG-LIN-013 |
easy | MCQ | A |
| 6 | ALG-LIN-017 |
medium | Numeric Entry | 14 |
| 7 | ALG-LIN-018 |
medium | MCQ | E |
| 8 | ALG-LIN-021 |
medium | MCQ | A |
| 9 | ALG-LIN-024 |
medium | QC | B |
| 10 | ALG-LIN-027 |
medium | Numeric Entry | 7 |
| 11 | ALG-LIN-030 |
medium | MCQ | B |
| 12 | ALG-LIN-033 |
medium | Select all that apply | A, D |
| 13 | ALG-LIN-038 |
medium | Select all that apply | B, C, E |
| 14 | ALG-LIN-034 |
hard | MCQ | A |
| 15 | ALG-LIN-040 |
hard | Numeric Entry | 5 |
| 16 | ALG-LIN-041 |
hard | QC | C |
| 17 | ALG-LIN-042 |
hard | MCQ | B |
| 18 | ALG-LIN-043 |
hard | Numeric Entry | 3 |
| 19 | ALG-LIN-044 |
hard | MCQ | A |
| 20 | ALG-LIN-045 |
hard | QC | D |
| 21 | ALG-LIN-049 |
hard | QC | B |
| 22 | ALG-LIN-050 |
hard | MCQ | D |
| 23 | ALG-LIN-051 |
hard | Numeric Entry | -12 |
| 24 | ALG-LIN-052 |
hard | MCQ | B |
| 25 | ALG-LIN-053 |
hard | Select all that apply | A, D |
| 26 | ALG-LIN-039 |
extreme hard | MCQ | C |
| 27 | ALG-LIN-054 |
extreme hard | MCQ | E |
| 28 | ALG-LIN-055 |
extreme hard | QC | D |
| 29 | ALG-LIN-056 |
extreme hard | Numeric Entry | 50 |
| 30 | ALG-LIN-057 |
extreme hard | Select all that apply | A, C |
Part C — Worked Solutions
Q1 · ALG-LIN-004 — Answer: B
Step 1 — solve for x. Subtract 4 from both sides: x = 10 - 4 = 6. Step 2 — compare. Column A = 6, Column B = 7, so Column B is greater. Check: substitute back into the original equation. 6 + 4 = 10. Correct. Answer: B.
Trap. Adding 4 instead of subtracting it gives x = 14, which makes Column A look greater and produces choice A. The operation must undo the +4, so it is subtraction.
Q2 · ALG-LIN-005 — Answer: B
Step 1 — move the constant. Subtract 8 from both sides: -3x = 2 - 8 = -6. Step 2 — divide by the coefficient of x, which is -3: x = (-6)/(-3) = 2. Check: substitute into the original equation. 8 - 3(2) = 8 - 6 = 2. Correct. Answer: B.
Trap. Losing a sign: writing 3x = 2 - 8 = -6 and then x = -6/3 = -2, choice A. The coefficient is -3, not 3, so the division must carry the minus too. Choice E, -6, is stopping at Step 1, -3x = -6, and reporting the right-hand side as x; choice D, 6, is the same halt one tidy-up later, at 3x = 6. Choice C, -2/3, comes from subtracting 8 from the left side only, leaving -3x = 2, and then dividing 2 by -3.
Q3 · ALG-LIN-011 — Answer: 17
Step 1 — clear the denominator. Multiply both sides by 4: x + 3 = 20. Step 2 — subtract 3 from both sides: x = 17. Check: substitute into the original equation. (17 + 3)/4 = 20/4 = 5. Correct. Answer: 17.
Trap. Dividing by 4 instead of multiplying, which gives x + 3 = 5/4 and then x = -7/4. The 4 is already dividing the left side, so the undoing operation is multiplication. A second error is subtracting the 3 before clearing the fraction, giving x/4 = 2 and x = 8.
Q4 · ALG-LIN-012 — Answer: B, D
An operation preserves the solution only if it is applied to both sides in full, and to every term on each side. A. FALSE - both sides were meant to be doubled, but only the 3x term was. Solving 6x - 5 = 14 gives 6x = 19, so x = 19/6. B. TRUE - x was added to both sides. Solving 4x - 5 = 7 + x gives 3x = 12, so x = 4. C. FALSE - only the left side was divided by 2. Solving (3x - 5)/2 = 7 gives 3x - 5 = 14, so 3x = 19 and x = 19/3. D. TRUE - 3x - 5 = 7 is linear and the coefficient of x is 3, which is not 0, so it has exactly one solution. E. FALSE - the signs on the left were flipped but the right side was left alone. Solving 5 - 3x = 7 gives -3x = 2, so x = -2/3. Check: substitute x = 4 into the original equation 3x - 5 = 7. Then 3(4) - 5 = 12 - 5 = 7. Correct. Answer: B and D.
Trap. Selecting A or C because the equation still 'looks like' the original. A doubles only the 3x term and gives x = 19/6; C divides only the left side and gives x = 19/3. Every term on both sides must receive the identical operation.
Q5 · ALG-LIN-013 — Answer: A
Step 1 — divide both sides by 3, since 3 multiplies the whole bracket: x + 4 = 7. Step 2 — subtract 4 from both sides: x = 3. Alternative route: distribute first. 3x + 12 = 21, so 3x = 9 and x = 3. Check: substitute into the original equation. 3(3 + 4) = 3(7) = 21. Correct. Answer: A.
Trap. Cancelling the 3 on the left without dividing the right, giving x + 4 = 21 and x = 17, which is choice E. Choice B, 7, is the value of x + 4 rather than of x, and choice C, 9, is the value of 3x. Choice D, 11, comes from distributing to 3x + 12 = 21 and then moving the 12 without changing its sign: 3x = 21 + 12 = 33.
Q6 · ALG-LIN-017 — Answer: 14
Step 1 — clear both denominators at once by multiplying both sides by 6, the LCM of 3 and 2: 2(2x - 1) = 3(x + 4). Step 2 — distribute on each side: 4x - 2 = 3x + 12. Step 3 — collect: subtract 3x from both sides, giving x - 2 = 12, so x = 14. Check: substitute into the original equation. (2(14) - 1)/3 = 27/3 = 9, and (14 + 4)/2 = 18/2 = 9. Correct. Answer: 14.
Trap. Cross-multiplying but distributing only to the first term, writing 4x - 1 = 3x + 4 and getting x = 5. The multiplier must reach every term inside the bracket.
Q7 · ALG-LIN-018 — Answer: E
Step 1 — distribute the 3: 3(2x - 4) = 6x - 12. Step 2 — distribute the -2 across both terms: -2(x + 1) = -2x - 2. Step 3 — combine: 6x - 12 - 2x - 2 = 4x - 14. Step 4 — solve: 4x - 14 = 14, so 4x = 28 and x = 7. Check: substitute into the original equation. 3(2(7) - 4) - 2(7 + 1) = 3(10) - 2(8) = 30 - 16 = 14. Correct. Answer: E.
Trap. Writing -2(x + 1) as -2x + 2, which flips the sign of the 1 instead of carrying the minus onto it: that gives 6x - 12 - 2x + 2 = 4x - 10 = 14 and x = 6, choice C. Choice D, 25/4, is the neighbouring slip -2x + 1, giving 4x - 11 = 14. Choice B, 5, comes from distributing the 3 to 2x only, 6x - 4; choice A, 4, is both bracket slips at once, 6x - 4 - 2x + 2 = 4x - 2 = 14.
Q8 · ALG-LIN-021 — Answer: A
Step 1 — translate in the order the operations happen. Let the number be n. 'Increased by 12' gives n + 12; 'then multiplied by 3' multiplies that whole result, giving 3(n + 12). Step 2 — write the equation: 3(n + 12) = 51. Step 3 — divide both sides by 3: n + 12 = 17. Step 4 — subtract 12 from both sides: n = 5. Check: substitute into the original wording. 5 increased by 12 is 17, and 17 multiplied by 3 is 51. Correct. Answer: A.
Trap. Reversing the order and writing 3n + 12 = 51, which gives n = 13, choice B; the multiplication applies to the whole result, so the bracket is required. Choice C, 15, multiplies only the 12, giving n + 36 = 51. Choice D, 17, is the value of n + 12 rather than of n. Choice E, 29, divides 51 by 3 correctly and then adds 12 instead of subtracting it.
Q9 · ALG-LIN-024 — Answer: B
Step 1 — notice that the comparison does not need the value of x. Whatever x is, x + 1 is exactly 1 more than x, so Column B is greater. Step 2 — confirm that x is in fact a single fixed number, so the comparison is well defined. 5x - 3 = 3x + 9 gives 2x = 12, so x = 6. Step 3 — compare: Column A = 6, Column B = 7. Check: substitute into the original equation. 5(6) - 3 = 27 and 3(6) + 9 = 27. Correct. Answer: B.
Trap. Solving for x, getting 6, and then comparing 6 against the literal expression 'x + 1' as though the answer depended on the sign of x. It does not: x + 1 exceeds x for every real x. Answering D because a variable appears in Column B is the error this question is built to catch.
Q10 · ALG-LIN-027 — Answer: 7
Step 1 — express everything in one unknown. The length is 3 times the width, so l = 3w. Step 2 — substitute into the formula: 56 = 2(3w) + 2w. Step 3 — simplify: 56 = 6w + 2w = 8w. Step 4 — divide by 8: w = 7. Check: substitute into the original formula. w = 7 gives l = 21, and P = 2(21) + 2(7) = 42 + 14 = 56. Correct. Answer: 7.
Trap. Substituting w = 3l instead of l = 3w, which reverses the relationship and gives 56 = 2l + 6l = 8l, so l = 7 and w = 21 - a rectangle whose 'width' is three times its length. A second error is dropping one of the 2s and solving 56 = 3w + w = 4w, giving w = 14.
Q11 · ALG-LIN-030 — Answer: B
Step 1 — treat a, b and c as if they were ordinary numbers; the steps are the same as for a numerical equation. Step 2 — add b to both sides: ax = c + b. Step 3 — divide both sides by a, which is allowed because a is not 0: x = (b + c)/a. Check: substitute back into the original equation. a[(b + c)/a] - b = (b + c) - b = c. Correct. Answer: B.
Trap. Moving the -b across the equals sign without changing its sign, giving x = (c - b)/a, which is choice A. Choice C is that same sign error with the two terms also swapped. Choice D, c/a - b, divides only the c term by a and leaves the b outside the division. Choice E, c - b/a, keeps the wrong sign on b as in A and then divides only that term by a, leaving c undivided.
Q12 · ALG-LIN-033 — Answer: A, D
Translate first. 'Three more than a number n' is n + 3. 'The number decreased by 9' is n - 9, and 'twice as much as' that quantity doubles the whole quantity, so it is 2(n - 9). A. TRUE - n + 3 = 2(n - 9) is the direct translation. Distributing gives n + 3 = 2n - 18, so n = 21. B. FALSE - the bracket has been dropped, so only n was doubled and the 9 was not. Solving n + 3 = 2n - 9 gives n = 12. C. FALSE - 'three more than n' is n + 3, not n - 3. Solving n - 3 = 2(n - 9) gives n - 3 = 2n - 18, so n = 15. D. TRUE - n = 21, and 21 = 3 * 7, so n is a multiple of 7. E. FALSE - n = 21, which is greater than 20. Only the two error values, 12 and 15, are less than 20. Check: substitute n = 21 into the original statement. Three more than 21 is 24. The number decreased by 9 is 21 - 9 = 12, and twice as much as that is 24. Correct. Answer: A and D.
Trap. Dropping the bracket and writing n + 3 = 2n - 9, which doubles the number but not the 9; that gives n = 12, and the student who does it also picks E. Reading 'three more than n' as n - 3 gives n = 15, also less than 20.
Q13 · ALG-LIN-038 — Answer: B, C, E
Test x = 2 in each equation, and check that x = 2 is permitted before accepting it. A. FALSE - cancelling gives (x - 2)(x + 2)/(x - 2) = x + 2, so the equation looks like x + 2 = 4 and x = 2. But that cancellation divided by (x - 2), which is 0 at x = 2. Substituting x = 2 into the original left side gives 0/0, which is undefined, so x = 2 is not a solution; the equation has no solution at all. B. TRUE - 3(2 - 2) + 5 = 3(0) + 5 = 5. Correct. C. TRUE - (2 + 6)/(2 + 2) = 8/4 = 2, and the denominator 2 + 2 = 4 is not 0, so the value is permitted. D. FALSE - 3(2) - 2(2 + 1) = 6 - 6 = 0, not 4. Solving properly: 3x - 2x - 2 = 4, so x = 6. E. TRUE - (2 + 4)/3 + (2 - 4)/2 = 6/3 + (-2)/2 = 2 - 1 = 1. Correct. Check: clear the fractions in E to confirm. Multiplying every term by 6 gives 2(x + 4) + 3(x - 4) = 6, so 5x - 4 = 6 and x = 2, matching the substitution. Answer: B, C and E.
Trap. Accepting A because cancelling (x - 2) leaves x + 2 = 4, which gives x = 2. Dividing by (x - 2) is only legal when x is not 2, so that cancellation destroys the very value it produces - the original equation is undefined at x = 2. Selecting D comes from writing -2(x + 1) as -2x + 2, which turns the equation into x + 2 = 4 and makes x = 2 look right; the true solution is x = 6.
Q14 · ALG-LIN-034 — Answer: A
Step 1 — solve the first formula for r. Multiply both sides by (1 - r), which is not 0: S(1 - r) = a. Step 2 — divide both sides by S, which is allowed because S is not 0: 1 - r = a/S. Step 3 — isolate r. Subtract 1 from both sides: -r = a/S - 1. Multiply by -1, changing the sign of every term: r = 1 - a/S = (S - a)/S. Step 4 — build the denominator of T: 1 + r = 1 + (S - a)/S = (S + S - a)/S = (2S - a)/S. Step 5 — substitute into T = a/(1 + r). Dividing by a fraction multiplies by its reciprocal: T = a * S/(2S - a) = aS/(2S - a). Check: take S = 4 and a = 3. Then r = (4 - 3)/4 = 1/4, and the original formula gives a/(1 - r) = 3/(1 - 1/4) = 3/(3/4) = 4 = S, so that value of r is right. Now T = a/(1 + r) = 3/(5/4) = 12/5, and aS/(2S - a) = (3)(4)/(8 - 3) = 12/5. Correct. Answer: A.
Trap. Stopping at Step 3 with -r and reporting r = a/S - 1. Then 1 + r = a/S and T = a/(a/S) = S, which is choice B. Choice C comes from computing 1 + r as 1 + a/S instead of 1 + (1 - a/S); choice D is the correct work with the final compound fraction inverted; choice E multiplies a by (1 + r) instead of dividing by it.
Q15 · ALG-LIN-040 — Answer: 5
Step 1 — notice that the equation only looks quadratic. It is a difference of two squares, and the squared terms will cancel. Step 2 — use A^2 - B^2 = (A - B)(A + B) with A = 2x + 5 and B = 2x - 3. A - B = (2x + 5) - (2x - 3) = 8, a constant. A + B = (2x + 5) + (2x - 3) = 4x + 2. Step 3 — the equation becomes 8(4x + 2) = 176, which is linear. Step 4 — divide both sides by 8: 4x + 2 = 22. Step 5 — solve: 4x = 20, so x = 5. Long route, same answer: expanding gives (4x^2 + 20x + 25) - (4x^2 - 12x + 9) = 32x + 16, and 32x + 16 = 176 gives 32x = 160 and x = 5. Check: substitute into the original equation. (2(5) + 5)^2 - (2(5) - 3)^2 = 15^2 - 7^2 = 225 - 49 = 176. Correct. Answer: 5.
Trap. Treating the equation as quadratic and reaching for the quadratic formula. The x^2 terms cancel, so no quadratic ever appears. The two live arithmetic errors: expanding (2x - 3)^2 as 4x^2 - 9, which leaves 20x + 34 = 176 and x = 7.1, and subtracting only the first term of the second square, which leaves 8x + 34 = 176 and x = 17.75.
Q16 · ALG-LIN-041 — Answer: C
Step 1 — collect the variable terms: px - qx = 3 - 5, so (p - q)x = -2. Step 2 — read the cases. If p - q is not 0, this has exactly one solution, x = -2/(p - q). So 'no solution' forces p - q = 0. Step 3 — check that p = q really does give no solution. The equation becomes 0 = -2, a contradiction, so there is indeed no x. (Had the constants been equal too, this would instead be the infinitely-many case.) Step 4 — therefore p = q and the two columns are equal. Check: take p = q = 7. Then 7x + 5 = 7x + 3 gives 5 = 3, impossible, so there is no solution. Correct. Answer: C.
Trap. Answering D on the grounds that p and q are unknown. The no-solution condition pins the relationship exactly: the coefficients of x must match while the constants differ, so p = q even though neither value is known. Answering A or B comes from assuming the side with the larger constant needs the larger coefficient.
Q17 · ALG-LIN-042 — Answer: B
Step 1 — name one unknown. Let B be Ben's age now. Step 2 — translate the first sentence. '4 years more than twice Ben's age' is 2B + 4, so Jian's age now is 2B + 4. Step 3 — translate the second sentence. Six years ago Jian was (2B + 4) - 6 = 2B - 2 and Ben was B - 6, and Jian's age then was three times Ben's age then: 2B - 2 = 3(B - 6). Step 4 — distribute the 3 to both terms in the bracket: 2B - 2 = 3B - 18. Collect: 18 - 2 = 3B - 2B, so B = 16, and Jian is 2(16) + 4 = 36. Step 5 — the second stage needs its own unknown. Let t be the number of years from now until the stated ratio holds. In t years Jian is 36 + t and Ben is 16 + t; both ages grow, so t must be added on both sides: 36 + t = 2(16 + t). Step 6 — distribute and collect: 36 + t = 32 + 2t, so 36 - 32 = 2t - t and t = 4. Step 7 — the question asks for the number of years, not for either age at that time. Check: substitute into the original wording. Jian is 36 and Ben is 16 now, and 36 is 4 more than twice 16. Six years ago Jian was 30 and Ben was 10, and 30 = 3 * 10. In 4 years Jian is 40 and Ben is 20, and 40 = 2 * 20. Correct. Answer: B.
Trap. Choice E, 40, is Jian's age when the ratio holds, and choice D, 20, is Ben's age then; the question asks for the waiting time, not an age. Choice C, 16, stops at Ben's age now, the answer to the first stage only. Choice A, 2, comes from ageing Ben but not Jian in the second stage, solving 36 = 2(16 + t); a gap of years applies to both people at once. The first stage has its own live error: subtracting 6 from Jian's side only, 2B - 2 = 3B - 6, gives B = 4 and a Jian who is younger than his own past self.
Q18 · ALG-LIN-043 — Answer: 3
Step 1 — solve the first equation for x, treating n as a number. Distribute: 3x - 3n = 2x + 5. Collect: x = 3n + 5. Step 2 — solve the second equation for x: x = 20 - 2n. Step 3 — 'the same solution' means these two expressions for x are the same number: 3n + 5 = 20 - 2n. Step 4 — collect: 3n + 2n = 20 - 5, so 5n = 15 and n = 3. Check: substitute n = 3 into both original equations. The first becomes 3(x - 3) = 2x + 5, that is 3x - 9 = 2x + 5, so x = 14. The second becomes x + 6 = 20, so x = 14. The solutions agree. Correct. Answer: 3.
Trap. Reporting 14, the common solution, instead of n. The question asks for the constant, not for x. The other live error is distributing the 3 to x only, writing 3x - n = 2x + 5; that gives x = n + 5, then n + 5 = 20 - 2n and n = 5, which fails the check because the two equations then have solutions 10 and 10 only by coincidence of the wrong first step.
Q19 · ALG-LIN-044 — Answer: A
Step 1 — cross-multiply: c(x - a) = b(x + a). Step 2 — distribute on both sides: cx - ca = bx + ba. Step 3 — collect the x terms on the left and the constants on the right, changing the sign of each term that moves: cx - bx = ba + ca. Step 4 — factor each side: x(c - b) = a(b + c). Step 5 — divide by (c - b), which is nonzero because b is not equal to c: x = a(b + c)/(c - b). Check: substitute back into the original equation with a = 1, b = 2, c = 3. Then x = 1(2 + 3)/(3 - 2) = 5, and (x - a)/b = (5 - 1)/2 = 2 while (x + a)/c = (5 + 1)/3 = 2. Correct. Answer: A.
Trap. Cross-multiplying the wrong pairs, b(x - a) = c(x + a), which gives x(b - c) = a(b + c) and so x = a(c + b)/(b - c), choice D: the correct magnitude with the denominator reversed in sign. Choice C, -a, comes from failing to change the sign of -ca when it moves right: cx - bx = ba - ca gives x(c - b) = -a(c - b), so x = -a. Choice B divides by (b + c) instead of by (c - b), exchanging the two factors. Choice E is a false shortcut: the numerators differ by 2a and the denominators by c - b, and those differences are divided one by the other.
Q20 · ALG-LIN-045 — Answer: D
Step 1 — clear the bracket once, keeping both letters: 4 - cx - 2c = k. Step 2 — isolate x: -cx = k - 4 + 2c, so x = (4 - 2c - k)/c. Step 3 — compare by difference rather than by computing either column. Column A - Column B = [(4 - 2c - 1) - (4 - 2c - 7)]/c = 6/c. Step 4 — read that difference. It is never 0, since c is never 0, so the columns are never equal. But its sign is the sign of c, and the stem fixes only that c is not 0. Step 5 — both signs are permitted. If c > 0 the coefficient of x after the bracket is cleared is -c, a negative number, so dividing by it reverses the order and the larger k gives the smaller x: Column A is greater. If c < 0 that coefficient is positive, the order is preserved, and Column B is greater. Check: substitute back into the original equation. With c = 3, k = 1 gives 4 - 3(x + 2) = 1, so -3x = 3 and x = -1; k = 7 gives 4 - 3(x + 2) = 7, so -3x = 9 and x = -3; here Column A is greater. With c = -3, k = 1 gives 4 + 3(x + 2) = 1, so 3x = -9 and x = -3; k = 7 gives 4 + 3(x + 2) = 7, so 3x = -3 and x = -1; here Column B is greater. Two permitted values of c give opposite orders. Correct. Answer: D.
Trap. Answering A is the trained reflex from the fixed-coefficient version of this equation: clear the bracket, see a minus in front of x, and conclude that the larger k must give the smaller x. That conclusion depends on c being positive, which the stem never says. Answering B is the untrained reflex, that a larger number on the right forces a larger x, and it happens to be right only when c is negative. Answering C confuses 'the difference does not depend on k' with 'the difference is 0'; the difference is 6/c, which is never 0.
Q21 · ALG-LIN-049 — Answer: B
Step 1 — solve for x: x = (n^2 + 2n)/(n + 1). The division is legal because n is positive, so n + 1 is not 0. Step 2 — do not compute; compare instead. Column B squared would be (n + 1)^2 = n^2 + 2n + 1. Step 3 — so (n + 1)(n + 1) = n^2 + 2n + 1, which is exactly 1 more than the right side of the given equation. Step 4 — therefore (n + 1)x = n^2 + 2n < (n + 1)(n + 1). Dividing both sides by the positive number n + 1 gives x < n + 1, for every positive integer n. Check: substitute n = 3 into the original equation. 4x = 9 + 6 = 15, so x = 3.75, and n + 1 = 4. Indeed 3.75 < 4. Correct. Answer: B.
Trap. Seeing n^2 + 2n and reading it as n(n + 2), then cancelling loosely to conclude x = n + 1 and answering C. The factorisation is n(n + 2), not (n + 1)^2, and it falls one short of (n + 1)^2, so x is always just below n + 1. Answering D because n is unknown misses that the gap has the same sign for every n.
Q22 · ALG-LIN-050 — Answer: D
Step 1 — expand the left side: 2(3x - a) + 4 = 6x - 2a + 4. Step 2 — the equation is 6x - 2a + 4 = 6x + b. Step 3 — subtract 6x from both sides. The variable disappears, leaving -2a + 4 = b. Step 4 — for infinitely many solutions this leftover statement must be true for every x, and it is a statement about constants only, so it must simply hold: b = 4 - 2a. Step 5 — the question asks for b + 2a, not for a or b separately: b + 2a = (4 - 2a) + 2a = 4. Check: substitute a = 1, so b = 2, into the original equation. 2(3x - 1) + 4 = 6x + 2 on the left and 6x + 2 on the right, an identity. And b + 2a = 2 + 2 = 4. Correct. Answer: D.
Trap. Distributing the 2 onto the +4 that sits outside the bracket, giving 6x - 2a + 8 = 6x + b and b + 2a = 8, which is choice E. Choice B, -4, comes from reading -2a + 4 = b as b = -2a - 4; choice A is both errors together; choice C comes from assuming an identity forces a = b = 0.
Q23 · ALG-LIN-051 — Answer: -12
Step 1 — multiply both sides by (x - 4). This step is legal only when x is not 4, so x = 4 can never be a solution of the original equation no matter what the algebra returns. 3x + c = 5(x - 4). Step 2 — distribute and collect: 3x + c = 5x - 20, so 2x = c + 20 and x = (c + 20)/2. Step 3 — for every c the algebra hands back exactly one candidate. The only way the original equation can have no solution is for that candidate to be the forbidden value: (c + 20)/2 = 4. Step 4 — solve: c + 20 = 8, so c = -12. Check: substitute c = -12 into the original equation. (3x - 12)/(x - 4) = 3(x - 4)/(x - 4) = 3 for every permitted x, and 3 is never equal to 5, so the equation has no solution. Correct. Answer: -12.
Trap. Concluding that no value of c can fail, because x = (c + 20)/2 always comes out of the algebra. Clearing the denominator multiplies both sides by an expression containing the variable, and that is only legal away from x = 4; when the candidate root lands exactly on 4, the step has manufactured a root the original equation does not have. Reporting 4, the excluded value, instead of c is the other live error.
Q24 · ALG-LIN-052 — Answer: B
Step 1 — solve the equation for x, treating k as a number. Collect the x terms: 4x - 2x = 18 - k, so 2x = 18 - k. Step 2 — isolate x: x = (18 - k)/2. Step 3 — impose the two conditions separately. For x to be an integer, 18 - k must be even; 18 is even, so k must be even. For x to be positive, 18 - k > 0, that is k < 18. Step 4 — list the k that satisfy both inside 1 <= k <= 30: 2, 4, 6, 8, 10, 12, 14, 16. Note that k = 18 is excluded, because it gives x = 0 and 0 is not positive. Step 5 — add them. Pairing the ends, 2 + 16 = 18, 4 + 14 = 18, 6 + 12 = 18, 8 + 10 = 18, which is 4 pairs of 18: the sum is 72. Check: substitute the two endpoints back into the original equation. k = 2 gives 4x + 2 = 2x + 18, so 2x = 16 and x = 8, a positive integer. k = 16 gives 4x + 16 = 2x + 18, so 2x = 2 and x = 1, a positive integer. And k = 18 gives 4x + 18 = 2x + 18, so x = 0, correctly excluded. Correct. Answer: B.
Trap. Including k = 18 and answering 90. It gives x = 0, and 0 is not a positive integer. Choice A, 36, sums the solutions x instead of the values of k, which is not what was asked. Choice D, 153, drops the parity condition and sums every k from 1 to 17; choice E, 240, drops the positivity condition and sums every even k from 2 to 30.
Q25 · ALG-LIN-053 — Answer: A, D
Step 1 — rearrange once: (p - r)x = s - q. Everything follows from whether p - r is 0. If p - r is not 0, there is exactly one solution, x = (s - q)/(p - r), whatever the constants do. If p - r = 0, the equation reads 0 = s - q: no solution when q is not equal to s, infinitely many when q = s. A. TRUE - a real number has a multiplicative inverse exactly when it is not 0, since 0 times anything is 0 and can never be 1. So the condition says p - r is not 0, which is precisely the criterion from Step 1. The inverse is even the tool used: multiplying (p - r)x = s - q by 1/(p - r) is what produces the single value of x. B. FALSE - this is the no-solution case. Example: 2x + 1 = 2x + 5 gives 0 = 4. C. FALSE - it says nothing about the coefficients. Example: 2x + 1 = 2x + 5 satisfies q not equal to s yet has no solution. D. TRUE - substitute p = 2r into p - r: 2r - r = r, and r is not 0, so p - r is not 0 and there is exactly one solution. E. FALSE - p + r = 0 gives p - r = p - (-p) = 2p, which is not 0 only when p is not 0. The condition as stated allows p = r = 0, and then the equation is q = s: no solution if q is not equal to s, infinitely many if q = s. Check: take D with r = 3, so p = 6, and q = 1, s = 7. Then 6x + 1 = 3x + 7 gives 3x = 6 and x = 2, exactly one solution. Take A with p = 4, r = -1, q = 0, s = 5: p - r = 5, whose multiplicative inverse is 1/5, and 4x + 0 = -x + 5 gives 5x = 5 and x = 1, exactly one solution. Take E with p = r = 0, q = 1, s = 1: the equation is 1 = 1, true for every x. Correct. Answer: A and D.
Trap. Option A is the criterion itself in disguise, and the disguise is the work: 'has a multiplicative inverse' has to be turned into 'is not 0' before it can be matched against (p - r)x = s - q. Passing over it because it is phrased about a number rather than about the equation costs a correct selection. Selecting C because different constants feel as though they should force a unique answer is the mirror error: they do not, since 2x + 1 = 2x + 5 has q not equal to s and no solution at all. Selecting E is the missing-zero-case error: p + r = 0 does force p - r = 2p, but 2p is 0 when p = 0, and p = r = 0 leaves the equation with either none or infinitely many solutions.
Q26 · ALG-LIN-039 — Answer: C
Step 1 — clear the denominator: 2x + 4 = k(x - 1) = kx - k. Step 2 — collect the x terms: 2x - kx = -k - 4, so x(2 - k) = -(k + 4). Step 3 — solve, noting k = 2 must be set aside: if k = 2 the equation reads 0 = -6, a contradiction with no solution at all. For k not equal to 2, x = (k + 4)/(k - 2). Step 4 — make the divisibility visible by splitting the fraction: x = (k - 2 + 6)/(k - 2) = 1 + 6/(k - 2). Step 5 — x is an integer exactly when (k - 2) divides 6, so k - 2 is one of -6, -3, -2, -1, 1, 2, 3, 6. Step 6 — that is 8 values of k: k = -4, -1, 0, 1, 3, 4, 5, 8. None of them makes x equal to 1, since 6/(k - 2) is never 0. Check: substitute k = 5 into the original equation. x = 1 + 6/3 = 3, and (2(3) + 4)/(3 - 1) = 10/2 = 5 = k. Correct. Answer: C.
Trap. Counting only the positive divisors of 6, which gives k - 2 in {1, 2, 3, 6} and the answer 4, choice A. Divisors run both ways. Choice B, 6, comes from demanding that (k - 2) divide 4 rather than 6, by reading x = (k + 4)/(k - 2) without the split; choice D, 9, comes from getting the eight values and then also counting k = 2, which in fact gives no solution.
Q27 · ALG-LIN-054 — Answer: E
Step 1 — only one letter is ever unknown at a time. F is given as 170, and employee X has both gross and net pay given, so X pins down r by itself: 1200 - r*1200 - 170 = 850. Step 2 — solve that one equation: 1030 - 1200r = 850, so 1200r = 180 and r = 0.15. Write the surviving fraction of gross pay as 1 - r = 0.85, so every employee satisfies N = 0.85G - 170. Step 3 — the turn. Z's net pay is not handed over as a number; it is handed over as a fraction of the very quantity being solved for. '80 percent of Z's gross pay' is 0.80G, so the unknown stands on both sides of the equation: 0.85G - 170 = 0.80G. Step 4 — collect the G terms on one side and the constant on the other: 0.85G - 0.80G = 170, so 0.05G = 170. Only five hundredths of the gross pay is left to carry the whole fixed deduction, which is why the answer is so much larger than X's pay. Step 5 — divide: G = 170/0.05 = 3400. Check: substitute G = 3400 into the original formula with r = 0.15 and F = 170. 3400 - 0.15(3400) - 170 = 3400 - 510 - 170 = 2720, and 2720/3400 = 0.80, exactly 80 percent as required. The same r and F reproduce X: 1200 - 0.15(1200) - 170 = 1200 - 180 - 170 = 850. Correct. Answer: E.
Trap. Choice D, $2,720, is Z's net pay - the value of both sides of the Step 3 equation - not the gross pay asked for. Choice C, $850, comes from treating the whole 20 percent shortfall as the fixed deduction and solving 170 = 0.20G, which ignores that the tax is doing most of that work. Choice B, $680, comes from taxing the pay after the deduction, solving 0.85G - 170 = 0.80(G - 170); the formula taxes gross pay, and the deduction comes off afterwards. Choice A, $340, is the decimal slip 170/0.5 in place of 170/0.05. The deeper trap is looking for a second employee's numbers to work with: everything after Step 2 comes from Z alone.
Q28 · ALG-LIN-055 — Answer: D
Step 1 — collect the given equation: 5b - 3b = 7a - 3a + 12, so 2b = 4a + 12 and b = 2a + 6. This is one relationship between two letters; a is still free. Step 2 — substitute b = 2a + 6 into both parts of Column A. Numerator: a + 2b = a + 2(2a + 6) = 5a + 12. Denominator: a + b = a + 2a + 6 = 3a + 6. Step 3 — note what the stem's condition buys. a + b is not 0 means 3a + 6 is not 0, that is a is not -2, so the division is safe. Nothing else pins a down. Step 4 — compare by subtracting, not by cross-multiplying, because the sign of the denominator 3a + 6 is unknown and cross-multiplying by a negative would flip the comparison unnoticed: (5a + 12)/(3a + 6) - 5/3 = [3(5a + 12) - 5(3a + 6)]/[3(3a + 6)] = (15a + 36 - 15a - 30)/(9a + 18) = 6/(9a + 18). Step 5 — read that difference. The numerator 6 is fixed and positive, so the sign of the difference is the sign of 9a + 18, that is the sign of a + 2. It is never 0, so the columns are never equal, but a > -2 makes Column A greater and a < -2 makes Column B greater. Both are permitted, so the relationship cannot be determined. Check: substitute two permitted pairs into the original equation and into Column A. Take a = 0, b = 6: left = 0 + 30 = 30 and right = 0 + 18 + 12 = 30, and (0 + 12)/(0 + 6) = 2, which is greater than 5/3. Take a = -3, b = 0: left = -9 + 0 = -9 and right = -21 + 0 + 12 = -9, and (-3 + 0)/(-3 + 0) = 1, which is less than 5/3. Correct. Answer: D.
Trap. Answering C is the trap the constant 12 is there to set. Without it the relationship would be b = 2a, the numerator and denominator would both carry a factor of a, that factor would cancel, and Column A would be 5/3 for every a. The 12 breaks that cancellation: b = 2a + 6 leaves (5a + 12)/(3a + 6), whose value moves with a. Answering A or B comes from testing a single value of a - a = 0 gives 2 and suggests A, a = -3 gives 1 and suggests B - and stopping there. The one value worth hunting for is a = -2, where the denominator would vanish; the stem excludes it, which is exactly the hint that a is otherwise free on both sides of it.
Q29 · ALG-LIN-056 — Answer: 50
Step 1 — name one unknown. The white count is given, so the only unknown is the green count: let g be the number of green tokens in the original collection. The original total is g + 30. Step 2 — translate the removal, and keep hold of the fact that taking tokens out shrinks the total as well as one colour. What remains is g - 12 green tokens out of g + 30 - 12 = g + 18 tokens in all: g - 12 = (1/4)(g + 18). Step 3 — clear the fraction by multiplying both sides by 4: 4g - 48 = g + 18. Collect: 3g = 66, so g = 22. The reduced collection is 10 green and 30 white, 40 tokens in all. Step 4 — the second stage brings its own single unknown. Let n be the number of green tokens added. Adding green tokens raises the green count and the total by the same n, so n appears in the numerator and the denominator at once: (10 + n)/(40 + n) = 2/5. Step 5 — clear the fraction by cross-multiplying: 5(10 + n) = 2(40 + n), so 50 + 5n = 80 + 2n. Collect: 3n = 30 and n = 10. Step 6 — the question asks for the size of the collection at that point, not for n and not for the green count: 40 + 10 = 50. Check: substitute back into the original wording. The collection starts as 22 green and 30 white, 52 in all. Remove 12 green: 10 green out of 40 remaining, and 10/40 = 1/4. Add green until greens are 2/5: adding 10 gives 20 green out of 50, and 20/50 = 2/5. Correct. Answer: 50.
Trap. Three of the four quantities in play are not the one asked for: 52 is the original total, 10 is the number added (and also the green count before adding), and 20 is the green count at the end. The live algebraic error is in Step 4: writing (10 + n)/40 = 2/5, as though adding tokens changed the green count but not the total. That gives n = 6 and a collection of 46, which fails the check, because 16 green out of 46 is not 2/5. Step 2 carries the same error in reverse: writing g - 12 = (1/4)(g + 30) forgets that removing green tokens shrinks the total too.
Q30 · ALG-LIN-057 — Answer: A, C
Step 1 — substitute x = 5: (10 + a)/(5 - b) = 3. This is legal only because x is not equal to b, so 5 - b is not 0. Step 2 — clear the denominator: 10 + a = 3(5 - b) = 15 - 3b. Step 3 — rearrange: a + 3b = 5, equivalently a = 5 - 3b. That single relation is everything the stem gives; a and b are not individually determined. A. TRUE - a + 3b = 5 is exactly Step 3. B. FALSE - 2b - a = 10 together with a = 5 - 3b gives 2b - 5 + 3b = 10, so 5b = 15 and b = 3. That is one particular case, not a must. C. TRUE - from Step 3, a - 2 = (5 - 3b) - 2 = 3 - 3b = 3(1 - b). Since b is an integer, a - 2 is 3 times an integer. D. FALSE - b = 5 is impossible. Clearing the denominator in general gives 2x + a = 3x - 3b, so x = a + 3b; with a + 3b = 5 the only root is x = 5. If b were 5, that root would equal b, which the stem excludes, and the equation would have no permitted solution. E. FALSE - a + b = 5 with a = 5 - 3b gives 5 - 3b + b = 5, so 2b = 0 and b = 0. It holds in one case only. Check: substitute a test pair back into the original equation (2x + a)/(x - b) = 3. Take b = 1, so a = 2. Then (2(5) + 2)/(5 - 1) = 12/4 = 3. Correct, and a + 3b = 2 + 3 = 5 and a - 2 = 0 = 3 * 0, while a + b = 3, not 5. Answer: A and C.
Trap. Selecting E because a + b = 5 resembles the correct a + 3b = 5; it holds only when b = 0. Selecting D because the stem calls a and b integers and says nothing more about b - but b = 5 would place the excluded value x = b exactly at the root x = 5, leaving the equation with no permitted solution.