Sutton SET Mathematics Mock Paper 8

C

Elimination shortcut: the answer must be greater than 41 367 and even. That rules out A because it is too small and B because it is odd. Keeping the smallest possible opening digits 41 leaves 3, 6 and 8; the smallest valid ending gives 41 368.

D swaps the final 6 and 8, making a larger valid number. E puts 3 in the thousands place, so it is larger than necessary. A starts with the smallest digits without respecting the lower limit, and B simply copies the lower limit without checking parity.

A

Elimination shortcut: an answer near 70 needs the opening 8 + 4 to be multiplied by 6. Only A does that without reducing the multiplier. Fully, (8 + 4) x 6 - 2 = 12 x 6 - 2 = 70.

B makes the multiplier only 6 - 2 and gives 24. C also uses 6 - 2, giving 48. D keeps 4 x 6 together and gives 30. E puts the subtraction inside the final bracket but still multiplies 4 x 6 first, so it also gives 30.

E

Elimination shortcut: the two used parts total a little more than one half, so the remainder must be a little less than one half. Of the options, 11/24 has that size. Fully, 5/12 = 10/24 and 1/8 = 3/24, so 13/24 is used and 24/24 - 13/24 = 11/24 remains.

A copies the fraction used for rolls. B subtracts 3/24 from 10/24 instead of adding the used parts. C assumes the loaves use one half, then subtracts the 1/8 used for rolls. D ignores the dough used for rolls.

B

Elimination shortcut: sand is 5 of the 7 equal parts, so the used sand must be more than half of 84 kg but less than 84 kg. After adding the unused 6 kg, 66 kg is the only sensible option. Fully, one part is 84 divided by 7 = 12 kg, so 5 parts are 60 kg and 60 + 6 = 66 kg was delivered.

A is the 2-part cement amount. C is the whole mixture used and forgets that only part is sand. D adds the spare sand to the whole mixture, and E multiplies 84 by 5 then divides by 2, reversing the role of the total and ratio parts.

D

Elimination shortcut: four pieces use just under 2 m, so just under 1 m should remain. That points to 97 cm, not 0.97 cm or 9.7 cm. Convert 2 m 85 cm to 285 cm, then calculate 285 - 4 x 47 = 285 - 188 = 97 cm.

A and B have the right digits but convert centimetres by factors of 100 and 10 unnecessarily. C gives the length of one cut piece. E subtracts only one 47 cm piece from 285 cm.

B

Elimination shortcut: the three given angles total 266 degrees, so the missing angle must be a little less than 100 degrees to reach 360 degrees. This rules out C, D and E. Fully, 360 - 266 = 94 degrees.

A results from losing a carried ten when adding the given angles. C copies one given angle. D subtracts the 104 degree angle from 270 but ignores the 72 degree angle. E gives the sum of the known angles instead of the missing one.

D

Elimination shortcut: removing a value far above the old mean must make the mean lower than 12, and the remaining total is 60 - 28 = 32. Dividing 32 by 4 gives 8 minutes.

A divides the 12-minute mean by the number left. B divides the remaining total by the original five services. C subtracts 5 from the old mean because there were five services. E gives the remaining total without dividing by four.

A

Elimination shortcut: the product must end in 6 because 8 x 7 ends in 6. Only 9176 has that last digit. Fully, 248 x 30 = 7440 and 248 x 7 = 1736, giving 7440 + 1736 = 9176.

B is 248 x 38, one extra section. C treats the 7 sections as 10 sections. D gives 248 x 70 and forgets the 30 sections. E adds an unnecessary zero to the correct product.

E

Elimination shortcut: 27/40 is greater than one half but less than three quarters, so A, B and C are too small. Multiply top and bottom by 25: 27/40 = 675/1000 = 0.675.

A moves the decimal point one place too far left. B treats the numerator as hundredths and ignores the denominator. C copies the denominator as tenths. D truncates the thousandths digit instead of writing the exact value.

C

Elimination shortcut: the missing vertex must share x = 5 with the top-right vertex and y = 1 with the bottom-left vertex. Only (5, 1) does both.

A changes the sign of the right-hand x-coordinate. B swaps the roles of the x and y values. D changes the sign of the required y-coordinate, and E reverses the two positive coordinates instead of matching the rectangle's rows and columns.

D

Elimination shortcut: XC is 100 - 10 = 90 and IV is 5 - 1 = 4, so the answer must be just above 90. Only 94 fits. Thus XCIV = 90 + 4 = 94.

A reads X as 60 when combining the symbols. B treats XC as 80. C subtracts IV from XC. E adds I before V as 5 rather than recognising IV as 4, and also treats XC as 100.

C

Elimination shortcut: the whole trip lasts just under 100 minutes, so after an 18-minute wait the moving time must be just over 80 minutes. From 09:47 to 11:26 is 99 minutes, and 99 - 18 = 81 minutes.

A subtracts the 18-minute wait twice. B treats 47 to 26 as a difference of 21 minutes and combines it incorrectly with the hours. D includes the wait. E adds the wait to the whole elapsed time.

B

Elimination shortcut: 30 rolls are 5/3 as many as 18, so the flour must be between 450 g and 900 g. Only 750 g fits. One roll needs 450 divided by 18 = 25 g, and 30 x 25 = 750 g.

A is the flour for the extra 12 rolls only. C doubles the recipe as if 30 were 36. D adds the required 750 g to the original 450 g. E multiplies 450 by 30 without first dividing by 18.

A

Elimination shortcut: the two unknown angles total 360 - 210 = 150 degrees, so the smaller one must be less than half of 150 degrees. Only 60 degrees fits. Five ratio parts total 150 degrees, one part is 30 degrees and two parts are 60 degrees.

B is the larger 3-part angle. C treats 2 : 3 as the fraction 2/3 of 150. D gives the total of both unknown angles. E adds the 210 degree angle to the smaller angle.

E

Elimination shortcut: 10 boxes hold 240 markers and 11 hold 264, both too few. Boxes must be whole, so 11.9 is impossible. Twelve boxes hold 288 markers, enough for all 286.

A copies the remainder after comparing 286 with 12 x 24 = 288. B estimates 286 as 240. C rounds 286 divided by 24 down despite 22 markers remaining. D writes an approximate quotient without rounding up to a whole box.

C

Elimination shortcut: one quarter withdraws, leaving 240. Three eighths is a little less than half, so the answer must be a little less than 120. This points to 90. Fully, 240 divided by 8 x 3 = 90.

A finds 3/8 of the 80 entrants who withdrew. B gives the number who withdrew. D finds 3/8 of the original 320 entrants and ignores the withdrawal. E gives everyone remaining before the relay choice.

A

Elimination shortcut: there are 180 - 112 = 68 children. Contactless children number 137 - 89 = 48, so the cash-paying children are 68 - 48 = 20.

B is the number of adults paying cash, 112 - 89. C is the total number paying cash, 180 - 137. D is the number of children paying contactlessly. E is the total number of children before separating payment types.

E

Elimination shortcut: all five options are two-digit multiples of 6, so check which divide 144 and then obey the word greatest. Since 144 divided by 72 = 2, 72 is a factor and it is the greatest listed value.

A, B, C and D are all factors of 144 and multiples of 6, but each ignores the instruction to choose the greatest one. They are tempting valid examples rather than the greatest valid example.

D

Elimination shortcut: the answer must be less than the enclosing area of 9 x 7 = 63 square metres but much more than the 12 square metre cut-out. Subtract the cut-out: 63 - 4 x 3 = 63 - 12 = 51 square metres.

A gives only the removed area. B adds the four stated lengths. C uses 7 x 7 for the whole platform, copying the shorter outer side as both dimensions and ignoring the cut-out. E gives the whole rectangle without removing the corner.

B

Elimination shortcut: 231 divided by 7 is just over 30, so the missing digit must be small. Since 33 x 7 = 231, the digit is 3.

A makes 31 x 7 = 217. C makes 35 x 7 = 245. D makes 37 x 7 = 259, and E makes 39 x 7 = 273. Each wrong option can be rejected by checking its product.

A

Elimination shortcut: taking nearly 2 kg from 4 1/2 kg must leave a little under 3 kg. Only 2 3/4 kg has the right size. Regroup 4 1/2 as 3 6/4, then 3 6/4 - 1 3/4 = 2 3/4 kg.

B subtracts the whole numbers but adds the fractional parts. C subtracts only 1/4 kg. D adds the fractional parts while subtracting the whole numbers incorrectly. E adds both mixed numbers instead of subtracting.

E

Elimination shortcut: four separate cubes have 4 x 6 = 24 faces. There are three joins, and each join hides two faces, so six faces are hidden. Therefore 24 - 6 = 18 faces remain visible.

A counts the joins. B gives the faces on one cube. C counts four faces on each of the three joins rather than the outside. D subtracts only two hidden faces per end cube and misses the middle joins.

D

Elimination shortcut: before any is used, there are 35 kg brown and 15 kg green. Brown falls to 25 kg while green stays 15 kg, so brown is still the larger part. This rules out A, B and C. The ratio 25 : 15 simplifies to 5 : 3.

A reverses the original ratio. B reverses the correct final ratio. C assumes using 10 kg makes the two amounts equal. E keeps the original ratio even though only brown compost changes.

C

Elimination shortcut: the full capacity is even, and subtracting 28 keeps it even. The passenger total must also be a little below 500, which points to 478. Fully, 5 x 76 = 380, 3 x 42 = 126, and 380 + 126 - 28 = 478.

A counts only the single-decker seats. B subtracts the empty seats from the double-decker seats but omits the other buses. D is the full capacity before empty seats are removed. E adds the empty seats to capacity.

B

Elimination shortcut: 0.375 is between one quarter and one half, so C, D and E are too large. Write it as 375/1000 and divide top and bottom by 125 to get 3/8.

A reads 0.375 as 3 hundredths. C drops the first decimal digit and treats 0.375 as 0.75. D truncates to 0.37 before converting. E uses hundredths instead of thousandths and produces 3.75.

B

Elimination shortcut: rising 14 from -6 crosses zero and reaches 8; going down 3 must finish below 8 but above zero. Only level 5 fits. Fully, -6 + 14 - 3 = 5.

A treats both moves as downward. C stops after the rise. D adds the final 3 instead of subtracting it. E adds all three magnitudes and ignores the negative starting level.

C

Elimination shortcut: the mean must lie between 5 kg and 9 kg, so A, B and E can be rejected. The total is 2 x 5 + 3 x 7 + 1 x 9 = 40 kg. Divide by 6 to get 6 2/3 kg.

A copies the greatest frequency. B gives the smallest recorded mass. D gives the mode, not the mean. E gives the total without dividing by the six days.

E

Elimination shortcut: the third interior angle is obtuse because the two 32 degree angles total only 64 degrees, so its neighbouring exterior angle must be acute. Fully, the third interior angle is 180 - 32 - 32 = 116 degrees. Angles on a straight line total 180 degrees, so the exterior angle is 180 - 116 = 64 degrees.

A halves one base angle. B copies one base angle. C subtracts 64 from 116 and gives the difference between the interior and exterior angles. D calculates 90 - 32, treating the third angle as a right angle.

D

Elimination shortcut: the total is 24 + 120 = 144 litres, or 144 000 cubic centimetres. The base area is 120 x 40 = 4800 square centimetres, and 144 000 divided by 4800 = 30 cm. A quick check is 4800 x 30 = 144 000.

A converts the 144-litre total to only 14 400 cubic centimetres. B finds the depth of the first 24 litres only. C finds the depth of the added 120 litres only. E copies the width as the depth.

A

Elimination shortcut: 48 groups are two groups fewer than 50 groups, so the amount removed is only 2 x 25, not another product near 1200. Therefore ? = 50.

B removes four groups of 25 instead of two. C gives 48 x 25, the left side of the equation. D gives 50 x 25 before the subtraction. E doubles 50 x 25 rather than finding the difference between the two products.

C

Fifteen per cent is 3/20. If 3 parts are 360 kg, one part is 120 kg and 20 parts are 2400 kg.

A finds 15% of 360 kg, treating the used amount as the whole. B subtracts 15% from the correct full amount and gives the unused mass. D adds the used 360 kg to the full amount. E treats 1% as 360 kg and multiplies by 100.

D

The two rows contain 12 squares and the right column contains 5. The two top-right squares have been counted twice, so 12 + 5 - 2 = 15 squares are shaded. The fraction is 15/30 = 1/2.

A counts only the right column. B counts only the two complete rows. C subtracts three overlap squares instead of two. E counts all 12 row squares and all 5 column squares without removing the overlap.

A

One lap is 2.8 divided by 7 = 0.4 km. Twelve laps cover 12 x 0.4 = 4.8 km.

B copies the original number of laps as a distance. C doubles the correct distance. D treats each lap as 1 km. E multiplies 2.8 by 12 without first finding the length of one lap.

E

Thirty-two frames need 32 x 7 = 224 bolts. The store has 216, so it needs 224 - 216 = 8 more bolts.

A calculates only the one bolt needed to turn the remainder of 6 into the next complete frame. B gives the number of frames beyond the 30 complete frames currently possible, not the bolts needed. C is the remainder when 216 is divided by 7. D adds one full frame's bolts without using the existing remainder.

B

A rhombus has four equal sides and two pairs of parallel sides. Excluding right angles makes the description unambiguous.

A kite need not have opposite sides parallel. A rectangle does not need all four sides equal. A square has right angles, which the question excludes. A trapezium has only one pair of parallel sides in the definition used here.

D

Convert 7.5 litres to 7500 ml. After the spill, 7500 - 300 = 7200 ml remains. Then 7200 divided by 300 = 24 bottles.

A converts 7200 ml to litres and stops. B gives the remaining litres. C removes the spilled 300 ml twice. E divides the original 7500 ml by 300 and ignores the spill.

B

The increase is 0.4 m. Compare it with the original 1.6 m: 0.4/1.6 = 1/4 = 25%.

A compares the increase with the new 2.0 m. C copies the 0.4 m increase as 40%. D treats any improvement as doubling. E gives the new distance as a percentage of the original rather than the increase.

C

List multiples until they match: multiples of 6 are 6, 12, 18, 24; multiples of 8 are 8, 16, 24. The first common multiple is 24 days.

A and B give one watering interval only. D adds 6 and 8 and then multiplies by 3. E multiplies 6 by 8, finding a common multiple but not the first one.

A

Each course starts with 1458 divided by 27 = 54 bricks. Set aside 6 to leave 54 - 6 = 48 bricks between the ends.

B stops before setting aside the end bricks. C adds 6 instead of subtracting it. D subtracts 6 only once from the whole delivery. E gives the full delivery without sharing or setting any aside.

E

Find the increases, not the cumulative totals: 350 m, 430 m, 300 m, 440 m and 450 m. The greatest increase is 450 m from 8 to 10 minutes.

A uses the first cumulative reading. B is 430 m, C is 300 m and D is 440 m. D is tempting because 1520 is a large cumulative total, but its interval increase is 10 m less than the final one.

E

The radius is half the 18 cm diameter, so each square side is 9 cm. Its perimeter is 4 x 9 = 36 cm.

A halves the radius again. B gives one side. C gives two sides, or copies the diameter. D counts only three sides. None completes all four sides except E.

A

Find 3/5 of 4/7 by multiplying: 3/5 x 4/7 = 12/35. This is already in simplest form.

B puts the bean numerator over the denominator for all beds. C gives all vegetable beds. D gives the bean fraction of vegetable beds only. E inverts 5/7 after mixing the two denominators.

D

Multiply the plan length by the scale: 8.4 x 2.5 = 8.4 x 2 + 8.4 x 1/2 = 16.8 + 4.2 = 21 m.

A divides 8.4 by 2.5. B subtracts the scale number 2.5 from 8.4. C adds 8.4 and 2.5 instead of multiplying. E converts metres to decimetres unnecessarily and becomes ten times too large.

B

Undo the operations in reverse: add 17 to get 112, then divide by 4 to get 28. Checking gives 28 x 4 - 17 = 95.

A subtracts 17 from 95 before dividing, undoing subtraction in the wrong direction. C also subtracts 17, then divides by 2 instead of 4. D stops after adding 17. E multiplies 112 by 4 instead of dividing.

C

The length is 84 divided by 7 = 12 m. The full perimeter is 2 x (12 + 7) = 38 m. Leave out the 3 m opening: 38 - 3 = 35 m.

A gives the missing length. B gives length plus width, only half the perimeter. D includes fencing across the gate. E subtracts the 3 m opening from the area rather than from the perimeter.

A

Use twentieths: 3/10 = 6/20 and 1/4 = 5/20. Altogether 11/20 = 55% is used for loaves and rolls, leaving 9/20 = 45% for buns.

B gives the combined percentage already used. C ignores the rolls and leaves 70%. D ignores the sourdough loaves and leaves 75%. E adds 30% and 75% instead of subtracting both used portions from 100%.

E

The swap changes 672 418 to 612 478. Subtracting gives 672 418 - 612 478 = 59 940.

A subtracts the digit values 7 - 1 and attaches one zero. B loses a factor of 10 from the correct decrease. C treats the new value of 7 as 600 instead of 70. D transposes the final two digits while regrouping in the subtraction.

B

The horizontal gap is 8 - (-6) = 14 squares, so halfway is 7 squares from either stop. Moving 7 right from -6 reaches 1. The y-coordinate stays 4, so the midpoint is (1, 4).

A averages the sizes 6 and 8 but keeps the negative sign from the first stop. C halves only the positive coordinate 8. D gives the half-gap of 7 as the coordinate. E gives the full horizontal gap as the coordinate.

C

Convert the total length to centimetres: 18 x 240 = 4320 cm. Since 57 x 75 = 4275 and 58 x 75 = 4350, there is enough for 57 complete shelves but not 58.

A gives one shelf per plank. B allows exactly three shelves from each plank and ignores that joining lets offcuts combine. D rounds up despite being 30 cm short. E divides by 7.5 rather than 75 after converting to centimetres.

D

Four batches use 4 x 0.35 = 1.40 kg. Subtract from the starting mass: 2.40 - 1.40 = 1.00 kg.

A copies the amount in one batch. B misreads 0.35 as 0.45 before multiplying by four. C subtracts five lots of 0.35 kg. E treats each batch as 0.25 kg, using only 1 kg altogether.