GSEB Class 8 Maths Solutions Chapter 8 Comparing Quantities InText Questions

Try These (Page 119)

Question 1. In a primary school, the parents were asked about the number of hours they spend per day in helping their children to do homework. There were 90 parents who helped for \( \frac{1}{2} \) hour to \( 1\frac{1}{2} \) hours given in the adjoining figure; 20% helped for more than \( 1\frac{1}{2} \) hours per day; 30% helped for \( \frac{1}{2} \) hour to \( 1\frac{1}{2} \) hours; 50% did not help at all. Using this, answer the following:
Answer:
(1) We know that 30% of the total surveyed parents corresponds to 90 parents. To find the total number of surveyed parents, we can set up an equation:
Let 'x' be the total number of surveyed parents.
\( 30\% \text{ of } x = 90 \)
\( \frac{30}{100} \times x = 90 \)
\( x = \frac{90 \times 100}{30} \)
\( x = 3 \times 100 \)
\( x = 300 \)
Therefore, the total number of surveyed parents is 300.

(2) The number of parents who did not help their children accounts for 50% of the total surveyed parents.
Number of parents who did not help \( = 50\% \text{ of } 300 \)
\( = \frac{50}{100} \times 300 \)
\( = 50 \times 3 \)
\( = 150 \)
So, 150 parents did not help with homework.

(3) The percentage of surveyed parents who helped their children for more than \( 1\frac{1}{2} \) hours is 20%.
Number of parents who helped for more than \( 1\frac{1}{2} \) hours \( = 20\% \text{ of } 300 \)
\( = \frac{20}{100} \times 300 \)
\( = 20 \times 3 \)
\( = 60 \)
Thus, 60 parents offered more than \( 1\frac{1}{2} \) hours of help.
In simple words: First, we figured out the total number of parents by using the 30% who helped. Then, we calculated how many didn't help and how many helped for a longer time based on the given percentages.

Exam Tip: When given a percentage corresponding to a specific number, use that information to calculate the total amount before finding other proportional parts.

Try These (Page 121)

Question 1. A shop gives 20% discount. What would the sale price of each of these be?
(a) A dress marked at Rs.120
(b) A pair of shoes marked at Rs.750
(c) A bag marked at Rs.250
Answer:
(a) For the dress:
Marked price \( = \text{Rs. } 120 \)
Discount rate \( = 20\% \)
Discount amount \( = 20\% \text{ of Rs. } 120 \)
\( = \frac{20}{100} \times 120 \)
\( = 2 \times 12 \)
\( = \text{Rs. } 24 \)
Sale price \( = \text{Marked price } - \text{ Discount} \)
\( = \text{Rs. } 120 - \text{ Rs. } 24 \)
\( = \text{Rs. } 96 \)

(b) For the pair of shoes:
Marked price \( = \text{Rs. } 750 \)
Discount rate \( = 20\% \)
Discount amount \( = 20\% \text{ of Rs. } 750 \)
\( = \frac{20}{100} \times 750 \)
\( = 2 \times 75 \)
\( = \text{Rs. } 150 \)
Sale price \( = \text{Marked price } - \text{ Discount} \)
\( = \text{Rs. } 750 - \text{ Rs. } 150 \)
\( = \text{Rs. } 600 \)

(c) For the bag:
Marked price \( = \text{Rs. } 250 \)
Discount rate \( = 20\% \)
Discount amount \( = 20\% \text{ of Rs. } 250 \)
\( = \frac{20}{100} \times 250 \)
\( = 2 \times 25 \)
\( = \text{Rs. } 50 \)
Sale price \( = \text{Marked price } - \text{ Discount} \)
\( = \text{Rs. } 250 - \text{ Rs. } 50 \)
\( = \text{Rs. } 200 \)
In simple words: To find the sale price, first calculate 20% of the original marked price for each item to get the discount amount. Then, subtract this discount from the original price.

Exam Tip: Remember that discount is always calculated on the marked price. The sale price is obtained by subtracting the discount from the marked price.

Question 2. A table marked at Rs.15,000 is available for 14,400. Find the discount given and the discount per cent?
Answer:
Marked price of the table \( = \text{Rs. } 15000 \)
Sale price of the table \( = \text{Rs. } 14400 \)
Discount given \( = \text{Marked price } - \text{ Sale price} \)
\( = \text{Rs. } 15000 - \text{ Rs. } 14400 \)
\( = \text{Rs. } 600 \)

Discount per cent \( = \frac{\text{Discount}}{\text{Marked price}} \times 100\% \)
\( = \frac{600}{15000} \times 100\% \)
\( = \frac{6}{150} \times 100\% \)
\( = \frac{1}{25} \times 100\% \)
\( = 4\% \)
In simple words: We first find how much money was taken off the price, which is the discount. Then, we use this discount to calculate what percentage it is of the original price.

Exam Tip: Always express the discount percentage with respect to the marked price, not the sale price.

Question 3. An almirah is sold at Rs.5,225 after allowing a discount of 5%. Find its marked price?
Answer:
Sale price of the almirah \( = \text{Rs. } 5225 \)
Discount rate \( = 5\% \)
We know that Discount \( = 5\% \text{ of Marked price} \)
Let the Marked Price be \( MP \).
So, \( MP - \text{Discount} = \text{Sale price} \)
\( MP - 5\% \text{ of } MP = \text{Rs. } 5225 \)
\( MP \left( 1 - \frac{5}{100} \right) = \text{Rs. } 5225 \)
\( MP \left( \frac{100 - 5}{100} \right) = \text{Rs. } 5225 \)
\( MP \left( \frac{95}{100} \right) = \text{Rs. } 5225 \)
\( MP = \frac{5225 \times 100}{95} \)
\( MP = 55 \times 100 \)
\( MP = \text{Rs. } 5500 \)
The marked price of the almirah is Rs.5500.
In simple words: Since the sale price is after a 5% discount, it means the sale price is 95% of the original marked price. We use this to calculate the full original marked price.

Exam Tip: If a discount of x% is given, the sale price is (100-x)% of the marked price. Use this relationship to find the marked price when sale price and discount percentage are known.

Try These (Page 123)

Question 1. Find selling price (SP) if a profit of 5% is made on
(a) a cycle of Rs.700 with Rs.50 as overhead charge.
(b) a lawn mower bought at Rs.1150 with Rs.50 as transportation charges.
(c) a fan bought for Rs.560 and expenses of Rs.40 made on its repairs.
Answer:
(a) For the cycle:
Total cost price (CP) \( = \text{Purchase price } + \text{ Overhead expenses} \)
\( = \text{Rs. } 700 + \text{ Rs. } 50 \)
\( = \text{Rs. } 750 \)
Profit \( = 5\% \text{ of CP} \)
\( = 5\% \text{ of Rs. } 750 \)
\( = \frac{5}{100} \times 750 \)
\( = \frac{3750}{100} \)
\( = \text{Rs. } 37.50 \)
Selling price (SP) \( = \text{CP } + \text{ Profit} \)
\( = \text{Rs. } 750 + \text{ Rs. } 37.50 \)
\( = \text{Rs. } 787.50 \)

(b) For the lawn mower:
Total cost price (CP) \( = \text{Purchase price } + \text{ Transportation charges} \)
\( = \text{Rs. } 1150 + \text{ Rs. } 50 \)
\( = \text{Rs. } 1200 \)
Profit \( = 5\% \text{ of CP} \)
\( = 5\% \text{ of Rs. } 1200 \)
\( = \frac{5}{100} \times 1200 \)
\( = 5 \times 12 \)
\( = \text{Rs. } 60 \)
Selling price (SP) \( = \text{CP } + \text{ Profit} \)
\( = \text{Rs. } 1200 + \text{ Rs. } 60 \)
\( = \text{Rs. } 1260 \)

(c) For the fan:
Total cost price (CP) \( = \text{Purchase price } + \text{ Repair expenses} \)
\( = \text{Rs. } 560 + \text{ Rs. } 40 \)
\( = \text{Rs. } 600 \)
Profit \( = 5\% \text{ of CP} \)
\( = 5\% \text{ of Rs. } 600 \)
\( = \frac{5}{100} \times 600 \)
\( = 5 \times 6 \)
\( = \text{Rs. } 30 \)
Selling price (SP) \( = \text{CP } + \text{ Profit} \)
\( = \text{Rs. } 600 + \text{ Rs. } 30 \)
\( = \text{Rs. } 630 \)
In simple words: First, calculate the total cost by adding overheads or expenses to the buying price. Then, find 5% of this total cost for the profit. Finally, add the profit to the total cost to get the selling price.

Exam Tip: Remember that overhead expenses (transport, repairs, etc.) are always added to the cost price before calculating profit or loss.

Question 2. A shopkeeper bought two TV sets at Rs.10,000 each. He sold one at a profit 10% and the other at a loss of 10%. Find whether he made an overall profit or loss?
Answer:
Cost Price (CP) of each TV set \( = \text{Rs. } 10000 \)
Total Cost Price (CP) of both TV sets \( = 2 \times \text{Rs. } 10000 \)
\( = \text{Rs. } 20000 \)

For the 1st TV set (sold at 10% profit):
Profit rate \( = 10\% \)
Profit amount \( = 10\% \text{ of Rs. } 10000 \)
\( = \frac{10}{100} \times 10000 \)
\( = \text{Rs. } 1000 \)
Selling Price (SP) of 1st TV \( = \text{CP } + \text{ Profit} \)
\( = \text{Rs. } 10000 + \text{ Rs. } 1000 \)
\( = \text{Rs. } 11000 \)

For the 2nd TV set (sold at 10% loss):
Loss rate \( = 10\% \)
Loss amount \( = 10\% \text{ of Rs. } 10000 \)
\( = \frac{10}{100} \times 10000 \)
\( = \text{Rs. } 1000 \)
Selling Price (SP) of 2nd TV \( = \text{CP } - \text{ Loss} \)
\( = \text{Rs. } 10000 - \text{ Rs. } 1000 \)
\( = \text{Rs. } 9000 \)

Total Selling Price (SP) of both TV sets \( = \text{SP of 1st TV } + \text{ SP of 2nd TV} \)
\( = \text{Rs. } 11000 + \text{ Rs. } 9000 \)
\( = \text{Rs. } 20000 \)

Since Total Cost Price \( = \text{Rs. } 20000 \) and Total Selling Price \( = \text{Rs. } 20000 \), there is no overall profit or loss in this transaction.
In simple words: We find the selling price for each TV separately, one with profit and one with loss. After adding up both selling prices, we compare it with the total cost price of both TVs. In this situation, the total money gained is the same as the total money spent.

Exam Tip: When an item is sold at the same percentage profit and loss, remember that the calculations must be based on the individual cost prices. In cases where CP is the same, equal percentage profit and loss will result in no overall profit or loss.

Try These (Page 124)

Question 1. Find the buying price of each of the following when 5% ST is added on the purchase of
(a) A towel at Rs.50 each
(b) Two bars of soap at Rs.35 each
(c) 5 kg of flour at 15 per kg
Answer:
(a) For a towel:
Cost of the towel \( = \text{Rs. } 50 \)
Sales tax \( = 5\% \text{ of Rs. } 50 \)
\( = \frac{5}{100} \times 50 \)
\( = \text{Rs. } 2.50 \)
Buying price of the towel \( = \text{Cost } + \text{ Sales tax} \)
\( = \text{Rs. } 50 + \text{ Rs. } 2.50 \)
\( = \text{Rs. } 52.50 \)

(b) For two bars of soap:
Cost of one soap bar \( = \text{Rs. } 35 \)
Cost of two soap bars \( = 2 \times \text{Rs. } 35 \)
\( = \text{Rs. } 70 \)
Sales tax \( = 5\% \text{ of Rs. } 70 \)
\( = \frac{5}{100} \times 70 \)
\( = \text{Rs. } 3.50 \)
Buying price of the two soap bars \( = \text{Cost } + \text{ Sales tax} \)
\( = \text{Rs. } 70 + \text{ Rs. } 3.50 \)
\( = \text{Rs. } 73.50 \)

(c) For 5 kg of flour:
Cost of flour per kg \( = \text{Rs. } 15 \)
Cost of 5 kg of flour \( = 5 \times \text{Rs. } 15 \)
\( = \text{Rs. } 75 \)
Sales tax \( = 5\% \text{ of Rs. } 75 \)
\( = \frac{5}{100} \times 75 \)
\( = \frac{375}{100} \)
\( = \text{Rs. } 3.75 \)
Buying price of 5 kg of flour \( = \text{Cost } + \text{ Sales tax} \)
\( = \text{Rs. } 75 + \text{ Rs. } 3.75 \)
\( = \text{Rs. } 78.75 \)
In simple words: For each item, first figure out its total cost. Then, calculate 5% of that cost for the sales tax. Finally, add the sales tax to the original cost to get the final buying price.

Exam Tip: Remember to calculate sales tax on the actual purchase price of the goods, which is the base amount before tax is added.

Question 2. If 8% VAT is included in the prices, find the original price of
(a) A TV bought for Rs.13,500
(b) A shampoo bottle bought for Rs.180
Answer:
(a) For the TV:
Cost of TV including VAT \( = \text{Rs. } 13500 \)
Rate of VAT \( = 8\% \)
Let the original price be \( x \).
Price including VAT \( = x + 8\% \text{ of } x \)
\( = x + \frac{8}{100}x \)
\( = x \left( 1 + \frac{8}{100} \right) \)
\( = x \left( \frac{100+8}{100} \right) \)
\( = x \left( \frac{108}{100} \right) \)
So, \( \frac{108}{100}x = \text{Rs. } 13500 \)
\( x = \frac{13500 \times 100}{108} \)
\( x = 125 \times 100 \)
\( x = \text{Rs. } 12500 \)
The original price of the TV was Rs.12500.

(b) For the shampoo bottle:
Cost of shampoo bottle including VAT \( = \text{Rs. } 180 \)
Rate of VAT \( = 8\% \)
Let the original price be \( x \).
Price including VAT \( = x + 8\% \text{ of } x \)
\( = x \left( \frac{108}{100} \right) \)
So, \( \frac{108}{100}x = \text{Rs. } 180 \)
\( x = \frac{180 \times 100}{108} \)
\( x = \frac{18000}{108} \)
\( x = \frac{500}{3} \)
\( x = \text{Rs. } 166 \frac{2}{3} \)
The original price of the shampoo bottle was Rs. \( 166 \frac{2}{3} \).
In simple words: If VAT is included, the current price is 108% of the original price. To find the original price, divide the given price by 108 and then multiply by 100.

Exam Tip: When VAT is included in the price, consider the given price as \( (100 + \text{VAT}\%) \) of the original price. This helps in working backward to find the original amount.

Try These (Page 125)

Question 1. Two times a number is a 100% increase in the number. If we take half the number, what would be the decrease in per cent?
Answer:
Let the original number be \( x \).
If we take half the number, it becomes \( \frac{x}{2} \).
Decrease in the number \( = \text{Original number } - \text{ Half the number} \)
\( = x - \frac{x}{2} \)
\( = \frac{x}{2} \)
Decrease percentage \( = \frac{\text{Decrease}}{\text{Original number}} \times 100\% \)
\( = \frac{\frac{x}{2}}{x} \times 100\% \)
\( = \frac{1}{2} \times 100\% \)
\( = 50\% \)
The decrease in percentage would be 50%.
In simple words: If you start with a number and then cut it in half, the amount it went down by is exactly half of the original number. So, it's a 50% decrease.

Exam Tip: Understand that a 100% increase means doubling the number, and halving the number implies a 50% decrease from the original amount.

Question 2. By what percent is Rs.2,000 less than Rs.2,400? Is it the same as the percent by which Rs.2,400 is more than Rs.2,000?
Answer:
Case I: Percentage by which Rs.2,000 is less than Rs.2,400.
Here, the base amount for comparison is Rs.2,400.
Difference \( = \text{Rs. } 2400 - \text{ Rs. } 2000 \)
\( = \text{Rs. } 400 \)
Percentage decrease \( = \frac{\text{Difference}}{\text{Base amount}} \times 100\% \)
\( = \frac{400}{2400} \times 100\% \)
\( = \frac{1}{6} \times 100\% \)
\( = \frac{50}{3}\% \)
\( = 16\frac{2}{3}\% \)

Case II: Percentage by which Rs.2,400 is more than Rs.2,000.
Here, the base amount for comparison is Rs.2,000.
Difference \( = \text{Rs. } 2400 - \text{ Rs. } 2000 \)
\( = \text{Rs. } 400 \)
Percentage increase \( = \frac{\text{Difference}}{\text{Base amount}} \times 100\% \)
\( = \frac{400}{2000} \times 100\% \)
\( = \frac{1}{5} \times 100\% \)
\( = 20\% \)

Comparing the two percentages, \( 16\frac{2}{3}\% \) is not the same as \( 20\% \).
Therefore, the two percentages are not the same.
In simple words: To find how much less Rs.2,000 is from Rs.2,400, we compare the difference to Rs.2,400. To find how much more Rs.2,400 is than Rs.2,000, we compare the difference to Rs.2,000. Since the base numbers are different, the percentages are also different.

Exam Tip: When calculating percentage increase or decrease, always identify the correct base value against which the change is being measured. This base value typically represents the original or reference amount.

Try These (Page 126)

Question 1. Find interest and amount to be paid on Rs.15000 at 5% per annum after 2 years?
Answer:
Given:
Principal (P) \( = \text{Rs. } 15000 \)
Rate of interest (R) \( = 5\% \) per annum
Time (T) \( = 2 \) years

Simple Interest (SI) \( = \frac{P \times R \times T}{100} \)
\( = \frac{15000 \times 5 \times 2}{100} \)
\( = 150 \times 5 \times 2 \)
\( = 150 \times 10 \)
\( = \text{Rs. } 1500 \)

Amount (A) \( = \text{Principal } + \text{ Simple Interest} \)
\( = \text{Rs. } 15000 + \text{ Rs. } 1500 \)
\( = \text{Rs. } 16500 \)
The interest earned is Rs.1500, and the total amount to be paid after 2 years is Rs.16500.
In simple words: We calculate the simple interest by multiplying the principal, rate, and time, then dividing by 100. The total amount is found by adding this interest to the original principal.

Exam Tip: For simple interest problems, carefully identify the principal, rate, and time. Ensure the rate and time units are consistent (e.g., rate per annum and time in years).

Note:
I. When interest is calculated on the amount of previous year then it is called compound interest (CI).
II. Unlike simple interest, the formula for compound interest, i.e., \( A = P \left[ 1 + \frac{R}{100} \right]^n \) gives us the amount directly.

Try These (Page 129)

Question 1. Find CI on a sum of Rs.8000 for 2 years at 5% per annum compounded annually?
Answer:
Given:
Principal (P) \( = \text{Rs. } 8000 \)
Rate (R) \( = 5\% \) p.a.
Time (T) \( = 2 \) years

Since the interest is compounded annually, we use the formula for Amount (A) with compound interest:
\( A = P \left[ 1 + \frac{R}{100} \right]^n \)
Here, \( n = T = 2 \) years.
\( A = 8000 \left[ 1 + \frac{5}{100} \right]^2 \)
\( A = 8000 \left[ \frac{100+5}{100} \right]^2 \)
\( A = 8000 \left[ \frac{105}{100} \right]^2 \)
\( A = 8000 \left[ \frac{21}{20} \right]^2 \)
\( A = 8000 \times \frac{21}{20} \times \frac{21}{20} \)
\( A = 8000 \times \frac{441}{400} \)
\( A = 20 \times 441 \)
\( A = \text{Rs. } 8820 \)

Now, Compound Interest (CI) \( = \text{Amount } - \text{ Principal} \)
\( = \text{Rs. } 8820 - \text{ Rs. } 8000 \)
\( = \text{Rs. } 820 \)
The compound interest on the sum is Rs.820.
In simple words: We calculate the total amount after two years using the compound interest formula, which means interest is added each year to the previous balance. Then, we subtract the original principal from this total amount to find the compound interest earned.

Exam Tip: For compound interest, ensure you use the correct formula \( A = P (1 + R/100)^n \) and subtract the principal from the final amount to get the CI. Pay close attention to whether compounding is annual, semi-annual, or quarterly.

Try These (Page 130)

Question 1. Find the time period and rate for each?
1. A sum taken for \( 1\frac{1}{2} \) years at 8% per annum is compounded half yearly.
2. A sum taken for 2 years at 4% per annum compounded half yearly.
Answer:
1. When a sum is taken for \( 1\frac{1}{2} \) years at 8% per annum, compounded half-yearly:
Time period (n) \( = 1\frac{1}{2} \text{ years} = \frac{3}{2} \text{ years} \)
Since it's compounded half-yearly, there are 2 compounding periods per year. So, the total number of periods \( = \frac{3}{2} \times 2 = 3 \) half-years.
Rate (R) per half-year \( = \frac{8\%}{2} = 4\% \text{ per half year} \).

2. When a sum is taken for 2 years at 4% per annum, compounded half-yearly:
Time period (n) \( = 2 \text{ years} \)
Since it's compounded half-yearly, there are 2 compounding periods per year. So, the total number of periods \( = 2 \times 2 = 4 \) half-years.
Rate (R) per half-year \( = \frac{4\%}{2} = 2\% \text{ per half year} \).
In simple words: When interest is compounded half-yearly, you double the number of years to get the total number of compounding periods and halve the annual interest rate to get the rate for each period.

Exam Tip: When interest is compounded semi-annually, remember to multiply the number of years by 2 for 'n' and divide the annual rate by 2 for 'R'.

Try These (Page 131)

Question 1. Find the amount to be paid
1. At the end of 2 years on Rs.2,400 at 5% per annum compounded annually.
2. At the end of 1 year on Rs.1,800 at 8% per annum compounded quarterly.
Answer:
1. For Rs.2,400 at 5% p.a. compounded annually for 2 years:
Principal (P) \( = \text{Rs. } 2400 \)
Rate (R) \( = 5\% \) p.a.
Time (T) \( = 2 \) years
Number of compounding periods (n) \( = 2 \)
Amount (A) \( = P \left[ 1 + \frac{R}{100} \right]^n \)
\( A = 2400 \left[ 1 + \frac{5}{100} \right]^2 \)
\( A = 2400 \left[ \frac{105}{100} \right]^2 \)
\( A = 2400 \left[ \frac{21}{20} \right]^2 \)
\( A = 2400 \times \frac{21}{20} \times \frac{21}{20} \)
\( A = 6 \times 21 \times 21 \)
\( A = 6 \times 441 \)
\( A = \text{Rs. } 2646 \)

2. For Rs.1,800 at 8% p.a. compounded quarterly for 1 year:
Principal (P) \( = \text{Rs. } 1800 \)
Annual Rate \( = 8\% \)
Rate per quarter (R) \( = \frac{8\%}{4} = 2\% \)
Time (T) \( = 1 \) year
Number of compounding periods (n) \( = 1 \times 4 = 4 \) quarters
Amount (A) \( = P \left[ 1 + \frac{R}{100} \right]^n \)
\( A = 1800 \left[ 1 + \frac{2}{100} \right]^4 \)
\( A = 1800 \left[ \frac{102}{100} \right]^4 \)
\( A = 1800 \left[ \frac{51}{50} \right]^4 \)
\( A = 1800 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} \)
\( A = \text{Rs. } 1948.38 \) (approximately)
In simple words: For annually compounded interest, use the given annual rate and number of years. For quarterly compounded interest, divide the annual rate by four and multiply the number of years by four to find the rate and periods per quarter. Then, apply the compound interest formula to get the final amount.

Exam Tip: Be careful with the compounding frequency. For quarterly compounding, divide the annual rate by 4 and multiply the time in years by 4 to get the rate per period and total number of periods (n).

Try These (Page 133)

Question 1. A machinery worth Rs.10,500 depreciated by 5%. Find its value after one year?
Answer:
Given:
Present value of machinery (P) \( = \text{Rs. } 10500 \)
Rate of depreciation (R) \( = 5\% \) p.a.
Time (T) \( = 1 \) year
Number of periods (n) \( = 1 \)

For depreciation, the formula for the value after 'n' years is:
\( A = P \left[ 1 - \frac{R}{100} \right]^n \)
\( A = 10500 \left[ 1 - \frac{5}{100} \right]^1 \)
\( A = 10500 \left[ \frac{100 - 5}{100} \right] \)
\( A = 10500 \left[ \frac{95}{100} \right] \)
\( A = 105 \times 95 \)
\( A = \text{Rs. } 9975 \)
The value of the machinery after one year will be Rs.9975.
In simple words: Depreciation means the value goes down. We subtract the depreciation percentage from 100%, then multiply this new percentage by the original value to find the new, reduced value after one year.

Exam Tip: For depreciation problems, use the formula \( A = P (1 - R/100)^n \) where the rate 'R' is subtracted, indicating a decrease in value over time.

Question 2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%?
Answer:
Given:
Present population (P) \( = 12 \text{ lakh} = 12,00,000 \)
Rate of increase (R) \( = 4\% \) p.a.
Time (T) \( = 2 \) years
Number of periods (n) \( = 2 \)

The formula for population growth (similar to compound interest for an increase) is:
\( A = P \left[ 1 + \frac{R}{100} \right]^n \)
\( A = 12,00,000 \left[ 1 + \frac{4}{100} \right]^2 \)
\( A = 12,00,000 \left[ \frac{104}{100} \right]^2 \)
\( A = 12,00,000 \left[ \frac{26}{25} \right]^2 \)
\( A = 12,00,000 \times \frac{26}{25} \times \frac{26}{25} \)
\( A = \frac{1200000 \times 676}{625} \)
\( A = 1920 \times 676 \)
\( A = 12,97,920 \)
The population of the town will be 12,97,920 after 2 years.
In simple words: To find the population after an increase, we use a formula like compound interest. We add the percentage increase to 100%, then multiply the current population by this new percentage for each year.

Exam Tip: Population growth problems use the compound interest formula with a positive rate, as the population increases over time. Ensure consistent units and accurate calculations.