Scholarship Paper 2 (ACER Level 2/3 form)

6 ÷ 1.5 = 4 and 10⁹ ÷ 10⁴ = 10⁵, so the answer is 4 × 10⁵. (4.5 × 10⁵ subtracts 1.5 from 6 instead of dividing; 9 × 10¹³ multiplies the two numbers.)

135/360 = 3/8, and 3/8 of 120 is 45. (30 treats the bus sector as a quarter; 60 treats it as a half — the angle says it is neither.)

The minute hand is 180° round from 12. The hour hand is 2½ hour-marks round: 2.5 × 30° = 75°. The angle between them is 180° − 75° = 105. (120 leaves the hour hand on the 2; 75 is just the hour hand's own position.)

100 = 14 × 7 + 2, so after 14 whole weeks it is Wednesday again, and 2 more days gives Friday. (Thursday counts today as the first of the 100 days.)

4 × 2¹⁰ = 2² × 2¹⁰ = 2¹², so n = 12. (40 multiplies the index by 4; 14 adds the 4 to the index; 20 adds two of the indices.)

The exterior angles add to 360°, so there are 360 ÷ 24 = 15 sides. (24 takes the angle for the number of sides; 12 divides 360 by 30, the exterior angle of a 12-gon.)

One printer prints 600 ÷ 3 ÷ 4 = 50 pages a minute, so 5 printers print 250 a minute. 1500 ÷ 250 = 6 minutes. (10 scales up the pages but forgets the extra printers; 4 assumes the time stays the same.)

Juice: 30% of 2 L = 0.6 L, and 80% of 3 L = 2.4 L, so 3 L of juice in 5 L: 60%. (55% averages 30 and 80, ignoring that there is more of the stronger drink.)

Four folds make 2 × 2 × 2 × 2 = 16 equal sections, and 16 sections have 15 lines between them. (16 counts the sections, not the creases; 31 and 32 come from five folds.)

Total = 5 × 38 = 190 mm. The bars show 35 + 50 + 20 + 45 = 150 mm, so May had 190 − 150 = 40 mm. (38 is the mean, not the missing value.)

Values 1–2 are 0, values 3–9 are 1, values 10–13 are 2. The 10th and 11th are both 2, so the median is 2. (1 is the mode, 1.95 is the mean, and 4 is the median of the "matches" row, which is the wrong list.)

From C, P is 1 right and 3 up. After a quarter turn clockwise that becomes 3 right and 1 down: from C(1, 1) that is (4, 0). ((−2, 2) turns anticlockwise; (0, −2) is a half turn; (4, −2) turns about the origin instead of C.)

30 more guests cost $480, so each guest is $16 and the fee is 460 − 20 × 16 = $140. 30 guests cost 140 + 30 × 16 = $620. ($690 scales $460 up by 30/20, counting the fee 1½ times; $700 is halfway between the two prices, but 30 is not halfway between 20 and 50.)

A is 50 and B is 70: an increase of 20 on 50, which is 20/50 = 40%. (The bar for B looks three times as tall because the axis starts at 40 — that gives the false 200%; 20% is the difference read as a percentage.)

Year 10 bus = 40 − 9 − 17 = 14. Bus total = 18 + 14 = 32. So the probability is 14/32 = 7/16. (7/40 divides by all 80 students; 7/20 divides by the 40 in Year 10; 7/9 compares with Year 9's bus riders only.)

The gap is 360° − 90° − 120° = 150°. Its exterior angle is 180° − 150° = 30°, and 360° ÷ 30° = 12 sides. (8 sides gives 135°, which fits with a square and another octagon, not with a hexagon.)

With 5 on the table, 4 and 2 stand up as walls. 3 folds round from 4 to make a third wall, and 6 folds round from 2 to make the fourth. That leaves 1 as the lid, opposite 5. (3 and 6 look far from 5 on the net, but they end up as walls next to it.)

The surrounding rectangle runs from x = 1 to 7 and y = 1 to 7: 6 × 6 = 36. The corner triangles are ½ × 6 × 2 = 6, ½ × 3 × 4 = 6 and ½ × 3 × 6 = 9. 36 − 21 = 15. (18 is half the rectangle, which is only right when a side of the triangle lies along the rectangle.)

She rides 20 km out and 20 km back: 40 km in 5 hours, which is 8 km/h. (4 uses only the 20 km out; 10 leaves out the two half-hour stops; 5 makes both mistakes.)

Prime totals: 1+1 = 2, 1+2 = 3, 1+4 = 5, 2+1 = 3, 2+3 = 5, 3+2 = 5, 3+4 = 7 — that is 7 of the 12 pairs, 7/12. (1/2 forgets that 2 is prime.)

A is on top in every view, so B, C, D and E are the four faces around A (and F is opposite A). Going round: B is next to C, C next to D, D next to E — so the ring is B, C, D, E and the face opposite D is B. (C and E are seen beside D; F is opposite A, not D.)

From B to C is 1 right and 3 up. Doing the same from A(1, 1) gives D = (2, 4). ((10, 6) makes a parallelogram ABDC with BC as a diagonal; (0, −2) makes one with AB as a diagonal — neither has the corners in the order ABCD.)

The 5 × 4 faces give 3 × 2 = 6 each, the 5 × 3 faces give 3 × 1 = 3 each, the 4 × 3 faces give 2 × 1 = 2 each. Total 2 × (6 + 3 + 2) = 22. (24 is the count with exactly two painted faces, along the edges; 8 is the corners; 6 is the unpainted ones inside.)

1.5 × 10¹¹ ÷ (3 × 10⁸) = 0.5 × 10³ = 500 seconds = 8 minutes 20 seconds. (50 seconds slips a power of ten; 5 minutes reads 500 seconds as 5 hundred-seconds.)

2⁶⁰ = (2³)²⁰ = 8²⁰ and 3⁴⁰ = (3²)²⁰ = 9²⁰. Comparing 5²⁰, 8²⁰ and 9²⁰ gives 5²⁰, 2⁶⁰, 3⁴⁰. (Ordering by the index alone puts 2⁶⁰ last; ordering by the base puts 5²⁰ last.)

Folding keeps angles, so before the fold FG also made 65° with the bottom edge on the right of F. The angles at F along the straight bottom edge are x, 65° and 65°, so x = 180° − 130° = 50°. (115° is 180° − 65°, which uses only one of the two 65° angles; 65° assumes x equals the marked angle.)

2¹⁰ × 5¹³ = (2 × 5)¹⁰ × 5³ = 125 × 10¹⁰: the digits 125 followed by ten zeros, 13 digits. (11 counts only 10¹⁰; 23 adds the indices.)

The middle square is ½ of the board, so the four shaded corners make the other ½. The smallest square is ½ of ½ = ¼ of the board. Shaded: ½ + ¼ = 3/4. (2/3 counts the three regions as if they were equal in size.)

A route is 4 rights and 3 ups in some order: 35 routes. A to P (2 right, 1 up) has 3 routes; P to B (2 right, 2 up) has 6; so 3 × 6 = 18 go through P. 35 − 18 = 17. (18 is the routes that DO pass through P; 26 adds 3 + 6 instead of multiplying.)

Since 3w = 2l, take w = 2k and l = 3k. The large rectangle is 3w = 6k wide and l + w = 5k tall, so its perimeter is 22k = 66 and k = 3: w = 6, l = 9. The large rectangle is 18 by 15, area 270 cm². (54 cm² is one small rectangle; 330 cm² is 5 × 66.)

The numbers are a, b, 7, 8, 8 with a + b = 30 − 23 = 7 and both below 7 (if b were 7, then 7 would appear as often as 8). The smallest a can be is 1 (with b = 6), so the range is 8 − 1 = 7. (8 takes a = 0 and b = 7, which is neither positive nor allowed by the mode.)

P goes 240 km in 3 hours: 80 km/h. Q goes 200 km in 2 hours: 100 km/h. At 11:00 the gap is 40 km, and Q closes it at 100 − 80 = 20 km/h, taking 2 hours: 1:00 pm. (12:00 noon closes the gap at 40 km/h, as if P had stopped at 11:00 and Q were twice as fast.)

The diagonal of the 12 by 8 rectangle goes through the grid corners (3, 2), (6, 4) and (9, 6), so it is four copies of the 3 by 2 diagonal, end to end: 4 × 4 = 16 squares. (20 is 12 + 8, which ignores that the line crosses a vertical and a horizontal grid line at the same moment at those corners.)

The columns are 3, 1, 2 tall and the rows 2, 3, 1. The tallest each pile can be is the smaller of its column and row heights: column 3 gives 2 + 3 + 1 = 6, column 1 gives 1 + 1 + 1 = 3, column 2 gives 2 + 2 + 1 = 5. Total 14, and these piles do give both views. (18 copies the front view into every row, which makes the side view wrong; 12 adds the cubes in the two views.)

Both dice 4 or less: 4 × 4 = 16 pairs. Both 3 or less: 3 × 3 = 9 pairs. So 16 − 9 = 7 pairs have 4 as the larger number: 7/36. (2/9 = 8/36 counts the double 4 twice; 1/6 leaves it out.)