15

8000

90000

70169308 is divisible by 11.

Numbers after commas are as follows: 9, 05, 00, 041.

We have, 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + (769 x 5) x 5 x 218  [\(\because\)25 = 5 x 5]
                                                              = 3845 x 5 x 782 + 3845 x 5 x 218
                                                              = 3845 x 5 x (782 + 218) [taking 3845 x 5 as common terms]
                                                              = 3845 x 5 X 1000 = 19225 x 1000 = 19225000.

We have, 125 x 40 x 8 x 25 = (125 x 40) x (8 x 25)
                                         = 5000 x 200 = 1000000

The predecessor of given numbers are as follows:

Here 280.64 + 25.5 + 28 can be written as

280.64 + 25.5 + 28 = 334.14

\(12\frac{4}{5}\)

we have, \(\frac{7}{12}-\frac{4}{15}\) = \(\frac{35-16}{60}\)
[ ∵ LCM of 12 and 5 is 60]
= \(\frac{19}{60}\)

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.

We draw the following table using tally marks from given information.

Numbers	Tally marks	Number of times
1		7
2		6
3		5
4	I I I I	4
5		11
6		7

Given, length of the table-top = 2 m
and breadth of the table-top = 1m 50 cm = 1m + 50 cm
\(=1m+50\times \frac { 1 }{ 100 } m\quad \left[ \because \quad 1cm=\frac { 1 }{ 100 } m \right] \)
\(=1m+\frac { 50 }{ 100 } m\quad =1m+0.50m=1.50m\)
∴  Area of the table-top = Length \(\times\)Breadth
= 2 m \(\times\)1.50 m =3sq m
Hence, the area of the table-top is 3 sq m

Fraction\(\frac { 10 }{ 3 } \)is the simplest form of given fraction

Number of CDs bought by Kristin = 3
Number of CDs received as gifts = 5
∴ Total number of CDs = 3 + 5 = 8
Hence, fraction of CDs bought by Kristin
=  \(\frac { Number\ of\ CDs\ bought\ by\ Kristin }{ Total\ number\ of\ CDs } =\frac { 3 }{ 8 } \) 
and fraction of CDs, she received as gifts
= \(\frac { Number\ of\ CDs\ recieved\ as\ gift }{ Total\ number\ of\ CDs } =\frac { 5 }{ 8 } \)

Given population is 2538473.
Rounded off to nearest thousands = 2538000

0°C to 10°C = +10° C -4° C to 8° C = +12°C (max.)
-15°C to -8°C = +7°C -7°C to 0°C = +7°C

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°.
                    

(a) \(2:3=\frac{2}{3}\ and\ 4:5=\frac{4}{5}\)
2: 3 and 4: 5 are not equal, therefore ratios are not in proportion.
(b) \(25 g:200g=\frac{25}{200}=\frac{1}{8}=1:8\)
\(6kg:48kg=\frac{6}{48}=\frac{1}{8}=1:8\)
Hence, they form a proportion.
(c) \(440m:2km=\frac{440}{2000}[\because1\ km=1000\ m]\)
\(=\frac{11}{50}=11:50\)
\(55\ cm:3 m=\frac{55}{300}=\frac{11}{60}=11:60\)
So, they are not in proportion.
(d) \(200 mL:2.5 L=\frac{200}{2500}=\frac{2}{25}=2:25\)
\(Rs.4:Rs.50=\frac{4}{50}=\frac{2}{25}=2:25\)
So, they are in proportion.

In order to get the missing number, we consider the fact that18 = 6 X 3 i.e. when we divide 18 by 3, we get 6.
So,to get the missing number of second ratio, 15 must also be divided by 3.
Then, we have 15 \(\div\) 3 = 5
Hence, the second ratio is \(\frac{5}{6}\)
i.e \(\frac{15}{18}=\frac{5}{6}\)................(i)
Similarly,to get third ratio, we multiply both terms of secondratio by 2. [\(\because\) 5 X 2 = 10]
i.e \(\frac{5}{6}=\frac{5\times2}{6\times2}=\frac{10}{12}=10:12\) [multiplying numerator and denominator by 2]
\(\therefore \frac{5}{6}=\frac{10}{12}\)...............(ii)
Hence, the third ratio is 10/12.
Now, to get the fourth ratio, we consider the fact that 30 = 6 X 5 i.e. when we divide 30 by 6, we get 5.
So, In second ratio we muItipIy by 5,i.e. \(\frac{5}{6}=\frac{5\times5}{6\times5}=\frac{25}{30}\)
\(\therefore \frac{5}{6}=\frac{25}{30}\)..................(iii)
From Eqs. (i), (ii) and (iii), we have

Here, from the above relation, we can say that all these are equivalent ratios.

(A) According to the question, 100 students can be represented by = 1symbol 
\(\therefore\) 1 student can be represented by \(\frac{1}{100}\) symbols
Now, for year 1996,
400 students can be represented by \(\frac{400\times 1}{100}\)
                        = 4 symbols
For year 1998,
535 students can be represented by \(\frac{535\times 1}{100}\)
                       = 5 complete symbols + 35 students
                       = 5 complete symbols and 1 incomplete symbol
For year 2000,
472 students can be represented by \(\frac{472\times 1}{100}\)
                      = 4 complete symbols + 72 students
                      = 4 complete symbols and 1 incomplete symbol
For year 2002,
600 students can be represented by \(\frac{600\times 1}{100}\)
                     = 6 complete symbols
For year 2004,
623 students can be represented by \(\frac{623\times 1}{600}\)
6 complete symbol + 23 students = = 6 complete symbols and 1 incomplete symbol
Hence, the required pictograph of given data is shown below

(a) 6 symbols represent total number of students in the year 2002.
(b) 5 complete symbols and 1 incomplete symbol represent total number of students for the year 1998.
(B) According to the question, 50 students can be represented by 1symbol
\(\therefore\) 1 student can be represented by \(\frac{1}{50}\) symbols
Now,  for year 1996,
400 students can be represented by \(\frac{400\times 1}{50}\) symbols
                          = \(\frac{400}{50}\)
                          = 8 complete symbols.
For year 1998,
535 students can be represented by \(\frac{535\times 1}{50}\) 
                         = \(\frac{535}{50}\) symbols
                         = 10 complete symbols + 1 incomplete symbol
For year 2000,
472 students can be represented by \(\frac{472\times 1}{50}\) symbols
                         = \(\frac{472}{50}\) 
                         = 9 complete symbols + 1 incomplete symbol
For year 2002,
600 students can be represented by \(\frac{600\times 1}{50}\) symbols
                         = \(\frac{600}{50}\)
                         = 12 complete symbol
For year 2004,
623 students can be represented by \(\frac{623\times 1}{50}\) symbols
                         = \(\frac{623}{50}\)
                         = 12 complete symbols + 1 incomplete symbol
Hence, the required pictograph of given data is shown below

Therefore, we observe that pictograph B is more informative because it gives better approximation.