Sutton SET Mathematics Mock Paper 7

A

Eliminate 80 000 immediately because a halfway number must lie between 38 950 and 41 050. The two totals are equally spaced around 40 000: 40 000 - 38 950 = 1050 and 41 050 - 40 000 = 1050.

The answer is 40 000. The tempting 40 050 comes from looking only at the last three digits. 39 000 and 41 000 round the given totals instead of finding their midpoint, while 80 000 is roughly their sum.

D

Eliminate 1663 on rough size: multiplying the whole opening subtraction by 18 would make a huge answer, but the brackets do not tell you to do that. Work inside the brackets first: 13 + 5 = 18. Then 4 x 18 = 72, so 96 - 72 + 7 = 24 + 7 = 31.

The answer is 31. The tempting 24 stops before adding 7. The 85 subtracts 18 without multiplying it by 4, 175 adds 72 instead of subtracting it, and 1663 comes from working blindly from left to right.

B

Eliminate 3/5 because that is what remains after the morning only, and eliminate 1/5 because that is the afternoon sale, not the closing stock. After the morning, 3/5 remains. The afternoon keeps 2/3 of that remainder, so 2/3 x 3/5 = 2/5 is left.

The answer is 2/5. The tempting 4/15 comes from adding 2/5 and 1/3 as though both sales were fractions of the original stock. The 1/3 copies the afternoon fraction, and 3/5 stops a step early.

E

Four ratio parts represent 36 pears, so one part is 36 divided by 4 = 9. The trader sells 7 x 9 = 63 apples. Add the 15 unsold apples: 63 + 15 = 78 apples at the start.

The tempting 63 is the number of apples sold, with the unsold apples forgotten. The 51 adds 15 to the 36 pears, 72 doubles the pear total, and 84 uses 12 as one ratio part by dividing 36 by 3 instead of 4.

C

Eliminate 4.23 m because a route longer than 3 km cannot be only a few metres. Convert 3 km 480 m to 3480 m, then add the detour: 3480 + 750 = 4230 m.

The answer is 4230 m. The tempting 4.23 is the correct number of kilometres with the wrong unit. The 3480 omits the detour, 3750 omits the original 480 m, and 1230 adds only 480 m and 750 m while losing the 3 km.

B

Eliminate any row whose total is not 360 degrees, because every quadrilateral has an interior angle total of 360 degrees. Option B gives 48 + 102 + 96 + 114 = 360 degrees.

The totals in A and C are both 350 degrees, so each is 10 degrees short. The totals in D and E are both 370 degrees, so each is 10 degrees too large. Those are tempting if one addition is rounded or a carried ten is lost.

A

Eliminate 486 pounds because that is all the income before any charge, and eliminate 671 pounds because paying charges cannot increase the money. The income is 18 x 27 = 486 pounds. The stall charges are 18 x 5 = 90 pounds, so 486 - 90 - 95 = 301 pounds.

The answer is 301 pounds. The tempting 391 subtracts the licence fee but forgets the per-stall charge. The 396 subtracts the per-stall charge but forgets the licence, while 486 stops before either deduction and 671 adds both charges.

D

Eliminate 0.61 and 3/5 because both are below 62%: they are 61% and 60%. Also eliminate 63% because it is above 5/8. Since 5/8 = 0.625, the number must lie strictly between 0.620 and 0.625. Only 0.621 does.

The tempting 0.625 equals the upper boundary, but the question says less than 5/8. The 63% crosses that boundary, while 0.61 and 3/5 cross the lower boundary.

C

Eliminate every number below 75 500 because it is nearer 75 000 than 76 000. Exactly halfway, 75 500 rounds up to 76 000, so it is the smallest possible number.

The tempting 76 499 is the largest whole number that rounds to 76 000, not the smallest. The 76 500 rounds to 77 000, while 75 000 and 75 499 round to 75 000.

E

Eliminate 3.50 cm because two strips that are each longer than a metre cannot leave only a few centimetres. Convert 2.4 m to 240 cm. Then 240 + 175 - 65 = 350 cm.

The answer is 350 cm. The tempting 3.50 is the right length in metres but carries the wrong unit. The 110 uses only 175 - 65, the 415 is the total before cutting, and 3500 multiplies by 100 once too many.

D

Point P is 4 squares to the left of x = 1, so its reflection is 4 squares to the right of that line, at (5, 5). Moving 2 squares down changes only the second coordinate, giving (5, 3).

The tempting (5, 7) moves up instead of down. The point (-3, 3) makes the translation but forgets the reflection, (3, 5) treats x = 1 as the y axis and forgets the final move, and (-5, 3) reflects in the wrong direction.

B

Eliminate 207 because it is the number left before sharing, not the number on one shelf. There are 7 x 48 = 336 books, then 336 - 129 = 207 remain. Share them: 207 divided by 9 = 23.

The answer is 23. The tempting 37 shares all 336 books and ignores the loans. The 14 comes from sharing only the 129 lent books and discarding the remainder, 207 stops before division, and 465 adds the lent books instead of subtracting them.

E

Eliminate 3/4 m because five pieces that long would use 3 3/4 m and leave nothing to set aside. The cord left is 3 3/4 - 1/2 = 3 1/4 = 13/4 m. Dividing by 5 gives 13/4 divided by 5 = 13/20 m.

The answer is 13/20 m. The tempting 13/4 m is the whole remaining length before it is shared. The 3/20 subtracts only the numerators, 3/5 divides the whole-number part by 5 and loses the quarter, and 3/4 ignores the set-aside cord.

A

The younger group has 7 - 5 = 2 more ratio parts than the older group. Those extra 2 parts are the 8 absent scouts, so one part is 4 scouts. The full troop has 5 + 7 = 12 parts, and 12 x 4 = 48 scouts.

The tempting 20 is the older group and 28 is the younger group. The 40 counts ten parts instead of twelve, while 56 multiplies the seven younger parts by the eight absent scouts instead of finding one part.

C

Eliminate every scalene option because equal angles in a triangle stand opposite equal sides, so the triangle is isosceles. The third angle is 180 - 45 - 45 = 90 degrees, making it right-angled as well.

The answer is a right-angled isosceles triangle. The obtuse option wrongly treats the two 45 degree angles as leaving more than 90 degrees. An equilateral triangle would have three 60 degree angles, and both scalene options ignore the equal pair.

E

Eliminate 8 and 15 degrees because the median of seven ordered values is the fourth value, not either end. Expand the frequencies in order: 8, 8, 10, 12, 12, 12, 15. The fourth value is 12 degrees Celsius.

The tempting 11 degrees is the mean, found from a total of 77 divided by 7. The 10 is the third value, reached by stopping one place before the middle, while 8 and 15 are the lowest and highest values.

C

Eliminate 4.5 because the question says the secret is a whole number. Undo the steps in reverse order: 81 divided by 3 = 27, then 27 - 9 = 18, then 18 divided by 2 = 9.

The answer is 9. The tempting 18 stops before undoing the doubling. The 6 divides by 3 instead of 2 at the final step, 4.5 halves once too often, and 36 subtracts 9 before undoing the final tripling, which reverses the steps in the wrong order.

A

Use elimination through the digit-sum test. A multiple of 9 has digits that add to a multiple of 9. The known digits total 4 + 8 = 12, so the missing digit must take the total to 18. That digit is 6.

The answer is 6 because 4 + 6 + 8 = 18. The tempting 3 makes a total of 15, 4 makes 16, 7 makes 19 and 9 makes 21. None of those totals is divisible by 9.

B

Line up the decimal points. First, 2.750 + 0.608 = 3.358. Then 3.358 - 1.900 = 1.458 mm.

The tempting 3.358 mm is the amount before evaporation. The 1.45 drops the final 0.008 by treating 0.608 as 0.600, 14.58 moves the decimal point, and 0.242 subtracts both 1.9 and 0.608 from 2.75 instead of adding the shower first.

D

Eliminate 75 hours 25 minutes because subtracting used time must leave less than the original 55 hours 35 minutes. Convert 2 days 7 hours to 55 hours. Regroup 55 hours 35 minutes as 54 hours 95 minutes, then subtract 19 hours 50 minutes to get 35 hours 45 minutes.

The tempting 35 hours 85 minutes subtracts without regrouping and leaves an impossible number of minutes. The 36 hours 15 minutes subtracts 35 from 50 instead of regrouping, 11 hours 45 minutes forgets the first full day, and 75 hours 25 minutes adds the times.

A

Twenty-seven small cubes make a 3 by 3 by 3 cube. On each of the 12 edges, the one middle cube has exactly two painted faces. The corner cubes have three painted faces, so the answer is 12.

The tempting 8 counts the corner cubes, and 6 copies the number of large faces. The 24 counts two painted faces for each edge cube instead of counting cubes, while 26 counts every small cube except the unpainted centre.

E

Eliminate 323 because that is a number of pegs, not packs. The tents need 5 x 18 x 4 = 360 pegs. After using the 37 stored pegs, 323 must be bought. Twelve packs hold only 300 pegs, so one more pack is needed: 13 packs.

The tempting 12 rounds 323 divided by 25 down and leaves tents without pegs. The 15 ignores the stored pegs, 2 divides the 37 stored pegs by the pack size, and 323 stops before converting pegs into packs.

D

Eliminate 256 and 320 because the smaller wheel must turn more often to cover the same distance. Five circumference parts x 320 turns gives the journey length in matching units. Divide by the rear wheel's 4 parts: 5 x 320 divided by 4 = 400 turns.

The answer is 400. The tempting 256 multiplies by 4/5 in the wrong direction. The 64 divides by 5 only, 320 assumes equal wheels, and 1600 multiplies by 5 without dividing by 4.

C

Find 3/7 of 7/12 by multiplying: 3/7 x 7/12 = 21/84. Cancel the sevens, or simplify 21/84, to get 3/12 = 1/4.

The tempting 3/7 is the fraction of fiction shelves, not of all shelves. The 3/19 and 10/19 come from adding numerators and denominators, while 7/36 multiplies by 1/3 and forgets that the mystery fraction is 3/7.

B

Eliminate 11 500 litres because 11 500 is a number of cubic centimetres, not litres. The box volume is 40 x 25 x 18 = 18 000 cubic centimetres. Empty space is 18 000 - 6500 = 11 500 cubic centimetres. Since 1000 cubic centimetres is 1 litre, this is 11.5 litres.

The tempting 18 litres is the whole box before the documents are added. The 6.5 litres is the occupied space, 24.5 litres adds occupied space to capacity, and 11 500 litres forgets the final conversion.

C

First find the audiobooks: 120 - 37 - 28 - 19 = 36. Then compare fiction with audiobooks: 37 - 36 = 1. One more fiction book was borrowed.

The tempting 36 is the number of audiobooks, not the difference. The 9 compares fiction with information books, 19 copies the graphic novel count, and 65 adds fiction and information instead of finding the missing group and comparing it.

D

Eliminate 8400 because swapping the digits does not remove both place values. The swap changes 508 407 to 504 807. Subtract: 508 407 - 504 807 = 3600.

The tempting 400 and 800 use one digit's new value without comparing the whole numbers. The 4000 subtracts the digits 8 - 4 and attaches three zeros, while 8400 adds the old place values instead of finding the decrease.

B

Eliminate four options by the last digit. The units calculation is 7 x 8 = 56, so the product must end in 6. Only 19 056 does. To check fully, use 397 x (50 - 2) = 19 850 - 794 = 19 056.

The tempting 19 850 stops after multiplying by 50. The 20 644 adds 794 instead of subtracting it, 19 200 replaces 397 by 400 and treats an estimate as exact, and 1905.6 introduces a decimal point that was not in either whole-number factor.

A

Eliminate 79.50 pounds and 88.50 pounds because a 12% reduction is greater than the 4.50 pound delivery charge, so the final amount must be below 75 pounds. Ten per cent of 75 pounds is 7.50 pounds and 2% is 1.50 pounds, so the reduction is 9 pounds. Pay 75 - 9 + 4.50 = 70.50 pounds.

The tempting 66 pounds is the reduced price before delivery. The 70.05 misaligns the decimal digits when adding 4.50, 79.50 adds delivery without applying the reduction, and 88.50 adds both the reduction and the delivery charge.

E

Angles around the centre total 360 degrees. The six triangles are equal, so each central angle is 360 divided by 6 = 60 degrees.

The tempting 120 degrees is an interior angle of a regular hexagon, not one of the six angles at its centre. The 30 halves the correct angle, while 45 and 90 divide the turn into eight or four equal parts instead of six.

B

The middle strip is 1/3 of the flag. Three quarters of that strip is shaded, so the shaded fraction is 3/4 x 1/3 = 3/12 = 1/4.

The tempting 3/4 describes the shaded part of the middle strip only, and 1/3 describes the whole middle strip. The 1/12 counts one triangle instead of three, while 3/7 adds the three strips and four triangles as though all seven parts were equal.

C

Eliminate 18.8 km and 180.8 km on rough size: at this scale, 4 cm represents 1 km, so just over 7 cm is under 2 km before the detour. Calculate 7.2 x 25 000 = 180 000 cm = 1.8 km. Add 0.8 km to get 2.6 km.

The tempting 1.8 km is the mapped route before the detour. The 1.0 km subtracts the detour, 18.8 km misses a conversion factor of 10, and 180.8 km treats 180 000 cm as 180 km.

A

She reads 36 x 4 = 144 pages first, then 58 x 3 = 174 pages. That is 144 + 174 = 318 pages read. Subtract from the collection: 1250 - 318 = 932 pages unread.

The tempting 318 is the total read, not the amount left. The 1106 subtracts only the first four days, 1076 subtracts only the next three days, and 1568 adds the pages read to the starting total.

D

Compare each numerator with half its denominator. The gaps from 1/2 are: 7/15 is 1/30 away, 9/20 is 1/20 away, 11/24 is 1/24 away, 13/25 is 1/50 away, and 17/30 is 1/15 away. The smallest gap is 1/50, so 13/25 is closest.

The tempting 11/24 has a numerator only 1 below 12, but its pieces are larger than fiftieths. Choosing 7/15, 9/20 or 17/30 usually comes from comparing numerator gaps without allowing for different denominator sizes.

E

Eliminate 0.03 cm and 3 cm on rough size: 120 litres spread over a base smaller than one square metre will be much deeper than a few centimetres. Convert 120 litres to 120 000 cubic centimetres. The base area is 80 x 50 = 4000 square centimetres, so the depth is 120 000 divided by 4000 = 30 cm.

The tempting 0.03 cm gives 0.03 metres but labels it centimetres. The 3 cm misses a factor of 10, 300 cm adds one, and 24 cm divides 120 000 by 5000 after adding the base dimensions instead of multiplying them correctly.

C

Rounding to the nearest hundred sends everything from 4550 up to 4649 to 4600. Exactly halfway, 4550, rounds up, so it is the smallest one that works.

4549 is the trap, and it is only one away: it is nearer to 4500, so it rounds down.

4500 rounds to itself. 4551 does round to 4600 but it is not the smallest. 4600 is the answer to the rounding, not the number you started with.

C

A parallelogram matches itself after a half turn, so its rotational symmetry is order 2. A sloping, general parallelogram has no line of symmetry, so it fits both parts of the description.

A square has rotational symmetry of order 4 and four symmetry lines. A rectangle and a rhombus each have symmetry lines, while an isosceles trapezium has one symmetry line but does not usually match itself after a half turn.

E

Eliminate 145 and 152 because those are numbers of oranges, not bags. Each crate starts with 2736 divided by 18 = 152 oranges. Remove 7 to leave 145, then 145 divided by 5 = 29 full bags.

The answer is 29. The tempting 30 divides the original 152 oranges by 5 and rounds down, ignoring the damaged fruit. The 145 stops before bagging, 152 stops before either later step, and 7 copies the number removed.

B

Three decimal places means 0.425 = 425/1000. Divide top and bottom by 25 to get 17/40. Seventeen and forty have no common factor greater than 1, so this is simplest form.

The tempting 425/1000 has the correct value but is not simplified. The 17/400 divides the numerator and denominator by different amounts, 42/5 treats 0.425 as 8.4, and 4 1/4 reads the digits as four and a quarter.

A

From 06:20 to 09:05 is 2 hours 45 minutes, or 165 minutes. This contains 165 divided by 15 = 11 intervals. There is a recording at both ends, so 11 intervals need 12 recordings.

The tempting 11 counts the gaps but forgets the starting recording. The 10 omits both endpoints, 15 copies the interval length, and 165 gives the duration in minutes instead of the number of recordings.

E

Eliminate 120 because a 90 degree sector is only one quarter of the circle, not three quarters. Since 90/360 = 1/4, the sector represents 160 divided by 4 = 40 children's books. Subtract the 15 needing repair: 40 - 15 = 25.

The tempting 40 is the sector total before repairs are removed. The 15 copies the repair count, 55 adds instead of subtracting, and 120 counts all the books outside the sector.

B

Eliminate 20 because that is the original blue group, not the red group. The red side stays at 3 parts in both ratios, so the part size can stay the same. Blue falls from 5 parts to 2 parts, a drop of 3 parts. Those 3 parts equal 12 badges, so one part is 4 and the red group is 3 x 4 = 12.

The tempting 4 is one ratio part and 8 is the blue group after the giveaway. The 20 is the blue group before it, while 32 adds the original red and blue groups.

C

The bicycles cost 4 x 13.50 = 54 pounds. The helmets cost 3 x 4.25 = 12.75 pounds. Add them and then use the voucher: 54 + 12.75 - 8 = 58.75 pounds.

The tempting 54 pounds is the bicycle cost only. The 66.75 pounds is the total before the voucher, 74.75 pounds adds the voucher instead of subtracting it, and 46 pounds subtracts the voucher from the bicycles but forgets the helmets.

A

Children make up 35% of 240, which is 84. That leaves 240 - 84 = 156 adults. One quarter of 156 is 156 divided by 4 = 39 adults using a computer.

The tempting 84 is the number of children and 156 is the number of adults before finding a quarter. The 60 is a quarter of all visitors, while 21 is a quarter of the children rather than the adults.

D

Use elimination by counting two short lists. Ten labels end in 7: 7, 17, up to 97. Ten labels have 7 in the tens place: 70 to 79. That seems to make 20, but 77 is in both lists and must be counted only once, giving 19.

The tempting 20 double-counts 77. The 10 counts only one digit position, 18 removes 77 twice instead of once, and 9 misses either 7 itself or one number from the block 70 to 79.

C

The number of sides is 72 divided by 9 = 8, so the shape is a regular octagon. A regular polygon has one line of symmetry for each side, so it has 8 lines of symmetry.

The tempting 9 copies the side length and 72 copies the perimeter. The 4 halves the number of sides, while 6 comes from dividing the perimeter by 12 instead of by the given side length.

A

The whole courtyard area is 18 x 12 = 216 square metres. A 1 m path on both sides leaves a central rectangle 16 m by 10 m, with area 160 square metres. The path area is 216 - 160 = 56 square metres.

The tempting 160 square metres is the planted centre and 216 square metres is the whole courtyard. The 60 is the courtyard perimeter, while 29 comes from shrinking each dimension by only 1 m, using 17 x 11, instead of removing 1 m at both ends.

B

Enclose the L-shape in a rectangle from x = 1 to x = 9 and y = 1 to y = 7. Its area is 8 x 6 = 48 square units. The missing top-right rectangle is 5 units wide and 4 units high, so its area is 20. Therefore the L-shape area is 48 - 20 = 28 square units.

The tempting 20 is the cut-out area and 48 is the enclosing rectangle. The 68 adds the cut-out instead of subtracting it, while 32 subtracts only 16 square units after using 4 rather than 5 for the cut-out width.

E

Use elimination before doing the long multiplication. The answer is roughly 170 x 24 divided by 6, which is about 680 trays. Also, after division by 6, multiplying the correct option back by 6 must end in 4 because the unbroken total does. Of the sensible-sized options, 689 passes that last-digit check. Fully: 173 x 24 = 4152, then 4152 - 18 = 4134, and 4134 divided by 6 = 689.

The tempting 692 divides all 4152 jars before removing breakages. The 674 subtracts 18 after converting jars into trays, 4134 is the number of unbroken jars rather than trays, and 18 copies the breakage count.

D

The original range is 28 - 18 = 10 minutes. Adding 22 keeps the smallest and largest values unchanged, so the range stays 10. It also makes 22 occur twice, more often than any other time, so 22 becomes the mode.

The tempting 18 or 24 would keep the range but would make that chosen number the mode instead. Adding 12 or 30 would change the range to 16 or 12 minutes, and neither would make 22 the mode.