Given, ratio = 3 : 5, firstly multiply and divide the numerator and denominator with the same number.
\(\frac{3}{5}=\frac{3\times2}{5\times2}=\frac{6}{10}\ and\ \frac{3}{5}=\frac{3\times3}{5\times3}=\frac{9}{15},\) hence two equivalent rati.os of 3: 5 are 6 : 10 and 9: 15.

\(\because\) Distance covered by Rohan in 1 h = 12 km and distance covered by Rakesh in 1 h = 8 km
\(The\ required\ ratio=\frac{Distance\ covered\ by\ Rohan}{Distance\ covered\ by\ Rakesh}=\frac{12}{8}\)
Now, divide numerator and denominator both by the HCF of 12 and 8 i.e. 4, \(\frac{12\div4}{8\div4}\ \ [\because HCF\ of\ 12\ and\ 8\ is\ 4]\)
\(=\frac{3}{2}=3:2\)

Ratio=3:11

4: 14 and 6 : 21

We have,\(45:60=\frac{45}{60}\) ,in simplest form \(=\frac{3}{4}=3:4\)
Now, \(36:48=\frac{36}{48}\),in simplest form \(=\frac{3}{4}=3:4\)
So, the ratio 45:60 and 36:48 are in proportion i.e. 45:60::36:48

Product of means = Product of extremes

Required area to yield 360 quintals wheat = 8 hectares of land
\(\therefore\) To yield 1 quintal wheat, required area \(=\frac{8}{360}\) hectares
\(\therefore\) To yield 540 quintals wheat, required area \(=540\times\frac{8}{360}\) hectares \(\frac{3\times8}{2}\) hectares = 12 hectares of land.

2 cm to 4 cm \(=\frac{2}{400}=1:200[\because\ 1\ m=100 cm]\)

Given,30% of Zuben's house area= 40% of Seema's house area
\(\therefore\)Required ratio of Zuben's house to Seema's house \(=\frac{4}{3}=4:3\)

Yes! we can think of some more examples where we have to divide a number of things in some ratio. 
Sum of the ratios = 2 + 3 = 5
Total money = Rs. 150
\(\therefore\) Money got by one brother
= \(\frac { 2 }{ 5 } \times 150\) = Rs. 60
Money got by another brother
= \(\frac { 3 }{ 5 } \times 150\) = Rs. 90

Given, Ravi walks 6 km in an hour i.e. distance covered by Ravi in one hour = 6 km.
and Roshan walks 4 km in an hour i.e. distance covered by Roshan in one hour = 4 km
\(\therefore\) Ratio of distance covered by Ravi to the distance covered by Roshan \(=\frac{Distance\ covered\ by\ Ravi\ in\ one\ hour}{Distance\ covered\ by\ Roshan\ in\ one\ hour}\)
\(=\frac{6\ km}{4\ km}=\frac{6}{4}=\frac{3}{2}=3:2\)
Hence, the required ratio is 3 : 2.

Given, number of holidays in one year = 73
We know that, number of days in one year = 365
\(\therefore\) Ratio of number of holidays to the number of days in one year \(=\frac {Number\ of\ holidays}{Total\ number\ of\ days}=\frac {73}{365}\)
\(=\frac{73\div73}{365\div73}=\frac{1}{5}=1:5\) [\(\because\)HCF of 73 and 365 = 73]
Hence, the required ratio is 1 : 5.

Here, we have to convert L into mL.
We know that,1L = 1000 mL
\(\therefore\)2 L=2 X 1000 mL = 2000 mL
\(\therefore\ Required\ ratio=\frac{500\ mL}{2000\ mL}=\frac{500}{2000}=\frac{5}{20}\)
[dividing numerator and denominator by 100]
\(Ratio=\frac{1}{4}=1:4\)

3:4

We have, 1: 5 and 3: 15
Here, \(3:15=\frac{3}{15}=\frac{3\div3}{15\div3}=\frac{1}{5}=1:5\) [dividing numerator and denominator both by 3]
i.e. 1 : 5 and 3 : 15 are in proportion.
Thus, the proper form is 1 : 5 :: 3: 15.

8:9

Quantity of burfi = 18 kg
Given, Khoya : Sugar = 7: 2
Total = 7 + 2 = 9
Quantity of khoya \(=\frac{18\times7}{9}=14\ kg\)
So, Reshma used 14 kg khoya.

School start at 7 : 00 am and gets over at 12 : 30 pm
Total time the student spend at shcool
= 5 h 30 min = (5 x 60 + 30) m = 330 min
Break time = 9: 50 am to 10 : 10 am = 20 min
\(\therefore\) Required ratio \(=\frac{20}{330}=2:33\)

Ratio of sides of triangle = 2 : 3 : 5
Total perimeter = 70
Total of the ratio of sides = 2 + 3 + 5 =10
\(\therefore\) Length of the longest side = \(70\times\frac{5}{10}=35\ cm\)

\(\frac { 40 }{ 5 } =\frac { 8 }{ 1 } \)
\(\frac { 42 }{ 6 } =\frac { 7 }{ 1 } \)
\(\because\) 8 > 7
\(\therefore\) First shopkeeper is costlier
\(\therefore\) Tomatoes must be purchased from the second shopkeeper.
The comparison by taking difference will not help him decide as the weights of tomatoes  purchased from the two shopkeepers are different.
 

Given, number of girls = 20 and number of boys = 15
\(\therefore\) Total number of students in the class = 20 +15 = 35
(a) Ratio of number of girls to the number of boys \(=\frac{Number\ of\ girls}{Number\ of\ boys}=\frac{20}{15}\)
\(=\frac{20\div5}{15\div5}=\frac{4}{3}=4:3\) [ \(\because\)HCFof 20 and 15=5]
(b) Ratio of number of girls to the total number of students in the class \(=\frac{Number\ of\ girls}{Total\ number\ of\ students}=\frac{20}{35}\)
\(=\frac{20\div5}{35\div5}=\frac{4}{7}\) [ \(\because\) HCF of 20 and 35 = 5]

Given, number of teachers = 102
and number of students = 3300
\(\therefore\)  Required ratio of teachers to the number of students \(=\frac{Number\ of\ teachers}{Number\ of\ students}=\frac{102}{3300}\)
\(\because\) 102=2x3x17and 3300=2x3x2x5x5x11
\(\therefore\) HCF of 102 and 3300=2x3=6
\(Ratio=\frac{102\div6}{3300\div6}=\frac{17}{55}=17:550\)

255 m

Total number of members = 100
Number of members, who play carom = 20
Number of members, who play table tennis = 24
Number of members, who play badminton = 16
Number of members, who do not play any game
=100 -(20 + 24 + 16)=100 - 60 = 40
(a) Ratio of members of carom to the members of table tennis \(=20:24=\frac{20}{24}=5:6\)
(b) Ratio of members of badminton to the members of carom \(=16:20=\frac{16}{20}=\frac{4}{5}=4:5\)
(c) Ratio of table tennis to the member of those, who play badminton \(=24:16=\frac{24}{16}=\frac{3}{2}=3:2\)
(d) Ratio of badminton to the number of those, who do not play any game \(=16:40=\frac{16}{40}=\frac{2}{5}=2:5\)

Given, milk needed for making cake = 1cup
and flour needed for making cake = \(2\frac{1}{2}\) cups
\(=\frac{5}{2}\) cups
Then, total amount needed = Milk + Flour
\(=(1+\frac{5}{2})\ cups=\frac{7}{2}\ cups\)
So, \(\frac{7}{2}\) cups of milk and flour needed to make cake for 6 persons.
Let the needed amount of cups of milk and flour to make cake for 8 persons = x [where, x is a multiple of cups]
By ratio and proportion law,\(\frac{\frac{7}{2}}{6}=\frac{x}{8}\)
\(\Rightarrow\frac{7}{2}\times8=6\times x\Rightarrow x=\frac{28}{6}\)
\(\Rightarrow x=\frac{14}{3}\) [dividing by 2
Hence, the cups needed for 8 persons is \(\frac{14}{3}\).

\(\frac { 150 }{ 120 } =\frac { 15 }{ 12 } =\frac { 5 }{ 4 } \)= 5:4

\(\frac { 6 }{ 2 } =\frac { 3 }{ 1 } \)=3:1

10:5=\(\frac { 10 }{ 5 } =\frac { 10\div 5 }{ 5\div 5 } =\frac { 2 }{ 1 } \)
\(20\div 10=\frac { 20 }{ 10 } =\frac { 20\div 10 }{ 10\div 10 } =\frac { 2 }{ 1 } \)
\(30\div 15=\frac { 30 }{ 15 } =\frac { 30\div 15 }{ 15\div 15 } =\frac { 2 }{ 1 } \)=2:1

100-60=40
40:100 = \(\frac { 40 }{ 100 } =\frac { 40\div 20 }{ 100\div 20 } =\frac { 2 }{ 5 } \)= 2:5

\(\frac { 5 }{ 80 } \times 32=2\)

1 year = 12 months
4000x12 = 48000

\(\frac { 8000 }{ 8 } \) = 1000