Ratio

                  Ratio

Ratio, Proportion and Variation forms an interesting lecture. The basic fundamentals of this lecture are extensively used in solving many questions in chapters like Work & Time, Time, Speed & Distance etc. You are likely to find quite a few questions in the actual exams, based on these concepts.

Ratio

The comparison between two quantities of the same kind of unit is the Ratio of one quantity to another. Two quantities of different Ratio kinds cannot be compared. Thus, there is no relation between 20 rupees and 20 men. The ratio of a and b is usually written as a : b or a/b .

Antecedent and Consequent

In the ratio a: b, a is called the antecedent (the first term) and b is called the consequent (the second term).

Imp.                       

The ratio of two numbers a and b, written as a : b, is the fraction a/b provided b # 0. Thus

a:b= a/b,b≠0.Ifa=b≠0,the ratio is 1:1 or 1/1=1.

ü  The ratio of 4 to 6 =4: 6 =4/ 6 =2/ 3 .

ü  2/3 : 4/5 = 2/3 = 5/6

                 4/5

ü 5x : 3y  =     5x    =  20x

                 3y/4   3y

 Commensurable Ratio

It is the ratio of two fractions or any two quantities which can be expressed exactly by the ratio of 2 integers.

e.g. ad : bc. For example, the ratio of 10 m to 40 m is 10 : 40 or 1 : 4, which is the ratio of two integers, so these are the commensurable quantities.

Incommensurable Ratio

It is the ratio of two fractions or any two quantities in which one or both of the terms is a surd quantity. No two integers can be found which will exactly measure their ratio i.e. cannot be exactly expressed by any 2 integers. For example, the ratio of a side of a square to its diagonal is 1 : √2 which cannot be expressed as a ratio of two integers. Thus, 1 and √2 are incommensurable quantities.

      

Composition of Ratios

Compounded Ratio

When two or more ratios are multiplied term wise, the ratio thus obtained is called their compounded ratio. For the ratios a : b and c : d, the compounded ratio is ac : bd.

El. What is the compounded ratio, for 2 : 3 and 4 : 5?

Sol. The compounded ratio is 2 x 4 : 3 x 5 or 8 : 15.

Duplicate Ratio

It is the compounded ratio of two equal ratios. Thus the duplicate ratio of a : b is a²/b² or a² : b².

E2.   Find the duplicate ratio of 4 : 5.

Sol. The duplicate ratio of 4 : 5 is 16 : 25.

E3.   If a : b is the duplicate ratio of a + x : b + x, then show that x²= ab.

Sol,   a    = (a +x )²   ,   a  =    a² + 2ax + x2 

   b       (b+x)²        b        b²+2bx+x²

=ab²+2abx+ax²=a²b+2abx+bx²

= ax² – bx²= a²b-ab² = x²(a-b)= ab(a-b)

So, x² = ab

Triplicate Ratio

It is the compounded ratio of three equal ratios. Thus the triplicate ratio of a : b is a³/b³ or a³ : b3.

E4.   Find the triplicate ratio of 4: 5.

Sol. The triplicate ratio of 4 : 5 is 64 : 125.

Sub-duplicate Ratio

For any ratio a : b, its sub-duplicate ratio is defined as √a : √b

E5.   Find the sub-duplicate ratio 16: 25.

Sol. Sub-duplicate ratio of 16 : 25 is √16 : √25, i.e. 4 : 5.

Sub-triplicate Ratio

For any ratio a : b, its sub-triplicate ratio is defined as ³√a : ³√b .

E6.   Find the sub-triplicate ratio of 27 : 64.

Sol. Sub-triplicate ratio of 27 : 64 is ³ √27 : ³ √64 , i.e. 3 : 4.

Reciprocal Ratio

For the ratio a : b, a, b # 0, the ratio 1/a :1/ b which is same as b : a is called its reciprocal ratio.

Equivalent Ratios

We can get the equivalent fraction of a certain fraction by either multiplying or dividing both the numerator and denominator of the given fraction by the same number. In the same way, we get the equivalent ratio of a certain ratio too. The equivalent ratios of a : b are 2a : 2b, 3a : 3b, 5a : 5b etc.

E7. Find two equivalent ratios of 4 : 5.

Sol. We can express 4 : 5 as a fraction, that is 4/5.

4    =   4x2  =       4x3   or     4    ₌    8   =   12

5         5x2            5x3                5       10       15

Or,4:5=8:10= 12:15.

So, 8 : 10 and 12 : 15 are equivalent ratios of 4 : 5.

Comparing Ratios

To compare the two ratios, we express them as fractions and then compare.

E8. Which is greater 3 : 4 or 4: 5?

Sol. 3 : 4 = ¾ and 4:5 = 4/5 .

3      =  3  x 5 = 15

4           4 x 5    20

4      =  4  x 4 = 16          

5          5  x 4    20

Since, 16   > 15       or,    4   >  3

           20      20              5        4

So,4:5 > 3:4.

4

Important

A ratio is said to be in its simplest form if the HCF of the antecedent and the consequent is 1.

E9. Divide 2400 in the ratio 3 : 5.

Sol. The first part is 3 units and the second part is 5 units. The total of both the parts = 3 units + 5 units = 8 units. Here, 8 units = 2400,

So, 1 unit = 2400/8 = 300.

The first part = 3 units = 3 x 300 = 900.

The second part = 5 units = 5 x 300 = 1500.

E10.A sum of money is divided between Vinod and Lokesh in the ratio of 3 : 7. Vinod gets Rs.240. What does Lokesh get?

Sol. Vinod gets 3 units = Rs.240.

So, 1 unit = 240/3 = 80.

Therefore, 7 units = 7 x 80 = 560. Thus, Lokesh gets Rs.560.

Important

a : b = ka : kb where k is a constant. a:b=a/k:b/k.

a : b  ≠ a+k: b+k, where a ≠ b. a : b ≠  a² : b², where a ≠  b.

a : b : c = X : Y : Z is equivalent to a   =b=  c

            X  Y  Z

a  : b > X : Y if aY > bX.  a : b < X : Y if aY < bX.

   If a  = c = e  =…. ,k  then   a+c+e+...  = k.
       b     d     f                                 b+d+f+...

Also note that: k = a /b = Xc  / Xd = -5e/-5f.

Each ratio  =   (a+Xc-5e)  =  k.               

                         (b+Xd-5f)

Here, we have randomly taken X and —5.

You can take any factor.

If a / b >1 or a > b, then  a+X  <   a    where a, b, X are natural

            b+X   b

numbers.

If a / b <1 or a < b, then  a+X  >   a    where a, b, X are natural

            b+X   b

numbers.

a² / b² , a³ / b³ , √a / √b , ³√a / ³√b    are respectively called duplicate, triplicate , sub-duplicate and sub-triplicate ratios of a/b.

Ell.  Find the compounded ratio of (a + X) : (a — X),

      (a² + X²)  : (a + X)², and (a² — X²)² : (a⁴ — X⁴).

Sol. a + X      x    a² + X²     x     (a² – X²)²

       a – X           (a + X)²             (a⁴ – X⁴)

=  a + X      x          a² + X²           x            (a² – X²)(a² – X²)

    a – X              (a + X)(a+X)                     (a² – X²)(a² + X²)

=     a² – X²             =        a² – X²            =    1 .

   ( a- X )(a+X )              a² – X²

Ratios from homogeneous equations

The ratios of x : y : z from the equations

a_l x + b_l y + c_l z = 0 and a₂x + b₂y + c₂z = 0 is given by

x : y : z = (b₁c₂ — b₂c_l) : (c₁a₂ — c₂a_l): (a₁b₂ — a₂b₁).

E12.If 4x — 2y — 7z = 0 and x + y — 4z = 0, then find the ratio of x : y : z..

 (8) -(-7) : (-7) - (-16) : (4) - (-2)

 Or 15   :  9  :  6

 x:y:z=15:9:6=5:3:2.

E13.If ax + cy + bz = 0, cx + by + az = 0 and bx + ay + cz = 0, then show that a³ + b³ + c³ = 3abc. -

Sol. Let ax+cy+bz=0...(1)      cx+by+az=0...(2)

From (1)&(2)

ac—b²: bc—a²: ab—c²

let       X           =          y           =             z             =          k

       ac – b²              bc – a²                ab – a²

Then x = k(ac — b²),  y = k(bc — a²),  z = k(ab — c²)  Putting them in equation bx + ay + cz = 0 .•. bk(ac—b²) + ak(bc—a²) + ck(ab—c²) = 0. Or abc-b³+abc-a³+abc-c³= 0.

.'. 3abc = a³ + b³ + C³.

Problems leading to the application of ratios

E14.The ratio of the number of boys to the number of girls in a school of 1638 is 5 : 2. If the number of girls increased by 60, then what must be the decrease in the number of boys to make the new ratio of boys to girls as 4 : 3?

Sol. Number of boys =5/7 x 1638 =1170.

Number of girls =2 /7x 1638 =468.

Number of girls after increase = 468 + 60 = 528.

Total number of boys = 528 x4/ 3 = 704.

The number of boys to be decreased = 1170 — 704 = 466.

E15.A bag contains an equal number of one rupee, 50 paise and 25 paise coins respectively. If the total value is Rs.35, then how many coins of each type are there?

     Sol. Here number of each type of coin is the same. Hence, we may write    

 Number of each type of coin =

     Total amount

Sum of value of each coin

Number of each type of coin =                           35

                     1+0.5 +0.25       

= 20 coins of each type.

E16.The sum of the squares of three numbers is 532 and the ratio of the first to the second as also of the second to the third is 3 : 2. What is the second number?

 Sol.     first number  =    3   x   3   =    9

        Second number      2        3         6

       [Make the second number same in both the ratios; i.e. 6]

and Second number = 3 x 2   =   6

         Third number     2 2   4

First : Second : Third = 9 : 6: 4.

(9x)² + (6x)² + (4x)² = 532            133x² = 532 = x = ±2 Second number is 6x = ±12 .

E17.A bag contains one rupee, fifty paise, twenty five paise and ten paise coins in the proportion 1: 3 : 5 : 7. If the total amount is Rs.22.25, then find the number of coins of each

kind.

Sol. Let the number of coins of one rupee, fifty paise, twenty five paise and ten paise be x, 3x, 5x, 7x respectively.

Now, value of one 50 paise coin in rupee = 1/2 .

Value of one 25 paise coin in rupee = 1/4

'      Value of one 10 paise coin in rupee = 1/10

        Number of coins x Value of coin in rupees = Amount in rupees

       [(x x1) +  (3x   x ½) + (5x  x  ¼) + (7x   x  1/10)] = Rs.22.25

89x =22.25   

x=5. 20

Number of 1 rupee coins = 5 x 1= 5 Number of 50 paise coins = 3x = 15. Number of 25 paise coins = 5x = 25. Number of 10 paise coins = 7x = 35.

E18.If a : b = 3:4 and b : c = 6:13, then find a : b : c.

Sol. The best way to solve such questions is to make b common in the two ratios.

Thus, we can write a : b = 9 : 12, and b : c = 12 : 26. Now that b is equal in both the ratios, we can write the same as       

a           b        c

      9          12

   12     26

Thus, we can write a : b : c = 9 : 12 : 26.

Using formula directly, we can get

a : b : c = (3 x 6) : (4 x 6) : (4 x 13)   =   9 : 12 : 26.

E19.Given a : b = 4 : 5, b : c = 20 : 11, c : d = 6 : 7.
Find a : b : c: d.

Sol. Here we proceed step by step. First of all, we try to get a : b : c. This can be done by making b common in the two ratios a : b and b : c. To do this, let us multiply the first ratio by 4. So a : b becomes 16 : 20.

Since b : c = 20 : 11, a : b : c can be directly obtained as a : b : c = 16 : 20 : 11. Now making c common between the two ratios a : b : c and c : d. To do this, we multiply the first ratio (a : b : c) throughout by 6 and the second ratio (c : d) throughout by 11.

       Hence a : b : c = (16 : 20 : 11) x 6 = 96 :120 : 66 and c:d=(6:7)x 11=66:77.

= Final answer is a : b : c : d = 96 : 120: 66 : 77.

 Using formula directly, we can get a : b : c : d

= (4 x 20 x 6) : (5 x 20 x 6) : (5 x 11 x 6) :  (5 x 11 x 7) = 480: 600 : 330 : 385

     =96:120:66:77.

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