SAT Math — Module 1 challenge

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 insight is invariant subtraction for the linear system combined with circle substitution and number-of-solutions exclusion. Subtracting the equations gives $(2k+2-2k-1)x+(k+4-k-3)y=5$, so $x+y=5$. With $y=5-x$ in $(x-1{)}^2+(y-4{)}^2=8$, we get $(x-1{)}^2+(1-x{)}^2=8$, so $2(x-1{)}^2=8$ and $x=3$ or $x=-1$, giving $(3,2)$ and $(-1,6)$. Using $(3,2)$ in $(2k+1)x+(k+3)y=4k+9$ gives $8k+9=4k+9$, so $k=0$. Using $(-1,6)$ gives $4k+17=4k+9$, impossible, and $k=2$ makes the determinant $-k+2$ zero with parallel distinct lines, so only $k=0$ yields a system solution on the circle.

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 $c=2$.

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 turn the story into a partition equation and then use divisibility inside a strict bound. Write $p=2f$ for the number who pass and $a$ for the number absent, so $2f+f+a=32$ and hence $3f=32-a$. Fewer than 5 means $0\le a\le 4$, so $28\le 3f\le 32$. But $3f$ must be a multiple of 3, and the only multiple of 3 from 28 to 32 is 30, so $f=10$ with $a=2$. The counts 20 passes, 10 fails, and 2 absents total 32 and meet every condition, so the only possible value is 10.

Bound the median with the total. Order the five distinct integers $a<b<m<d<e$; the range gives $e=a+6$ and the mean gives a total of 50. Distinctness inside the interval from $a$ to $a+6$ gives $b\le m-1$, $d\le a+5$ and $a\le m-2$, so the total is at most $3a+2m+10\le 5m+4$; if $m\le 9$ that is at most 49, too small. Likewise $b\ge a+1$, $d\ge m+1$ and $a\ge m-4$ give a total of at least $5m-4$; if $m\ge 11$ that is at least 51, too large. Hence the median is 10, as in the set 7,8,10,12,13, which has total 50 and range 6.

The insight is to regroup the cost as a multiple of the item count plus a notebook surcharge and then check endpoints with the nonstrict bounds. With $p$ pens, the cost is $4n+3p=3(n+p)+n$, which is at most 35, while $n+p$ is at least 10. Hence $30+n\le 3(n+p)+n\le 35$, giving $n\le 5$, while $n\ge 0$ by definition. Every value occurs: $n$ notebooks together with $10-n$ pens uses exactly 10 items and costs $30+n$ dollars, which is at most 35 for each of $n=0,1,2,3,4,5$, so the attainable values are exactly 0 through 5.

The shortcut is to solve the linear pair once in terms of $m$ and convert the circle condition into a quadratic bound. From $y=3x-7$, $x+2(3x-7)=m$ gives $x=(m+14)\mathrm{/}7$ and $y=(3m-7)\mathrm{/}7$. Then $x^2+y^2<49$ becomes $(m+14{)}^2+(3m-7{)}^2<2401$, or $10m^2-14m+245<2401$, so $5m^2-7m-1078<0$. Factoring gives $(m+14)(5m-77)<0$, so $-14<m<77\mathrm{/}5=15.4$. The greatest integer strictly below 15.4 is 15, and $m=15$ indeed gives a point inside the circle. Trying integer $m$ values one by one would take far too long.

The insight is symmetry: equal angles placed in vertically opposite corners force OE and OF to form a single straight line. Vertical angles give angle AOD equal to angle BOC, call it $x$. Then the remainders are angle EOD $=x-50$ and angle FOC $=x-50$, so those remainders are equal. Angle AOC is supplementary to angle AOD, so it measures $180-x$. Then angle AOF, which goes from OA through OC to OF, measures $(180-x)+(x-50)=130$. Adding angle AOE gives $50+130=180$, so points E, O, and F are collinear and angle EOF measures $180^{\circ }$. Trying to find $x$ first stalls because the crossing angle was never given.

The insight is an extremal bound plus an explicit construction to show the bound is sharp. The total must be $8\times 11=88$, a one-step equation. To have many numbers above 14, make everything as small as the rules allow: numbers above 14 cost at least $15,16,17,18,19$ and numbers below cost at least $1,2,3$. With five numbers above 14, the cheapest distinct list is $1,2,3,15,16,17,18,19$, whose sum is 91, already above 88, so five or more is impossible. With four, $1+2+3+4+15+16+17+18=76$, leaving 12 to add while staying distinct and above 14; $1,2,3,4,15,16,17,30$ totals 88 with four numbers above 14. So the maximum is 4.

The insight is to turn each speed into a lap time with a common unit and take the least common multiple by prime factorization. Since 22 ft/s equals 15 mi/h, 33 ft/s equals 22.5 mi/h. For 3 miles the lap times are 3 divided by 22.5 hours equals 8 minutes, 3 divided by 30 hours equals 6 minutes, and 3 divided by 18 hours equals 10 minutes. The joint return is the least common multiple of 8, 6, and 10, whose prime factors give $2^3$ times 3 times 5 equals 120 minutes. Listing multiples to 120 would take far longer than factoring.

The insight is to combine the commutativity of successive multipliers with the asymmetry of intermediate divisibility. Both orders multiply the start by $6\mathrm{/}5$ and $5\mathrm{/}4$, so both end at $3\mathrm{/}2$ times the start, but the middle year differs. The 20 percent then 25 percent order needs the start to be a multiple of 5 and the middle to be a multiple of 4, forcing the start to be a multiple of 10. The reverse order needs the start to be a multiple of 4 with the middle automatically a multiple of 5, forcing only a multiple of 4. A start working for both must be a common multiple of 10 and 4, hence a multiple of 20. Below 50 the possibilities are 20 and 40, so the largest is 40, giving yearly counts 40, 48, 60 in one order and 40, 50, 60 in the other.

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 let the first equation bound the search: with $x$ and $y$ positive integers, $3x+5y=64$ leaves only finitely many pairs, and divisibility picks them out. Since $64-3x$ must be a positive multiple of 5, $x$ is 3 mod 5, so $(x,y)$ is $(3,11)$, $(8,8)$, $(13,5)$, or $(18,2)$. From $7x+py=92$, $p=(92-7x)\mathrm{/}y$. The first three pairs give $71\mathrm{/}11$, $36\mathrm{/}8$, and $1\mathrm{/}5$, none an integer, while $(18,2)$ gives $(92-126)\mathrm{/}2=-17$. A student who sees the finite list finishes without a calculator; one who solves parametrically faces messy fractions. Thus $p=-17$.

The insight is to add the six orthogonal views instead of counting cubes. The top and bottom each see the 3 by 4 footprint, giving 12 and 12. The front and back each see $4+3+1=8$. The left and right each see $3+2+1=6$. Adding gives $12+12+8+8+6+6=52$. Counting cube by cube forces tracking dozens of hidden glued faces and usually misses the bottom or double counts the steps.

The insight is to square the sum of the legs instead of solving for each leg, combining the linear perimeter with the Pythagorean relation. Let the legs be $a$ and $b$. Then $a+b+9=20$, so $a+b=11$. By the Pythagorean theorem $a^2+b^2=81$. Squaring the sum gives $(a+b{)}^2=a^2+2ab+b^2$, so $121=81+2ab$. Thus $2ab=40$ and $ab=20$. The area is $ab\mathrm{/}2$, which equals 10 square inches. Trying to find $a$ and $b$ separately leads to an ugly quadratic with irrational roots, while the area follows at once.

Square what is asked instead of solving. Let $T=\sqrt{r}+\sqrt{s}$ with $T>0$. Then $T^2=r+s+2\sqrt{rs}$. By Vieta from $x^2-11x+16=0$, $r+s=11$ and $rs=16$, so $\sqrt{rs}=4$ and $T^2=11+8=19$, giving $T=\sqrt{19}$. Solving gives the messy pair $(11\pm \sqrt{57})\mathrm{/}2$ whose square roots do not simplify individually, so the direct route stalls while squaring first finishes in seconds.

The insight is to drop perpendiculars from $P$ to the sides to create four right triangles whose squares cancel into an opposite-corner invariant. Write $P=(x,y)$ in a coordinate placement with $A=(0,0)$, $B=(W,0)$, $C=(W,H)$, $D=(0,H)$. Then $P{A}^2=x^2+y^2$, $P{B}^2=(W-x{)}^2+y^2$, $P{C}^2=(W-x{)}^2+(H-y{)}^2$, $P{D}^2=x^2+(H-y{)}^2$. Adding opposite pairs gives $P{A}^2+P{C}^2=x^2+y^2+(W-x{)}^2+(H-y{)}^2=P{B}^2+P{D}^2$. Hence $2^2+9^2=6^2+P{D}^2$, so $4+81-36=49$ and $PD=7$ after taking the positive root. No coordinate of $P$ or side of the rectangle is ever needed.

The key insight is that averaging passes through a linear function, so the mean bonus is the formula applied to the mean hours. First compute the overall mean hours by weighting: $(15\cdot 40+10\cdot 60)\mathrm{/}25=1200\mathrm{/}25=48$. Let the hours be $H_i$ and bonuses $B_i=m{H}_i+b$. Then the mean bonus is $(1\mathrm{/}25)\sum B_i=m(1\mathrm{/}25)\sum H_i+b=m\cdot 48+b$. But an employee with 48 hours gets 720, so $m\cdot 48+b=720$. Hence the mean bonus is 720 without ever finding $m$ and $b$, which cannot be found from one point.

The insight is to invert to tanks per hour and clear denominators to a factorable quadratic, then keep only the admissible positive root. Let the faster take $n$ hours, so the slower takes $n+2$. Then $1\mathrm{/}n+1\mathrm{/}(n+2)=1\mathrm{/}2.4=5\mathrm{/}12$. Clearing gives $12(2n+2)=5n(n+2)$, or $5n^2-14n-24=0$, which factors as $(5n+6)(n-4)=0$. The positive root is $n=4$, so the slower takes 6 hours. A student who adds or averages times never forms the reciprocal equation and is stuck with decimals.

The insight is that integrality of a fractional-slope line is a divisibility condition modulo the denominator. The slope is $(11.5-7)\mathrm{/}(8-2)=4.5\mathrm{/}6=3\mathrm{/}4$, so $f(x)=(3\mathrm{/}4)x+11\mathrm{/}2$ since $7-(3\mathrm{/}4)\cdot 2=11\mathrm{/}2$. Then $f(x)$ is an integer exactly when $(3x+22)\mathrm{/}4$ is an integer, which happens when $3x+22$ is a multiple of 4. Since $3x+22\equiv 3x+2$ mod 4, this needs $3x\equiv 2$ mod 4, so $x\equiv 2$ mod 4. In 0 to 100 these are $2,6,\dots ,98$, which is $(98-2)\mathrm{/}4+1=25$ values. Checking all 101 values by hand is long, but the congruence is short.

The insight is to avoid solving for the sides and instead recover their sum with the identity $(a+b{)}^2=a^2+b^2+2ab$, then use the fact that any corresponding lengths share the same linear scale. Let the true sides be $a$ and $b$ in feet. Then $ab=1200$ and $a^2+b^2=50^2=2500$ by the Pythagorean theorem, so $(a+b{)}^2=2500+2400=4900$ and $a+b=70$, giving a true perimeter of 140 feet. The drawing diagonal of 2.5 cm corresponds to 50 feet, so the linear factor is $2.5\mathrm{/}50=1\mathrm{/}20$ cm per foot. Hence the drawing perimeter is $140\mathrm{/}20=7$ cm, a short computation once the sum is seen.