\(\frac { 2 }{ 5 } +\frac { 1 }{ 6 } =\frac { 17 }{ 30 } \)

HCF of 91, 112 and 49 = 7

yes

We have, 10824,
\(\begin{matrix} 1 \\ \downarrow \\ O \end{matrix}\begin{matrix} 0 \\ \downarrow \\ E \end{matrix}\begin{matrix} 8 \\ \downarrow \\ O \end{matrix}\begin{matrix} 2 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\)
Sum of digits at odd places from right = 4 +8 + 1= 13
Sum of digits at even places from right = 2 + 0 = 2
Now, difference = 13 - 2 = 11
So, 10824 is divisible by 11.

Here, -8 > -13. So, -13 is on the left of -8. Hence, if we are at -8 on number line, then move to the left from -8 to reach at -13.

Production of milk in one day = 75678 L                                  75678
Supply to milk depot in one day = 67689 L                             -67689
So, the milk left in the dairy = (75678 - 67689) = 7989 L           7989

Numbers after commas are as follows: 73, 75, 307.

We have, (1063 x 127) - (1063 x 27) = 1063 (127 - 27)= 1063 x 100 = 106300 [taking 1063 as common term].

We know that, one day = 24 h
\(\therefore\) The required fraction = \(\frac { 15\ h }{ 24\ h } =\frac { 15 }{ 14 } =\frac { 5 }{ 8 } \)

To find 6 - 2
Let us start from 6 and make 2 equal jumps to the left of 6. Each jump is equal to 1 unit. Now, we reach at 4.

\(\therefore\)  6 - 2 = 4.

(a) Factors of 8 are: 1,2,4 and 8
Factors of 12 are: 1,2,3,4,6 and 12
Factors of 20 are: 1,2,4,5, 10 and 20
Hence, common factors of 8, 12 and 20 are 1, 2 and 4.
(b) Factors of9 are 1,3 and 9
Factors of 15 are 1, 3, 5 and 15
Factors of 21 are 1,3, 7 and 21
Hence, common factors of 9, 15 and 21 are 1 and 3.

Increasing order of these interger is -14 < -13 < - 12 < - 11 < - 10 < - 9, because we are moving to the right from -15 to -8.

Rs 313.86

We have \(\Box{ } -\frac { 5 }{ 8 } =\frac { 1 }{ 4 } \)
Here\(\frac { 5 }{ 8 } \) is subtracted from missing fraction to get\(\frac { 1 }{ 4 } .\) This means addition of \(\frac { 5 }{ 8 } \) and \(\frac { 1 }{ 4 } \) gives the missing fraction.
∴ Missing fraction = \(\frac { 1 }{ 4 } +\frac { 5 }{ 8 } \)
⇒ \(\frac { 1 }{ 4 } =\frac { 1\times 2 }{ 4\times 2 } =\frac { 2 }{ 8 } \)
Now, \(\frac { 1 }{ 4 } +\frac { 5 }{ 8 } =\frac { 2 }{ 8 } +\frac { 5 }{ 8 } =\frac { 2+5 }{ 8 } =\frac { 7 }{ 8 } \)
Hence  \(\boxed { \frac { 7 }{ 8 } } -\frac { 5 }{ 8 } =\frac { 1 }{ 4 } \)

In figures (a) and (c), there ate only two lines of symmetry i. e. horizontal line of symmetry and vertical line of symmetry.

But for figure (b), we can draw more than two lines of symmetry as shown below.

Given, cost of 7 m of cloth = Rs. 294
\(\therefore\) Cost of 1 m of cloth \(=Rs.\frac{294}{7}=Rs.42\)
\(\therefore\) Cost of 5 m of cloth = Rs.42 X 5 = Rs. 210
Hence, the cost of 5 m of cloth is Rs. 210.

The perimeter of the given figure is 28 cm

(a) 24 persons
(b) R brand.

Simplest form of the fraction \(\frac { 84 }{ 98 } \) is \(\frac { 6 }{ 7 } \)

We know that, all given fractions are greater than zero but less than or equal to 1. So, they will lie between 0 and 1. So, we have to divide the gap between 0 and 1 into ten equal parts, then each part shows\(\frac { 1 }{ 10 } \) and 5 parts show\(\frac { 5 }{ 10 } \), 10 parts show\(\frac { 10 }{ 10 } \) Also, 0 shows \(\frac { 0 }{ 10 } \)

\(\left[ also,\frac { 0 }{ 5 } =0 \right] \)
Hence, points A, B, C and D respectively represent the fractions \(\frac { 0 }{ 10 } ,\frac { 1 }{ 10 } ,\frac { 5 }{ 10 } \)and \(\frac { 10 }{ 10 } .\)

Here, given number is 60 i.e. LX.

\(2\frac{3}{20}\)

To find the line of symmetry of angle 75°, we use the following steps
Step I Draw \(\bar{OB}\) of any length.
                  
Step II Place the centre of the protractor at O and the zero edge along \(\bar{AB}\).
                   
Step III Start with zero near B, mark point C at 75°.
                    
Step IV Join OC. \(\angle\)BOC is the required' angle of measure 75°.
                   
Step V With O as centre and using compasses, draw an arc that cuts both rays of \(\angle\)O at P and Q.
                     
Step VI With P as centre, draw (in the interior of \(\angle\)O an arc whose radius is more than half of the length of PQ).
Step VII With the same radius and with Q as centre, draw another arc in the interior of \(\angle\)O. Let the two arcs intersect at D.
                   
Step VIII Join OD then \(\bar{OD}\) is the required bisector of \(\angle\)O. i.e. OD is the line of symmetry of an angle of measure 75°.
                    

To draw a circle, of diameter 6.1 cm, we use the following steps:
Step I Draw a line segment \(\bar{PQ}\) of length 6.1 cm.
                  
Step II With P as centre, using compasses, draw an arc of a circle (here, we can draw a circle also) with radius more than half of the length of \(\bar{PQ}\) .
Step III With the same radius and with Q as centre, draw another circle using compasses. Let it cut the previous circle at M and N.
                       
Step IV Now, join \(\bar{MN}\) . It cuts \(\bar{PQ}\) at O.
Therefore, MN is the perpendicular bisector of \(\bar{PQ}\) and O is the mid-point of \(\bar{PQ}\). Now, with O as centre and OP or OQ as radius, draw a circle.
Thus, it is a circle whose diameter is the line segment \(\bar{PQ}\).
Hence, the circle PMQN is the required circle.

To draw the 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 'age-group' and along vertical line mark 'number of persons'.
(ii) Now, take scale of 1 unit length = 20000 along the vertical line and then mark the corresponding values.
Also, the heights of bars for various groups are as follows:

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

(a) From bar graph, we see that the lengths of bars for age group 30-44 and 45-59 are same, so age group (30 - 44) and (45 - 59) have the same population.
(b) Persons having age 60 or above are called senior citizens.
\(\therefore\) Number of senior citizens in the town
          = Number of persons of age group (60 - 74) + Number of persons of age 75 and above
          = 80000 + 40000 = 120000.

In the given pictograph, 1 picture = 2 bulbs
Now, number of bulbs sold on Monday = 6 pictures
                                                              = 6 x 2 = 12bulbs
Number of bulbs sold on Tuesday = 8 x 2 = 16 bulbs
Number of bulbs sold on Wednesday = 4 x 2 = 8 bulbs
Number of bulbs sold on Thursday = 5 x 2 = 10 bulbs
Number of bulbs sold on Friday = 7 x 2 = 14 bulbs
Number of bulbs sold on Saturday = 4 x 2 = 8 bulbs
Number of bulbs sold on Sunday = 9 x 2 = 18 bulbs
(a) Number of bulbs sold on Friday = 7 x 2 = 14 bulbs
(b) Maximum number of bulbs were sold on Sunday i.e. 18 bulbs.
(c) The same number of bulbs were sold on Wednesday and Saturday i.e. 8 bulbs.
(d) The minimum number of bulbs were sold on Wednesday and Saturday i.e. 8 bulbs.
(e) Total number of bulbs sold in a week
                      = 12 + 16 + 8 + 10 + 14 + 8 + 18 = 86
Now, number of cartons which can hold 9 bulbs = 1 and number of carton which can hold 1 bulb = \(\frac{1}{9}\)
\(\therefore\) Number of cartons which can hold 86 bulbs = \(\frac{1\times 86}{9}\)
                     = \(\frac{86}{9}=9\frac{5}{9}=10\)
Hence, 10 cartons were needed in the given week.

Given, total money = Rs.10,00,000
Money spent on a colour TV = Rs. 16,580
Money spent on a motorcycle = Rs. 45,890
and money spent on a flat = Rs. 8,70,000
Total amount spent = 16,580 + 45,890 + 8,70,000
= Rs. 9,32,470
∴ Money left with him = 10,00,000 - 9,32,470
= Rs.67,530
Hence, Rs. 67,530 was left with him.

We know that, all the negative integers are less than zero and they have '-' sign.