SAT Math Tips and Strategies, 24 Explained Examples

The answer is $83$.

1. Group $9a + 6b$ as $3(3a + 2b)$, so the target is $3(3a + 2b) - 4$.
2. Substitute the known whole expression to get $3(29) - 4$.
3. Calculate $87 - 4 = 83$. The individual values of $a$ and $b$ are not needed.

As a check, expanding $3(3a + 2b) - 4$ gives exactly the requested expression, so every pair allowed by the equation gives $3(29) - 4 = 83$. For example, the pair $(a, b) = (9, 1)$ satisfies the equation and produces $83$.

Watch for this. A common mistake is to multiply only one term, or to find one pair of values for $a$ and $b$ and assume those values are the only ones.

The video then compares a separate target that matches only in part, $9a + 4b$. The pairs $(9, 1)$ and $(7, 4)$ both satisfy $3a + 2b = 29$, yet they give $85$ and $79$, so a partial match does not fix one value. Watch this comparison.

The answer is $27$ liters.

1. Convert $2.5$ minutes to $150$ seconds.
2. Multiply $180$ milliliters per second by $150$ seconds to get $27{,}000$ milliliters.
3. Convert $27{,}000$ milliliters to $27$ liters.

In one line, with every unit canceling except liters,

As a check, $180$ milliliters per second is $0.18$ liters per second, and $0.18 \times 150 = 27$. Another way, one minute delivers $10.8$ liters, and $2.5 \times 10.8 = 27$.

Watch for this. A common mistake is to multiply $180$ by $2.5$, which treats minutes as seconds, or to keep a count in milliliters with a liter label. The multiplication also relies on the constant rate, since a changing rate would need information about how it changes.

The video then applies the same unit check to a speed, converting $72$ kilometers per hour to meters per second. Multiplying by $\frac{1{,}000 \text{ m}}{1 \text{ km}}$ removes kilometers, and multiplying by $\frac{1 \text{ h}}{3{,}600 \text{ s}}$ removes hours, which gives $20$ meters per second. Watch this comparison.

The answer is $12$ batteries.

1. Let $n$ be the number of batteries, a nonnegative integer. The total mass gives $7 + 4n \le 58$.
2. Subtract $7$ and divide by $4$ to get $n \le \frac{51}{4} = 12.75$.
3. The greatest whole number allowed is $12$.
4. Twelve batteries give $55$ kilograms, and thirteen give $59$ kilograms, which exceeds the limit.

As a check, $7 + 4(12) = 55 \le 58$, while $7 + 4(13) = 59 > 58$. These neighboring integers establish the maximum.

Watch for this. A common mistake is to round $12.75$ to $13$ without checking the inequality, or to forget the mass of the rack.

The video then keeps the boundary $12.75$ and compares two requests on separate number lines. For a maximum with $n \le 12.75$, the count $12$ works and $13$ is too large. For a minimum with $n \ge 12.75$, the count $12$ is too small and $13$ is the first whole number that works. The decimal is the same but the allowed side has changed, so no single rule to round up or down would work. Watch this comparison.

The answer is $7$.

1. Subtract the second equation from the first, $(4x + 3y) - (4x - y) = 38 - 10$.
2. This gives $4y = 28$, so $y = 7$.
3. For a check, the second equation then gives $x = \frac{17}{4}$, which satisfies both equations.

As a check, at $\left(\frac{17}{4}, 7\right)$ the left sides are $17 + 21 = 38$ and $17 - 7 = 10$. The two lines have different slopes, so the system has exactly one solution.

Watch for this. A common mistake is to write $3y - y$ instead of $3y - (-y)$ when subtracting the second equation.

The video then uses the same coefficient scan to choose addition. In the separate system $3u + 2v = 11$ and $-3u + 5v = 17$, the coefficients of $u$ are opposites, so adding the complete equations gives $7v = 28$ and $v = 4$. Substitution gives $u = 1$, and both equations check. Watch this comparison.

The answer is (C), $7$ centimeters.

1. For a proposed width $w$, the length is $w + 3$ and the area is $w(w + 3)$.
2. The widths $5$, $6$, $7$ and $10$ give the areas $40$, $54$, $70$ and $130$.
3. Only the width $7$ gives the required area, and its length is $10$.

As a check, $7(7 + 3) = 70$ and the two side lengths differ by $3$. All four choices were tested, and exactly one satisfies the area.

Choice (D), $10$ centimeters, is the length of the correct rectangle, but the question asks for the width, and a width of $10$ would give an area of $130$ square centimeters. Choices (A) and (B) give areas of $40$ and $54$ square centimeters, which are too small.

The video then shows when the direction of the tests can guide you. Because the width is positive, both factors $w$ and $w + 3$ grow as $w$ grows, so the area grows too. An area below $70$ says to try a larger width, and an area above $70$ says to try a smaller one. That is a mathematical reason to choose the next test, while the position of a choice in the list is not. Watch this comparison.

The answer is no. At $x = 1$ the two sides are $9$ and $5$, and the correct expansion is $x^2 + 4x + 4$.

1. At $x = 0$ both proposed sides equal $4$, so this value does not expose the error.
2. At $x = 1$ the left side is $(1 + 2)^2 = 9$ while the right side is $1^2 + 4 = 5$, which disproves the claim that the equation holds for every real $x$.
3. Expand $(x + 2)(x + 2)$ to get $x^2 + 2x + 2x + 4 = x^2 + 4x + 4$.

As a check, the difference between the two proposed sides is $4x$. It is zero at $x = 0$ but not at $x = 1$, which is why the first test missed the error.

Watch for this. A common mistake is to treat agreement at one input as proof of an identity.

The video then shows why $0$ and $1$ can be weak tests. Suppose someone claims that $x^2 \ge x$ for every real $x$. Both $0$ and $1$ satisfy it, but $\frac{1}{2}$ is an allowed input whose square, $\frac{1}{4}$, is less than $\frac{1}{2}$. That single fraction disproves the claim. Watch this comparison.

The answer is that the expression is defined for every real $x$ except $4$, and on that domain it equals $x + 4$.

1. Record the condition $x - 4 \ne 0$, so $x \ne 4$.
2. Factor the numerator as $x^2 - 16 = (x - 4)(x + 4)$.
3. Cancel the common nonzero factor to get $x + 4$, while keeping $x \ne 4$.
4. The graph of the original expression has a hole at $(4, 8)$. The polynomial $x + 4$ on its own is defined for every real number.

As a check, at $x = 6$ both expressions equal $10$, since $\frac{36 - 16}{6 - 4} = \frac{20}{2} = 10$. At $x = 4$ the original denominator is zero, even though $x + 4$ would give $8$.

Watch for this. A common mistake is to include $x = 4$ after canceling because the simpler polynomial has a value there.

The video then compares a fraction with no common factor, $\frac{x + 4}{x + 2}$. At $x = 2$ it equals $\frac{6}{4} = \frac{3}{2}$, but erasing the $x$ terms would leave $\frac{4}{2} = 2$, a different value. Only whole nonzero factors cancel, never terms inside a sum. Watch this comparison.

The answer is $p = 3$.

1. Rewrite the right side as $2(p - 3)$.
2. At $p = 3$ the equation becomes $0x = 0$, which is true for every real $x$.
3. For $p \ne 3$, dividing by $p - 3$ is valid and gives $x = 2$, a single solution.

As a check, at $p = 3$ both sides are zero whatever $x$ is. For every other value of $p$, the equation $(p - 3)(x - 2) = 0$ forces $x = 2$. No value of $p$ produces $0x$ equal to a nonzero constant in this equation.

Watch for this. A common mistake is to divide right away and conclude that $x = 2$ is the only possibility for every value of $p$. A zero coefficient alone does not promise infinitely many solutions either, since $0x = 5$ has none.

The video then sums up the three cases for an equation $Ax = B$. If $A \ne 0$, there is exactly one solution, $x = \frac{B}{A}$. If $A = 0$ and $B = 0$, every real $x$ works. If $A = 0$ but $B \ne 0$, no $x$ works. Watch this comparison.

The answer is $x = 2$.

1. The right side must be nonnegative, so $x \ge -1$. This also keeps the radicand nonnegative.
2. Square both sides to get $x + 7 = (x + 1)^2$, or $x^2 + x - 6 = 0$.
3. Factor as $(x + 3)(x - 2) = 0$ to get the candidates $x = 2$ and $x = -3$.
4. The original equation holds at $2$, since both sides equal $3$. At $-3$ it gives $2 = -2$, so reject $-3$.

As a check, the condition $x \ge -1$ already excludes $-3$, and direct substitution confirms $2$ and rejects $-3$. The extra candidate appears because $2$ and $-2$ have the same square.

Watch for this. A common mistake is to keep every root of the squared equation as a solution of the original equation.

The video then shows a case where the sign check settles everything at once. The equation $\sqrt{z} = -4$ has no real solution, because a real square root is never negative. Squaring would give $z = 16$, but then the left side is $4$, not $-4$. Watch this comparison.

The answer is $5$, reached when $x = 6$.

1. Rewrite $x^2 - 12x + 41$ as $(x - 6)^2 + 5$.
2. A square is never negative, so the expression is at least $5$.
3. At $x = 6$ the square is zero, so the bound $5$ is reached.

As a check, expanding $(x - 6)^2 + 5$ returns $x^2 - 12x + 41$, and substituting $x = 6$ gives $36 - 72 + 41 = 5$.

Watch for this. A common mistake is to report $x = 6$ when the question asks for the minimum value, or to take the constant $41$ as the minimum.

The video then restricts the inputs to $x \le 4$. The square is still nonnegative, but $x = 6$ is no longer allowed. Every allowed input is at least $2$ units from $6$, so $(x - 6)^2 \ge 4$, and the new minimum is $9$, reached at $x = 4$. Watch this comparison.

The answer is $h = 4$.

1. Rewrite the quadratic as $y = (x + 1)^2 + 4$.
2. Intersections satisfy $(x + 1)^2 = h - 4$.
3. There are no real intersections if $h < 4$, one if $h = 4$ and two if $h > 4$.
4. So $h = 4$, and the common point is $(-1, 4)$.

As a check, at $h = 4$ the equation $(x + 1)^2 = 0$ has the single root $-1$. The discriminant of $x^2 + 2x + (5 - h) = 0$ is $4h - 16$, which is zero exactly when $h = 4$.

Watch for this. A common mistake is to count points from a graph that is cut off, or to assume that a line that looks like it touches the curve gives exactly one real solution.

The video then tests what a picture can hide, using the level $y = 5$. The equation $(x + 1)^2 = 1$ gives the two inputs $-2$ and $0$. A drawing that starts at $x = -1$ would show only the second point, but that window does not remove the first intersection. Watch this comparison.

The answer is $125$ percent.

1. An increase of $3$ in $t$ multiplies $N$ by $\frac{3}{2}$.
2. Six hours contain two such intervals, so the multiplier is $\left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25$.
3. The relative increase is $2.25 - 1 = 1.25$, or $125$ percent.

As a check, for every $t$ the ratio is $\frac{N(t + 6)}{N(t)} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$. For example, $N(0) = 240$ and $N(6) = 540$, and $\frac{540 - 240}{240} = 1.25$.

Watch for this. A common mistake is to read $50$ percent as an hourly rate, or to report the final amount as a percentage when the question asks for the increase. A growth factor of $2.25$ means an increase of $125$ percent, not $225$ percent.

The video then applies the same model over nine hours. The exponent changes by $\frac{9}{3} = 3$, so the multiplier is $\left(\frac{3}{2}\right)^3 = \frac{27}{8}$, an increase of $\frac{19}{8}$ of the starting amount, or $237.5$ percent. Adding three copies of $50$ percent would miss that each increase acts on an amount that has already grown. Watch this comparison.

The answer is $132$ dollars.

1. The sale price is $70$ percent of the original price $P$, so $0.70P = 92.40$.
2. Divide by $0.70$ to get $P = 132$.
3. Check that $30$ percent of $132$ is $39.60$ and $132 - 39.60 = 92.40$.

Watch for this. A common mistake is to add $30$ percent of $92.40$ to the sale price. That gives $92.40 \times 1.30 = 120.12$, which uses the wrong base, and a $30$ percent discount on $120.12$ does not return $92.40$.

The video then reverses an increase with the same rule. An amount becomes $84$ after a $20$ percent increase, so $1.2P = 84$ and $P = 70$. As a check, $20$ percent of $70$ is $14$, and $70 + 14 = 84$. Watch this comparison.

The answer is $0$ percent, so the final amount equals the original amount.

1. Let the starting amount be $P > 0$. After the increase it is $1.25P$.
2. The decrease leaves $0.80(1.25P) = P$.
3. The combined multiplier is $1$, so the overall change is zero.
4. The multipliers are reciprocals, since $1.25 \times 0.80 = 1$, or $\frac{5}{4} \cdot \frac{4}{5} = 1$.

As a check, with $P = 80$ the amount becomes $100$ and then returns to $80$. Equal percentages do not cancel in general, since a $10$ percent increase followed by a $10$ percent decrease has multiplier $1.10 \times 0.90 = 0.99$, a $1$ percent decrease.

Watch for this. A common mistake is to subtract the percentages and claim a $5$ percent increase, or to assume that any equal increase and decrease cancel.

The video then separates the net change from the path. Taking the two changes in the other order also gives a product of $1$. Starting at $80$, a $20$ percent decrease gives $64$, and a $25$ percent increase of $64$ adds $16$, back to $80$. The middle amount differs from the first path. The order does not matter here because only percentage multipliers are involved, and that can change when fixed charges, limits or other rules come between them. Watch this comparison.

The answer is $8$ minutes.

1. Recover the totals, $8 \cdot 5 = 40$ minutes and $12 \cdot 10 = 120$ minutes.
2. The combined total is $160$ minutes over $20$ batches.
3. The combined mean is $\frac{160}{20} = 8$ minutes.

As a check, with weights $\frac{8}{20}$ and $\frac{12}{20}$ the mean is $\frac{8}{20} \cdot 5 + \frac{12}{20} \cdot 10 = 8$. The result lies between $5$ and $10$ and closer to the mean of the larger group.

Watch for this. A common mistake is to average $5$ and $10$ to get $7.5$, which gives the smaller and the larger group equal weight.

The video then swaps the group sizes while keeping the two means. Twelve batches at a mean of $5$ minutes contribute $60$ minutes, and eight batches at a mean of $10$ minutes contribute $80$ minutes, so the combined mean is $\frac{140}{20} = 7$ minutes. The means are still $5$ and $10$, but the combined mean changed from $8$ to $7$. Watch this comparison.

The answer is $\frac{8}{29}$.

1. The condition restricts attention to the walkers, $8 + 21 = 29$ students.
2. Among those walkers, $8$ belong to Robotics.
3. The conditional probability is $\frac{8}{29}$.

As a check, the four counts add to $50$, and the formula gives

Watch for this. A common mistake is to divide by all $50$ students or by all $20$ Robotics members. The fraction $\frac{8}{50}$ is the probability of walking and being in Robotics without the condition, and $\frac{8}{20}$ conditions on Robotics instead.

The video then checks the denominator by saying the probability as a sentence. Out of the $29$ walkers, $8$ are in Robotics, which is $\frac{8}{29}$. Out of the $20$ Robotics students, $8$ walk, which is $\frac{8}{20}$ and answers a different question with the same favorable count. Watch this comparison.

The answer is that the median increases from $7$ to $17$ and the standard deviation does not change.

1. The ordered values of B are $13$, $15$, $17$, $19$, $21$, so the middle value changes from $7$ to $17$.
2. Both means also move by $10$, from $7$ to $17$, and the distances from each mean stay $-4$, $-2$, $0$, $2$, $4$.
3. Because these deviations are unchanged, the standard deviation is unchanged, whichever standard definition is used for both sets.

As a check, both lists have range $8$. Their population variances both equal $\frac{16 + 4 + 0 + 4 + 16}{5} = 8$, and their sample variances are equal too.

Watch for this. A common mistake is to assume that every statistic increases by $10$ when each value increases by $10$.

The video then multiplies every value of A by $-2$. The new mean and median are both $-14$, and each deviation is $-2$ times the old one. The order reverses, but the sizes of the distances double, so the standard deviation doubles and does not become negative. Watch this comparison.

The answer is that random assignment supports the investigation of a causal effect, while volunteering alone does not make the group a representative sample of all students.

1. Entry into the study was by volunteering, not by a stated random sample of all students.
2. The treatments were given by random assignment.
3. Random assignment helps compare the two treatment groups, while the way participants entered limits generalization.
4. A completed analysis would still have to assess chance variation and how the study was carried out. The design alone gives no numerical treatment effect.

As a check, the prompt states random assignment but not random sampling, and it gives no outcome data, so no claim that the reminders worked is made.

Watch for this. A common mistake is to treat random assignment as if the participants were a random sample of the population, or to assume that any observed difference proves a cause.

The video then turns the design around. A study that randomly samples students from a defined school population but only observes their existing habits can support estimates for that population. It has not assigned the habits at random, so other factors could explain differences between groups, and sampling alone would not establish a causal effect. Watch this comparison.

The answer is $243$ cubic centimeters.

1. The area ratio is $\frac{243}{108} = \frac{9}{4}$, so the positive length factor is $k = \frac{3}{2}$.
2. The volume factor is $k^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$.
3. Multiply to get $72 \cdot \frac{27}{8} = 243$ cubic centimeters.
4. One possible smaller prism measures $3$ by $4$ by $6$, which confirms that the given area and volume can occur together.

As a check, a $3$ by $4$ by $6$ prism has surface area $108$ and volume $72$. Scaling its edges to $4.5$, $6$ and $9$ gives surface area $243$ and volume $243$. The two equal numbers have different units and come from different scale factors.

Watch for this. A common mistake is to multiply the volume by the area ratio, which gives $72 \cdot \frac{9}{4} = 162$ cubic centimeters and uses only two dimensions of change.

The video then doubles only the height of the $3$ by $4$ by $6$ prism. The new prism is $3$ by $4$ by $12$, and its volume doubles from $72$ to $144$ rather than multiplying by $8$, because only one dimension changed. The cube rule applies only when all three directions scale together. Watch this comparison.

The answer is the center $(4, -3)$ and the radius $6$.

1. Group $x^2 - 8x$ and $y^2 + 6y$.
2. Add $16$ and $9$ to both sides to get $(x - 4)^2 + (y + 3)^2 = 36$.
3. The center is $(4, -3)$, and the radius is $\sqrt{36} = 6$.

As a check, expanding the completed squares and subtracting $25$ gives back the original equation. The point $(10, -3)$ is $6$ units from the center and satisfies $100 + 9 - 80 - 18 = 11$.

Watch for this. A common mistake is to read the signs of the center straight from the expanded terms, or to report $36$ as the radius.

The video then adds a form check. If the completed equation is written as $2\left[(x - 4)^2 + (y + 3)^2\right] = 72$, first divide the whole equation by $2$ to get $36$ on the right. The radius is still $6$, not $\sqrt{72}$. Watch this comparison.

The answer is $0$.

1. The hypotenuse is $AB = \sqrt{7^2 + 24^2} = 25$.
2. Relative to angle $A$, the opposite side is $BC = 24$, so $\sin(\theta) = \frac{24}{25}$.
3. Relative to angle $B$, the adjacent leg is $BC = 24$, so $\cos(\phi) = \frac{24}{25}$.
4. The two ratios are equal, so the difference is $0$.

As a check, $7^2 + 24^2 = 625 = 25^2$. The two acute angles are complementary, which gives a second reason that $\sin(\theta) = \cos(\phi)$.

Watch for this. A common mistake is to keep the same opposite and adjacent labels after switching the reference angle, or to change which side is the hypotenuse.

The video then points out a shorter route. Before any length is found, $\sin(\theta) = \frac{BC}{AB}$ and $\cos(\phi) = \frac{BC}{AB}$, so the difference is zero without evaluating either fraction. Watch this comparison.

The answer is $7$ meters.

1. Let the width be $w > 0$ and the length $2w$.
2. The area equation is $2w^2 = 98$, so $w^2 = 49$ and the candidates are $7$ and $-7$.
3. A width must be positive, so $w = 7$ and the length is $14$.
4. Check that $7 \cdot 14 = 98$ square meters and $14 = 2 \cdot 7$.

Watch for this. A common mistake is to keep the negative root as a width, or to give $14$ because it is also a side of the field.

The video then warns against inventing a restriction. A temperature or a change in a balance can be negative unless the problem rules it out, and the equation $z^2 = 49$ for any real $z$ keeps both $7$ and $-7$. The field excludes $-7$ only because a width is a positive length. Watch this comparison.

The answer is $26.7$ centimeters.

1. Use $C = 2\pi r$ with the exact radius, so $C = 2\pi\sqrt{18} = 6\pi\sqrt{2}$.
2. Evaluate the final expression, which is about $26.6573$ centimeters.
3. Round the circumference to $26.7$ centimeters.

As a check, the exact value lies between $26.65$ and $26.75$, so its nearest tenth is $26.7$.

Watch for this. A common mistake is to round an intermediate quantity early and expect the final rounding to stay accurate.

The video then compares this with rounding the radius first. A radius of $4.2$ gives a circumference of about $26.3894$, which rounds to $26.4$ and changes the requested result. The formula is the same, but the early rounding changed its input. Watch this comparison.

The answer is $2$. One useful note is “smaller root, roots with product $10$ and sum $7$.”

1. Record both the requested quantity and a useful relationship, such as the smaller root and roots with product $10$ and sum $7$.
2. The numbers $2$ and $5$ have that sum and product, so $(x - 2)(x - 5) = 0$.
3. The roots are $2$ and $5$, and the smaller one is $2$.
4. Substitute $2$ into the original equation to check, $4 - 14 + 10 = 0$.

As a check, expanding $(x - 2)(x - 5)$ gives $x^2 - 7x + 10$. The note is one example of a useful strategy, not the only correct response.

Watch for this. A common mistake is to return to an unlabeled calculation, or to find a valid root without checking whether the question asks for the smaller or the larger one. No single time limit for setting a problem aside suits every problem or every student.

The video then shows that a note helps even before you know the method. For this example, “smaller root, try a factor pair” keeps the target and the next move. For the earlier fraction with denominator $x - 4$, a note such as “exclude $4$, factor the numerator” keeps a different kind of risk in view. Watch this comparison.