Sutton SET Mathematics Mock Paper 5

D

A number is divisible by 9 when its digits have a total divisible by 9. The known digits total 4 + 7 + 5 + 3 = 19. Adding 8 gives 27, so the missing digit is 8.

Elimination shortcut: the total must reach the next multiple of 9 after 19, which is 27. You only need the difference, 8, rather than testing every whole number.

Adding 1, 3, 5 or 9 gives 20, 22, 24 or 28. None of those totals is divisible by 9. A common trap is to choose 9 simply because the question mentions divisibility by 9.

B

The levels add 6 x 38 = 228 points. Priya then has 250 + 228 = 478 points. Spending 47 leaves 478 - 47 = 431 points.

Elimination shortcut: 6 x 38 ends in 8, so the total after adding 250 also ends in 8. Subtracting 47 must give an answer ending in 1. Only 431 does.

478 is the total before the tool is bought. 275 adds 6 and 38 instead of multiplying them, 525 adds the tool cost instead of subtracting it, and 1512 multiplies 6 by the starting score and then adds points incorrectly.

C

After 3/8 is sent away, 5/8 remains. The pie maker takes 2/5 of that remainder: 2/5 x 5/8 = 1/4 of the whole crop. Therefore 1 - 3/8 - 1/4 = 3/8 is left.

Elimination shortcut: the second amount sent is 2/5 of 5/8, which cancels quickly to 1/4. The farm sends 3/8 + 2/8 = 5/8 altogether, leaving 3/8.

1/4 is the amount sent for pies, while 3/20 multiplies 3/8 by 2/5 even though the second fraction applies to what remains. 7/40 subtracts 3/8 and 2/5 as if both were fractions of the whole. 13/40 subtracts only their product from one.

E

Each roll is 235 cm, so four rolls measure 4 x 235 = 940 cm. The workers use 680 cm. The length left is 940 - 680 = 260 cm.

Elimination shortcut: the starting length is just under 10 m and about 7 m is used, so roughly 3 m remains. Only 260 cm is close to that rough size.

680 cm is the amount used. 25 cm uses only one roll and mishandles the subtraction, 160 cm treats each roll as 210 cm, and 2600 cm has the right digits but is ten times too large.

A

A right angle is 90 degrees. The ratio has 2 + 3 = 5 equal parts, so one part is 90 divided by 5 = 18 degrees. The larger angle is 3 x 18 = 54 degrees.

Elimination shortcut: the larger angle must be more than half of 90 degrees but less than 90 degrees. This rules out 18, 30, 36 and 45 degrees immediately.

18 degrees is one ratio part, 36 degrees is the smaller angle, and 30 degrees comes from dividing 90 by the larger ratio number. Choosing 45 degrees ignores that the two ratio parts are unequal.

B

There are 7 ratio parts, so each part is 49 divided by 7 = 7 kg. Wholemeal starts at 35 kg and rye at 14 kg. After 7 kg of wholemeal is used, the amounts are 28 kg and 14 kg. Their simplest ratio is 2 : 1.

Elimination shortcut: only wholemeal is removed, so the first part of the original 5 : 2 ratio must become smaller, but wholemeal still outweighs rye. This rules out 5 : 2, 1 : 2 and the unchanged-style answer 3 : 2 before simplification is checked.

28 : 14 shows the correct amounts but is not simplified. 3 : 2 subtracts 2 ratio parts from the first number instead of subtracting 7 kg from the actual amount, and 1 : 2 reverses the two flours.

C

The four shortfalls are 128 - 119 = 9, 145 - 137 = 8, 96 - 91 = 5 and 132 - 121 = 11. Add them: 9 + 8 + 5 + 11 = 33.

Elimination shortcut: add the planned total and completed total by their final digits. The difference must end in 3 because 501 - 468 does. Of the sensible shortfall-sized options, only 33 ends in 3.

8 and 11 are single workers' shortfalls. 501 is the total planned and 468 is the total completed, so both stop before finding the difference.

D

Moving 7 squares right changes the first coordinate: -5 + 7 = 2, giving (2, 2). Reflection in the x-axis keeps the first coordinate and changes the sign of the second, giving (2, -2).

Elimination shortcut: a move right must make the first coordinate positive, so remove every option beginning with a negative number. The reflection must then make the second coordinate negative, leaving (2, -2).

(2, 2) forgets the reflection. (-12, -2) moves left before reflecting. (-2, 2) and (-2, -2) subtract 5 from 7 without following the direction on the grid.

B

Numbers rounding to 36 600 to the nearest hundred run from 36 550 to 36 649. Every number in that interval also rounds to 37 000 to the nearest thousand. Only 36 624 is in the interval.

Elimination shortcut: use the stricter nearest-hundred clue first. The number must begin 36 6 and its last two digits must be below 50. That leaves 36 624 without needing to round every option twice.

36 524 rounds to 36 500, while 36 650 rounds to 36 700. 37 024 rounds to 37 000 to the nearest hundred, and 37 600 rounds to 38 000 to the nearest thousand.

A

She cuts 18 x 12 = 216 postcards. After rejecting 29, she has 187. Dividing by 5 gives 37 complete packs with 2 postcards left, so she can make 37 complete packs.

Elimination shortcut: a complete pack count must be a little below 200 divided by 5, so it must be just below 40. Also, 187 cannot fill 38 packs because those need 190 postcards. This settles 37 quickly.

38 rounds up and would need too many postcards. 187 is the number of usable postcards, 43 ignores the rejected cards before dividing, and 935 multiplies by 5 instead of dividing.

E

Write all three amounts as decimals: 3/4 = 0.75 and 30% = 0.30. The amount becomes 0.75 - 0.30 + 0.125 = 0.575.

Elimination shortcut: after using 0.30, the tank holds 0.45. Adding 0.125 must give a result a little above one half but still below 3/4. Only 0.575 fits that range.

0.45 stops before ink is added. 0.325 subtracts the added ink, 0.875 ignores the ink used, and 1.175 adds all the quantities instead of subtracting the amount used.

D

Add the three dimensions: 48 + 27 + 11 = 86 cm. The box is 90 - 86 = 4 cm below the maximum.

Elimination shortcut: 48 + 27 is already 75, and adding 11 puts the box in the high eighties. Its spare allowance must therefore be only a few centimetres. Only 4 cm has the right rough size.

15 cm forgets to include the height. 42 cm subtracts only the length from the limit, 86 cm is the dimension total before comparison, and 176 cm adds the limit to the dimensions.

B

The four diameters total 4 x 18 = 72 cm. Four discs have only three gaps between them, adding 3 x 3 = 9 cm. The row is 72 + 9 = 81 cm long.

Elimination shortcut: the discs alone cover 72 cm, and the gaps add less than 12 cm. The answer must be between 72 cm and 84 cm, so 81 cm is the only possibility.

72 cm forgets the gaps. 84 cm counts four gaps, 45 cm uses the 9 cm radius instead of the 18 cm diameter, and 87 cm counts five gaps.

E

Work inside the brackets first: 24 + 16 = 40. Multiply next: 40 x 3 = 120. Finally, 180 - 120 = 60.

420 adds 120 to 180. 444 adds 180 and 24 before multiplying, while 132 subtracts 16 x 3 but ignores the 24. 140 works out the bracket but forgets to multiply it by 3.

B

Use twentieths first: 2/5 = 8/20 and 3/4 = 15/20. Their total is 23/20, so half of that is 23/40.

Elimination shortcut: the halfway value must be greater than 2/5 but less than 3/4. This immediately removes 5/18, 23/20 and 7/20. Then compare the two remaining choices: 7/10 is much closer to 3/4, while 23/40 is equally spaced from both ends.

5/18 adds the original numerators and denominators. 23/20 adds the two fractions but does not halve the total. 7/20 finds the difference, and 7/10 is the average of the two numerators over a copied denominator rather than the average of the fractions.

C

Seven ratio parts represent 42 bottles, so one part is 6. There are originally 3 x 6 = 18 cyan bottles. After 12 arrive there are 30 cyan bottles, giving 30 : 42. Divide both parts by 6 to get 5 : 7.

3 : 7 leaves the ratio unchanged. 7 : 5 reverses the colours. 15 : 21 divides both amounts by 2 but does not simplify fully, and 4 : 7 treats the extra 12 bottles as one ratio part instead of two.

E

The named groups contain 4 + 5 + 3 = 12 batches, so 3 batches remain. The cracked pots total 4 x 0 + 5 x 1 + 3 x 2 + 3 x 3 = 0 + 5 + 6 + 9 = 20.

Elimination shortcut: 15 is the number of batches, not the number of cracked pots. The total must be at least 5 + 6 + 9 = 20 from the three non-zero groups, which removes every option below 20; 36 has counted categories and frequencies incorrectly.

11 omits the final three batches, 9 counts only their cracked pots, and 36 multiplies the number of batches already named by the final value of 3.

A

The parts are 600 000 + 400 + 170 + 12. The last three parts total 582, so the whole number is 600 582.

Elimination shortcut: there are no thousands or ten thousands, so the number must start 600. Also, 17 tens and 12 ones together make 182, and adding 4 hundreds makes the final three digits 582. This leaves 600 582.

604 182 puts the 4 hundreds into the thousands column. 600 482 forgets to exchange 10 of the 17 tens for a hundred, while 600 522 mishandles the 12 ones. 604 712 simply places the spoken quantities side by side.

D

Angles in a quadrilateral total 360 degrees. The two known angles total 110 + 70 = 180 degrees, leaving 180 degrees. Split that equally: 180 divided by 2 = 90 degrees.

Elimination shortcut: the two known angles already make a straight angle of 180 degrees. The two equal angles must make the other 180 degrees, so each must be a right angle. Only 90 degrees fits.

70 and 110 copy a known angle. 80 subtracts 110 from 360 and then divides by 3, treating all remaining angles as equal. 180 is the total of the two unknown angles, not the size of each one.

C

The delivery contains 23 x 48 = 1104 bolts. After 275 are used, 1104 - 275 = 829 bolts remain.

Elimination shortcut: 23 x 48 ends in 4. Subtracting a number ending in 5 gives a result ending in 9. Only 829 and 1379 end in 9, and the answer must be smaller than 1104, so it is 829.

275 is the number used and 1104 is the starting total. 1379 adds the used bolts instead of subtracting them, while 804 subtracts 300 and forgets to add back 25.

A

Convert 2 1/3 to the improper fraction 7/3. Dividing 7/3 by 7 is the same as multiplying by 1/7, so 7/3 x 1/7 = 1/3 metre.

3 m turns the final fraction upside down. 2/21 divides only the fraction part by 7 and loses the two whole metres. 16 1/3 multiplies the mixed number by 7, and 7/3 is the whole cord before it is shared.

C

The whole sheet has area 32 x 24 = 768 square cm. The picture has area 26 x 18 = 468 square cm. The unprinted area is 768 - 468 = 300 square cm.

Elimination shortcut: both products end in 8, so their difference must end in 0. Only 300 does. You can then confirm it by finding the two areas.

468 square cm is the printed area and 768 square cm is the whole sheet. 36 square cm multiplies the differences in the side lengths, and 112 square cm is the perimeter of the outer sheet.

E

Four half-shaded squares contain the same shaded area as two whole squares. The shaded area is therefore 6 + 2 = 8 small squares out of 16. Since 8/16 simplifies to 1/2, the answer is 1/2.

Elimination shortcut: pair the four half-squares to make two whole shaded squares. Six wholes plus two wholes is exactly half of the 16-square design, so there is no need to add separate fractions.

3/8 counts only the six fully shaded squares. 5/8 and 10/16 count every half-shaded square as fully shaded. 4/10 uses the numbers of half-shaded and shaded squares as a new fraction instead of comparing shaded area with total area.

C

One standard task is worth 325 divided by 5 = 65 points. Forty per cent of 65 is 26, so the challenge task is worth 65 + 26 = 91 points.

26 is only the increase, and 65 is the standard value before the increase. 105 adds 40 points instead of 40%, while 130 doubles the standard value instead of increasing it by 40%.

A

The post office receives 18 x 25 = 450 stamps. It sells 167 + 96 = 263 stamps. The number remaining is 450 - 263 = 187.

Elimination shortcut: the sales total ends in 3, so subtracting it from 450 must give an answer ending in 7. Only 187 and 617 do, and a remainder cannot be greater than the starting 450, so 187 is forced.

283 subtracts only the first sale, 354 subtracts only the afternoon sale, 617 adds the morning sale to the starting stock, and 450 ignores both sales.

D

Twenty per cent of 50 cm is 10 cm, so the increased width is 60 cm. Twenty per cent of 60 cm is 12 cm. Decreasing 60 cm by 12 cm gives 48 cm.

Elimination shortcut: the 20% decrease is taken from 60 cm, so it removes 12 cm, which is more than the 10 cm first added. The final width must therefore be slightly less than 50 cm. Only 48 cm fits.

40 cm takes both changes as decreases. 60 cm stops after the increase, 50 cm assumes equal percentages cancel even though they act on different amounts, and 52 cm adds the 2 cm difference instead of subtracting it.

B

Angles around a point total 360 degrees. Remove the stated angle: 360 - 38 = 322 degrees for the other three angles together.

Elimination shortcut: removing an acute angle from a full turn leaves more than 270 degrees. Only 322 degrees is large enough.

38 degrees copies the given angle, and 76 degrees doubles it. 142 degrees finds the angle next to 38 degrees, while 180 degrees counts only one side of a straight line rather than all three remaining angles.

B

The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. There are exactly eight.

Elimination shortcut: list factor pairs only until they meet. For 24 they are 1 and 24, 2 and 12, 3 and 8, 4 and 6. Four pairs give eight factors.

18 has six factors: 1, 2, 3, 6, 9 and 18. The number 25 has three, 27 has four, and 32 has six. A common error is to count a square number's repeated middle factor twice.

D

Six boxes fit along the 60 cm length, four along the 40 cm width and three along the 30 cm height. Therefore 6 x 4 x 3 = 72 boxes fit.

Elimination shortcut: each crate dimension is divided by 10, so the answer is 6 x 4 x 3. This is much quicker than finding both volumes in cubic centimetres.

130 adds the three crate dimensions. 24 counts only one layer, while 7200 finds a mixture of area and height. 72 000 is the crate's volume in cubic centimetres but does not divide by the volume of one box.

A

The mornings account for 3 x 46 = 138 crates and the afternoons for 2 x 58 = 116 crates. The total is 254 crates over 5 sessions, so the mean is 254 divided by 5 = 50.8.

Elimination shortcut: the combined mean must lie between 46 and 58, and it must be closer to 46 because there are three morning sessions but only two afternoon sessions. This removes 254, 46, 58 and also 52, which is exactly halfway.

52 averages the two given means without allowing for the different numbers of sessions. 254 is the total, while 46 and 58 each ignore one part of the data.

B

To make 5 pieces from one rod, the carpenter needs 4 cuts. For 8 separate rods, she makes 8 x 4 = 32 cuts.

13 adds the numbers of rods and pieces. 40 counts pieces rather than cuts. 39 subtracts one cut from the total number of pieces, as if all eight rods began as one long rod, and 64 assumes two cuts are needed for each piece.

E

Multiply both amounts by 10 to remove the decimals. Then 7.2 divided by 0.3 has the same answer as 72 divided by 3, which is 24 pots.

Elimination shortcut: each pot holds less than 1 kg, so there must be more than 7.2 pots. Also, 0.3 kg is close to one third of a kilogram, so the answer should be close to 3 x 7.2, which is about 22. Only 24 is sensible.

2.4 divides 7.2 by 3 but ignores that the divisor is 0.3. 21.6 multiplies the two visible whole-number digits, 240 moves the decimal by one place too many, and 7.5 adds instead of dividing.

D

There are 28 divided by 4 = 7 exchange groups. Each group gives 3 patterned cards, so Tariq receives 7 x 3 = 21 new cards. Adding the 5 he already has gives 26 patterned cards.

21 counts only the new patterned cards. 33 adds all 28 blank cards to the 5 patterned cards without exchanging them. 37 reverses the exchange relationship, and 12 multiplies the two exchange numbers without using the 28 cards.

C

A full turn is 360 degrees. A regular octagon has 8 equal steps around its centre, so one step is 360 divided by 8 = 45 degrees.

Elimination shortcut: moving by one vertex is less than a quarter turn because the octagon has more than four vertices. It is also one eighth of a full turn, so 45 degrees is the only suitable option.

22.5 degrees halves the correct one-step turn. 90 and 135 degrees move two or three vertex positions, while 360 degrees returns every vertex to its own starting position after a full turn.

C

The first 6 has value 6. The final 6 has value 0.006. Their difference is 6.000 - 0.006 = 5.994.

6.006 adds the two values instead of finding their difference. 0.994 treats the units digit as 1, while 5.94 treats the final 6 as hundredths. 6.6 treats it as tenths.

E

Use 200 as a friendly number. Then 198 x 47 = 200 x 47 - 2 x 47 = 9400 - 94 = 9306. Also, 202 x 53 = 200 x 53 + 2 x 53 = 10 600 + 106 = 10 706. Their total is 20 012.

Elimination shortcut: the first product ends in 6 because 8 x 7 ends in 6. The second also ends in 6 because 2 x 3 ends in 6. Their sum must end in 2, which leaves 10 012, 40 012 and 20 012. A rough estimate of 200 x (47 + 53) is 20 000, so only 20 012 has the right size.

20 000 assumes the adjustments around 200 cancel exactly. 10 012 loses one block of 10 000, 40 012 doubles the estimate, and 19 988 reverses both adjustments around 200.

A

March has 31 days. Three days after 28 March is 31 March, and the fourth day after is 1 April. That leaves 13 more days, so the seventeenth day after 28 March is 14 April.

13 April counts 28 March as day one even though the wording says after it is sent. 15 April takes four days to reach the end of March and then adds all 13 remaining days after 1 April. 14 May changes the month twice, and 31 March counts only three days.

B

Three fifths of 30 is 18, so there are 18 girls. Two thirds of 18 is 12, so 12 girls play a brass instrument.

That leaves 18 - 12 = 6 girls who do not, which is B.

You can rule options out before finishing. The answer has to be smaller than the number of girls, so 18 and 20 are impossible whatever the second step gives.

12 is the number who DO play, so it stops one step short. 18 is all the girls. 4 comes from taking a third of 12 rather than a third of 18.

D

The horizontal length is 5 - (-3) = 8 units. The vertical length is 4 - (-2) = 6 units. The perimeter is 2 x (8 + 6) = 28 units.

14 adds one horizontal and one vertical side but does not go all the way round. 48 multiplies 8 by 6 and finds the area, while 16 and 20 double only one pair of sides and then add incorrectly.

E

Batch P has 5%, batch R has 3/60 = 5%, and batch S has 8/160 = 5%. Batch Q has 6/150 = 2/50 = 4%, so batch Q has the smallest proportion.

Batch P is tempting because 5 is a small count, but its batch is also smaller than Q. Batches R and S each have the same 5% rate as P. The proportions are not all equal because Q's 4% is lower.

C

Nine lots of 3.40 pounds cost 9 x 3.40 pounds = 30.60 pounds. Subtract the 5 pound voucher: 30.60 pounds - 5 pounds = 25.60 pounds.

25 pounds drops the 60 pence. 30.60 pounds ignores the voucher, while 35.60 pounds adds it. 1.60 pounds subtracts 5 pounds from the price of one parcel instead of from the whole order.

A

The first job uses 4 x 15 = 60 printer-minutes to make 360 labels, so one printer-minute makes 6 labels. The second job also uses 6 x 10 = 60 printer-minutes, so it makes the same 360 labels.

240 adjusts for the shorter time but not the extra printers. 540 adjusts for six printers but not the shorter time. 900 multiplies both visible changes, and 60 is the number of printer-minutes rather than labels.

B

The water volume is 50 x 40 x 30 = 60 000 cubic centimetres. Divide by 1000 because each litre occupies 1000 cubic centimetres: 60 000 divided by 1000 = 60 litres.

Elimination shortcut: 50 x 40 makes 2000, so each centimetre of depth holds 2 litres. A depth of 30 cm therefore holds 60 litres, without writing 60 000 first.

600 litres divides by 100 instead of 1000. 120 litres doubles the answer, 60 000 litres keeps the cubic-centimetre number but changes only the unit label, and 20 litres uses the base area but forgets the 30 cm depth.

C

Equal sides that sit next to each other rather than opposite each other is the definition of a kite, and a kite has exactly one line of symmetry, down its long diagonal.

A rhombus is the tempting answer, because its four equal sides do include two pairs of adjacent ones. It is ruled out by the counting: a rhombus has two lines of symmetry and two pairs of equal opposite angles, and the question says exactly one of each.

A parallelogram and a rectangle have their equal sides opposite each other, not adjacent. A trapezium need not have any equal sides at all.

A

The difference between 25% and 10% is 15% of the whole print run. Therefore 15% represents 18 posters. Five per cent represents 6 posters, so 100% represents 20 x 6 = 120 posters.

72 treats the 18-poster difference as 25% of the total. 180 treats it as 10%, while 270 multiplies 18 by 15 instead of scaling from 15% to 100%. 18 is only the difference between the two groups.

E

Each term is double the previous term, then add 1: 8 x 2 + 1 = 17, 17 x 2 + 1 = 35, and 35 x 2 + 1 = 71. Therefore the next term is 71 x 2 + 1 = 143.

79 adds 8 to the last term. 107 adds the last two differences, while 134 doubles 67 after misreading the pattern. 142 doubles 71 but forgets to add 1.

B

The first crate already uses one of the 48 numbers, so there are 47 steps from the first crate to the last. Calculate 156 + 47 = 203.

202 takes only 46 steps. 204 adds all 48 as if the first crate came before the count began. 108 subtracts 48, and 7488 multiplies the starting number by the number of crates.

D

After the proof, 2.400 - 0.375 = 2.025 kg remains. The main run uses 3/5, so 2/5 remains. One fifth of 2.025 kg is 0.405 kg, and two fifths is 0.81 kg.

1.215 kg is the 3/5 used in the main run, not the amount left. 0.96 kg takes 2/5 of the starting 2.4 kg and forgets the proof. 2.025 kg stops after the proof, while 1.41 kg subtracts 3/5 as though it meant 0.6 kg.

E

The tools weigh 8 x 85 = 680 g. With the 62 g box, the parcel weighs 742 g. It can take 750 - 742 = 8 g more.

70 g forgets the mass of the box. 742 g is the parcel's current mass, and 688 g subtracts the box mass from the limit but ignores the tools. 132 g adds the limit gap for the tools to the box mass instead of finding the final small allowance.

A

Follow the pair of instructions one complete round at a time. The displays are 5 x 3 - 4 = 11, then 11 x 3 - 4 = 29, then 29 x 3 - 4 = 83.

29 stops after two rounds. 131 multiplies by 3 three times and subtracts 4 only once. 99 subtracts 4 after the first multiplication but not after the later ones, while 187 adds 4 after each multiplication instead of subtracting it.