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Given:

Simple interest = Rs. 4944

Time = 2 years

The rate percent is 15 times of the time

Rate% = 30

Concept used:

C.I. = P{(1 + R/100)T – 1}

S.I. = (P × R × T)/100

S.I. → Simple interest

P → Principal

T → Time

R → Rate%

C.I. → Compound interest

Calculations:

The rate percent is 15 times of the time

Rate% = 2 × 15 = 30%

S.I. = (P × R × T)/100

⇒ 4944 = (P × 30 × 2)/100

⇒ P = Rs. 8240

C.I. = P{(1 + R/100)T – 1}

⇒ 8240{(1 + 30/100)² – 1}

⇒ 8240{(13/10) × (13/10) – 1}

⇒ 8240 × 69/100

⇒ Rs. 5685.6

∴ The compound interest at the same rate of interest at same sum is Rs. 5685.6

Given:

For 3 years, the difference between simple interest and compound interest = Rs. 616

Rate = 8%

For compound interest, time = 2 years.

Concept used:

C.I. = P{(1 + R/100 )^T – 1}

D = P × (R/100)²(300 + R)/100

S.I. → Simple interest

P → Principal

T → Time

R → Rate%

C.I. → Compound interest

D → Difference between S.I. and C.I. for 3 years difference,

Calculations:

D = P × (R/100)2(300 + R)/100

⇒ 616 = P × (8/100)²(300 + 8)/100

⇒ 616 = P × (64/10000) × (308/100)

P = (616 × 10000 × 100)/(308 × 64)

⇒  Rs. 31,250

At the same sum,

C.I. = P{(1 + R/100 )T – 1}

⇒ 31250{(1 + 8/100)² – 1}

⇒ 31250 × {(108/100) × (108/100) – 1}

⇒ 31250 × {(11664/10000) – 1}

⇒ 31250 × (1664/10000)

⇒ Rs. 5200 

∴ The compound interest is Rs. 5200

Given:

S.I. = Rs.1210

C.I. = Rs.1331

Time = 2 years

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 and the rate be R.

So, 1210 = (P × R × 2)/100

⇒ PR = 1210 × 50      ----(i)

Similarly, 1331 = P(1 + R/100)² - P

⇒ 1331 = P(100 + R)²/(100)² - P

⇒ 1331 = P(10000 + R² + 200R - 10000)/10000

⇒ 1331 = PR(R + 200)/10000 

⇒ 1331 × 10000 = 1210 × 50(R + 200) (by putting the value of PR from (i))

⇒ 11 × 20 = R + 200

⇒ R = 20%

∴ The rate of interest per annum is 20%.

Given:

Increased rate = 5%

Time = 5 year

In 5 year Rs. 1300 amounts to be Rs. 1560.

Concept:

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

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

Simple interest = Amount – Principle

Calculation:

Simple interest = 1560 – 1300 = Rs. 260 

Rate = (260 × 100)/(1300 × 5) = 4%

New rate = 5% + 4% = 9%

Simple interest = (1300 × 9 × 5)/100 = 585 Rs.

Amount = Simple interest + Principle

⇒ 585 + 1300

Given:

The difference between CI and SI for 2 years = 6250

Principal = 25000

Concept used:

The difference between CI and SI for 2 years = P(R/100)²

Calculation:

25000 × (R/100)² = 6250

⇒ R²/100 = 25

⇒ R² = 2500

⇒ R = √2500

⇒ R = 50%

∴ Rate of interest is 50%

GIVEN:

Manvendra paid Rs. 4800 as interest on a loan he took 5 years ago at 8% rate of simple interest.

FORMULA USED:

Simple Interest = PRT ÷ 100

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

CALCULATION:

Applying the formula:

4800 = (P × 8 × 5)/100

Solution:

Given: Sum Manish borrows = 54000

Amount repaid by Manish at the end of second year = 32340

Rate of interest = 10%

Concept used: Amount at the end of second year = Amount at the end of first year {1 + (rate/100)}

Calculation: Compound interest for first year = (54000 × 10)/100

⇒ 5400

Amount at the end of first year = 54000 + 5400

⇒ 59400

Let the amount repaid be Rs x

Then, the sum at the beginning of the second year = 59400 – x

Amount at the end of second year = (59400 – x) + [{(59400 – x) × 10}/100]

⇒ (59400 – x) + [{(59400 – x) × 10}/100] = 32340

⇒ 59400 – x + 5940 – 0.1x = 32340

⇒ 65340 – 1.1x = 32340

⇒ 1.1x = 65340 – 32340

⇒ 1.1x = 33000

⇒ x = 30000

∴ Amount repaid by Manish at the end of first year = 30000

Given:

P = Rs. 7500

R = 10%

N = 3 Years

Amount paid back at end of year 1 = Rs. 2800

Amount paid back at end of year 2 = Rs. 1500

Formula used:

A = P × {1 + (R / 100)}^N

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

A = P + I

Calculation:

Amount to be paid at end of 1st year = 7500 × {1 + (10 / 100)}

⇒ A = 8250

She pays Rs. 2800 at end of 1st year

⇒ Amount left = 8250 – 2800

⇒ Amount left = 5450

Amount at end of 2^nd year = 5450 × {1 + (10 / 100)}

⇒ Amount at end of 2^nd year = 5995

She paid Rs. 1500 at the end of 2^nd year

⇒ Amount left to pay = (5995 – 1500)

⇒ Amount left to pay at end of 2^nd year = Rs. 4495

Amount at end of 3^rd year = 4495 × {1 + (10 / 100)}

⇒ Amount at end of 3^rd year = 4944.5

Given:

Total amount = Rs. 21,840

Time = 3 years

Rate = 12% per annum at SI

Formula used:

For installment at SI,

A = nx + {(n – 1) + (n – 2) +(n – 3) + …}Rx/100

where,

 A = Total amount or Debt.

x = value of each installment

n = time for installment

R = Rate (%)

Calculations:

A = nx + {(n – 1) + (n – 2) +(n – 3) + …}Rx/100

⇒ 21840 = 3x + (2 + 1)12x/100

⇒ 21840 = 3x + (36x/100)

⇒ 21840 = 336x/100

⇒ 21840/336 = x/100

⇒ 65 = x/100

⇒ 6500 = x

∴ Annual installment is Rs. 6500

Given:

Sum = Rs. 1550

Rate = 5% and 8%

Time = 5 yrs

SI = Rs. 500

Formula Used:

S.I = P × r × t/100

Calculations:

Let the sum lent at 5% be P

Sum lent at 8% = (1550 - P)

Then, (P × 5 × 5)/100 + [(1550 - P) × 8 × 5] /100 = 500

⇒ 25P – 40P + 1550 × 40 = 50000

⇒ -15P + 62000 = 50000

⇒ -15P = 50000 – 62000

⇒ -15P = -12000

⇒ P = 800

Sum lent at 8% = 1550 – 800 = 750

required ratio = 800 : 750 = 16 : 15.

∴ The ratio of money lent at 5% to that at 8% is 16 : 15.

Given: 

Simple interest for 2 years at 6% per annum is 180.

Concept: 

SI = (P × R × T)/100

A = P(1 + r/100)ⁿ

Here, N is the number of terms in the year.

Calculation: 

The simple interest is 

⇒ 180 = (P × 6 × 2)/100

⇒ P = 1500 rupee

Now, When interest is compounded annually, then the amount is 

⇒ 1500(1 + 6/100)² 

⇒ 1500 × 106/100 × 106/100

⇒ 1685.40

∴ when interest is compounded annually then the amount is 1685.40 rupee.

Given:

Interest received after 2 years = Rs. 15300

Interest received after 3 years = Rs. 16830

Sum invested on compound interest = Rs. 40000

Concept Used:

When a sum is compounded half yearly, its rate percent gets halved and time gets doubled.

Formula Used:

r% = ((C.I.3 – C.I.2)/C.I.2) × 100

When a sum is compounded half yearly,

C.I. = P × (1 + (r)/100)T – 1)

where r% → rate percent per annum at which the sum was invested.

C.I.2 and C.I.3 → Compound interests after 2 and 3 years respectively.

C.I. → Compound Interest, Principal → P, Time period → T.

Calculations:

Let compound interest after 2 and 3 years be C.I.2 and C.I.3 respectively.

r% = ((C.I.3 – C.I.2)/C.I.2) × 100

⇒ r% = ((16830 – 15300)/15300) × 100

⇒ r% = (1530/15300) × 100

⇒ r% = 10%

C.I. = P × (1 + (r)/100)T – 1)

⇒ C.I. = Rs. 40000 × (1 + (10/100)1 – 1)

⇒ C.I. = Rs. 4000

∴ The rate percent at which the sum was invested is 10% per annum compound interest received when Rs. 40000 are compounded for 1 year is Rs. 4000.

GIVEN:

Principal (P)= Rs1200, Rate(R) = 10%, Time(N) = 2 years

FORMULA USED:

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

CALCULATION:

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

⇒ A = 1200(1 + 10/100)²

⇒ A = 1200(1 + 1/10)¹⁰

⇒ A = 1200 × 11/10 × 11/10

⇒ A = 12 × 121

⇒ A = Rs1452

⇒ Amount = Principle + Interest

⇒ Interest = Amount - Principle

⇒ Interest =Rs1452 - Rs1200

Given:

Anjali borrowed Rs. 25,000 at the rate of 10% per annum under compound interest.

Amount repaid after 1 year = Rs. 15,000

Formula used:

Amount after n years under compound interest = P[1 + (r/100)]^T

Calculation:

Amount that she needs to pay at the end of one year = 25000 [1 + (10/100)]

⇒ Rs. 27500

She repaid Rs. 15,000 at the end of one year

⇒ Rs. (27500 – 15000) = Rs. 12,500 = Principal for second year

Amount that he needs to pay to clear = 12500 (1 + 10/100)

⇒ Rs. 13750

∴ The required amount is Rs. 13,750

Given:

A and B borrow equal sum of money for 3 years at the rate of 5% for SI and CI respectively

Difference between interest of A and B is 427

Formula Used:

SI = (P × R × T)/100

CI = P (1 + R/100)^t - P

Calculation:

SI of A = (P × 3 × 5)/100 = 3P/20    ____(i)

CI of B = P (1 + 5\100)³ - P 

⇒ P ×  \(\frac{{21}}{{20}} × \frac{{21}}{{20}} × \frac{{21}}{{20}}\) - P

⇒ \(\frac{{9261P}}{{8000}}\) - 1

⇒ 1261P/8000    ____(ii)

According to question,

⇒ \(\frac{{1261P}}{{8000}} - \frac{{3P}}{{20}}\) = 427

⇒ 61P/8000 = 427

⇒ P = (427 × 8000)/61

⇒ P = Rs.56000

Interest Paid by B = 56000 (1 + 5\100)3 - P 

⇒ 56000 × \(\frac{{21}}{{20}} × \frac{{21}}{{20}} × \frac{{21}}{{20}}\)  - P

⇒ 56000 × (9261/8000) - P

⇒ 64827 - P

⇒ 64827 - 56000 = Rs.8827

∴ The sum borrowed is Rs.56,000 and interest paid by B is Rs.8827.

Given:

Principal = Rs. 300

Rate of interest = 10%

time = 2 years

Formula used:

Compound interest = Principal[1 + (rate/100)]^time - Principal

Calculation:

Compound Interest for 2 years

⇒ 300[1 + (10/100)]² - 300

⇒ 363 - 300

⇒ 63

∴ The compound interest for 2 years is Rs. 63.

Given:

Rate of interest = 10% 

At the end of second year, money paid by her = Rs. 12474 

Concept used:

A = P × (1 + R/100)^T 

Where,

A → Amount 

P → Principal 

R → Rate of interest 

T → Time 

Calculations:

Let the principal be 100x 

Money after completion of first year = 100x × (1 + 10/100)¹ 

⇒ 100x × 110/100 = 110x 

She borrowed the same principal in second year 

So principal for second year = 110x + 100x = 210x 

Money after completion of second year = 210x × (1 + 10/100)1 

⇒ 231x = 12474 

⇒ x = 54 

Money borrowed each year =100x = 5400

∴ Money borrowed each year is Rs. 5400 

Given:

Principal Amount = Rs.12200

Rate of interest = 25%

Three equal annual instalment means (Time) = 3 years

Formula:

A = P × (1 + R/100)ⁿ

Here,

A = Amount

P = Principal Amount

R = Rate of interest

n = time in year

I_A = I × (1 + R/100)^t

Here,

I_A = Instalment Amount

I = Instalment

R = Rate of interest

t = (n – 1)year, (n – 2)year, ……..

Here, n = time in years

Calculation:

We know that –

A = P × (1 + R/100)ⁿ  ……. (1)

Put all the given values in equation (1)

A = 12200 × (1 + 25/100)³

⇒ 12200 × (1 + 1/4)³

⇒ 12200 × (5/4)³

⇒ 12200 × 125/64

Now,

Let the instalment be Rs.y

Instalment Amount (I_A) for three years means = (n – 1)year  +  (n – 2)year +(n – 3)year

We know that –

I_A for (n – 1)year = I × (1 + R/100)ⁿ ….. (2)

Put all the given values in equation (2) then we get

I_A for (n – 1)year = y × (1 + 25/100)²

⇒ y × (1 + 1/4)²

⇒ y × (5/4)²

⇒ 25y/16

Similarly,

I_A for (n – 2)year = y × (1 + 25/100)¹

⇒ y × (1 + ¼)

⇒ y × 5/4

⇒ 5y/4

Now,

I_A for (n – 3)year = y × (1 + 25/100)⁰

⇒ y × 1

⇒ y

Instalment Amount (I_A) for three years = 25y/16 + 5y/4 + y

⇒ (25y +20y + 16y)/16

⇒ 61y/16

Now,

We equate the Amount (A) & Instalment Amount (I_A)

12200 × 125/64 = 61y/16

⇒ 200 × 125/4 = y

⇒ y = 50 × 125

⇒ y = 6250

∴ The Value of each instalment will be Rs.6250

Given:
Simple interest more than 1% the simple interest increased = Rs. 240

Time = 2 years

Concept used:

S.I. = (P × R × T)/100

Calculation:

S.I. = (P × R × 2)/100      ----(1)

S.I. = [P × (R + 1) × 2]/100      ----(2)

Equation (1) subtract from equation (2)

[P × (R + 1) × 2]/100 – (P × R × 2)/100 = 240

⇒ P[(R + 1) × 2 – (R × 2)]/100 = 240

⇒ P(2R + 2 – 2R) = 240 × 100 

⇒ P × 2 = 24000

⇒ P = 24000/2

⇒ P = 12000

∴ The principal lends on simple interest for two years is Rs. 12000.

Given

Amount = 2(principal)

Formula

Simple interest = (p × r × t)/100

Where p, r and t represents principal, rate of interest and time

Calculation

Simple interest = Amount – Principal

⇒  Rs. 2P – P

⇒  P

⇒  P = ( P × r × 5)/100

⇒  r = 20 %

Alternative method

If a sum of money becomes ‘a’ times in ‘t’ years at simple interest, then formula for calculating rate of interest will be as follows:

Rate of interest = 100(a – 1)/t

Solution:

a= 2, t = 5

Rate of interest = 100(2 – 1)/5

Given

Principal = Rs. 1500, Time = 12 months and Rate = 12% per annum

Formula used

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

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

Concept

When compound quarterly is given, then time is multiply by four and rate is divided by four.

Calculation

Rate = 12%/4 = 3% per annum

Time = 12 months or 1 year

⇒ 1 × 4

⇒ 4

A = P(1 + r/100)t 

⇒ A = 1500(1 + 3/100)⁴

⇒ A = 1500(103/100)⁴

⇒ A = Rs. 1688.26

∴ The amount is Rs. 1688.26.

Given:

Principal: Rs 48,000

Rate = 8 % per annum

Time = 9 months

Formula used:

Amount = P [1 + r %]^t

Calculation:

Rate = (8/4) = 2 % per annum

Time = 9 months = 3 cycles

Amount = Rs 48,000 [1 + (2/100)]³

 ⇒ 48,000 (102/100)³

 ⇒ Rs 50,937.984

Compound interest = Amount – Principal

 ⇒ Rs 50,937.984 – Rs 48,000

 ⇒ Rs 2937.984

∴ The compound interest is Rs. 2937.984

Given:

Principal = ₹ 16000

Rate = 20%

Time = 6 months = 1/2 year

Interest is compounded quarterly 

Concept Used:

If interest is compounded quarterly means interest is calculated in every three months that is 4 times in a year.

We can simply convert this problem into a normal compound interest problem by multiplying the time by 4 and dividing the rate by 4 

Formula Used:

Amount = Principal[1 + (Rate/100)]Time

Amount = Principal + Interest

Calculation:

New rate = 20%/4 = 5%

New time = 1/2 × 4 = 2 years

Amount = ₹ 16000[1 + 5/100]²

⇒ ₹ 16000[1 + 1/20]²

⇒ ₹ 16000[21/20]²

⇒ ₹ 16000[441/400]

So, Principal + Interest = ₹ 17640

⇒ Interest = ₹ 17640 – ₹ 16000

⇒ Interest = ₹ 1640

∴ The compound interest on ₹ 16000 at the rate of 20% per annum for 6 months if the interest is compounded quarterly is ₹ 1640 

Given:

Amount of a bike (P) = Rs. 84000

Rate of interest = 10%

Formula used:

P × {1 + (R/100)}² = I × {1 + (R/100)} + I

Where p = principal

R = rate percentage at compound interest

I = each installment

Calculation:

84000 × {1 + (10/100)}² = I × {1 + (10/100)} + I

⇒ 84000 × (11/10) × (11/10) = I × (11/10) + I

⇒ 840 × 11 × 11 = (21/10) × I

⇒ I = (840 × 10 × 11 × 11)/21

⇒ I = 48400

∴ The value of each installment is Rs. 48,400

Formula used:

SI = P × R × T/100

Where P = Principal, R = Rate and T = Time

Calculation:

Let the principal be Rs. X.

Now, the simple interest for first one and a half

year = [P × 8 × (3/2)/100]

⇒ 12P/100

The Simple interest for next 6 months = [P × 10 × (1/2)/100]

⇒ 5P/100

The Simple interest for next 10 months = [P × 12 × (10/12)/100]

⇒ 10P/100

The Simple interest for the remaining 15 months = [P × 4 × (15/12)/100]

⇒ 5P/100

According to Question,

(12P/100) + (5P/100) + (10P/100) + (5P/100) = 22400.32

⇒ 32P/100 = 22400.32

⇒ P = 22400.32 × (100/32)

⇒ Rs. 70,001

∴ The principal will be Rs. 70,001

Given:

A sum of money gets doubled in 10 years at simple interest(SI)

Formula used:

SI = \(\dfrac{P \times R \times T}{100}\)

Amount = P + SI

P = Principal

R = Rate of interest

T = Time

Calculation:

According to the question:

Amount after 10 years = 2P

Hence, the SI in 10 years = P

⇒ P = \(\dfrac{P \times R \times 10}{100}\)

⇒ R = 10

∴ Rate of interest = 10%

Given:

Principal = Rs 3200

Rate of interest = 10% per annum

Amount = Rs 3362

Concept used:

Amount = P × [1 + (r/100)]^t

But here, the amount is calculated quarterly. Then the amount obtained will be:

Amount = P × [1 + r/(n × 100)]^nt

Where n is the number of terms, here n = 4.

Calculation:

3362 = 3200 × [1 + 10/(4 × 100]^4t

⇒ 3362/3200 = (41/40)^4t

⇒ 1681/1600 = (41/40)4t 

⇒ (41/40)² = (41/40)^4t

On comparing powers,

⇒ 2 = 4t 

⇒ t = 1/2 

∴ The time period is 1/2 year.

Given:

Time = 10 years

Money triple in 10 years.

Formula used:

Simple interest = (P × R × T)/100

Where P → Principal

T → Time

R → Rate

Calculation:

Let the principal be 'x'.

So, Amount = 3x

Simple interest = 2x

Simple interest = (P × R × T)/100

⇒ 2x = (x × R × 10)/100

⇒ R = (2 × 100)/10

⇒ R = 20%

∴ The annual rate of interest is 20%.

Given data:

Time period t = 4 years 

After the four years  Amount = 2p

Formulae used :

Amount = p (1 + r/100 )^t

Calculations:

Let the principal be 'p'

Rate percent be 'r'

2p = p (1 + r/100)⁴

⇒ 2 = (1 + r/100)⁴

⇒ 2^1/4 = 1 + r/100

⇒ 2^1/4 = (100 + r)/100

⇒ 1.19 × 100  = 100 + r

⇒ 119 = 100 + r

⇒ r = 119 - 100

⇒ r = 19 %

∴ The rate of interest is 19%

Given:

Amount (A₁) = Rs.3136

Time = 2years

Amount (A₂) = Rs.3512.32

Time = 3 years

By using Ratio method

Rate = 12% = 3/25

25               ∶          28

25               ∶          28

25               ∶          27                       (for 3^rd year = (3/25) × (2/3) = 2/25)

15625        ∶         21168

Now,

15625 = 2500

1 = 2500/15625

1 = 0.16

21168 = 0.16 × 21168

⇒ 3386.88

≈ 3387

∴ The New amount will be Rs.3387

Given:

Time = 3 years

Amount ∶ Principal = 7 ∶ 4 

Formula Used:

Interest = Amount – Principal

Simple interest = Principal × Interest Rate × Time /100

Calculation:

Amount/Principal = 7/4

So, Principal = 4x

And, Amount = 7x

Interest = Amount – Principal

⇒ 3x = 7x - 4x

Simple interest = Principal × Interest Rate × Time /100

⇒ 3x = 4x × R × 3/100

⇒ R = 25% p.a.

∴ The rate of interest is 25% p.a.

Given:

A sum of money placed at compound interest doubles itself in 8 years.

Formula used:

A = P(1 + R/100)n

where A = amount 

p = principal

R = rate 

n =  time

Calculations:

Let P be Rs.1, then A = Rs.2

A = P(1 + R/100)n

⇒ 2 =  1(1 + R/100)⁸

⇒ 4 = 1(1 + R/100)¹⁶ (by squaring both sides)

comparing this to A = P(1 + R/100)n 

we get n =  16 years

∴ the correct answer is 16 years.

Given:

Time = 6 years

Formula used:

SI = (P × R × T)/100

Calculation:

The sum becomes 15 times of itself. So, the SI = 15P - P = 14P

⇒ 14P = (P × R × 6)/100

⇒ R = 233.33%

Given:

Principal = Rs. 3000

Rate of interest = 5%

Time = 2 years

Formula used:

S.I = PRT/100

Calculation:

S.I = Rs. (3000 × 5 × 2)/100

⇒ Rs. 300

∴ The required simple interest is Rs. 300

Solution:

Formula Used:

SI = Amount – Principal

SI = P × T × R /100

Calculations:

Let the principal be Rs. x.

Amount = Rs. 132

SI = 132 – x      ----(1)

Also SI = x × 2 × 5 /100      ----(2)

Equating equation (1) and (2)

⇒ x × 2 × 5 /100 = 132 – x

⇒ 11x = 1320

⇒ x = 120

If the rate becomes two times i.e. 10% per annum,

⇒ SI = 120 × 3.5 × 10 /100

⇒ SI = 42

⇒ Amount = 120 + 42 = Rs. 162

Hence it amounts Rs. 162 at the rate 10% for 3.5 years.

Given: 

Sum of Money becomes 3 times

Rate of simple interest = 5% per annum

Formula Used;

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

Amount =  Principal + Simple interest

Calculation:

Let the principal be Rs A

Then after tripling Amount will become 3A

Simple interest = 3A - A = 2A

2A = (A × 5 × time)/100

⇒ Time = (2A × 100)/(A × 5) = 40 years

∴ The Time period is 40 years.

Given:

Rate of interest = 5%

Time period in years = 3 years

Formula Used:

We know that, SI = (P × R × T)/100

Where, P = Principal

R = rate of interest

T = Time period in years

Calculations:

Let the initial rate be R% and the increased rate is (R + 5)%

According to question;

\(\frac{{{\rm{P}} × 3 × \left( {{\rm{R\;}} + {\rm{\;}}5} \right)}}{{100}} - \frac{{{\rm{P}} × 3 × {\rm{R}}}}{{100}} = 4500\)        ---(1)

P × 3/100 × (R + 5 - R) = 4500

P × 3/100 × (5) = 4500

P = 4500/15 × 100

⇒ P = Rs. 30000

∴ The principal amount is Rs. 30000

Given:

Difference in simple interest of 4 years and 3 years = Rs. 42

Rate of interest = 5%

Formula used:

SI = (P × R × T)/100

Calculation:

Simple interest for 1 year = Difference in simple interest of 4 years and 3 years

⇒ Simple interest for 1 year = Rs, 42

⇒ (P × R × T)/100 = 42

⇒ (P × 5 × 1)/100 = 42

⇒ P = 42 × 100/5 = 42 × 20 = 840

∴ The sum is Rs. 840

Given:

First Amount = Rs. 756

Time = 2 years

Second Amount = Rs. 873

Time = \(3\frac{1}{2}\) years = 7/2 years

Formula Used:

Simple Interest = Amount – Principal 

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

Calculation:

Let the principal be 'x'. 

Simple Interest = Amount - Principal 

⇒ 756 – x

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

Rate = (S.I × 100)/ P × t

⇒ Rate = ((756 – x) × 100)/ (x × 2)

Simple Interest = Amount - Principal 

⇒ 873 – x

Rate = ((873 – x) × 100)/ (x × 7/2)

According to the Question,

((756 – x) × 100)/ (x × 2) = ((873 – x) × 100)/ (x × 7/2)

⇒ ((756 – x) × 100)/2 = 2((873 –- x) × 100)/ 7

⇒ 7((756 – x) × 100)= 4((873 – x) × 100)

⇒ 529200 – 700x = 349200 – 400x

⇒ 300x = 180000

⇒ x = 600

Rate = ((756 – x) × 100)/ (x × 2)

⇒ ((756 – 600) × 100)/(600 × 2)

⇒ (156 × 100)/ 1200

⇒ 13%

∴ The rate of interest per annum is 13%.

Given:

I = 20% of P (S.I.)

N = 4 years

For C.I. P = 5000, N = 2 Years

Formula used:

A = P × {1 + (R / 100)}^N

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

I = PRN / 100

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

A = P - I

Calculation:

Here, I = (P × R × 4) / 100

⇒ 0.20P = 4PR / 100

⇒ 20 = 4 × R

⇒ R = 5%

Now, calculating C.I

⇒ A = 5000 × {1 + (5 / 100)}²

⇒ A = 5000 × (21 × 21) / (20 × 20)

⇒ A = 12.5 × 21 × 21

⇒ A = 5512.5

C.I. = 5512.5 – 5000

⇒ C.I. = 512.5

∴ Required compound interest is Rs. 512.5

Given:

The sum at the same simple interest becomes amount to Rs 457 in 5 years and Rs 574 in 10 years.

Concept Used:

Sum = Amount - Interest

Calculation:

The amount after 5 years is Rs. 457 and sum after 10 years is Rs. 574

The simple interest in 5 years is (574 - 457) = 117

Sum = 457 - 117

⇒ 340

∴ The value of the sum is Rs. 340.

Given:

Principal = Rs. 40,000

Time = 2 years

Rate = 4% per annum

Formula used:

SI = (P × R × T)/100

Where,

SI = Simple interest

R = Rate

T = Time

Calculations:

SI = (P × R × T)/100

⇒ SI = (40,000 × 4 × 2)/100

⇒ SI = 3,200

∴ The simple interest after 2 years is Rs. 3,200

Given 

Rahul invest certain amount for is 3 years 

Shyam invest same amount for is 12 years 

Formula Used 

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

Calculation 

According to question,

⇒ Principal and rate is same 

Simple interest depend on time period, 

Ratio of time = Ratio of Simple interest 

Ratio of time (Rahul : Shyam)

⇒ 3 : 12 = 1 : 4 

Ratio of Simple interest (Rahul : Shyam)

⇒ 1 : 4 

∴ The Ratio of Simple interest  for Rahul : Shyam is 1 : 4 

Given - 

principal = Rs. 9,000

Invested in three parts at 3%, 4% and 6% on simple interest 

Formula used - 

S.I = (principal × rate × time)/100

Solution - 

Let the amount be x, y and z respectively.

⇒ s.i = (x × 3 × 1/100) = (y × 4 × 1/100) = (z × 6 × 1/100)

⇒ x : y : z = (1/3) : (1/4) : (1/6)

⇒ x : y : z = 4 : 3 : 2

⇒ the amount invested at 6% = (2 × 9000/9) = Rs. 2000

∴ amount invested at 6% is Rs, 2.000.

Given:

⇒ The investment ratio of Sumit and Sanket =  5  :   6      ----(1)

⇒ The ratio of rate offered = 6  :  11      ----(2)

⇒ The interest earned by Sumit = Interest earned by Sanket – Rs. 3600      ----(3)

Formula used:

⇒ Simple interest = (Principle × Rate × Time)/100      ----(4)

Calculation:

As the interest was simple interest, taking the ratio of their interest we get,                     

                             Sumit                :             Sanket

⇒ Interest       (5×6×2)/100           :         (6×11×2)/100

⇒ Interest                   5y               :                 11y

So the difference between their interests is 6y and this equals the given interest of Rs. 3600

⇒ 6y = Rs. 3600

⇒ 1y = Rs. 600

⇒ So, the interest earned by Sumit = 5 × 600 = Rs. 3000 

Given:

Sum = Rs.2800

Time = 2 years

Rate of interest = 15%

Formula used:

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

Calculations:

CI = 2800[(1 + (15/100))² - 1]

⇒ 2800 × [((100 + 15)/100)² - 1] = 2800 × [(115/100)² - 1]

⇒ 2800(1.3225 - 1) = 2800(0.3225)

⇒ 903

∴ The compound interest = 903

Given:

Principal = Rs. 8000 

Rate = 5%

Time = 3 years.

Formula used:

C.I = P[(1 + R/100)N – 1]

Calculation:

C.I = P[(1 + R/100)^N – 1]

⇒ C.I = 8000[(1 + 5/100)³ – 1]

⇒ C.I = 8000 × [(21/20)³  – 1]

⇒ C.I = 8000 × [(9261 – 8000)/8000]

⇒ C.I = (8000 × 1261/8000)

⇒ C.I = Rs. 1261

∴  The compound interest is  Rs. 1261

Given:

Rate = 10% per annum

Maturity Amount after one year = ₹ 13,230

Interest is compounded half yearly

Concept Used:

For half yearly compound interest just double the time and half the rate then the question is general compound interest question.

Formula Used:

Amount = Principal[1 + Rate/100]^Time

Calculation: 

Rate = 10%/2 = 5% 

Time = 1 × 2 = 2 year

Amount = Principal[1 + Rate/100]Time

⇒ ₹ 13,230 = Principal[1 + 5/100]²

⇒ ₹ 13,230 = Principal[1 +1/20]²

⇒ ₹ 13,230 = Principal[21/20]²

⇒ ₹ 13,230 = (441/400)Principal

⇒ Principal = ₹ (13230 × 400)/441

⇒ Principal = ₹ 5292000/441

⇒ Principal = ₹ 12,000

∴ The sum is ₹ 12,000

Given:

Rahul lent a certain amount of money to his friend at the rate of 6% per annum for 1^st 5 years simple interest

8% per annum simple interest for next 7 years

12% per annum simple interest for the period beyond 12 years

Formula used:

Simple Interest = {Sum (P) × rate of interest (r) × time (t)}/100

Calculation:

Let the certain amount of money be Rs. x

S.I at 6% rate for 5 years = 30 × x/100

S.I at 8% rate for 7 years = 56 × x/100

S.I at 12% rate for (15 – 12) years i.e. for 3 years = 36 × x/100

According to the question

⇒ 30 × x/100 + 56 × x/100 + 36 × x/100 = 8540

⇒ 122 × x = 854000

⇒ x = 7000

∴ The amount of money Rahul lent to his friend was Rs. 7000

Given:

Two friends Amit and Rahul invested an equal amount of sum Rs. P in two different schemes.

Simple interest offered by the two schemes were at the rate 9% and 6% respectively and at the end of 2 years

Difference between the total amount received by Amit and Rahul was Rs. 2166.

Formula used:

Simple Interest = {Sum (P) × rate of interest (r) × time (t)}/100

Amount (A) = Principal (P) + Simple interest (S.I)

Calculation:

After 2 years, Rahul’s amount from first scheme

⇒ P + (P × 9 × 2)/100 = 118P/100

Rahul’s amount from second scheme

⇒ P + (P × 6 × 2)/100 = 112P/100

Difference between the amounts received by Rahul from both schemes

118P/100 – 112P/100 = 2166

⇒ 6P = 2166 × 100

⇒ P = 36100

∴ The value of P is Rs. 36100