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Given

The sum invested at 8% p.a. amounts to Rs. 20,280 at the end of one year, when the interest is compounded half-yearly.

Formula used

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

A = Amount

P = Principal

R = Rate

T = Time

Calculation

Let Principal be Rs.x

⇒ Time is compounded half yearly so, rate is 8/2 = 4%

⇒ x(1 + 4/100)² = 20280

⇒ x(104/100)² = 20280

⇒ x(26/25)² = 20280

⇒ x = (20280 × 25 × 25)/(26 × 26)

⇒ x = Rs.18750

⇒ Amount = 18750(1 + 8/100)²

⇒ A = (18750 × 27 × 27)/(25 × 25)

⇒ Amount = Rs.21870

⇒ CI = 21870 - 18750

∴ Compound interest for 2 years at the same rate is Rs.3120

Given:

Principal (P) = Rs. 15625

Time (T) = 3 years

Amount (A) = Rs. 21,952

Concept used:

Amount = P(1 + R/100)T

Calculation:

Let the rate of interest per annum be R.

21952 = 15625(1 + R/100)³

⇒ 21952/15625 = (1 + R/100)3

⇒ ∛(21952/15625) = 1 + R/100

⇒ 28/25 = 1 + R/100

⇒ 28/25 – 1 = R/100

⇒ (28 – 25)/25 = R/100

⇒ 3/25 = R/100

⇒ R = (3/25) × 100

⇒ R = 3 × 4

⇒ R = 12%

∴ The rate per cent per annum is 12%.

Given:

The sum = Rs 12,000

Time = \(1 \frac{1}{2}\) years

Rate = 10% p.a.

Formula used:

A = P(1 + R/100)^t

Here, A, P, R and t are the Amount, Principal, Rate and time respectively

Concept used:

When compounded half-yearly then,

Rate is half and time is doubled

Calculation:

Rate = 10%/2 = 5% and Time = \(1 \frac{1}{2}\) × 2 = 3 half yearly

Now, A = P(1 + R/100)t

⇒ A = 12000(1 + 5/100)³

⇒ A = 12000 × 21/20 × 21/20 × 21/20

⇒ A = 13891.5

∴ The total amounts to be paid is Rs 13891.50

Given:

A sum = Rs. 12,000

Rate = 10% per annum

And time = 1.5 years.

Formula used:

Amount = Principal{1 + (R/100)}ⁿ

Where n = time in years and R = rate percentage/annum

Compund interest = Amount - Principal

If time = n years, rate = R% and interest compounded half-yearly

Then time = 2n  and rate = (R/2)%

Calculation:

Interest compounded half-yearly:

Then time = 2 × 1.5

⇒ 3 years

And the rate = 10/2

⇒ 5% 

Amount = 12000 × {1 + (5/100)}³

⇒ 12000 × {1 + (1/20)}³

⇒ 12000 × {(20 + 1)/20}³

⇒ 12000 × (21/20)³

⇒ 12000 × (9261/8000)

⇒ 12 × 9261/8

⇒ 1,11,132/8

⇒ Rs. 13891.5

Compound interest = amount - principal

⇒ 13891.5 - 12000

⇒ Rs. 1891.5

∴ The compound interest is Rs. 1891.5

Given:

A worker borrowed some money on simple interest at the rate of 6% p. a. for the first 3 years, at the rate of 9% p. a. for the next 3 years, and at the rate of 14% p. a. for the period beyond 6 years. If he paid a total interest of Rs.6,490 at the end of 7 years.

Concept used:

Simple interest 

Calculation:

Let the principal be x = 100

Si at the rate of 6% p. a. for 3 years 

⇒ 6 × 3 × x = 18x

Si at the rate of 9% p. a. for 3 years

⇒ 9 × 3 × x = 27x

Si at the rate of 14% p. a. for 1 year

⇒ 14 × x = 14x

As per the question,

⇒ 18x + 27x + 14x = 6490

⇒ 59x = 6490

⇒ x = 110

∴ Principal = Rs.11000

Given:

Principle is 25000

Rate is 4%

Time is 1 year and CI is calculated Half Yearly

Formula Used:

Compound Interest (CI) = P × (1 + R/100)ⁿ

Calculation: 

CI is calculated Half Yearly so Rate = 4/2 = 2 %

⇒ Time becomes 2 years

⇒ CI in 2 years = 25000 × (1 + 2/100)² = 26010

Given:

Principal, p = Rs. 50,000

Time = 1 years

Compounded half yearly, n = 2

Rate, r = 10/2 = 5%

Formula used:

Amount = Principal × [1 + r/100]ⁿ 

Amount = Principal + Interest

Calculation:

Amount = 50,000 × [1 + 5/100]²

⇒ Amount = 50000 × 21/20 × 21/20 

⇒ Amount = 125 × 21 × 21

⇒ Amount = Rs. 55,125

Amount = Principal + Interest

⇒ 55,125 = 50,000 + Interest

⇒ Amount of interest = Rs. 5,125

∴ The amount of interest is Rs. 5,125.

Given:

Principal(P) = Rs. 6400

Rate of Interest(r)  = 24%

Time =  1.5 years compounded half-yearly.

Formula used:

If the interest is compounded half yearly, the rate of interest will be halved and time will be doubled.

CI = P [1 + (R/100)]ⁿ - P

Calculation:

CI = 6400 [1 + (12/100)]³ - 6400

⇒ 8991.54 – 6400 = 2591.54

On Solving, we get

CI = Rs. 2591.54

∴ The compound interest is Rs. 2591.54

Given:

Sum = Rs. 8000

Rate of interest = 30% compounded half yearly

Interest = Rs. 4167

Formula:

If the interest is compounded half yearly, the rate of interest will be halved and time will be doubled.

CI = P [1 + (R/100)]ⁿ – P

Calculation:

CI = P [1 + (R/100)]ⁿ - P

4167 = 8000 [1 + (15/100)]ⁿ - 8000

(1.15)ⁿ = 12167/8000

N = 3

Hence, time = 3/2 = 1.5 years

∴ The required time is 1.5 years

Given:

Sum = Rs. 4000

Rate = 22% compounded half yearly

Time = 1.5 years

Formula:

If the interest is compounded half yearly, the rate of interest will be halved and time will be doubled.

CI = P [1 + (R/100)]ⁿ – P

Calculation:

CI = P [1 + (R/100)]ⁿ – P

CI = 4000 [1 + (11/100)]³ – 4000

⇒ 5470.52 – 4000 = 1470.5

∴ The compound interest is Rs. 1470.5

Given:

Sum = 2200

Rate of interest = 12%

Number of years = 2 years compounded half yearly

Concept:

In compound interest half yearly means the rate of interest is get halved

Formula used:

Amount = P[(1 + {(R/2) × 1/100})²ⁿ]

Calculations:

Amount = 2200[(1 + {(12/2) × 1/100})^{2 × 2})

⇒ Amount = 2200 (1 + 6/100)⁴

⇒ Amount = 2200 × 106 × 106 × 106 × 106/(100 × 100 × 100 × 100)

⇒ Amount = 2777.449 ≈ 2777.45

∴ The amount after 2 years is Rs. 2777045

22Given:

A certain sum amounts to Rs. 15,500 in 2 years at 12% p.a. simple interest. Next time is 1.5 year and the rate of interest is 10% p.a. and interest compounded half yearly.

Concept Used:

Interest = Principal - amount

If p be the principal and the rate of interest is r then simple interest in t year is ptr/100

If p be the principal, r be the rate of interest, the interest compounded half yearly then amount after t year will be p(1 + r/200)^2t

Calculation:

Let the principal be p

Simple interest get in 2 years at 12% p.a. simple interest is (2 × 12p)/100

⇒ 24p/100

Amount = 24p/100 + p

⇒ 124p/100

Accordingly,

124p/100 = 15500

⇒ p = 12500

The principal is 12500

Total time is 1.5 year 

The rate of interest is 10% p.a.

Interest compounded half yearly

Amount after 1.5 years will be 12500 × (1 + 10/200)^{1.5 × 2}

⇒ 12500 × (21/20)³

⇒ 14470.31

⇒ 14470

∴ The amount will be 14470 after 1.5 year

Given:

Present fare per person = Rs. 1000, total fare charged by the train = Rs. 2500000,

The fare of a royal train between to stations increases at the rate of 25% per month at simple interest,

The number of passenger decreases at the rate of 10% per annum compounded monthly

Formula used:

S.I = P × R × T/100

\({\rm{A}} = {\rm{P\;}}{\left( {1 - {\rm{\;}}\frac{{\rm{r}}}{{100}}} \right)^2},\;where\;A = final\;amount\;and\;p = initial\;amount\;of\;water\)

Calculation:

Present number of passenger in the train = 2500000/1000 = 2500

Fare per person in the 2nd month = 1000 + 1000 × 25 × 2/100 = Rs. 1500

Number of passenger in the 2nd month = 2500 (1 - 10/100)² = 2025

Given:

A Simple Interest on a particular sum is 60% of the principle in 5 years.

Let P be the principle and R% is the rate of interest p.a.

Formula Used:

Simple Interest ( S.I ) = ( P× R× T )/ 100

Where P = Principle, R = Rate of interest, T = time period

Compound Interest ( C.I ) = P(1 + R/100)^T – P

Where P = Principle, R = Rate of interest, T = time period

Calculations:

⇒ As from the given formula

  S.I for 5 years = 60%× P = P× R%× 5

⇒ R% = 60% / 5 = 12%      ____( 1 )

As 12% = ( +3/25) = 28/25

⇒ C.I = 6250( 1 + 12/100 )² – 6250

(∵ Principle is given = Rs 6250 )

⇒ C.I = 6250 (28/25)² – 6250 = 6250× 784/625 – 6250

( As 28² = 784, 25² = 625 )

C.I = 7840 – 6250 = Rs 1590

Given 

Increase in salary after 6 years = 60% 

Formula Used 

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

simple Interest = (principal × rate × time)/100

Calculation 

Let the salary at 1st year be 100x 

Increased in salary = 60% of 100x = 60x 

⇒ 60x = (100x × 6 × R)/100  

⇒ R = 10% 

Compound Interest after 3 years = P(1 + r/100)^t - P

⇒ 12000(1 + 10/100)³ - 12000

⇒ 12000 × (11/10)³ - 12000

⇒ 15972 - 12000 = Rs. 3972

∴ The required answer is Rs 3972

Given:

Time = 2 years

Sum = Rs.1250

Rate = 4%

Formula Used:

R = √(D/P) × 100

where, 

R = Rate

P = Sum

D = CI – SI

Calculation:

R = √(D/P) × 100

⇒ 4 = √(D/1250) × 100

⇒ 4/100 = √(D/1250)

Squarring on both sides,

⇒ 16/10,000 = D/1250

⇒ D = (16 × 1250)/10,000

⇒ D = 2

∴ Difference between CI and SI is Rs.2.

Given:

Sum = Rs. 320

Amount = Rs. 405

Rate of interest = 12.5%

Formula Used:

If a^m = aⁿ, then

m = n

Amount = P × (1 + r%/100)^t

⇒ P → Principal, r% → Rate of interest per annum, t → Time period

Calculations:

Let the time period be t years.

Amount = P × (1 + r%/100)t

⇒ 405 = 320 × (1 + 12.5/100)^t

⇒ 405/320 = (9/8)^t

⇒ 81/64 = (9/8)² = (9/8)^t

⇒ t = 2 years

∴ The time in which a sum of Rs. 320 amount to Rs. 405 is 2 years.

Given:

An amount becomes twice in 15 years

Formula used:

S.I = PRT/100

Where, P = Principal

R = Rate of interest

T = Time period

Calculation:

Let the principal be P

Amount after 15 years = 2P

SI = (2P – P)

⇒ P

Now, according to the question

(P × R × 15)/100 = P

⇒ 15R/100 = 1

⇒ R = 20/3%

∴ The rate of interest is 20/3%

Given:

Rate = 7%

For simple interest time = 3 years

Simple interest = Rs. 7,455

For compound interest time = 2 years

Rate of interest for compound interest = 4%

Concept used:

C.I. = P {(1 + R/100 )T – 1}

S.I. = (P × R × T)/100

Where,

C.I → Compound interest

S.I. → Simple interest

P →  Principal

T → Time

R → Rate%

Calculations:

S.I. = (P × R × T)/100

⇒ 7455 = (P × 7 × 3)/100

⇒ P = 745500/21

⇒ P = 35500

C.I. = 35500 × {(1 + 4/100)² – 1}

⇒ 35500 × [{(26/25) × (26/25)} – 1]

⇒ 35500 × (676 – 625)/625

⇒ 35500 × 51/625

⇒ 2896.8

∴ The compound interest is Rs. 2896.8

Given:

Ram saves Rs.60,000 at the beginning of each year and puts the money in a bank.

Rate of compound interest = 10% annually

Time = 4 year

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:

First year deposit in bank Rs.60,000

After 1 year Amount = 60000 × (1 + 10/100)¹

After 2 year Amount = 60000 × (1 + 10/100)²

After 3 year Amount = 60000 × (1 + 10/100)³

After 4 year Amount = 60000 × (1 + 10/100)⁴

Total amount after 4 year = 60000 × 1.1 × {1 + 1.1 + (1.1)² + (1.1)³}

⇒ 66000 × (1 + 1.1 + 1.21 + 1.331)

⇒ 66000 × 4.641

⇒ Rs.3,06,306

∴ Total saving of Ram at the end of 4 years is Rs.3,06,306.

Given :

Amount = 169

time = 2 year 

Rate% = 4% 

Formula Used :

\(Amount = P\;{\left( {1 + \frac{r}{{100}}} \right)^t}\)

Where P is the principal, r is the rate of interest and t be the time period

Calculation:

Accordingly,

\(169 = P\;{\left( {1 + \frac{4}{{100}}} \right)^2}\)

P = (169 × 25× 25)/26× 26

P = 156.25

∴ The required amount is Rs. 156.25.

Given:

Amount in second year = 4624 Rs.

Amount in Third year = 4913 Rs.

Formula used 

\(Amount = P\;{\left( {1 + \frac{r}{{100}}} \right)^t}\)

Where P is the principal, r is the rate of interest and t be the time

Solution:

Accordingly,

\(4624 = \;{P\left( {1 + \frac{r}{{100}}} \right)^2}\)      ----(i)

\(4913 = P\;{\left( {1 + \frac{r}{{100}}} \right)^3}\)      ----(ii)

Divide eq (ii) by (i)

\(\frac{{4913}}{{4624}} = {\left( {1 + \frac{r}{{100}}} \right)^1}\)      ----(iii)

\(r = \frac{{\left( {4913 - 4624} \right) × 100}}{{4624}} \)

r = 6.25%

6.25% = 1/16

from Eq (i) 

4624 = P× (17/16)× (17/16)

⇒ P = 4096

∴ The sum of money is Rs. 4096

Given:

A certain sum of money becomes 7 times of itself in 18 years if invested at the simple interest.

Formula used:

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

Calculation:

A sum of money becomes 7 times of itself in 18 years

Let the sum be x

The amount after 18 years will be 7x.

Interest = 7x – x = 6x

S I = (p × r × t)/100

6x = (x × r × 18)/100

r = 100/3%

If the rate of interest divided by 100 is applied on a sum of Rs. 54000 for three years under compound interest,

R = 100/3%

According to the question,

Amount = 54000 (1 + 1/3)³

⇒ 23000 × 4 × 4 × 4

⇒ Rs. 128000

∴ The amount after 4 years is Rs. 128,000

Given:

A sum of amount becomes 7 times in 20 years at simple interest

Formula used:

S.I = PRT/100

Amount = Principal + S.I

Calculation:

Let the principal be Rs. x

Amount = Rs. 7x

S.I = Rs. (7x – x) 

⇒ Rs. 6x

Now, according to the question

(x × R × 20)/100 = 6x

⇒ R = 6 × 5%

⇒ 30%

∴ The rate of simple interest is 30%

Given:

Money becomes Rs. 1200 in 4 years

Money becomes Rs. 1500 in 7 years 

Calculation:

According to question

Difference between the money of 7 years and 4 years = 1500 - 1200

⇒ Rs. 300

So, Interest of 3 years = Rs. 300

Now interest of one year = 300/3 = Rs. 100

⇒ Interest of 4 years = 4 × 100 

⇒ Rs. 400

Principal amount = 1200 - 400

⇒ Rs. 800

∴ The Principal amount is Rs. 800

Given:

A = Rs. 3900

N = 5 years

P = Rs. 1500

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:

I = 3900 – 1500

⇒ I = Rs. 2400

Accordingly,

2400 = (1500 × R × 5) / 100

⇒ 2400 × 100 / (1500 × 5) = R

⇒ R = 32%

Now, rate of interest is reduced by 2%

⇒ New rate of interest is (32 – 2) = 30%

I = (1500 × 30 × 5) / 100

⇒ I = 2250

⇒ A = 1500 + 2250

Given:

Time = 4 years

Rate = 12.5%

Formula used:

Amount = Simple interest + Principal

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:

Let, the principal is P

SI = (P x R x T)/100

⇒ P - 400 = (P × 4 × 12.5)/100

⇒  P - 400 = P × 4 × (1/8)

⇒  P - 400 = P/2

⇒  2P - 800 = P

⇒  P = 800 Rs.

∴ The sum is 800 Rs.

Given:

The difference between the simple interest of 3 years at 8% and that of 2 years at 9% rate of interest is Rs. 96.

Concept Used:

SI = (P × R × T)/100

Calculation:

Let, the principal be P

Interest in first case is (3P × 8)/100 = 24P/100

Interest in 2nd case is (2P × 9)/100 = 18P/100

Accordingly,

24P/100 - 18P/100 = 96

⇒ 6P/100 = 96

⇒ P = 1600

∴ The required principal is Rs. 1600.

Given:

C.I. – S.I. = Rs. 27

R = 5%

N = 2 years

Formula used:

In case of compound interest

A = P × {1 + (R / 100)}^N

Where P = Principal amount, R = Rate of interest in %, N = Number of years, A = Amount

In case of simple interest

I = PRN / 100

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

A = P + I

Calculation:

Here, C.I. – S.I. = 27

Also, C.I. = P × {1 + (R / 100)}^N – P

And S.I = PRN/100

Accordingly,

⇒ P × {1 + (R / 100)}N – P – (PRN / 100) = 27

⇒ P{(1 + R/100)^N - 1 - (RN/100)} = 27

⇒ P{(1 + 5/100)² - 1 - (5 × 2)/100} = 27

⇒ P × {441/400 - 1 - 1/10} = 27

⇒ P × (1/400) = 27

⇒ P = 27 × 400

⇒ P = 10800

Given:

C.I = Rs 618

T = 2 years

R = 6 % per annum

Formula used:

C.I = P[(1 + r %)^t – 1]

 618 = P[(1 + 6 %)² -1] ⇒ 618 = P[(106/100)² – 1]

⇒ 618 = P [(11236/10000) – 1 ]

⇒ 618 = P [(11236 – 10,000)/10,000]

⇒ 618 = P [ 1236/10,000]

⇒ [(618 × 10,000)/1236] = P

⇒ 5000 = P

∴ The principal is Rs. 5000

Given:

C.I. - S.I. = Rs.108.3

Rate (R) = 19%

Time = 2 years

Formula used:

The difference between S.I. and C.I. for 2 years = PR2/(100)2

Where P → Principal

R → Rate

Calculations:

According to the question,

The difference between the compound interest and simple interest for 2 years = PR2/(100)2

⇒ 108.3 = Z × 19 × 19/10000

⇒ Z = Rs.3000

∴ The value of Z is Rs.3000.

Given:

C.I. - S.I. = Rs.1445

Rate (R) = 17%

Time = 2 years

Formula used:

The difference between S.I. and C.I. for 2 years = PR2/(100)2

Where P → Principal

R → Rate

Calculations:

According to the question,

The difference between the compound interest and simple interest for 2 years = PR²/(100)²

⇒ 1445 = P × 17 × 17/10000

⇒ P = Rs.50000

∴ The value of P is Rs.50000.

Given:

Principal = Rs.12500

Time = 1 year

Actual Rate = 4%

But due to half-yearly

Rate = 2 percent

Formula:

C.I – S.I for 2 years = (P × R²)/100 × 100

Calculation:

Note: Due to half-yearly we use formula C.I – S.I for 2 year

We know that –

C.I – S.I for 2 years = (P × R²)/100 × 100        ……………   (1)

Now,

Put all the given values in equation (1) then we get

C.I – S.I for 2 years = (12500 × 2²)/100 × 100

⇒ (12500 × 4)/100 × 100

⇒ (125 × 4)/100

⇒ 125/25

⇒ 5

∴ The Difference between C.I and SI for 1 year is Rs. 5

Concept used:

In the case of the compound interest, Interest is earned on the interest every year.

And, In case of simple interest, Interest is only earned on the principal only and interest is the same in every year.

Calculation:

From the above statement we come to know that Statement I is correct and Statement II is wrong

∴ The correct answer is option 3

Given:

P = Rs. 5000

R = 25%

n = 2

t = 1.5 years

Formula used:

A = P × {1 + (R / n)}^nt

Where P = Principal amount, R = Rate of interest in %, n = number of times interest applied per year and t = time period

Calculation:

Here A = 10240 × {1 + (25 / 100 × 2)}^{2 × 1.5}

⇒ A = 10240 × {1 + (1 / 8)}³

⇒ A = 10240 × (9 × 9 × 9) / (8 × 8 × 8)

⇒ A = 20 × 729

⇒ A = 14580

Given:

P = Rs. 3800

R = 20%

n = 2

t = 1 year

Formula used:

A = P × {1 + (R / 100n)}^nt

Where P = Principal amount, R = Rate of interest in %, n = number of times interest applied per year and t = time period

Calculation:

Scheme 1:

Here, interest compounded yearly

⇒ A = 3800 × {1 + (20 / 100)}

⇒ A = 4560

Scheme 2:

Interest compounded half yearly

⇒ A = 3800 × {1 + (20 / 100 × 2)}²

⇒ A = 3800 × {(11 × 11) / (10 × 10)}

⇒ A = 3800 × (121 / 100)

⇒ A = 4598

Required difference = 4598 – 4560

Given:

Difference of S.I. on certain sum of money lent for 3 years and 5 years is Rs. 980. 

Rate of interest = 7% p.a.

Concepts used:

S.I. = P × R × T/100

Where,

P → Principal

R → Rate of interest per annum

T → Time (in years)

Calculation:

S.I. = P × R × T/100

Where, P = Principal, R = rate of interest per annum, T = time (in years)

S.I. for 3 years = P × 7 × (3/100)

⇒ 21P/100

S.I. for 5 years = P × 7 × (5/100)

⇒ 35P/100

The difference of S.I. on a certain sum of money lent for 3 years and 5 years is Rs. 980.

(35P/100) – (21P/100) = Rs. 980

⇒ 14P/100 = Rs. 980

⇒ P = Rs. 980 × (100/14)

⇒ P = Rs. 7,000.

∴ The sum of money lent is equal to Rs. 7,000.

Given:

A sum of money lent at S.I. amounts to Rs. 6,500 after 2 years and amounts to Rs. 9,000 after 7 years at same rate of interest

Concepts used:

S.I. = P × R × T/100

Where, P = Principal, R = Rate of interest per annum, T = Time (in years)

Calculation:

Amount after 2 years = Rs. 6500

Amount after 7 years = Rs. 9000

S.I of 5 years = Rs. (9000 – 6500)

⇒ Rs. 2500

S.I of 1 year = Rs. 2500/5

⇒ Rs. 500

S.I of 2 years = Rs. (500 × 2)

⇒ Rs. 1000

Principal = Rs. (6500 – 1000)

⇒ Rs. 5500

Rate of interest = (500 × 100)/5500%

⇒ 100/11%

⇒ 9.09%

∴ The rate of interest is 9.09%

Given:

The difference between simple interest and compound interest on the principal = Rs. 8

Rate of interest = 4% p.a

Number of years = 2 years

Formula used:

SI = PTR/100

CI = P[(1 + r/100)n - 1]

Calculations:

Let principal be 'p'

SI = p × 4 × 2/100

⇒ SI = 8p/100

CI = p[(1 + 4/100)2 - 1]

⇒ CI = p[(1 + 1/25)2 - 1]

⇒ CI = p[(26/25)2 - 1]

⇒ CI = p(1.0816 - 1)

⇒ CI = 0.0816p

Difference between SI and CI = 0.0816p - 0.08p

⇒ 0.0016p = 8

⇒ p = 8/0.016

⇒ p = 5000

∴ The Principal is Rs 5000

Given:

Selling price of 5 tulips = Rs. 56

Gain% = 40%

Calculation:

SP of 1 tulip = 56/5

⇒ Let CP of 1 tulip be x

⇒ According to Q

⇒ X × 140/100 = 56/5

⇒ X = Rs 8

⇒ CP of 20 Tulips = 20 × 8 =160

∴ 20 tulips would cost him Rs. 160

Given:

Principal = Rs 35,800

Rate of interest, R = 15%

Time period, T = 4 years

Concept used:

Total amount = Principal, P + Simple Interest, SI

SI = PRT/100

Calculation:

SI = (35,800 × 15 × 4)/100 = 21,480

Total amount = 35,800 + 21,480 = Rs 57,280

∴ Total amount obtained at the end of four years is Rs 57,280.

Given:

Rate of interest = 20%

Years = 2

Principal amount = 8000

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:

Here, 40% of amount is compounded

⇒ P for C.I. = (40 / 100) × 8000

⇒ P for C.I. = 3200

⇒ 3200 × (1 + (20 / 100))²

⇒ 3200 × (6 / 5)²

⇒ 3200 × 36 / 25

⇒ 4608

⇒ Interest earned = 4608 – 3200

⇒ I = 1408

Now,

⇒ 60% of amount = 8000 – 3200

⇒ P for S.I. = 4800

⇒ 4800 × 20 × 2 / 100

⇒ I = 1920

∴ Total interest earned is Rs. 3328

Given:

The ratio of 3 years of C.I. and S.I. of one year on a certain sum of money is 4.75 ∶ 1.

Formula used:

(x + y)³ = x³ + y³ + 3xy(x + y)

S.I. = (P × R × T)/100 

A = P(1 + R/100)n

Where A → amount

n → time

S.I. → Simple Interest

P → Principal

R → Rate 

T → Time

Calculations:

Let the sum of money be Rs. P. and rate of interest be R% p.a.

Compound interest for 3 years = P[(1 + R/100)³ - 1]

⇒ P[1 + 3 × 1 × (R/100)(1 + R/100) + (R/100)³ -1]

⇒ P(R/100)(3 + 3 × R/100 + R²/10000)

Simple interest for one year = (P × R × 1)/100

According to the question,

⇒ {P(R/100)(3 + 3 × R/100 + R2/10000)}/(P × R × 1)/100 = 4.75/1.

⇒ 3 + 3 × (R/100) + R²/10000 = 4.75

⇒ R2/10000 + 3 × (R/100) - 1.75 = 0

⇒ R² + 300R - 17500 = 0

⇒ R2 + 350R - 50R - 17500 = 0

⇒ R(R + 350) - 50(R + 350) = 0

⇒ (R + 350)(R - 50) = 0

Thus R = - 350% (not valid) , or R = 50%

∴ The rate is 50%.

Given:

Time = 2 years 

Rate (R) = 10%

C.I. = Rs.630

Formula used:

S.I. = (P × R × T)/100 

A = P(1 + R/100)n

Where A → amount

n → time

S.I. → Simple Interest

P → Principal

R → Rate 

T → Time

Calculations:

Let the principal be Rs. P.

C.I. = P(1 + 10/100)² - P

⇒ 630 = P{(121/100) - 1}

⇒ P = Rs.3000

Now

S.I. = (3000 × 10 × 2)/100

⇒ S.I. = Rs.600

∴ The simple interest is Rs.600.

Given:

Loan amount = Rs. 2,60,200

Rate of interest = 4% pa

Formula used:

Amount = Principal × [1 + r/100]n 

Calculations:

Let the first part be x

The second part = (260200 – x)

According to the question,

x(1 + 4/100)⁷ = (260200 – x)(1 + 4/100)⁹

⇒ x/(260200 – x) = (1 + 4/100)⁹/(1 + 4/100)⁷

⇒ x/(260200 – x) = 26/25 × 26/25

⇒ x/(260200 – x) = 676/625

⇒ 625x = 676 × 260200 – 676x

⇒ 1301x = 676 × 260200

⇒ x = (676 × 260200)/1301

⇒ x = 135200

Second part = 260200 – x

⇒ 260200 – 135200

⇒ 125000

Given-

Principal = 24

Amount = 56

Rate = 16%

Concept Used-

Interest = Amount - Principal

Interest = Principal × Rate × Time/100

Calculation- 

Interest = 56 - 24 

⇒ 32

According to Question -

Interest = Principal × Rate × Time/100

⇒ 32 = 24 × 16 × Time/100

⇒ Time = 25/3 years

∴ Required time = 8 years 4 months

Given:

Compound interest for the second year = Rs. 464

Rate (R) = 16%

Formula used:

A = P(1 + R/100)n

Where A → amount

P → principal

R → rate 

n → time

Calculations:

Let the principal be Rs. P.

According to the question,

Amount after 1st year = P × 29/25 = 29P/25

Amount after 2nd year = (29P/25) × (29/25) = (841P/625)

So, Compound interest for the second year = (841P/625) – 29P/25

⇒ (841P – 725P)/625

⇒ 116P/625

But Compound interest for the second year = Rs. 464

Thus (116P/625) = 464

⇒ P = 2500

∴ The Principal is Rs. 2500.

Given:

Amount (A) = Rs. 297.91

Principal (P) = Rs. 156.25

Rate (R) = k%

Time (n) = 3 years

Formula used:

A = P(1 + R/100)n

Where A → amount

P → principal

R → rate 

n → time

Calculations:

A = P(1 + R/100)n

⇒ 297.91 = 156.25(1 + R/100)³

⇒ (1 + R/100)3 = 297.91/156.25

⇒ (1 + R/100)3 = (31/25)³

⇒ (1 + R/100) = 31/25

⇒ (R/100) = 6/25

⇒ R = 24%

∴ The value of k is 24%.

Given:

⇒ Principle = Rs.200

⇒ Rate of interest = 10%

⇒ Time period = 1 year

Formula:

Let P = Principle, R = Rate of interest and N = time period

⇒ Simple interest = PNR/100

⇒ Compound interest = P(1 + (R/2)/100)²ⁿ - P

Calculation:

⇒ Simple Interest after a year = (200 × 10 × 1)/100 = Rs.20

⇒ Compound interest after a year compounded semi annually = 200(1 + 5/100)² - 200

= Rs.20.5

∴ Required difference between interests = 20.5 - 20 = Rs.0.5 = 50 paise.

Given:

Rate = 11%

Time = 3 years

Simple interest(S.I) = Rs. 9570

Formula used:

Simple interest = (P × R × T)/100

Calculation:

⇒ P = (S.I × 100)/(R × T)

⇒ P = (9570 × 100)/(11 × 3)

⇒ P = 29000

∴ The principal amount borrowed by him was Rs. 29000.