Work through the levels in order. Type your answer and press Check — open the worked solution if you get stuck. Answers like -6, 3/4 or 0.25 are typed exactly as shown. A ⭐ Bonus Challenge at the end is extra credit — skip it if the rest was hard work.
Level 1 · Warm-up
1 Two consecutive positive integers multiply to 156. What is the larger of the two?
Show worked solution
n2 + n − 156 = 0, so (n + 13)(n − 12) = 0. The integers are positive, so n = 12 and the larger is 13. Check: 12 × 13 = 156 ✓. −12 is the larger of the negative pair −13 and −12, which "positive" rules out; 12 is the smaller; 14 adds 2, as for odd or even numbers.
2 A rectangle’s length is 7 cm more than its width, and its area is 78 cm2. Find its width and length.
Show worked solution
w2 + 7w − 78 = 0, so (w + 13)(w − 6) = 0. w = −13 is rejected because a width cannot be negative. Width 6 cm, length 13 cm. Check: 6 × 13 = 78 ✓.
3 The square of a positive number is 30 more than the number itself. Find the number.
Show worked solution
n2 − n − 30 = 0, so (n − 6)(n + 5) = 0 and n = 6 or n = −5. The number is positive, so it is 6. Check: 36 = 6 + 30 ✓.
Level 2 · Core
4 A 12 m by 8 m pool has a paved border x m wide all the way round it. The border has the same area as the pool. How wide is the border?
Show worked solution
(12 + 2x)(8 + 2x) − 96 = 96, so 4x2 + 40x − 96 = 0, x2 + 10x − 24 = 0 and (x + 12)(x − 2) = 0. A width cannot be negative, so x = 2 m. Check: 16 × 12 = 192 = 96 + 96 ✓. 4 m puts the border on one side only, (12 + x)(8 + x) − 96 = 96; 12 m keeps the rejected root without its sign; 24 m does both.
5 Two consecutive even integers multiply to 168. Find both possible values of the smaller integer.
Show worked solution
n(n + 2) = 168, so n2 + 2n − 168 = 0 and (n + 14)(n − 12) = 0. Smaller root −14, larger root 12. Both work: 12 × 14 = 168 and (−14) × (−12) = 168.
6 A triangle’s height is 6 cm less than its base, and its area is 56 cm2. Find the base, in cm.
Show worked solution
½b(b − 6) = 56, so b2 − 6b − 112 = 0 and (b − 14)(b + 8) = 0. A base cannot be negative, so b = 14. Check: height 8 and ½ × 14 × 8 = 56 ✓.
Level 3 · Challenge
7 A 9 cm by 5 cm photo sits in a frame x cm wide all the way round. Photo and frame together cover 165 cm2.
1 (9 + x)(5 + x) = 165 describes it.
2 x2 + 7x − 30 = 0
3 The frame is 3 cm wide.
Which statements are correct?
1 (9 + x)(5 + x) = 165 describes it.
2 x2 + 7x − 30 = 0
3 The frame is 3 cm wide.
Which statements are correct?
Show worked solution
The outer size is (9 + 2x) by (5 + 2x), not (9 + x) by (5 + x) (1 ✗). (9 + 2x)(5 + 2x) = 165 gives 4x2 + 28x − 120 = 0, and ÷ 4 gives x2 + 7x − 30 = 0 (2 ✓). (x + 10)(x − 3) = 0, and −10 is rejected, so x = 3 (3 ✓). Check: 15 × 11 = 165. 2 and 3.
8 Two numbers differ by 5, and their squares add to 73. Find both possible values of the smaller number.
Show worked solution
x2 + (x + 5)2 = 73, so 2x2 + 10x − 48 = 0, x2 + 5x − 24 = 0 and (x + 8)(x − 3) = 0. Smaller root −8, larger root 3. Both work: 3 and 8 give 9 + 64 = 73, and −8 and −3 give 64 + 9 = 73.
9 A rectangle has a perimeter of 34 cm and a diagonal of 13 cm. Find its area, in cm2.
Show worked solution
x2 + (17 − x)2 = 169, so 2x2 − 34x + 120 = 0, x2 − 17x + 60 = 0 and (x − 5)(x − 12) = 0. Both roots give the same rectangle, 5 cm by 12 cm, so the area is 60 cm2. Check: 2(5 + 12) = 34 and 52 + 122 = 169 = 132 ✓.