Ratio and Proportion are commonly asked questions in aptitude test and competitive exams. These problems require special arithmetic calculation ability to solve it quickly. Questions from these category could be so tricky, so you must equipped yourself with all the shortcut methods and fast calculation is one of the plus for ratio and proportion problems. Let’s start with some formula and definitions :

**Ratio** : The ratio of two quantities a and b in the same units , is the fraction a/b and we write it as a : b.

**Proportion** : The equality of ratios is called proportion.

=> a : b : : c : d <=> (b x c) = (a x d)

Product of means (b,c) = Product of extremes (c,d)

**Componendo and Dividendo** : If a / b = c / d , then (a + b) / (a – b) = (c + d) / (c – d)

You would have studied these definitions and formulas many times, let’s start with solving some problems :

**Problem 1** : **If a : b = 5 : 9 and b : c = 4 : 7, find a : b : c**

**Solution 1** : Actually its very tricky question if you do not know how to solve it. In this kind of problems , first see the common term and b is the common term in here. So the trick in here is to make the value of b common in both the ratio and that could be done by multiplying the values with the corresponding ratio values i.e.

a : b = 5 : 9 ( Multiply it by 4 which is the ratio value of b in b : c)

a : b = 5 *4 / 9*4 => 20 : 36

Similarly, b : c = 4 : 7 (Multiply it by 9 which the ratio value of b in a : b)

b : c = 4*9 / 7*9 => 36 : 63

So, now the value of b has been equated in both the ratio , so we can finally conclude

Therefore , a : b : c = 20 : 36 : 63

**Problem 2** : *If A : B = (1/2) : (3/8) , B : C = (1/3) : (5/9) and C : D = (5/6) : (3/4), then find A : B : C : D*

**Solution 2** : So now here 4 terms are given instead of 3 like the previous problem ,

Now to solve this problem we need to first find out A : B : C using the method we used in the previous problem

Therefore , A : B = (1/2 * 1/3 ) : (3/8 * 1/3) = (1/6) : (1/8) and B : C = (1/3 * 3/8) : (5/9 * 3/8) = (1/8) : (5/24)

Hence , A : B : C = (1/6) : (1/8) : (5/24)

Since C : D = (5/6) : (3/4) , So C is the common term in here ,

Therefore, A : B : C = (1/6 * 5/6) : (1/8 * 5/6) : (5/24 * 5/6) = 5/36 : 5/48 : 25/144

Hence C : D = (5/6 * 5/24) : (3/4 * 5/24) = 25/144 : 15/96

Therefore, A : B : C : D = 5/36 : 5/48 : 25/144 : 15/96 = 1/6 : 1/8 : 5/24 : 3/16 (Take 5/6 as common)

Multiply the numerators with the LCM of the denominators (48)

Thus A : B : C : D = 8 : 6 : 10 : 9

**Problem 3** : **A bag contains 50 paise, 25 paise and 10 paise coins in the ratio 5 : 9 : 4, amounting to Rs 206. Find the number of coins of each type.**

**Solution 3** : Let the number of coins of 50p , 25p and 10p be 5x, 9x and 4x

Therefore, (5x/2) + (9x/4) + (4x/10) = 206 => x = 40

Hence, No of 50 p coins = 5x = 5 * 40 = 200; 25 p coins = 9x = 9*40= 360 and 10 p coins = 4x = 4*40=160

**Problem 4** : *In a mixture of is 60 liters , the ratio of milk and water is 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is ?*

**Solution 4** : Quantity of milk = 60 * (2/3) = 40 liters,

Thus quantity of water = 60 – 40 = 20 liters

New ratio of milk and water is 1 : 2

Let quantity of water to be added further is x liters.

There ratio of milk to water = 40 / (20+x)

Hence, 40 / (20+x) = 1/2 => x = 60

There quantity of water to be added is 60 liters.

**Problem 5** : **Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy 15 times as heavy as water ?**

**Solution 5** : Given Gold (G) = 19 times Water (W) = 19W

Therefore Copper (C) = 9W

New alloy = 15W

Lets mix 1 gm of Gold with x gm of Copper , which would result in (x + 1)gm of alloy

1 gm of Gold + x gm of Copper = (x+1) gm of alloy

19W + (9W)* x = (x+1)* 15W

=> x = 2/3

Therefore ratio Gold to Copper in the alloy = 1 : (2/3) => 3 : 2

**Problem 6** : **Based on Componendo and Dividendo **

**If p = 4xy / (x + y), find the value of [ ( (p + 2x) / (p – 2x) ) + ( (p + 2y) / (p – 2y) ) ]**

**Solution 6** : Given p = 4xy / (x + y) ,

Therefore p / 2x = 2y / (x+y)

Apply Componendo and Dividendo,

p+2x / p-2x = 2y+x+y / 2y-x-y => p+2x / p-2x = 3y + x / y-x

Similarly, p+2y / p-2y = 3x + y / x-y

Add both the equations, answer will be 2.

**Problem 7** : **The ratio of incomes of A and B is 5 : 4 and the ratio of their expenditures is 3 : 2. If at the end of the year, each saves Rs 1600, then what is the income of A ?**

**Solution 7** : Suppose the income of A and B is 5x and 4x respectively and their expenditures is 3y and 2y respectively.

Therefore , 5x – 3y = 1600 and 4x – 2y = 1600 , On solving both the equations ,

x = 800 , y = 800

Thus income of A = 5x = 5*800 = Rs 4000

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