SAT Math — Module 2 challenge

The insight is a two-hour block plus remainder case work for who finishes. The six-hour pipe does $1\mathrm{/}6$ tank per hour and the four-hour pipe does $1\mathrm{/}4$, so one cycle of each gives $5\mathrm{/}12$ in 2 hours. Two cycles give $10\mathrm{/}12$ in 4 hours, leaving $2\mathrm{/}12=1\mathrm{/}6$, exactly one hour of the starting pipe, so 5 hours total. Treating the pipes as simultaneous gives 2.4, averaging gives 4.8, and rounding to full cycles gives 6.

The insight is to scale percentages to a concrete hypothetical total and then reverse the conditional denominator. Imagine 1000 students. Then 400 are seventh-graders and 600 are eighth-graders. Bus riders among seventh-graders are $400\times 3\mathrm{/}10=120$, while bus riders among eighth-graders are $600\times 1\mathrm{/}2=300$. Total bus riders are $120+300=420$. Among those 420 bus riders, 120 are seventh-graders, so the probability is ${\displaystyle \frac{120}{420}}={\displaystyle \frac{2}{7}}$ after dividing by 60. A student who only reads the 3 out of 10 rate misses the different base sizes.

The insight is to extract perfect squares and then use the bound $n<0$ to replace absolute value correctly. Since $\sqrt{18n^2}=\mathrm{\mid }n\mathrm{\mid }\sqrt{18}=\mathrm{\mid }n\mathrm{\mid }\cdot 3\sqrt{2}$ and $\sqrt{8n^2}=\mathrm{\mid }n\mathrm{\mid }\cdot 2\sqrt{2}$, and $\mathrm{\mid }n\mathrm{\mid }=-n$ for negative $n$, the first term is $2(-n)3\sqrt{2}=-6n\sqrt{2}$ and the second is $3(-n)2\sqrt{2}=-6n\sqrt{2}$, so $E=-12n\sqrt{2}$, which is positive as a sum of principal roots must be. Adding radicands directly or dropping roots misses the perfect-square structure and gives unlike or rootless results.

The insight is an enclosing-box re-cut that avoids slant heights, followed by quadratic rescaling. The triangle fits in the 6 by 6 box from 0 to 6 in each direction, whose area is 36 square centimeters. The three outside right triangles have areas 6 times 2 over 2, which is 6, 4 times 4 over 2, which is 8, and 2 times 6 over 2, which is 6, totaling 20. So the drawing area is 36 minus 20, which is 16. Trying base times height on slant sides would need square roots, while the box uses only whole numbers. Linear scale 4 gives area factor 16, so the true area is 16 times 16, which is 256 square meters.

The insight is zero-product factorization for the linear system combined with median-ordering case analysis distinguishing mean. Subtracting gives $(k-2)(x-y+2)=0$; $k=2$ gives $5x+y=14$ infinite, so for unique solution $y=x+2$ and $x=(k+4)\mathrm{/}3$, $y=(k+10)\mathrm{/}3$. Comparing $x$, $y$, $k$ gives median $x$ for $k<2$, median $k$ for $2<k<5$, median $y$ for $k>5$. Setting median 3 gives $x=3$, so $k=5$ outside $k<2; $y=3$, so $k=-1$ outside $k>5; $k=3$ inside $2<k<5$. Hence only $k=3$ works, while mean $(5k+14)\mathrm{/}9=3$ would give $13\mathrm{/}5$.

Insight: reframe probability bounds as a compound inequality and use exponent growth with reversal. Let $r=32-2^k$ and $p=r\mathrm{/}30$, so $0<p<1$ gives $0<32-2^k<30$. From $0<32-2^k$ get $2^k<32$, so $k<5$ and $k\le 4$ since $2^4=16$ and $2^5=32$. From $32-2^k<30$ get $-2^k<-2$, so $2^k>2$ after multiplying by $-1$ and flipping, giving $k>1$ and $k\ge 2$ since $2^1=2$ and $2^2=4$. With $k\ge 0$ this leaves $k=2$, 3, and 4 with red counts 28, 24, and 16, all strictly between 0 and 30. Trying each $k$ without the bound structure is longer.

The key is that only the change in the total matters, with percents as fractions, plus the range as a sum-difference system. The original total is $5(34)=170$ and the new total is $5(36)=180$, a rise of 10. Let the smallest be $s$ and the greatest be $g$. The 60 percent rise adds $(3\mathrm{/}5)s$ and the 20 percent cut subtracts $(1\mathrm{/}5)g$, so $(3\mathrm{/}5)s-(1\mathrm{/}5)g=10$, or $3s-g=50$. The range gives $g=s+10$. Substituting yields $3s-(s+10)=50$, so $2s=60$ and $s=30$. Hence $g=40$. The middle three do not change and never enter the change equation.

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 a tuned weighted combination plus an integer slack impossibility. Note $(3\mathrm{/}8)(3x+y)+(7\mathrm{/}8)(x+3y)=2x+3y$. Since $3x+y\le 20$ and $x+3y\le 24$, we get $2x+3y\le (60+168)\mathrm{/}8=28.5$, so any integer objective is at most 28. Write $s1=20-3x-y$ and $s2=24-x-3y$, both nonnegative integers. Then $28.5-(2x+3y)=(3s1+7s2)\mathrm{/}8$. If the objective were 28, then $3s1+7s2=4$, which has no nonnegative integer solution. Hence 28 is impossible. The pair $x=3$ and $y=7$ satisfies both constraints and gives $6+21=27$, so the greatest possible value is 27.

The key insight is adding to bound the sum plus a square bound for the product. Adding gives $3(x+y)=(x+2y)+(2x+y)\le 40$, so $x+y\le 13$ for integers since $3\times 14=42$. From $(x-y{)}^2\ge 0$ we have $xy\le ((x+y)\mathrm{/}2{)}^2$, and with integer sum at most 13 the product is at most $6\times 7=42$, with smaller sums giving at most 36. Hence area $xy\mathrm{/}2$ is at most 21. The pair $x=6$ and $y=7$ satisfies $6+14=20$ and $12+7=19$ and gives area $42\mathrm{/}2=21$, so the greatest possible area is 21.

The insight is to count the complement instead of listing numbers containing 7. There are $9\times 10\times 10=900$ three-digit numbers. Numbers with no 7 have 8 choices for the hundreds digit (1 to 9 except 7) and 9 choices each for the tens and ones (0 to 9 except 7), so $8\times 9\times 9=648$. Thus numbers with at least one 7 are $900-648=252$. Letting $252\mathrm{/}(252+n)=1\mathrm{/}3$ gives $756=252+n$, so $n=504$. Listing all 900 numbers would take far too long.

The key is a cheapest-list bound plus a construction attaining it, since an angle has a positive complement exactly when it is acute under $90^{\circ }$. Four distinct angles with no complement need at least $90+91+92+93=366^{\circ }$, using $90^{\circ }$ as the smallest non-acute integer and distinctness to force the increase, which already exceeds $360^{\circ }$, so four is impossible. Three is attainable, for example $1^{\circ }+91^{\circ }+92^{\circ }+176^{\circ }=360^{\circ }$, with three distinct angles at least $90^{\circ }$ and one acute angle, all distinct. Hence the greatest possible number without a complement is three.

The insight is to recover the hidden red count from the without-replacement product by using consecutive-integer factoring, then use complementary counting for blue. Total pairs are 13 times 12 divided by 2, which is 78. Favorable red pairs are 78 times $6\mathrm{/}13$, which is 36. Since 36 equals 9 times 8 divided by 2, there are 9 red marbles and 13 minus 9, which is 4, blue marbles. Blue pairs are 4 times 3 divided by 2, which is 6. So the blue probability is 6 divided by 78, which is $1\mathrm{/}13$.

The insight is to combine a counting re-frame for new pairs with percent re-basing. With $n$ students originally there are $n(n-1)\mathrm{/}2$ pairs. The newcomer pairs with each of the $n$ old students, so exactly $n$ new handshakes appear. The percent increase is therefore $n$ divided by $n(n-1)\mathrm{/}2$, which simplifies to $2\mathrm{/}(n-1)$. Setting $2\mathrm{/}(n-1)=1\mathrm{/}5$ gives $n-1=10$, so $n=11$. Indeed 55 pairs become 66, an increase of 11, which is 20 percent of 55.

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 insight is a case split for intersecting lines combined with a divisibility filter on the parameter. Opposite angles are equal while adjacent angles are supplementary, so either $5x+50=9x+46$ or $5x+50+9x+46=180$. The first gives $4x=4$ and $x=1$, which is not a multiple of 3. The second gives $14x+96=180$, so $14x=84$ and $x=6$, which is a multiple of 3. Then the two angles are $5(6)+50=80^{\circ }$ and $9(6)+46=100^{\circ }$, so the smaller is $80^{\circ }$. A solver who assumes one position without checking the multiple condition fails.

The key combines turning the mean into a total with a Pythagorean difference of squares. The five total $5(10)=50$ and the three known total $8+12+14=34$, so the other two, call them $p$ and $q$, satisfy $p+q=16$. Together with 8 they form a right triangle. If 8 were the hypotenuse then $p^2+q^2=64$ would combine with $(p+q{)}^2=256$ to force $pq=96$, which has no integer pair summing to 16. So 8 is a leg and $q^2-p^2=64$, which factors as $(q-p)(q+p)=64$. Since $q+p=16$, we get $q-p=4$. Solving $p+q=16$ and $q-p=4$ gives $p=6$ and $q=10$. The full list is 6, 8, 10, 12, 14, whose smallest is 6.

The key is that the two groups overlap in the two middle numbers, so adding their totals double counts the median pair, and an even-count median is their average. The total of all eight is $8(20)=160$. The five smallest total $5(16)=80$ and the five largest total $5(26)=130$, together 210. The excess $210-160=50$ is the sum of the fourth and fifth numbers counted twice. For eight numbers the median is the average of those two, so $50\mathrm{/}2=25$. No listing is needed once the overlap is seen.

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 to add the perimeter equations and read off the border area as width times average perimeter without ever finding the length and width. Let the outer garden be $l$ by $w$ and the inner hole be $l-6$ by $w-6$ since 3 feet are removed from each side. Then the outer perimeter is $2l+2w$ and the inner is $2l+2w-24$, so their sum is $4l+4w-24=120$ and $l+w=36$. The path area is $lw-(l-6)(w-6)=6(l+w)-36=216-36=180$. Trying to find $l$ and $w$ separately is impossible from one sum, a failing route, while the combined quantity finishes in seconds.

The key is to re-base percents that refer to different wholes and then balance the two counts. Let the numbers of boys and girls be $x$ and $y$. Walkers total $0.20x+0.80y$, and girls who walk total $0.80y$. Since half the walkers are girls, $0.80y=0.50(0.20x+0.80y)$, so $0.80y=0.10x+0.40y$ and $0.40y=0.10x$, giving $x=4y$. Hence girls are one out of five students, which is $20\mathrm{\%}$. A student who treats every percent as a share of the same whole is pushed into equating the walker share with the school share directly.

The insight is to convert first and then use total distance over total time rather than any average of speeds. Since 15 mi/h equals 22 ft/s, 22 ft/s is 15 mi/h and 44 ft/s is 30 mi/h. For one mile, the time is one quarter divided by 15 plus three quarters divided by 30, which is $1\mathrm{/}60$ plus $1\mathrm{/}40$ equals $1\mathrm{/}24$ hours. Dividing distance by time gives 24 mi/h. Weighting matters because three quarters of the distance is fast, while a plain mean of speeds either ignores the weights or ignores the harmonic form.