Chapter 10 — The Challenge Set

Answer: A

Explanation

Two straight lines fail to meet in exactly one situation: they are parallel and they are not the same line. Both halves matter here, and the second half is the whole question.

Parallel first. Two equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ describe parallel lines when the coefficients are in proportion:

$$\frac{k}{9} = \frac{4}{k} \quad\Longrightarrow\quad k^2 = 36 \quad\Longrightarrow\quad k = 6 \text{ or } k = -6$$

Now test each one, which is the step that gets skipped.

If $k = 6$ the system is $6x + 4y = 10$ and $9x + 6y = 15\mathclose{.}$ Multiply the first by $\tfrac{3}{2}\mathclose{:}$ $9x + 6y = 15\mathclose{.}$ That is the second equation exactly. The two equations are the same line, so every point on it is a solution — infinitely many, not none.

If $k = -6$ the system is $-6x + 4y = 10$ and $9x - 6y = 15\mathclose{.}$ Multiply the first by $-\tfrac{3}{2}\mathclose{:}$ $9x - 6y = -15\mathclose{.}$ The left sides now agree and the right sides do not, $-15 \ne 15\mathclose{.}$ Parallel, distinct, no solution.

So $k = -6\mathclose{,}$ option A.

Why each wrong option is wrong:

• B, $0$ — took both roots and added them, reading "what is the value of $k$" as "the sum of the possible values". Only one value survives the test, so there is nothing to sum.
• C, $6$ — solved $k^2 = 36$ and took the positive root. This is the coincident case: it gives infinitely many solutions, which is the opposite of what was asked. Proportional coefficients only make lines parallel; whether they are the same line depends on the constants on the right, and nothing about $k^2 = 36$ checks those.
• D, $36$ — solved as far as $k^2 = 36$ and stopped. A very common way to lose a hard question: the algebra was right and the last step was never taken.

Takeaway: "No solution" is two conditions, not one — parallel and distinct. Whenever a proportionality gives you two candidate values, substitute both back and look at the constants. One of them is usually the coincident case, and it is usually the one that looks nicer.

Answer: B

Explanation

"Intersect at exactly one point" means the equation you get by setting the two expressions equal has exactly one solution.

$$x^2 + kx + 9 = 3x + 5 \quad\Longrightarrow\quad x^2 + (k - 3)x + 4 = 0$$

A quadratic has exactly one solution when its discriminant is zero:

$$(k-3)^2 - 4(1)(4) = 0 \quad\Longrightarrow\quad (k-3)^2 = 16 \quad\Longrightarrow\quad k - 3 = \pm 4$$

So $k = 7$ or $k = -1\mathclose{,}$ and their sum is $6$ — option B.

Note what happened to the constant: the $9$ and the $5$ combined into a $4$ before the discriminant was taken. Taking the discriminant of the original $x^2 + kx + 9$ is the single most common way to lose this question.

Why each wrong option is wrong:

• A, $7$ — found both values correctly and then reported only $k = 7\mathclose{.}$ The question asks for the sum, and a question that asks for a sum is telling you there is more than one value.
• C, $-1$ — the same slip the other way round.
• D, $-6$ — took $k - 3 = \pm 4$ and wrote $k = -3 \pm 4\mathclose{,}$ giving $1$ and $-7\mathclose{,}$ which sum to $-6\mathclose{.}$ The $3$ moves across the equals sign and changes sign as it goes.

Takeaway: Tangency is a discriminant question in disguise. Set the two expressions equal, collect everything on one side first, and only then reach for $b^2 - 4ac$ — using the coefficients of the combined equation, never the original curve.

Answer: A

Explanation

Divide one reading by the other. The $a$ cancels, which is the whole reason to start there:

$$\frac{f(5)}{f(2)} = \frac{ab^5}{ab^2} = b^3 = \frac{486}{18} = 27$$

So $b = 3\mathclose{.}$ Now use either reading to get $a\mathclose{:}$

$$f(2) = a \cdot 3^2 = 9a = 18 \quad\Longrightarrow\quad a = 2$$

And $f(0) = a \cdot b^0 = a \cdot 1 = 2\mathclose{,}$ option A.

The last line is the point of the question. In any model of the form $ab^x\mathclose{,}$ $f(0)$ is $a$ — the starting value — because $b^0 = 1\mathclose{.}$

Why each wrong option is wrong:

• B, $3$ — reported the growth factor $b$ instead of the initial value $a\mathclose{.}$ These are the two constants in the model and it is easy to hand back the wrong one under time pressure.
• C, $6$ — computed $a \cdot b = 6\mathclose{,}$ which is $f(1)\mathclose{,}$ not $f(0)\mathclose{.}$ Off by one step down the sequence.
• D, $27$ — stopped at $b^3 = 27$ without taking the cube root, and without going on to find $a$ at all.

Takeaway: For $f(x) = ab^x\mathclose{,}$ dividing two readings kills $a$ and hands you $b\mathclose{;}$ dividing the exponents tells you what power of $b$ you have. And $f(0) = a$ always — if a question asks for the value at zero, it is asking for $a\mathclose{.}$

Section 2 — Answers That Are Not Solutions

Squaring both sides, multiplying by a denominator and taking a root can all create values that satisfy the equation you wrote but not the equation you started with. On a hard question that extra value is always offered as an option.

Answer: A

Explanation

Two absolute values are equal when the things inside are equal, or when one is the negative of the other. Both cases have to be worked.

Case 1: $2x - 5 = x + 4\mathclose{,}$ so $x = 9\mathclose{.}$

Case 2: $2x - 5 = -(x + 4) = -x - 4\mathclose{,}$ so $3x = 1$ and $x = \tfrac{1}{3}\mathclose{.}$

Both check: at $x = 9\mathclose{,}$ $|13| = |13|\mathclose{;}$ at $x = \tfrac{1}{3}\mathclose{,}$ $\left|-\tfrac{13}{3}\right| = \left|\tfrac{13}{3}\right|\mathclose{.}$ Neither is extraneous here, which is worth noticing — with absolute values on both sides both cases usually survive, unlike the one-sided kind.

The sum is $9 + \tfrac{1}{3} = \tfrac{28}{3}\mathclose{,}$ option A.

Why each wrong option is wrong:

• B, $9$ — solved the first case and stopped. A single bar equation almost always has two solutions; a question asking for their sum is confirming it.
• C, $\frac{26}{3}$ — found both and then subtracted rather than added.
• D, $\frac{1}{3}$ — solved the second case only.

Takeaway: $|A| = |B|$ gives two equations, $A = B$ and $A = -B\mathclose{.}$ Write both down before solving either — the second is the one that gets forgotten, and it is the one with the awkward fraction.

Answer: B

Explanation

Square both sides. That is the only way to get at $x\mathclose{,}$ and it is also what creates the problem.

$$x + 7 = (x - 5)^2 = x^2 - 10x + 25 \quad\Longrightarrow\quad x^2 - 11x + 18 = 0$$

$$(x - 2)(x - 9) = 0 \quad\Longrightarrow\quad x = 2 \text{ or } x = 9$$

Now check both in the original equation, which squaring has quietly changed.

At $x = 2\mathclose{:}$ the left side is $\sqrt{9} = 3\mathclose{;}$ the right side is $2 - 5 = -3\mathclose{.}$ A square root sign means the non-negative root, so the left side can never be negative. $3 \ne -3\mathclose{,}$ and $x = 2$ is not a solution of the equation that was asked.

At $x = 9\mathclose{:}$ left side $\sqrt{16} = 4\mathclose{,}$ right side $9 - 5 = 4\mathclose{.}$ ✓

So $x = 9\mathclose{,}$ option B.

Why each wrong option is wrong:

• A, $2$ — the extraneous root. It solves the squared equation perfectly and fails the original. Squaring turns $-3$ into $9$ just as it turns $3$ into $9\mathclose{,}$ so it cannot tell the two apart; only substituting back can.
• C, $11$ — added the two roots. There is only one solution, so there is nothing to add.
• D, $18$ — multiplied them, which is the constant term of the quadratic and not a solution of anything.

Takeaway: Every time you square both sides you must substitute back. The extraneous root appears exactly when the two sides had opposite signs, and it will always be one of the options.

Section 3 — Two Ideas at Once

The rest of the chapter is questions that need something from one topic and something from another. There is no signal in the wording about which two.

Answer: C

Explanation

Get the circle into standard form by completing the square in both variables:

$$x^2 - 10x + y^2 + 6y = -18$$ $$(x - 5)^2 - 25 + (y + 3)^2 - 9 = -18$$ $$(x - 5)^2 + (y + 3)^2 = 16$$

So the centre is $(5, -3)$ and the radius is $\sqrt{16} = 4\mathclose{.}$

The distance from the origin to the centre is

$$\sqrt{5^2 + (-3)^2} = \sqrt{34} \approx 5.83$$

That is more than the radius, so the origin lies outside the circle. The nearest point of the circle sits on the line joining the origin to the centre, one radius back from the centre:

$$\sqrt{34} - 4$$

which is option C, about $1.83\mathclose{.}$

Why each wrong option is wrong:

• B, $\sqrt{34}$ — the distance to the centre. The centre is not on the circle; every point of the circle is $4$ units away from it.
• A, $\sqrt{34} + 4$ — went one radius past the centre instead of stopping short of it. That is the furthest point of the circle from the origin, not the nearest.
• D, $4$ — the radius, which is the distance from the centre to the circle, not from the origin.

Takeaway: For a point outside a circle, nearest distance is $d_{\text{centre}} - r$ and furthest is $d_{\text{centre}} + r\mathclose{.}$ Complete the square first — a general-form equation tells you nothing until you do.

Answer: B

Explanation

This is the one fact the question rests on: areas of similar figures scale as the square of the length ratio. So if areas are in the ratio $9 : 25\mathclose{,}$ lengths are in the ratio $\sqrt{9} : \sqrt{25} = 3 : 5\mathclose{.}$

Perimeter is a length, so it scales by $\tfrac{5}{3}\mathclose{:}$

$$24 \times \frac{5}{3} = 40$$

option B.

Why each wrong option is wrong:

• A, $14.4$ — used $\tfrac{3}{5}$ instead of $\tfrac{5}{3}$ and made the larger triangle smaller than the smaller one. Always sanity-check the direction.
• C, $\frac{200}{3}$ — scaled the perimeter by the area ratio $\tfrac{25}{9}\mathclose{,}$ giving $\tfrac{200}{3} \approx 66.7\mathclose{.}$ This is the mistake the question is built to catch: the ratio you are given is not the ratio you may apply.
• D, $120$ — multiplied by $5$ and never divided by $3\mathclose{,}$ using one half of the ratio as though it were the whole scale factor.

Takeaway: Lengths scale by $k\mathclose{,}$ areas by $k^2\mathclose{,}$ volumes by $k^3\mathclose{.}$ Given an area ratio, take the square root before you touch any length.

Answer: D

Explanation

The identity to reach for is $\sin(\theta) = \cos(90^\circ - \theta)\mathclose{:}$ sine and cosine agree exactly when their angles are complementary, adding to $90°\mathclose{.}$

So the two angles here must add to $90\mathclose{:}$

$$x + (3x - 10) = 90 \quad\Longrightarrow\quad 4x - 10 = 90 \quad\Longrightarrow\quad 4x = 100 \quad\Longrightarrow\quad x = 25$$

Check: $\sin(25°) = \cos(65°)\mathclose{,}$ and $3(25) - 10 = 65\mathclose{.}$ ✓ Option $25\mathclose{.}$

Why each wrong option is wrong:

• B, $20$ — wrote $x + 3x + 10 = 90\mathclose{,}$ flipping the sign of the $-10$ when moving it. Solving gives $x = 20\mathclose{,}$ and $3(20) - 10 = 50\mathclose{,}$ but $20 + 50 \ne 90$ — the check would have caught it.
• C, $22.5$ — dropped the $-10$ and solved $4x = 90\mathclose{,}$ giving $22.5\mathclose{.}$
• A, $5$ — set the two angles equal, $x = 3x - 10\mathclose{.}$ Equal angles give equal sines; a sine equals a cosine when the angles add to $90°\mathclose{.}$

Takeaway: $\sin A = \cos B$ means $A + B = 90\mathclose{.}$ It does not mean $A = B\mathclose{,}$ and it does not mean $A + B = 180\mathclose{.}$ Write the sum equation immediately and then just solve a linear equation.

Answer: C

Explanation

Composition is not commutative, and this question is about how far apart the two orders are.

$$f(g(x)) = (x + 3)^2 = x^2 + 6x + 9$$ $$g(f(x)) = x^2 + 3$$

Set them equal:

$$x^2 + 6x + 9 = x^2 + 3$$

The $x^2$ cancels — which is why a question that looks quadratic is linear:

$$6x + 9 = 3 \quad\Longrightarrow\quad 6x = -6 \quad\Longrightarrow\quad x = -1$$

Check: $f(g(-1)) = f(2) = 4$ and $g(f(-1)) = g(1) = 4\mathclose{.}$ ✓ Option $-1\mathclose{.}$

Why each wrong option is wrong:

• A, $-3$ — solved $g(x) = 0\mathclose{.}$ A natural thing to do with a composition question and not what was asked.
• B, $-2$ — expanded $(x + 3)^2$ as $x^2 + 3x + 9\mathclose{,}$ halving the middle term. The middle term of $(x + a)^2$ is $2ax\mathclose{,}$ not $ax\mathclose{.}$
• D, $1$ — moved the $9$ across as $+9$ and solved $6x = 6\mathclose{.}$ The $9$ is on the left, so subtracting it from both sides makes the right side $3 - 9 = -6\mathclose{.}$

Takeaway: Write out both compositions in full before comparing them. When the outer function is a square, expand it properly — $(x+3)^2$ has a $6x\mathclose{,}$ and losing it turns a solvable question into a different one.

Answer: C

Explanation

The two percentages are taken of different amounts, which is the entire question. Work with a multiplier for each step rather than adding and subtracting percentages.

Take the original price as $100\mathclose{.}$

An increase of $20\%$ multiplies by $1.20\mathclose{:}$ $100 \times 1.20 = 120\mathclose{.}$

A decrease of $20\%$ multiplies by $0.80$ — and it is applied to $120\mathclose{,}$ not to $100\mathclose{:}$ $120 \times 0.80 = 96\mathclose{.}$

So the final price is $96\%$ of the original, option C. The item is $4\%$ cheaper than it started, even though the two percentages were equal.

The reason is that $20\%$ of $120$ is $24\mathclose{,}$ while $20\%$ of $100$ is only $20\mathclose{.}$ The decrease is taken from a bigger number, so it is bigger.

Why each wrong option is wrong:

• B, $100$ — assumed $+20\%$ and $-20\%$ cancel. They only would if both were percentages of the same amount, and they never are in a sequence like this.
• A, $104$ — got the size of the change right and its direction wrong. A rise followed by an equal fall always leaves you below where you started, never above.
• D, $80$ — applied the decrease to the original price and ignored the increase entirely.

Takeaway: Chain percentage changes as multipliers: $1.20 \times 0.80 = 0.96\mathclose{.}$ Never add or subtract the percentages themselves, and remember the answer to "up then down by the same percent" is always a net loss.

Answer: C

Explanation

The factor theorem says $(x - r)$ divides $p(x)$ exactly when $p(r) = 0\mathclose{.}$ Two factors give two equations, and two equations are what two unknowns need.

From $(x - 1)\mathclose{,}$ substitute $x = 1\mathclose{:}$

$$2 + a - 5 + b = 0 \quad\Longrightarrow\quad a + b = 3$$

From $(x + 2)\mathclose{,}$ substitute $x = -2$ — note the root is $-2\mathclose{,}$ not $2\mathclose{:}$

$$2(-8) + 4a + 10 + b = 0 \quad\Longrightarrow\quad 4a + b = 6$$

Subtract the first from the second: $3a = 3\mathclose{,}$ so $a = 1$ and then $b = 2\mathclose{.}$

$a + b = 3\mathclose{,}$ option C.

There is a shortcut worth seeing: the first equation is $a + b = 3\mathclose{,}$ so the question was answered before the second factor was ever used. The second factor is there to make you do work you did not need.

Why each wrong option is wrong:

• A, $1$ — solved the system correctly and handed back $a\mathclose{.}$
• B, $2$ — handed back $b\mathclose{.}$
• D, $6$ — reported $4a + b\mathclose{,}$ the left side of the second equation, rather than $a + b\mathclose{.}$

Takeaway: $(x + 2)$ is a factor when $p(-2) = 0\mathclose{.}$ The sign flips, and it is the most reliable way to lose this question. And read what is actually being asked — sometimes the first line of working already contains it.

Answer: B

Explanation

Go back to the total. A mean is a total divided by a count, and the count has not changed — only the total has.

$$\text{original total} = 20 \times 72 = 1440$$

One score of $35$ has to come out and one of $75$ has to go in, so the total changes by $75 - 35 = 40\mathclose{:}$

$$\text{correct total} = 1440 - 35 + 75 = 1480$$

$$\text{correct mean} = \frac{1480}{20} = 74$$

option B. Equivalently, the mean rises by $\tfrac{40}{20} = 2\mathclose{.}$

Why each wrong option is wrong:

• A, $72$ — assumed correcting one value cannot move the mean. It moves it by the change divided by the count.
• C, $75.75$ — added the new score without removing the old one, giving a total of $1515$ over $20$ scores. There are still only $20$ scores; one of them changed value.
• D, $76$ — divided the change of $40$ by $10$ instead of $20\mathclose{.}$ The count is the number of scores, not half of it.

Takeaway: For any "one value was recorded wrongly" question, convert the mean to a total, adjust the total by the difference, and divide by the unchanged count.

Answer: B

Explanation

"Selected at random from those who passed" changes the population. You are no longer choosing from $100$ candidates — you are choosing from the $73$ who passed. That number is the denominator.

Of those $73\mathclose{,}$ the ones who sat in the morning number $45\mathclose{.}$ So

$$P(\text{morning} \mid \text{passed}) = \frac{45}{73}$$

option B, about $0.616\mathclose{.}$

The reliable method: find the phrase that restricts who is being chosen from, take that total as the denominator, and count within it.

Why each wrong option is wrong:

• A, $\frac{9}{20}$ — $\tfrac{45}{100}$ is the probability of picking someone who is both a morning candidate and a pass, out of everyone. It ignores the restriction entirely.
• C, $\frac{73}{100}$ — $\tfrac{73}{100}$ is simply the probability of passing.
• D, $\frac{3}{4}$ — $\tfrac{45}{60}$ is $P(\text{passed} \mid \text{morning})\mathclose{,}$ the chance a morning candidate passed. Same cell, different total, and it answers the question the other way round. Conditional probabilities are not symmetric, and this is the standard trap.

Takeaway: In "given that", the given event supplies the denominator. Underline the conditioning phrase, write its total down first, then look for the numerator inside it.

Answer: D

Explanation

The bases are different, so nothing can be compared yet. Both are powers of $2\mathclose{,}$ which is the way in:

$$4^{x+1} = (2^2)^{x+1} = 2^{2x+2} \qquad 8^{x-1} = (2^3)^{x-1} = 2^{3x-3}$$

Now the bases match, so the exponents must be equal:

$$2x + 2 = 3x - 3 \quad\Longrightarrow\quad x = 5$$

Check: $4^6 = 4096$ and $8^4 = 4096\mathclose{.}$ ✓ Option $5\mathclose{.}$

Why each wrong option is wrong:

• A, $-1$ — distributed the $3$ as $3x + 3$ instead of $3x - 3\mathclose{.}$ The exponent is $x - 1\mathclose{,}$ so both of its terms are multiplied by $3\mathclose{,}$ sign included.
• B, $2$ — rewrote $8$ as $2^3$ but left the $4$ alone, comparing $x + 1$ with $3(x - 1)\mathclose{.}$ Both sides have to be converted before exponents can be equated.
• C, $3$ — treated the bases as multipliers and solved $4(x+1) = 8(x-1)\mathclose{,}$ reaching $x = 3\mathclose{.}$ An exponent is not a coefficient.

Takeaway: Same base first, then equate exponents. Write $(2^2)^{x+1} = 2^{2(x+1)}$ out in full — the bracket is what stops the $2$ being forgotten.

Answer: B

Explanation

Adding all three equations is far faster than substituting, and it is the move the question is built around. Each variable appears in exactly two of the three equations, so adding gives

$$2(x + y + z) = 7 + 9 + 12 = 28 \quad\Longrightarrow\quad x + y + z = 14$$

Now each variable is the total minus the equation that does not contain it. The equation without $x$ is $y + z = 9\mathclose{,}$ so

$$x = 14 - 9 = 5$$

option B. The same trick gives $y = 14 - 12 = 2$ and $z = 14 - 7 = 7\mathclose{,}$ and $5 + 2 = 7$ checks out. ✓

Why each wrong option is wrong:

• A, $2$ — subtracted the wrong equation and produced $y\mathclose{.}$
• C, $7$ — produced $z\mathclose{.}$ Note $7$ also appears in the question as the first right-hand side, which makes it feel familiar.
• D, $14$ — stopped at $x + y + z = 14\mathclose{.}$

Takeaway: When every variable appears the same number of times, add all the equations at once. Each unknown is then the grand total minus the equation it is missing from, with no substitution at all.

Answer: D

Explanation

A parabola opening upwards takes its minimum at the vertex, which sits at

$$x = -\frac{b}{2a} = -\frac{-6}{2} = 3$$

The minimum value is the function evaluated there:

$$f(3) = 3^2 - 6(3) + c = 9 - 18 + c = c - 9$$

Setting that equal to $4$ gives $c = 13\mathclose{,}$ option D.

Check by completing the square: $x^2 - 6x + 13 = (x - 3)^2 + 4\mathclose{,}$ whose least value is $4$ at $x = 3\mathclose{.}$ ✓

Why each wrong option is wrong:

• A, $-5$ — wrote $f(3) = c + 9\mathclose{,}$ losing a sign when combining $9 - 18\mathclose{.}$
• B, $1$ — worked out $f(3)$ as $9 - 6 + c\mathclose{,}$ forgetting to multiply the 6 by 3.
• C, $4$ — took $c$ to be the minimum value. $c$ is the $y$-intercept, $f(0)$ — the two coincide only when the vertex sits on the $y$-axis.

Takeaway: Vertex $x$ from $-\tfrac{b}{2a}\mathclose{,}$ then substitute for the vertex $y\mathclose{.}$ Completing the square gives both at once and is worth doing as a check: $(x - p)^2 + q$ has minimum $q$ at $x = p\mathclose{.}$

Answer: D

Explanation

Rates add and subtract; times do not. Convert each device to a rate first — the fraction of the tank it handles in one hour.

The tap fills $\tfrac{1}{6}$ of the tank per hour; the drain removes $\tfrac{1}{9}$ per hour. They oppose each other, so the net rate is a subtraction:

$$\frac{1}{6} - \frac{1}{9} = \frac{3}{18} - \frac{2}{18} = \frac{1}{18}$$

The tank gains $\tfrac{1}{18}$ of itself each hour, so filling all of it takes

$$1 \div \frac{1}{18} = 18 \text{ hours}$$

option D. It is slower than the tap alone, which it has to be — a sanity check worth ten seconds.

Why each wrong option is wrong:

• A, $3$ — subtracted the two times. The same error in the other direction.
• B, $3.6$ — added the rates instead of subtracting. That is the answer to "two taps both filling", and it gives $3.6$ hours — faster than either device alone, which is impossible when one of them empties the tank.
• C, $15$ — added the two times. Times cannot be added; only rates can.

Takeaway: Turn every "takes $n$ hours" into "$\tfrac{1}{n}$ per hour" before doing anything else. Combine the rates, invert once at the end, and check the answer falls on the sensible side of the faster device.

Answer: C

Explanation

The parabola opens upwards, so it lies above the axis everywhere exactly when it never touches the axis — that is, when it has no real roots. No real roots means a negative discriminant:

$$b^2 - 4ac < 0 \quad\Longrightarrow\quad 36 - 4c < 0 \quad\Longrightarrow\quad c > 9$$

The inequality is strict, so $c = 9$ is not allowed: at $c = 9$ the expression is $(x - 3)^2\mathclose{,}$ which equals $0$ at $x = 3$ and so is not $> 0$ for all $x\mathclose{.}$

The least integer strictly greater than $9$ is $10\mathclose{,}$ option C.

Why each wrong option is wrong:

• B, $9$ — solved $36 - 4c = 0$ and took the boundary. The boundary is the single value that fails, because it puts a repeated root exactly on the axis.
• A, $3$ — took a square root that was not called for, reaching $3\mathclose{.}$
• D, $36$ — computed $b^2 = 36$ and reported it.

Takeaway: "Positive for all $x$" with an upward parabola means discriminant $< 0\mathclose{,}$ strictly. Whenever an inequality is strict, test the boundary separately — it is offered as an option precisely because it so nearly works.

Answer: C

Explanation

Two areas, subtracted.

The square has area $10^2 = 100\mathclose{.}$ A quarter circle of radius $10$ is one fourth of a full circle of that radius:

$$\frac{1}{4}\pi r^2 = \frac{1}{4}\pi(10)^2 = 25\pi$$

What is left is

$$100 - 25\pi$$

option C, about $100 - 78.5 = 21.5\mathclose{.}$

The sanity check matters here: the answer must be positive and smaller than the square. Any option that comes out negative can be discarded on sight.

Why each wrong option is wrong:

• A, $100 - 100 \pi$ — subtracted a whole circle of radius $10\mathclose{,}$ area $100\pi \approx 314\mathclose{,}$ from a square of area $100\mathclose{.}$ That is roughly $-214\mathclose{:}$ a negative area, impossible and visible without any calculation at all.
• B, $100 - 50 \pi$ — used a half circle. The shape is bounded by two sides of the square meeting at a corner, and a corner is a right angle — one quarter of a full turn.
• D, $25 \pi$ — gave the quarter circle itself: the region removed, not the region left.

Takeaway: A sector's fraction is its angle over $360°\mathclose{,}$ and a square's corner is $90°\mathclose{,}$ so a quarter. On any shaded-region question, check the sign and the rough size of your answer against the picture before choosing.

Answer: A

Explanation

Write a model for each tank as a function of the time $t$ in minutes. One is falling, so its rate is negative; the other is rising.

$$A(t) = 240 - 15t \qquad B(t) = 60 + 5t$$

Set them equal:

$$240 - 15t = 60 + 5t \quad\Longrightarrow\quad 180 = 20t \quad\Longrightarrow\quad t = 9$$

option A. Check: $A(9) = 240 - 135 = 105$ and $B(9) = 60 + 45 = 105\mathclose{.}$ ✓

The $20$ is the important number. The tanks close the $180$-litre gap between them at a combined $20$ litres per minute, because one is losing while the other gains — the rates add, even though the signs in the two models differ.

Why each wrong option is wrong:

• B, $12$ — divided $240$ by $20\mathclose{,}$ ignoring the $60$ litres already in tank B. The gap to close is $240 - 60 = 180\mathclose{,}$ not $240\mathclose{.}$
• C, $18$ — used $15 - 5 = 10$ for the combined rate, giving $18$ minutes. Subtracting is right when both tanks move the same way; here they move towards each other, so the gap closes at the sum of the rates.
• D, $105$ — solved correctly and then reported the shared volume, $105$ litres, instead of the time. Both numbers are in the working; only one was asked for.

Takeaway: Two linear models meet where their expressions are equal. Watch the sign of each rate as you build them, and when you have finished, read the question again to see whether it wants the time or the value.