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Given:

Amount in one year = 650

Amount in two years = 676

Concept used:

Amount = P(1 + R/100)^T

Calculation:

Let the principal be x.

For one year,

⇒ x(1 + R/100)¹ = 650      ----(1)

For two years,

⇒ x(1 + R/100)² = 676      ----(2)

Dividing equation (2) and (1),

[x(1 + R/100)²]/[x(1 + R/100)¹] = 676/650

⇒ (1 + R/100) = 26/25

From equation (1),

⇒ x(26/25) = 650

⇒ x = (650 × 25)/26

⇒ x = 25 × 25

⇒ x = 625

∴ The principle is Rs. 625.

Given:

A= Rs. 1155

R = 5% and 10%

N = 2 years

Formula used:

I = PRN / 100

Where P = Principal amount, R = Rate of interest in %, N = Number of years, I = Interest earned

A = P + I    

Where A = Amount received

Calculation:

A = P + I

⇒ A = P + (PRN / 100)

⇒ A = P (1 + RN / 100)

⇒ A = P (1 + (5 × 1 / 100)) × (1 + (10 × 1 / 100)) [∵ Interest was 5% and 10% in 1^st and 2^nd year]

⇒ 1155 = P × (21 / 20) × (11 / 10)

⇒ P = (1155 × 20 × 10) / (21 × 11)

⇒ P = 5 × 20 × 10

⇒ P = Rs. 1000

∴ Principal amount invested Rs. 1000.

Short Trick

If we take 100 as the principle.

After the first year the principle increased to 105.

And after second year, the amount increased to 115.5.

If he received 1155 as amount.

Then, 1 unit is 10

Then, 100 unit is 1000.

Given:

Amount in 4 years = Rs 1237.50

Amount in 6 years = Rs 1443.75

Formula used:

SI = (P × R × T)/100

A = P + SI

Here, A, P, R, T and SI are amount, principal, rate, time and simple interest respectively

Concept used:

SI is equal every year

Calculation:

The amount in 4 years = Rs 1237.50

The amount in 6 years = Rs 1443.75

SI for 2 years = 1443.75 – 1237.50 = 825/4

Then, SI for 4 years = 825/4 × 2 = 825/2

Principal = Amount for 4 years – SI for 4 years 

⇒ 1237.5 – 825/2

⇒ 825

∴ The principal is Rs 825

Given:

Interest on the second part of money = Rs. 6,000

Rate of interest is 10% and 20%

Formula used:

S.I = (P × R × T)/100

Here, S.I, P, R and T is simple interest, Principle, Rate and Time respectively

Calculation:

Let the investment in two banks be 2x and 3x

S.I = (3x × 20 × 1)/100

⇒ 6,000 = 3x/5

⇒ x = 10,000

Now, Investment of A = 20,000

S.I = (20,000 × 10 × 1)/100 = 2,000

∴ The interest in the first part is Rs. 2,000

Given:

Amount = Rs. 900

Time = 4 years

Rate of interest = 5%

Formula used:

Amount = Principal + Interest

S.I = PRT/100

Calculation:

S.I = (P × 4 × 5)/100

⇒ P/5

Now, according to question

⇒ P + (P/5) = Rs. 900

⇒ 6P/5 = Rs. 900

⇒ P = Rs. 150 × 5

⇒ Rs. 750

∴ The sum is Rs. 750

GIVEN:

For the first 2 years, r = 4% per annum, t = 2 years

next 4 years, r = 6% per annum, t = 4 years

next 3 years, r = 8% per annum, t = 3 years

 FORMULAE USED:

SI = (p × r × t)/100

CALCULATION:

 the simple interest for the first 2 years = (p × 4 × 2)/100 = 2p/25

the simple interest for the next 4 years = (p × 6 × 4)/100 = 6p/25

And then, for the 

 the simple interest for the next 3 years = (p × 8 × 3)/100 = 6p/25

⇒ Total interest = (2p/25) + (6p/25) + (6p/25) = 14p/25

So, 14p/25 = 4480

⇒ p = Rs. 8000

Hence, the principal amount is Rs. 8000

Given:

Principal = 48400

Rate (R) = 4% pa

Time (T) = 6 yr

Calculations:

In case 1 :

SI = (P × R × T) / 100

SI = (48400 × 4 × 6) / 100

SI = 11616

Amount = Principal + SI

⇒ Amount = 48400 + 11616

⇒ Amount = 60016

In case 2:

SI = (60016 × 4 × 6) /100

⇒ SI = 14403.84

∴ The Simple interest after the last 6 years is Rs. 14403.84.

Given:

rate of interest per annum for the first two years is 4%.

The rate of interest for next four years is 6% and for the next three years is 8%.

Formula used:

simple interest = (P × R × T)/100

Calculation:

Total simple interest for 9 years = 1120

⇒ P × [(2 × 4) + (4 × 6) + (3 × 8)]\100 = 1120

⇒ P × 56/100 = 1120

⇒ P = 2000 Rs.

∴ The sum is 2000 Rs.

Given:

Principal (P₁) = 5000, Rate (R₁) = 12%, Time (T) = 4 years

Principal (P₂) = x, Rate (R₂) = 10%, Time (T) = 4 years

Formula Used:

Simple Interest (I) = (P × R × T)/100

Where, 

P → Principal

R → Rate 

T → Time

Calculations:

Simple interest gained from 5000 = (5000 × 12 × 4)/100 = 2400

Let the other Principal be x.

S.I. gained = (4800 – 2400) = 2400

⇒ (x × 10 × 4)/100 = 2400

⇒ x = Rs. 6000

∴ The correct answer is Rs. 6000. 

Given:

P = Rs. 800 each

N = 3 years

I₂ – I₁ = Rs. 48

Formula used:

I = PRN / 100

Where P = Principal amount, R = Rate of interest in %, N = Number of years, I = Interest earned

A = P + I    

Where A = Amount received

Calculation:

I₂ – I₁ = Rs. 48

⇒ (PR₂N/100) – (PR₁N / 100) = 48

⇒ PN(R₂ – R₁) / 100 = 48

⇒ 800 × 3 (R2 – R1) / 100 = 48

⇒ R₂ – R₁ = (48 × 100) / (800 × 3)

⇒ R₂ – R₁ = 16 × 100 / 800

⇒ R₂ – R₁ = 2%

Given:

The rate of interest on certain sum of money at 10% for 5 years and 6 years is differing by Rs. 75

Formula used:

Simple interest = (P × R × T)/100

Amount = Principal + Interest

Calculation:

Interest for one year is Rs. 75 and we know that the simple interest is calculated on initial sum for each year

75 = (P × 10 × 1)/100

⇒ P = Rs. 750

∴ The sum is Rs. 750.

Given:

Principal = Rs. 1800

Time = 5 years

Rate of interest = 8%

Formula used:

Simple Interest = (Principal × time × rate)/100

Calculation:

Simple Interest earned

⇒ (1800 × 5 × 8)/100

⇒ 720

∴ The per annum interest is Rs. 720. 

Given:

simple interest= Rs.1,400

time = 4 years

Formula Used:

SI = P × T × R /100

Calculation:

Let the Rate = x

We know that,

SI = P × T × R /100

⇒ 672 = 1400 × 4 × x /100 + 1400 × 2 × x /100

⇒ 672 = 56x + 28x 

⇒ 672 = 84x

⇒ x = 672/84

⇒ x = 8

∴ Rate = 8%

Given:

Amount = 3/2 of the sum

Time = 3 years

Formula used:

Simple interest = (P × R × T)/100

Amount = Principal + Interest

Calculation:

Let the sum of money be P

Amount after 3 years will be 3/2P

Now, by using the formula

3/2P – P = (PR × 3)/100

⇒ R = 50/3 = 16\(\frac{2}{3}\)%

∴ The rate of interest is 16\(\frac{2}{3}\)%. 

Given:

R = 30%

Formula used:

Simple interest = PRT/100

Amount = Principal + Simple interest

Calculation:

Let principal be X Rs. and time be taken T years to become it 16X.

⇒ P = X Rs. and A = 16X Rs.

⇒ S.I. = 16X - X = 15X

⇒ 15X = PRT/100

⇒ 15X = (X × 30 × T)/100

⇒ 100 = 2T

⇒ T = 50 years.

∴ In 50 years a sum of money will become sixteen times.

Given:

Time = 6 years

Formula used:

SI = (P x R x T)/100,

Amount = SI + P

where,

P → Principal,

R → Rate of Interest,

T → Time Period of the Loan/deposit in years,

S.I. → Simple Interest

Calculation:

Let, the P = x, A = 2x

⇒ SI = A - P

⇒ SI = 2x - x = x

⇒ x = (x x R x 6)/100

⇒ R = 50/3%

In 10 times Let, P = x, A = 10x

⇒ SI = 10x - x = 9x

⇒ SI = 9x

⇒  9x = (x x 50 x T)/100 × 3

⇒ T = 54 years

∴ It will take 54 years to become 10 times of itself.

Given:

A sum of money becomes 3 times itself in 5 years.

Concept used:

Simple interest, SI = P × R × T/100

Where P is principal, R is the rate of interest and T is the time period.

Amount, A = P + SI

Calculation:

Amount, 3P = P + P ×  R × 5/100

2P = 5PR/100

Or, 40%

Amount = 5P = P + 40% × P × T/100

4P = 4PT/10

Or, T = 10 years 

∴ Time taken for amount to become 5 times itself is 10 years.

Given:

Principle = 19000

Time = 2 years

Rate = 40%

Condition = Semi annually 

Concept used:

The interest compounded is Semi-annually ( half yearly )

New rate = Rate/2 

New time = Time × 2

Formula used:

C.P = P{(1 + R/100)^T – 1}

Where, 

C.P → Compound interest 

P → Principal 

R → Rate 

T → Time 

Calculations:

New rate = 40/2 = 20%

New time = 2 × 2 = 4

C.P = P{(1 + R/100)T – 1}

⇒ 19000{(1 + 20/100)⁴ –  1}

⇒ 19000{(1 + 1/5)⁴ – 1}

⇒ 19000 {(6/5)⁴ – 1}

⇒ 19000 {(6/5 × 6/5 × 6/5 × 6/5) – 1}

⇒ 19000{(1296/625) – 1}

⇒ 19000(1296 – 625)/625

⇒ 19000 × (671/625)

⇒ 30.4 × 671

⇒ Rs. 20,398.4

∴ The compound interest is Rs. 20,398.4

Given: 

A certain sum of money becomes 25 times of itself in a span of 50 years on simple interest.

Formula:

r = (100 × SI)/(P × t)

SI = A – P

Where, 

SI → Simple interest 

A → Amount 

P → Principal 

r → rate 

t → time

Calculation:

Let the sum of money be P

Amount becomes 25 times = 25P

SI = 25P – P

⇒ 24P

r = (100 × 24P)/(P × 50)

⇒ 2400P/50P

⇒ 2400/50

⇒ 48

∴ The rate of interest is 48%

Given:

At a certain rate of interest, a sum of 3600 would fetches an interest 43.56 less when put under simple interest for 2 years than when put under compound interest for 2 years

Formula:

S.I = (principal × rate × time)/100

Interest under compound interest = P[1 + (R/100)]^T – P

Calculation:

According to the question,

⇒ [3600 × {1 + (R/100)}² – 3600] – [(3600 × R × 2)/100] = 43.56

⇒ 3600 (1 + (R/100)}² – 72R = 43.56 + 3600

⇒ R = 11%

Hence, rate is 11%

Given:

A certain sum of money at compound interest turns 2 times itself in 9 years. 

Concept used:

A = P(1 + R/100)T

Where,

A → Amount

P → Principal 

R → Rate of interest 

T → Time

Calculations:

A = P(1 + R/100)T 

According to the question

⇒ 2P = P(1 + R/100)⁹ 

⇒ 2 = (1 + R/100)⁹      ----(1)

⇒ 32P = P(1 +  R/100)^T 

⇒ 32 = (1 + R/100)^T

⇒ 2⁵ = (1 + R/100)^T 

From eq. (1)

⇒ {(1 + R/100)⁹}⁵ = (1 + R/100)^T

⇒ (1 + R/100)⁴⁵ = (1 + R/100)^T

⇒ T = 45 years.

∴ The required time is 45 years 

Given:

An amount under simple interest fetches an interest 6 times of itself in 40 years.

The same rate of interest is applied on an amount of Rs. 71890 for two years under compound interest.

Formula Used:

Amount after n years under compound interest = p × (1 + r/100)ⁿ

Compound Interest = p × [1 + (r/100)]ⁿ – p [Where p is Principal, r is interest rate, n is time period]

Calculation:

Let the sum be x.

And Interest = 6x

S I = (p × r × t)/100

⇒ 6x = (x × r × 40)/100

⇒ r = 15

For second case,

Interest = Rs. [71890 × (1 + 15/100)² – 71890] = Rs. [95074.525 – 71890] = Rs. 23184.525

∴ The interest is Rs. 23184.525

Given:

Principal = Rs. P

Rate = 4r %

Time = 2 years

Formula used:

A = P [1 + (r/100)]^t

Where A, P, t and r is amount, principal, time and rate respectively

Calculation:

A = P [(1 + 4r/100]²

⇒ P [(1 + r/25)]²

∴ The amount received is P [(1 + r/25)]2

Given:

Rs. 12000 becomes double of itself in 5 years when it is compounded.

Formula used:

Amount(A) = P[1 + (R/100)]^T

P = Principal

R = Rate of interest

T = Time

Calculation:

P = Rs. 12000

According to the question:

2P = P[1 + (R/100)]⁵

⇒ 2 = [1 + (R/100)]⁵

If it is deposited for 20 years, then amount will be given by:

A = P[1 + (R/100)]²⁰

⇒ A = 12000[1 + (R/100)]^{5 × 4}

⇒ A = 12000(2)⁴

⇒ A = 12000 × 16 = 192000

∴ The amount will become Rs. 192000 in 20 years

Given:

A sum becomes Rs. 33868.8 after 2 years under compound interest.

The same amount at 10% compounded annually becomes 35937 after 3 years.

Formula Used:

In compound interest amount = P (1 + r /100)ⁿ [where principal be P and rate be r]

Calculation:

35937 = P (1 + 10/100)² [Where principal be P]

⇒ P = 27000

⇒ 27000 (1 + r /100)² = 33868.8 [Where rate be r]

⇒ R = 12%

∴ The initial rate is 12%

Given:

The rate of interest = 10% per annum

Formula used:

\(CI\ =\ P\ × [{(1\ +\ {R\over 100})^T\ -\ 1}]\)     (Where CI = Compound interest, P = The Principle, R = The rate, and T = Time)

Calculation:

Let us assume the sum is X

⇒ \(640\ =\ X\ × [{(1\ +\ {10\over 100})^1\ -\ 1}]\)

⇒ \(640\ =\ X\ × [{( {11\over 10}) -\ 1}]\)

⇒ 640 = X/10

⇒ X = 6400

⇒ The compounded interest on the same sum in 2 years on the same rate = \(\ 6400\ × [{(1\ +\ {10\over 100})^2\ -\ 1}]\)

⇒ \(\ 6400\ × [{( {121\over 100}) -\ 1}]\)

⇒ The compound interest in two years = 6400 × \({21\over 100}\) = 1344

⇒ The compounded interest on only second-year = (The interest in two years) - (The interest in the first year) = 1344 - 640 = 704

∴ The required result will be 704.

Given:

The principle = 4000 and Time = 2 years

Formula used:

\(CA\ =\ P\ \times {(1\ +\ {R\ \over 100})^T}\)

(Where CA = Compounded amount, P = The principle, R = The rate of interest and T = Time)

Calculation:

Let us assume the rate of interest be R

⇒ \(5760\ =\ 4000\ \times {(1\ +\ {R\over 100})^2}\)

⇒ \({5760\over4000} =\ {(1\ +\ {R\over 100})^2} \)

⇒ \(1.44 =\ {(1\ +\ {R\over 100})^2}\)

⇒ \(1.2\ =\ {1\ +\ {R\over 100}}\)

⇒ R = 20%

∴ The required result will be 20%.

GIVEN:

principal = Rs. 84500

interest rate = 8% per annum

time = 1 year 6 months

compounded half-yearly

CONCEPT:
for half-yearly, the time period will be doubled and the rate will be halved 

FORMULA USED:
\(A=P(1+\frac{R}{100})^t\)

CALCULATION:

R = 4%

Time period (t) = 3 years

\(A =84500(1+\frac{4}{100})^3 \)

\(\\A=84500(1.04)^3\)

\(\\A\approx95051\)

\( \\interest =95051 - 84500=Rs10551\)

ALTERNATE SOLUTION:

R = 4%

T = 3 years

\(4\% =\frac{1}{25} \)

\(\\\therefore P = 25\space ratio , A= 26\space ratio\)

\(\\for\space3years\)

\(\\\Rightarrow\space P=(25)^3=15625 \space ratio\)

\( \\A=(26)^3 = 17576\space ratio\)

\(\\interest =17576-15625=1951\space ratio \)

\(\\15625\space ratio=Rs. 84500 \)

\(\\\therefore 1951\space ratio \approx Rs. 10551\)

Given:

P = Rs. 13000

r = 20%

t = 6 months

Formula:

If interest compounded quarterly, then

r = 20/4 = 5%

t = 6/12 × 4 = 2 years

Calculation: 

CI = P [(1 + r/100)t – 1]

⇒ CI = 13000[(1 + 5/100)2 – 1]

⇒ CI = 13000 × [105/100 × 105/100 – 1]

⇒ CI = 13000 × (11025 – 10000)/10000

⇒ CI = 13 × 1025/10

⇒ CI = Rs. 1332.5

The maturity amount = 13000 +  1332.5 = Rs. 14332.5

Given:

Principal = Rs. 13,000

Time = 1 year

Rate of interest = 2%

Formula used:

S.I = PRT/100

Calculation:

S.I = (Rs. 13000 × 1 × 2)/100

⇒ Rs. 260

Amount = Principal + S.I

⇒ Rs. 13,000 + Rs. 260

⇒ Rs. 13,260

∴ The required amount is Rs. 13,260

Given:

Principal = Rs. 450

Time = 2.5 years

Rate of interest = 3%

Formula used:

S.I = PRT/100

Calculation:

S.I = Rs. (450 × 2.5 × 3)/100

⇒ Rs. 33.75

∴ The required S.I is Rs. 33.75

Given:

A shopkeeper takes a loan (Principal) = Rs. 54000

Time = 6 years

Rate = 6%

Concept used:

Simple interest = (Principal × Rate × Time)/100

Calculation:

Simple interest = (Principal × Rate × Time)/100

⇒ (54,000 × 6 × 6)/100

⇒ 540 × 36

⇒ Rs. 19,440

∴ The simple interest is Rs. 19,440

Given

Amount = Rs. 416

Time = 1 year 4 months

Rate of interest = 3%

Formula used

SI = (P × R × T)/100

Amount(A) = P + SI

SI → Simple interest, P → Principal, R → Rate, T → Time

Calculation

Time = 1 year 4 months = 4/3 years

A = P + (P × R × T)/100

⇒ 416 = P + (P × 3 × 4/3)/100

⇒ 41600 = 100P + 4P

⇒ 104 P = 41600

⇒ P = Rs. 400

Given:

Amita borrowed ( Principal ) = 7500

Rate = 8%

Time = 3 years

She lent it to Anju,

Rate = 12%

Time = 3 years

Concept used:

Simple interest = (Principal × Rate × Time)/100

Calculations:

Simple interest in 3 years

⇒ (7500 × 12 × 3)/100 – (7500 × 8 × 3)/100

⇒ 2700 – 1800 = 900

Profit in 3 years = Rs. 900

∴ The profit is Rs. 900

Given:

Amount after 3 years is Rs 50400.

Amount after 6 years is Rs 241920.

Formula Used:

Amount = \(P × {\left( {1 + \frac{R}{{100}}} \right)^n}\)

CI = Amount - Principal = \(P × {\left( {1 + \frac{R}{{100}}} \right)^n}\) - P = \(P × \left\{ {{{\left( {1 + \frac{R}{{100}}} \right)}^n} - 1} \right\}\)

Calculation:

Let the sum be Rs x.

The rate of interest be R%.

According to the question,

Amount for 3 years = 50400

⇒ \(x × {\left( {1 + \frac{R}{{100}}} \right)^3}\) = 50400 ....(I)

Amount for 6 years = 241920

⇒ \(x × {\left( {1 + \frac{R}{{100}}} \right)^6}\) = 241920 ....(II)

On dividing (II) ÷  (I)

⇒ \({\left( {1 + \frac{R}{{100}}} \right)^3}\) = 241920/50400

⇒ \({\left( {1 + \frac{R}{{100}}} \right)^3}\) = 4.8 ....(III)

On putting the value of equation (III) in(I)

⇒ x × 4.8 = 50400

⇒ x = 10500

Therefore, the certain amount of sum is Rs 10500.

Given:

Time taken for the sum of money to become 4 times its present value = 10 years

Formula Used:

When interest is compounded annually, the amount received is obtained as:

Amount = P × [1 + (R/100)]ⁿ

where P = Principal,

R = Rate of interest for compounding principal annually,

n = time period (in years)

Calculation:

∵ The sum becomes 4 times its initial value after 10 years,

We get the equation for amount as follows:

4P = P × [1 + (R/100)]¹⁰

⇒ 4 = [1 + (R/100)]¹⁰      ----(i)

Let's assume that the time taken by the sum to become 16 times its value = x years

Now, we get the equation for amount as follows:

16P = P × [1 + (R/100)]^x

⇒ 16 = [1 + (R/100)]^x

⇒ 4² = [1 + (R/100)]^x      ----(ii)

From equation (i), we get:

⇒ 42 = {[1 + (R/100)]¹⁰}²      ----(iii)

On combining eqautions (ii) and (iii), we get:

[1 + (R/100)]x = {[1 + (R/100)]10}2

⇒ x = 10 × 2 

⇒ x = 20

∴ The sum will become 16 times its initial amount after a period of 20 years.

Given:

Rate of interest = 17%

Time = 2 years

The difference between CI and SI = Rs. 433.50

Formula used:

Simple interest = P × R% × T

Calculation:

Let the principal be Rs. x.

SI = P × R% × T

⇒ SI = x × 17% × 2

⇒ SI = x × 17/100 × 2

⇒ SI = 34x/100

⇒ SI = 0.34x

In the case of CI:

1st year CI = P × R% × T

⇒ 1st year CI = x × 17% × 1

⇒ 1st year CI = 17x/100

⇒ 1st year CI = 0.17x

Amount after 1st year = P + CI 

⇒ Amount = x + 0.17x

⇒ Amount = 1.17x

2nd year CI = 1.17x × 17% × 1

⇒ 2nd year CI = 0.1989x

Total CI = 1st year CI + 2nd year CI

⇒ CI = 0.17x + 0.1989x

⇒ CI = 0.3689x

Required difference = CI – SI

⇒ 433.50 = 0.3689x – 0.34x

⇒ 433.50 = 0.0289x

⇒ x = 433.50/0.0289

⇒ x = Rs. 15000

∴ The sum is Rs. 15000.

Given:

Ram borrow Rs.50,000 at 5% annually compound interest

At the end of each year he pay back Rs.2100

Clear debit in 3 years

Concept:

Compound interest is interest on interest.

A = P × (1 + R/100)^T

A = Final amount

P = Initial amount

R = Rate of interest

T = Time

Calculation:

At end of 1 year amount = 50000 × (1 + 5/100)

⇒ 50000 × (105/100)

⇒ Rs.52,500

After payback Rs.2,100 remaining amount for pay at the end of 1 year = 52500 – 2100

⇒ Rs.50,400

At end of 2 year amount = 50400 × (1 + 5/100)

⇒ 50400 × (105/100)

⇒ Rs.52,920

After payback Rs.2,100 remaining amount for pay at the end of 2 year = 52920 – 2100

⇒ Rs.50,820

At end of 3 year amount = 50820 × (1 + 5/100)

⇒ 50820 × (105/100)

⇒ Rs.53,361

∴ Rs.53,361 should he pay at the end of the third year to clear his debit.

Given:

Compound interest: P = 10000, R = 5%, n = 2

Simple Interest: R = 10%, n = 2

Formula used:

Simple interest: (P × R × T) / 100

Compound interest: Amount = P (1 + (R/100))ⁿ

Where, n = Number of years, P = Principal Amount, R = Rate of Interest, A = Amount

Calculations:

Find interest from amount compounded

⇒ A = 10000 × (1 + (5 / 100)²

⇒ A = 10000 × (1 + (1 / 20))²

⇒ A = 10000 × 21 × 21 / 400

⇒ A = 25 × 21²

⇒ A = 11,025

⇒ I = 11025 – 10,000

⇒ I = 1025

Now, interest earned by simple interest

⇒ I = 1025 – 350

⇒ I = 675

⇒ 675 = P × 10 × 2 / 100

⇒ 67500 / 20 = P

⇒ P = 3375

Given:

I = (1/16) × P

R = T

Formula used:

S.I. = P × R × T/100

Calculation:

According to given

⇒ S.I. = P × R × R/100

⇒ 1/16 = R²/100

⇒ R² = 100/16

⇒ R = 10/4

⇒ R = 2.5%

Given:

Amount becomes double in 12 years

Formula Used:

Simple Interest = (P × r × t)/100

Amount = P + Simple Interest

Where P → Principal

r → Rate of interest

t → Time

Calculation:

Let principal be 100

Amount = P + Simple Interest

⇒ 2 × 100 = 100 + Simple Interest

⇒ Simple Interest = 100

Simple Interest = (P × r × t)/100

⇒ 100 = (100 × r × 12)/100

⇒ r = 100/12

⇒ r = 25/3

\( \Rightarrow {\rm{r}} = 8\frac{1}{3}\;\% \)

∴ The rate of interest is \(8\frac{1}{3}\;\% .\)

Concept used:

R = (I × 100)/(P × T)

A = P + I

Where R = rate of interest, P = principal, I = interest

T = time and A = Amount

Calculation:

Let P = 100 then A = 300

I = 300 - 100 = 200

R = (200 × 100)/(100 × 12.5)

⇒ R = 200/12.5

∴ Rate of interest is 16%

Given:

Principle = 10950 Rs.

Rate = 8%

time = 220 days

Formula used:

Amount =  P + SI

SI = (P x R x T)/100,
Where P = Principal, R = Rate of Interest, T = Time Period of the Loan/deposit in years, SI = Simple Interest

Calculations:

SI = (10950 × 8 × 220)/(365 × 100)

⇒ SI = 528 Rs.

⇒ Amount = 528 + 10950

⇒ Amount = 11478 Rs.

∴ The amount is Rs.11478 

Given:

Amount obtained at a sum for 4 years at r % = SI on the same sum at r % for 14 years

Formula used ;

Amount = Principal + SI

\(SI = {P\times r\times t \over 100}\)

Calculation :

Let the principal be P

According to the question

The amount for 4 years at r % = SI for 14 years at r %

P + SI for 4 years at r % = SI for 14 years at r %

⇒ P + \({P\times r\times t \over 100}\) = \({P\times r\times T \over 100}\)

P + \({P\times r\times 4 \over 100}\) = \({P\times r\times 14 \over 100}\)

⇒ P = (14Pr/100) - (4Pr/100)

⇒ P = Pr/10

⇒ r = 10

∴ The value of 'r' is 10 %

Concept used:

R = (I × 100)/(P × T)

A = P + I

Where R = rate of interest, I = interest, P = principal

T = time and A = amount

Calculation:

25/16 times that means if P is 16 than A is 25.

P = 16, A = 25

I = 25 -16 = 9

r = (9 × 100)/(16 × 40)

⇒ r = 90/64 = 45/32 = 1.4% approx