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GIVEN: Principal = Rs. 720

Amount after 3/2 years = Rs. 882

We KNOW that,

A = P + (P × R × T)/100

⇒ 882 = 720 + (720 × R × 3/2)/100

⇒ 16200 = 720 × R × 3/2

⇒ R = 16200/(720 × 1.5)

⇒ R = 15%

New Principal = Rs. 800, new Amount = Rs. 1040

⇒ 1040 = 800 + (800 × 15 × t)/100

⇒ 24000 = 800 × 15 × t

⇒ t = 2 years

∴ the correct option is 4)

Principal = 2500 & Amount = 8100

As we KNOW, A = P (1 + r/100)²

⇒ (1 + r/100)² = 81/25

⇒ 1 + r/100 = 9/5

⇒ r = 80%

Amount after 4 years

A = 2500 (1 + 80/100)⁴

⇒ A = 2500 × 9/5 × 9/5 × 9/5 × 9/5

Let the principal be Rs. P.

Simple INTEREST = (Principal × Rate × Time)/100 = (P × 10 × 2)/100 = 0.2P = 4000

∴ P = Rs. 20000

C’I at the same rate for same period = Principal{(1 + Rate%)^Time - 1}

= 20000 {(1 + 0.1)² - 1} = 20000{1.21 - 1} = 4200

Difference in interests = Rs. (4200 - 4000) = Rs. 200

On a SUM of Rs. P at a RATE of R% p.a. after n years,

COMPOUND interest = P[(1 + R/100)ⁿ - 1]

LET the sum INVESTED be Rs. P

⇒ CI = 4160 = P[(1 + 8/100)² - 1]

⇒ 4160 = P[0.1664]

⇒ P = 4160/0.1664 = Rs. 25000

∴ The sum invested is Rs. 25000

AMOUNT under Simple Interest (A) = P [1 + (TR/100)]----(1)

Given Principal (P) = Rs. 1200, Amount (A) = Rs.1740, TIME (T) = 3 years

Substituting all the values in Equation 1, we get

1740 = 1200 [1 + (3R/100)]

⇒ [1 + (3R/100)] = (1740/1200)

⇒ (3R/100) = 1.45 - 1

⇒ Rate of Interest (R) = (0.45 × 100)/3 = 45/3 = 15%

Rate of Interest after an increase of 3% = 15% + 3% = 18%

Amount at the NEW rate of Interest = 1200 [1 + (3 × 18)/100)]

⇒ 1200 [1 + (54/100)] = 1200 × (1 + 0.54) = 1200 × (1.54) = Rs. 1848

$(A = P{\left( {1 + \frac{R}{{n \times 100}}} \RIGHT)^{n \times t}})$ 

Where, A is AMOUNT, P is sum, r is RATE%, t is time in years, n is number of times INTEREST is compounded annually.

Given, certain sum, invested with interest of 12% per annum, compounded semiannually amounts to Rs. 7022.50 at the end of 1 year

$(\Rightarrow 7022.50 = P{\left( {1 + \frac{{12}}{{200}}} \right)^2})$ 

⇒ 7022.50 = 1.06² P

⇒ P = Rs. 6250 

Time = 1 + 73/365 = 1 + 1/5 = 6/5 yrs.

Now,

Simple INTEREST = (Principal × RATE × time)/100 = (12500 × 8 × 6/5)/100 = Rs. 1200

For annually, principal (P) = Rs. 25000, rate (R) = 20% and time (T) = 1 year

For semi-annually, principal (P) = Rs. 25000, semi-annually rate (R) = 20%/2 = 10%

And time (T) for semi-annually = 1 × 2 = 2 half year

COMPOUND interest (CI, annually) = 25000(1 + 20/100) - 25000

⇒ CI (annually) = Rs. 5000

And Compound interest (CI, semi-annually) = 25000(1 + 10/100)² - 25000

⇒ CI (semi-annually) = (25000 × 11/10 × 11/10) - 25000

⇒ CI (semi-annually) = Rs. 5250

⇒ Let, Principal be RS. x then Amount = Rs. 8x, Rate = r% and TIME (n) = 3 years.

⇒ A = P × [1+ (r/100)]ⁿ

⇒ 8x = x × [1 +(r/100)]³

⇒ 2³ = [1+ (r/100)]³

⇒ 2 = 1 + (r/100) ------ 1

⇒ Amount becomes 16x then

⇒ A = P × [1+ (r/100)]ⁿ

⇒ 32X = x × [1+ (r/100)]ⁿ

⇒ putting the value from equation 1 we get

⇒ 32 = 2ⁿ

⇒ 2⁵ = 2ⁿ

Let M be the sum of money.

Simple interest, SI = P × R × T/100

where

P = principal amount

R = rate of interest and

T = time.

? TOTAL interest = Rs. 624

$(\Rightarrow \frac{{\frac{{2M}}{3} \times 6 \times 6}}{{100}} + \frac{{\frac{M}{3} \times 4 \times 6}}{{100}} = 624 \Rightarrow M = 1950)$

A amount is DOUBLED when it GETS the same amount as interest.

So if we have to MAKE money 16 times then its interest would be 15 times.

Time taken to 1 time interest = 5 YEARS

<P class="MsoNormal" style="margin-left:0in;text-indent:0in;mso-list:none">Let the sum be Rs. ‘P’

Given, R = 12%, Time = 2 YEARS

On a sum of Rs. P at R% per annum in n years,

When compounded annually, CI = P[(1 + R/100)ⁿ - 1]

⇒$ CI = P × [(1 + 12/100)² - 1] = 0.2544P

?$ Simple interest = (PRINCIPAL × Rate × Time)/100

⇒$ SI = (P × 12 × 2)/100 = 0.24P

Now,

Difference = CI - SI = Rs. 360

⇒$ 0.2544P - 0.24P = 360

⇒$ P = 360/0.0144 = Rs. 25000

Let the sum DEPOSITED at 8% interest be Rs. x

Snehashish deposited two parts of a sum of Rs. 28000 in different cooperative banks at the rates of 8% per annum and 14% per annum respectively.

So, the sum deposited at 14% interest = Rs. (28000 – x)

In one year he got Rs. 3200 as the total interest.

So, we can write now,

(x × 1 × 8%) + [(28000 – x) × 1 × 14%] = 3200

⇒ 8x + 392000 – 14X = 320000

⇒ 6x = 72000

⇒ x = 12000

We know that, FORMULA for SIMPLE interest:

$(S.I. = \frac{{P \times R \times T}}{{100}})$

Where,

S.I. = Simple Interest

P= principal

T = Time

R = Rate of interest

Interest PAYABLE to A = 3500 × (10/100) × 3 = 1050

Interest payable to B = 3500 × (11.5/100) × 3 = 1207.5

The basic amount in both transactions is same. So, difference between interests in two transactions is the gain.

LET the rate of interest be r% and principle be y

Interest can be given as 3y – y = 2y

5 YEARS 4 MONTHS can be written as 5 + 4/12 years = 16/3 years, using this in simple interest formula

Using simple interest formula

⇒ 2y = (y × r × 16/3) /100

⇒ r = 600/16 = 37.5

∴ The rate of interest is 37.5%

As per the given DATA,

Principal = Rs. 13500

Rate R = 12.5%

Time T = 2 years

We know that simple INTEREST (SI) = (P × T × R)/100

⇒ SI = (13500 × 12.5 × 2)/100 = Rs. 3375

We know that Amount = Simple interest + principal

⇒ Amount = 3375 + 13500 = Rs. 16875

∴ MATURITY VALUE = Rs. 16,875

Solution : 

Let 'x' be the part of money which is lent at 24% rate of interest and '(17500 -x)' be the remaining part which is lent at 10% rate of interest.

GIVEN: N = 5years

Simple Interest 1 = a = (5 × x × 24)/100 

a = 120x/100 = 1.2x

Simple Interest 2 = b = (5 × 17500-x × 10)/100 

b = (17500-x)/2 = 8750 - x/2 

Given: a + b = 13300

1.2x + 8750 - x/2 = 13300

(2.4x -x)/2 = 13300 - 8750 

1.4x /2 = 4550

1.4x = 9100

x = 9100/1.4

x = Rs.6500

(17500 - 6500) = 11,000

x : (17500 -x) = 6500 /11000

= 13:22

So, the correct option is 4). 13:22

LET the principal amount be ‘X’

∴ According to the 1^st given CONDITION,

⇒ 1000 = (x × 4 × 5)/100

∴ x = 5000

Let the PRINCIPAL or sum be Rs. x

3 installments of Rs. 300 for 3 years give Amount = x(1 + 25/100)³

⇒ x(1 + 25/100)³ = 300(1 + 25/100)² + 300(1 + 25/100)¹ + 300(1 + 25/100)⁰

⇒ x(1.25)³ = 300(1.25)² + 300(1.25) + 300

⇒ x = 1143.75/(1.25)³

Given,

Time period, T = 15 years

Rate of interest for JATIN, R₁ = 13%

Rate of interest for Gabru, R₂ = 15%

Let the original amount of loan TAKEN by Jatin from the bank be Rs. x.

We KNOW the FORMULA for simple interest-

SI = (P × T × R)/100

Where,

P = Principal amount

T = Time period

R = Rate of interest

Profit = Rs. 1470

⇒ (P × T × R₂)/100 – (P × T × R₁)/100 = 1470

⇒ (x × 15 × 15/100) – (x × 13 × 15/100) = 1470

⇒ x = Rs. 4900

A bank offers 10% COMPOUND interest per half year. A customer deposits Rs. 2000 each on 1^st January and 1^st July of a year.

Compound interest for FIRST 6 months = Rs. 2000 × [1 + (10/100)] = Rs. 2200

So, interest = Rs. 2200 - 2000 = Rs. 200

Principal amount for next 6 months = Rs. 2000 + 2200 = Rs. 4200

So, the compound interest for LAST 6 months = Rs. 4200 × [1 + (10/100)] = Rs. 4620

So, interest = Rs. 4620 - 4200 = Rs. 420

<P>SIMPLE INTEREST = PRT/100

Here, R = 5%

T = 4 years

Total AMOUNT = P + I = 12000

⇒ I = 12000 - P

⇒ I = P × 5 × 4/100

⇒ 12000 - P = 20P/100

⇒ 12000 – P = 0.2P

⇒ 12000 = 1.2P

∴ P = 12000/1.2 = Rs. 10,000

S.I = (P × R × T)/100

Where, P = principal

R = rate of interest

T = TIME in years

When T = 15, S.I = 9P – P = 8P

⇒ 8P = (P × R × 15)/100

⇒ R = 800/15 = 160/3

When P becomes 33 times that is 33P, S.I = 33P – P = 32P, T = ?

⇒ 32P = (P × 160/3 × T)/100

We know that:

Amount:

$(A = \LEFT[ {P{{\left( {1 + \frac{r}{{100}}} \right)}^t}} \right])$ 

Where, A = amount INCLUDING interest

P = principal

r = rate of interest

t = time period

Now, ACCORDING to the question,

Amount after 7^th year-

$(= P{\left( {1 + \frac{{15}}{{100}}} \right)^7})$ 

Amount after 8^th year-

$(= P{\left( {1 + \frac{{15}}{{100}}} \right)^8} = P\left[ {{{\left( {1 + \frac{{15}}{{100}}} \right)}^7} \times {{\left( {1 + \frac{{15}}{{100}}} \right)}^1}} \right])$ 

∴ required ratio-

$(= \frac{{P{{\left( {1 + \frac{{15}}{{100}}} \right)}^7}}}{{P\left[ {{{\left( {1 + \frac{{15}}{{100}}} \right)}^7} \times {{\left( {1 + \frac{{15}}{{100}}} \right)}^1}} \right]}} = \frac{1}{{1 + \frac{{15}}{{100}}}} = \frac{{100}}{{115}} = \frac{{20}}{{23}} = 20:23)$ 

<P>S.I = AMOUNT – Principal

= 75375 – 45000

= RS 30375

$(S.I.\; = \frac{{P \times R \times T}}{{100}})$

Where,

S.I. = Simple Interest

P= Principal

T = Time

R = Rate of interest

∴ $(30375\; = \frac{{45000 \times R \times 9}}{{100}})$

⇒ $(R = \frac{{30375 \times 100}}{{45000 \times 9}})$

A bank offers 5% compound interest per half year. A customer deposits RS. 4800 each on 1^st JANUARY and 1^st July of a year.

Compound interest for first 6 months = Rs. 4800 × [(1 + (5/200)) - 1] = Rs. 120

Principal amount for next 6 months = Rs. 4800 + 4800 + 120 = Rs. 9720

So, the compound interest for last 6 months = Rs. 9720 × [(1 + (5/200)) - 1] = Rs. 243

⇒ Students between 20 and 30 = 55 – 42 = 13

⇒ Students between 30 and 40 = 42 – 26 = 16

Let the rate of interest be r%.

SIMPLE interest earned on RS. 25000 in 2 years = 25000 × 0.16 = Rs. 4000

⇒ 4000 = (25000 × r × 2)/100

⇒ r = 8%

∴ COMPOUND interest in 2 years = 25000 × [1 + 8/100]² - 25000

Given that, rate of compound interest = 25% and amount of the FIRST installment is Rs.X

We know that, COMPOUNDED amount,

A = P[1 + r/100]ⁿ

Here A = Amount, P = Principal, r = Rate

P = x/[1 + 25/100]¹

⇒ $(A\; = \;P{\left. {\left( {1\; + \;\frac{r}{{100}}} \right.} \right)^t})$

where

P = principal 

r = rate of INTEREST

t = time

⇒ A = 5000(1 + 4/100)²

On solving we get,

⇒ A = 5408

⇒ CI = A - P = 5408 - 5000 = Rs. 408

⇒ SI = CI/2

⇒ Simple interest = 408/2 = Rs. 204

⇒ SI = (p × R × T)/100

p = pricipal

R = rate of interest

T = time

On solving we get

⇒ 204 = (p × 4 × 3)/100 

⇒ P = Rs. 1700

Here we will use the usual METHOD of computing interest rates.

AMOUNT = 72000{1 + (8/100)}{1 + (9/100) }{1 + (10/100) }

On computing we will get the amount Rs. 93234.240.

∴ The amount OBTAINED at the end of 3 YEARS = Rs. 93234.240

Let n be NUMBER of half years

Amount = 8000 + 2648 = 10648

From the problem statement

⇒ 10648 = 8000(1 + 0.2/2)ⁿ

⇒ 1.331 = (1.1)ⁿ

⇒ n = 3

Simple interest for 3^rd year, the duration is only 1 year

Let the rate of interest be 100r

⇒ Simple interest = (PRINCIPAL × rate × 1)/100

⇒ 1750 = Principal × r----(1)

Also,

⇒ Compound interest = Principal × [(1 + Rate/100)^TIME - 1]

⇒ 3622.5 = Principal × [(1 + r)² - 1]

⇒ 3622.5 = Principal × [r² + 2r]

⇒ 3622.5 = Principal × r × (r + 2)----(2)

Dividing equation (2) by (1), we GET,

⇒ 3622.5/1750 = r + 2

⇒ r = 0.07

The formula for amount payable at annual compound interest, INCLUDING principal sum, is: $(A = P{\left( {1 + \frac{r}{{100 \times N}}} \right)^{nt}})$

Where:

A = Amount

P = the principal INVESTMENT amount (the initial deposit or loan amount)

r = the annual interest rate (decimal)

n = the NUMBER of times that interest is compounded PER year

t = the number of years the money is invested or borrowed for

Here n = 1

Let the rate be r%

2508.8 = 2000 (1 + r/100)²

A = P[1 + (R/100)]ⁿ, where A = AMOUNT after ADDING compound Interest, R = rate of interest and n = NUMBER of years

⇒ 12321 = 10000[1 + (R/100)]²

⇒ (111/100)² = [1 + (R/100)]²

⇒ [1 + (11/100)]² = [1 + (R/100)]²

⇒ R = 11%

∴ Rate of compound interest = 11%

<P>Rs. 1800 is INVESTED at 4% per annum simple interest.

The interest is added to the PRINCIPAL after every 5yr.

Formula of Simple interest = (P × R × T)/100 where P is the principal, R is the RATE of interest and T is the TIME period.

So, the simple interest in 5 years

= Rs. 1800 × 5 × (4/100) = Rs. 360.

New principal amount = Rs. 1800 + 360 = Rs. 2160

So, the interest in next duration = Rs. 2448 - 2160 = Rs. 288

Let the x years needed to have interest Rs. 288.

Now we can write,

2160 × x × (4/100) = 288

⇒ x = 28800/8640

⇒ x = 10/3

Let the principle be P.

Compounded AMOUNT = P{1 + (11/100)}² = 1.2321P

Compound interest = 1.2321P – P = 0.2321P

According to the question,

⇒ 0.2321P = 1160.5

⇒ P = 5000

Now,

Let ‘x’ be the amount Rahul borrowed

∴ Interest paid by Rahul after 2 years = (x × 2 × R)

∴ According to the given condition,

⇒ 2xr = 4000

∴ XR = 2000----(1)

Interest to be paid by Ramesh

According to the given conditions,

$(x \times \left( {1 + 2r + {r^2} - 1} \right) = 4200)$

$(\therefore xr \times \left( {r + 2} \right) = 4200)$----(2)

From equations (1) and (2) we GET.

(r + 2) = 4200/2000

∴ r = 0.1

Here, For the FIRST YEAR: Principal, p = RS. 20000; time period, t = 1 year; r = 10%

∴ Amount at the END of first year = p + prt/100 = 20000 + (20000 × 10 × 1)/100 = Rs. 22000

This will be principal amount for the second year, where r = 12%

∴ Amount at the end of second year = p + prt/100 = 22000 + (22000 × 12 × 1)/100 = Rs. 24640

GIVEN, PRINCIPAL P = RS. 10,000 and Time period, T= 2 years

Let the rate of INTEREST per annum be R%.

Formula used:

Simple interest, S.I = P × R × T

Compound interest, C.I = P × [(1 + R)^T - 1]

∴ S.I = 10000 × R × 2

= Rs. 20000R

And, C.I = 10000 × [(1 + R)² – 1]

= Rs. (10000R² + 20000R)

Given that, C.I – S.I = Rs. 25

∴ (10000R² + 20000R) – 20000R = 25

⇒ R² = 0.0025

⇒ R = 0.05

∴ R% = 0.05 × 100= 5%

GIVEN that principal P = Rs. 49,780

Rate R = 7%

NUMBER of investments = 3

⇒ Value of each INSTALLMENT = $(\FRAC{P}{{\left( {\frac{{100}}{{100\; + \;R}}} \right)\; + \;{{(\frac{{100}}{{100\; + \;R}})}^2}\; + \;{{\left( {\frac{{100}}{{100\; + \;R}}} \right)}^3}}}\; = \;\frac{{49,780}}{{\left( {\frac{{100}}{{100\; + \;7}}} \right)\; + \;{{(\frac{{100}}{{100\; + \;7}})}^2}\; + \;{{\left( {\frac{{100}}{{100\; + \;7}}} \right)}^3}}})$ = 49,780/2.62 = Rs. 19,000

Let the number be xy

x × y = 18----(1)

⇒ 10X + y + 63 = 10y + x

⇒ 9x + 63 = 9y

x + 7 = y----(2)

From (1) and (2) we get

⇒ x(x + 7) = 18

⇒ x² + 7X – 18 = 0

(x + 9)(x – 2) = 0 ⇒ x = 2 as it can’t be negative

⇒ 2Y = 18 ⇒ y = 9

<P>LET the PRINCIPAL be P

⇒ A = P(1 + r/100n)^nt

⇒ 7727.104 = P(1 + 12/100)³

⇒ P = 7727.104/(1 + 12/100)³

⇒ P = 7727.104/(1.12)³

<P>? SIMPLE INTEREST = (Principal × Rate × Time)/100

Let the principal and the rate be Rs. P and R% respectively

Simple interest after 3 years = 3PR/100 = 840

⇒ PR/100 = 840/3 = 280

We KNOW that, Compounded amount, $(A\; = \;P{\left[ {1 + \FRAC{r}{{100}}} \right]^t})$

Where, P = Principal amount, r = % rate of interest, t = time in years

Given, a sum of money compounded annually comes 1.96 times of itself in 2 years

$(P{\left[ {1 + \frac{r}{{100}}} \right]^2} = 1.96\;P)$ 

$(\RIGHTARROW {\left[ {1 + \frac{r}{{100}}} \right]^2} = 1.96)$ 

$(\Rightarrow \frac{{100\; + \;r}}{{100}}\; = \;1.4)$ 

⇒ 100 + r = 140

<P>We KNOW the formula for simple interest -

$(\Rightarrow \;S.I.\; = \frac{{P\; \times \;R\; \times \;T}}{{100}})$

SI = Simple interest

P = Principal

R = Rate of interest

T = Time period

P = Rs. 5000

Amount at the end of first year = 5000+(5000 × 10 × 1)/100 = Rs. 5500

Amount he paid back = Rs. 1500

For the second year:

P = 5500 – 1500 = Rs. 4000

Amount = 4000+(4000 × 10 × 1)/100 = Rs. 4400

For the third year

P = 4400 – 1500 = Rs. 2900

Amount payable at the end of third year = 2900+(2900 × 10 × 1)/100 = Rs. 3190

Formula for compound INTEREST:-

$(A = P{\left( {1 + \FRAC{R}{{100}}} \right)^T})$, CI = A – P

Where CI is compound interest, A is amount, P is principal, R is rate, T is time

Given, Sum of money BECOMES 8 times in 3 years.

∴ A = 8P

$(\Rightarrow 8{\rm{P}} = P{\left( {1 + \frac{R}{{100}}} \right)^3})$

$(\Rightarrow {\rm{\;}}8 = {\left( {1 + \frac{R}{{100}}} \right)^3})$

$(\Rightarrow {2^3} = {\left( {1 + \frac{R}{{100}}} \right)^3})$

⇒ 2 = 1 + R/100

⇒ R = 100 %

Let the time be T in which sum of money becomes 16 times of itself.

∴ A = 16P

$(\Rightarrow 16{\rm{P\;}} = P{\left( {1 + \frac{R}{{100}}} \right)^T})$

$(\Rightarrow 16{\rm{\;}} = {\left( {1 + \frac{{100}}{{100}}} \right)^T})$

⇒ 2⁴ = 2^T

By property of indices ; a^b = a^c, ⇒ b = c

Let money is Rs. P and RATE is R%.

S.I. = Amount - Money = 77P/50 - P = 27P/50

⇒ 27P/50 = (P × R × 18)/(100 × 5)

The difference between C.I. and S.I. for 2 YEARS = S.I. on the INTEREST of first year

⇒ S.I. for first year = RS. 400

⇒ The S.I. on interest of first year = C.I. – S.I. = 900 – 800 = Rs. 100

⇒ Interest on Rs.400 for 1 year = Rs. 100

⇒ Interest on Rs.100 for 1 year = x

⇒ i.e, x = [(100 × 100)/400] = Rs. 25

⇒ Thus the rate of interest = 25%

$(\Rightarrow {\rm{Principal\;AMOUNT}} = {\rm{}}\frac{{\left( {100 \times {\rm{S}}.{\rm{I}}} \right){\rm{\;}}}}{{RT}})$

⇒ Principal amount = (100 × 800)/(2 × 25)

⇒ Principal amount = Rs. 1600

⇒ S.I. for 3 years = (1600 × 3 × 25)/100

⇒ S.I. for 3 years = Rs. 1200

⇒ C.I. for 3 years = 1600[(1 + 0.25) ³ – 1]

⇒ C.I. for 3 years = Rs. 1525

⇒ Difference between C.I. and S.I. for 3 years = 1525 – 1200 = Rs. 325

<P>Let principal be P and RATE be r

⇒ TIME = r

⇒ r² = 100/7

⇒ r = 3.78%

∴ The rate percentage is 3.78%