Compound Interest And Uniform Rate Of Increase Or Decrease

We know that amount after 3 years

Where p = principal, r = rate of interest and n = time = 3 years

From given, p = ₹5000 and r = 8.5%

So, the amount after 3 years

= ₹6386.44

We know that amount after 3 years

Where p = principal, r = rate of interest and n = time = 3 years

From given, p = ₹5000 and r = 8%

So, the amount after 3 years

= ₹6298.56

We know that amount after 2 years

Where p = principal, r = rate of interest and n = time = 2 years

From given, p = ₹2000 and r = 6%

So, the amount after 2 years

= ₹2247.2

∴ Compound interest for 2 years = ₹2247.2 - ₹2000

= ₹247.2

We know that amount after 3 years

Where p = principal, r = rate of interest and n = time = 3 years

From given, p = ₹30000 and r = 9%

So, the amount after 3 years

= ₹38850.87

We know that if the rate of compound interest is r% per annum and the interest is compounded half yearly or if the number of phase of compound interest in a year is 2 then the amount in n years

Where p = principal, r = rate of compound interest and n = time interval

From given, p = ₹80000, r = 5% and n = years years

∴ We get amount for years

= ₹86151.25

Let principal = ₹ x

At the rate of 8% compound interest per the annum the amount for 2 years

∴ Compound interest for 2 years

By the condition,

∴ x = ₹17704.32692

∴ Principal is ₹17704.32692

Let principal = ₹x

At the rate of 10% compound interest per the annum the amount for 3 years

∴ Compound interest for 3 years

By the condition,

∴ x = ₹8000

∴ Principal is ₹8000

Let principal = ₹x

At the rate of 9% compound interest per the annum the amount for 2 years

= ₹1.1881x

∴ Compound interest for 2 years = 1.1881x – x

By the condition, 1.1881x – x = 29702.50

⇒ 0.1881x = 29702.50

∴ x = ₹157908.0276

∴ Principal is ₹157908.0276

Let principal = ₹x

At the rate of 8% compound interest per the annum the amount for 3 years

= ₹1.259x

∴ Compound interest for 3 years =₹ (1.259x – x)

By the condition, 1.259x – x = 31492.80

⇒ 0.259x = 31492.80

∴ x = 121593.82

∴ Principal is ₹121593.82

Let principal = ₹x

∴ Simple interest at 7.5% for 2 years

Compound interest at 7.5% for 2 years

∴ Compound interest for 2 years

By the condition,

∴ x = ₹2133333.33

∴ Principal is ₹2133333.33

Let principal = ₹x

∴ Simple interest at 5% for 3 years

= ₹0.15x

At the rate of 5% compound interest per the annum the amount for 3 years

= ₹1.157x

∴ Compound interest for 3 years = 1.157x – x = 0.157x

By the condition, 0.157x – 0.15x = 10000

⇒ 0.007x = 10000

∴ x = 1428571.42

∴ Principal is ₹1428571.42

Let principal = ₹x

∴ Simple interest at 9% for 2 years

= ₹0.18x

At the rate of 9% compound interest per the annum the amount for 2 years

= ₹1.1881x

∴ Compound interest for 2 years = ₹1.18x – x = 0.1881x

By the condition, 0.1881x – 0.18x = 129.60

⇒ 0.0081x = 129.60

∴ x = ₹16000

∴ Principal is ₹16000

Let principal = ₹x

∴ Simple interest at 10% for 3 years

At the rate of 10% compound interest per the annum the amount for 3 years

∴ Compound interest for 3 years

By the condition,

∴ x = 30000

∴ Principal is ₹30000

We know that if principal = p and rate of compound interest per annum for first and 2^nd years are r₁% and r₂%,

Then we get the amount for 2 years

From given p = ₹6000, r₁ = 7% and r₂ = 8%

∴ Amount for 2 years

= ₹6933.6

∴ Compound interest for 2 years = ₹6933.6 - ₹6000

= ₹933.6

We know that if principal = p and rate of compound interest per annum for first and 2^nd years are r₁% and r₂%,

Then we get the amount for 2 years

From given p = ₹5000, r₁ = 5% and r₂ = 6%

∴ Amount for 2 years

= ₹5565

∴ Compound interest for 2 years = ₹5565 - ₹5000

= ₹565

Let the principal = ₹p and the rate of interest per annum = r%

∴ The interest for 1 year at the rate of simple interest r% per annum

Again the amount at the rate of compound interest r% per annum for 2 years

∴ Compound interest for 2 years

By the condition, … (1)

… (2)

Dividing equation (2) by (1),

⇒ 1/2 r = 102 – 100

⇒ 1/2 r = 2

⇒ r = 2 × 2

∴ r = 4

From (1),

⇒ p = 50 × 25

∴ p = 1250

∴ Principal is ₹1250 and rate of interest is 4%.

Let the principal = ₹p and the rate of interest per annum = r%

∴ The interest for 2 years at the rate of simple interest r% per annum

Again the amount at the rate of compound interest r% per annum for 2 years

∴ Compound interest for 2 years

By the condition, … (1)

… (2)

Dividing equation (2) by (1),

⇒ r = 206 – 200

∴ r = 6

From (1),

∴ p = 70000

∴ Principal is ₹70000 and rate of interest is 6%.

We know that if the rate of compound interest is r% per annum and the interest is compounded half yearly, the number of phase of compound interest in a year is 2, then the amount for n years

From given, p = ₹6000, r = 8 and n = 1

∴ Amount for 1 year

= ₹6489.6

∴ Compound interest for 1 year = ₹6489.6 - ₹6000

= ₹489.6

We know that if the rate of compound interest is r% per annum and the interest is compounded quarterly, the number of phase of compound interest in a year is 4, then the amount for n years

From given, p = ₹6250, r = 10 and years

∴ Amount for 9 months or 3/4 year

= ₹6730.56

∴ Compound interest for 9 months = ₹6730.56 - ₹6250

= ₹480.56

Let the rate of compound interest per annum be r% per annum.

We know that the amount at rate of r% compound interest per annum for 2 years

By the condition,

∴ r = 8

∴ Rate of interest per annum is 8%.

Let it be in ‘n’ years ₹40000 in compound interest at the rate of 8% per annum will amount to ₹46656.

∴ Amount for n years

By the condition,

∴ n = 2

∴ In the rate of compound interest if 8% per annum ₹40000 will amount to ₹46656 in 2 years.

Let the rate of compound interest per annum be r% per annum.

We know that the amount at rate of r% compound interest per annum for 2 years

By the condition,

∴ r = 10

∴ Rate of interest per annum is 10%.

Let it be in ‘n’ years ₹50000 in compound interest at the rate of 10% per annum will amount to ₹60500.

∴ Amount for n years

By the condition,

∴ n = 2

∴ In the rate of compound interest if 10% per annum ₹50000 will amount to ₹60500 in 2 years.

Let it be in ‘n’ years ₹300000 in compound interest at the rate of 10% per annum will amount to ₹399300.

∴ Amount for n years

By the condition,

∴ n = 3

∴ In the rate of compound interest if 10% per annum ₹300000 will amount to ₹399300 in 3 years.

We know that if the rate of compound interest is r% per annum and the interest is compounded half yearly, the number of phase of compound interest in a year is 2, then the amount for n years

From given, p = ₹1600, r = 10 and n = year years

∴ Amount for year

= ₹1852.2

∴ Compound interest for year = ₹1852.2 - ₹1600

= ₹252.2

We know that population after n year

Where p = present population, r% = rate of increase in population in one year and n = number of years.

Given p = 10000, r% = 3% and n = 2 years

⇒ Population after 2 years

= 10609

∴ Population after 2 years of village of Pahalanpur will be 10609.

We know that population after n year

Where p = present population, r% = rate of increase in population in one year and n = number of years.

Given p = 80000000, r% = 2% and n = 3 years

⇒ Population after 3 years

= 84896640

∴ Population after 3 years of the state will be 84896640.

We know that after depreciation, the price of machine after n years

Where p = present price of machine, r% = rate of decrease in a year and n = number of years.

Given p = ₹100000, r% = 10% and n = 3 years

⇒ Price of machine after 3 years

= ₹72900

∴ Price of the machine after 3 years is ₹72900.

Let the number of students readmitted 2 years before be x.

∴ At present the number of readmitted students

By the condition,

∴ x = 3200

∴ The number of students readmitted 2 years before was 3200.

Let the number of street accidents 3 years before in the district be x.

∴ At present the number of street accidents in that district

By the condition,

∴ x = 12000

∴ The number of street accidents 3 years before in the district was 12000.

We know that population after n year

Where p = present population, r% = rate of increase in population in one year and n = number of years.

Given p = 406 quintals = 40600 kg [∵ 1 quintal = 100 kg], r% = 10% and n = 3 years

⇒ Production of fishes after 3 years

= 54038.6 kg

∴ Production of fishes after 3 years will be 54038.6 kg or 504.386 quintals.

Let the height of the tree 2 years before be x.

∴ At present the height of the tree

By the condition,

∴ x = 16.66 metre

∴ The height of the tree 2 years before was 16.6 metre.

The amount the family had to spend for electric bill 3 years before was ₹4000.

Let amount that the family have to spend to pay the electric bill in the present year be x.

∴ At present the amount that has to be paid

By the condition,

∴ x = ₹3429.5

∴ The amount that the family have to spend to pay the electric bill in the present year is ₹3429.5.

We know that after reduction, the weight after n years

Where p = present weight, r% = rate of decrease in a year and n = number of years.

Given p = 80 kg, r% = 10% and n = 3 years

⇒ Weight after 3 years

= 58.32 kg

∴ Savan Babu’s weight after 3 years is 58.32 kg.

Let the sum of number of students in all M.S.K 3 years before be x.

∴ At present the sum of number of students

By the condition,

∴ x = 3000

∴ The sum of number of students 3 years before in all the M.S.K in the district was 3000.

The number of farmers using fertilizers and insecticides 3 years before was 3000.

Let present number such farmers in the village be x.

∴ At present the number of such farmers

By the condition,

∴ x = 1536

∴ The number of such farmers in that village in the present year is 1536.

We know that after depreciation, the price of machine after n years

Where p = present price of machine, r% = rate of decrease in a year and n = number of years.

Given p = ₹18000, r% = 10% and n = 3 years

⇒ Price of machine after 3 years

= ₹13122

∴ Price of the machine after 3 years is ₹13122.

We know that after reduction, the population after n years

Where p = present population, r% = rate of decrease one year and n = number of years.

Given p = 1200 families r% = 75% and n = 2 years

⇒ Number of families without electricity after 2 years

= 75

∴ The number of families without electricity after 2 years is 75.

The number of users of cold drink in the town 3 years before was 80000.

Let present number of users of cold drink be x.

∴ At present the number of users of cold drink

By the condition,

∴ x = 33750

∴ The number of users of cold drink in the present year is 33750.

Let the number of smokers in the city 3 years before be x.

∴ At present the number of smokers in the city

By the condition,

∴ x = 41289.43

∴ The number of smokers in the city before 3 years was 41289.43.

In case of compound interest, the time period and the rate of compound interest are constant. So the rate of compound interest per annum is equal.

In compound interest, after a definite period of time, the interest is accrued to the principal to get the new principal.

So, the principal changes every year.

We know that population after n year

Where p = present population, r% = rate of increase in population in one year and n = number of years.

Given r% = 2r%

⇒ Population after n years

We know that after depreciation, the price of machine after n years

Where p = present price of machine, r% = rate of decrease in a year and n = number of years.

Given p = ₹2p and r% = 2r%

⇒ Price of machine after n years

Let the rate of compound interest per annum be r% per annum.

We know that the amount at rate of r% compound interest per annum for 2 years

By the condition,

∴ r = 10

∴ Rate of interest per annum is 10%.

(i) False

The compound interest is always higher than simple interest.

(ii) True

The principal in compound interest changes each time interest is compounded as the interest earned is added to the principal amount.

(i) The compound interest and simple interest for one year at the fixed rate of interest on fixed sum of money are unequal.

In simple interest, the interest is calculated on principal only.

While in compound interest, the interest accrued is added to the principal to get the new principal.

(ii) If some things are increased by fixed rate with respect to time, that is uniform rate of growth.

Examples are increase in population, increase in the number of learners, etc.

(iii) If some things are decreased by fixed rate with respect to time this is uniform rate of decreases or depreciation.

Examples are fall of efficiency of machine, decrease valuations of old building and furniture, etc.

Let the rate of compound interest per annum be r% per annum.

We know that the amount at rate of r% compound interest per annum for 2 years

By the condition,

∴ r = 5

∴ Rate of interest per annum is 5%.

Let A = amount, P = principal, r = rate of interest compounded per annum and n = number of years.

Let it be in ‘n’ years A in compound interest will amount to 2A.

∴ Amount for n years = A (1 + r)ⁿ

By the condition, A (1 + r)ⁿ = 2A

⇒ (1 + r)ⁿ = 2A/ A

⇒ (1 + r)ⁿ = 2

⇒ (1 + r) = 2^1/n … (1)

Let the amount become 4 times in t years.

⇒ 4A = A(1 + r)^t

⇒ (1 + r)^t = 4

⇒ (1 + r)^t = 2²

⇒ (1 + r) = 2^2/t… (2)

From (1) and (2),

⇒ 2^1/n = 2^2/t

∴ t = 2n

∴ The sum of money will become four times in 2n years.

Let principal = ₹x

At the rate of 5% compound interest per the annum the amount for 2 years

∴ Compound interest for 2 years

By the condition,

∴ x = ₹6000

∴ Principal is ₹6000

Let the Price of Machine at present = Rs x

Price of machine depreciates at the rate of r%

Therefore, Price of machine before one year was =

Price of machine before n years =

Given : The present population = p

Let previous year’s population be x

Then,

As the population increases by rate r%

And population before n years was