1.3 SPM Practice (Short Questions)


Question 1:
Round off each of the following numbers to the number of significant figures indicated in brackets.
(a) 80616 (3 s. f.)
(b) 60932 (3 s. f.)
(c) 0.4783 (2 s. f.)
(d) 3.047 (3 s. f.)
(e) 0.00567 (2 s. f.)
(f) 0.05086 (3 s. f.)

Solution:

(a) 80600
(b) 60900
(c) 0.48
(d) 3.05
(e) 0.0057
(f) 0.0509


Question 2:
Express each of the following as a single numbers.
(a) 8.565 × 10-5
(b) 1.304 × 105
(c) 6.754 × 10-6
(d) 1.0352 × 104

Solution
:

(a) 0.00008565
(b) 130400
(c) 0.000006754
(d) 10352


Question 3:
Express each of the following in standard form.
(a) 376510
(b) 47865400
(c) 0.000507
(d) 0.00006408

Solution
:

(a) 3.7651 × 105
(b) 4.78654 × 107
(c) 5.07 × 10-4
(d) 6.408 × 10-5


Question 4:
1.3 × 1015 + 3.2 × 1014

Solution:

1.3 × 1015 + 3.2 × 1014
= 1.3 × 1015 + 0.32 × 1015
= (1.3 +0. 32) × 1015
= 1.62 × 1015


Question 5:
0.0000036 – 2.1 × 10-7

Solution:

0.0000036 – 2.1 × 10-7
= 3.6 × 10-6 – 2.1 × 10-7
= 3.6 × 10-6 – 0.21 × 10-6
= (3.6 - 0.21) × 10-6
= 3.39 × 10-6

1.2 Standard Forms

1.2.1 Standard Form (Part 1)
1. Standard form is a way of writing very large numbers or very small numbers in the form of A × 10n, where  1 A < 10  and n is a positive or a negative integer.

Example 1:
Express each of the following numbers in standard form.
(a) 7244
(b) 32567
(c) 750000
(d) 0.65
(e) 0.0428
(f) 0.000369

Solution:
(a) 7244 = 7.244 × 1000 = 7.244 × 103

(b) 32567 = 3.2567 × 10000 = 3.2567 × 104

(c) 750000 = 7.5 × 100000 = 7.5 × 105

(d) 0.65
= 6.5 × 1 10
= 6.5 × 10-1

(e) 0.0428
= 4.28 × 1 100
= 4.28 × 10-2

(f) 0.000369
= 3.69 × 1 10000
= 3.69 × 10-4


Example 2:
Express each of the following numbers in standard form.
(a) 63.4
(b) 2738
(c) 23000
(d) 428000000
(e) 0.0063
(f) 0.000000038

Solution
:

(a) 63.4 = 6.34 × 10

(b) 2738 = 2.738 × 1000 = 2.738 × 10³

(c) 23000 = 2.3 × 10000 = 2.3 × 104

(d) 428000000 = 4.28 × 100000000 = 4.28 × 108

(e) 0.0063
= 6.3 × 1 1000
= 6.3 × 10-3
 
(f) 0.000000038
= 3.8 × 1 100000000
= 3.8 × 10-8
 

1.1 Significant Figures


1.1.1 Significant Figures (Part 1)
1. Significant figures are the relevant digits in an integer or a decimal number which has been rounded up to a value according to a degree of accuracy.

2. In rounding off positive numbers to a given number of significant figures, the significance of zero is shown as below.

(a)
All non-zero digits in a number are significant figures (s. f.).
Example:
(i) 568 (3 s. f.)
(ii) 36.97 (4 s. f.)

(b)
All zeroes between non-zero digits are significant.
Example:
(i) 7001 (4 s. f.)
(ii) 3.04 (3 s. f.)
(iii) 22.054 (5 s. f.)

(c)
All zeroes that lie on the right of a non- zero digit in a decimal are significant.
Example:
(i) 0.70 (2 s. f.)
(ii) 4.500 (4 s. f.)
(iii) 3.00 (3 s. f.)

(d)
Zeroes that lie on the left of a non-zero digit in a decimal are not significant.
Example:
(i) 0.05 (1 s. f.)
(ii) 0.0040 (2 s. f.)
(iii) 0.07040 (4 s. f.)

(e)
Zeroes at the end of a whole number are to be considered as non significant unless stated otherwise.
Example
(i) 40 (1 s. f.)
(ii) 3670 (3 s. f.)
(iii) 704200 (4 s. f.)


Example 1:
Round off the following numbers correct to three significant figures.
(a) 246 = 246 (3 s. f.)
(b) 2463 = 2460 (3 s. f.)
(c) 24632 = 24600 (3 s. f.)
(d) 0.00745 = 0.00745 (3 s. f.)
(e) 0.007453 = 0.00745 (3 s. f.)
(f) 0.007455 = 0.00746 (3 s. f.)
(g) 0.007403 = 0.00740 (3 s. f.)

Example 2:

Round off each of the following numbers to the number of significant figures indicated in brackets.
(a) 3548 (2 s. f.)
(b) 0.5089 (3 s. f.)
(c) 33.028 (1 s. f.)
(d) 0.40055 (3 s. f.)
(e) 0.681 (2 s. f.)
(f) 38.97 (3 s. f.)

Solution:

(a) 3500 (2 s. f.)
(b) 0.509 (3 s. f.)
(c) 30 (1 s. f.)
(d) 0.401 (3 s. f.)
(e) 0.68 (2 s. f.)
(f) 39.0 (3 s. f.)

1.3 SPM Practice (Short Questions)


Question 9:
3.17× 10 8 1.20× 10 9 =

Solution:
3.17× 10 8 1.20× 10 9 =3.17× 10 8 ( 0.120× 10 1 × 10 9 ) =3.17× 10 8 0.120× 10 8 =( 3.170.120 )× 10 8 =3.05× 10 8


Question 10:
Express  0.096 ( 2× 10 3 ) 3  in standard form.

Solution:
0.096 ( 2× 10 3 ) 3 = 0.096 8× 10 9 =1.2× 10 11


Question 11:
Express 0.0000643.5× 10 6  in standard form.

Solution:
0.0000643.5× 10 6 =6.4× 10 5 3.5× 10 6 =6.4× 10 5 0.35× 10 5 =( 6.40.35 )× 10 5 =6.05× 10 5


Question 12:
200.7× 10 11  is written as 2.007× 10 b  in the standard form. State the value of b.

Solution:
200.7× 10 11 =2.007× 10 2 × 10 11                   =2.007× 10 13                 b=13


Question 13:
Find the product of 0.1985 and 0.5.
Round off the answer correct to two significant figures.

Solution:
0.1985 × 0.5
= 0.09925
= 0.10 (two significant figures)

1.3 SPM Practice (Short Questions)


Question 14:
The area of a rectangular public parking area is 7.2 km2 . Its width is 2400 m. The length, in m, of the parking area is

Solution:
Area = Length × Width Length × 2.400 km = 7.2 k m 2 Length= 7.2 2.4            =3 km            =3000 m            =3× 10 3  m


Question 15:
It is given that 30 solid metal cylinders each with a radius of 70 cm and a height of 300 cm, are melted to make 40 identical solid spheres.
Find the volume, in cm3 , of each solid sphere.

Solution:
Volume of 40 spheres = Volume of 30 cylinders =30× 22 7 × 70 2 ×300 =138 600 000  cm 3 Volume of 1 sphere = 138 600 000 40 =3 465 000  cm 3 =3.465× 10 6  cm 3 =3.47× 10 6  cm 3