SSAT Upper Quantitative II challenge

The insight is to treat the added mix as unknown and eliminate it to get a divisibility condition, then minimize. Let the start be $4k$ and $3k$ and let $t$ red marbles be added, so $12-t$ green are added. Then $(4k+t)\mathrm{/}(3k+12-t)=7\mathrm{/}5$, so $20k+5t=21k+84-7t$, giving $k=12t-84$. Positivity forces $k>0$, so $t>7$, leaving $t=8,9,10,11,12$ with $k=12,24,36,48,60$ and original totals 84,168,252,336,420. The least is 84, from 48 and 36 plus 8 and 4 to reach 56 and 40. Without eliminating the split, solvers must guess both the start and the split.

The insight is inclusion-exclusion plus seeing the union. Each quarter has area $\pi \times 12^2\mathrm{/}4=36\pi$, so the sum is $72\pi$. The diagonal $BD$ splits the square into triangles $ABD$ and $BCD$. Every point of triangle $ABD$ is at most 12 from $A$ because the farthest vertices $B$ and $D$ are exactly 12 away, so that triangle lies in the first quarter; similarly the other triangle lies in the second quarter. Hence the union of the quarters is exactly the square of area 144. Therefore the overlap equals sum minus union, $72\pi -144$. Adding sectors without subtracting the square greatly overcounts.

The insight is to replace displacement by effective base area and then enforce the rim capacity as a bound. The brick footprint is $6\times 6=36$, so without overflow the water would stand at $800\mathrm{/}(100-36)=800\mathrm{/}64=12.5$ inches. Because 12.5 exceeds the 12-inch rim, some water must leave. At the rim the submerged brick occupies $36\times 12=432$ cubic inches, leaving $10\times 10\times 12-432=1200-432=768$ cubic inches for water. The tank started with 800, so $800-768=32$ cubic inches overflow.

The key is turning nearest-percent rounding into a double inequality and eliminating small totals in groups. Reporting 18 percent means the true share lies from 17.5 percent inclusive to 18.5 percent exclusive. For totals through 5 the whole interval lies below 1, so no positive count fits. For totals 6 through 10 the interval lies strictly between 1 and 2 because 0.175 times the total exceeds 1 while 0.185 times the total stays below 2, so no integer fits. With total 11 the interval runs from 1.925 to 2.035 and contains 2, realized by 2 over 11, so 11 is least. Checking each total by decimal division would be longer.

The insight is to compress alternating work into repeating two-hour cycles and then track whose turn handles the remainder. The hourly fractions are $1\mathrm{/}8$ and $1\mathrm{/}12$, so each two-hour cycle fills $1\mathrm{/}8+1\mathrm{/}12=5\mathrm{/}24$. Four cycles fill $20\mathrm{/}24=5\mathrm{/}6$ in 8 hours, leaving $1\mathrm{/}6$. The next turn belongs to the first pipe, adding $1\mathrm{/}8$ to reach $23\mathrm{/}24$ in 9 hours and leaving $1\mathrm{/}24$. The second pipe needs $(1\mathrm{/}24)\mathrm{/}(1\mathrm{/}12)=1\mathrm{/}2$ hour, so the total is $19\mathrm{/}2$ hours. Adding alternating fractions hour by hour is long, while one division locates the four full cycles immediately.

The insight is an extremal bound: for a fixed base the area is largest when the altitude is largest, and an altitude cannot exceed the slant side that contains it. Write the area with base $AB$, so the area equals one half times 5 times $h$, where $h$ is the distance from $C$ to the line through $AB$. Since $h$ is a leg of a right triangle with hypotenuse $AC=7$, $h$ is at most 7, with equality only when $AC$ is perpendicular to $AB$. Hence the largest area occurs for a right angle at $A$. Then $BC$ is the hypotenuse of legs 5 and 7, so $BC$ equals $\sqrt{25+49}$, which is $\sqrt{74}$. Any other position tilts $AC$ and shortens $h$.

The insight is to combine without-replacement probability counts with denominator divisibility to pin the total before factoring. Let total be $t$ with $r$ red and $b$ blue, $r+b=t<10$. Then $r(r-1)\mathrm{/}t(t-1)=2\mathrm{/}7$ and $b(b-1)\mathrm{/}t(t-1)=1\mathrm{/}7$, so $t(t-1)$ is a multiple of 7. For $t<10$, consecutive-product multiples of 7 occur only at $t=7$ giving 42 and $t=8$ giving 56. With $t=7$, $r(r-1)=12$ factors as $4\cdot 3$ giving $r=4$, and $b(b-1)=6$ gives $b=3$ with $4+3=7$. With $t=8$, $r(r-1)=16$ has no integer root. The student who uses single draws gets 2, one who swaps colors gets 3, and one who uses complements or totals gets larger values.

The insight is to group the geometric progression into triples and use remainders modulo seven instead of adding twenty-one powers. Since $1+2+4=7$, factor each block as $2^{3k}+2^{3k+1}+2^{3k+2}=2^{3k}(1+2+4)=7\cdot 2^{3k}$, which is a multiple of 7. With powers 0 through 20 there are exactly seven such triples, so $S$ is a sum of multiples of 7. Equivalently $S+1=2^{21}=(2^3{)}^7$ is 1 mod 7 because 8 is 1 mod 7, so $S$ is 0 mod 7. The remainder is therefore 0.

The insight is that integrality repeats with the reduced denominator of the slope rather than requiring all 43 checks. The rise is $41-11=30$ and the run is $46-4=42$, so slope $30\mathrm{/}42=5\mathrm{/}7$ in lowest terms. Starting at $(4,11)$, the line gives $y-11=(5\mathrm{/}7)(x-4)$, which is integral exactly when $x-4$ is a multiple of 7. Between 4 and 46 inclusive those values are 4, 11, 18, 25, 32, 39, and 46, a total of 7. Checking every $x$ would mean 43 substitutions, while the step observation finishes the count at once.

The insight is surround-and-subtract coordinate area combined with prism volume as base area times height. The enclosing 5 by 4 rectangle has area 20. The three outside right triangles have areas 2.5, 6, and 2, totaling 10.5. Subtracting leaves base area 9.5. A right prism has volume base area times height, so 9.5 times 4 equals 38 cubic centimeters.

The insight is that each inner minus flips every sign inside it, so working outward turns nesting into alternation. Indeed $50-(49-X)=50-49+X$, and expanding $X$ the same way gives $50-49+48-47+\cdots +2-1$. Now pair neighbors: $(50-49)+(48-47)+\cdots +(2-1)$, where each pair equals 1. There are 50 numbers hence 25 such pairs, so the value is 25. Evaluating from the innermost parentheses outward would need 49 separate subtractions, while the reframe plus count finishes immediately.

The insight is to read each absolute value as distance and pair symmetrically around the median. For any $n$, $\mathrm{\mid }n-1\mathrm{\mid }+\mathrm{\mid }n-17\mathrm{\mid }$ is the sum of distances to 1 and 17, which is at least 16, with equality whenever $n$ lies between them. Similarly $\mathrm{\mid }n-5\mathrm{\mid }+\mathrm{\mid }n-13\mathrm{\mid }$ is at least 8, with equality between 5 and 13, and $\mathrm{\mid }n-9\mathrm{\mid }$ is at least 0, with equality at 9. Adding gives at least $16+8+0=24$. The value $n=9$ lies in both intervals and hits 9, so it attains $8+4+0+4+8=24$. Testing integers one by one is the long failing route.

The insight is to express the intercepts through the slope and turn positivity into a divisibility condition. Write the slope as the integer $m<0$; then $y-18=m(x-12)$, so the $x$-intercept satisfies $-18=m(x_0-12)$, hence $x_0=12-18\mathrm{/}m$. Since $x_0$ must be a positive integer and $m$ is negative, $m$ must be a negative divisor of 18. The $y$-intercept $y_0=18-12m$ is then automatically a positive integer. The negative divisors of 18 are $-1,-2,-3,-6,-9,-18$, six lines, with $x$-intercepts 30,21,18,15,14,13. Testing all negative slopes one by one would be much longer.

The insight is central symmetry, not area subtraction with a case split. A half-turn about the rectangle center $(5,3)$ swaps the two halves of the rectangle, so any line cutting area in half must pass through that fixed center; otherwise rotation would produce a different line with swapped areas. Hence the line is $y-3=2(x-5)$, or $y=2x-7$. Setting $y=0$ gives $2x=7$, so $x=7\mathrm{/}2$. This also shows the line meets the bottom at $7\mathrm{/}2$ and the top at $13\mathrm{/}2$, both inside the sides, so no side-case analysis is needed. Using an edge midpoint instead of the center gives each of the other values.

The insight is to group the points by parallel lines and prove that cross pairs can never have slope 1, then count combinations. The first four points lie on $y=x$ and the last four on $y=x+3$, both of slope 1. Within each group any two points determine slope 1, giving $6+6=12$ favorable pairs. For a cross pair $(a,a)$ and $(b,b+3)$, the slope is $(b+3-a)\mathrm{/}(b-a)=1+3\mathrm{/}(b-a)$, which cannot equal 1 when $b\mathrm{\ne }a$, and is undefined when $b=a$. Hence no cross pair works. There are $8\times 7\mathrm{/}2=28$ total pairs, so the probability is $12\mathrm{/}28=3\mathrm{/}7$. Computing all 28 slopes one by one would take far too long.

The insight is to combine the three-dimensional Pythagorean extension with symmetric squaring of the dimension sum to get surface without edges. Let edges be $l,w,h$. Base diagonal squared is $l^2+w^2$, adding $h^2$ gives space diagonal squared $l^2+w^2+h^2=81$. Also $(l+w+h{)}^2=l^2+w^2+h^2+2P$, so $225=81+2P$ where $P$ is half the surface. Thus $2P=144$ and surface 144. Edges stay undetermined, so solving for them fails. The student who squares only one part reports a square, one who halves again reports half surface, and one who adds reports the sum.

The insight is to reduce the slope to lowest terms to see the smallest integer step. The run is $15-(-6)=21$ and the rise is $18-4=14$, so slope is $14\mathrm{/}21=2\mathrm{/}3$. Hence integer points advance by multiples of 3 horizontally and 2 vertically. Starting at $(-6,4)$, points are $(-6+3t,4+2t)$ for $t=0,\dots ,7$ since $21\mathrm{/}3=7$ steps, giving $7+1=8$ points including both ends. Checking all 22 integer $x$ values one by one is the long route, while the reduced step counts them at once.

The key is to write the series sum and then restrict the factors by parity. If the first term is $a$, the sum is $n(2a+n-1)\mathrm{/}2=105$, so $n(2a+n-1)=210$. Since their sum $2a+2n-1$ is odd, the two factors have opposite parity, and since $2a+n-1$ exceeds $n$, we get $n$ squared below 210, so $n$ is at most 14. Testing divisors of 210 below 14 with opposite parity to the cofactor gives runs such as 15 through 20, 12 through 18, and 6 through 15, while $n=14$ gives cofactor 15 and $a=1$, namely 1 through 14, which indeed sums to 105. The value $n=15$ would force $a=0$, which is not positive, so 14 is greatest.

The insight is to use the arithmetic series average together with divisor parity and a size bound. Let there be $k$ terms starting at $a$, so $188=ka+k(k-1)\mathrm{/}2$ and $a=188\mathrm{/}k-(k-1)\mathrm{/}2$ must be a positive integer. Also $2\cdot 188=k(2a+k-1)$ with $2a-1\ge 1$, so $k$ divides 376 and $k<\sqrt{376}<20$. Write $376=8\cdot 47$ with 47 prime. If $k$ is even then $376\mathrm{/}k$ must be odd, so $k$ must contain $2^3$, leaving only 8 or 16 below 20. Checking gives $188\mathrm{/}8-7\mathrm{/}2=20$ valid and $188\mathrm{/}16-15\mathrm{/}2$ non-integral. Odd $k$ would need $376\mathrm{/}k$ even, but the only odd divisors below 20 fail the integrality test. The student who only averages tries every $k$ by long division, while one who picks the complementary factor takes 47.

The key is the telescoping difference for consecutive products combined with a strict rational inequality. Since 1 over k times k plus 1 equals 1 over k minus 1 over k plus 1, all interior terms cancel and the sum through n collapses to 1 minus 1 over n plus 1, which equals n over n plus 1. Requiring more than 95 percent means n over n plus 1 exceeds 19 over 20, so 20 times n exceeds 19 times n plus 19 and n exceeds 19. The value 19 gives exactly 19 over 20, so strict inequality forces the least integer 20. Adding nine fractions with common denominators would be much longer.

The insight is to write each percent change as a reduced fraction and force divisibility through coprime denominators, then use the price interval to pin down one value. A $12\mathrm{\%}$ discount multiplies by $88\mathrm{/}100=22\mathrm{/}25$, so the sale price $22\times$ original$\mathrm{/}25$ is integral only if the original cents are a multiple of 25. An $18\mathrm{\%}$ tax multiplies the sale price by $118\mathrm{/}100=59\mathrm{/}50$, so the total is original$\times 22\times 59\mathrm{/}(25\times 50)=$original$\times 649\mathrm{/}625$. Since $649=11\times 59$ shares no prime factor with $625=5^4$, the original must be a multiple of 625 cents. Between 500 and 1000 cents the only multiple is 625 cents, so the original price is 6.25 dollars.

The insight is a bound plus a construction. Let the sides be $a$ and $b$ with $a^2+b^2=100$. Since $(a-b{)}^2\ge 0$, we have $a^2+b^2\ge 2ab$, so $100\ge 2ab$. Then $(a+b{)}^2=a^2+b^2+2ab\le 100+100=200$, so $a+b\le \sqrt{200}=10\sqrt{2}$ and the perimeter $2(a+b)$ is at most $20\sqrt{2}$. This bound is attainable: $a=b=5\sqrt{2}$ gives $a^2+b^2=50+50=100$ and perimeter $20\sqrt{2}$. A student who only tries integer guesses such as 6 and 8 never proves maximality.

The insight is that frequencies come from area, then cumulative counts locate the median, then pushing values to interval tops maximizes it. Widths are 10, 10, 20, and 10, so areas are proportional to 20, 30, 40, and 30, which sum to 120 and therefore equal the exact counts. Ordered scores run 1-20 in the first interval, 21-50 in the second, 51-90 in the third, and 91-120 in the fourth. For 120 scores the median is the average of the 60th and 61st values, both in 20-39. The largest attainable value puts those scores at 39 and 39, giving median 39.

The insight is to convert speeds to times with the inverse proportion for equal distances and then use an extremal bound with an explicit construction. Write each leg distance as $d$. Total time is $10d\mathrm{/}60=d\mathrm{/}6$. A leg over 90 uses less than $d\mathrm{/}90$ time, while any leg uses at most $d\mathrm{/}30$ time because speed is at least 30. With $k$ fast legs, total time is less than $kd\mathrm{/}90+(10-k)d\mathrm{/}30=d(1\mathrm{/}3-k\mathrm{/}45)$. Since this must exceed $d\mathrm{/}6$, we get $1\mathrm{/}3-k\mathrm{/}45>1\mathrm{/}6$, so $k<7.5$ and $k\le 7$. The bound is attainable: drive 7 legs at 105 miles per hour and 3 legs at 30 miles per hour, giving $7\mathrm{/}105+3\mathrm{/}30=1\mathrm{/}15+1\mathrm{/}10=1\mathrm{/}6$ of $d$, so 7 can occur.

The insight is to turn rounded percents into intervals for integer counts and then find the least total with a construction. For $n$ students with $a$ apple votes, $26.5\le 100a\mathrm{/}n<27.5$, and similarly $32.5\le 100b\mathrm{/}n<33.5$ and $39.5\le 100c\mathrm{/}n<40.5$. For the cherry interval to hold an integer, $n$ below 15 leaves only $n=5$ with 2 and $n=10$ with 4, but neither admits an apple count since 1.325 to 1.375 and 2.65 to 2.75 hold no integer. Thus no total below 15 works, which is the bound. For $n=15$, the counts 4, 5, and 6 give 26.67%, 33.33%, and 40%, rounding to 27, 33, and 40, which is the construction, so 15 is attainable and minimal.