Qiyas GAT Practice Test.

Answer: B
Let the two groups be 5 parts and 7 parts, so the total is 12 parts. Since 12 parts represent 36 students, 1 part represents 36 ÷ 12 = 3 students. Group revision students = 7 × 3 = 21. The percentage is 21 ÷ 36 × 100 = 58.333...%, which is 58⅓%. Check: 58⅓% of 36 is 21, so the result matches the ratio. A likely mistake is to use 7 ÷ 5 instead of 7 ÷ 12; the base must be the whole group, not only the individual-revision group.

Answer: B
First find the reduced price: 15% of 240 = 36, so the price becomes 240 − 36 = 204. Then increase this by 10%: 10% of 204 = 20.4, so the final price is 204 + 20.4 = 224.4. Compared with the original 240, the decrease is 240 − 224.4 = 15.6. The percentage decrease is 15.6 ÷ 240 × 100 = 6.5%. Check using multipliers: 240 × 0.85 × 1.10 = 224.4, confirming the same result. A common mistake is to subtract 15% and add 10% directly to get a 5% decrease, but the 10% increase is applied to the reduced price, not the original price.

Answer: C
The original ratio has 4 + 1 = 5 parts. Since the mixture is 25 liters, each part is 25 ÷ 5 = 5 liters. Water = 4 × 5 = 20 liters, and syrup = 1 × 5 = 5 liters. Syrup does not change when only water is added. For the new ratio 6:1, 5 liters of syrup must match 6 × 5 = 30 liters of water. The mixture already has 20 liters of water, so the amount of water to add is 30 − 20 = 10 liters. Check: after adding 10 liters, water:syrup = 30:5 = 6:1. A likely mistake is to make the total mixture 6 + 1 = 7 parts without keeping the syrup amount fixed.

Answer: C
Let the tens digit be a and the ones digit be b. The original number is 10a + b, and the reversed number is 10b + a. The digit sum gives a + b = 11. The reversed number is 27 less than the original, so 10b + a = 10a + b − 27. Rearranging gives 9b − 9a = −27, so b − a = −3, or a − b = 3. Now solve a + b = 11 and a − b = 3. Adding the equations gives 2a = 14, so a = 7. Then b = 4. The original number is 74. Check: 7 + 4 = 11, and the reverse 47 is 27 less than 74. A likely mistake is to give 47, which is the reversed number, not the original.

Answer: C
Total forms = 180. Online forms = 60% of 180 = 108. Paper forms = 180 − 108 = 72. Online forms needing correction = 1/6 of 108 = 18, so complete online forms = 108 − 18 = 90. Paper forms needing correction = 1/4 of 72 = 18, so complete paper forms = 72 − 18 = 54. Total complete forms = 90 + 54 = 144. Each box holds 18 complete forms, so boxes filled = 144 ÷ 18 = 8. Check: 8 boxes hold 8 × 18 = 144 complete forms. A likely mistake is to subtract only one correction group or to apply 25% to all 180 forms instead of only the paper forms.

Answer: B
Compare consecutive ratios. The first number doubles each time: 2, 4, 8, 16, so the next first number is 32. The second number triples each time: 3, 9, 27, 81, so the next second number is 243. Therefore, the next ratio is 32:243. Check: from 16:81 to 32:243, the first term is multiplied by 2 and the second by 3, matching the earlier pattern. A likely mistake is to multiply both terms by the same number, but the displayed pattern uses different multipliers for the two positions.

Answer: C
Let the total number of students be T. Mathematics students are 40% of T, so the remaining students are 60% of T. Physics students are 30% of the remaining 60%, which is 0.30 × 0.60T = 0.18T. Students choosing neither are the remaining part after mathematics and physics: T − 0.40T − 0.18T = 0.42T. We are told 0.42T = 84, so T = 84 ÷ 0.42 = 200. Check: 40% of 200 = 80; remaining = 120; 30% of 120 = 36; neither = 200 − 80 − 36 = 84. A likely mistake is to take 30% of the whole group instead of 30% of the remaining students.

Answer: C
Let the original quantity be x. After a 20% increase, it becomes 1.20x. Then a 20% decrease leaves 80% of that amount: 0.80 × 1.20x = 0.96x. The final value is 288, so 0.96x = 288. Thus x = 288 ÷ 0.96 = 300. Check: 20% increase of 300 gives 360; 20% decrease of 360 is 72; 360 − 72 = 288. A likely mistake is to think a 20% increase and a 20% decrease cancel exactly. They do not, because the second percentage is taken from the increased value.

Answer: A
First identify the relationship in the first pair. An embargo is a restriction placed on trade. Now apply the same relationship: a quarantine is a restriction placed on contact or movement between people. The exact matching answer is “contact.” A likely mistake is to choose “illness,” because quarantine may be used during illness, but illness is the reason for the restriction, not the activity being restricted.

Answer: C
We need one combined ratio for Amal, Badr, and Camil. Amal:Badr = 3:4. Badr:Camil = 2:5. Make Badr’s parts match. In the first ratio, Badr has 4 parts. In the second, Badr has 2 parts, so multiply the second ratio by 2: Badr:Camil = 4:10. Therefore Amal:Badr:Camil = 3:4:10. Camil receives 10 parts and Amal receives 3 parts, so the difference is 7 parts. This difference equals 140 riyals, so 1 part = 140 ÷ 7 = 20 riyals. Total parts = 3 + 4 + 10 = 17, so the total budget is 17 × 20 = 340 riyals. Check: Amal = 60, Badr = 80, Camil = 200; Camil − Amal = 140. A likely mistake is to join the ratios as 3:4:5 without making Badr’s value consistent.

Answer: C
Machine A produces 3/5 = 60% of all parts. Machine B produces the remaining 40%. The overall defect rate is a weighted average: from Machine A, 60% × 4% = 0.60 × 0.04 = 0.024 = 2.4% of all parts. From Machine B, 40% × 7% = 0.40 × 0.07 = 0.028 = 2.8% of all parts. Total defect rate = 2.4% + 2.8% = 5.2%. If 78 parts are 5.2% of the total, then total parts = 78 ÷ 0.052 = 1500. Check: Machine A makes 900 parts, with 36 defects; Machine B makes 600 parts, with 42 defects; 36 + 42 = 78. A likely mistake is to average 4% and 7% as 5.5%, ignoring that the machines produce different amounts.

Answer: B
Let the initial amounts be A = 7x and B = 5x. Transferring 20% of Tank A means moving 0.20 × 7x = 1.4x from A to B. After the transfer, Tank A has 7x − 1.4x = 5.6x, and Tank B has 5x + 1.4x = 6.4x. Then 10 liters are added to Tank A, so the final amounts are A = 5.6x + 10 and B = 6.4x. The final ratio is 3:2, so (5.6x + 10) ÷ 6.4x = 3 ÷ 2. Cross-multiply: 2(5.6x + 10) = 3(6.4x), so 11.2x + 20 = 19.2x. Thus 20 = 8x, so x = 2.5. The initial total is 7x + 5x = 12x = 12 × 2.5 = 30 liters. Check: initially A = 17.5 and B = 12.5. Transfer 3.5 liters: A = 14, B = 16. Add 10 to A: A = 24, B = 16, giving 24:16 = 3:2. A likely mistake is to calculate 20% of the total water instead of 20% of Tank A only.

Answer: B
Let a be the price of one adult ticket and s be the price of one student ticket. The information gives 3a + 5s = 205 and 2a + 3s = 130. To eliminate a, multiply the first equation by 2: 6a + 10s = 410. Multiply the second equation by 3: 6a + 9s = 390. Subtracting gives s = 20. Substitute into 2a + 3s = 130: 2a + 60 = 130, so 2a = 70 and a = 35. The required cost is 4a + 2s = 4(35) + 2(20) = 140 + 40 = 180. Check: 3(35) + 5(20) = 205 and 2(35) + 3(20) = 130, so the prices fit. A likely mistake is to subtract the original equations directly and get a + 2s = 75, which is not enough to find the final cost.

Answer: C
During the first 4 minutes, only draining happens, so the amount lost is 12 × 4 = 48 liters. The container then has 180 − 48 = 132 liters. After that, water leaves at 12 liters per minute and enters at 5 liters per minute, so the net loss is 12 − 5 = 7 liters per minute. To go from 132 liters to 70 liters, the container must lose 132 − 70 = 62 liters. At 7 liters per minute, this takes 62/7 = 8 6/7 minutes after the pump starts. Total time from the start is 4 + 8 6/7 = 12 6/7 minutes. Check: over 12 6/7 minutes the drain removes 154 2/7 liters, while the pump adds 44 2/7 liters, for a net loss of 110 liters; 180 − 110 = 70. A likely mistake is to use the 7-liter net rate from the beginning, forgetting that the pump starts only after 4 minutes.

Answer: B
Let the smallest even integer be x. Then the three consecutive even integers are x, x + 2, and x + 4. Their sum plus twice the smallest is x + (x + 2) + (x + 4) + 2x = 5x + 6. Set this equal to 206: 5x + 6 = 206. Then 5x = 200, so x = 40. The middle integer is x + 2 = 42. Check: the integers are 40, 42, and 44; their sum is 126, and twice the smallest is 80; 126 + 80 = 206. A likely mistake is to answer 40 because it is the first value found, but the question asks for the middle integer.

Answer: C
Let P be the original tag price. A 15% discount means the customer first pays 85% of the tag price, so the discounted price is 0.85P. Then the coupon subtracts SAR 18, giving 0.85P − 18 = 152. Add 18 to both sides: 0.85P = 170. Divide by 0.85: P = 170 ÷ 0.85 = 200. Check: 15% of 200 is 30, so the price after discount is 170; after the SAR 18 coupon, the payment is 152. A likely mistake is to subtract 15% from 152 instead of applying the discount to the original unknown price.

Answer: B
Work with rates in tanks per hour. Pipe A fills 1/6 of the tank per hour, and Pipe B fills 1/9 of the tank per hour. The drain removes 1/18 of the tank per hour. The net filling rate is 1/6 + 1/9 − 1/18. Using denominator 18, this is 3/18 + 2/18 − 1/18 = 4/18 = 2/9 of a tank per hour. Time equals total work divided by rate: 1 ÷ (2/9) = 9/2 = 4.5 hours. Check: in 4.5 hours, the net amount filled is (2/9)(4.5) = 1 full tank. A likely mistake is to add the drain rate instead of subtracting it.

Answer: B
The total cost of the almonds is 12 × 30 = 360. The total cost of the walnuts is 45x. The total mass is 12 + x, and the average cost is 36, so (360 + 45x)/(12 + x) = 36. Multiply both sides by 12 + x: 360 + 45x = 36(12 + x) = 432 + 36x. Subtract 36x from both sides: 360 + 9x = 432. Subtract 360: 9x = 72, so x = 8. Check: total cost is 360 + 45(8) = 720 and total mass is 20 kg; 720 ÷ 20 = 36. A likely mistake is to average 30 and 45 directly to get 37.5, ignoring the different masses.

Answer: C
Let the width be w meters. Then the length is w + 5 meters. The area is w(w + 5) = 84, so w^2 + 5w − 84 = 0. Factor: w^2 + 5w − 84 = (w + 12)(w − 7). Thus w = 7 or w = −12. A negative width is impossible, so w = 7. The length is 12. The perimeter is 2(length + width) = 2(12 + 7) = 38 meters. Check: 7 × 12 = 84, and the length is 5 more than the width. A likely mistake is to stop at the dimensions and answer 12 or 7 instead of calculating the perimeter.

Answer: C
We need one consistent ratio for all three shares. Amal:Bader = 3:4. Also Bader:Careem = 2:5. To make Bader's number match, rewrite Bader:Careem by multiplying both parts by 2, giving 4:10. Now Amal:Bader:Careem = 3:4:10. The total number of parts is 3 + 4 + 10 = 17. Since the total is SAR 510, each part is 510 ÷ 17 = 30. Careem receives 10 parts, so Careem's share is 10 × 30 = 300. Check: Amal gets 90 and Bader gets 120, so Amal:Bader = 90:120 = 3:4; Bader:Careem = 120:300 = 2:5. A likely mistake is to combine the ratios as 3:4:5, but that would not preserve Bader:Careem = 2:5.

Answer: B
Let f be the fixed starting fee and r be the charge per kilometer. The two fares give f + 8r = 34 and f + 13r = 51.50. Subtract the first equation from the second: 5r = 17.50, so r = 3.50. Substitute into f + 8r = 34: f + 28 = 34, so f = 6. For a 20 km trip, the distance charge before discount is 20 × 3.50 = 70. A 10% discount on the distance charge reduces it by 7, leaving 63. Add the starting fee: 63 + 6 = 69. Check: without promotion, the 20 km fare would be 6 + 70 = 76, and only the distance part is discounted. A likely mistake is to apply the 10% discount to the entire SAR 76, which would not follow the condition in the question.

Answer: B
The two parallel sides differ by 18 − 10 = 8 m. Because the left endpoints are joined by the perpendicular side, the horizontal offset occurs on the slanted side. The perpendicular side equals the height, so it is 6 m. The slanted side forms a right triangle with horizontal leg 8 m and vertical leg 6 m. Its length is √(8² + 6²) = √(64 + 36) = √100 = 10 m. The perimeter is 18 + 10 + 6 + 10 = 44 m. Check: all four outer sides have been counted exactly once. A likely mistake is to add only the two bases and the height, forgetting the slanted side.

Answer: A
A flight plan guides a pilot’s intended route and actions during a flight. The matching idea for a teacher is not simply a book or classroom, but a planned guide for teaching. A teacher follows a lesson plan. “Textbook” is tempting because teachers may use it, but it is a resource, not the same kind of planned course of action. “Exam” measures learning rather than guiding instruction.

Answer: B
First find the area of the rectangle: 16 × 10 = 160 cm². The semicircle has diameter 10 cm, so its radius is 5 cm. The area of the full circle would be 3.14 × 5² = 3.14 × 25 = 78.5 cm². The semicircle area is half of that: 78.5 ÷ 2 = 39.25 cm². Remaining area = 160 − 39.25 = 120.75 cm². Check: the answer must be less than 160 cm² but not by a full circle’s area. A likely mistake is using 10 cm as the radius instead of the diameter.

Answer: C
The interior angles of a quadrilateral sum to 360°. So (3x + 10) + (2x + 20) + 90 + (x + 60) = 360. Combine like terms: 6x + 180 = 360. Then 6x = 180, so x = 30. Now compute the angles: 3x + 10 = 100°, 2x + 20 = 80°, the given angle is 90°, and x + 60 = 90°. Two angles are 90°, but neither is the largest. The largest angle is 100°. Check: 100 + 80 + 90 + 90 = 360, so the values are consistent. A likely mistake is to stop at x = 30 and choose 30° instead of finding the requested angle.

Answer: A
Convert the drawing dimensions to real dimensions. The rectangle is 8 × 4 = 32 m by 5 × 4 = 20 m, so its real area is 32 × 20 = 640 m². The fountain’s drawn radius is 1 cm, so its real radius is 1 × 4 = 4 m. Its area is 3.14 × 4² = 3.14 × 16 = 50.24 m². The remaining area is 640 − 50.24 = 589.76 m². Check: scaling lengths by 4 makes areas scale by 16, not by 4. A likely mistake is to subtract the drawing-circle area directly from the real courtyard area.

Answer: B
The cube displaces a volume of water equal to its own volume. The cube’s volume is 30³ = 30 × 30 × 30 = 27,000 cm³. The tank’s base area is 90 × 40 = 3,600 cm². Water-level rise = displaced volume ÷ base area = 27,000 ÷ 3,600 = 7.5 cm. Check: base area × rise = 3,600 × 7.5 = 27,000 cm³, matching the cube volume. A likely mistake is to divide by the tank perimeter instead of the base area.

Answer: B
The height-to-shadow ratio is the same for vertical objects at the same time of day. For the pole, the ratio is 3 ÷ 4.5 = 2/3. The building height is therefore (2/3) × 27 = 18 m. The total height up to the antenna top is (2/3) × 30.75 = 20.5 m. So the antenna height is 20.5 − 18 = 2.5 m. Check: the antenna adds 30.75 − 27 = 3.75 m of shadow, and (2/3) × 3.75 = 2.5 m. A likely mistake is to report the total height, 20.5 m, instead of the antenna height.

Answer: B
Find the length of each outer side. From (0,0) to (12,0) is 12 m. From (12,0) to (12,5) is 5 m. From (12,5) to (7,5) is 5 m. From (7,5) to (7,9) is 4 m. From (7,9) to (0,9) is 7 m. From (0,9) back to (0,0) is 9 m. The full perimeter is 12 + 5 + 5 + 4 + 7 + 9 = 42 m. Since a 2 m gate is not fenced, fence needed = 42 − 2 = 40 m. Check: the two sides around the inward corner are still part of the outer boundary. A likely mistake is to treat the shape as a 12 by 9 rectangle and ignore the notch.

Answer: C
Figure n has rows of n, n − 1, n − 2, and so on down to 1 square, all aligned on the left. Count only the outside boundary. The top edge contributes n unit sides, and the left edge contributes n unit sides. From the top-right corner down to the bottom-left, the staircase boundary consists of n vertical unit sides and n horizontal unit sides: for example, it goes down once, then alternates left and down until the final bottom horizontal side is included. Thus the total perimeter in unit sides is n + n + n + n = 4n. For Figure 7, that is 4 × 7 = 28 unit sides. Each unit side is enlarged to 3 cm, so the perimeter is 28 × 3 = 84 cm. Check with small figures: Figure 1 has perimeter 4 unit sides, and Figure 2 has perimeter 8 unit sides, matching 4n. A likely mistake is to count the number of squares instead of the outside boundary.

Answer: C
The diagonals bisect each other, so half of the 24 cm diagonal is 12 cm and half of the 10 cm diagonal is 5 cm. These half-diagonals meet at a right angle, forming a right triangle whose hypotenuse is one side of the rhombus. The side length is √(12² + 5²) = √(144 + 25) = √169 = 13 cm. A rhombus has four equal sides, so the perimeter is 4 × 13 = 52 cm. Check: the 5-12-13 right triangle fits the half-diagonals. A likely mistake is to add the diagonals, getting 34, but diagonals are not the outer sides.

Answer: B
The cylinder’s radius is half the diameter: 14 ÷ 2 = 7 cm. Its full volume is πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 1,540 cm³. It is filled to 3/5, so the water volume is (3/5) × 1,540 = 924 cm³. The rectangular container’s base area is 22 × 7 = 154 cm². Water height = volume ÷ base area = 924 ÷ 154 = 6 cm. Check: 154 × 6 = 924 cm³, the same water volume. A likely mistake is to use 14 cm as the radius, which would make the cylinder volume four times too large.

Answer: A
A scalpel is a specialized tool used by a surgeon. We need a pair in which the first item is a specialized tool used by the second person. “Chisel : sculptor” fits: a chisel is a tool a sculptor may use. “Canvas : painter” is tempting, but a canvas is mainly a surface or medium, not a tool like a scalpel. “Court : judge” gives a place and person, while “medicine : patient” gives treatment and receiver.

Answer: B
First find the five-day total: 240 + 180 + 210 + 270 + 300 = 1200 visitors. Wednesday and Thursday together had 270 + 300 = 570 visitors. The required percentage is 570 ÷ 1200 × 100 = 47.5%. A likely mistake is to divide 570 by only the two-day total, but the question asks for the percentage of the five-day total.

Answer: B
Use a weighted average because the groups have different sizes. Science total points: 48 × 76 = 3648. Humanities total points: 32 × 82 = 2624. Business total points: 20 × 70 = 1400. Total points = 3648 + 2624 + 1400 = 7672. Total students = 48 + 32 + 20 = 100. Overall average = 7672 ÷ 100 = 76.72. A common error is to average 76, 82, and 70 directly, which ignores the different group sizes.

Answer: C
Calculate revenue and cost separately. Notebook revenue = 250 × 18 = 4500; notebook cost = 250 × 11 = 2750. Calculator revenue: 70 calculators at 75 gives 70 × 75 = 5250, and 20 discounted calculators at 65 gives 20 × 65 = 1300, so calculator revenue = 6550. Calculator cost = 90 × 52 = 4680. Backpack revenue = 60 × 120 = 7200; backpack cost = 60 × 82 = 4920. Total revenue = 4500 + 6550 + 7200 = 18250. Total cost = 2750 + 4680 + 4920 = 12350. Profit = 18250 − 12350 = 5900 riyals. A likely mistake is to apply the calculator discount to all 90 calculators instead of only 20.

Answer: A
Compute the weekly totals. Week 1 total = 120 + 45 + 35 = 200 kg. Week 2 total = 135 + 50 + 40 = 225 kg. Week 3 total = 150 + 60 + 42 = 252 kg. Week 4 total = 140 + 80 + 38 = 258 kg, so Week 4 is the greatest. From Week 3 to Week 4, plastic increased by 20 kg, while paper decreased by 10 kg and metal decreased by 4 kg. The net change is +20 − 10 − 4 = +6 kg. So Week 4 exceeded Week 3 by 6 kg mainly because of the plastic increase. A likely mistake is to notice that paper is largest in most weeks and assume it drove the final increase, but paper actually fell in Week 4.

Answer: C
Look at the stated repeating increases: +3, +6, +9, +6. From Day 9, the next increase is the first in the cycle, +3, so Day 10 = 54 + 3 = 57. Day 11 then adds +6, giving 57 + 6 = 63. Day 12 then adds +9, giving 63 + 9 = 72. A likely mistake is to keep adding 6 because several differences are 6, but the full four-step cycle must be followed.

Answer: B
Find each day’s total and sandwich fraction. Sunday total = 180 + 120 + 60 = 360, so sandwich fraction = 120/360 = 1/3. Monday total = 150 + 160 + 70 = 380, so sandwich fraction = 160/380 = 8/19. Tuesday total = 170 + 140 + 90 = 400, so sandwich fraction = 140/400 = 7/20. Compare the values: 1/3 ≈ 0.333, 8/19 ≈ 0.421, and 7/20 = 0.35. The greatest is Monday, 8/19. A common error is to choose Tuesday because its total is largest, but the question asks for the fraction, not the total number of meals.

Answer: C
First find the total travel minutes for the original 80 students. Walkers: 40 × 12 = 480 minutes. Bus students: 30 × 28 = 840 minutes. Car students: 10 × 18 = 180 minutes. Original total = 480 + 840 + 180 = 1500 minutes. After adding 20 cyclists, there are 100 students with average 19.5 minutes, so the new total is 100 × 19.5 = 1950 minutes. Cyclists’ total minutes = 1950 − 1500 = 450. Cyclists’ average = 450 ÷ 20 = 22.5 minutes. A likely mistake is to average the four group averages directly, but the cyclist average is unknown and the group sizes matter.

Answer: B
The total number of students remains 24 because one left and one joined. Update the frequencies: score 60 becomes 2 students; score 90 becomes 1 student; all other frequencies stay the same. For 24 students, the median is the average of the 12th and 13th ordered scores. Cumulative positions after the change: 60 covers positions 1–2; 65 covers 3–6; 70 covers 7–11; 75 covers 12–17. Therefore both the 12th and 13th scores are 75, so the new median is 75. A likely mistake is to average 60 and 90 because those are the changed scores, but the median depends on the middle positions after ordering all scores.

Answer: B
Start with 1200 liters. From 8:00 to 10:00, 2 hours of adding 150 liters per hour gives 2 × 150 = 300 liters, so the tank reaches 1500 liters. From 10:00 to 11:00, 220 liters are used: 1500 − 220 = 1280. From 11:00 to 1:00, the net gain per hour is 90 − 30 = 60 liters. Over 2 hours, that is 2 × 60 = 120 liters, so the tank reaches 1400 liters. From 1:00 to 2:00, 160 liters are drained: 1400 − 160 = 1240 liters. A likely mistake is to add 90 liters per hour without subtracting the leak.

Answer: A
First convert last year’s percentages into amounts. Rent = 30% of 240,000 = 72,000. Salaries = 45% of 240,000 = 108,000. Supplies = 15% of 240,000 = 36,000. Events = 10% of 240,000 = 24,000. Apply this year’s changes: rent becomes 72,000 × 1.10 = 79,200; salaries stay 108,000; supplies become 36,000 × 0.80 = 28,800; events become 24,000 × 2 = 48,000. This year’s total = 79,200 + 108,000 + 28,800 + 48,000 = 264,000. Salary fraction = 108,000/264,000 = 108/264 = 9/22. A likely mistake is to keep salaries at 45%, but the total budget changed.

Answer: A
A preface comes before the main text of a book and introduces or prepares for it. An overture is an introductory musical piece that comes before a larger staged musical work. Therefore, the best parallel is “opera.” A common mistake is “chapter,” because a preface appears near chapters, but an overture is not a section of a chapter-like item; it introduces the larger performance.

Answer: A
Camouflage is used to prevent or reduce detection. Now find what insulation prevents or reduces. Insulation reduces heat loss or unwanted heat transfer. The closest answer is “heat loss.” A likely mistake is “warmth,” because insulation may keep something warm, but it does so by reducing heat loss; the relationship asks for what is prevented, not what is preserved.

Answer: A
An archive is a place where records are stored and preserved. A reservoir is a place where water is stored. So the answer must be “water.” A tempting error is “dam,” because dams may create reservoirs, but a dam is not what a reservoir stores. Another error is “river,” which may feed a reservoir but is not the stored contents in the same direct way.

Answer: A
Concise and wordy are opposites: concise means expressed in few words, while wordy means using too many words. Therefore, candid should be paired with its opposite. Candid means honest or direct, so the opposite is “evasive.” A likely mistake is “brief,” because brief is related to concise, not to candid. Another mistake is “frank,” which is a synonym of candid, not an opposite.

Answer: A
The passage says some people expected the craft to disappear, but that did not happen. The older net-makers inspected the machine’s output and repaired unusual tears the machine could not recognize. The final sentence states that the machine changed routine work while increasing the value of experienced judgment. So the machine and experienced net-makers are complementary: the machine handles routine production, while people provide judgment for difficult cases. A likely mistake is to say the machine replaced them completely, but the passage explicitly says the most respected workers still used their experience.

Answer: A
Each pair connects a creator or specialist with the kind of product or representation that person makes. A scribe produces or copies a manuscript; a composer creates a symphony; an architect creates a blueprint. A cartographer makes a map. “Compass” is tempting because cartographers may use one, but the pattern asks for the product, not the tool. “Voyage” is an activity, and “mountain” may appear on a map but is not produced by the cartographer.

Answer: A
A filter removes impurities from a liquid, air flow, or similar mixture. Apply the same relationship to an editor: an editor removes or corrects errors from writing. Therefore, the best answer is “errors.” “Writer” is tempting because an editor works with writers, but that is not the removed item. “Paper” is the medium, and “ink” is part of writing materials, not the flaw being removed.

Answer: A
The survey favored later hours, but it represented only university students. The gate records gave wider evidence about actual use and showed two important patterns: student use dropped after 10:30 p.m., while younger pupils used the library in the late afternoon. The manager’s decision combined both pieces of evidence: a small evening extension plus more late-afternoon staffing. Therefore, the best conclusion is that the manager did not rely on the survey alone because it missed some users and did not fully match actual attendance. A likely mistake is to say the survey was useless; the passage says it influenced the one-hour extension, but it was incomplete.

Answer: A
The passage presents a disagreement: shop owners dislike the short-term appearance, while the project leader argues for longer-term protection and environmental benefits. The council continues the project but asks for signs, showing it accepts the project’s purpose while recognizing public misunderstanding. The central idea is not merely that cafés lost a view or that signs are decorative. It is that an unpopular-looking restoration may have practical benefits that need to be explained. A likely mistake is to focus only on tourism, but tourism is just one of several benefits mentioned.

Answer: B
Let the number of novel boxes be n. Then the number of science boxes is 5 - n. Novel boxes hold 18 books each, and science boxes hold 12 books each, so the total is 18n + 12(5 - n). This equals 78. Simplify: 18n + 60 - 12n = 78, so 6n + 60 = 78. Subtract 60: 6n = 18. Divide by 6: n = 3. Check: 3 novel boxes hold 54 books, and 2 science boxes hold 24 books; 54 + 24 = 78. The friend’s closeness idea is not exact reasoning.

Answer: A
Both groups had the same tasks and found the same spelling errors, so the difference was not spelling ability. The key difference was the order: Group One handled small errors first and then treated the article as almost finished; Group Two examined the argument first and suggested more structural changes. Therefore, the teacher concluded that task order affects attention. A likely mistake is to claim Group Two was simply better, but the passage does not say the groups differed in ability.

Answer: A
The clock is broken, but people still use it as a meeting place. The fruit seller explains that its broken state keeps people nearby, talking and sometimes buying. The fresh paint and benches show that the market maintains the clock as a landmark, even without repairing its timekeeping function. Thus, the clock’s value is social and commercial rather than practical as a clock. A likely mistake is to infer neglect, but fresh paint and benches show attention, not neglect.

Answer: A
The quiet corridor reduced noise complaints, so the original rule partly worked. However, it created a new problem: students gathered at the entrance to finish conversations, causing delays and blocking the staircase. The discussion area gave students another place to talk away from exam rooms. The result confirms the reason: the corridor cleared while complaints stayed low. A likely mistake is to say the discussion area replaced the quiet rule; the passage says complaints remained low, meaning the quiet purpose continued.

Answer: A
The phrase is explained by the sentences that follow it. The app neatly places tasks on a calendar and sends reminders, but it cannot judge importance or difficulty. Students who follow the notifications may work in a tidy order without making wise priorities. Therefore, the phrase means the app supports organization better than decision-making. A likely mistake is to think the app is described as slow or ugly, but the review praises its clean design and fast reminders.

Answer: A
Mariam’s comparison shows that limited space or time forces careful selection. The club test supports her view: most members found the two-minute speech harder because they had to decide what to omit. So she would agree that a short speech can require more careful planning than a longer one. A likely mistake is to assume shorter always means easier; the passage directly challenges that assumption.

Answer: A
The first committee member judges success too broadly, expecting walkers all day. The second member uses the sun pattern to interpret the behavior: walkers shift to the planted side after noon because that is when the shade matters most. The decision to plant a second row suggests the first row helped but did not solve the whole-day shade problem. Therefore, the most reasonable evaluation is partial success with a need for expansion. A likely mistake is to call it a total failure, but afternoon use increased on the planted side.

Answer: A
The guide contrasts decorated or golden objects with plain, repaired pots. Repeated repairs show that the pots stayed useful in daily life. Because the guide wants visitors to understand ordinary meals, these practical objects provide stronger evidence than rare display pieces. Therefore, she emphasizes the repaired plain pots because they reveal everyday use. A likely mistake is to think expensive objects are always more informative; the passage says golden bowls were often for display, not daily meals.