Sutton SET Mathematics Mock Paper 9

A

Elimination shortcut: the number must begin 40 to be as small as possible, and its last two digits must form a multiple of 4. This removes C because 98 is not divisible by 4, and D and E are already too large in their first three digits.

With 4, 0 and 3 fixed first, test the remaining digits in increasing arrangements. The smallest ending that makes the last two digits divisible by 4 is 896, giving 403 896.

B is divisible by 4 but is larger because its final three digits are 968. C fails the divisibility test, while D puts 3 before 0 and E places large digits too early.

B

Elimination shortcut: 18 x 24 ends in 2, and subtracting 35 must give an answer ending in 7. Only 397 has that last digit.

Fully, 18 x 24 = 432, then 432 - 35 = 397 usable sheets.

362 subtracts 35 twice, 432 forgets the discarded sheets, 467 adds 35, and 15 120 multiplies the number produced by the discarded amount.

C

Elimination shortcut: the halfway value must be strictly between 35% and 60%, so A and D are boundary values and E is far too large. This leaves B or C.

Write 35% as 7/20 and 3/5 as 12/20. Their sum is 19/20, and half of that is 19/40.

A copies the lower boundary, B rounds 35% down to 30% before averaging, D copies the upper boundary, and E adds the two boundaries without dividing by 2.

D

Elimination shortcut: the original total has 2 + 5 = 7 equal parts, so it must be a multiple of 7. Only 21 fits.

The extra 6 adults account for the rise from 2 parts to 4 parts. Two parts equal 6, so one part is 3. The original total is 7 x 3 = 21.

A is one part, B copies the number joining, C is the original child group, and E is the total after the 6 adults arrive.

E

Elimination shortcut: the final amount must be larger than both 1350 ml already present and the 1260 ml added. Only 1390 ml and 2610 ml survive that size check, and a sum of about 1.4 litres and 1.3 litres must be about 2.7 litres.

Convert 1.35 litres to 1350 ml. The bottles add 3 x 420 = 1260 ml, so the container holds 1350 + 1260 = 2610 ml.

A finds the difference between 1350 and 1260, B gives only the added water, C gives only the starting water, and D is the empty space in a 4 litre container, although no 4 litre capacity was asked for.

B

Elimination shortcut: a convex interior angle must be less than 180 degrees, so C, D and E cannot be the missing angle. The known angles already total more than 360 degrees, so A is the trap from using a quadrilateral total.

A pentagon's interior angles total 540 degrees. The four known angles total 88 + 104 + 117 + 96 = 405 degrees, so the missing angle is 540 - 405 = 135 degrees.

A uses 360 degrees and ignores the negative result, C copies a triangle's total, D is the sum of the known angles, and E is the total for the whole pentagon.

C

Elimination shortcut: after setting aside 17, there are 1358 programmes. Since 28 x 50 = 1400, the answer is just under 50 boxes, leaving only 48 or 49.

Now 28 x 48 = 1344, leaving 14 programmes, while 28 x 49 = 1372 is too many. Therefore 48 complete boxes are filled.

A copies the proof count, B copies the box size, D rounds the division up and would require too many programmes, and E gives the number of programmes available rather than the number of boxes.

D

Elimination shortcut: the increase is 75% - 60% = 15% of the run. If 15% is 18 pages, the whole must be several times 18 and only 120 gives exactly 15% as 18.

Five per cent is 18 divided by 3 = 6 pages, so 100% is 20 x 6 = 120 pages.

A mistakes the increase for the whole, B is the original 60%, C is the final 75%, and E treats 18 as 6% rather than 15%.

E

Elimination shortcut: a number rounding to 64 000 can be at most 64 499. To make it as large as possible without repeating 6 and 4, choose the largest available hundreds digit, 3, then the largest remaining tens and units digits, 9 and 8.

This forms 64 398, which has distinct digits and rounds down to 64 000.

A and C repeat 0, B is valid but begins 63 and is smaller, and D swaps the final 9 and 8 into the smaller order.

A

Elimination shortcut: the full session is a little over 2 1/2 hours, and removing 25 minutes leaves a little over 2 hours. Only 126 minutes has the right size.

From 22:47 to midnight is 73 minutes, and midnight to 01:18 is 78 minutes. The full session is 151 minutes, so observations last 151 - 25 = 126 minutes.

B includes the cloud break, C adds the break, D adds 2 hours 47 minutes to 1 hour 18 minutes, and E subtracts the clock displays as if they were ordinary base-ten numbers.

C

Elimination shortcut: the final left move makes the x-coordinate 4 - 5 = -1, so A and E are impossible. The reflection puts the point below y = -2, and the upward move cannot take it to positive 5, removing D.

The point is 5 squares above y = -2, so its reflection is 5 squares below at (4, -7). Translating gives (4 - 5, -7 + 2) = (-1, -5).

A reflects the x-coordinate as well, B moves down instead of up after reflecting, D reflects in the x-axis rather than y = -2, and E moves right instead of left.

D

Elimination shortcut: the capacity is about 300 + 170 = 470, so after removing 47 the answer should be a little over 420. Only 427 fits that rough size.

The two blocks contain 9 x 34 = 306 and 6 x 28 = 168 places. Total capacity is 474, and 474 - 47 = 427 people are seated.

A subtracts the two block capacities, B and C give only one block, and E adds the 47 empty places to the capacity.

E

Elimination shortcut: each of A to D is the total obtained by leaving out one positive amount, so the complete total must be larger than 2.275 litres. Only E can include all four amounts.

Line up the decimal points: 0.125 + 0.480 + 0.705 + 1.090 = 2.400 litres, or 2.4 litres.

A omits 1.09 litres, B omits 0.705 litres, C omits 0.48 litres, and D omits 0.125 litres.

A

Elimination shortcut: the cyan amount does not change, so its 5 parts keep the same size. Magenta rises by exactly 1 part, from 3 parts to 4, and that rise is the 12 ml added. One part is therefore 12 ml.

The original magenta amount is 3 x 12 = 36 ml, the new magenta amount is 48 ml, the cyan amount is 60 ml, and the original total is 96 ml. Those are exactly what B, C, D and E mistakenly report instead of one part.

B

Elimination shortcut: the two equal base angles together equal the exterior angle of 124 degrees, so each must be half of 124. This points straight to 62 degrees.

Equally, the interior vertex angle is 180 - 124 = 56 degrees. The base angles total 180 - 56 = 124 degrees, and 124 divided by 2 is 62 degrees.

A halves the 56 degree vertex angle, C treats two 56 degree angles as known and subtracts them from 180, D forgets to halve the exterior angle, and E copies the triangle's angle total.

D

Elimination shortcut: the four known readings average 41 cm because they total 164 cm. To raise the mean of all five to 42 cm, the missing reading must be above 42 cm. Of the sensible readings, 46 cm is the only one that can do this.

The required total is 5 x 42 = 210 cm. The known total is 35 + 39 + 44 + 46 = 164 cm, leaving 210 - 164 = 46 cm.

A is the range of the known readings, B copies the mean, C copies one listed reading, and E gives the required total rather than the missing reading.

E

Elimination shortcut: 14 x 6 and 9 x 4 both end in even digits, so their sum is even. Adding 5 must make a total ending in 5, and only 125 does.

The cast needs 14 x 6 = 84 pages and the crew needs 9 x 4 = 36 pages. Including spares gives 84 + 36 + 5 = 125 pages.

A counts only crew pages, B only cast pages, C omits the spares, and D treats five spare pages as one spare page.

A

Elimination shortcut: a common factor cannot be greater than the smaller number, so D and E are impossible. Also 99 does not divide 132 and 132 does not divide 198, leaving 66.

132 = 2 x 66 and 198 = 3 x 66, so 66 is a factor of both and is the greatest one.

B is half of 198 only, C is the whole smaller number but not a factor of 198, D copies the larger number, and E adds the two numbers.

B

Elimination shortcut: 37.5% is 3/8, so slightly more than one third of 320 is used. About 200 remain, and adding 28 should give a little over 220. Only 228 fits.

One eighth of 320 is 40, so 3/8 is 120. There are 320 - 120 = 200 left, then 200 + 28 = 228.

A subtracts the 28 new dressings, C subtracts only 28 from the original stock, D adds 28 but ignores those used, and E adds both the used and new dressings to 320.

C

Elimination shortcut: the contents are three quarters of the total mass, so the packaging should be somewhat under 1 kg. Only 850 g has that size and the requested unit.

Convert 3.6 kg to 3600 g and 2 kg 750 g to 2750 g. The packaging mass is 3600 - 2750 = 850 g.

A has the right amount in kilograms but labels it grams, B divides by 10 once too often, D copies the contents mass, and E adds the contents and total masses.

E

Elimination shortcut: there are 10 vertices, but a diagonal cannot join the chosen vertex to itself or to either of its 2 neighbours. Remove those 3 vertices: 10 - 3 = 7.

A subtracts 7 rather than 3, B halves the eight other vertices, C simply halves 10, and D wrongly excludes the opposite vertex as well as the two neighbours.

A

Elimination shortcut: 24 x 100 = 2400 and 24 x 30 = 720, so the quotient is a little under 130. Only 126 m is possible.

Indeed, 24 x 126 = 24 x 100 + 24 x 20 + 24 x 6 = 2400 + 480 + 144 = 3024.

B divides by 6, the sum of the divisor's digits, C divides only by 4, D divides only by 2, and E multiplies by 24 instead of dividing.

B

Elimination shortcut: one width plus one height is half the perimeter, 33 m. The width is the larger share but less than 33 m, leaving only 21 m.

There are 7 + 4 = 11 parts in 33 m, so one part is 3 m. The width is 7 x 3 = 21 m.

A is the height, C is width plus height, D doubles the width, and E copies the whole perimeter.

C

Elimination shortcut: if 3/8 is 90, then 1/8 is 30 and the full run is 240. Five eighths remain, which is 150, so 20% of the remainder must be 30.

A finds 20% of the 90 already identified, B finds 10% of the whole run, D finds 20% of the whole run, and E stops at the number remaining.

D

Elimination shortcut: the whole floor is just under 100 square metres and the unpainted square is only 9 square metres, so the painted area must be in the high eighties. Only 87 square metres fits.

The floor area is 12 x 8 = 96 square metres. The trapdoor area is 3 x 3 = 9 square metres, leaving 96 - 9 = 87 square metres.

A gives only the trapdoor, B gives the floor perimeter, C treats the trapdoor as a 3 m by 8 m strip, and E forgets to remove the trapdoor.

A

Elimination shortcut: the minimum stays 14, while the maximum falls from 37 to 24. The decrease in the range is therefore the same as the fall in the maximum: 37 - 24 = 13.

Fully, the old range is 37 - 14 = 23 and the new range is 24 - 14 = 10. It decreases by 23 - 10 = 13.

B copies the minimum, C is the old range, D is the new maximum, and E is the removed score.

B

Elimination shortcut: read the subtractive pairs. CM is 1000 - 100 = 900, XL is 50 - 10 = 40, and IV is 5 - 1 = 4. Their sum must end in 44, leaving 944 or 1044; CM settles it as 944.

A subtracts C twice in CM, C reverses IV into VI, D reads CM as 1000, and E adds every symbol separately instead of using the three subtractive pairs.

C

Elimination shortcut: 286 is just under 300 and 37 is just under 40, so the product should be a little under 12 000. Also 6 x 7 ends in 2. Only 10 582 has both the right size and last digit.

Fully, 286 x 30 = 8580 and 286 x 7 = 2002. Adding gives 10 582.

A and B are the two partial products left uncombined, D replaces 37 by 40 and treats the estimate as exact, and E adds an extra zero to the correct product.

D

Twelve and a half per cent is 1/8. One eighth of 1.6 m is 0.2 m, so the depth becomes 1.6 + 0.2 - 0.15 = 1.65 m.

A is only the net rise, B subtracts the percentage rise, C applies only the fall, and E adds the 0.15 m fall instead of subtracting it.

E

A full turn around any polygon is 360 degrees. With turns of 45 degrees, the number of sides is 360 divided by 45 = 8, so the polygon is a regular octagon.

The shapes in A, B, C and D would complete only 3, 4, 5 or 6 of the required eight 45 degree turns, giving totals of 135, 180, 225 or 270 degrees instead of 360 degrees.

B

The two complete columns contain 2 x 4 = 8 shaded squares. Three more are shaded, giving 11 out of 24, or 11/24.

A counts only the two full columns, C assumes there is one extra in all four rows, D counts five extras instead of three, and E reports the fraction in the four columns outside the first two before considering any actual shading.

C

The press produces 1260 divided by 18 = 70 pages per minute. It runs for 40 - 4 = 36 minutes, so it produces 70 x 36 = 2520 pages.

A gives one minute's output, B gives only the original 18-minute output, D ignores the pause, and E doubles the correct output because 36 minutes is twice 18 minutes and then doubles once more.

D

Nine pairs are 18 gloves, leaving 900 - 18 = 882. Each kit receives 882 divided by 18 = 49 gloves, then one is used, leaving 48.

A copies the number of pairs, B gives the number of gloves kept at the desk, C removes a pair rather than one glove from each kit, and E stops before the practice glove is used.

E

Multiply both numbers by 100: 2.4 divided by 0.08 is the same as 240 divided by 8, which is 30.

A reverses the division and truncates 1/30, B and C place the decimal point two places or one place too far left, and D rounds 0.08 to 0.1 and uses that estimate as an exact divisor.

A

The original perimeter is 2 x (14 + 9) = 46 cm. Cutting from a corner removes lengths of 5 cm and 3 cm from the outside but adds new cut edges of exactly 5 cm and 3 cm, so the perimeter stays 46 cm.

B adds the two 3 cm edges without removing the old edge, C does the same with the two 5 cm edges, D adds the whole 16 cm perimeter of the cut-out, and E gives the remaining area, 126 - 15, in the wrong kind of unit.

C

The last two digits are 28, so divisibility by 4 is already satisfied. For divisibility by 9, the digit sum 4 + ? + 2 + 8 = 14 + ? must be a multiple of 9. Only ? = 4 makes 18.

A, B, D and E make digit sums of 16, 17, 20 and 22, so each fails the divisibility-by-9 condition even though the ending 28 passes the divisibility-by-4 condition.

D

The diagonals of every rectangle are equal in length. A sloping general parallelogram can have diagonals of different lengths, so D distinguishes the rectangle.

A is guaranteed for a rhombus but not a rectangle, B is true of every parallelogram and therefore does not distinguish one, C is true for a square but not every rectangle, and E is not guaranteed even for a square.

E

Pair the outside numbers: 37 + 63 = 100, 38 + 62 = 100, and so on. There are 13 such pairs, totalling 1300, with the middle number 50 left. The sum is 1350.

A is the count of numbers, B is the middle number, C is one pair, and D forgets to add the unpaired middle number.

A

Seven tenths multiplied by eight hundredths is 56 thousandths, which is 0.056.

B has one decimal place too few, C adds the two decimals instead of multiplying, D moves the decimal point two places too far right, and E ignores both decimal place values.

B

The base area is 30 x 20 = 600 square centimetres. Height is volume divided by base area: 18 000 divided by 600 = 30 cm.

A divides by 100 once too often, C divides by 30 and then by 10 instead of 20, D divides by only one base dimension, and E multiplies the volume by the base area.

D

Of the 65 people who pass, 31 are adults, so 65 - 31 = 34 children pass.

A is the number of adults who fail, B is the number of children who fail, C copies the adults who pass, and E is the total number of children whether they pass or fail.

E

The difference between 7 parts and 2 parts is 5 parts, and these equal 25 m. One part is 5 m. The total is 2 + 3 + 7 = 12 parts, so it is 12 x 5 = 60 m.

A, B and D are the individual short, middle and long stretches. C copies the difference between the longest and shortest instead of adding all three.

A

The four numbers have an average of 110 divided by 4 = 27.5, so they lie equally around 27.5: 26, 27, 28 and 29. The largest is 29.

B divides the total by 2, C subtracts 4 from the total, D copies the total, and E multiplies the total by the number of terms.

B

Use denominator 24: 5/6 = 20/24, 1/4 = 6/24 and 1/8 = 3/24. Therefore 20/24 - 6/24 + 3/24 = 17/24, which is already in simplest form.

A stops after the subtraction, C converts 1/8 incorrectly as 5/24, D treats 1/4 as 1/24, and E adds all three fractions instead of subtracting the quarter.

C

From 400 to 499, the tens and units must not be 4, giving 9 x 9 = 81 numbers. In each of the 200s and 300s, the 4 can be in the tens or units place, giving 9 + 9 = 18 numbers per hundred. The total is 81 + 18 + 18 = 117.

A counts only one of the 200s or 300s, B counts both but omits the 400s, D also includes 244 and 344 by double-counting the two possible digit positions, and E counts all 100 numbers in the 400s even when another 4 appears.

E

A triangular prism has two matching triangular ends joined by three rectangular faces, so E is exact.

A describes a tetrahedron, B a square-based pyramid, C a pentagonal prism, and D a cube.

A

Nine marks make 8 gaps. Their total length is 8 x 25 = 200 mm, which is 20 cm.

B counts 9 gaps before converting, C gives one 25 mm interval but labels it centimetres, D has the correct number of millimetres with the wrong unit, and E both counts 9 gaps and keeps the millimetre number under a centimetre label.

B

The horizontal base is 6 - (-2) = 8 units and the perpendicular height is 6 - 1 = 5 units. The area is 1/2 x 8 x 5 = 20 square units.

A gives only the height, C gives the 8 by 5 enclosing rectangle without halving, D uses the upper y-coordinate 6 as the height, and E uses that wrong height and doubles instead of halving.

C

Both products contain groups of 489. The difference between 63 groups and 58 groups is 5 groups, so the answer is 489 x 5 = 2445.

A gives only the difference between the multipliers, B gives one group, D is the smaller product, and E is the larger product.

D

The changes are a rise of 0.6 m, a fall of 0.3 m and a rise of 0.6 m. Their total is 0.6 + 0.3 + 0.6 = 1.5 m.

A gives only the fall, B gives one rise, C gives the net change from first to last, and E copies the final reading.