Sutton SET Mathematics Mock Paper 10

C

Elimination shortcut: the known digits add to 5 + 7 + 2 = 14. A multiple of 9 needs a digit sum that is a multiple of 9, so only 4 makes the sum 18. The number is 5472, which does round down to 5000.

A and B produce digit sums of 14 and 16, so neither number is divisible by 9. D and E produce numbers at least 5500, so they round to 6000; they also miss the divisibility test.

E

Elimination shortcut: the bracket is 12, so 144 divided by 12 is 12. The multiplication contributes 72, and adding 18 gives 90.

A stops after 18 + 12. B omits the opening 18. C works from left to right, using (18 + 144) divided by 12 x 6. D ignores the brackets and uses 18 + 144 divided by 5 + 7 x 6.

B

Elimination shortcut: 5/8 remains after the morning, and the afternoon leaves 3/5 of that remainder. So 3/5 x 5/8 = 3/8 is left. You do not need to find the afternoon sale separately.

A multiplies the two fractions sold. C copies the afternoon fraction as though it referred to the original harvest. D adds 3/8 and 2/5 as though both were original-harvest fractions. E stops after the morning.

D

Elimination shortcut: the answer must be a little over 2 litres, so A, B and E have impossible sizes. Convert 4.5 litres to 4500 ml and 1.275 litres to 1275 ml. Then 4500 - 850 - 1275 = 2375 ml.

A has the right numerical amount in litres but labels it as millilitres. B copies the first amount used. C is the total used, not the remainder. E adds the used solution to the starting amount.

A

Elimination shortcut: an exterior angle equals the two opposite interior angles together. Those two base angles are equal, so each is 124 divided by 2 = 62 degrees.

B subtracts the top interior angle, 56 degrees, from 124 instead of halving the exterior angle. C assumes a right angle. D is 180 - 62 and confuses a supplementary angle with a base angle. E copies the exterior angle.

D

Elimination shortcut: white exceeds yellow by 7 - 5 = 2 ratio parts. Those 2 parts are the 18 pots sold, so one part is 9. The original total is 15 parts, or 15 x 9 = 135 pots.

A is only the number sold. B and C are the original yellow and white groups. E adds the 18 sold pots to 135 even though 135 is already the original total.

B

Elimination shortcut: a missing value in a group with mean 24 should be near 24, so only 26 has a sensible size. Fully, the required total is 6 x 24 = 144 kg. The five known loads total 118 kg, leaving 144 - 118 = 26 kg.

A copies the number of loads. C is the known total, and D is the required total before the missing value is found. E adds 118 and 144 instead of finding their difference.

E

Elimination shortcut: 340 is 85% of the original. Since 5% must be a whole twentieth, divide 340 by 17 to get 5% = 20 pears. Therefore 100% = 20 x 20 = 400 pears.

A is 15% of 340. B takes 15% off 340 again. C treats the remaining amount as the original. D adds 15% of 340 rather than restoring 15% of the unknown original.

A

Elimination shortcut: a number is divisible by 4 when its last two digits form a multiple of 4. Start with the smallest allowed thousands digit, 3, then the smallest hundreds digit, 1. The ending 48 works, giving 3148.

B is divisible by 4 but is larger because its tens digit is 8. C has the smaller available hundreds arrangement after a different choice, but 18 is not divisible by 4. D abandons the smallest possible thousands digit. E puts the digits in descending order and is odd.

C

Elimination shortcut: two matching triangular ends joined by three rectangular faces describe a triangular prism. It also has 6 vertices and 9 edges, so every clue agrees.

A has 6 rectangular faces and 8 vertices. B has 5 vertices and only one square base. D has 4 triangular faces, 4 vertices and 6 edges. E has curved surfaces and no vertices.

E

Elimination shortcut: the run lasts more than an hour because it passes 23:47, but less than two hours. From 22:47 to midnight is 73 minutes, then another 18 minutes gives 91 minutes.

A counts only the time after midnight. B subtracts the displayed minute numbers. C counts from 23:47 and loses the first hour. D treats an hour as 100 minutes, using 100 - 47 + 18.

A

Undo the operations in reverse: 220 divided by 2 is 110, then 110 - 18 = 92, and 92 divided by 4 = 23.

B divides by 2 instead of undoing the multiplication by 4. C divides the final result by 4 and ignores the addition and final doubling. D stops before the last reverse step. E stops after undoing only the final doubling.

D

Elimination shortcut: after setting aside half a metre, a little over 2 m remains, so six pieces must each be a little over 1/3 m. Calculate 2 3/4 - 1/2 = 2 1/4 = 9/4. Then 9/4 divided by 6 = 9/24 = 3/8 m.

A divides the set-aside 1/2 m by 6. B ignores the 3/4 m before dividing. C divides only the 2 whole metres by 6. E is the remaining length before it is divided.

C

Elimination shortcut: A to M moves 6 squares right and 6 squares down. The same move again reaches B, so B is at (1 + 6, -2 - 6) = (7, -8).

A reverses both directions. B swaps the coordinates of M. D makes the horizontal move but loses the vertical change. E uses only part of each displacement instead of repeating the complete A-to-M move.

B

Elimination shortcut: 18 cm x 25 = 450 cm, or 4.5 m. Shortening this by 50 cm, which is 0.5 m, gives 4 m.

A has the right digits after the subtraction but divides by 10 once too many. C is the original full-size length before shortening. D multiplies the 50 cm change by the scale. E converts 450 cm to metres in the wrong direction.

B

Elimination shortcut: cutting a rectangle from a corner removes 4 m and 3 m from the outside but adds new inside edges of the same two lengths. The perimeter stays 2 x (12 + 9) = 42 m.

A subtracts the removed lengths once from the original perimeter. C adds the removed 3 m twice. D adds both 4 m and 3 m to the original perimeter. E gives the original area, 12 x 9, but labels it as a length.

D

Elimination shortcut: the middle of five equally spaced numbers is their mean. Since 370 divided by 5 = 74, the numbers are 70, 72, 74, 76 and 78. The largest is 78.

A is the smallest, B is the second number and C is the middle number. E treats consecutive even numbers as though the gap were 3, reaching 80 instead of 78.

A

Elimination shortcut: subtracting almost one tenth from just over 0.5 must give just over 0.4. Write 0.09 as 0.090, then 0.506 - 0.090 = 0.416 mm.

B subtracts 0.01. C makes no change. D adds 0.009 after misaligning the decimal places. E adds 0.09 instead of subtracting it.

E

Elimination shortcut: the amount added exceeds the amount used by 0.06 of the capacity, so the final level must be 0.06 above the starting 0.75. Thus 0.75 + 0.18 - 0.12 = 0.81.

A gives only the net change. B treats both later amounts as subtractions. C subtracts the added amount and adds the used amount. D assumes the two changes cancel exactly.

C

Adjacent angles in a rhombus total 180 degrees, so each angle beside the 58 degree angle is 180 - 58 = 122 degrees. Opposite angles are equal, making 122 degrees the largest.

A copies the smaller angle. B assumes every rhombus has right angles. D adds 58 to 90. E finds the reflex angle 360 - 58 rather than an interior angle of the rhombus.

A

Elimination shortcut: the full volume is 80 x 50 x 30 = 120 000 cubic centimetres, or 120 litres. Three quarters is 90 litres, and 90 - 15 = 75 litres remain.

B is the amount before draining. C adds the 15 litres. D is the full capacity. E adds the drained amount to the full capacity.

C

Elimination shortcut: 486 - 37 = 449 labels are needed. Seventeen rolls give only 425 labels, while 18 rolls give 450, so 18 is the smallest sufficient number.

A rounds down and leaves 24 labels missing. B is 449 divided by 25 but a fraction of a roll cannot be opened. D ignores the 37 labels already held and rounds 486 divided by 25 up. E is the number of labels needed, not rolls.

B

Elimination shortcut: the first group provides 4 x 3 = 12 worker-hours, so the rate is 18 divided by 12 = 1.5 kits per worker-hour. The second group provides 6 x 5 = 30 worker-hours, making 30 x 1.5 = 45 kits.

A changes only the number of workers. C multiplies the original 18 by the original 3 hours. D multiplies by the new 5 hours without allowing for the original time. E changes both numbers by multiplying but never divides by the original 4 workers and 3 hours.

D

Elimination shortcut: the rise is 12 percentage points, but the question compares that rise with the original 30%. Since 12 divided by 30 = 0.4, the share has risen by 40% of its original size.

A loses a zero when converting 0.4 to a percentage. B gives the percentage-point increase. C copies the original share. E adds 30 and 42 instead of comparing their difference with 30.

E

Elimination shortcut: both team totals are even, and subtracting 17 makes the answer odd. It must also exceed 700, so only 739 fits. Fully, 14 x 36 = 504, 9 x 28 = 252, and 504 + 252 - 17 = 739.

A copies the rejected number. B is only the second team's total. C subtracts the rejected apples from the first team only and omits the second team. D is only the first team's total.

D

Angles at a point total 360 degrees. The two right angles use 180 degrees, leaving 180 degrees in 2 + 3 = 5 parts. One part is 36 degrees, so the larger angle is 3 x 36 = 108 degrees.

A is one ratio part. B is the smaller angle. C assumes the remaining angles are equal. E is the combined total of the two remaining angles.

E

Elimination shortcut: 18 squared is 324, and 324 is even with digit sum 9, so it is divisible by both 2 and 3 and therefore by 6. The neighbouring squares are 17 squared = 289 and 19 squared = 361, outside the range.

A, B and C are multiples of 6 but are not square numbers. D is neither divisible by 3 nor a square number; it is included for checking only the final zero.

C

Elimination shortcut: 15% of 320 is 48, leaving 272 usable bulbs. The non-tulip share is 5/8, a little over half of 272, so 170 is the only sensible option. Indeed, 272 divided by 8 x 5 = 170.

A is the 3/8 tulip share of the usable bulbs. B takes 3/8 of all 320 bulbs. D takes 5/8 of all bulbs and ignores the damage. E stops after finding the usable total.

B

Elimination shortcut: 27, 28, 29 and 30 September are the first 4 days. That leaves 15 more days, so the nineteenth and final day is 15 October.

A counts only 18 days because it excludes one endpoint. C counts 20 days by adding 19 after the opening date. D transfers the duration directly to an October date. E keeps the day number and merely changes the month.

A

Test the two remainders. Since 63 is a multiple of 7, 67 leaves 4. Since 65 is a multiple of 5, 67 leaves 2, so 67 meets both conditions.

B leaves 2 after division by both 7 and 5. C leaves 4 after division by 7 but 4 after division by 5. D leaves 0 after division by 7. E leaves 0 after division by 7 and 4 after division by 5.

C

The bottom strip is 1/4 = 5/20 of the rectangle. Each of the five parts in the top strip is 1/20 of the rectangle, so its three shaded parts make 3/20. Altogether, 5/20 + 3/20 = 8/20 = 2/5.

A counts only the shaded pieces in the top strip. B counts only the bottom strip. D adds shaded numerators and denominators, using (3 + 1)/(5 + 4). E describes the shaded fraction of the top strip alone.

A

In the metal to glass ratio, 7 parts are 42, so metal is 10 x 6 = 60. In the wooden to metal ratio, 5 parts are 60, so wooden is 3 x 12 = 36.

B copies the glass count. C is the metal count. D adds wooden and metal pieces. E is the total number of all three materials.

E

Thirty-five per cent of 480 is 168 reserved wristbands. Morning lessons use 3/7 of 168, which is 72. Therefore 480 - 72 = 408 wristbands are not reserved for morning lessons.

A copies the percentage. B is the number reserved for morning lessons. C is the total reserved for all lessons. D is the number not reserved for any lesson, so it wrongly excludes wristbands reserved for later lessons.

B

In the units column, exchange one ten so 14 - 7 = 7. In the tens column, the reduced missing digit must exchange one hundred before subtracting 6. With a missing digit of 2, this gives 11 - 6 = 5, and the hundreds check is 7 - 2 = 5.

A forgets that the borrowed ten reduces the missing digit. C forgets the first borrow but allows for the second. D copies the units digit of the top number. E copies a digit from the answer instead of completing the exchanges.

D

The tile has 96 divided by 12 = 8 sides. Exterior turns around any polygon total 360 degrees, so each equal turn is 360 divided by 8 = 45 degrees.

A is the number of sides. B copies the side length. C divides 360 by 12 instead of by 8. E is the interior angle, 180 - 45, rather than the exterior turn.

E

Elimination shortcut: the first 6 is worth 60 000 and the second is worth 0.06. From hundredths to ten-thousands is six place-value moves, so the value is multiplied by 10 six times: 1 000 000 times.

A, B, C and D count only two, three, four or five of the six place-value moves. Each misses one or more factors of 10.

B

Elimination shortcut: the time is less than 2 hours, so the distance must be less than 9.6 km but greater than 4.8 km. Forty-five minutes is 3/4 hour, and 3/4 of 4.8 km is 3.6 km. The total is 4.8 + 3.6 = 8.4 km.

A counts only the first hour. C rounds the time up to 2 hours. D removes the decimal point from the correct distance. E multiplies the hourly distance by 45 as though 45 minutes meant 45 hours.

D

Follow each complete pair: 3 becomes 20, then 20 becomes 54, then 54 becomes 122. The displayed number is 122.

A adds 7 three times and doubles only once. B reverses every pair, doubling before adding. C adds 7 once and then doubles three times. E carries out a fourth pair after reaching 122.

C

After growing, the height is 120% of the original. Keeping 85% after trimming gives 85% of 120%, which is 102% of the original. Therefore 51 cm is 102%, so 1% is 0.5 cm and 100% is 50 cm.

A treats 15% as 15 cm and subtracts it from 51. B takes 15% off the final height again. D assumes the two changes cancel. E adds 20% to the final height instead of reversing both changes.

A

The first 8 trays contain 8 x 6 = 48 damaged plants and the next 12 contain 12 x 4 = 48. That is 96 across 20 trays, so the mean is 96 divided by 20 = 4.8.

B takes the simple mean of 6 and 4, ignoring that the groups have different sizes. C adds the two means. D gives the tray count. E is the total number of damaged plants before division.

B

The difference 11 - 3 = 8 ratio parts represents 64 seedlings, so one part is 8. The healthy group is 11 x 8 = 88 seedlings.

A is the damaged group. C is the total group. D multiplies the 3 damaged parts by the difference of 64. E multiplies 11 by 64 without first finding one part.

D

Before bruised pears are removed, 15 x 48 divided by 9 = 80 bags could be filled. Since 1/8 are bruised, 7/8 are good, so the number of bags is 7/8 of 80 = 70.

A copies the denominator of the bruised fraction. B is the number of bags the 90 bruised pears would fill. C divides the 630 good pears by 10 instead of 9. E ignores the bruised pears.

E

The joined cuboid measures 10 cm by 2 cm by 2 cm. Its surface area is two 10 x 2 faces in each of two directions, plus two 2 x 2 ends: 40 + 40 + 8 = 88 square centimetres.

A gives the row length but labels it as area. B gives the volume, 10 x 2 x 2. C is the surface area of a row only 6 cm long, so it forgets two cubes. D counts the four long faces but omits both end faces.

A

Convert 62.5% to 5/8. In twenty-fourths, 5/8 = 15/24 and 7/12 = 14/24, so the difference is 1/24.

B uses 8 + 12 = 20 as a denominator. C finds the correct numerator gap but uses 12 instead of the common denominator 24. D doubles that denominator error. E gives 62.5% as a fraction but does not subtract 7/12.

C

The lowest common multiple of 6, 8 and 9 is 72. The multiples around 200 are 144, 216 and 288, so the smallest one above 200 is 216.

A is divisible by 6 and 9 but not 8. B and D are divisible by 8 but not by both 6 and 9. E is a common multiple but is not the smallest one above 200.

A

Convert 4.05 kg to 4050 g. The kits together weigh 4050 - 450 = 3600 g, and 3600 divided by 12 = 300 g per kit.

B divides the total mass by 12 without removing the box. C copies the box mass. D is the combined kit mass before sharing. E is the total mass of box and kits in grams.

C

The midpoint of -6 and 4 is -1, while the shared y-coordinate stays 3. Moving 5 squares down changes the y-coordinate to 3 - 5 = -2, so C is (-1, -2).

A halves the horizontal distance of 10 but uses it as the x-coordinate. B subtracts only 1 vertically. D loses the negative sign on the midpoint. E adds 5 to both midpoint coordinates instead of moving only down.

E

Each tray is 307 - 297 = 10 g short. Across 84 trays, the shortfall is 84 x 10 = 840 g. There is no need to calculate either large total.

A is the shortfall on one tray. B is one gram per tray. C assumes a 2 g shortfall per tray. D copies the actual mass on one tray.

D

The sector represents 18/90 = 1/5 of the crates. One fifth of the 360 degrees in a full circle is 72 degrees.

A copies the crate count as an angle. B gives the percentage, 20%, but labels it in degrees. C copies the number of crates outside a different comparison. E gives the angle of the other four fifths of the chart.

B

With six values, the median is the mean of the third and fourth values. The added time must sit between 35 and 41, making the middle pair 35 and the new value. Their total must be 2 x 37 = 74, so the new value is 74 - 35 = 39 minutes.

A would make the median 35. C would leave the middle pair as 35 and 41, giving 38. D is the required total of the middle pair, not the added value. E is the total of the original five times.