First we move 5 steps to the left of 0 reaching - 5 and then from this point we move 3 steps to the
right. We reach the point - 2. Thus, (- 5) + 3 = - 2.

Given, ribbon bought by Sarita = \(\frac { 2 }{ 5 } \)m and ribbon bought by Lalitha =\(\frac { 3 }{ 4 } \) m
∴ Total length of the ribbon bought by them =\(\frac { 2 }{ 5 } +\frac { 3 }{ 4 } \)
Now, LCM of 5 and 4 = 20
∴  \(\frac { 2 }{ 5 } +\frac { 3 }{ 4 } =\frac { 2\times 4 }{ 4\times 5 } +\frac { 3\times 5 }{ 4\times 5 } \)=\(\frac { 8 }{ 10 } +\frac { 15 }{ 20 } =\frac { 8+15 }{ 20 } =\frac { 23 }{ 20 } \)
Hence, total length of the ribbon bought by them is\(\frac { 23 }{ 20 } \) m.

\(\frac { 2 }{ 3 } +\frac { 3 }{ 4 } +\frac { 1 }{ 2 } =\frac { 23 }{ 12 } \)

Thus, 25.65 + 9.005 + 3.7 = 38.355

We have \(\Box { } -\frac { 3 }{ 21 } =\frac { 5 }{ 21 } \)
Here\(\frac { 3 }{ 21 } \) is subtracted from missing fraction to get \(\frac { 5 }{ 21 } .\)
So, we add \(\frac { 3 }{ 21 } \)to \(\frac { 5 }{ 21 } \)get missing fraction
∴ Missing fraction =\(\frac { 3 }{ 21 } +\frac { 5 }{ 21 } =\frac { 3+5 }{ 21 } =\frac { 8 }{ 21 } \)
Thus, \(\boxed { \frac { 8 }{ 21 } } -\frac { 3 }{ 21 } =\frac { 5 }{ 21 } \)

\(\frac { 6 }{ 14 } ,\frac { 9 }{ 21 } ,\frac { 12 }{ 28 } ,\frac { 15 }{ 35 } \)

(-7) - 8 - (-25) = (-7) + (-8) + 25   [∵ additive inverse of -25 is 25]
= -15 + 25 = -15 + 15 + 10   [∵ 25=15+10]
=0+10=10  [∵ -15+15=0]

Decimals in words are given below:
We have, 5.008 = Five point zero zero eight

We have, (37) + (-2) + (-65) + (-8)
= 37 - (2 + 65 + 8) = 37 - 75
=37 -37 -38    [∵ 75 = 37 + 38 ⇒ - 75 = - 37 - 38]
= 0 - 38 = - 38  [∵ -37+37=0]

Integers between -15 and -5 are -14, -13:-12, -11, -10, - 9, - 8, - 7, - 6.

\(200+60+5+\frac { 1 }{ 10 } =265.1\)

We know that, total minutes in an hour = 60 min
∴   Required field fraction = \(\frac { 40\quad min }{ 60\quad min } =\frac { 40 }{ 60 } =\frac { 2 }{ 3 } \)

We draw the following table using given information

To draw a bar graph, we will use the following steps:
(i) Firstly, draw two perpendicular lines, one is horizontal and one is vertical. Along the horizontal line, mark 'days' and along the vertical line mark 'number of books sold'.
(ii) Now, take scale of 1 unit length = 5 books, along the vertical line and then mark the corresponding values.
Also, the heights of the bars for various days are as follows:

(iii) Drawbars of equal width and of height calculated in Step (ii) on the horizontal line with equal spacing (or gap) between them.
Thus, we get the following bar graph.

The given decimals can be written as
(a) 0.29=\(0+\frac { 2 }{ 10 } +\frac { 9 }{ 100 } \)
(b) 2.08=\(2+\frac { 0 }{ 10 } +\frac { 8 }{ 100 } \)
(c) 19.60=10+9+\(\frac { 6 }{ 10 } +\frac { 0 }{ 100 } \)
(d) 14832=100+40+8+\(\frac { 3 }{ 10 } +\frac { 2 }{ 100 } \)
(e) 200.812=200+00+0+\(\frac { 8 }{ 10 } +\frac { 1 }{ 100 } +\frac { 2 }{ 1000 } \)
Now, the place value table is given below:

The integers lying between -2 and 2 are -1, 0 and 1. So, there are 3 integers.

The predecessor of -1 = -1 - 1 = -2
\(\therefore\)  The required predecessor is -2.

The additive inverse of 14 is -14.

Increasing order of these integers is - 29 < - 28 < - 27 < - 26 < - 25 < - 24, because we are moving to the right from -30 to -23.

Distance travelled by bus = \(3\frac{1}{2}\) km
Nasir walked to reach town = \(1\frac{1}{8}\) km
∴ Total distance = \(3\frac{1}{2}+1\frac{1}{8}=\frac{7}{2}+\frac{9}{8}\)
= \(\frac{28+9}{8}\) [ ∵ LCM of 2 and 8 is 8]
=\(\frac{37}{8}=4\frac{5}{8}\)km

2.876 kg - 0.316 kg
i.e.   
\(\quad 2.876\\-0.316\\ \_\_\_\_\_\_\_\_\\ \ Rs\ 2.560 kg\\ \_\_\_\_\_\_\_\_\)

-17,-15,-5,-2,0,2,3,5