RRB Averages, Ratio and Proportion

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Theory

‘Averages, Ratio and Proportion’ is an important concept in Quantitative Aptitude. It is applicable in many arithmetic questions directly or indirectly. Average of a group of numbers can be obatined by the sum of numbers which are to be  divided by the quantity of number. Ratio is used to compare the quantities having the same units. It is represented by ‘:’. Proportion is used to represent the relationship between ratios. It is represented by ‘::’. Let’s assume two fractions are $$\frac{P}{Q}$$ and $$\frac{R}{S}$$. So it can be represented as P:Q and R:S.

P:Q :: R:S

$$\frac{P}{Q} = \frac{R}{S}$$

There are many properties in this topic out of which dividendo and componendo are used very frequently.

If P:Q :: R:S, then as per dividendo and componendo property, it will be (P+Q) : (P-Q) :: (R+S) : (R-S).

$$\frac{P+Q}{P-Q} = \frac{R+S}{R-S}$$

Formula

Average = $$\frac{\text{Sum of data}}{\text{Number of data}}$$

Solved Example

Q) If E:F is 2:5 and F:G is 4:3, then find out E:G.

Sol.

E:F ⇒ 2:5 Eq.(i)

F:G ⇒ 4:3 Eq.(ii)

From Eq.(i) and Eq.(ii).

E:F:G ⇒ 8:20:15

So E:G ⇒ 8:15

Formula


$$\frac{A}{B} = \frac{B-M}{M-A}$$

The above given is the formula for mixture and alligations.

 If any mixture initially is in pure form and when a fixed amount of mixture is taken out in a repeated manner and replaced with water, then the following formula can be used.

quantity of pure liquid after n number of time taken out mixture = $$\text{original mixture} \times(1- \frac{\text{fix amount of mixture which is taken out}}{\text{original mixture}})^{n}$$

Solved Example

Q) A shopkeeper has two varieties of Rice which cost price is Rs. 16 and Rs. 21 per kg. He mixed a certain quantity of both and formed a mixture of 100 kg and selling the mixture at the rate of Rs. 18 per kg. In this way, he neither gets profit nor loss. Find out the quantity of Rice in the mixture whose rate is Rs. 21 per kg.

Sol.

So the ratio of cheaper to dearer quantity is 3:2 respectively.

Quantity of Rice in the mixture whose rate is Rs. 21 per kg = (2/5) of 100 = 40 kg

Solved Example

Q) Find out the average of the numbers 20, 35, 38 and 25.

Sol. Average = $$\frac{20+35+38+25}{4}$$

= $$\frac{118}{4}$$

= 29.5

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