Ratio and Proportions

Important Facts:

Ratio: The ratio of two qualities a and b in the same units,
is the fraction a/b and we write it as a:b. In the ratio, a:b, we 
call ‘a’ as the first term of antecedent and b, the second 
term consequent.
Ex: The ratio 5:9 represents 5/9 with antecedent=5 ,consequent=9

Rule: The multiplication or division of each term of 9 ratio
by the same non-zero number does not affect the ratio.

Proportion: The equality of two ratios is called proportion.
If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in
proportion. Here a and b are called extremes, while b and c are 
called mean terms.
     Product of means=product of extremes
     Thus, a:b::c:d  =>   (b*c)=(a*d) 
    
Fourth proportional: If a:b::c:d, then d is called the fourth
proportional to a,b and c.

Third proportional: If a:b::b:c, then c is called third 
proportional to a and b.

Mean proportional: Mean proportional between a and b is SQRT(a*b).

Comparision of Ratios:

We say that (a:b)>(c:d)   =>   (a/b)>(c/d)

Compounded ratio: The compounded ratio of the ratios (a:b),
(c:d),(e:f) is (ace:bdf).

Duplicate Ratio: If (a:b) is (a2: b2 )

Sub-duplicate ratio of (a:b) is (SQRT(a):SQRT(b))

Triplicate ratio of (a:b) is  (a3: b3 )

Sub-triplicate ratio of (a:b) is (a1/3: b1/3 ).
If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d)  (componend o and
dividend o)
 
Variation:

we say that x is directly proportional to y, if x=ky for some
constant k and we write.
We say that x is inversely proportional to y, if xy=k for some
constant  and we write.

X is inversely proportional to y. 
    If a/b=c/d=e/f=g/h=k then 
   k=(a+c+e+g)/(b+d+f+h)
         If a1/b1,a2/b2, a3/b3..............an/bn are unequal
fractions then the ratio.
a1+a2+a3+..........an/(b1+b2+b3+...............bn) lies between the 
lowest & the highest of the three fractions.

                                                

Simple Problems 

1.If a:b =5:9 and b:c=4:7 Find a:b:c?

Sol:  a:b=5:9 and b:c=4:7=4*9/4:9*4/9=9:63/9
         a:b:c=5:9:63/9=20:36:63

2.Find the fourth proportion to 4,9,12

Sol:   d is the fourth proportion to a,b,c
            a:b=c:d
            4:9=12:x
             4x=9*12=>x=27

3.Find third proportion to 16,36

Sol: If a:b=b:c then c is the third proportion to a,b
           16:36=36:x
            16x=36*36
              x=81

4.Find mean proportion between 0.08 and 0.18

Sol:  mean proportion between a and b=square root of ab
         mean proportion =square root of 0.08*0.18=0.12
	   
5.If  a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d is

Sol:     a:b=2:3 and b:c=4:5=4*3/4:5*3/4=3:15/4
            c:d=6:7=6*15/24:7*15/24=15/4:35/8
            a:b:c:d=2:3:15/4:35/8=16:24:30:35
	    
6.2A=3B=4C then A:B:C?

Sol:    let 2A=3B=4C=k  then
         A=k/2,   B=k/3,    C=k/4
          A:B:C=k/2:k/3:k/4=6:4:3
	      
7.15% of x=20% of y then x:y is

Sol:      (15/100)*x=(20/100)*y
                       3x=4y
                       x:y=4:3
		       
8.a/3=b/4=c/7 then (a+b+c)/c=

Sol:    let a/3=b/4=c/7=k
        (a+b+c)/c=(3k+4k+7k)/7k=2

9.Rs 3650 is divided among 4 engineers, 3 MBA’s and 5 CA’s
such that 3 CA’s get as much as  2 MBA’s and 3 Eng’s as
much as 2 CA’s .Find the share of an MBA.

Sol:     4E+3M+5C=3650
         3C=2M,   that is M=1.5C
         3E=2C     that is E=.66 C
       Then, (4*0.66C)+(3*1.5C)+5C=3650
                  C=3650/12.166
                   C=300
        M=1.5 and C=450.
                                                
Difficule Problems   

1.Three containers A,B and C are having mixtures of milk and
water in the ratio of 1:5 and 3:5 and 5:7 respectively. If
the capacities of the containers are in the ratio of all the
three containers are in the ratio 5:4:5, find the ratio of 
milk to water, if the mixtures of all the three containers 
are mixed together.

Sol: Assume that there are 500,400 and 500 liters respectively 
     in the 3 containers.
Then ,we have, 83.33, 150 and 208.33 liters of milk in each of
the three containers.
Thus, the total milk is 441.66 liters. Hence, the amount of
water in the mixture is 1400-441.66=958.33liters.
Hence, the ratio of milk to water is 
   441.66:958.33 => 53:115(using division by .3333)
  The calculation thought process should be 
        (441*2+2):(958*3+1)=1325:2875
         Dividing by 25   => 53:115.                          

2.A certain number of one rupee,fifty parse and twenty five 
paise coins are in the ratio of 2:5:3:4, add up to Rs 210.
How many 50 paise coins were there?

Sol: the ratio of 2.5:3:4 can be written as 5:6:8
let us assume that there are 5 one rupee coins,6 fifty 
paise coins and 8 twenty-five paise coins in all.
        their value=(5*1)+(6*.50)+(8*.25)=5+3+2=Rs 10
    If the total is Rs 10,number of 50 paise coins are 6.
if the total is Rs 210, number of 50 paise coins would be
             210*6/10=126.

3.The incomes of A and B are in the ratio of 4:3 and their 
expenditure are in the ratio of 2:1 . if each one saves 
Rs 1000,what are their incomes?

Sol:    Ratio of incomes of A and B=4:3
      Ratio of expenditures of A and B=2:1
Amount of money saved by A=Amount of money saved by B=Rs 1000
   let the incomes of A and B be 4x and 3x respectively
   let the expense of A and B be 2y and 1yrespectively
Amount of money saved by A=(income-expenditure)=4x-2y=Rs 1000
   Amount of money saved by B=3x-y=Rs 1000 
  this can be even written as 6x-2y=Rs 2000
         now solve 1 and 3 to get
                x=Rs 500    
  therefore income of A=4x=4*500=Rs 2000
         income of B=3x=3*500=Rs 1500
			 
4.A sum of Rs 1162 is divided among A,B and C. Such that 4
times A's share share is equal to 5 times B's share and 7 
times C's share . What is the share of C?

Sol: 4 times of A's share =5 times of B's share=7 times of
 C's share=1
    therefore , the ratio of their share =1/4:1/5:1/7
          LCM of 4,5,7=140
     therefore, ¼:1/5:1/7=35:28:20
   the ratio now can be written as 35:28:20 
   therefore C's share=(20/83)*1162=20*14=Rs 280.
		 
5.The ratio of the present ages of saritha and her mother is 
2:9, mother's age at the time of saritha's birth was 28 years,
what is saritha's present age?

Sol:  ratio of ages of saritha and her mother =2:9
     let the present age of saritha be 2x years.              
     then the mother's present age would be 9x years
            Difference in their ages =28 years
                    9x-2x=28 years
                      7x=28=>x=4
        therefore saritha's age =2*4=8 years