Aller au contenu principal
Ecolint Campus des NationsMathématiques
← Dossiers de révision

Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 10 · Exponents and Indices

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.32
2.1
3.x7x^7
4.x5x^5
5.x6x^6
6.4x64 x^6
7.18\dfrac{1}{8}
8.3×1033 \times 10^3
9.4×10−34 \times 10^{-3}
10.25 000
Silver
11.12x712 x^7
12.5x45 x^4
13.x9x^9
14.x3x^3
15.x−4x^{-4}
16.6×1096 \times 10^9
17.4×1044 \times 10^4
18.x5x^5
19.3×1043 \times 10^4 (30 000 times)
20.9x49 x^4
Gold
21.5
22.3
23.4
24.14\dfrac{1}{4}
25.4x24 x^2
26.3.5×1053.5 \times 10^5
27.14\dfrac{1}{4}
28.x=5x = 5
29.116\dfrac{1}{16}
30.x2/3x^{2/3}
Platinum
31.x=4x = 4
32.x=2x = 2
33.4x74 x^7
34.≈3.33×105\approx 3.33 \times 10^5 times (≈ 333 000)
35.x4y6\dfrac{x^4}{y^6}
36.8
37.1.8×10111.8 \times 10^{11} m
38.x=2x = 2
39.4x6y44 x^6 y^4
40.x=64x = 64

Pack B — Answers

Bronze
1.81
2.1
3.x7x^7
4.x6x^6
5.x12x^{12}
6.25x425 x^4
7.125\dfrac{1}{25}
8.4.5×1044.5 \times 10^4
9.7×10−57 \times 10^{-5}
10.6 300
Silver
11.10x710 x^7
12.6x66 x^6
13.x8x^8
14.x3x^3
15.x−6x^{-6}
16.2×1062 \times 10^6
17.3×1033 \times 10^3
18.x3x^3
19.3×1043 \times 10^4 (30 000 times)
20.−8x9-8 x^9
Gold
21.9
22.4
23.9
24.17\dfrac{1}{7}
25.27x227 x^2
26.4.2×1064.2 \times 10^6
27.19\dfrac{1}{9}
28.x=7x = 7
29.827\dfrac{8}{27}
30.x3/5x^{3/5}
Platinum
31.x=5x = 5
32.x=3x = 3
33.9x29 x^2
34.≈3.33×105\approx 3.33 \times 10^5 times (≈ 333 000)
35.x6y8\dfrac{x^6}{y^8}
36.27
37.6×10106 \times 10^{10} m
38.x=3x = 3
39.9x6y69 x^6 y^6
40.x=125x = 125

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 1, 2, 4, 8, 16 (b) gn=2n−1g_n = 2^{n-1} (c) ≈2.15×109\approx 2.15 \times 10^9
Full working
(a) Doubling sequence: 1, 2, 4, 8, 16.

(b) Geometric with u1=1u_1 = 1, r=2r = 2. So gn=1⋅2n−1=2n−1g_n = 1 \cdot 2^{n-1} = 2^{n-1}.

(c) g32=231=2,147,483,648≈2.15×109g_{32} = 2^{31} = 2{,}147{,}483{,}648 \approx \mathbf{2.15 \times 10^9}.
2Problem 2
Answer
(a) 1.2 times (b) Yes, just (c) ≈1.1×10−3\approx 1.1 \times 10^{-3} m³ (= 1.1 L)
Full working
(a) 9×10−67.5×10−6=97.5=1.2\frac{9 \times 10^{-6}}{7.5 \times 10^{-6}} = \frac{9}{7.5} = 1.2.

(b) The capillary (9 μm) is wider than the cell (7.5 μm), so yes — barely.

(c) Cell radius r=3.75×10−6r = 3.75 \times 10^{-6} m. Volume per cell: V=43π(3.75×10−6)3=43π⋅52.73×10−18≈2.21×10−16V = \frac{4}{3}\pi (3.75 \times 10^{-6})^3 = \frac{4}{3}\pi \cdot 52.73 \times 10^{-18} \approx 2.21 \times 10^{-16} m³. Total: 5×1012×2.21×10−16≈1.1×10−35 \times 10^{12} \times 2.21 \times 10^{-16} \approx 1.1 \times 10^{-3} m³ — about **1.1 litres**, which matches real blood volume reasonably.
3Problem 3
Answer
See working.
Full working
(a) By the division law: x5x5=x5−5=x0\frac{x^5}{x^5} = x^{5-5} = x^0. But x5x5=1\frac{x^5}{x^5} = 1 (any non-zero quantity divided by itself). So x0=1x^0 = 1.

(b) Using the division law: x3x5=x3−5=x−2\frac{x^3}{x^5} = x^{3-5} = x^{-2}. Directly: x3x5=1x5−3=1x2\frac{x^3}{x^5} = \frac{1}{x^{5-3}} = \frac{1}{x^2}. Combining: x−2=1x2x^{-2} = \frac{1}{x^2}. This justifies the negative index rule.

(c) Using the power rule: (x1/2)2=x1/2⋅2=x1=x(x^{1/2})^2 = x^{1/2 \cdot 2} = x^1 = x. So x1/2x^{1/2} is the number which squared gives xx, i.e. the square root: x1/2=xx^{1/2} = \sqrt{x}.
4Problem 4
Answer
(a) Pn=4.5×104⋅1.03nP_n = 4.5 \times 10^4 \cdot 1.03^n (b) ≈6.05×104\approx 6.05 \times 10^4 (c) 15 years
Full working
(a) Compound growth multiplier 1.03 per year. Pn=4.5×104⋅1.03nP_n = 4.5 \times 10^4 \cdot 1.03^n.

(b) After 10 years: P10=4.5×104⋅1.0310=4.5×104⋅1.3439≈6.05×104P_{10} = 4.5 \times 10^4 \cdot 1.03^{10} = 4.5 \times 10^4 \cdot 1.3439 \approx 6.05 \times 10^4.

(c) Solve 4.5×104⋅1.03n>7×1044.5 \times 10^4 \cdot 1.03^n > 7 \times 10^4, i.e. 1.03n>74.5=1.5561.03^n > \frac{7}{4.5} = 1.556. Test: 1.0314≈1.5131.03^{14} \approx 1.513 (no); 1.0315≈1.5581.03^{15} \approx 1.558 (yes). So after **15 years** (i.e. in 2035).
5Problem 5
Answer
6×10−3<150<3×10−2<1.5×10−1<0.46 \times 10^{-3} < \dfrac{1}{50} < 3 \times 10^{-2} < 1.5 \times 10^{-1} < 0.4
Full working
Convert each to standard form or decimal:
- 3×10−2=0.033 \times 10^{-2} = 0.03
- 0.4=0.40.4 = 0.4
- 1.5×10−1=0.151.5 \times 10^{-1} = 0.15
- 150=0.02\frac{1}{50} = 0.02
- 6×10−3=0.0066 \times 10^{-3} = 0.006

Ordered: 0.006<0.02<0.03<0.15<0.40.006 < 0.02 < 0.03 < 0.15 < 0.4, i.e. 6×10−3<150<3×10−2<1.5×10−1<0.46 \times 10^{-3} < \frac{1}{50} < 3 \times 10^{-2} < 1.5 \times 10^{-1} < 0.4.
6Problem 6
Answer
(a) 40, 20, 10, 5 (b) M=80⋅(12)nM = 80 \cdot \left(\tfrac{1}{2}\right)^n (c) After 35 days (mass 0.625 g)
Full working
(a) Each 5-day period halves: 80 → 40 → 20 → 10 → 5.

(b) M=80⋅(12)n=802nM = 80 \cdot \left(\frac{1}{2}\right)^n = \frac{80}{2^n} after nn half-lives (=5n= 5n days).

(c) Need 802n<1\frac{80}{2^n} < 1, i.e. 2n>802^n > 80. 26=64<80<128=272^6 = 64 < 80 < 128 = 2^7. So n=7n = 7 half-lives = **35 days**: M=80128=0.625M = \frac{80}{128} = 0.625 g.
7Problem 7
Answer
(a) x3x^3 (b) x3x^3 (c) x2x^2 (d) x=81x = 81
Full working
(a) x6=(x6)1/2=x6/2=x3\sqrt{x^6} = (x^6)^{1/2} = x^{6/2} = x^3.

(b) x93=(x9)1/3=x9/3=x3\sqrt[3]{x^9} = (x^9)^{1/3} = x^{9/3} = x^3.

(c) x5x1/2=x5/2x1/2=x5/2−1/2=x2\frac{\sqrt{x^5}}{x^{1/2}} = \frac{x^{5/2}}{x^{1/2}} = x^{5/2 - 1/2} = x^2.

(d) x1/2=9⇒x=92=81x^{1/2} = 9 \Rightarrow x = 9^2 = 81.
8Problem 8
Answer
(a) 1.5×10291.5 \times 10^{29} grains (b) 1.3×10171.3 \times 10^{17} atoms
Full working
(a) Number of grains = 6×10244×10−5=64×1024−(−5)=1.5×1029\frac{6 \times 10^{24}}{4 \times 10^{-5}} = \frac{6}{4} \times 10^{24-(-5)} = 1.5 \times 10^{29}.

(b) Number = 1.3×1071×10−10=1.3×1017\frac{1.3 \times 10^7}{1 \times 10^{-10}} = 1.3 \times 10^{17}.
9Problem 9
Answer
All four are wrong. See working for corrections.
Full working
(a) **Wrong.** Power of a power *multiplies* indices: (x3)2=x3×2=x6(x^3)^2 = x^{3 \times 2} = x^6.

(b) **Wrong.** Adding *like terms* — the indices don't combine: x3+x3=2x3x^3 + x^3 = 2x^3.

(c) **Wrong.** Any non-zero number to the power 0 is 1: x0=1x^0 = 1 (for x≠0x \neq 0).

(d) **Wrong.** Power applies to *both* the coefficient and the variable: (2x)3=23⋅x3=8x3(2x)^3 = 2^3 \cdot x^3 = 8 x^3.
10Problem 10
Answer
(a) 56\dfrac{5}{6} (b) 15\dfrac{1}{5} (c) Not in general
Full working
(a) a−1+b−1=12+13=36+26=56a^{-1} + b^{-1} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}.

(b) (a+b)−1=(2+3)−1=5−1=15(a + b)^{-1} = (2 + 3)^{-1} = 5^{-1} = \frac{1}{5}.

(c) 56≠15\frac{5}{6} \neq \frac{1}{5}, so (a+b)−1≠a−1+b−1(a+b)^{-1} \neq a^{-1} + b^{-1} in general. This is a common misconception: the negative index does **not** distribute over a sum. The general rule for fractions: 1a+1b=a+bab\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}, not 1a+b\frac{1}{a+b}.
11Problem 11
Answer
x=5x = 5
Full working
Write both sides with base 2. RHS: 4x−1=(22)x−1=22(x−1)=22x−24^{x-1} = (2^2)^{x-1} = 2^{2(x-1)} = 2^{2x-2}.

Equation: 2x+3=22x−22^{x+3} = 2^{2x-2}. Same base, so indices equal: x+3=2x−2x + 3 = 2x - 2. Rearranging: 5=x5 = x. So **x=5x = 5**.

Check: LHS = 28=2562^8 = 256; RHS = 44=2564^4 = 256 ✓.
12Problem 12
Answer
(a) See working (b) 4.02×10134.02 \times 10^{13} km (c) ≈42,500\approx 42{,}500 years
Full working
(a) Distance = speed × time = (3×108)×(3.154×107)=9.462×1015(3 \times 10^8) \times (3.154 \times 10^7) = 9.462 \times 10^{15} m ≈ **9.46×10159.46 \times 10^{15} m** ✓.

(b) Distance to Proxima Centauri = 4.25×9.46×10154.25 \times 9.46 \times 10^{15} m = 40.205×101540.205 \times 10^{15} m = 4.02×10164.02 \times 10^{16} m. Convert to km (÷ 1000): 4.02×10134.02 \times 10^{13} km.

(c) Time = 4.02×1016 m3×104 m/s=1.34×1012\frac{4.02 \times 10^{16} \text{ m}}{3 \times 10^4 \text{ m/s}} = 1.34 \times 10^{12} s. In years: 1.34×10123.154×107≈4.25×104=42,500\frac{1.34 \times 10^{12}}{3.154 \times 10^7} \approx 4.25 \times 10^4 = \mathbf{42{,}500} years.