Comparing Fractions

Rewrite both over \(24\). \(\tfrac{5}{8}=\tfrac{5\times 3}{8\times 3}=\tfrac{15}{24}\) and \(\tfrac{7}{12}=\tfrac{7\times 2}{12\times 2}=\tfrac{14}{24}\), so the north crew leads with \(15\) pieces to \(14\). The gap is \(\tfrac{15}{24}-\tfrac{14}{24}=\tfrac{1}{24}\), already in simplest form. North is ahead by \(\boxed{1/24}\) of the path. Comparing \(5\) against \(7\) directly does not work, since an eighth of the path and a twelfth of the path are different lengths.

Over the shared bottom \(56\), \(\tfrac{4}{7}=\tfrac{4\times 8}{56}=\tfrac{32}{56}\) and \(\tfrac{5}{8}=\tfrac{7\times 5}{56}=\tfrac{35}{56}\). Since \(35\gt 32\), \(\tfrac{5}{8}\) is larger, and the gap is \(\tfrac{35}{56}-\tfrac{32}{56}=\tfrac{3}{56}\), already in simplest form. So the gap is \(\boxed{3/56}\). Those two tops are exactly the diagonal products, which is why the diagonals settle the order without writing the common denominator at all.

Cross-multiply. \(7\times 11=77\) belongs to \(\tfrac{7}{9}\) and \(9\times 8=72\) belongs to \(\tfrac{8}{11}\), so \(\tfrac{7}{9}\) is higher. Those products are the numerators over \(9\times 11=99\), so the gap is \(\tfrac{77}{99}-\tfrac{72}{99}=\tfrac{5}{99}\). Since \(99=9\times 11\) shares nothing with \(5\), that is simplest form. The left marker beats the right by \(\boxed{5/99}\) of the route.

Both tops are \(6\), and an eleventh is a fatter piece than a thirteenth, so \(\tfrac{6}{11}\) is the larger share. For the gap use \(11\times 13=143\), where \(\tfrac{6}{11}=\tfrac{78}{143}\) and \(\tfrac{6}{13}=\tfrac{66}{143}\), so \(\tfrac{78}{143}-\tfrac{66}{143}=\tfrac{12}{143}\), already in simplest form. The lead is \(\boxed{12/143}\). The denominators alone settled the direction, and the common bottom was only needed to measure the gap.

With the tops matched, \(\tfrac{4}{15}=\tfrac{12}{45}\) and \(\tfrac{3}{10}=\tfrac{12}{40}\). Both hold \(12\) pieces, and cutting into \(40\) parts makes bigger pieces than cutting into \(45\), so the red ribbon \(\tfrac{3}{10}\) is the larger share. Over \(30\), \(\tfrac{3}{10}=\tfrac{9}{30}\) and \(\tfrac{4}{15}=\tfrac{8}{30}\), so the gap is \(\tfrac{9}{30}-\tfrac{8}{30}=\tfrac{1}{30}\). Red leads by \(\boxed{1/30}\) of a roll. With matched tops the smaller denominator wins, the reverse of the matched-bottom rule.

Each tank is one piece short of full, \(\tfrac{1}{9}\) for the first and \(\tfrac{1}{12}\) for the second. A twelfth is a thinner sliver than a ninth, so tank 2 has less empty room and \(\tfrac{11}{12}\gt\tfrac{8}{9}\). Both missing pieces are measured down from the same full mark, so the fill levels differ by \(\tfrac{1}{9}-\tfrac{1}{12}=\tfrac{4}{36}-\tfrac{3}{36}=\tfrac{1}{36}\). Tank 2 is fuller by \(\boxed{1/36}\) of a tank.

The top bar has \(\tfrac{1}{14}\) left to fill and the bottom has \(\tfrac{1}{15}\). A fifteenth is the smaller piece, so the \(\tfrac{14}{15}\) bar is closer to done. Both leftovers are measured down from the same finish line, so the distance between the bars is \(\tfrac{1}{14}-\tfrac{1}{15}=\tfrac{15-14}{14\times 15}=\tfrac{1}{210}\). The bars are \(\boxed{1/210}\) apart. Subtracting the unit fractions this way makes the big denominator appear once, at the end, instead of at the start.

A score beats \(\tfrac{1}{2}\) when its top is more than half its bottom. \(\tfrac{5}{9}\) wins since \(5\gt 4\tfrac{1}{2}\). \(\tfrac{4}{9}\) loses. \(\tfrac{7}{15}\) loses since \(7\lt 7\tfrac{1}{2}\). \(\tfrac{9}{16}\) wins since \(9\gt 8\). \(\tfrac{11}{20}\) wins since \(11\gt 10\). \(\tfrac{8}{17}\) loses since \(8\lt 8\tfrac{1}{2}\). That is \(\boxed{3}\) qualifying scores. Half of a whole split into \(b\) parts is \(\tfrac{b}{2}\) parts, so one comparison per score does the job with no common denominators.

Twice \(7\) is \(14\), which clears \(13\), so \(\tfrac{7}{13}\) sits above half. Twice \(9\) is \(18\), short of \(20\), so \(\tfrac{9}{20}\) sits below half. Phone 1 is the more charged one. Cross-multiplying agrees, \(7\times 20=140\) against \(9\times 13=117\), and those are the numerators over \(13\times 20=260\), so the gap is \(\tfrac{140}{260}-\tfrac{117}{260}=\tfrac{23}{260}\). Since \(23\) is prime and \(260=2^2\times 5\times 13\), that is lowest terms. Phone 1 leads by \(\boxed{23/260}\).

Compare the sizes over \(40\). \(\tfrac{3}{5}=\tfrac{24}{40}\) and \(\tfrac{5}{8}=\tfrac{25}{40}\), so \(-\tfrac{5}{8}\) dips deeper below \(0\) and \(-\tfrac{3}{5}\) is the larger, warmer value. The distance between the readings is the same as the distance between the sizes, \(\tfrac{25}{40}-\tfrac{24}{40}=\tfrac{1}{40}\), already in simplest form. The two mornings are \(\boxed{1/40}\) of a degree apart. Below \(0\) the bigger size is the smaller value, so compare as positives and then flip the verdict.

Priya is \(\tfrac{1}{12}\) short of the finish and Mara is \(\tfrac{1}{6}\) short, and the smaller leftover means farther along, so Priya is ahead of Mara. For Dev, cross-multiply \(\tfrac{7}{9}\) against Mara's \(\tfrac{5}{6}\), giving \(7\times 6=42\) and \(5\times 9=45\), so Dev trails Mara and therefore Priya too. The order from least to most is Dev, Mara, Priya, so the gap from leader to last is \(\tfrac{11}{12}-\tfrac{7}{9}=\tfrac{33}{36}-\tfrac{28}{36}=\tfrac{5}{36}\), already in simplest form. That is \(\boxed{5/36}\).

The rate \(\tfrac{98}{99}\) is short by \(\tfrac{1}{99}\) and \(\tfrac{100}{101}\) is short by \(\tfrac{1}{101}\). The bigger miss belongs to \(\tfrac{98}{99}\), so \(\tfrac{100}{101}\) is the larger rate. The gap between the rates is the gap between the misses, \(\tfrac{1}{99}-\tfrac{1}{101}=\tfrac{101}{9999}-\tfrac{99}{9999}=\tfrac{2}{9999}\), and \(9999=9\times 11\times 101\) is odd, so that is lowest terms. The rates differ by \(\boxed{2/9999}\).

Since \(4\) and \(7\) share no factor, the shared denominator is \(4\times 7=28\). Rosa has \(\tfrac{3}{4}=\tfrac{21}{28}\) and Ben has \(\tfrac{5}{7}=\tfrac{20}{28}\), so Rosa has painted more, and the gap is \(\tfrac{21}{28}-\tfrac{20}{28}=\tfrac{1}{28}\), already in simplest form. Rosa leads by \(\boxed{1/28}\) of a wall.

The diagonal products are \(8\times 10=80\) for \(\tfrac{8}{11}\) and \(11\times 7=77\) for \(\tfrac{7}{10}\), so \(\tfrac{8}{11}\) is larger. Those same products are the numerators over \(11\times 10=110\), so the gap is \(\tfrac{80}{110}-\tfrac{77}{110}=\tfrac{3}{110}\), already in simplest form. The larger beats the smaller by \(\boxed{3/110}\).

Set the diagonal products equal, \(6\times 20=k\times 15\), so \(120=15k\) and \(k=120\div 15=\boxed{8}\). Quick check, \(\tfrac{6}{8}\) and \(\tfrac{15}{20}\) both reduce to \(\tfrac{3}{4}\).

Both tops are \(8\), and thirteenths are bigger pieces than fifteenths, so \(\tfrac{8}{13}\gt\tfrac{8}{15}\) and the person with the 13-strand ribbon has more. Over \(15\times 13=195\), \(\tfrac{8}{13}=\tfrac{120}{195}\) and \(\tfrac{8}{15}=\tfrac{104}{195}\), so the gap is \(\tfrac{120}{195}-\tfrac{104}{195}=\tfrac{16}{195}\). Since \(16=2\times 2\times 2\times 2\) and \(195=3\times 5\times 13\), that is already in simplest form, so the gap is \(\boxed{16/195}\).

Both matched forms have a top of \(18\), and \(80\) parts are bigger than \(105\) parts, so \(\tfrac{18}{80}\gt\tfrac{18}{105}\) and the second recipe, \(\tfrac{9}{40}\), uses more. Both \(40\) and \(35\) divide \(280\), so \(\tfrac{9}{40}=\tfrac{63}{280}\) and \(\tfrac{6}{35}=\tfrac{48}{280}\), and the gap is \(\tfrac{63}{280}-\tfrac{48}{280}=\tfrac{15}{280}=\tfrac{3}{56}\) after cancelling a \(5\). The second recipe uses \(\boxed{3/56}\) more of the jar.

Double the top and compare it with the bottom. \(\tfrac{7}{12}\) gives \(14\gt 12\), a win. \(\tfrac{5}{11}\) gives \(10\lt 11\), a loss. \(\tfrac{10}{19}\) gives \(20\gt 19\), a win. \(\tfrac{6}{13}\) gives \(12\lt 13\), a loss. \(\tfrac{9}{17}\) gives \(18\gt 17\), a win. That is \(\boxed{3}\) scores above \(\tfrac{1}{2}\). Doubling works because \(\tfrac{1}{2}\) is exactly the case where the top equals half the bottom.

Twice \(8\) is \(16\), which clears \(15\), so \(\tfrac{8}{15}\gt\tfrac{1}{2}\). Twice \(7\) is \(14\), short of \(16\), so \(\tfrac{7}{16}\lt\tfrac{1}{2}\). Green is the farther checkpoint. For the gap use \(15\times 16=240\), where \(\tfrac{8}{15}=\tfrac{128}{240}\) and \(\tfrac{7}{16}=\tfrac{105}{240}\), so \(\tfrac{128}{240}-\tfrac{105}{240}=\tfrac{23}{240}\). Since \(23\) is prime and does not divide \(240\), the gap is \(\boxed{23/240}\).

Cup 1 leaves \(\tfrac{1}{10}\) empty and cup 2 leaves \(\tfrac{1}{11}\). The bigger bottom makes the smaller piece, so cup 2 has less empty space and \(\tfrac{10}{11}\) is the fuller one. Both gaps are measured down from the same brim, so the levels differ by \(\tfrac{1}{10}-\tfrac{1}{11}=\tfrac{11}{110}-\tfrac{10}{110}=\tfrac{1}{110}\). The two levels are \(\boxed{1/110}\) of a cup apart.

Survey 1 is \(\tfrac{1}{50}\) short of done and survey 2 is \(\tfrac{1}{51}\) short. A fifty-first is the smaller leftover, so survey 2 is further along. Both leftovers are measured from the same finished whole, so the completion fractions differ by \(\tfrac{1}{50}-\tfrac{1}{51}=\tfrac{51}{2550}-\tfrac{50}{2550}=\tfrac{1}{2550}\), already in simplest form. Survey 2 leads by \(\boxed{1/2550}\).

Compare the sizes over \(9\times 7=63\). \(\tfrac{5}{9}=\tfrac{35}{63}\) and \(\tfrac{4}{7}=\tfrac{36}{63}\), so \(\tfrac{4}{7}\) is the bigger size, which puts \(-\tfrac{4}{7}\) farther below \(0\) and makes \(-\tfrac{5}{9}\) the warmer reading. The distance between them is the difference of the sizes, \(\tfrac{36}{63}-\tfrac{35}{63}=\tfrac{1}{63}\). The readings are \(\boxed{1/63}\) apart.

All four denominators divide \(72\). Rewriting, \(\tfrac{7}{12}=\tfrac{42}{72}\), \(\tfrac{5}{8}=\tfrac{45}{72}\), \(\tfrac{11}{18}=\tfrac{44}{72}\), and \(\tfrac{2}{3}=\tfrac{48}{72}\). The pieces are now the same size, so the biggest numerator wins. That is \(48\), which came from \(\boxed{2/3}\).

Over \(12\), the leftover sand is \(\tfrac{3}{4}-\tfrac{1}{3}=\tfrac{9}{12}-\tfrac{4}{12}=\tfrac{5}{12}\) of a scoop. Half a scoop is \(\tfrac{6}{12}\), so what remains is a little less than half, short by \(\tfrac{6}{12}-\tfrac{5}{12}=\tfrac{1}{12}\). The sand misses the halfway mark by \(\boxed{1/12}\) of a scoop.

Both \(16\) and \(20\) divide \(80\). Maya has \(\tfrac{13}{16}=\tfrac{65}{80}\) and Theo has \(\tfrac{17}{20}=\tfrac{68}{80}\), so Theo leads. The gap is \(\tfrac{68}{80}-\tfrac{65}{80}=\tfrac{3}{80}\), and since \(80=2\times 2\times 2\times 2\times 5\) has no factor of \(3\), that is already in simplest form. Theo is ahead by \(\boxed{3/80}\).

Each mosaic is one tile from done, \(\tfrac{1}{124}\) missing for the first and \(\tfrac{1}{200}\) for the second. The bigger denominator makes the smaller hole, so mosaic 2 is the more complete one. The completions differ by the difference of the holes. Here \(124=4\times 31\) and \(200=8\times 25\) share only a factor of \(4\), so the least common denominator is \(\tfrac{124\times 200}{4}=6200\), and \(\tfrac{1}{124}-\tfrac{1}{200}=\tfrac{50}{6200}-\tfrac{31}{6200}=\tfrac{19}{6200}\), lowest terms since \(19\) is prime. Mosaic 2 leads by \(\boxed{19/6200}\).