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Life Mathematics

Word Problems on Percentages

Sukhiram is the sarpanch of a village and is very good at mathematics. The current population of his village is 18550. From the record of past years, he observed that the population of the village is increasing at the rate of 4% per year. On the basis of the current growth rate, he wants to know the population of the village after 2 years. How can he do so?

Actually, it is a problem of growth (or appreciation). If he knows the formula, then he can easily find out what he wants to know.

Let us understand the formula that should be used here.

The growth per unit time is called the rate of growth.

Currently, the population of the village is increasing at the rate of 4% every year.

This means that in a year the population increases by 4% of the previous population. Hence, the rate of growth is 4% and is denoted by r.

∴ r = 4%

If V0 is the current measure of quantity and V is the measure of quantity after n years, then

This is the relation between the current measure of a quantity and the measure of a quantity after n years.

On using this formula, we can find the population of the village after two years. Let us see how.

In the given example,

r = 4%

V0 = 18550

n = 2

On substituting these values in equation (i), we obtain

The population of the village cannot be in decimals. Hence,

V = 20064 (approximately)

In our daily lives, we come across several things that grow. For example: Population of a city or village, prices of goods, height or weight of children etc.

On the other hand, the value of machines or vehicles decreases with time.

The decrease in value per unit time is called rate of depreciation.

If the rate of depreciation is constant, then we have the following formula:

Here, r is the rate of depreciation per year; V0 is the current value and V is the value after n years.

When a quantity increases in the first year at the rate of, then decreases at the rate of in the second year and then increases at the rate of in the third year, then the formula is

Here, V0 is the initial measure and V is the quantity after 3 years.

Let us now solve some problems to understand these formulae better.

Example 1

Priya bought a diamond necklace worth Rs 300000. The value of the necklace appreciates by 6% every year. What will be the value of the necklace after 3 years?

Solution:

The value of the necklace after three years can be calculated by using the compound interest formula.

P = Rs 300000

Rate of appreciation, R = 6% p.a.

Therefore, the value of the necklace after 3 years

Thus, the value of the necklace after 3 years will be Rs 357304.80.

Example 2

There were 5000 students in a school in the year 2004. In the year 2005, the number of students increased by 5% of the number of students in the previous year. In the year 2006, the number of students decreased by 12% of the number of students in the previous year. How many students did the school have in the year 2006?

Solution:

V0 = 5000

Rate of appreciation in the year 2005, r1 = 5%

Rate of depreciation in the year 2006, r2 = 12%

Number of students in the year 2006 is given by

Therefore, the number of students in the year 2006

Thus, the number of students in the school in the year 2006 is 4620.

Suppose you go to a grocery shop to buy eggs. If the shopkeeper is selling 2 eggs for Rs 4, then what amount is required to buy 5 eggs?

We can solve this problem using the unitary method.

First, let us determine the cost of 1 egg.

It is given that cost of 2 eggs = Rs 4

Cost of 1 egg =

Hence, cost of 5 eggs = 5 × Rs 2 = Rs 10

What do we observe in the above example?

One simple thing that we observe is that as the number of eggs increases, its cost also increases. Such situations are examples of direct variation or direct proportion. In our day-to-day lives, we come across various such situations. For example, if a car moves with constant speed, then the distance covered by it is in direct proportion with the time taken to cover the distance.

Direct variation or Direct proportion can be defined as follows:

When two variable quantities increase or decrease simultaneously such that their ratio remains unchanged, then it is an example of direct variation and the quantities are said to be in direct proportion. It is said that one variable "varies directly" with the other.

Let us look at an example of direct proportion in this video.

Writing direct variation or direct proportion in the form of symbols:

If there is direct variation between two variables x and y, then it is represented as:

x α y  

It is read as "x varies directly as y" or "x is directly proportional to y".  

Also, in direct proportion or direct variation, the ratio of variables is always a constant value.

Thus,  , where k is known as the constant of proportionality.

So, it can be concluded that:

If x α y, then , where k is a constant. 

or

If , then  x α y, where k is a constant. 

 

is the equation of direct proportion.

Now, suppose variables x and y are in direct proportion. If y1 and y2 are the values of y corresponding to the respective values x1 and x2 of x, then .

Therefore, we can write the equation as follows:

Now, let us consider the following situation to check whether the variables involved in it are in direct proportion or not.

Arnab goes to a stationery shop to buy some pens. If each pen costs Rs 5, then what amount is required to buy 4 such pens? Also, determine the amount that Arnab need to pay to buy 10 such pens?

It is given that cost of 1 pen = Rs 5

Hence, cost of 4 pens = 4 × Rs 5 = Rs 20

Similarly, cost of 10 pens = 10 × Rs 5 = Rs 50

This information can be represented in the tabular form, where the number of pens is denoted by variable x and their corresponding cost is denoted by variable y, as

Number of pens: x

1

4

10

Cost (Rs): y

5

20

50

If we observe the ratio of the corresponding values of x and y, then we see that

We, thus, observe that as the number of pens (x) increases, their cost (y) also increases in such a manner that their ratio remains constant, say k. Thus, in this case, the value of k is.

So, this is an example of direct variation.

Hence, we say that x and y are in direct proportion, if

Thus, to check whether the variables x and y are in direct proportion, we need to find the ratio for their corresponding values. If this ratio remains constant, then the variables are in direct proportion, otherwise they are not.

Let us now discuss some examples based on this concept.

Example 1: 

The scale of a map is given as 1:10000. The distance between two buildings in a city on the map is 5 cm. What is the actual distance between the two buildings?

Solution:

Given situation is an example of direct variation.

So, it can be said that the distance on the map and the actual distance between the buildings are in direct proportion.

Let the actual distance between the buildings be x.

Thus, the actual distance between the buildings is 500 m.

Example 2: 

If 1 kg 600 g of rice is sufficient for 20 people, then what quantity of rice will be sufficient for 27 people? Also calculate for how many people 2 kg 400 g of rice will be sufficient.

Solution:

Let x kg of rice be sufficient for 27 people and 2 kg 400 g of rice be sufficient for y number of people.

1 kg 600 g = 1.6 kg

2 kg 400 g = 2.4 kg

The given information can be represented by the following table:

Quantity of rice (kg): x

1.6

x

2.4

Number of people: y

20

27

y

In the given case, quantity of rice increases and decreases as the number of people increases and decreases respectively. So, this is an example of direct variation.  

So, it can be said that the quantity of rice and the number of people are in direct proportion.

Hence, 2 kg 160 g of rice will be sufficient for 27 people and 2 kg 400 g of rice will be sufficient for 30 people.

Example 3:

A car travels at a constant speed of 35 km/h. How far can it travel in 15 minutes?

Solution:

Speed of car = 35 km/h

This means that the car travels 35 km in 60 minutes.

Let the car travel x km in 15 minutes.

The given information can be represented by the following table:

Distance covered by the car (km): x

35

x

Time taken by the car (min): y

60

15

In the given case, the time taken by the car increases and decreases as the dista…

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