# Selina Concise Solutions for Chapter 4 Decimal Fractions (Decimals) Class 7 ICSE Mathematics

### Exercise 4 (A)

1. Convert the following into fractions in their lowest terms :
(i) 3.75
(ii) 0.5
(iii) 2.04
(iv) 0.65
(v) 2.405
(vi) 0.085
(vii) 8.025

2. Convert into decimal fractions

(ii) 79/100
(iii) 37/10,000
(iv) 7543/104
(v) 3/4

3. Write the number of decimal places in :
(i) 0.4762
(ii) 7.00349
(iii) 8235.403
(iv) 35.4
(v) 2.608
(vi) 0.000879

(i) In 0.4762, there are four places.
(ii) In 7.00349, there are five places.
(iii) In 8235.403, there are three places.
(iv) In 35.4, there is one place.
(v) In 2.608, there are three places.
(vi) In 0.000879, there are six places.

4. Write the following decimals as word statements :
(i) 0.4,0.9,0.1
(ii) 1.9, 4.4, 7.5
(iii) 0.02, 0.56, 13.06
(iv) 0.005,0.207, 111.519
(v) 0.8, 0.08, 0.008, 0.0008
(vi) 256.1, 10.22, 0.634

(i) 0.4 = zero point four, 0.9 = zero point nine, 0.1 = zero point one.
(ii) 1 .9 = one point nine, 4.4 = four point four, 7.5 = seven point five.
(iii) 0.02 = zero point zero two, 0.56 = zero point five six, 13.06 = thirteen point zero six.
(iv) 0.005 = zero point zero zero five, 0.207 = zero point two zero seven, 111.519 = one
hundred eleven point five one nine.
(v) 0.8 = zero point eight, 0.08 = zero point zero eight, 0.008 = zero point zero zero
eight, 0.0008 = zero point zero zero zero eight
(vi) 256.1 = Two hundred fifty six point one, 10.22 = Ten point two two, 0.634 = zero
point six three four.

5. Convert the given fractions into like fractions:
(i) 0.5,3.62,43.987 and 232.0037
(ii) 215.78, 33.0006, 530.3 and 0.03569
(i) 0.5, 3.62, 43.987 and 232.0037
In these decimals, the greatest places of decimal is 4
∴0.5 = 0.5000
3.62 = 3.6200
43.987 = 43.9870
232.0037 = 232.0037

(ii) 215.78, 33.0006, 530.3 and 0.03569
In these decimals, the greatest places of decimal is 5
∴ 215.78 = 215.78000
33.0006 = 33.00060
530.3 = 530.30000
0.03569 = 0.03569

### Exercise 4 (B)

(i) 0.5 and 0.37 (ii) 3.8 and 8.7
(iii) 0.02, 0.008 and 0.309
(iv) 0. 4136, 0. 3195 and 0.52
(v) 9.25, 3.4 and 6.666
(vi) 3.007, 0.587 and 18.341
(vii) 0.2, 0.02 and 2.0002
(viii) 6. 08, 60.8, 0.608 and 0.0608
(ix) 29.03, 0.0003, 0.3 and 7.2
(x) 3.4, 2.025, 9.36 and 3.6221

2. Subtract the first number from the second :
(i) 5.4, 9.8
(ii) 0.16, 4.3
(iii) 0.82, 8.6
(v) 2.237, 9.425
(vi) 41 .03, 59.46
(vii) 3.92. 26.86
(viii) 4.73, 8.5
(ix) 12.63, 36.2
(x) 0.845, 3.71

3. Simplify :
(i) 28.796 -13.42 – 2.555
(ii) 93.354 – 62.82 – 13.045
(iii) 36 – 18.59 – 3.2
(iv) 86 + 16.95 – 3.0042
(v) 32.8 – 13 – 10.725 +3.517
(vi) 4000 – 30.51 – 753.101 – 69.43
(vii) 0.1835 + 163.2005 – 25.9 – 100
(viii) 38.00 – 30 + 200.200 – 0.230
(ix) 555.555 + 55.555 – 5.55 – 0.555
(i) 28.796 - 13.42 -2.555
= 28.976 - (13.42 + 2.555)
= 28.796 - 15.975
= 12.821

(ii) 93.354 - 62.82 - 13.045
= 93.354 - (62.82 + 13.045)
= 93.354 - 75.865
= 17.489

(iii) 36 - 18.59 - 3.2
= 36 -(18.59 + 3.2)
= 36 - 21.79
= 14.21

(iv) 86 + 16.95 - 3.0042
= 102.95 - 3.0042 = 99.9458

(v) 32.8 - 13 - 10.725 + 3.517
= (32.8 + 3.517) - (13 + 10.725)
= 36.317 - 23.725 = 12.592

(vi) 4000 - 30.51 - 753.101 - 69.43
= 4000 - (30.51 + 753.101 + 69.43)
= 4000 - 853.041

(vii) 0.1835 + 163.2005 - 25.9 - 100
= (0.1835 + 163.2005) - (25.9 + 100)
= 163.3840 - 125.9 = 37.484

(viii) 38.00 - 30 + 200.200 - 0.230
= (38.00 + 200.200) -(30 + 0.230)
= 238.200 - 30.230
= 207.970 = 207.97

(ix) 555.555 + 55.555 - 5.55 - 0.555
= (555.555 + 55.555) - (5.55 + 0.555)
= 611.110 - 6.105 = 605.005

4. Find the difference between 6.85 and 0.685.
Difference between  6.85 and 0.685
= 6.85 - 0.685
= 6.165

5. Take out the sum of 19.38 and 56.025 then subtract it from 200.111  .
Sum of 19.38 + 56.025 = 75.405

6. Add 13.95 and 1.003; and from the result, subtract the sum of 2.794 and 6.2  .
Sum of 13.95 and 1.003
= 13.95 + 1.003  = 14.953

7. What should be added to 39.587 to give 80.375 ?
Sum = 80.375
Given number = 39.587
∴ The number which is to be added
= 80.375 - 39.587 = 40.788

8. What should be subtracted from 100 to give 19.29 ?
Sum = 100
The number = 19.29
∴ The number which is to be subtracted
= 100 - 19.29 = 80.71

9. What is the excess of 584.29 over 213.95 ?
Total = 584.29
Given number = 213.95
Required difference = 584.29 - 213.95 = 370.34

10. Evaluate :
(i) (5.4 - 0.8) + (2.97 - 1.462)
(ii) (6.25 + 0.36) - (17.2 - 8.97)
(iii) 9.004 + (3 - 2.462)
(iv) 879.4 - (87.94 - 8.794)

11. What is the excess of 75 over 48.29 ?
Excess of 75 over 48.29

12. If A = 237.98 and B = 83.47 .
Find :
(i) A - B
(ii) B - A .
(i) A - B
A = 237.98
B = 83.47
⇒ A - B = 154.51

(ii) B -A
= 83.47 - 237.98
= -154.51

13. The cost of one kg of sugar increases from ₹28.47 to  ₹32.65. Find the increase in cost.
Initial cost of sugar = ₹ 28.47
Increase cost of sugar = ₹ 32.65
∴ Increase of sugar in cost = ₹5.18

### Exercise 4 (C)

1. Multiply:
(i) 0.87 by 10
(ii) 2.948 by 100
(iii) 6.4 by 1000
(iv) 5.8 by 4
(v) 16.32 by 28
(vi) 5. 037 by 8
(vi) 4.6 by 2.1
(viii) 0.568 by 6.4
(i) 0.87 × 10 = 8.7
(ii) 2.948 × 100 = 294.8
(iii) 6.4 × 1000 = 6400
(iv) 5.8 × 4 = 23.2
(v) 16.32 × 32 = 456.96
(vi) 5.037 × 8 = 40.296
(vii) 4.6 × 2.1 = 9.66
(viii) 0.568 × 6.4 = 3.6352

2. Multiply each number by 10, 100, 1000 :
(i) 0.5
(ii) 0.112
(iii) 4.8
(iv) 0.0359
(v) 16.27
(vi) 234.8
(i) 0.5 × 10 = 5,
⇒ 0.5× 100 = 50,
⇒ 0.5 × 1000 = 500

(ii) 0.112 × 10 = 1.12,
⇒ 0.112 × 100 =11.2,
⇒ 0.112 × 1000 = 112

(iii) 4.8 × 10 = 48,
⇒ 4.8 × 100 = 480,
⇒ 4.8 × 1000 = 4800

(iv) 0.0359 × 10 = 0.359,
⇒ 0.0359 × 100 = 3.59,
⇒ 0.0359 × 1000 = 35-9

(v) 16.27 × 10 = 162.7,
⇒ 16.27 × 100 = 1627,
⇒ 16.27 × 1000 = 16270

(vi) 234.8 × 10 = 2348,
⇒ 234.8 × 100 = 23480,
⇒ 234.8 × 1000 = 234800

3. Evaluate:
(i) 5.897 × 2.3
(ii) 0.894 × 87
(iii) 0.01 × 0.001
(iv) 0.84 × 2.2 × 4
(v) 4.75 × 0.08 × 3
(vi) 2.4 × 3.5 × 4.8
(vii) 0.8 × 1.2 × 0.25
(viii) 0.3 × 0.03 × 0.003
(ix) 12.003 × (0.2)5

4. Divide :
(i) 54.9 by 10
(ii) 7.8 by 100
(iii) 324.76 by 1000
(iv) 12.8 by 4
(v) 27.918 by 9
(vi) 4.672 by 8
(vii) 4.32 by 1.2
(viii) 7.644 by 1.4
(ix) 4.8432 by 0.08
(i) 54.9 ÷ 10 = 5.49
(ii) 7.8 ÷ 100 = 0.078
(iii) 324.76 ÷ 1000 = 0.32476
(iv) 12.8 ÷ 4 = 3.2
(v) 27.918 ÷ 9 = 3.102
(vi) 4.672÷ 8 = 0.584
(vii) 4.32 ÷ 1.2
= 432 ÷ 120
= 3.6
(viii) 7.644 ÷ 1.4
= 7644 ÷ 1400
=5.46
(ix) 4.8432 ÷ 0.08
= 48432 ➗ 800
= 60.54

5. Divide each of the given numbers by 10, 100, 1000 and 10000
(i) 2.1
(ii) 8.64
(iii) 5.01
(iv) 0.0906
(v) 0.125
(vi) 111.11
(vii) 0.848 × 3
(viii) 4.906 × (0.2)2
(ix)  (1.2)2 × (0.9)2
(i) 2.1 ÷ 10 = 0.21
2.1 ÷ 100 = 0.021
2.1 ÷ 1000 = 0.0021
2.1 ÷ 10000 = 0.00021

(ii) 8.64 ÷ 10 = 0.864
8.64 ÷100 = 0.0864
8.64 ÷ 1000 = 0.00864
8.64 ÷ 10000 = 0.000864

(iii) 5.01 ÷ 10 = 0.501,
5.01 ÷ 100 = 0.0501
5.01 ÷ 1000 = 0.00501
5.01 ÷ 10000 = 0.000501

(iv) 0.0906 ÷ 10 = 0.00906
0.0906 ÷ 100 = 0.000906
0.0906 ÷ 1000 = 0.0000906
0.0906 ÷ 10000 = 0.00000906

(v) 0.125 ÷ 10 = 0.0125
0.125 ÷ 100 = 0.00125
0.125 ÷ 1000 = 0.000125
0.125 ÷ 10000 = 0.0000125

(vi) 111.11 ÷ 10 = 11.111
111.11 ÷ 100 = 1.1111
111.11 ÷ 1000 = 0.11111
111.11 ÷ 10000 = 0.011111

(vii) 0.848 × 3 = 2.544
2.544 ÷10 = 0.2544
2.544 ÷ 100 = 0.02544
2.544 ÷ 1000 = 0.002544
2.544 ÷ 10000 = 0.0002544

(viii) 4.906 × (0.2)2 = 4.906 × 0.2 × 0.2
= 4.906 × 0.04 = 0.19624
= 0.19624 ÷ 10 = 0.019624
= 0.19624 ÷ 100 = 0.0019624
= 0.19624 ÷ 1000 = 0.00019624
= 0.19624 ÷ 10000 = 0.000019624

(ix) (1.2)2 × (0.9)2 = 1.2×1.2×0.9×0.9 = 1.44×0.81 = 1.1664
1.1664 ÷ 10 = 0.11664
1.1664 ÷ 100 = 0.011664
1.1664 ÷ 1000 = 0.0011664
1.1664 ÷ 10000 = 0.00011664

6. Evaluate :
(i) 9.75 ÷ 5
(ii) 4.4064
÷ 4
(iii) 27.69
÷ 30
(iv) 19.25
÷ 25
(v) 20.64
÷ 16
(vi) 3.204
÷ 9
(vii) 0.125
÷ 25
(viii) 0.14616
÷ 72
(ix) 0.6227
÷ 1300
(x) 257.894
÷ 0.169
(xi) 6.3
÷ (0.3)3

(xi) 6.3 ÷ (0.3)3
= 6.3 ÷ (0.3 × 0.3)
= 6.3 ÷ (0.09)
= 630 ÷ 09
= 630 ÷ 9
= 70

7. Evaluate:
(i) 4.3 × 0.52 × 0.3
(ii) 3.2 × 2.5 × 0.7
(iii) 0.8 × 1.5 × 0.6
(iv) 0.3 × 0.3 × 0.3
(v) 1.2 × 1.2 × 0.4
(vi) 0.4 × 0.04 × 0.004
(vii) 0.5 × 0.6 × 0.7
(Viii) 0.5 × 0.06 × 0.007
(i) 4.3 × 0.52 × 0.3

(Sum of decimal places = 1 + 2+ 1 = 4)
∴ 4.3 × 0.52 × 0.3 = 0.6708

(ii) 3.2 × 2.5 × 0.7

(Sum of decimal places = 1+1+1 = 3)
∴ 3.2 × 2.5 × 0.7 = 5.600 or 5.6

(iii) 0.8 × 1.5 × 0.6
(Sum of decimal places = 1+1+1 = 3)
∴ 0.8 × 1.5 × 0.6 = 0.720 or 0.72

(iv) 0.3 × 0.3 × 0.3

(Sum of decimal places = 1+1+1 = 3)
∴ 0.3 × 0.3 × 0.3 = 0.027

(v)  1.2 × 1.2 × 0.4

(vi) 0.4 × 0.04 × 0.004
(Sum of decimal places = 1+2+3= 6)
∴ 0.4 × 0.04 × 0.004 = 0.000064

(vii) 0.5 × 0.6 × 0.7

(Sum of decimal places = 1 + 1+ 1 = 3)
∴ 0.5 × 0.6 × 0.7 = 0.210 or 0.21

(viii) 0.5 × 0.06 × 0.007

(Sum of decimal places = 1+ 2+ 3 = 5)
∴ 0.5 × 0.06 × 0.007 = 0.00021

8. Evaluate:
(i) (0.9)2
(ii) (0.6)2 × 0.5
(iii) 0.3 × (0.5)2
(iv) (0.4)3
(v) (0.2)3   × 5
(vi) (0.2)3   × 0.05
(i) (0.9)2
⇒ 0.9 × 0.9 = 0.81
(Sum of decimal places 1 + 1= 2)

(ii) (0.6)2  ×  0.5
⇒ 0.6 ×  0.6 × 0.5
⇒ 0.36 × 0.5 = 0.180 or 0.18
(Sum of decimal places = 1 + 1 + 1 = 3)

(iii) 0.3 × (0.5)2
⇒ 0.3 × 0.5 × 0.5
⇒ 0.3 × 0.25 = 0.075
(Sum of decimal places 1 + 1 + 1 = 3)

(iv) (0.4)3
⇒ 0.4 × 0.4 × 0.4
⇒ 0.16 × 0.4 = 0.064
(Sum of decimal places 1 + 1 + 1 = 3)

(v) (0.2)3   × 5
⇒ 0.2 × 0.2 × 0.2 × 5
⇒ 0.08 × 5 = 0.40 or 0.4
(Sum of decimal places 1 + 1 + 1 = 3)

(vi) (0.2)3   × 0.05
⇒ 0.2 × 0.2 × 0.2 × 0.05
⇒ 0.008 × 0.05 = 0.00040
(Sum of decimal places = 5)

9. Find the cost of 36.75 kg wheat at the rate of  ₹12.80 per kg.
Total weight of wheat  = 36.75 kg
Cost of 1 kg of wheat = ₹ 12.80
∴ Cost of 36.75 kg of wheat
= 36.75 × 12.80 = ₹ 470.40

10. The cost of a pen is ₹ 56.15. Find the cost of 16 such pens.
Cost of one pen = ₹ 56.15
∴ Cost of 16 pens
= ₹ 56.15 × 16 = ₹ 898.40

11. Evaluate:
(i) 0.0072 ÷ 0.06
(ii) 0.621 ÷ 0.3
(iii) 0.0532 ÷ 0.005
(iv) 0.01162 ÷ 0.14
(v) (7.5 × 40.4) ÷ 25
(vi) 2.1 ÷ (0.1 × 0.1)
(i)  0.0072 ÷ 0.06

12. Fifteen identical articles weigh 31.50 kg. Find the weight of each article.
Weight of 15 articles = 31.50 kg
∴ Weight of one article
= 31.50- 15 = 2.1 kg

13. The product of two numbers is 211.2. If one of these two numbers is 16.5, find the
other number.
The product of two numbers = 211.2
one number = 16.5
∴ Second number = 211.2 ÷ 16.5
= (211.2 × 10)/(16.5×10
= 2112/165 = 12.8

14. One dozen identical articles cost ₹45.96. Find the cost of each article.
∴ Weight of one dozen articles = ₹45.96
One dozen = 12
∴ Cost of one article = 45.96 + 12 = ₹3.83

### Exercise 4 (D)

1. Find whether the given division forms a terminating decimal or a non-terminating
decimal:
(i) 3 ÷ 8
(ii)8 ÷ 3
(iii) 6÷ 5
(iv) 5 ÷ 6
(v) 12.5 ÷ 4
(vi) 23 ÷ 0.7
(vii) 42 ÷ 9
(viii) 0.56 ÷ 0.11
(i) 3 ÷ 8 = 0.375
Hence it is terminating decimal.

(ii) 8 ÷ 3 = 2.666....
Hence it is non - terminating decimal.

(iii) 6 ÷ 5 = 1.2
Hence it is terminating decimal.

(iv) 5 ÷ 6 = 0.8333.....
Hence it is non - terminating decimal .

(v) 12.5 ÷ 4 = 3.125
Hence it is terminating decimal.

(vi) 23 ÷ 0.7
= 230 ÷ 7
= 32.8571428.......
Hence it is non - terminating decimal.

(vii) 42 ÷ 9 = 4.666.....
Hence it is non - terminating decimal.

(viii) 0.56 ÷ 0.11
= 56 ÷ 11
= 5.0909.....
Hence it is non - terminating decimal .

2. Express as recurring decimals :

(ii) 10/11
(iii) 5/6
(iv) 2/13
(v) 1/9
(vi) 17/90
(vii) 5/18
(viii) 7/12

3. Convert into vulgar fraction :

(i) 0. 3

(i) 0. 8

(i) 4. 4

(i) 23. 7

4. Convert into vulgar fraction :

(i) 0. 35

(i) 2. 23

(i) 1. 28

(i) 5. 234

5. Convert into vulgar fraction :

(i) 0. 37

(i) 0. 245

(i) 0. 685

(i) 0. 442

### Exercise 4 (E)

1. Round off:
(i) 0 .07, 0.112, 3.59, 9.489 to the nearest tenths.
(ii) 0.627, 100.479, 0 065 and 0.024 to the nearest hundredths.
(iii) 4.83,0.86,451 .943 and 9.08 to the nearest whole number.
(i) 0.07 = 0.1,
0.112 = 0.1
3.59 = 3.6,
9.489 = 9.5

(ii) 0.627 = 0.63,
100.479 = 100.48
0.065 = 0.07,
0.024 = 0.02

(iii) 4.83 = 5,
0.86= 1,
451.943 = 452
9.08 = 9

2. Simplify, and write your answers correct to the nearest hundredths :
(i) 18 .35 × 1.2
(ii) 62.89 × 0.02
(i) 18. 32 × 1.2 = 22.02

(ii) 62.89 × 0.02 = 1.2578 = 1.26

3. Write the number of significant figures (digits) in :
(i) 35.06
(ii) 0.35
(iii) 7.0068
(iv) 19 .0
(v) 0.0062
(vi) 0 4.2 × 0.6
(vii) 0.08 × 25
(viii) 3.6 ÷ 0.12 .
(i) 35.06 : In this significant figures i.e. digits are 4
(ii) In 0.35, significant figures are 2
(iii) In 7.0068, significant figures are 5
(iv) In 19.0, significant figures are 3
(v) In 0.0062, significant figures are 2
(vi) In 4.2 × 0.6 = 2.52, significant figure are 3
(vii) In 008 × 25 = 2.00 = 2 significant figure is 1
(viii) In 3.6 ÷ 0 .12 or 360 ÷ 12 = 30, significant figure are 2.

4. Write :
(i) 35.869,0 008426,4.952 and 382.7, correct lo three significant figures.
(ii) 60.974. 2.8753, 0.001789 and 400.04, correct to four significant figures.
(iii) 14.29462, 19.2, 46356.82 and 69, correct to five significant figures.
(i) Correct to three significant figures are
35.869 → 35.9
0.008426 → 0.00843
4. 952 → 4.95
382.7 →383

(ii) Correct to four significant figures
60.974 → 60.97
2. 8753 → 2.875
0.001789 → 0.001789
400.04 → 400.0

(iii) Correct to five significant figures
14.29462 → 14.295
19.2 → 19.200
46356.82 → 46357
69 → 69.000

### Exercise 4 (F)

1. The weight of an object is 3 .06 kg. Find the total weight of 48 similar objects.
Weight of one object = 3.06 kg
∴ Weight of 48 objects  = 3.06 × 48 = 146.88 kg

2. Find die cost of 17.5 m cloth at the rate of Rs. 112.50 per metre.
Cost of 1 metre cloth = Rs. 112.50
∴ Cost of 17.5 m cloth
= Rs. 112.50 × 17.5
= Rs. 1968.750
= 1968.75

3.  One kilogramme of oil costs Rs. 73.40 . Find the cost of 9.75 kilogramme of the oil.
Cost of 1 kg oil = Rs. 73.40
∴ Cost of 9.75 kg oil = Rs. 73.40 × 9.75
= Rs. 715.6500
= Rs. 715.65

4. Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.
Weight of 8 objects = 51.2 kg
∴ Weight of 1 object = 51.2/8 kg = 6.4 kg

5. 18.5 m of cloth costs Rs. 666. Find the cost of 3.8 m cloth.
Cost of 18.5 m cloth = Rs. 666
Cost of 1 m cloth = Rs. 666 ÷18.5 and cost of 3.8 m cloth
= Rs. (666 ÷18.5)× 3.8
= Rs. (6660 ÷ 185)× 3.8
= Rs. 36×3.8
= Rs. 136.80

6. Find die value of:
(i) 0.5 of Rs. 7.60 + 1.62 of Rs. 30
(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg
(iii) 6.25 of 8.4 – 4.7 of 3.24
(iv) 0.98 of 235 – 0 .09 of 3.2
(i) 0.5 of  Rs. 7.60 + 1.62 of Rs. 30
= Rs. 3.80 + Rs. 48.60
= Rs. 52.40

(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg
= 16.79 kg  + 0.432 kg = 17.222 kg

(iii) 6.25 of 8.4 - 4.7 of 3.24
= 52.500 - 15.228 = 37.272

(iv) 0.98 of 235 - 0.09 of 3.2 = 230.30 - 0.288  = 230.012

7. Evaluate:
(i) 5.6 – 1 .5 of 3.4
(ii) 4.8 ÷ 0.04 of 5
(iii) 0.72 of 80 + 0.2
(iv) 0.72 ÷ 80 of 0.2
(v) 6.45 + (3.9 – 1.75)
(vi) 0.12 of(0.104 – 0.02)+ 0.36×0.5
(i) 5.6  - 1.5 of 3.4
= 5.6 - 5.1

= 5.6 - 5.1 = 0.5

(ii) 4.8 ÷ 0.04 of 5
= 4.8 ÷ 0.20
= 4.8 ÷ 0.2
= 48 ÷ 2
= 24

(iii) 0.72 of 80 ÷ 0.2
= 57.60 ÷ 0.2
= 57.6 ÷ 0.2
= 576 ÷ 2
= 288

(iv) 0.72 ÷ 80 of 0.2
= 0.72 ÷ 16.0
= 0.72 ÷ 16
=72 ÷ 1600

(v) 6.45 ÷ (3.9 - 1.75)
= 6.45 ÷ (3.90 - 1.75)
= 6.45 ÷ 2.15
= 645 ÷ 215
= 3

(vi) 0.12 of (0.104 - 0.02) + 0.36×0.5
= 0.12 of 0.084 + 0.36×0.5
= 0.01008 + 0.180
= 0.19008