ISEE Upper Math challenge

The idea is to measure the base along a horizontal side, where the two slanted lines are easy to compare, rather than vertically between them. On $y=1$ the lines are at $x=-1$ and $x=4$, a horizontal side of length 5; on $y=5$ they are at $x=1$ and $x=6$, again length 5. The height between the horizontal sides is $5-1=4$. The area is base times height, $5\cdot 4=20$.

The insight is to use the triangle inequality that the sum of the two smaller sides must exceed the largest, so only triples with a small sum can fail and the rest need no check. There are 7 numbers, so total triples are 7 times 6 times 5 divided by 6, which is 35. For a triple to fail, the two smallest must sum to at most the largest, which is at most 10. The only pairs with sum at most 10 are 4 and 5 with sum 9 and 4 and 6 with sum 10. With 4 and 5, the largest can be 9 or 10, giving 2 failing triples. With 4 and 6, the largest can only be 10, giving 1 more. So 3 triples fail and 35 minus 3 is 32, so the probability is $32\mathrm{/}35$.

The insight is to count powers by bounding instead of listing and to recognize the overlap as sixth powers from the least common multiple of 2 and 3. Squares need $n$ times $n$ at most 500. Since 22 times 22 equals 484 and 23 times 23 equals 529, there are 22 squares. Cubes need $n$ times $n$ times $n$ at most 500. Since 7 times 7 times 7 equals 343 and 8 times 8 times 8 equals 512, there are 7 cubes. Both means sixth powers, since a number that is both must have exponent a multiple of 2 and 3. Since 2 to the sixth equals 64 and 3 to the sixth equals 729, there are 2 such numbers, namely 1 and 64. By inclusion, 22 plus 7 minus 2 equals 27.

The insight is to treat the metal as the outer solid minus the inner air and then to remember that area scales with the square of the radius so a square root is required. The outer solid is $\pi$ times 9 squared times 4, or $324\pi$, so the inner air is $324\pi$ minus $180\pi$, or $144\pi$. Hence inner radius squared times 4 equals 144, inner radius squared equals 36, inner radius equals 6, and inner diameter equals 12. A linear proportion using 180 over 324 or forgetting to double the radius gives a short wrong number, while the subtraction plus square root finishes quickly.

Think like a tutor: use the line itself. The line through $(0,u)$ and $(v,0)$ satisfies ${\displaystyle \frac{x}{v}}+{\displaystyle \frac{y}{u}}=1$, which is equal slopes in disguise. Substituting $(k,k)$ gives ${\displaystyle \frac{k}{v}}+{\displaystyle \frac{k}{u}}=1$. Keep the reciprocal together instead of clearing everything: ${\displaystyle \frac{k}{u}}=1-{\displaystyle \frac{k}{v}}={\displaystyle \frac{v-k}{v}}$. Invert once to get ${\displaystyle \frac{u}{k}}={\displaystyle \frac{v}{v-k}}$, so $u={\displaystyle \frac{vk}{v-k}}$. Clearing denominators first and chasing $u$ through each term is long and sign prone, while one common denominator plus one inversion is short. With $v=6$ and $k=2$, this gives $u=3$.

The insight is a parity re-frame. The product is even unless both numbers are odd, so $36-3\cdot 3=27$ outcomes have an even product. A sum is even when both numbers are even or both are odd, but both odd makes the product odd. So among the 27 outcomes with an even product, the sum is even exactly when both numbers are even, which happens in $3\cdot 3=9$ outcomes. The probability is ${\displaystyle \frac{9}{27}}={\displaystyle \frac{1}{3}}$.

The insight is to rewrite chained division as one fraction and then minimize its denominator. The quotient $a÷b$ equals ${\displaystyle \frac{a}{b}}$, and parentheses decide whether each later divisor lands in the numerator or denominator, while the second number 3 must stay in the denominator. To maximize, keep only 3 below and flip 4 and 5 above, giving ${\displaystyle \frac{60\cdot 4\cdot 5}{3}}={\displaystyle \frac{1200}{3}}=400$. The left-to-right evaluation gives 1, flipping only $3÷4$ gives 16, and flipping only $4÷5$ gives 25, all smaller. Trying all five parenthesizations with fractions is far longer than the one extremal fraction.

The insight is to recover attempt totals from made and percent before any averaging, then use total makes over total attempts. Ana attempted 9 divided by 0.9 equals 10 in June and 12 divided by 0.4 equals 30 in July, so 21 makes on 40 attempts for 52.5 percent. Ben attempted 21 divided by 0.7 equals 30 in June and 2 divided by 0.2 equals 10 in July, so 23 makes on 40 attempts for 57.5 percent. Ben is higher despite trailing in each month on percent because his makes concentrate in the larger sample, so averaging percents reverses the order.

The insight is to count by extremes first, choosing the smallest and largest to fix the range and then counting middle choices, instead of listing all triples. Total triples are 8 times 7 times 6 divided by 6, which is 56. For range 5, the smallest can be 1, 2 or 3 with largest 5 more, giving 3 endpoint pairs. Between each such pair there are 4 integers for the middle number, since 5 minus 1 is 4. So favorable triples are 3 times 4, which is 12. Thus the probability is 12 divided by 56, which is $3\mathrm{/}14$.

The insight is to translate each purchase into a linear equation and then notice an invariant in the total rather than solving the system. Let $a$, $b$, $o$ be the prices, so $2a+b+o=1.90$ and $a+2b+o=2.00$ and $a+b+2o=2.10$. Adding all three word equations counts each fruit exactly four times, so $4(a+b+o)=1.90+2.00+2.10=6.00$. Dividing by 4 gives $a+b+o=1.50$ in one step. Solving the three-by-three system by elimination would require clearing decimals and many steps without a calculator, while the total makes the arithmetic trivial.

The insight is to use the overlap as a bridge between the two fractions and then force integrality with the neither bound. Let the overlap be $t$. Then tennis is $3t\mathrm{/}2$ and basketball is $2t$, so tennis needs $t$ even and the union is $5t\mathrm{/}2$. With 30 total and fewer than 5 neither, the union exceeds 25 but is at most 30. So 25 is less than $5t\mathrm{/}2$ and $5t\mathrm{/}2$ is at most 30, giving 10 less than $t$ and $t$ at most 12. Evenness leaves $t=12$, realized by 18 tennis, 24 basketball, union 30.

The insight is to regroup before computing and use difference of squares. Exponents come before subtraction, but adjacent pairs can be grouped as $(100^2-99^2)$ through $(2^2-1^2)$ without changing the value. Each pair factors as $(a-b)(a+b)$ with $a-b=1$, so it equals $a+b$. Hence the whole expression equals $(100+99)+(98+97)$ through $(2+1)$, which is the sum $100+99$ down to 1. That sum pairs to 50 pairs averaging 101, totaling 5050. Computing 100 squares directly would be very long, while regroup plus factoring is short.

The insight is that two of the three triangle inequalities are automatic, leaving one counting problem solved by complement. With sides $a,b,10$ and $1\le a,b\le 10$, both $a+10>b$ and $b+10>a$ always hold, so a non-degenerate triangle needs only $a+b>10$ with strict inequality. Counting directly is tedious, so count failures $a+b\le 10$: for sums $s=2$ through 10 there are $s-1$ pairs, totaling $1+2+\cdots +9=45$. Out of $10\cdot 10=100$ ordered pairs, $100-45=55$ succeed, so the probability is $55\mathrm{/}100=11\mathrm{/}20$. The boundary $a+b=10$ is degenerate and must be counted as failure.

The insight is to read $\mathrm{\mid }x-3\mathrm{\mid }+\mathrm{\mid }x+5\mathrm{\mid }$ as the sum of the distances from $x$ to 3 and to $-5$ on the number line. For any $x$ between $-5$ and 3, the two distances add to exactly 8. Each step beyond either end adds 2 to the total. The total can be at most 12, so we can go 2 units past each end. This gives $-7\le x\le 5$. (Algebra agrees: for $x\ge 3$, $2x+2\le 12$ gives $x\le 5$; for $x\le -5$, $-2x-2\le 12$ gives $x\ge -7$.) The integers from $-7$ to 5 number 13.

The insight is to read negative exponents as reciprocals and work with their sum rather than solving for $a$ and $b$. Since $a^{-1}=1\mathrm{/}a$, adding gives $1\mathrm{/}a+1\mathrm{/}b=(a+b)\mathrm{/}ab=7\mathrm{/}5$. Squaring that sum gives $1\mathrm{/}{a}^2+2\mathrm{/}ab+1\mathrm{/}{b}^2=49\mathrm{/}25$, so subtracting the cross term $2\mathrm{/}5=10\mathrm{/}25$ leaves $1\mathrm{/}{a}^2+1\mathrm{/}{b}^2=39\mathrm{/}25$, which is $a^{-2}+b^{-2}$. Solving $t^2-7t+5=0$ for $a$ and $b$ needs messy square roots of 29 and squaring their reciprocals without a calculator, while the symmetric computation uses only 7 and 5.

The insight is that a remainder statement hides a divisor statement plus an integer ceiling for a strict inequality. If $200=qk+6$ then $qk=194$, so $k$ divides 194 and $k$ is greater than 6. Since $194=2$ times 97, its positive divisors are $1,2,97,194$. The largest one greater than 6 is 194, so $K=194$. Then $3n>194$ gives $n$ greater than $194\mathrm{/}3$, which is 64 with remainder 2. The least integer greater than that is 65, because 3 times 64 is 192, still too small, while 3 times 65 is 195.

The insight is that iteration soon cycles, so 100 steps reduce to a remainder modulo the period. Compute directly: starting from 2, the next value is $(1+2)\mathrm{/}(1-2)=3\mathrm{/}(-1)=-3$, then $(1-3)\mathrm{/}(1+3)=(-2)\mathrm{/}4=-1\mathrm{/}2$, then $(1-1\mathrm{/}2)\mathrm{/}(1+1\mathrm{/}2)=(1\mathrm{/}2)\mathrm{/}(3\mathrm{/}2)=1\mathrm{/}3$, then $(1+1\mathrm{/}3)\mathrm{/}(1-1\mathrm{/}3)=(4\mathrm{/}3)\mathrm{/}(2\mathrm{/}3)=2$, back to the start. Hence values repeat every 4 steps. Since $100=4\times 25$ leaves remainder 0, the 100th value equals the starting value 2.

Think like a tutor: add rates, not times. Pipe A fills ${\displaystyle \frac{1}{a}}$ tank per hour, Pipe B fills ${\displaystyle \frac{1}{b}}$, Pipe C fills ${\displaystyle \frac{1}{c}}$, and together they fill ${\displaystyle \frac{1}{t}}$ per hour, so ${\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}={\displaystyle \frac{1}{t}}$. Combine the known parts first: ${\displaystyle \frac{1}{c}}={\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}={\displaystyle \frac{ab-at-bt}{abt}}$. Invert only at the end to get $c={\displaystyle \frac{abt}{ab-at-bt}}$. Multiplying by $abct$ at the start and then trying to collect $c$ from three terms is long and sign prone, but one denominator plus one inversion is quick. With $a=4$, $b=4$, and $t=1$, this gives $c=2$.

Two angles formed by two intersecting lines are either vertical, and so equal, or adjacent, and so supplementary, so test both cases. If they are vertical, $3x+10=2x+30$ gives $x=20$, so each angle is $70^{\circ }$ and its complement is $20^{\circ }$, which equals $x$. If they are adjacent, $(3x+10)+(2x+30)=180$ gives $x=28$, so the angles are $94^{\circ }$ and $86^{\circ }$, and the acute one has complement $4^{\circ }$, not $28^{\circ }$. Only the vertical case meets every condition, so the acute angle measures $70^{\circ }$.

The key insight is that equal travel times make distances proportional to speeds, turning perpendicular motion into a scaled right triangle solved by factoring. The distance gap 20 over 2 hours means speeds differ by 10, so write them as $v$ and $v+10$ with distances $2v$ and $2v+20$. Perpendicular directions give $(2v{)}^2+(2v+20{)}^2=100^2$, which simplifies to $v^2+10v-1200=0$. Factoring gives $(v+40)(v-30)=0$, so the slower speed is 30 and the faster is 40 miles per hour.

The insight is to factor and cancel before multiplying out, then enforce the domain instead of accepting every quadratic root. Since $x^2-4=(x-2)(x+2)$, for $x$ not equal to 2 the fraction simplifies to $(x+2)(x-3)$. Setting $(x+2)(x-3)=-4$ gives $x^2-x-6=-4$, or $x^2-x-2=0$, which factors as $(x-2)(x+1)=0$. The value 2 makes the original denominator zero so it cannot satisfy the equation, leaving $-1$ as the only solution. Checking gives $(-3)(-4)\mathrm{/}(-3)=-4$. Expanding the cubic directly hides the hole.

Think of $\mathrm{\mid }x-a\mathrm{\mid }$ as distance on the number line. Here $\mathrm{\mid }x-2\mathrm{\mid }+\mathrm{\mid }x-9\mathrm{\mid }$ is the sum of distances from $x$ to 2 and to 9, whose minimum is the distance 7 between 2 and 9, attained for every $x$ between them. The allowance is 10, so there are 3 extra units beyond the minimum, which by symmetry extend equally to the two sides, 1.5 to the left of 2 and 1.5 to the right of 9. Hence the graphed set is $0.5\le x\le 10.5$. The integers there are $1,2,\dots ,10$, which is 10 integers, counted directly with no casework on three absolute-value regions.

The insight is to split by the sign of the run and then bound numerator and denominator separately to their extremes. Write slope as $(y-4)\mathrm{/}(x-4)$ with $x\mathrm{\ne }4$. If $x>4$ the denominator is positive, so the fraction is largest when the numerator is largest and the denominator smallest: $4\mathrm{/}1=4$ at $(5,8)$. If $x<4$ the denominator is negative, so the largest value needs a negative numerator: $(-4)\mathrm{/}(-1)=4$ at $(3,0)$. Checking only the four corners gives 1, which misses these edge points.

The idea is to use the angle sum of the quadrilateral and never look for the angles one at a time. The angles add to $360^{\circ }$, so $A+B=360^{\circ }-150^{\circ }=210^{\circ }$. In triangle $APB$ the angles at $A$ and $B$ are ${\displaystyle \frac{A}{2}}$ and ${\displaystyle \frac{B}{2}}$, which total $105^{\circ }$. So angle $APB=180^{\circ }-105^{\circ }=75^{\circ }$, which is exactly half of $C+D$.

The key insight is that each removal deviation from 30 must be balanced over $n-1$ survivors, and subtracting eliminates the unknown total. Let the total be $S=30n$ and the removed numbers be $a$ and $b$. Then $(S-a)\mathrm{/}(n-1)=29$ gives $a=30n-29n+29=n+29$, and $(S-b)\mathrm{/}(n-1)=31$ gives $b=31-n$. Hence $a-b=2n-2$. Since the difference is 12, $2n-2=12$ so $n=7$. The larger removed number is the one whose deletion lowers the mean, $a=7+29=36$, while the smaller is 24.

Think like a tutor: never solve for the sides. Let the sides be $l$ and $w$. Perimeter gives a sum, $l+w={\displaystyle \frac{P}{2}}$, and Pythagoras gives $l^2+w^2=d^2$. The clever move is the square of a sum, $(l+w{)}^2=l^2+2lw+w^2$. Let $A=lw$. Then ${\displaystyle \frac{P^2}{4}}=d^2+2A$, so $2A={\displaystyle \frac{P^2}{4}}-d^2$ and $A={\displaystyle \frac{P^2-4d^2}{8}}$. Trying to find $l$ and $w$ separately forces a messy quadratic, but treating the product as one block finishes quickly. With 3,4,5, $P=14$ and $d=5$ give 12.

The idea is to set two slopes equal and then keep both roots of the resulting square. The slope from $(1,1)$ to $(3,k)$ is ${\displaystyle \frac{k-1}{2}}$, and from $(1,1)$ to $(k,9)$ it is ${\displaystyle \frac{8}{k-1}}$. Equal slopes give $(k-1{)}^2=16$, so $k-1=4$ or $k-1=-4$, giving $k=5$ or $k=-3$. Both work: for $k=5$ the slopes are 2 and 2; for $k=-3$ they are $-2$ and $-2$. The sum of the possible values is $5+(-3)=2$.

The key insight is to treat the graph as two separate meetings, one before and one after Car A turns, and discard any algebraic intersection outside its time interval. Car A outbound is $d=60t$ for $0\le t\le 2$ and return is $d=240-60t$ for $2\le t\le 4$, while Car B is $d=120-20t$. Outbound $60t=120-20t$ gives $t=1.5$ and $d=90$. Return $240-60t=120-20t$ gives $40t=120$, so $t=3$, which lies in both $[2,4]$ and $[0,6]$, giving $d=60$. The turnaround gap at $t=2$ is 40 and Car B is then at 80, so those numbers describe positions at the turn, not meetings.

The insight is to factor the bases into twos and fives and pair them into tens before counting digits. Since $4^7=(2^2{)}^7=2^{14}$ and $25^5=(5^2{)}^5=5^{10}$, the product is $2^{14}\times 5^{10}=2^4\times 10^{10}=16\times 10^{10}$. Renormalizing $16\times 10^{10}=1.6\times 10^{11}$ gives scientific notation with $n=11$. Multiplying out 16384 by 9765625 is far too long without a calculator, while pairing twos with fives leaves only 16 times a power of ten.

Write the angles as $12a$ and $16b$ with positive integers $a$ and $b$. Supplementary means $12a+16b=180$, so $3a+4b=45$. Since $3a$ and 45 are multiples of 3, $b$ must be a multiple of 3: $b=3$ gives $a=11$ (angles $132^{\circ }$ and $48^{\circ }$), $b=6$ gives $a=7$ ($84^{\circ }$ and $96^{\circ }$), and $b=9$ gives $a=3$ ($36^{\circ }$ and $144^{\circ }$). The acute members are $48^{\circ }$, $84^{\circ }$ and $36^{\circ }$, so the largest is $84^{\circ }$. A multiple of 16 such as $80^{\circ }$ fails because its supplement, $100^{\circ }$, is not a multiple of 12.

The insight is to name the counts: if $m$ copies are added and $n$ copies are subtracted, the result is $m(x^2-2x+3)-n(2x^2+x-4)=(m-2n)x^2+(-2m-n)x+(3m+4n)$. No $x^2$ term means $m=2n$. The $x$ term gives $-2m-n=-20$, so $-5n=-20$, $n=4$ and $m=8$. The constant term is $3(8)+4(4)=24+16=40$. Note that subtracting $-4$ adds 4 for each subtracted copy.

The insight is that multiplying out makes the squares cancel: $(n+3)(n+5)=n^2+8n+15$ and $(n+1)(n+2)=n^2+3n+2$, so the difference is just $5n+13$. Now $5n+13=4(n+3)+(n+1)$, so it is divisible by 4 exactly when $n+1$ is, that is, when $n$ leaves remainder 3 on division by 4. Those $n$ are $3,7,11,\dots ,47$, which is $(47-3)\mathrm{/}4+1=12$ integers.

The idea is extremal optimization under a fixed sum with a strict inequality and a divisibility restriction. Because O is on a straight line the two outer angles plus the right angle make 180 degrees, so the two unknown angles sum to 90 degrees. Write the smaller as s and the larger as 90 minus s. The condition that the larger exceeds twice the smaller gives 90 minus s greater than two s, so 90 is greater than three s and s is less than 30. To make the larger as small as possible the smaller must be as large as possible, since they sum to a constant. The largest multiple of 7 below 30 is 28, giving 62 for the larger, which indeed exceeds twice 28. Using equality, ignoring divisibility, or reversing the extremal direction gives the common wrong values.

The key is to recognize the numerator as a square in three variables and cancel the common factor, using positivity to ensure the denominator is not zero. Expanding $(2a+3b+c{)}^2$ gives $4a^2+9b^2+c^2+12ab+4ac+6bc$, which matches the numerator exactly, so the quotient equals $2a+3b+c$ because $2a+3b+c$ is positive and therefore nonzero for positive $a$, $b$, and $c$. Trying polynomial division without first regrouping into the square would be long, while the recognition makes the cancellation one step.

The insight is mirror construction for palindromes plus a case split when the mirror falls short, finished by a strict integer step. Mirroring the first two digits of 3271 gives 3223, which reads the same forward and backward but is still below 3271. So the leading pair must increase from 32 to 33, giving 3333, and no 32 palindrome exceeds 3271 while 3333 is the smallest 33 palindrome. Thus $P=3333$. Then $n-12>3333$ gives $n$ greater than 3345. The least integer greater than 3345 is 3346, because 3345 minus 12 equals 3333, which is not greater.

The insight is to translate the digit sum into linear equations that differ by which position repeats, then enforce digit bounds with leading-zero and distinctness. For patterns where the repeated digit is the hundreds, both 2 times $a$ plus $b$ equals 12 with $a$ nonzero and $b$ different from $a$. The solutions are $a$ equals 2, 3, 5 and 6 with $b$ equals 8, 6, 2 and 0, giving 4 numbers for each of the two such patterns. For the pattern where the tens and ones repeat, $a$ plus 2 times $b$ equals 12 gives $a$ equals 8, 6 and 2 with $b$ equals 2, 3 and 5, giving 3 numbers. The total is 4 plus 4 plus 3, which is 11.

The insight is prime factorization of a product of two binomial values combined with a positive-negative case split. Let $u=n+4$ and $v=n+10$, so $v-u=6$ and $uv$ is prime and positive. Hence $u$ and $v$ are both positive or both negative, and since their product is prime, the smaller in absolute value must be 1 or $-1$. With $v$ larger, only $u=1$, $v=7$ giving $n=-3$ and $uv=7$, or $u=-7$, $v=-1$ giving $n=-11$ and $uv=7$, are possible. Both give the prime 7, and every other integer $n$ gives a composite or nonpositive product. Trying values of $n$ one by one would never establish that only two work.

The insight is repetend scaling by 100 plus the symmetry of $ab$ with $ba$. Let $N=0.\overline{ab}$. Then $100N$ shifts the repetend two places, so $100N-N=99N$ equals the integer $10a+b$, giving $N=(10a+b)\mathrm{/}99$. Likewise $0.\overline{ba}=(10b+a)\mathrm{/}99$. Adding gives $11(a+b)\mathrm{/}99=(a+b)\mathrm{/}9$. Also $1.\bar{2}=1+2\mathrm{/}9=11\mathrm{/}9$. Hence $(a+b)\mathrm{/}9=11\mathrm{/}9$ and $a+b=11$. Testing digit pairs by long division would be very long, while the shift-and-subtract symmetry reduces the whole problem to one short equation.

The insight is that an even-count median of 85 that is not an observed score must be the average of neighboring values 80 and 90, which pins the unknown frequency through its own position. Let the unknown be $x$, so the total is 12 plus $x$. The last 80 is in position 4 plus $x$ and the first 90 is next, so the two middle positions must be those two, giving quantity 12 plus $x$ divided by 2 equals 4 plus $x$. Solving gives $x$ equals 4, and indeed 4 seventies, 4 eighties, 5 nineties, and 3 hundreds make the 8th 80 and 9th 90.

The insight is to reduce the slope to lowest terms to find the smallest integer step, then count multiples. The run is 24 and the rise is 18, so the slope is ${\displaystyle \frac{18}{24}}={\displaystyle \frac{3}{4}}$ in lowest terms. Any move between integer points must be a whole multiple of $(4,3)$, so the 24 run splits into $24\mathrm{/}4=6$ minimal steps, giving $6+1=7$ integer points including both ends.

Insight: negate the compound inequality instead of solving for $x$, then filter by remainders and pair symmetrically. Since $y=3x-7$, observe $-3x+7=-y$, so $-50\le -y<10$. Multiplying by $-1$ reverses both signs to give $50\ge y>-10$, so $-10<y\le 50$. Because $y=3x-7$ with integer $x$, $y$ leaves remainder 2 upon division by 3. Among $-9$ through 50, those values are $-7$, $-4$, up to 50, twenty terms with first $-7$ and last 50. Pairing first with last gives 20 times 43 divided by 2, which is 430. Solving for $x$ first and evaluating each $y$ separately is much longer.

The insight is to subtract the two polynomials. $A(x)-B(x)=(p-q)x+(q-p)=(p-q)(x-1)$. Since $p\mathrm{\ne }q$, this difference is zero only when $x=1$, so $r$ must be 1. Then $A(1)=1+p+q=10$, which gives $p+q=9$. Finally $A(2)+B(2)=(4+2p+q)+(4+2q+p)=8+3(p+q)=8+27=35$. The individual values of $p$ and $q$ are never needed.

The insight is to conserve water volume on the reduced base and first check the partial-submersion bound. Water volume is 8 by 6 by 2 equals 96. If the cube sticks out, water covers $48-16=32$ square inches, so depth would be $96\mathrm{/}32=3$, which is at most 4, so the assumption holds and 3 is correct. Using the full base for the total gives $(96+64)\mathrm{/}48=10\mathrm{/}3$, adding the rise on the reduced base to the old depth gives $2+2=4$, and dividing the total by the reduced base gives $160\mathrm{/}32=5$. The bound selects the correct case in seconds while ignoring the footprint fails.

The key insight is that overall rates are weighted by patient counts, not averaged, so Clinic B can lead each month yet trail overall for small $k$. Clinic A overall is $(10+36)\mathrm{/}(100+40)=46\mathrm{/}140=23\mathrm{/}70$. Clinic B overall is $(5+k)\mathrm{/}(40+k)$. Requiring $(5+k)\mathrm{/}(40+k)>23\mathrm{/}70$ gives $350+70k>920+23k$, so $47k>570$ and $k>12.12$, hence the least integer is 13. Indeed $17\mathrm{/}52$ at $k=12$ stays below $46\mathrm{/}140$, while $18\mathrm{/}53$ at $k=13$ passes it.