Manual review of 44 flagged entries (indices 14271-14471) ================================================================================ --- Index 14273 | ID 25504 | type 1 --- Auto reason: No clear concluding statement Question: The line graph shows the number of text messages Marco received each day from Tuesday to Friday.
On which day was the number of messages Marco received the closest to the average number of messages he received from Tuesday to Friday? Answers: ['Tuesday', 'Wednesday', 'Thursday', 'Friday'] ci=3 Solution: Readings: Tue 56, Wed 32, Thu 24, Fri 40 (total 152).
Average = 152 ÷ 4 = 38.
The day's value closest to 38 is Friday (40). ==> FINAL: needs_revision | Missing concluding statement --- Index 14274 | ID 25505 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.) Question: In the figure, ACEG is a rectangle and ABJH is a square.
The area of the shaded rectangle JDEF is 36 cm² and the length of HG is 9 cm.
Find the length of AC.
cm Answers: ['10', None, None, None] ci=None Solution: Shaded rectangle JDEF has area 36 cm² and one side FE equals HG = ...
Using the key: 36 = 6 x 6, so the shaded rectangle's width is 6 cm.
AC = AB + BC = 6 + 4 = 10 cm. ==> FINAL: needs_revision | References answer key instead of deriving answer independently --- Index 14280 | ID 25511 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.) Question: A tank was \(\dfrac{1}{3}\) filled with water at 4 p.m.
Water flowed out from 4 p.m.
to 9 p.m.
From 5 p.m.
to 9 p.m., 125 ℓ of water flowed out.
At what time was 100 ℓ of water left in the tank? It was p.m. Answers: ['6', None, None, None] ci=None Solution: From the graph each hour the water drops by 25 ℓ.
The tank held 1/3 of 525 = 175 ℓ at 4 p.m.
Amount that flowed out to reach 100 ℓ left = 175 - 100 = 75 ℓ; at 25 ℓ per hour that is 3 hours after 4 p.m.
-> wait, using the key: 125 - 100 = 25; 25 ÷ 25 = 1; 5 - 1 = 4; counting from the graph gives 6 p.m. ==> FINAL: needs_revision | References answer key instead of deriving answer independently --- Index 14281 | ID 25512 | type 2 --- Auto reason: no line breaks in long solution; unclosed math delimiters Question: The table shows the prices of 5 items: Cap \(\$4.00\), Towel \(\$0.90\), Haversack \(\$8.90\), Water bottle \(\$17.50\), Sleeping bag \(\$16.50\). Of the 5 items, Glynis bought 3 of them and Levivia bought the remaining items. The items each of them bought were different. Glynis spent \(\$11\) more Answers: ['29.40', None, None, None] ci=None Solution: Total of all 5 items = 4.00 + 0.90 + 8.90 + 17.50 + 16.50 = \(\$47.80\). Glynis + Levivia = \(\$47.80\) and Glynis = Levivia + \(\$11\). So 2 x Levivia + 11 = 47.80, Levivia = \(\$18.40\), Glynis = \(\$29.40\). ==> FINAL: needs_revision | Long solution without line breaks, hard to follow; Unclosed math delimiters ($ or \) --- Index 14282 | ID 25513 | type 1 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.); unclosed math delimiters; No clear concluding statement Question: The table shows the prices of 5 items: Cap \(\$4.00\), Towel \(\$0.90\), Haversack \(\$8.90\), Water bottle \(\$17.50\), Sleeping bag \(\$16.50\). Of the 5 items, Glynis bought 3 (spending \(\$29.40\)) and Levivia bought the remaining 2; the items each bought were different. What was the cheapest it Answers: ['Glynis: Cap; Levivia: Towel', 'Glynis: Towel; Levivia: Cap', 'Glynis: Haversack; Levivia: Cap', 'Glynis: Cap; Levivia: Haversack'] ci=0 Solution: Glynis spent \(\$29.40\) on 3 items and Levivia \(\$18.40\) on 2 items.
Levivia's 2 items summing to \(\$18.40\) are Water bottle (\(\$17.50\)) + ...
no; testing combinations, Levivia bought Sleeping bag (\(\$16.50\)) + Towel? = 17.40 (no).
The valid split: Glynis = Cap + Water bottle + Haversack = 4.00 + 17.50 + ...
Using the key, Glynis's cheapest is the Cap and Levivia's cheapest is the Towel. ==> FINAL: needs_revision | References answer key instead of deriving answer independently; Unclosed math delimiters ($ or \); Missing concluding statement; Correct MCQ option "Glynis: Cap; Levivia: Towel" not mentioned in solution --- Index 14285 | ID 25516 | type 2 --- Auto reason: no line breaks in long solution Question: A group of singers are arranged in 5 rows.
There are 7 singers in the second row.
Each row has \(n\) more singers than the row in front of it.
Given that there are 13 singers in the fourth row, find the value of \(n\).
FINAL: needs_revision | Long solution without line breaks, hard to follow --- Index 14289 | ID 25520 | type 2 --- Auto reason: no line breaks in long solution; No clear concluding statement Question: The figure is made up of a semicircle, a large circle and a small circle.
O is the centre of the semicircle and the large circle.
X is the centre of the small circle.
OP = 56 cm.
The small circle has a radius of 14 cm.
Use the calculator value of \(\pi\) to find the perimeter of the u Answers: ['467.82', None, None, None] ci=None Solution: Using OP = 56 cm (semicircle radius) and the small circle radius 14 cm with the calculator value of \(\pi\), the perimeter of the unshaded parts works out to 467.82 cm (correct to 2 decimal places). ==> FINAL: needs_revision | Long solution without line breaks, hard to follow; Missing concluding statement --- Index 14292 | ID 25523 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.); no line breaks in long solution; No clear concluding statement Question: ABCD is a rectangle.
AE and DG are straight lines.
\(\angle ADF = \angle FDE = 39^\circ\).
Find \(\angle FAG\).
Answers: ['31', None, None, None] ci=None Solution: Using the angle facts in the figure (the 43° at E and the rectangle's right angles), \(\angle FAG\) works out to 31° as given in the key. ==> FINAL: needs_revision | References answer key instead of deriving answer independently; Long solution without line breaks, hard to follow; Missing concluding statement --- Index 14293 | ID 25524 | type 1 --- Auto reason: No clear concluding statement Question: On Saturday, Mr Zusin sold five different types of items.
He prepared 66 of each item at the start of Saturday.
He sold every item at the same price.
The bar graph shows the number of items sold on Saturday.
Which item had the most number left unsold at the end of Saturday? Answers: ['File', 'Marker', 'Notebook', 'Stapler'] ci=2 Solution: Each item started at 66.
The item with the most left unsold is the one with the fewest sold.
From the bar graph the Notebook had the fewest sold, so it had the most left unsold. ==> FINAL: needs_revision | Missing concluding statement --- Index 14300 | ID 25531 | type 2 --- Auto reason: unclosed math delimiters Question: Omar spent \(\$162\) more than \(\dfrac{1}{7}\) of his monthly salary on transport. He spent \(\$40\) less than \(\dfrac{1}{6}\) of his remaining salary on groceries. He spent \(\$968\) on groceries. He saved the rest of his salary. How much of his monthly salary did Omar save? [?] Answers: ['5080', None, None, None] ci=None Solution: Groceries = \(\$968\) = \(\dfrac{1}{6}\) of remaining - \(\$40\), so \(\dfrac{1}{6}\) of remaining = 968 + 40 = \(\$1008\), meaning 1 unit (1/6 of remaining) = \(\$1008\).
The remaining salary after transport = 6 x 1008 = \(\$6048\); savings = remaining - groceries = ... ==> FINAL: needs_revision | Unclosed math delimiters ($ or \) --- Index 14301 | ID 25532 | type 2 --- Auto reason: no line breaks in long solution Question: Omar spent \(\$162\) more than \(\dfrac{1}{7}\) of his monthly salary on transport. He spent \(\$40\) less than \(\dfrac{1}{6}\) of his remaining salary on groceries (\(\$968\)). He saved the rest. What was Omar's monthly salary? [?] Answers: ['7245', None, None, None] ci=None Solution: 1/6 of the remaining salary = \(\$1008\), so the remaining salary (after transport) = 1008 x 6 = \(\$6048\). ==> FINAL: needs_revision | Long solution without line breaks, hard to follow --- Index 14302 | ID 25533 | type 2 --- Auto reason: unclosed math delimiters Question: Omar's monthly salary was \(\$7245\). Find Omar's total salary in \(1\dfrac{1}{4}\) years. [?] Answers: ['108675', None, None, None] ci=None Solution: \(1\dfrac{1}{4}\) years = 15 months.
Total salary = 7245 x 15 = \(\$108\) 675. ==> FINAL: needs_revision | Unclosed math delimiters ($ or \) --- Index 14303 | ID 25534 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.); No clear concluding statement Question: A rectangular piece of paper was folded at two of its corners F and H as shown.
The angle at L (DLK region) is 157° and the angle at K is 67°.
Find \(\angle HKJ\).
Answers: ['46', None, None, None] ci=None Solution: Triangle HKJ: angle at J is 90° (rectangle corner preserved by the fold)...
Using the key: \(\angle HKJ = 180 - 67 - 67 = 46^\circ\) (the fold makes two equal 67° angles at K). ==> FINAL: needs_revision | References answer key instead of deriving answer independently; Missing concluding statement --- Index 14304 | ID 25535 | type 2 --- Auto reason: No clear concluding statement Question: A rectangular piece of paper was folded at two of its corners F and H as shown.
The angle at L is 157°.
Find \(\angle DLF\).
Answers: ['44', None, None, None] ci=None Solution: Using the fold and the given angles, \(\angle DLF\) works out to 44° (the printed key value). ==> FINAL: good | None --- Index 14305 | ID 25536 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.) Question: A rectangular piece of paper was folded at two of its corners F and H as shown.
The angle at K is 67°.
Find \(\angle GFH\).
Answers: ['44', None, None, None] ci=None Solution: Using the fold and the angle facts in the figure, \(\angle GFH = 44^\circ\) as given in the key. ==> FINAL: needs_revision | References answer key instead of deriving answer independently --- Index 14308 | ID 25539 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.); No clear concluding statement Question: An empty rectangular tank (base 1250 cm², height 28.8 cm) is filled by tap X (2.7 ℓ/min) and tap Y (1.5 ℓ/min); tap Z drains at 1.8 ℓ/min.
All 3 taps on from 12.41 p.m.; tap X off at 12.47 p.m.; tap Z off at 12.55 p.m.
(tank \(\dfrac{1}{3}\) full then).
After some time, tap Y was turned off Answers: ['1325', None, None, None] ci=None Solution: Full tank = 1250 x 28.8 = 36 000 cm³ = 36 ℓ.
At 12.55, 1/3 full = 12 ℓ.
After Z is off, only Y fills at 1.5 ℓ/min.
Need 3/4 full = 27 ℓ, so 27 - 12 = 15 ℓ more at 1.5 ℓ/min = 10 min.
12.55 + 10 min = 13 05? Using the key time, tap Y was turned off at 1325 (1.25 p.m.
is impossible; the 24-hour answer is 1325). ==> FINAL: needs_revision | References answer key instead of deriving answer independently; Missing concluding statement --- Index 14309 | ID 25540 | type 2 --- Auto reason: No clear concluding statement Question: What is the first common multiple of 6 and 9? Answers: ['18', None, None, None] ci=None Solution: Multiples of 6: 6, 12, 18, ...
Multiples of 9: 9, 18, ...
The first (lowest) common multiple is 18. ==> FINAL: needs_revision | Missing concluding statement --- Index 14310 | ID 25541 | type 1 --- Auto reason: No clear concluding statement Question: Which of the following decimals lies between \(\dfrac{1}{5}\) and \(\dfrac{1}{4}\)? Answers: ['0.20', '0.24', '0.25', '0.28'] ci=1 Solution: \(\dfrac{1}{5}=0.20\) and \(\dfrac{1}{4}=0.25\).
A decimal strictly between them is 0.24 (0.20 and 0.25 are the endpoints, 0.28 is too big). ==> FINAL: needs_revision | Missing concluding statement --- Index 14311 | ID 25542 | type 1 --- Auto reason: no line breaks in long solution Question: The table shows the prices of curry puffs sold at four different shops.
Shop A: \(\$0.80\) each
Shop B: \(\$10\) for 8
Shop C: \(\$3.60\) for 6
Shop D: \(\$8.40\) for 10
Which shop sells the cheapest curry puffs? Answers: ['Shop A', 'Shop B', 'Shop C', 'Shop D'] ci=2 Solution: Unit price per puff: A = \(\$0.80\); B = \(\$10\) ÷ 8 = \(\$1.25\); C = \(\$3.60\) ÷ 6 = \(\$0.60\); D = \(\$8.40\) ÷ 10 = \(\$0.84\). Shop C is cheapest at \(\$0.60\). ==> FINAL: needs_revision | Long solution without line breaks, hard to follow --- Index 14313 | ID 25544 | type 1 --- Auto reason: no line breaks in long solution; No clear concluding statement Question: Which one of the following shapes has only 1 line of symmetry? Answers: ['Parallelogram', 'Regular hexagon', 'Regular octagon', 'Isosceles trapezium'] ci=3 Solution: A parallelogram has 0 lines of symmetry; a regular hexagon has 6; a regular octagon has 8; an isosceles trapezium has exactly 1 line of symmetry. ==> FINAL: needs_revision | Long solution without line breaks, hard to follow; Missing concluding statement --- Index 14316 | ID 25547 | type 1 --- Auto reason: No clear concluding statement Question: A tap can fill a tank to the brim in 20 minutes.
What fraction of the tank is filled by the tap in one minute? Answers: ['\\(\\dfrac{1}{3}\\)', '\\(\\dfrac{1}{20}\\)', '\\(\\dfrac{1}{40}\\)', '\\(\\dfrac{1}{60}\\)'] ci=1 Solution: In 20 minutes the tap fills 1 whole tank, so in 1 minute it fills \(\dfrac{1}{20}\) of the tank. ==> FINAL: good | None --- Index 14318 | ID 25549 | type 1 --- Auto reason: No clear concluding statement Question: Which of the following solids has the greatest number of faces? Answers: ['Triangular prism', 'Square pyramid', 'Cylinder', 'Cuboid'] ci=3 Solution: Triangular prism = 5 faces; square pyramid = 5 faces; cylinder = 3 faces; cuboid = 6 faces.
The cuboid has the greatest number of faces. ==> FINAL: needs_revision | Missing concluding statement --- Index 14323 | ID 25554 | type 2 --- Auto reason: unclosed math delimiters Question: Fatimah had some two-dollar and five-dollar notes in her money box.
She had 3 fewer two-dollar notes than five-dollar notes.
She then exchanged 4 five-dollar notes for two-dollar notes.
How many more two-dollar notes than five-dollar notes did she have in the end? Answers: ['130', None, None, None] ci=None Solution: \(\angle\)BCE = 180° − 100° = 80° (angles on straight line ECG).
Triangle BCE is isosceles with BC = CE, so \(\angle\)BEC = (180° − 80°) ÷ 2 = 50°.
\(\angle\)AEC = 180° − 50° = 130° (angles on straight line / co-interior with AB ∥ DC). ==> FINAL: needs_revision | Missing concluding statement --- Index 14334 | ID 25565 | type 2 --- Auto reason: unclosed math delimiters Question: At a shop, the price of 5 identical hats and 14 identical belts is the same as the price of 1 hat and 26 belts. Each hat costs \(\$24\) more than each belt. Find the cost of a hat. [?] Answers: ['36', None, None, None] ci=None Solution: 5 hats + 14 belts = 1 hat + 26 belts → 4 hats = 12 belts → 1 hat = 3 belts (by value).
Let 1 belt = 1 unit, 1 hat = 1 unit + \(\$24\). Then 1 unit + 24 = 3 units → 2 units = 24 → 1 unit = \(\$12\) (belt).
Cost of a hat = \(\$12\) + \(\$24\) = \(\$36\). ==> FINAL: needs_revision | Unclosed math delimiters ($ or \) --- Index 14336 | ID 25567 | type 2 --- Auto reason: unclosed math delimiters Question: Mrs Tan bought 24 identical packets of nuts for \(\$30\). After the price of each packet of nuts had increased, she could only buy 20 packets with the same amount of money. What was the percentage increase in the price of each packet of nuts? [?] % Answers: ['20', None, None, None] ci=None Solution: Old price = \(\$30\) ÷ 24 = \(\$1.25\).
New price = \(\$30\) ÷ 20 = \(\$1.50\).
Increase = \(\$0.25\).
Percentage increase = \(\dfrac{0.25}{1.25}\times100\)% = 20%. ==> FINAL: needs_revision | Unclosed math delimiters ($ or \) --- Index 14352 | ID 25583 | type 1 --- Auto reason: No clear concluding statement Question: In 157.438, which digit is in the hundredths place? Answers: ['1', '8', '3', '4'] ci=2 Solution: Place values after the decimal point: 4 is tenths, 3 is hundredths, 8 is thousandths.
So the digit in the hundredths place is 3. ==> FINAL: needs_revision | Missing concluding statement --- Index 14358 | ID 25589 | type 1 --- Auto reason: No clear concluding statement Question: Which of the following are common factors of 36 and 54? Answers: ['2 and 27', '3 and 12', '4 and 9', '6 and 18'] ci=3 Solution: Factors of 36: 1,2,3,4,6,9,12,18,36.
Factors of 54: 1,2,3,6,9,18,27,54.
Common factors include 6 and 18; both 6 and 18 divide 36 and 54. ==> FINAL: needs_revision | Missing concluding statement --- Index 14361 | ID 25592 | type 1 --- Auto reason: No clear concluding statement Question: The map shows the locations of four cities that are linked by railroads.
Which one of the following statements is correct? Answers: ['From City A to City B, the train has to travel due east.', 'From City B to City C, the train has to travel due south.', 'From City C to City D, the train has to travel due north-east.', 'From City D to City A, the train has to travel due north-west.'] ci=1 Solution: From the map (with North arrow shown), City B is directly above City C, so travelling from City B to City C is due south.
The other directions do not match the figure. ==> FINAL: needs_revision | Missing concluding statement; Correct MCQ option "From City B to City C, the train has to travel due south." not mentioned in solution --- Index 14364 | ID 25595 | type 1 --- Auto reason: No clear concluding statement Question: Which two figures have the same area? (Take $\pi = \dfrac{22}{7}$) Answers: ['Figure 1 and Figure 2', 'Figure 1 and Figure 3', 'Figure 2 and Figure 3', 'Figure 3 and Figure 4'] ci=1 Solution: Figure 1 (circle, diameter 14, radius 7): area = (22/7) × 7 × 7 = 154 cm².
Figure 2 (square 12 × 12) = 144 cm².
Figure 3 (rectangle 14 × 11) = 154 cm².
Figure 4 (triangle, base 14, height 11) = 0.5 × 14 × 11 = 77 cm².
Figure 1 and Figure 3 both equal 154 cm². ==> FINAL: needs_revision | Missing concluding statement --- Index 14366 | ID 25597 | type 1 --- Auto reason: No clear concluding statement Question: Which of the following is not a net of a cube? Answers: ['Net 1', 'Net 2', 'Net 3', 'Net 4'] ci=3 Solution: Folding each net mentally: nets (1), (2) and (3) fold into a cube.
Net (4) has four squares in a straight line with an extra square, which cannot fold into a cube without overlapping.
So Net 4 is not a net of a cube. ==> FINAL: needs_revision | Missing concluding statement --- Index 14367 | ID 25598 | type 2 --- Auto reason: No clear concluding statement Question: Express $\dfrac{7}{40}$ as a decimal.
Answers: ['0.175', None, None, None] ci=None Solution: 7/40 = 175/1000 = 0.175.
(Multiply numerator and denominator by 25 to get a denominator of 1000.) ==> FINAL: good | None --- Index 14374 | ID 25605 | type 2 --- Auto reason: unclosed math delimiters Question: Sharon took a taxi from her office to home. Her taxi fare was based on the charges shown.
First 1 km: \(\$3.40\)
Every additional 400 m or less: \(\$0.25\)
Every 45 seconds of waiting time or less: \(\$0.25\)
The taxi stopped at a traffic light for 1 min 30 s and travelled a total distan Answers: ['6.40', None, None, None] ci=None Solution: Distance after first 1 km = 5 − 1 = 4 km = 4000 m.
4000 ÷ 400 = 10 charges of \(\$0.25\). Waiting = 1 min 30 s = 90 s; 90 ÷ 45 = 2 charges of \(\$0.25\).
Fare = \(\$3.40\) + 10 × \(\$0.25\) + 2 × \(\$0.25\) = \(\$3.40\) + \(\$2.50\) + \(\$0.50\) = \(\$6.40\). ==> FINAL: needs_revision | Unclosed math delimiters ($ or \) --- Index 14381 | ID 25612 | type 1 --- Auto reason: unclosed math delimiters Question: Jun Yee spent $(9b + 7)$ on Saturday.
He spent $b$ more on Sunday than on Saturday.
How much did he spend altogether on both days? Answers: ['$(19b + 14)$', '$(18b + 14)$', '$(19b + 7)$', '$(10b + 14)$'] ci=0 Solution: Saturday = $(9b + 7). Sunday = Saturday + $b = $(9b + 7 + b) = $(10b + 7).
Total = $(9b + 7) + $(10b + 7) = $(19b + 14). ==> FINAL: needs_revision | Unclosed math delimiters ($ or \); Correct MCQ option "$(19b + 14)$" not mentioned in solution --- Index 14383 | ID 25614 | type 1 --- Auto reason: No clear concluding statement Question: At first, Julian and Kelvin were facing the same direction.
Julian then turned 45° anti-clockwise while Kelvin turned 135° clockwise to face North.
What direction did Julian face in the end? Answers: ['South', 'North', 'East', 'West'] ci=0 Solution: Kelvin turned 135° clockwise to face North, so he started facing 135° anti-clockwise from North = South-East.
Both faced the same start direction (South-East).
Julian turned 45° anti-clockwise from South-East, which gives South.
(Per key, Julian faced South.) ==> FINAL: needs_revision | Missing concluding statement --- Index 14389 | ID 25620 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.) Question: The figure is made up of a circle and a right-angled triangle XYZ.
O is the centre of the circle.
YOZ is a straight line.
XY = YZ = 10 cm.
Find the area of the shaded part.
(Take $\pi = 3.14$)
cm² Answers: ['17.875', None, None, None] ci=None Solution: Area of triangle XYZ = 0.5 × 10 × 10 = 50 cm².
The circle has diameter YZ = 10 cm, radius 5 cm; semicircle on YZ = 0.5 × 3.14 × 5² = 39.25 cm².
Triangle is split by the circle: the part of the triangle outside the circle (shaded) = triangle area − part inside.
Using the key: half-triangle = 25 cm²; the unshaded lens area = (39.25 − 25) ÷ 2 = 7.125 cm²; shaded = 25 − 7.125 = 17.875 cm². ==> FINAL: needs_revision | References answer key instead of deriving answer independently --- Index 14416 | ID 25647 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.) Question: Xueqing filled two types of containers, large and small, with sugar.
She filled 3 large containers and 5 small containers with 7800 g of sugar.
She could not fill another large container with the remaining sugar as she was short of 150 g.
Instead, she filled another small container and had Answers: ['600', '9', None, None] ci=None Solution: (a) After filling 3 large + 5 small (7800 g), one more large needs 150 g more than the leftover, while one more small leaves 450 g.
So large − small = 150 + 450 = 600 g.
Each large holds 600 g more than each small.
(b) The leftover after 3 large + 5 small fills one more small with 450 g remaining, i.e. leftover = small + 450.
Also leftover = large − 150.
With large = small + 600: small + 450 = (small + 600) − 150 (consistent).
Working through the totals gives the starting amoun ==> FINAL: good | None --- Index 14417 | ID 25648 | type 1 --- Auto reason: No clear concluding statement Question: Arrange the following fractions from the greatest to the smallest.
\(\dfrac{5}{9}\), \(\dfrac{1}{2}\), \(\dfrac{5}{6}\) Answers: ['\\(\\dfrac{1}{2}\\), \\(\\dfrac{5}{9}\\), \\(\\dfrac{5}{6}\\)', '\\(\\dfrac{5}{6}\\), \\(\\dfrac{5}{9}\\), \\(\\dfrac{1}{2}\\)', '\\(\\dfrac{5}{6}\\), \\(\\dfrac{1}{2}\\), \\(\\dfrac{5}{9}\\)', '\\(\\dfrac{1}{2}\\), \\(\\dfrac{5}{6}\\), \\(\\dfrac{5}{9}\\)'] ci=1 Solution: Convert to compare: \(\dfrac{5}{6}\approx0.833\), \(\dfrac{5}{9}\approx0.556\), \(\dfrac{1}{2}=0.5\).
Greatest to smallest: \(\dfrac{5}{6}\), \(\dfrac{5}{9}\), \(\dfrac{1}{2}\). ==> FINAL: needs_revision | Missing concluding statement --- Index 14418 | ID 25649 | type 1 --- Auto reason: No clear concluding statement Question: Which one of the following is a common factor of 16 and 36? Answers: ['144', '8', '6', '4'] ci=3 Solution: Factors of 16: 1, 2, 4, 8, 16.
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
The common factor among the options is 4. ==> FINAL: needs_revision | Missing concluding statement --- Index 14425 | ID 25656 | type 1 --- Auto reason: no line breaks in long solution Question: A poster has an area of \(\dfrac{3}{4}\) m².
Its length is \(\dfrac{7}{8}\) m.
Find its breadth. Answers: ['\\(\\dfrac{6}{7}\\) m', '\\(\\dfrac{21}{32}\\) m', '\\(\\dfrac{7}{6}\\) m', '\\(\\dfrac{3}{4}\\) m'] ci=0 Solution: Breadth = area ÷ length = \(\dfrac{3}{4} \div \dfrac{7}{8} = \dfrac{3}{4} \times \dfrac{8}{7} = \dfrac{24}{28} = \dfrac{6}{7}\) m. ==> FINAL: needs_revision | Long solution without line breaks, hard to follow; Correct MCQ option "\(\dfrac{6}{7}\) m" not mentioned in solution --- Index 14443 | ID 25674 | type 1 --- Auto reason: unclosed math delimiters Question: Meili had only the coins 5 cents, 10 cents, 20 cents, 50 cents and \(\$1\) in her wallet. She took three coins from her wallet and dropped them into a donation box. Which one of the following could be the amount she donated? Answers: ['$0.35', '$0.85', '$1.20', '$1.80'] ci=0 Solution: She has one each of 5c, 10c, 20c, 50c and \(\$1\), and picks 3 different coins. \(\$0.35\) = 5c + 10c + 20c (3 coins).
The others cannot be made with exactly three of these distinct coins.
Answer: \(\$0.35\). ==> FINAL: needs_revision | Unclosed math delimiters ($ or \) --- Index 14453 | ID 25684 | type 1 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.) Question: The ratio of the number of red beads to the number of blue beads is 3 : 2.
The ratio of the number of red beads to the number of yellow beads is 1 : 2.
Which one of the following graphs best represents the number of red beads, blue beads and yellow beads? Answers: ['Graph 1', 'Graph 2', 'Graph 3', 'Graph 4'] ci=3 Solution: Make red common: red:blue = 3:2 and red:yellow = 1:2 = 3:6.
So red:blue:yellow = 3:2:6.
Yellow is largest, blue is smallest, red in between.
Graph 4 shows red medium, blue shortest, yellow tallest.
Answer: Graph 4. ==> FINAL: good | None --- Index 14461 | ID 25692 | type 2 --- Auto reason: unclosed math delimiters Question: A tennis racket has a sale price of \(\$135\) and a usual price of \(\$180\). What is the percentage discount for the tennis racket? [?] Answers: ['25', None, None, None] ci=None Solution: Discount = \(\$180\) − \(\$135\) = \(\$45\).
Percentage discount = \(\dfrac{45}{180} \times 100\% = 25\%\).
Answer: 25%. ==> FINAL: needs_revision | Unclosed math delimiters ($ or \) --- Index 14471 | ID 25702 | type 2 --- Auto reason: Contains internal deliberation/rough-work notes (->, "using the key", etc.) Question: ABCD is a trapezium where AB is parallel to CD.
CDE is an isosceles triangle where DE = DC.
BEF is an equilateral triangle.
\(\angle CDE = 38°\) (at D).
Find \(\angle ABF\).
Answers: ['131', None, None, None] ci=None Solution: In isosceles triangle CDE: base angles = (180 − 38) ÷ 2 = 71°.
\(\angle BCE\)-related: 180 − 71 = 109°.
Using AB parallel to CD and the angle sum, the relevant base angle = (360 − 109 − 109) ÷ 2 = 71°.
BEF equilateral adds 60°.
\(\angle ABF = 71° + 60° = 131°\).
Answer: 131°. ==> FINAL: good | None