Math Section Strategy

Backsolving, picking numbers, estimation, calculator use, traps and pacing on ACT Math.

Worked Examples

<details> <summary><b>Example 1: Solve for x, then answer the real question</b></summary>

Question: If $4x - 7 = 21$, what is the value of $x + 3$?

Solution:

1. What is asked? The value of $x + 3$, not $x$.
2. $4x = 28$, so $x = 7$.
3. $x + 3 = 10$. ✓

Trap: The choices would include 7 (the value of $x$). A student who stops at step 2 picks it and loses an easy point.

</details> <details> <summary><b>Example 2: A multi-step problem with several tempting stopping points</b></summary>

Question: The length of a rectangle is 3 cm more than twice its width. The perimeter is 54 cm. What is the area of the rectangle, in square centimeters?

Solution:

1. What is asked? The area.
2. Let the width be $w$; the length is $2w + 3$.
3. Perimeter: $2(w + 2w + 3) = 54 \implies 3w + 3 = 27 \implies w = 8$.
4. Length $= 2(8) + 3 = 19$.
5. Area $= 8 \times 19 = 152$ square centimeters. ✓

Trap: 8 (the width), 19 (the length), and 54 (the perimeter) are all numbers you wrote down along the way. Only 152 answers the question.

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Worked Examples

<details> <summary><b>Example 1: Matching a radical answer with decimals</b></summary>

Question: Which of the following is equal to $\sqrt{72}$? The choices are $6\sqrt{2}$, $8\sqrt{3}$, $4\sqrt{3}$, and $2\sqrt{6}$.

Solution:

1. By hand: $72 = 36 \times 2$, so $\sqrt{72} = 6\sqrt{2}$. ✓
2. Calculator check: $\sqrt{72} \approx 8.485$. The choices evaluate to $6\sqrt{2} \approx 8.485$, $8\sqrt{3} \approx 13.86$, $4\sqrt{3} \approx 6.93$, and $2\sqrt{6} \approx 4.90$. Only $6\sqrt{2}$ matches.

Takeaway: If you forget how to simplify a radical, the decimal comparison still finds the answer in seconds.

</details> <details> <summary><b>Example 2: Letting the graph solve a system</b></summary>

Question: The graphs of $y = x^2 - 3$ and $y = 2x$ intersect at two points. What is the sum of the $x$-coordinates of those points?

Solution (algebra): Set them equal: $x^2 - 3 = 2x \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0$, so $x = 3$ or $x = -1$. The sum is $2$. ✓

Solution (graphing): Graph both equations and use the intersect feature twice: the points are $(-1, -2)$ and $(3, 6)$. Same sum, $2$.

Takeaway: If the quadratic does not factor nicely, the graph still gives the intersection points.

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Worked Examples

<details> <summary><b>Example 1: A ticket problem, solved by testing choices</b></summary>

Question: A theater sold 40 tickets for a total of 380 dollars. Adult tickets cost 12 dollars and student tickets cost 7 dollars. How many adult tickets were sold? Choices: 12, 16, 20, 24.

Solution:

1. Label: choice = number of adult tickets; students = 40 minus the choice.
2. Test 16 (a middle value): $16 \times 12 = 192$ and $24 \times 7 = 168$. Total 360, which is too low.
3. More adult tickets raise the total (each one adds 5 dollars over a student ticket), so go bigger.
4. Test 20: $20 \times 12 = 240$ and $20 \times 7 = 140$. Total 380. ✓

Answer: 20 adult tickets. Testing 16 also ruled out 12, since 12 would give an even smaller total.

</details> <details> <summary><b>Example 2: Backsolving a radical equation</b></summary>

Question: What value of $x$ satisfies $\sqrt{x + 7} = x - 5$? Choices: 2, 9, 11, 13.

Solution:

1. Test 9: $\sqrt{16} = 4$ and $9 - 5 = 4$. ✓
2. For comparison, test 2: $\sqrt{9} = 3$ but $2 - 5 = -3$. ✗ A square root is never negative.

Why it matters: Squaring both sides produces $x = 2$ and $x = 9$. A student who solves the quadratic and picks the first root chooses 2, an extraneous solution. Backsolving tests the original equation, so it never falls for this.

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Worked Examples

<details> <summary><b>Example 1: Variables in the choices</b></summary>

Question: Pens cost $p$ cents each. How many pens can be bought with $d$ dollars? Choices: $\frac{100d}{p}$, $\frac{dp}{100}$, $\frac{100p}{d}$, $\frac{d}{100p}$.

Solution:

1. Pick $p = 50$ cents and $d = 2$ dollars.
2. Target: 2 dollars is 200 cents, and $200 \div 50 = 4$ pens.
3. Plug in: $\frac{100(2)}{50} = 4$ ✓, $\frac{2(50)}{100} = 1$, $\frac{100(50)}{2} = 2500$, $\frac{2}{5000} = 0.0004$.

Answer: $\frac{100d}{p}$. Only one choice hit 4, so no second round is needed.

</details> <details> <summary><b>Example 2: A "must be true" question</b></summary>

Question: If $x < y < 0$, which of the following must be true? Choices: $xy < 0$, $\frac{x}{y} > 1$, $x + y > 0$, $x^2 < y^2$.

Solution:

1. Pick $x = -4$ and $y = -2$ (both negative, $x$ smaller).
2. $xy = 8$, not negative ✗. $\frac{x}{y} = 2 > 1$ ✓. $x + y = -6$, not positive ✗. $x^2 = 16$ and $y^2 = 4$, so $x^2 < y^2$ is false ✗.
3. Try another pair to confirm, $x = -3$, $y = -1$: $\frac{x}{y} = 3 > 1$ ✓.

Answer: $\frac{x}{y} > 1$. Since $x$ is farther from zero than $y$ and both are negative, the quotient is always positive and greater than 1.

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Worked Examples

<details> <summary><b>Example 1: Estimating instead of calculating</b></summary>

Question: What is 19.8% of 401? Choices: 7.94, 79.4, 321.6, 794.

Solution:

1. Round: 19.8% is about 20%, and 401 is about 400.
2. 20% of 400 is 80.
3. Only 79.4 is close to 80. ✓ (7.94 and 794 are decimal-point slips; 321.6 is the remaining 80.2% of 401, the part that is not taken.)

Time used: about 10 seconds, with no calculator.

</details> <details> <summary><b>Example 2: Eliminating with the triangle inequality</b></summary>

Question: Two sides of a triangle have lengths 7 and 10. Which of the following could be the perimeter? Choices: 19, 20, 27, 34.

Solution:

1. The third side $s$ must satisfy $10 - 7 < s < 10 + 7$, so $3 < s < 17$.
2. The perimeter is $17 + s$, so it must be between $20$ and $34$, not including either end.
3. 19 and 20 are too small (they need $s \le 3$); 34 needs $s = 17$, which makes a flat "triangle." Only 27 ($s = 10$) works. ✓

Takeaway: You did not need the exact third side; you only needed the bounds.

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Worked Examples

<details> <summary><b>Example 1: Drawing the missing diagram</b></summary>

Question: Two cyclists leave the same point. One rides 30 miles due north and the other rides 40 miles due east. How far apart are they, in miles?

Solution:

1. Draw it: an arrow up (30) and an arrow right (40) from one point. The distance between the riders is the segment joining the arrow tips.
2. The arrows meet at a right angle, so the distance is a hypotenuse: $\sqrt{30^2 + 40^2} = \sqrt{2500} = 50$. ✓
3. Recognize the 3-4-5 triangle scaled by 10 for a faster check.

Trap avoided: Without a diagram, many students add the distances (70), but the riders are not on one straight road.

</details> <details> <summary><b>Example 2: An extraneous solution in a rational equation</b></summary>

Question: How many real solutions does $\frac{x}{x - 2} = \frac{2}{x - 2} + 3$ have?

Solution:

1. Multiply every term by $x - 2$: $x = 2 + 3(x - 2) = 3x - 4$.
2. Solve: $2x = 4$, so $x = 2$.
3. Check in the original: $x = 2$ makes both denominators zero, so it is not allowed.

Answer: zero solutions. The equation has no solution, even though the algebra produced a number.

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Worked Examples

<details> <summary><b>Example 1: Recognize a structure, then solve for what is asked</b></summary>

Question: If $x^2 - y^2 = 24$ and $x - y = 4$, what is the value of $x$?

Solution:

1. Recognize the difference of squares: $x^2 - y^2 = (x + y)(x - y)$.
2. Substitute: $(x + y)(4) = 24$, so $x + y = 6$.
3. Add the equations $x + y = 6$ and $x - y = 4$: $2x = 10$, so $x = 5$. ✓
4. Check with a different method: $y = 1$, and $25 - 1 = 24$ ✓.

Strategies used: solving for an expression ($x + y$) and checking by a second method.

</details> <details> <summary><b>Example 2: Diagram plus backsolving</b></summary>

Question: The length of a rectangle is 4 inches more than its width, and its diagonal is 20 inches. What is the width, in inches? Choices: 8, 10, 12, 16.

Solution:

1. Draw the rectangle with its diagonal: a right triangle with legs $w$ and $w + 4$ and hypotenuse 20.
2. Backsolve with a middle choice, 12: legs 12 and 16, and $12^2 + 16^2 = 144 + 256 = 400 = 20^2$ ✓.
3. Pattern check: 12-16-20 is the 3-4-5 triangle scaled by 4.

Trap: 16 is the length, not the width. The algebraic route ($w^2 + 4w - 192 = 0$) gives the same answer but takes longer.

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