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Pass Publications GCSE Mathematics Higher 1
Number
EXERCISES
Topic 1 ∼ Bodmas
1. Evaluate the following:
(a) 5 + 9 × 3 (b) 15 − 2 × 3 (c) 7 + 8 × 7
(d) 9 − 9 × 0.1 (e) 30 − (13 × 3) (f) −15 + [17 × (−3)].2. Find the following results:
(a) 5(7 × 3 + 4) (b) 3[3 × (−2) + 8] (c) −7(6 × 9 − 60)
(d) 9(2 × 8 − 9) (e) [(−3) × 7] × (−5) (f) −6 × [(−9) × (−1)].3. (a) (16 + 1)(8 ÷ 4) (b)
(35 + 5
) (32 ÷ 3
)(c)
(72 − 3
) (42 ÷ 4
)(d)
(133 × 5
) (92 ÷ 9
)(e)
(122 − 44
) ÷ 10 (f)(142 − 96
) × 110.
4. (a)(1
5 + 110
) ÷ 35 (b) 1 1
10 ÷ (13
4 + 214
)(c) 21
7 ÷ 2 316 + 3
4
(d)(3
4 × 134
) × 5 (e)(23
4 + 114
) × (11
5 − 210
)(f)
134
234 − 11
2
.
5. (a) Multiply by 234 the sum of 11
4 and 234.
(b) Find the product of 315 and 5
8.
(c) Find the difference between 52 and 32.
(d) Find the quotient of 415 divided by the sum of 3
5 and 525.
(e) Subtract 134 from the product of 17
8 and 45.
(f) Divide the difference between 212 and 3
4 by 418.
2 Pass Publications GCSE Mathematics Higher
SolutionsTopic 1 Bodmas
1. (a) 5 + 9 × 3 = 5 + 27 = 32 (b) 15 − 2 × 3 = 15 − 6 = 9
(c) 7 + 8 × 7 = 7 + 56 = 63 (d) 9 − 9 × 0.1 = 9 − 0.9 = 8.1
(e) 30 − (13 × 3) = 30 − 39 = −9 (f) −15 +[17 × (−3)] = −15 − 51 = −66.
2. (a) 5(7 × 3 + 4) = 5(21 + 4) = 5(25) = 125
(b) 3 [3 × (−2) + 8] = 3{−6 + 8} = 3(2) = 6
(c) −7(6 × 9 − 60) = −7(54 − 60) = −7(−6) = 42
(d) 9(2 × 8 − 9) = 9(16 − 9) = 9(7) = 63
(e) [(−3) × 7] × (−5) = {−21} × (−5) = 105
(f) −6 × [(−9) × (−1)] = −6 × {9} = −54.
3. (a) (16 + 1)(8 ÷ 4) = (17)(2) = 34
(b) (35 + 5)(32 ÷ 3) = (243 + 5)(3) = 248 × 3 = 744
(c) (72 − 3)(42 ÷ 4) = (49 − 3)(4) = 46 × 4 = 184
(d) (132 × 5)(92 ÷ 9) = (169 × 5)(9) = 845 × 9 = 7605
(e) (122 − 44) ÷ 10 = (144 − 44) ÷ 10 = 100 ÷ 10 = 10
(f) (142 − 96) × 1
10= (196 − 96) × 1
10= 100 × 1
10= 10.
4. (a)
(1
5+ 1
10
)÷ 3
5=
(2
10+ 1
10
)÷ 3
5= 3
10÷ 3
5= 3
10× 5
3= 1
2
(b) 11
10÷
(1
3
4+ 2
1
4
)= 11
10÷
(7
4+ 9
4
)= 11
10÷ 16
4= 11
10× 1
4= 11
40
(c) 21
7÷ 2
3
16+ 3
4= 15
7÷ 35
16+ 3
4= 15
7× 16
35+ 3
4= 48
49+ 3
4= 192 + 147
196= 339
196
Pass Publications GCSE Mathematics Higher 3
(d)
(3
4× 1
3
4
)× 5 =
(3
4× 7
4
)× 5 = 21
16× 5 = 105
16
(e)
(2
3
4+ 1
1
4
)×
(1
1
5− 2
10
)=
(11
4+ 5
4
)×
(6
5− 2
10
)= 16
4×
(12
10− 2
10
)= 4
(f)1
3
4
23
4− 1
1
2
=7
411
4− 3
2
=7
411
4− 6
4
= 7
4× 4
5= 7
5.
5. (a) 23
4
(1
1
4+ 2
3
4
)= 11
4
(5
4+ 11
4
)= 11
4
(16
4
)= 11
(b) 31
5× 5
8= 16
5× 5
8= 2
(c) 52 − 32 = 25 − 9 = 16
(d)4
1
53
5+ 5
2
5
=21
53
5+ 27
5
=21
530
5
= 21
30= 7
10
(e) 17
8× 4
5− 1
3
4= 15
3
82× 4
5− 1
3
4= 3
2− 7
4= −1
4
(f)2
1
2− 3
4
41
8
=5
2− 3
433
8
=10
4− 3
433
8
=7
433
8
= 14
33.
4 Pass Publications GCSE Mathematics Higher
EXERCISES
Topic 2 ∼ Percentage
1. Find the values of the following:
(a) 5% of 100 kg (b) 7% of 28 m (c) 10% of £125
(d) 22% of 120 tonnes (e) 25% of 50 N (f) 35% of 5 A.
2. Find 33% of 250 kg.
3. What is 150% of £1 000?
4. Write 250 kg as a percentage of 25 kg.
5. Write 52 p as a percentage of £3.
6. The bus fares in outer London have increased from 70 p to £1. What is the percentageincrease?
7. The following items cost:
(a) £70 (b) £125
(c) £3.75 before a sale discount of 37.5%. How much is now the cost?
8. A T.V. costs £575 before V.A.T. of 17.5% is applied. How much do you pay withV.A.T.?
9. A car costs £13750 includingV.A.T. of 17.5%. How much is the cost of the car withoutV.A.T.?
10. If Pythagoras who lived 2600 years ago invested one drachma at 1% p.a., calculate:
(a) the simple interest and
(b) the compound interest earned today.
11. The formula Cn = C(1 + r
100
)n is used to work out the capital accumulated after n
years, Cn, at r% and C is the capital invested. If C = £1500, r = 5 and n = 10,calculate Cn.
12. If Cn = £14500, r = 7, n = 2 find C.
13. If C5 = C1.055, what is the ratio C5C
?
14. £5000 is invested at 2.65% p.a. for 5 years. How much is invested at the end ofyear 5?
15. A car is bought for £14995. The depreciation of the car is 15% per annum, what isthe value of the car after 3 years?
Pass Publications GCSE Mathematics Higher 5
16. A house was bought for £72500 on the 15th November 1985, and it costs £525000 onthe 14th November 2004. What is the average value of increase p.a.? If inflation onaverage is 5% during this period of time, what is the net gain or loss of this investment?
17. Increase the following amounts by 5%:
(a) £10 (b) £125 (c) £500 (d) £1250.
18. Decrease the following amounts by 8%:
(a) 50 kg (b) 25 m (c) 425 tonnes (d) £525.
19. Write down the percentages of the following fractions:
(a) 34 (b) 7
8 (c) 2 (d) 113 (e) 4
5 (f) 110 (g) 3
7 (h) 915.
20. Write down the fractions of the following percentages:
(a) 2% (b) 35% (c) 85% (d) 125%.
21. Change these fractions into decimals:
(a) 12 (b) 1
4 (c) 38 (d) 9
16 (e) 34 (f) 11
40 (g) 532
(h) 1920 (i) 7
16 (j) 339 (k) 34
5 (l) 425 (m) 85
9 (n) 725.
22. Change these fractions into recurring decimals:
(a) 13 (b) 1
18 (c) 411 (d) 4
3 (e) 416 (f) 43
7 (g) 323 (h) 2 4
13.
23. Change these decimals into fractions:
(a) 0.1 (b) 0.35 (c) 0.25 (d) 0.15
(e) 0.75 (f) 0.03 (g) 0.111 (h) 0.775.
24. Express the following as fractions:
(a) 5.05 (b) 0.375 (c) 0.125 (d) 6.9
(e) 8.75 (f) 5.125 (g) 3.68 (h) 0.003.
25. Write the following as percentages:
(a)3
5(b)
3
8(c)
27
40(d)
14
15(e) 0.98 (f) 0.001 (g) 2.17 (h) 0.33
26. Change the following percentages to fractions:
(a) 25% (b) 35% (c) 33% (d) 335%
(e) 5% (f) 4112% (g) 871
2% (h) 3912%.
6 Pass Publications GCSE Mathematics Higher
SolutionsTopic 2
1. (a)5
100× 100 kg = 5 kg (b)
7
100× 28 m = 1.96 m
(c)10
100× £125 = £12.50 (d)
22
100× 120 tonnes = 26.4 tonnes
(e)25
100× 50 N = 12.5 N (f)
35
100× 5 A = 1.75 A.
2.33
100× 250 kg = 82.5 kg . 3.
150
100× £1000 = £1500.
4.250 kg
25 kg× 100 = 1000%. 5.
52 p
300 p× 100 = 17.3% to 3 s.f..
6.30 p
70 p× 100 = 300
7= 42.9% to 3 s.f..
7. (a) £70 − 37.5
100£70 = £70 × 0.62.5 = £43.75
(b) £125 × 0.625 = £78.13 (c) £3.75 × 0.625 = £2.34.
8. £575 + 17.5
100× £575 = £575 × 1.175 = £675.63.
9.£13750
1.175= £11702.13.
10. (a) 0.01×2600 = 26 drachmae (b) 1×1.012600 = 1.7201718×1011 drachmae.
11. Cn = C(
1 + r
100
)n
C10 = £1500
(1 + 5
100
)10
= £2443.34.
12. £14500 = £C
(1 + 7
100
)2
£C = 14500
1.072 = £12664.86.
13.C5
C= 1.055 = 1.276 to 3 d.p. = 1.28 to 3 s.f..
14. C5 = £5000 × 1.02655 = £5698.56.
15. £14995 − 15
100£14995 = £14995 × 0.85
after one year £14995 × 0.853 = £9208.80 after 3 years.
Pass Publications GCSE Mathematics Higher 7
16. 525000 = 72500
(1 + 5
100
)19
525000
72500=
(1 + 5
100
)19
7.241379311
19 = 1 + 5
100(1.109823081 − 1)100 ⇒ r = 10.98% to 2 d.p.
72500 × 1.0519 = 183203.89 due to inflation
£525000 − £183203.89 due to inflation
£525000 − £183203.89 = 341796.1
= £342000 to 3 s.f. net gain.
17. (a) £10 × 1.05 = £10.50 (b) £125 × 1.05 = £131.25
(c) £500 × 1.05 = £525 (d) £1250 × 1.05 = £1312.50.
18. (a) 50 kg × 0.92 = 46 kg (b) 25 m × 0.92 = 23 m
(c) 425 tonnes × 0.92 = 391 tonnes (d) £525 × 0.92 = £483.
19. (a)3
4× 25
25= 75
100= 0.75% = 75% (b)
7
8× 12.5
12.5= 87.5
100= 87.5%
(c)2
1× 100
100= 200% (d) 1
1
3× 100
100= 400
3= 133.3̇%
(e)4
5× 20
20= 80
100= 80% (f)
1
10× 10
10= 10
100= 10%
(g)3
7= 0.428571428 = 42.9% to 3 s.f. (h)
9
15= 0.6 = 60%.
20. (a) 2% = 2
100= 1
50(b) 0.35 = 7
20
(c) 0.85 = 85
100= 17
20(d) 1.25 = 5
4.
8 Pass Publications GCSE Mathematics Higher
21. (a) 0.5 (b) 0.25 (c) 0.375 (d) 0.5625 (e) 0.75 (f) 0.275 (g) 0.15625
(h) 0.95 (i) 0.4375 (j) 3.6̇ (k) 3.8 (l) 4.4 (m) 8.5̇ (n) 7.4.
22. (a) 0.3̇3̇ (b) 0.05̇ (c) 0.3̇6̇ (d) 1.3̇
(e) 4.16̇ (f) 4.4̇2̇8̇5̇7̇1̇ (g) 3.6̇ (h) 2.3̇0̇7̇6̇9̇2̇.
23. (a)1
10(b) 0.35 =
35
100=
7
20(c) 0.25 =
1
4
(d) 0.15 =15
100= 3
20(e)
3
4(f)
3
100
(g)111
1000(h)
775
1000=
155
200=
31
40.
24. (a) 5.05 = 5 + 5
100= 5
1
20(b) 0.375 = 3
8(c)
1
8(d) 6 9
10 (e) 834 (f) 51
8
(g) 3 68100 = 334
50 = 31725 (h) 3
1000.
25. (a)3
5× 20
20= 60
100= 60%
(b)3
8× 12.5
12.5= 37.5
100= 37.5%
(c)27 × 2.5
40 × 2.5= 67.5
100= 67.5%
(d)14
15= 0.93̇ = 93.3̇% to 3 s.f .
(e) 98% (f) 0.1% (g) 217% (h) 33%.
26. (a) 25% = 25
100= 1
4(b)
35
100= 7
20
(c)33
100(d)
335
100= 3
35
100= 3
7
20
(e)5
100= 1
20(f)
41.5
100= 415
1000= 83
200.
(g)87.5
100= 875
1000= 175
200= 35
40= 7
8(h)
39.5
100= 395
1000= 79
200.
Pass Publications GCSE Mathematics Higher 9
EXERCISES
Topic 3 ∼ Decimals
1. Find the sum of the following decimal numbers:
3.56, 2.09, 0.065, 15.39, 1.70 without the use of a calculator.
2. Evaluate:
3.79 + 1.99 + 0.97 + 0.005 without the use of a calculator.
3. Determine the difference between 37.99 and 22.01, without the use of a calculator.
4. Evaluate:
25.09 − 16.98 without the use of a calculator.
5. Find the product 1.95 and 0.67, without the use of a calculator.
6. Evaluate:
15.69 × 0.87 without the use of a calculator.
7. Evaluate the quotient of 0.96 divided by 0.08, without the use of a calculator.
8. Evaluate:1.210.11 without the use of a calculator.
9. Write the following recurring decimals as fractions:
(a) 0.15̇ (b) 0.33̇ (c) 1.3̇5̇
(d) 6.87̇ (e) 0.9̇3̇ (f) 0.6̇6̇.
Show clearly all the steps.
10. Convert the following decimals to fractions:
(a) 0.607 (b) 0.65 (c) 1.35
(d) 2.75 (e) 0.00125 (f) 6.25.
11. Without the use of a calculator, add the following:
(a) 1.011 + 0.0003 + 2.046
(b) 2.003 + 0.003 + 1.003.
12. Subtract 1.999 from 2.000.
13. Multiply: (a) 0.97 × 0.369 (b) 3.767 × 1.001.
10 Pass Publications GCSE Mathematics Higher
14. Divide 4.769 by 0.0023.
15. 4.765×23.9 = 113.8835. Find the following answers without the use of a calculator:
(a) 476.5 × 0.239 (b) 0.4765 × 239
(c)1138835
239(d)
11388.35
4765.
16. Find the following recurring decimals as fractions:
(a) 0.6̇7̇ (b) 0.75̇ (c) 1.1̇1̇3̇.
17. Express 0.2̇3̇5̇ as ND
where N and D are integers.
18. Change the following decimals to fractions:
(a) 0.825 (b) 0.125 (c) 1.35 (d) 3.725.
*Difficult
Pass Publications GCSE Mathematics Higher 11
SolutionsTopic 3 ∼ Decimals1. 3.56 2. 3.79 3. 37.99 −
2.09 1.99 22.010.065 0.97 15.98
15.39 0.0051.70 6.755
22.805
4. 25.09 − 5. 1.95 × 6. 15.69 ×16.98 0.67 0.878.11 1365 10983
1170 125521.3065 13.6503
7.0.96
0.08= 96
8= 12. 8.
1.21
0.11= 121
11= 11.
9. (a) 0.15̇ = 0.15555 . . . = x
10x = 1.55555 . . .
x = 0.15555 . . .
9x = 1.4
x = 1.4
9= 14
90= 7
45(b) 0.3̇3̇ = 0.3333 . . . = x
10x = 3.3333 . . .−
x = 0.3333 . . .
9x = 3
x = 1
3(c) 1.3̇5̇ = 1.353535 . . . = x
100x = 135.3535 . . .−
x = 1.3535 . . .
99x = 134
x = 134
99
12 Pass Publications GCSE Mathematics Higher
(d) 6.87̇ = 6.8777 . . . = x
100x = 687.777 . . .−
10x = 68.777 . . .
90x = 619
x = 619
90
(e) 0.9̇3̇ = 0.939393 . . . = x
100x = 93.9393 . . .−
x = 0.9393 . . .
99x = 93
x = 93
99= 31
33
(f) 0.6̇6̇ = 0.666666 . . . = x
100x = 66.6666 . . .−
x = 0.6666 . . .
99x = 66
x = 66
99= 6
9= 2
3.
10. (a) 0.607 = 607
1000(b) 0.65 = 65
100= 13
20
(c) 1.35 = 135
100= 27
20(d) 2.75 = 275
100= 55
20= 2
3
4
(e) 0.00125 = 125
100000= 1
800(f) 6.25 = 625
100= 6
1
4.
11. (a) 1.011 (b) 2.0030.0003 + 0.003 +2.046 1.0033.0573 3.009
12. 2.000 −1.9990.001
Pass Publications GCSE Mathematics Higher 13
13. (a) 0.9700.369 ×8730
58202910
0.357930
(b) 3.7671.001 ×3767
000000003767
3.770767
14.4.769
0.0023= 47690
23= 2073.478 to 3 d.p.
2073.47826123
∣∣4769046 −1690161 −
8069 −110
92 −180161 −190184 −
6046 −14
15. 4.765 × 23.9 = 113.8835
(a) 476.5 × 0.239 = 4.765 × 100 × 23.9
100= 113.8835
(b) 0.476.5 × 239 = 4.765
10× 23.9 × 10 = 113.8835
(c)1138835
239= 113.8835 × 10000
23.9 × 10= 4.765 × 1000 = 4765
(d)11388.35
4765= 113.8835 × 100
4.765 × 1000= 23.9
10= 2.39.
14 Pass Publications GCSE Mathematics Higher
16. (a) 0.6̇7̇ = 0.676767 . . . = x
100x − x = 99x
67.6767 . . . − 0.6767 . . . = 67
99x = 67
x = 67
99
(b) 0.75̇ = 0.7555 . . . = x
100x = 75.55 . . .
100x − 10x = 90x = 75.55 . . . − 7.55 . . .
90x = 68
x = 68
90= 34
45
(c) 1.1̇1̇3̇ = 1.113113113 . . . = x
1000x = 1113.113113 . . .
1000x − x = 999x = 1112
x = 1112
999.
17. 0.2̇3̇5̇ = 0.235235235 . . . = x
1000x = 235.235235 . . .
x = 0.235235 . . .
999x = 235
x = 235
999.
18. (a) 0.825 = 825
1000= 165
200= 33
40
(b) 0.125 = 125
1000= 1
8
(c) 1.35 = 135
100= 27
20= 1
7
20
(d) 3.725 = 3725
1000= 149
40= 3
29
40.
Pass Publications GCSE Mathematics Higher 15
EXERCISES
Topic 4 ∼ Irrational numbers
1. The square roots of the prime numbers are irrational numbers. Are the squares ofirrational numbers always rational? Explain by giving an example.
2. Sketch five right angle triangles whose vertical sides are unequal.
(a) The hypotenuse is rational and so are the vertical sides.
(b) The hypotenuse is irrational and the vertical sides are rational.
(c) All three sides are irrational.
(d) The hypotenuse is rational and the vertical sides are irrational.
(e) The hypotenuse is irrational, one vertical side is rational and the other irrational.Insert the numbers on the diagrams that you have chosen.
3. Simplify the following surds:
(a)√
24 (b)√
12 (c)√
8 (d)√
18
(e)√
500 (f)√
45 (g)√
48 (h)√
1000
(i)√
125 (j)√
800 (k)√
300.
The term surd is derived from the word absurd.
4. Simplify the following surds:
(a) 3√
12 × 4√
3 (b)√
125 × 5√
5
(c)√
243 × √27 (d) 2
√60 × √
15.
5. Write down two irrational numbers between 3 and 4.
6. Write down two irrational numbers between 1 and 2.
7. Prove that√
3 is irrational.
8. Is a recurring number irrational?
*9. For GCSE π = 227 to 3 s.f. Can you find a better and closer to π fraction? Clue – use
the numbers 113355.*
10. Find two irrational numbers x and y which represent the sides of a rectangle such thatthe perimeter is rational.
16 Pass Publications GCSE Mathematics Higher
11. The sides of a rectangle are irrational and the area is rational. Find these sides if thearea of the rectangle is 50 m2.
*12. Find the value of(√
15 − 2√
3)2
.
*13. Find the value of(√
5 − √3) (√
5 + √3)
.
*14. 2 � 1
2 � 1
The above rectangle has irrational sides as shown.Determine (a) the perimeter (b) the area.
*15. If x = 2√
2 − 1, y = 2√
2 + 1 determine the value of xyx−y
.
16. Simplify
√125
20.
17. Simplify√
26(
2√
13 − 2√
2)
.
18. Simplify the quotient
(7 − √
2) (
7 + √2)
47.
19. Simplify the quotient
(6 − √
5) (
6 + √5)
√31
20. a and b are two positive irrational numbers. The sum and the product are rational.Express 1
a+ 1
bas a single fraction, explain why a+b
ab or aba+b
are always rational.
21. Find the value√
7 × √63.
22. Find the value of k if√
5 × √500 = k
√5.
23. Find the value of
√243 − √
3√12
.
Pass Publications GCSE Mathematics Higher 17
SolutionsTopic 4
1. π2 is an irrational number
π = 3.141592654 to 9 d.p.
π2 = 9.869604401 to 9 d.p..
2.
5 4
3
(a)
32 + 42 = 52
9 + 16 = 25
5 1
2
(b)
22 + 12 = (√
5)2
4 + 1 = 5
52
3
2(c) (√
3)2 +
(√5)2 =
(2√
2)2
3 + 5 = 8
7
(d)
2
3(√
2)2 +
(√7)2 = 32
2 + 7 = 9
3
(e)
2
7(√
3)2 + 22 =
(√7)2
3 + 4 = 7.
3. (a)√
24 = √2 × 2 × 2 × 3 = √
4√
2√
3 = 2√
6
(b)√
12 = √2 × 2 × 3 = √
4√
3 = 2√
3
(c)√
8 = √2 × 2 × 2 = √
2 × 2√
2 = 2√
2
(d)√
18 = √2 × 3 × 3 = √
2√
9 = 3√
2
(e)√
500 = √10 × 10 × 5 = √
100√
5 = 10√
5
(f)√
45 = √3 × 9 = √
9√
3 = 3√
3
18 Pass Publications GCSE Mathematics Higher
(g)√
48 = √3 × 16 = √
16√
3 = 4√
3
(h)√
1000 = √100
√10 = 10
√10
(i)√
125 = √25
√5 = 5
√5
(j)√
800 = √100
√2 × 2 × 2 = 10 × 2
√2 = 20
√2
(k)√
300 = √100
√3 = 10
√3.
4. (a) 3√
12 × 4√
3 = 12√
12 × 3 = 12√
36 = 12 × 6 = 72
(b)√
125 × 5√
5 = 5√
5 5√
5 = 125
(c)√
243 × √27 = √
3 × 81 × √27 = √
3 × 27 × 81 = √81 × 81 = 81
(d) 2√
60 × √15 = 2 × √
4 × 15 × √15 = 2 × 2 × 15 = 60.
5. 3,√
11,√
13, 4. 6. 1,√
2,√
3, 2.
7. If√
3 = a
bwhere a and b are integers and have common factor, squaring both sides
3 = a2
b2 a2 = 3b2
3b2 is a multiple of 3 and therefore odd. So a2 is an odd as well and so a is odd. Thismeans a = 3c, ∴ a2 = 9c2 9c2 − 3b2 hence 3c2 = b2.
8. 0.1̇2̇5̇ = 0.125125.125 . . . = x
1000x = 125.125125 . . .
x = 0.125125 . . .
999x = 125
x = 125
999= 0.1̇2̇5̇
Which is a rational number therefore a recurring number is not an irrational number.
9. π = 3.141592654 = 3.14 to 3 s.f.22
7= 3.142857143 = 3.14 to 3 s.f.
355
113= 3.14159292 = 3.141593 to 7 s.f.
π = 3.141592654 = 3.141593 to 7 s.f.
therefore355
113is equal to π to 7 s.f..
Pass Publications GCSE Mathematics Higher 19
10.
5� 3
9� 3
9� 3
5� 3
perimeter = 9 + √3 + 9 + √
3 + 5 − √3 + 5 − √
3 = 28 units.
11. (8� 14 ) m
(8� 14 ) m
(8 + √
14) (
8 − √14
)= 64 − 14 = 50 m2.
12.(√
15 − 2√
3)2
=(√
15)2 + 2
(√15
) (−2
√3)
+(−2
√3)2
= 15 − 4√
45 + 4 × 3 = 27 − 4√
45 = 27 − 12√
5.
13.(√
5 − √3) (√
5 + √3)
=(√
5)2 −
(√3)2 = 5 − 3 = 2.
14.(√
2 + 1)
2 + 2(√
2 − 1)
= 2√
2 + 2 + 2√
2 − 2
perimeter = 4√
2
area =(√
2 + 1) (√
2 − 1)
=(√
2)2 − 12 = 2 − 1 = 1.
15.xy
x − y=
(2√
2 − 1) (
2√
2 + 1)
(2√
2 − 1)
−(
2√
2 + 1)
=(
2√
2)2 − 12
−2= −7
2.
16.
√125
20=
√5 × 5 × 5
4 × 5= 5
2.
17.√
26(
2√
13 − 2√
2)
= √2√
13(
2√
13 − 2√
2)
= 26√
2 − 4√
13.
20 Pass Publications GCSE Mathematics Higher
18.
(7 − √
2) (
7 + √2)
47=
(7)2 −(√
2)2
47= 49 − 2
47= 47
47= 1.
19.
(6 − √
5) (
6 + √5)
√31
= 36 − 5√31
= 31√31
×√
31√31
= √31.
20.1
a+ 1
b= a + b
ab
a = 2 − √2, b = 2 + √
2 positive irrational numbers
a + b
ab=
(2 − √
2)
+(
2 + √2)
(2 − √
2) (
2 + √2)
= 4
4 − 2= 4
2= 2 ⇒ ab
a + b= 2
4= 1
2.
21.√
7 × √63 = √
7 × √7√
9 = 7 × 3 = 21.
22.√
5 × √500 = k
√5
√5 × √
5√
100 = 50
k = 50√5
×√
5√5
= 10√
5.
23.
√243 − √
3√12
=√
3√
81 − √3√
3√
4
=√
3(9 − 1)√3(2)
= 8
2= 4.
Pass Publications GCSE Mathematics Higher 21
EXERCISES
Topic 5 ∼ Significant figures and decimal places
1. Consider the number 3.141592654 which is correct to 10 significant figures. Writedown the number correct to:
(a) nine significant figures (b) seven significant figures
(c) five significant figures (d) three significant figures and
(e) one significant figure.
2. Consider the number 0.00125. Write down this number to 2 and 1 significant figures.
3. Write the following numbers correct to the approximation given in brackets:
(a) 0.12345 (3 s.f.) (b) 476.7 (3 s.f.)
(c) 46.9539 (4 s.f.) (d) 0.0098765 (5 s.f.)
(e) 0.0098765 (4 s.f.) (f) 0.0098765 (3 s.f.)
(g) 35 × 19 (3 s.f.) (h) 137 × 679 (4 s.f.)
(i) 37.5 × 139.65 (3 s.f.) (j) 479 × 0.012567 (3 s.f.)
(k) 29 × 39 × 767 (3 s.f.).
4. The mass of the earth is calculated to be 5.976 × 1024 kg to 4 significant figures. Writedown this mass to 3, 2 and 1 s.f.
5. The electronic mass is given as 9.109534 × 10−31 kg. Write down this to 3 s.f.
6. Round off to the nearest whole number the following:
(a) 9.3 (b) 7.56
(c) 7.499 (d) 999
(e) 4.75 × 3.76 (f) 0.37 × 6.99 × 37.
7. Round off the following to the nearest 10:
(a) £375 (b) 524 kg
(c) 35.95 mm (d) 10.99 m.
8. 125975 to:
(a) the nearest 10 (b) the nearest 100
(c) the nearest 1000 (d) the nearest 10000.
22 Pass Publications GCSE Mathematics Higher
SolutionsTopic 5
1. 3.141592654
(a) 3.14159265 (b) 3.141593
(c) 3.1416 (d) 3.14 (e) 3.
2. 0.0013 to 2 s.f.
0.001.
3. (a) 0.123 (b) 477 (c) 46.95
(d) 0.0098765 (e) 0.009877 (f) 0.00988
(g) 35 × 19 = 665 (h) 137 × 679 = 93020
(i) 37.5 × 139.65 = 5236.875 = 5240 to 3 s.f.
(j) 479 × 0.012567 = 6.019593 = 6.02 to 3 s.f.
(k) 29 × 39 × 767 = 867477 = 867000 to 3 s.f.
4. 5.976 × 1024 kg
5.98 × 1024 kg to 3 s.f.
6.0 × 1024 kg to 2 s.f.
6 × 1024 kg to 1 s.f.
5. 9.109534 × 10−31 kg
9.11 × 10−31 kg to 3 s.f.
6. (a) 9 (b) 8 (c) 7 (d) 999
(e) 4.75 × 3.76 = 17.86 (18)
(f) 0.37 × 6.99 × 37 = 95.6931 (96).
7. (a) £375 = £380 to the nearest 10 (b) 524 kg = 520 kg to the nearest 10
(c) 35.95 mm = 40 mm to the nearest 10 (d) 10.99 m = 10 m to the nearest 10.
8. (a) 125980 (b) 126000 (c) 126000 (d) 130000.
Pass Publications GCSE Mathematics Higher 23
EXERCISES
Topic 6 ∼ Upper and lower bounds
1. The mass of a young girl is 45 kg correct to the nearest kg. What are the upper andlower bounds of the mass?
2. The height of a basket ball player is 2.05 m to the nearest centimetre. What are theupper and lower bounds of the height?
3. Asif weighs a bag of potatoes. He records the mass as 1.1 kg. The mass is recorded tothe nearest tenth of a kilogram. What are the upper and lower bounds of the possiblemass?
4. A straight road is measured to the nearest 5 m, the road is 1000 m long. Find the actuallength.
5. A rectangular garden is 10 m × 6 m, the measurements are given to the nearest metre.Determine the upper and lower bounds of the sides and of the area.
6. The area of a triangle is given as 12× base × height
h
b
If h = 20 cm to the nearest centimetre and b = 10 cm to the nearest centimetre, findthe maximum and minimum areas.
7. In the diagram there are three shapes, a square, a rectangle and an equilateral triangle
3.5 4.5 1.5
3.51.5 1.5
2.5
(i) (ii) (iii)
All the sides are given in mm and to the nearest tenth of a mm.
(a) Find the longest and the shortest sides.
(b) Find the maximum and minimum perimeters.
24 Pass Publications GCSE Mathematics Higher
8. The sides are given to the nearest hundredth of a centimetre.
h
5.00 cm
5.00 cm5.00 cm
(a) Find the longest and the shortest height, h.
(b) Find the maximum and minimum area.
9. The volume of a metal block of copper is 50 cm3 to the nearest 1 cm3, the density ofcopper is 8.9 g/cm3 to the nearest tenth. Calculate:
(a) The upper bound of its mass.
(b) The lower bound of its mass.
Approximations
10. Write down the following numbers to the nearest thousand:
(a) 7399 (b) 90501 (c) 13099
(d) 656935 (e) 75555 (f) 2599.
11. Write down the following numbers to the nearest integer:
(a) 9.6 (b) 10.7 (c) 10.45 (d) 15.55
(e) 5.125 (f) 5.509 (g) 5.499 (h) 6.53
(i) 56.63 (j) 839.6 (k) 1023.5 (l) 15235.435.
12. Write down the following numbers to the nearest ten:
(a) 29 (b) 24 (c) 35 (d) 127 (e) 3,059
(f) 5,395 (g) 6,949 (h) 79,499 (i) 856.79 (j) 7,509.
13. Find the sum of 7,679, 9,699, 10,950, 15,999. Give your answer as an approximationto the nearest ten.
14. Find the quotients of the following numbers to the nearest unit:
(a) 769 by 29 (b) 76 by 23
(c) 623 by 24 (d) 87 by 4
(e) 425 by 23 (f) 576 by 26
(g) 975 by 22 (h) 699 by 28.
15. A book contains 21,976 words. Write down the number of words to the nearest
(i) ten (ii) hundred (iii) thousand.
Pass Publications GCSE Mathematics Higher 25
Estimations
16. Without the use of a calculator, workout the following estimations to the nearest ten:
(a)7.69 × 9.49
3.75(b)
(3.72 − 2.22
) × 103
10.49
(c)47.9 × 8.90
24.9.
17. Without the use of a calculator give the answer to 1 s.f. for the following:
(a)4.93 × 5.47
24.9(b)
5.79 × 7.79
4.75
(c)655 × 856
66 × 207.
18. Estimate the following quotients to 1 s.f.:
(a)7699 × 3333
8888(b)
279.5 ÷ 62.9
55.5 ÷ 14.9
(c)3.25 × 37.5 × 7.55
39.9.
19. (a) Use the calculator to work out the value of√
2.562+4.353.95−1.05 . Write down all the figures
on your calculator display.
(b) Give your answer to part (a) to an appropriate degree of accuracy.
20. Work out an estimate for the following quotients:
(a)48.0356
4.012 (b)27.975
7.035(c)
77.95
0.01.
21. Use the calculator to work out the value of 27.9×13.932.7−14.9
(a) Write down all the figures on your calculator display.
(b) Write your answer to part (a) to an appropriate degree of accuracy.
26 Pass Publications GCSE Mathematics Higher
SolutionsTopic 6 ∼ Estimations and Errors
1. 45.5 kg is the upper bound of the mass, 44.5 kg is the lower bound of the mass.
If m is the mass of the young girl
44.5 < m < 45.5
m = 45.499 this is 45 to 2 s.f.
m = 44.501 this is 45 to 2 s.f.
2. 2.055 m the upper bound of the height
2.045 m the lower bound of the height.
If h is the height of the basket ball player
204.5 cm < h < 205.5 cm
h = 205.499 cm which to the nearest cm is 205
h = 204.501 cm which to the nearest cm is 205.
3. 1.15 kg is the upper bound of the mass. 1.05 kg is the lowest bound of the mass.
4. The length of the road is between 997.5 m and 1002.5 m.
1000 m ± 2.5 m recorded to the nearest x, M ± 1
2x.
5.6 m
10 m
9.5 m < length < 10.5 m
5.5 m < width < 6.5 m
9.5 × 5.5 m2 < Area < 10.5 × 6.5 m2
52.25 m2 < A < 68.25 m2.
6. Area max = 20.5 × 10.5
2= 107.625 cm2 108 cm2 to 3 s.f.
Area min = 19.5 × 9.5
2= 92.625 cm2 92.6 cm2 to 3 s.f.
Pass Publications GCSE Mathematics Higher 27
7. (a) (i) 3.55 mm longest; 3.45 mm shortest side
(ii) 4.55 mm, 4.45 mm, 2.25 mm, 2.45 mm
(iii) 1.55 mm, 1.45 mm
(b) (i) 4 × 3.55 = 14.2 mm maximum perimeter
4 × 3.45 = 13.8 mm minimum perimeter
(ii) 2 × 4.55 + 2 × 2.55 = 14.2 mm
2 × 4.45 + 2 × 2.45 = 13.8 mm
(iii) 3 × 1.55 = 4.65 mm
3 × 1.45 = 4.35 mm.
8.
h
2.50cm
2.50cm
5.00 cm5.00 cm
(a) hmax = √5.0052 − 2.4952
= 4.33877863 = 4.34 cm to 3 s.f.
hmin =√
4.9952 − 2.5052 = 4.321458087
= 4.32 cm to 3 s.f.
(b) A = 1
2hb �⇒ Amax = 1
2hmaxbmax
= 1
24.33877863 × 5.005
= 10.85779352 = 10.9 cm2 to 3 s.f.
Amin = 1
2hmin × bmin
= 1
2× 4.321458087 × 4.995
= 10.79284157 = 10.8 cm2 to 3 s.f.
28 Pass Publications GCSE Mathematics Higher
9. Vcu = 50 cm3, ρcu = 8.9 g/cm3
(a) mmax = ρcuVcu = 50.5 × 8.95
= 451.975 g = 452 g to 3 s.f.
min = ρcuVcu = 49.5 × 8.85
= 438.075 = 438 g to 3 s.f.
Approximation.
10. (a) 7000 (b) 91000 (c) 13000
(d) 657000 (e) 76000 (f) 3000.
11. (a) 10 (b) 11 (c) 10 (d) 16 (e) 5 (f) 6
(g) 5 (h) 7 (i) 57 (j) 840 (k) 1024 (l) 15235
12. (a) 30 (b) 20 (c) 40 (d) 130 (e) 3060
(f) 5400 (g) 6950 (h) 79500 (i) 860 (j) 7510
13. 76799699
1095015999 +44327 44330 to the nearest 10
14. (a)769
29= 26.51724138 = 27 to the nearest unit
(b)79
23= 3.434782609 = 3 to the nearest unit
(c)623
24= 25.95833333 = 26 to the nearest unit
(d)87
4= 21.75 = 22 to the nearest unit
(e)425
23= 18.47876087 = 18 to the nearest unit
(f)576
26= 22.15384615 = 22 to the nearest unit
(g)975
22= 44.31818182 = 44 to the nearest unit
(h)699
28= 24.96428571 = 25 to the nearest unit
Pass Publications GCSE Mathematics Higher 29
15. (i) 21980 (ii) 22000 (iii) 22000
Estimations.
16. (a)7.69 × 9.49
3.75= 8 × 9
4= 18
(b)
(3.72 − 2.22
) × 103
10.49= 1.5 × 5.9 × 103
10.49
= 1.5 × 6 × 100
10= 900
10= 90
(c)47.9 × 8.90
24.9= 50 × 9
25= 18.
17. (a)4.93 × 5.47
24.5= 5 × 5
25= 1 to 1 s.f.
(b)5.79 × 7.79
24.9= 6 × 8
25= 48
25= 2 to 1 s.f.
(c)655 × 856
66 × 207=70010 ×8004
70 ×200= 40.
18. (a)7699 × 3333
8888= 7700 × 3000
9000= 7700
3= 2900 = 3000 to 1 s.f.
(b)279.5 ÷ 62.9
55.5 ÷ 14.9=
300
6060
15
= 5
4= 1 to 1 s.f.
(c)3.25 × 37.5 × 7.55
39.9= 3 × 40 × 7
40= 20 to 1 s.f.
19. (a)
√2.562 + 4.35
3.95 − 1.05=
√10.9036
2.9= 3.302059963
2.9= 1.138641367 = 1.14 to 3 s.f.
20. (a)48.0356
4.012 = 48
16= 3 (b)
27.975
7.035= 28
7= 4 (c)
77.95
0.01= 7795 = 7800.
21. (a)27.9 × 13.9
32.7 − 14.9= 387.81
17.8= 21.78707865 (b) 21.8 to 3 s.f.
30 Pass Publications GCSE Mathematics Higher
EXERCISES
Topic 7 ∼ Fractions
1.8
3− 5
3÷ 20
212. 1 −
(1
4+ 3
4× 4
5
)
3.1
2+ 1
3+ 1
44. 1
1
6+ 3
7+ 1
42
5. 1 −(
2
3+ 1
4
)6. 2
2
3+ 1
3
6+ 1
1
9
7.
(3 − 5
2
)÷
(1 + 1
5
)8. 3
2
3+ 2
3
5
9.7
4− 3
4× 8
910. 1
3
8+ 2
7
8
11.1
5− 6
5÷ 10
3512. 12
2
3+ 6
3
7
13.1
5− 3
5÷ 12
714.
1
8+ 2
8+ 3
8
15.
(2 + 3
4
)÷
(3
4− 5
)16.
1
2+ 2
3+ 3
4+ 4
5
17.26
7− 22
7× 10
3318. 2
1
4+ 3
1
8+ 4
1
16
19. 13
4÷ 7
420. 4
3
4− 1
3
8
21.16
5− 6
5÷ 10
2522. 2
15
16− 1
5
8
23. 23
8÷ 1
3
424. 2
3
9− 1
3
25. 33
4÷ 1
1
426.
43
64− 7
16
27. 23
8+ 1
3
428. 5
4
5− 2
3
5
29. 33
4− 1
1
430. 1
3
4+ 5
6− 1
3
Pass Publications GCSE Mathematics Higher 31
31.33
4 × 45
178 ÷ 3
8
32.3
11× 5
2
5
33.3
7− 6
7÷ 30
734.
1 − 35
310 − 1
5
35. 21
2+ 3
7
8÷ 3
1
436.
(3
4− 1
2
)÷
(3
1
2+ 2
1
4
)2
37. Find the sum of 234, 33
5 and 412.
38. Subtract 178 from the sum of 23
8 + 349.
39. Find the difference between 735 and 32
3.
40. From the sum of 338 and 21
4 subtract 3 116.
41. Subtract 238 from the difference between 4 3
16 and 1 432.
42. Find the product of 7 316 and 13
4.
43. 32
5÷ 2
1
1044. 2
3
5÷ 1
1
20
45. Simplify51
4 × 23
114 × 11
8
46.43
5
3 310 ÷ 1
10
47.41
4 × 23
14 × 23
5
48.
(1
4+ 1
3
)× 3
549.
145 ÷ 310
1523 × 13
5
50. Multiply the sum of 335 and 23
7 by 237.
32 Pass Publications GCSE Mathematics Higher
SolutionsTopic 7 ∼ Fractions
1.8
3− 5
3÷ 20
21= 8
3− 5
3× 21
20= 8
3− 7
4= 32
12− 21
12= 11
12.
2. 1 −(
1
4+ 3
4× 4
5
)= 1 −
(1
4+ 3
5
)= 1 − 5 + 12
20= 1 − 17
20= 3
20.
3.1
2+ 1
3+ 1
4= 6
12+ 4
12+ 3
12= 13
12= 1
1
12.
4. 11
6+ 3
7+ 1
42= 7
6+ 3
7+ 1
42= 7 × 7
6 × 7+ 3 × 6
7 × 6+ 1
42
= 49
42+ 18
42+ 1
42= 68
42= 1
26
42= 1
13
21.
5. 1 −(
2
3+ 1
4
)= 1 −
(2 × 4
3 × 4+ 3 × 1
4 × 3
)
= 1 −(
8
12+ 3
12
)= 1 − 11
12= 1
12.
6. 22
3+ 1
3
6+ 1
1
9= 8
3+ 9
6+ 10
9
= 8 × 6
3 × 6+ 9 × 3
6 × 3+ 10 × 2
9 × 2
= 48
18+ 27
18+ 20
18= 95
18+ 5
5
18.
7.
(3 − 5
2
)÷
(1 + 1
5
)=
(6
2− 5
2
)÷
(5
5+ 1
5
)
= 1
2÷ 6
5= 1
2× 5
6= 5
12.
8. 32
3+ 2
3
5= 11
3+ 13
5= 11 × 5
15+ 13 × 3
5 × 3
= 55
15+ 39
15= 94
15= 6
4
15.
Pass Publications GCSE Mathematics Higher 33
9.7
4− 3
4× 8
9= 7
4− 1 × 2
1 × 3= 7
4− 2
3= 7 × 3
4 × 3− 2 × 4
3 × 4= 21
12− 8
12= 13
12= 1
1
12.
10. 13
8+ 2
7
8= 11
8+ 23
8= 34
8= 4
2
8= 4
1
4.
11.1
5− 6
5÷ 10
35= 1
5− 6
5× 35
10= 1
5− 3 × 7
1 × 5= 1
5− 21
5= −20
5= −4.
12. 122
3+ 6
3
7= 38
3+ 45
7= 38 × 7
3 × 7+ 45 × 3
7 × 3= 266
21+ 135
21= 401
21= 19
2
21.
13.1
5− 3
5÷ 12
7= 1
5− 3
5× 7
12= 1
5− 1 × 7
5 × 4= 4 × 1
5 × 4− 7
20= 4 − 7
20= − 3
20.
14.1
8+ 2
8+ 3
8= 6
8= 3
4.
15.
(2 + 3
4
)÷
(3
4− 5
)=
(2 × 4
4+ 3
4
)÷
(3
4− 5 × 4
1 × 4
)
=(
8
4+ 3
4
)÷
(3
4− 20
4
)= 11
4÷ −17
4= 11
4× − 4
17= −11
17.
16.1
2+ 2
3+ 3
4+ 4
5
= 30
2 × 30+ 2 × 20
3 × 20+ 3 × 15
4 × 15+ 4 × 12
5 × 12
= 30
60+ 40
60+ 45
60+ 48
60= 163
60= 2
43
60.
17.26
7− 22
7× 10
33= 26
7− 2
7× 10
3= 26 × 3
7 × 3− 20
21= 78 − 20
21= 58
21= 2
16
21.
18. 21
4+ 3
1
8+ 4
1
16= 9
4+ 25
8+ 65
16
= 9 × 4
4 × 4+ 25 × 2
8 × 2+ 65
16= 36
16+ 50
16+ 65
16= 151
16= 9
7
16.
19. 13
4÷ 7
4= 7
4÷ 7
4= 7
4× 4
7= 1.
34 Pass Publications GCSE Mathematics Higher
20. 43
4− 1
3
8= 19
4− 11
8= 19 × 2
4 × 2− 11
8= 38
8− 11
8= 27
8= 3
3
8.
21.16
5− 6
5÷ 10
25= 16
5− 6
5× 25
10= 16
5− 3 × 5
1 × 5= 16
5− 15
5= 1
5.
22. 215
16− 1
5
8= 2
15
16− 1
10
16= 1
5
16.
23. 23
8÷ 1
3
4= 2
3
8÷ 1
6
8= 19
8÷ 14
8= 19
8× 8
14= 19
14= 1
5
14.
24. 23
9− 1
3= 21
9− 3
9= 18
9= 2.
25. 33
4÷ 1
1
4= 15
4÷ 5
4= 15
4× 4
5= 3.
26.43
64− 7
16= 43
64− 7 × 4
16 × 4= 43 − 28
64= 15
64.
27. 23
8+ 1
3
4= 2
3
8+ 1
6
8= 3
9
8= 4
1
8.
28. 54
5− 2
3
5= 29
5− 13
5= 16
5= 3
1
5.
29. 33
4− 1
1
4= 2
2
4= 2
1
2.
30. 13
4+ 5
6− 1
3= 7
4+ 5
6− 1
3= 7
4+ 5
6− 2
6
= 7
4+ 3
6= 7 × 3
4 × 3+ 3 × 2
6 × 2= 21 + 6
12= 27
12= 2
3
12= 2
1
4.
31.3
3
4× 4
5
17
8÷ 3
8
=15
4× 4
515
8× 8
3
= 3
5.
32.3
11× 5
2
5= 3
11× 27
5= 81
55= 1
26
55.
33.3
7− 6
7÷ 30
7= 3
7− 6
7× 7
30= 3
7− 1
5= 3 × 5
7 × 5− 1 × 7
5 × 7= 15 − 7
35= 8
35.
Pass Publications GCSE Mathematics Higher 35
34.1 − 3
53
10− 1
5
=5
5− 3
53
10− 2
10
=2
51
10
= 2
5× 10
1= 4.
35. 21
2+ 3
7
8÷ 3
1
4= 5
2+ 31
8÷ 13
4= 5
2+ 31
8× 4
13= 5
2+ 31 × 1
26
= 5 × 13
2 × 13+ 31
26= 65 + 31
26= 96
26= 3
18
26= 3
9
13.
36.
(3
4− 1
2
)÷
(3
1
2+ 2
1
4
)2
=(
3
4− 2
4
)÷
(7
2+ 9
4
)2
= 1
4÷
(14
4+ 9
4
)2
= 1
4÷
(23
4
)2
= 1
4× 16
232 = 4
529.
37. 23
4+ 3
3
5+ 4
1
2= 2 + 3 + 4 + 3
4+ 3
5+ 1
2= 9 + 3
4+ 3
5+ 2
4
= 9 + 5
4+ 3
5= 10 + 1
4+ 3
5= 10 + 5
20+ 12
20= 1017
20.
38.
(2
3
8+ 3
4
9
)− 1
7
8= 19
8+ 31
9− 15
8
= 31
9+ 4
8= 31
9+ 1
2= 62 + 9
18= 71
18= 3
17
18.
39. 73
5− 3
2
3= 38
5− 11
3= 38 × 3
5 × 3− 11 × 5
3 × 5= 114
15− 55
15= 59
15= 3
14
15.
40.
(3
3
8+ 2
1
4
)− 3
1
16=
(27
8+ 9
4
)− 49
16= 27
8+ 18
8− 49
16= 45
8− 49
16
= 45 × 2
8 × 2− 49
16= 90
16− 49
16= 41
16= 2
9
16.
41.
(4
3
16− 1
4
32
)− 2
3
8= 67
16− 36
32− 19
8
= 67 × 2
16 × 2− 36
32− 76
32= 134 − 36 − 76
32= 22
32= 11
16.
36 Pass Publications GCSE Mathematics Higher
42. 73
16× 1
3
4= 115
16× 7
4= 805
64= 12
37
64.
43. 32
5÷ 2
1
20= 17
5÷ 21
10= 17
5× 10
21= 34
21= 1
13
21.
44. 23
5÷ 1
1
20= 13
5÷ 21
20= 13
5× 20
21= 52
21= 2
10
21.
45.5
1
4× 2
3
11
4× 1
1
8
=21
4× 2
35
4× 9
8
=7
245
32
= 7
2× 32
45= 7 × 16
45= 112
45= 2
22
45.
46.4
3
5
33
10÷ 1
10
=23
533
10× 10
1
= 23
5 × 33= 23
165.
47.4
1
4× 2
31
4× 2
3
5
=17
4× 2
31
4× 13
5
=17
613
20
= 17
6× 20
13= 17 × 10
39= 170
39= 4
14
39.
48.
(1
4+ 1
3
)× 3
5=
(3
12+ 4
12
)× 3
5= 7
12× 3
5= 7
20.
49.1
4
5÷ 3
10
152
3× 1
3
5
=9
5÷ 55
152
3× 8
5
=9
5× 15
5516
15
= 27
55÷ 16
15
= 27
55× 15
16= 27 × 3
11 × 16= 81
176.
50.
(3
3
5+ 2
3
7
)× 2
3
7=
(18
5+ 17
7
)× 17
7= 18 × 7 + 17 × 5
35× 17
7
= 126 + 85
35× 17
7= 211 × 17
35 × 7= 3587
245= 14
157
245.
Pass Publications GCSE Mathematics Higher 37
EXERCISES
Topic 8 ∼ Sequences
1. Write down the 6th and 7th terms of the sequence 12, 22, 32, 42, 52, . . . , . . . .
2. Write down the 5th and 6th terms of the sequence, 3, 12, 27, 48, . . . , . . . .
3. Write down the 9th and 10th terms of the sequence 11,
12,
13, . . . .
4. Calculate the nth and 50th terms:
(a) 1, 4, 7, 10, 13, 16, . . .
(b) 1, 2, 4, 8, 16, 32, . . .
(c) 10, 8, 6, 4, 2, 0, −2, . . . .
5. 2 × 3, 3 × 4, 4 × 5, . . . , . . . . Find the next two terms.
6. Write the next three terms of the sequence 1, 1, 2, 3, 5, 8, . . . , . . . . . . . .
7. Write the next three terms of the sequence a, a, 2a, 3a, 5a, 8a, . . . .
8. 1, 12,
34,
58,
1116,
2132, . . . , . . . . Write the next term.
9. 10, 20, 15, 1712, 161
4, 1678, . . . . Write the next three terms.
10. Write down three terms of an arithmetic sequence.
11. Write down five terms of a geometric series.
12. Write down the geometric mean of the sequence 3, 9, 27.
13. What is the common ratio of the sequence 1, 13,
19,
127,
181, . . .
14. Write down two harmonic sequences.
15. Here are the first five terms of a sequence 1, 3, 6, 10, 15, . . . , . . . . Find the followingfour terms.
16. Find the 4th and 5th terms:
(a) 0.01, 0.0001, 0.000001, . . . , . . .
(b) 12,
34,
58, . . . , . . .
(c) 15, 11, 7, . . . , . . .
17. 1, 23,
35,
47,
59, . . . , . . . . Find the missing terms.
18. 2, 212, 22
3, 234, 24
5, . . . . What is the convergent limit?
19. Write down a divergent sequence.
20. Write down an oscillating sequence.
38 Pass Publications GCSE Mathematics Higher
SolutionsTopic 8 ∼ Sequences
1. 62, 72.
2. 3, 12, 27, 48, . . .
12 × 3, 22 × 3, 32 × 3, 42 × 3, 52 × 3, 62 × 3
∴ 75, 108.
3.1
9,
1
10since
1
nis the nth term
substitute n = 9 and n = 10.
4. (a) 1, 4, 7, 10, 13, 16, . . .
the differences are constant, 3, therefore the nth term is linear an + b
If n = 1, a + b = 1 . . . (1)
n = 2, 2a + b = 4 . . . (2)
(2) − (1) = a = 3, substitute in (1)
3 + b = 1 ⇒ b = −2
therefore 3n − 2 is the nth term.
If n = 50, 3 × 50 − 2 = 150 − 2 = 148
the nth and 50th terms are 3n − 2 and 148.(b) 1, 1×2, 1×2×2, 1×2×2×2, . . .
1, 2, 4, 8, . . . 2n−1
If n = 1, 21−1 = 20 = 1;
If n = 2, 22−1 = 2 etc
If n = 50, 249 is the 50th term.
(c) −2n + 12, −2(50) + 12 = −100 + 12 = −88.
5. 2 × 3, 3 × 4, 4 × 5, 5 × 6, 6 × 7.
6. 1, 1, 2, 3, 5, 8, 13, 21, 34 Fibonacci series
1 + 1 = 2, 1 + 2 = 3, 3 + 5 = 8, 5 + 8 = 13,
8 + 13 = 21, 13 + 21 = 34.
7. a, a, 2a, 3a, 5a, 8a, 13a, 21a, 34a Fibonacci series.
Pass Publications GCSE Mathematics Higher 39
8. 1,1
2,
3
4,
5
8,
11
16,
21
32
1 + 1
22
=3
22
= 3
41
2+ 3
42
=2
4+ 3
42
=5
42
= 5
83
4+ 5
82
=6
8+ 5
82
=11
82
= 11
165
8+ 11
162
=10
16+ 11
162
=21
162
= 21
3211
16+ 21
322
=22
32+ 21
322
= 43
64.
9. 10, 20, 15, 171
2, 16
1
2, 16
7
8, 16
9
16, 16
23
32, 16
41
6410 + 20
2= 30
2= 15
20 + 15
2= 35
2= 17
1
2
15 + 171
22
=32
1
22
= 161
4
171
2+ 16
1
42
=33
3
42
= 161
2+ 3
8= 16
7
8
161
4+ 16
7
82
=32 + 9
82
= 169
16
167
8+ 16
9
162
=32 + 23
162
= 16 + 23
32= 16
23
32
169
16+ 16
23
322
= 16 + 41
64= 16
41
64.
40 Pass Publications GCSE Mathematics Higher
10. 2, 4, 6 we add 2 to the first term, makes 4
we add 2 to the second term, makes 6.
11. 1, 1 × 1
3, 1 × 1
3× 1
3, 1 × 1
3× 1
3× 1
3, 1 × 1
3× 1
3× 1
3× 1
3, 1,
1
3,
1
9,
1
27,
1
81.
12. 3, 9, 27; a, b, c is a geometric series
b = √ac is the geometric mean
therefore√
3 × 27 = √81 = 9 is the geometric mean.
13.
1
31
=1
91
3
= 1
3= common ratio.
14.1
5,
1
8,
1
11; 5, 8, 11 form an arithmetic sequence
1
6,
1
10,
1
146, 10, 14 form an arithmetic sequence.
15. 1, 3, 6, 10, 15, 21, 28, 36, 45
1 + 2 = 3
3 + 3 = 6
6 + 4 = 10
10 + 5 = 15
15 + 6 = 21
21 + 7 = 28
28 + 8 = 36
36 + 9 = 45
The rule is, we addone more every timeafter the first one.
16. (a) 0.01, 00001, 0.000001, 0.00000001, 0.0000000001
we multiply each time by 0.01
(b)1
2,
3
4,
5
8,
11
16,
21
321
2+ 3
42
=2
4+ 3
42
= 5
85
8+ 3
42
= 11
16
5
8+ 11
162
= 21
32
Pass Publications GCSE Mathematics Higher 41
(c) 15 − 4 = 11, 11 − 4 = 7,
7 − 4 = 3, 3 − 4 = −1.
17.1
1,
2
3,
3
5,
4
7,
5
9,
6
11,
7
13
The rule is: we add one to the numerator and the denominator is an odd numberincreasing by 2.
18. 2, 21
2, 2
2
3, 2
3
4, 2
4
5, . . . 3
5
6,
6
7,
7
8,
8
9,
9
10, ....
n
n + 1→ 1.
19. 3, 9, 27, 81, ....3 × 3n−1
we multiply each term by 3 indefinitely.
20. 2, −2, 2, −2.
42 Pass Publications GCSE Mathematics Higher
EXERCISES
Topic 9 ∼ Standard form
For the following, calculators are not allowed:
1. Simplify the following indicial arithmetic expressions:
(a) 25 (b) 2−4 (c) 4−12 (d) 25−1
2
(e) 813 (f) 8−1
3 (g) 16−14 (h) 125
13 .
2. Find the values of the following root expressions:
(a)√
25 (b) 3√
125 (c) 3√
64 (d)√
256 (e) 5√
32
(f) 4√
256 (g) 8√
256 (h) 4√
81 (i) 5√
243.
3. Evaluate the following:
(a)(2−1
)−2(b)
(3−1
)2(c)
(52
)−2(d)
(102
)−3(e)
(10−2
)−1
(f)1(
3−1)−3 (g)
1
10−5(h) 10−3 (i)
(10−2
)3(j)
((10−1
)−2)−3
.
4. Evaluate the following:
(a)
√9
4(b) 3
√64
27(c)
√1
32(d)
√36 × 4 (e) 3
√125
243.
5. Simplify:
(a) 3√
3 − √3 + 5
√3 (b) 5
√5 + √
5 (c)√
5 × √5 × √
5 × √5
(d)√
8 (e)√
27 (f)√
125 (g)√
75
(h)√
45 (i)
√1
20(j)
√25
8(k)
√64
9.
6. Express√
27(√
3 − 1)
in terms of prime numbers.
7. Express√
8(
3 − √32
)in terms of prime numbers.
8. Write in standard form and to 3 s.f.:
(a) 0.0396 (b) 0.569 (c) 5.89 (d) 0.0000349
(e) 11000 (f) 23.695 (g) 6495000 (h) 73990
(i)10
0.0002(j)
100
0.00001(k)
1
0.005.
Pass Publications GCSE Mathematics Higher 43
9. Write in full the following standard index forms:
(a) 7.35 × 10−6 (b) 3.65 × 109 (c) 1.23 × 105
(d) 3.5 × 10−6 (e) 4.756 × 101 (f) 1.234 × 10−5
(g) 3.75 × 10−3 + 2.75 × 10−6 (h) 7.95 × 105 − 2.95 × 104
(i) 2.5 × 106 × 5 × 10−5
(j)1.25 × 107
6.25 × 10−6 (k)2.187 × 10−10
2.43 × 10−12 .
10. Write one billion and one trillion in standard form.
11. Find the smallest number of the following:
3.5 × 103, 3.55 × 103, 3.55 × 10, 3.5 × 10−3.
12. In physics, the electronic charge is given as 1.6021892 × 10−19 C. Write this numberin full and to 3 s.f.
13. In physics, the Avogadro number in mol−1 is given as 6.022045 × 1023. Write thisnumber in full and to 3 s.f.
14. In electronics, Boltzman’s constant in Jk−1 is given as 1.38062 × 10−23. Write thisnumber in full and to 3 s.f.
15. In electronics, the electronic mass in kg is given as 9.109534 × 10−31. Write thisnumber in full and to 3 s.f.
16. In physics, Planck’s constant in Js is given as 6.626176 × 10−34. Write this numberin full and to 3 s.f.
17. In physics, the gravitational constant in Nm2kg−2 is given as 0.00000000006672.Write this number in standard form to 3 s.f.
18. Work out:
(a)(3 × 105
) + (7 × 104
)(b) 4 × 10−6 − 3 × 10−7
(c)(5 × 104
) × (6 × 105
)(d)
8 × 106
4 × 10−12 .
19. The wealth of the UK is estimated to be 5 trillion pounds. Write this number in full.
20. The wealth of the world is estimated to be £1022. Write this out in full and also writeit in words.
21. The mass of the earth is 5.976 × 1024 kg and the volume is given as 43π (6800 km)3.
What is the average density of the earth in standard form to 3 s.f.
44 Pass Publications GCSE Mathematics Higher
SolutionsTopic 9 ∼ standard form
1. (a) 25 = 2 × 2 × 2 × 2 × 2 = 32 (b) 2−4 = 1
24 = 1
2 × 2 × 2 × 2= 1
16
(c) 4−12 = 1
412
= 1√4
= 1
2(d) 25−1
2 = 1
2512
= 1√25
= 1
5
(e) 813 = 3
√8 = 2 (f) 8−1
3 = 1
813
= 13√
8= 1
2
(g) 16−14 = 1
1614
= 14√
16= 1
2(h) 125
13 = 3
√125 = 5.
2. (a)√
25 = 5 (b) 3√
125 = 5 (c) 3√
64 = 4
(d)√
256 = 16 (e) 5√
32 = 2 (f) 4√
256 = 4
(g) 8√
256 = 8√
16 × 16
= 8√
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 8√
28 = 21 = 2
(h) 4√
81 = 3 (i) 5√
243 = 5√35 = 3
55 = 31 = 3.
3. (a)(2−1
)−2 = 22 = 4 (b)(3−1
)−2 = 3−2 = 1
32 = 1
9
(c)(52
)−2 = 5−4 = 1
54 = 1
625(d)
(102
)−3 = 10−6 = 0.000001
(e)(10−2
)−1 = 102 = 100 (f)1
(3−1)−3 = 1
33 = 1
27
(g)1
10−5= 105 = 100000 (h) 10−3 = 1
103 = 1
1000
(i)(10−2
)3 = 10−6 = 1
1000000(j)
((10−1
)−2)−3 = 10−6 = 1
1000000.
4. (a)
√9
4= 3
2(b) 3
√64
27= 4
3(c)
√1
32= 1
4√
2
(d)√
36 × 4 = √36
√4 = 6 × 2 = 12
(e) 3
√125
243= 3
√5 × 5 × 5
3 × 3 × 3 × 3 × 3= 5
3 3√
9.
Pass Publications GCSE Mathematics Higher 45
5. (a) 3√
3 − √3 + 5
√3 = 7
√3 (b) 5
√5 + √
5 = 6√
5
(c)√
5 × √5 × √
5 × √5 = 5 × 5 = 25 (d)
√8 = √
2 × 2 × 2 = 2√
2
(e)√
27 = √3 × 3 × 3 = 3
√3 (f)
√125 = √
5 × 5 × 5 = 5√
5
(g)√
75 = √5 × 5 × 3 = 5
√3 (h)
√45 = √
3 × 3 × 5 = 3√
5
(i)1√20
= 1√2 × 2 × 5
= 1
2√
5(j)
√25
8=
√5 × 5
2 × 2 × 2= 5
2√
2
(k)
√64
9= 8
3.
6.√
27(√
3 − 1)
= √3 × 3 × 3
(√3 − 1
)= 3
√3
(√3 − 1
)= 3
(3 − √
3)
.
7.√
8(
3 − √32
)= 3
√8 − √
8√
8√
4 = 3 × 2√
2 − 8 × 2 = 2(
3√
2 − 23)
.
8. (a) 0.0396 = 3.96 × 10−2 (b) 0.569 = 5.69 × 10−1
(c) 5.89 = 5.89 × 100 (d) 0.0000349 = 3.49 × 10−5
(e) 11000 = 1.10 × 104 (f) 23.695 = 2.37 × 101
(g) 6495000 = 6.50 × 106 (h) 73990 = 7.40 × 104
(i)10
0.0002= 100000
2= 5.00 × 104 (j)
100
0.00001= 10000000 = 1.00 × 107
(k)1
0.005= 1000
5= 200 = 2.00 × 102.
9. (a) 7.35 × 10−6 = 0.00000735 (b) 3.65 × 109 = 3650000000
(c) 1.23 × 105 = 123000 (d) 3.5 × 10−6 = 0.0000035
(e) 4.756 × 101 = 47.56 (f) 1.234 × 10−5 = 0.00001234
(g) 3.75 × 10−3 + 2.75 × 0−6 = 0.00375 + 0.00000275 = 0.00375275
(h) 7.95 × 105 − 2.95 × 104 = 795000 − 29500 = 765500
(i) 2.5 × 106 × 5 × 10−5 = 12.5 × 101 = 125
(j)1.25 × 107
6.25 × 10−6 = 1
5× 1013 = 0.2 × 1013
(k)2.187 × 10−10
2.43 × 10−12 = 2187 × 10−13
243 × 10−14 = 9 × 10−13+14 = 9 × 10 = 90.
46 Pass Publications GCSE Mathematics Higher
10. One billion = 1000000000 = 1.00 × 109
One trillion = 1000000000000 = 1.00 × 1012.
11. 3500, 3550, 35.5, 0.0035
3.5 × 10−3 is the smallest number.
12. 1.6021892 × 10−19
= 0.00000000000000000016021892 C
= 0.000000000000000000160 C to 3 s.f.
13. 6.022045 × 1023
= 602204500000000000000000 mol−1
= 602000000000000000000000 mol−1.
14. 1.38062 × 10−23
= 0.0000000000000000000000138 Jk−1.
15. 9.109534 × 10−31 kg
= 0.000000000000000000000000000000911kg.
16. 0.000000000000000000000000000000000663 Js.
17. 0.00000000006672 = 6.67 × 10−11 Nm2 Kg−2.
18. (a)(3 × 105
) + (7 × 104
) = 300000 + 70000 = 370000 = 3.7 × 105
(b) 4 × 10−6 − 3 × 10−7 = 0.000004 − 0.0000003 = 0.0000397 = 3.97 × 10−5
(c)(5 × 104
) × (6 × 105
) = 30 × 109 = 3.00 × 1010
(d)8 × 106
4 × 10−12 = 2 × 1018 = 2.00 × 1018.
19. £5000000000000000.
20. £10000000000000000000000
Ten million trillion.
21. ρ = m
V= 5.976 × 1024
4
3π(6800000)3
= 4537.276454 = 4.54 × 103 kg/m3.