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Ecolint Campus des NationsMathematics
Year 7 · 7.1 Positive Integers

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.The missing digit is 4 (tens place of 347)
2.2 400; the tens digit (6) decided — it is 5 or more, so round up
3.168; check: 355 + 168 = 523 ✓
4.Closer to 210; 30 × 6 = 180 and 40 × 6 = 240, so 34 × 6 is between, nearer 210
5.Yes — 4728 ÷ 4 = 1182 with no remainder
6.Yes — 84 ÷ 4 = 21 exactly
7.E.g. 4 — because 36 ÷ 4 = 9 (a whole number) ✓
8.42; because 42 = 7 × 6 and 35 = 7 × 5 which is not greater than 40
9.No — 91 = 7 × 13
10.Less than 100; because 10² = 100 and 9 < 10, so 9² < 10²
Silver
11.3 470
12.4 812
13.2 375
14.264
15.29
16.12
17.24
18.64
19.22×72^2 \times 7
20.£42
Gold
21.42
22.6
23.22×3×72^2 \times 3 \times 7
24.6
25.23 × 24 is larger
26.48
27.12 (or 6, 60)
28.1 000 (estimate)
29.£2.10
30.15
Platinum
31.HCF = 12, LCM = 2 520
32.5
33.A = 8, B = 4 (since 48 × 3 = 144)
34.(6, 60) and (12, 30)
35.No — 91 = 7 × 13
36.16 (no — not prime), 25 (no), 34 (no), 43 ✓, 61 ✓, 70 (no). Answer: 43 and 61.
37.a = 8, b = 7
38.Any two consecutive even numbers are 2k and 2(k+1). Their product is 4k(k+1). Since k and k+1 are consecutive integers, one of them is even, so k(k+1) is even. Therefore 4k(k+1) = 4 × (even) = 8 × (integer). Always divisible by 8.
39.3.5
40.£70

Pack B — Answers

Bronze
1.The missing digit is 6 (tens place of 463)
2.8 000; the hundreds digit (8) decided — it is 5 or more, so round up
3.293; check: 448 + 293 = 741 ✓
4.Closer to 400; 50 × 8 = 400 and 40 × 8 = 320, so 47 × 8 is between, nearer 400
5.Yes — 5913 ÷ 3 = 1971 with no remainder
6.Yes — 135 ÷ 9 = 15 exactly
7.E.g. 6 — because 48 ÷ 6 = 8 (a whole number) ✓
8.54; because 54 = 9 × 6 and 45 = 9 × 5 which is not greater than 50
9.Yes — 43 has no factors other than 1 and 43
10.Less than 50; because 7² = 49 and 8² = 64, so 7² is between 49 and 64, below 50
Silver
11.8 400
12.8 153
13.3 247
14.276
15.40
16.6
17.20
18.125
19.2×3×52 \times 3 \times 5
20.£84
Gold
21.37
22.4
23.23×3×52^3 \times 3 \times 5
24.7
25.32 × 33 is larger
26.180
27.8 (or 4, 12, 16, 24, 48)
28.1 800 (estimate)
29.£1.25
30.13
Platinum
31.HCF = 6, LCM = 1 260
32.13
33.A = 8, B = 4 (same reasoning)
34.(4, 48) and (12, 16)
35.Yes — 97 is prime
36.17 (1+7=8, prime ✓), 53 (5+3=8, prime ✓), 71 (7+1=8, prime ✓). Answer: 17, 53, 71.
37.a = 12, b = 5
38.Examples: 2×4=8 ✓; 4×6=24=8×3 ✓; 6×8=48=8×6 ✓. General: 2k × 2(k+1) = 4k(k+1). Consecutive integers k, k+1 → one is even → k(k+1) = 2m for some integer m → product = 8m. QED.
39.3
40.£51.50

Problem-solving — Worked Solutions

1Problem 1
Answer
10 lockers are open: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Full working
A locker ends up open if and only if it is toggled an **odd** number of times. Locker nn is toggled once for each of its factors. Most numbers have an even number of factors (they pair up: if dd divides nn, so does n÷dn \div d). The exception is **perfect squares**, where one factor equals its own pair (n\sqrt{n}), giving an odd total. The perfect squares from 1 to 100 are 1,4,9,16,25,36,49,64,81,1001, 4, 9, 16, 25, 36, 49, 64, 81, 100 — that is 121^2 through 10210^2. So exactly **10 lockers** are open.
2Problem 2
Answer
12 ways
Full working
Systematically vary the number of 20¢ coins (0, 1, or 2):

- **Two 20¢ coins** (40¢ used, 10¢ remaining): (1×10¢, 0×5¢) or (0×10¢, 2×5¢). **2 ways.**
- **One 20¢ coin** (20¢ used, 30¢ remaining from 5¢ and 10¢): (0×10, 6×5), (1×10, 4×5), (2×10, 2×5), (3×10, 0×5). **4 ways.**
- **No 20¢ coins** (50¢ from 5¢ and 10¢ only): (0×10, 10×5), (1×10, 8×5), (2×10, 6×5), (3×10, 4×5), (4×10, 2×5), (5×10, 0×5). **6 ways.**

Total: 2 + 4 + 6 = **12 ways.** Since the number of 20¢ coins is fixed in each group, the groups are mutually exclusive and exhaustive.
3Problem 3
Answer
(a) 1089 (b) 1089 (c) Always 1089 — see working.
Full working
(a) 731 reversed = 137. 731 − 137 = 594. 594 reversed = 495. 594 + 495 = **1089** ✓.

(b) 852 reversed = 258. 852 − 258 = 594. 594 reversed = 495. 594 + 495 = **1089** ✓.

(c) Write the number as 100a+10b+c100a + 10b + c. Its reverse is 100c+10b+a100c + 10b + a. Difference (assuming a>ca > c): (100a+10b+c)−(100c+10b+a)=99a−99c=99(a−c)(100a + 10b + c) − (100c + 10b + a) = 99a − 99c = 99(a − c). Since aa and cc are single digits with a>ca > c, we have a−c∈{2,3,4,5,6,7,8,9}a − c \in \{2, 3, 4, 5, 6, 7, 8, 9\}, giving 99(a−c)∈{198,297,396,495,594,693,792,891}99(a−c) \in \{198, 297, 396, 495, 594, 693, 792, 891\}. These are all 3-digit multiples of 99. For each: its digit-reversal is also in the list, and the sum of any of these with its reversal equals 1089. For example 198+891=1089198 + 891 = 1089, 297+792=1089297 + 792 = 1089, 396+693=1089396 + 693 = 1089, 495+495=990495 + 495 = 990 — wait, for a−c=5a − c = 5 we get 495; its reversal 594; 495+594=1089495 + 594 = 1089 ✓. The result is always **1089**.
4Problem 4
Answer
Saturday
Full working
January has 31 days; February has 28 days (non-leap year). From 1 January to 1 March is 31 + 28 = 59 days later. 59 ÷ 7 = 8 remainder 3. So 1 March is 3 days after Wednesday: +1 = Thursday, +2 = Friday, +3 = Saturday. **1st March is a Saturday.** (Note: teacher may wish to double-check with a real calendar for a specific year.)
5Problem 5
Answer
(a) 10 (b) 45 (c) n(n−1)2\dfrac{n(n-1)}{2} (d) 20 people
Full working
(a) Person 1 shakes hands with 4 others, person 2 with 3 remaining others (not counting person 1 again), and so on: 4+3+2+1=104 + 3 + 2 + 1 = \mathbf{10} handshakes.

(b) Same pattern: 9+8+7+6+5+4+3+2+1=459 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = \mathbf{45} handshakes.

(c) Each of the nn people shakes hands with (n−1)(n-1) others. That gives n(n−1)n(n-1) — but each handshake has been counted twice (once for each person). So the formula is n(n−1)2\dfrac{n(n-1)}{2}.

(d) Set n(n−1)2=190\dfrac{n(n-1)}{2} = 190, so n(n−1)=380n(n-1) = 380. We need two consecutive integers with product 380. Try: 19×20=38019 \times 20 = 380 ✓. So n=20n = \mathbf{20} people attended.
6Problem 6
Answer
(a) 72 minutes (b) 15 times (c) 2 times
Full working
(a) We need the LCM of 8, 12, and 18. Prime factorisations: 8=238 = 2^3, 12=22×312 = 2^2 \times 3, 18=2×3218 = 2 \times 3^2. LCM = 23×32=8×9=722^3 \times 3^2 = 8 \times 9 = \mathbf{72} minutes.

(b) In 120 minutes (2 hours), Bell A tolls at: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120. That is 120÷8=15120 \div 8 = \mathbf{15} times.

(c) Bell B tolls at: 12, 24, 36, 48, 60, 72 minutes. We remove times when B coincides with A or C. LCM(8, 12) = 24: B and A both toll at 24, 48, 72. LCM(12, 18) = 36: B and C both toll at 36, 72. Check each: 12 — is it a multiple of 8? 12÷8=1.512 ÷ 8 = 1.5 ✗; of 18? 12÷18<112 ÷ 18 < 1 ✗ → **B alone** ✓. 24 — multiple of 8 ✓ → with A. 36 — multiple of 18 ✓ → with C. 48 — multiple of 8 ✓ → with A. 60 — 60÷8=7.560 ÷ 8 = 7.5 ✗, 60÷18≈3.3360 ÷ 18 ≈ 3.33 ✗ → **B alone** ✓. 72 — all three. B tolls alone at minutes 12 and 60: **2 times**.
7Problem 7
Answer
A = 2, B = 7 (since 23 × 7 = 161)
Full working
For A3×B=161A3 \times B = 161, the two-digit number A3A3 must end in 3. Possibilities: 13, 23, 33, 43, 53, 63, 73, 83, 93. Divide 161 by each: 161÷13≈12.4161 \div 13 \approx 12.4 (no); 161÷23=7161 \div 23 = 7 ✓ (7 is a single digit). Check: 23×7=16123 \times 7 = 161 ✓. So A=2A = 2 and B=7B = 7.
8Problem 8
Answer
110
Full working
Look at the differences: 6−2=4, 12−6=6, 20−12=8, 30−20=10 — second differences are constant (2), so it is quadratic. Notice: 2 = 1×2, 6 = 2×3, 12 = 3×4, 20 = 4×5, 30 = 5×6. The pattern is n(n+1)n(n+1). For the 10th term: 10×11=11010 \times 11 = \mathbf{110}.
9Problem 9
Answer
126
Full working
Three-digit multiples of 7 start at 105. Check digit sums in order: 105 → 1+0+5=6 (no); 112 → 1+1+2=4 (no); 119 → 1+1+9=11 (no); 126 → 1+2+6=9 ✓. The smallest three-digit multiple of 7 with digit sum 9 is **126**. Check: 126 ÷ 7 = 18 ✓; digit sum = 1+2+6 = 9 ✓.
10Problem 10
Answer
Magic sum = 15. Grid (rows): [2, 7, 6], [9, 5, 1], [4, 3, 8].
Full working
The main diagonal (top-left to bottom-right) contains 2, 5, 8. These sum to 15, so the **magic sum = 15**.

Now fill the grid. Let the entries be:
2abc5def8\begin{array}{|c|c|c|}\hline 2 & a & b \\\hline c & 5 & d \\\hline e & f & 8 \\\hline\end{array}


Row 1: 2+a+b=15⇒a+b=132 + a + b = 15 \Rightarrow a + b = 13.
Row 3: e+f+8=15⇒e+f=7e + f + 8 = 15 \Rightarrow e + f = 7.
Col 1: 2+c+e=15⇒c+e=132 + c + e = 15 \Rightarrow c + e = 13.
Col 3: b+d+8=15⇒b+d=7b + d + 8 = 15 \Rightarrow b + d = 7.
Anti-diagonal (top-right to bottom-left): b+5+e=15⇒b+e=10b + 5 + e = 15 \Rightarrow b + e = 10.

From a+b=13a + b = 13 and b+d=7b + d = 7: subtract ⇒a−d=6\Rightarrow a - d = 6.
Col 2: a+5+f=15⇒a+f=10a + 5 + f = 15 \Rightarrow a + f = 10.
Row 2: c+5+d=15⇒c+d=10c + 5 + d = 15 \Rightarrow c + d = 10.

We know the standard 3×3 magic square using the integers 1–9 has magic sum 15. The only 3×3 magic square (up to rotation/reflection) containing 2, 5, 8 on the leading diagonal uses all of 1–9 exactly once. Testing: rows [2,7,6], [9,5,1], [4,3,8]. Verify: rows: 15,15,15 ✓; cols: 15,15,15 ✓; diagonals: 2+5+8=15, 6+5+4=15 ✓. Answer: **[2,7,6], [9,5,1], [4,3,8]**.
11Problem 11
Answer
30 m × 30 m = 900 m²
Full working
Let the length be ll and the width be ww. Perimeter: 2l+2w=1202l + 2w = 120, so l+w=60l + w = 60. Area A=l×wA = l \times w. Try values: if l=30,w=30l = 30, w = 30: A=900A = 900. If l=40,w=20l = 40, w = 20: A=800A = 800. If l=50,w=10l = 50, w = 10: A=500A = 500. The area is largest when the rectangle is a **square** (both sides equal). With l+w=60l + w = 60, the maximum is at l=w=30l = w = 30 giving A=900A = \mathbf{900} m².
12Problem 12
Answer
5:12 pm (the clock shows 11 hours 12 minutes have passed, so 5:12 pm)
Full working
From 6:00 am to 6:00 pm is 12 real hours. The clock loses 4 minutes every real hour, so in 12 hours it loses 12×4=4812 \times 4 = 48 minutes. The clock therefore shows 12 hours−48 minutes=11 hours and 12 minutes12 \text{ hours} - 48 \text{ minutes} = 11 \text{ hours and } 12 \text{ minutes} elapsed. Starting from 6:00 am + 11 h 12 min = **5:12 pm**. (The clock shows 5:12 pm even though the actual time is 6:00 pm.)