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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.

Banks provide four different types of accounts, which are as follows:

(1) Savings Bank Account

(2) Current Bank Account

(3) Fixed Deposit Account

(4) Recurring (or cumulative) Deposit Account

Now let us study about Recurring Deposit Account.

To encourage the habit of saving among low and middle income groups, banks and post offices provide Recurring Time Deposit schemes.

Under this scheme, an investor deposits a fixed amount every month for a fixed time period. This fixed time period is called maturity period.

After completing the maturity period, the investor gets the amount deposited with the interest. The total amount received by the investor is called maturity value.

In this scheme, the interest is compounded quarterly at a fixed rate. The rate of interest is revised from time to time.

 

To understand how the interest is calculated in a Recurring Deposit Scheme, let us look at the following video.

Now, let us solve some more examples to understand the concept better.

Example 1:

Suresh has a Recurring Deposit Account in a bank. He deposits Rs 1000 per month for one year. At the time of maturity, he gets Rs 12650. What is the rate of simple interest?

Solution:

It is given that:

Money deposited per month, p = Rs 1000

Maturity period, n = 12 months

We know that,

S.I.=p×nn+12×12×r100S.I.=1000×12×132×12×r100S.I.=65r

It is given that:

Amount of maturity = 12650

p × n + S.I. = 12650

1000 × 12 + 65r = 12650

65r = 12650 − 12000

65r = 650

r=65065=10

Thus, the rate of interest is =10% per annum.

Example 2:

Abhishek has a Recurring Deposit Account in a bank for 2 years at 10% p.a. simple interest. If he gets Rs 3750 as the interest after maturity period, then find the monthly installment and the amount of maturity.

Solution:

Let the monthly installment be Rs x.

It is given that r = 10%

n = (2 × 12) months = 24 months

We know that,

S.I.=p×nn+12×12×r1003750=x×24×252×12×10100x=3750×1025x=1500

∴ Monthly installment = Rs 1500

Amount of maturity = p × n + S.I.

= 1500 × 24 + 3750

= 39750

Thus, the monthly installment is Rs 1500 and amount of maturity is Rs 39,750.

In many cases, it is better if we represent increase or decrease in quantities in percentage terms rather than describing them numerically. 

Let us take the case where Manoj got a raise of Rs 5000 in his monthly salary. Now, we also have to consider his original salary in order to ascertain whether the raise was high or low. For example, if his original salary was Rs 10000, then this means that he got a raise of 50% in his monthly salary, which is quite high. However, let us suppose that his original monthly salary was Rs 50000. Now this would mean that Manoj got a raise of only 10%, which is quite low.

Thus, if we talk only about the increase or decrease in a quantity without referring to its original value, then it will cause confusion in some cases. However, this problem does not arise if we use percentages.

To understand the concept of percentage increase, look at the following video.

In the same way, we can find the percentage decrease, using the following formula.

Let us consider an example to understand percentage decrease.

In an athletic event, 60 students of a school participated last year. This year, the number of students of that school taking part is decreased by 5%. Find the number of students taking part in the athletic event this year.

Here, we have percentage decrease = 5%

Original number of students = 60

⇒ Decrease in number of students = 3

Thus, number of students taking part this year = 60 – 3 = 57

Thus, 57 students participated this year.

Let us solve some examples to understand the concept better.

Example 1:

The price of a toy was decreased by 20%. If this meant a decrease of Rs 125 in its price, then what were the original and the reduced prices of that toy?

Solution:

Let the original price of the toy to be x.

Percentage decrease in the price of the toy = 20

Decrease in the price of the toy = Rs 125

In this case,

Thus, original price of the toy = Rs 625

And, reduced price of the toy = Rs 625 − Rs 125 = Rs 500

Example 2:

The population of a city in 2002 was 2.5 crores. If it increased by 15% in a year, then what was the population in 2003?

Solution:

Original population of the city = 2.5 crores

Percentage increase in population = 15 %

Let x be the increase in population of the city.

In this case,

Thus, the population increased by 0.375 crores over the year.

Thus, population of the town in 2003 = (2.5 + 0.375) crores = 2.875 crores

 

Example 3:

The enrolment at a school increased from 1400 to 1500 in one year. What is the percentage increase in the enrolment?

Solution:

Increase in the enrolment = 1500 - 1400 = 100

Percentage increase in the enrolment

Thus, the enrolment is increased by 7.14%.

Example 4:

The selling price of a DVD player was dropped by 20% in one year. If the selling price of the DVD player is Rs 8000 now, then find the selling price of the DVD player one year before.

Solution:

Let x be the selling price of the DVD player before one year. The selling price of the DVD player was dropped by 20%. Therefore, now the selling price of the DVD is (100 - 20)% of x = 80% of x

∴ 80% of x = Rs 8000

Thus, the selling price of the DVD player before one year was Rs 10000.

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