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LET the sum BORROWED by Mohan = P

Total SI to be paid by Mohan to the bank = P × 3 × 1/100 = 3P/100 [give SI rate = 3%]

As GIVEN, rate of compound interest = 6% compounded HALF yearly

We can say that, rate of CI is 3% compounded annually

Total amount earned by Mohan by CI = P - P (1 + 3/100)² = P - P (103/100)²

⇒ P - (10609P/10000) = 609P/10000

Profit earned by Mohan = (total amount earned by CI) - (Total SI paid by Mohan to the bank) = 618

⇒ (609P/10000) - (3P/100) = 618

⇒ 309P/10000 = 618

⇒ P = 20000

∴ Amount borrowed by Mohan = Rs. 20000

Let RS. P be the principal.

According to the question,

(P × 23 × 2)/100 + P [(1 + 23/100)² - 1] = 2961

⇒ 0.46P + 0.5129P = 2961

⇒ 0.9729P = 2961

⇒ P = 2961/0.9729

So, SI for 1 YEAR = ((2961/0.9729) × 23 × 1)/100 = Rs. 700

Short TRICK:

Total percentage of SI in 2 years = 23 + 23 = 46%

And total percentage of CI in 2 years = 23 + 23 + (23 × 23)/100 = 51.29

According to question (CI + SI) = 46 + 51.29 = 97.29

⇒ 97.29 unit = 2961

For CI:

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

Where,

A is the amount at the end of time t,

P is the principal,

t is time,

r is rate

Let the annual INSTALLMENT be a.

For 1 YEAR,

$({\rm{a}} = {P_1} \times {\left( {\;1 + \frac{r}{{100}}\;} \right)^1})$

For 2 YEARS

$({\rm{a\;}} = {P_2} \times {\left( {\;1 + \frac{r}{{100}}\;} \right)^2})$

For 3 years

$({\rm{a}} = {P_3} \times {\left( {\;1 + \frac{r}{{100}}\;} \right)^3})$

GIVEN, P1 + P2 + P3 = 63840 and r = 8%

$(\Rightarrow \frac{a}{{1 + \frac{8}{{100}}}} + \frac{a}{{{{\left( {\;1 + \frac{8}{{100}}} \right)}^2}}} + \frac{a}{{\;{{\left( {1 + \frac{8}{{100}}} \right)}^3}}} = {\rm{\;}}63840)$

⇒ a × (2.577 ) = 63840

The interest rate is 10 paise PER RUPEE per month.

⇒ Total interest rate on Rs. 50 per month = 50 × 10 = 500 paise = Rs. 5(? Re. 1 = 100 paise)

⇒ SIMPLE interest on Rs. 50 for 6 MONTHS = 5 × 6 = Rs. 30

LET the sum of money invested be a and rate of interest be r

Time of INVESTMENT = 2 years 6 months = 2.5 years

Simple interest = (PRINCIPLE × time × rate)/100 = (a × r × 2.5)/100 = ar/40

According to the question,

Amount = (13/10) × Sum of money invested

⇒ (ar/40) + a = (13/10) × a

⇒ (r + 40)/40 = 13/10

⇒ r + 40 = 52

For CALCULATING simple interest, Time PERIOD = 2 years, Rate = 5%, simple interest = Rs. 50

Let Principal = P

We know, Simple interest = (P × R × T)/100

50 = (P × 5 × 2)/100

P = Rs. 500

Now, for calculating compound interest, Time period = 2 years, Let the rate = 5%, Principal = Rs. 500

We know, Amount = Principal (1 + Rate/100)ⁿ

And compound interest = Amount - Principal = 500 (1 + 5/100)² - 500 = Rs. 51.25

We know that,

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

Where R = rate

T = no. of years 

n = no. of times interest is compounded

P = principal

C.I = A - P

Where C.I = compound interest,

A = Amount 

Given, the bank offers 15% compound interest per half year. A customer deposits Rs 2400 each on 1st JANUARY and 1st July of a year

Thus, for SUM of money deposited on 1^st January, rate will be compounded TWICE while for sum of money deposited on 1^st July, rate is compounded once 

$(A = 2400{\left( {1 + \frac{{15}}{{200}}} \right)^{1 \times 2}} + 2400\left( {1 + \frac{{15}}{{100}}} \right))$

⇒ A = Rs. 5353.5

C.I = 5353.5 - 2400 - 2400 = Rs. 553.5

Given Sum(Principal) = Rs. 125000

Amount at the end compounded semiannually = Rs. 140608

RATE of interest per annum (r) = 8%

For semiannual compounding, the interest rate will be HALVED = (8/2)% = 4%

From the equation of COMPOUND interest,

Amount = Principal [1 + (r/100)]²ⁿ

⇒ 140608 = 125000(1 + (4/100))²ⁿ

⇒ 140608/125000 = (104/100)²ⁿ

⇒ (52/50)³ = (52/50)²ⁿ

As bases are equal, equalize the powers, n = 3/2 = 1.5

Formula for Amount under Simple Interest (A) = P [1 + (nr/100)]

⇒ A = 125000 [1 + (12/100)]

⇒ A = 125000 (1 + 0.12)

⇒ A = 125000 × 1.12

⇒ A = Rs. 140000

<P>LET the sum of money be Rs. x and the time required to amount to five times itself be t years.

So, the interest in t year should be Rs. 4x.

In case of simple interest, we KNOW,

(P × T × R)/100 = SI

Where, P = PRINCIPAL amount, T = Duration in years, i = Interest rate per year, SI = Total simple interest

Then,

x × t × 16% = 4x

⇒ t × (16/100) = 4

⇒ t = 400/16 = 25

We know the formula for compound interest –

CI = [P{1 + r/100}^t – 1]

Where,

CI = Compound interest

P = Principal

R = Rate of interest

T = TIME period

According to the QUESTION,

⇒ kP – P = [P {1 + r/100}⁶ – 1]

⇒ kP = P {1 + r/100}⁶----- (1)

∴ Total sum BECOMES after 24 year

⇒ P {1 + r/100}²⁴ = P [{1 + r/100}⁶]⁴

Using equation (1)

= Pk⁴

LET the sum = P

⇒ $(4P = P \TIMES {\left( {1 + \frac{R}{{100}}} \right)^2})$

$(\Rightarrow 4 = {\left( {1 + \frac{R}{{100}}} \right)^2})$

⇒ $(2 = 1 + 0.01R \Rightarrow R = 100)$(1)

When sum becomes 16 times:

⇒ $(16P = P \times {\left( {1 + \frac{R}{{100}}} \right)^n})$

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

⇒ $({2^4} = {2^n})$

Let us assume money invested on scheme X, Y and Z be Rs. x, y and z respectively

Also assuming interest GAIN by Dharmesh from each scheme be α

We know the formula for simple Interest,

SI = (P × R × T)/100

Where,

SI = Simple Interest

P = Principle

R = Rate of Interest

t = Time period

∴ Interest gain from scheme X = (x × 4 × 7)/100 = α

⇒ x = 100α/28

Similarly Interest gain from scheme Y = (y × 6 × 4)/100 = α

⇒ y = 100α/24

And Interest gain from scheme Z = (z × 7 × 3)/100 = α

⇒ z = 100α/21

∴ x : y : z = 100α/28 : 100α/24 : 100α/21

We know that,

A : B : C = nA : nB : nC

Where n is a constant

∴ By MULTIPLYING Right hand side RATIO by 42/25 we GET,

⇒ x : y : z = 6 : 7 : 8

Let the PRINCIPAL = Rs. 100

Now, S.I. on Rs. 100 = PTR/100

$(\Rightarrow \frac{{100 \times 5 \times 3}}{{100}} = {\rm{Rs}}.{\rm{}}15)$

Thus, S.I. on Rs. 100 = Rs. 15

$(\begin{ARRAY}{l} \Rightarrow {\rm{Now}},{\rm{\;C}}.{\rm{I}}.{\rm{\;on\;Rs}}.{\rm{\;}}100 = {\rm{P}}{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)^{\rm{t}}} - {\rm{P}}\\ \Rightarrow 100{\left( {1 + \frac{5}{{100}}} \right)^3} - 100 = {\rm{Rs}}.{\rm{\;}}15.7625 \end{array})$

⇒ C.I. on Rs. 100 = Rs. 15.7625

Now, the difference of C.I. and S.I. = 15.7625 – 15 = 0.7625

⇒ For Rs. 100 the difference in C.I. and S.I. is Rs. 0.7625

⇒ For Rs. X the difference in C.I. and S.I. is Rs. 366

$(\Rightarrow {\rm{X}} = \frac{{100 \times 366}}{{0.7625}} = {\rm{Rs}}.{\rm{\;}}48000)$

∴ On Rs. 48000, the difference of compound and simple interest, shall come out to be Rs. 366

Now formula for simple interest can be given as

⇒ SI = (P × T × R)/100

Where,

SI = Simple Interest

P = Principal amount

T = Time PERIOD

R = Rate of Interest

Now as interest is (1/36)^TH of the principal and the rate of interest and time period are equal the formula will become

⇒ P/36 = (P × R × R)/100

⇒ R² = 100/36 = 10²/6²

$(\begin{array}{L} \Rightarrow {\rm{R}} = 10/6 = 5/3 = 1\frac{2}{3}\% \\\therefore {\rm{Rate\;of\;interest}} = 1\frac{2}{3}\%\END{array})$

Let the savings in the banks A and B be X and y respectively

∴ According to the given CONDITIONS,

⇒ (x × 0.5 × r1) = (y × 0.5 × r2)

$(\therefore \FRAC{x}{y} = \frac{{r2}}{{r1}} = \frac{4}{5})$

Let P = principle, N = time and R = rate percent per annum

Then,

Simple interest = (P × N × R)/100

Given,

An AMOUNT becomes 9 times in 32 years when kept in a scheme of simple interest.

⇒ Amount = S.I + Principle

⇒ 9P = (P × 32 × R) /100 + P

⇒ 900P = 32PR + 100P

⇒ 800 = 32R

⇒ R = 800/32

⇒ R = 25%

Then,

Amount = S.I + Principle

⇒ 19P = (P × N × 25)/100 + P

⇒ 19P = NP/4 + P

⇒ 19 = N/4 + 1

⇒ 76 = N + 4

⇒ N = 76 - 4

⇒ N = 72

Principal Amount = Rs. 2000

Amount = Rs. 2420

Number of years = 2

Amount = P(1 + R/100)^t, where P is the principal, r is the rate of INTEREST and t is the time period

2420 = 2000 (1 + r/100)²

1.21 = 1.0201r²

$(\therefore r = \sqrt {\frac{{1.21}}{{1.0201}}} \approx 10\%)$

Let the money LENT be Rs. X

Now,

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

And,

SIMPLE Interest = (P × r × n)/100

Where, P = Principal Amount

r = RATE of interest (per annum)

n = Number of years

Hence,

X × [1 + 12.5/100]⁴ - X = 1200

X = 1994

Now,

Simple Interest = (1994 × 12.5 × 4)/100 = 997

Given, present height of the tree = 64 cm

The height of the tree increases by 1/8 th every year.

After 1^st year

Height of tree = 64 + 1/8 of 64

⇒ Height of tree = 64 + 8 = 72 cm

After 2^nd year

Height of tree = 72 + 1/8 of 72

⇒ Height of tree 72 + 9 = 81 cm

Another approach, it can be considered a CI problem:

PRINCIPAL (initial height) = 64 cm, rate = 1/8 × 100 % = 12.5 %, t= 2 years, A= final height

For CI:

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

Where,

A is the length at the end of time t,

P is the height at the starting of the time,

t is time in years.

r is growth rate in percent

⇒ Final height = $(64{\left( {1 + \frac{{12.5}}{{100}}} \right)^2})$

⇒ Final height = 64 × 1.125²

⇒ Final height = 64 × 1.265625 = 81 cm.

As we know-

$(CI\; = \;\left[ {P \times {{\left( {1 + \frac{R}{{100}}} \right)}^T}} \right] - P)$

And SI = (P × R × T)/100

Where, P = principal AMOUNT, R = % rate of interest, T = time period in years, CI = compound interest and SI = simple interest

Here, let the principal amount be Rs. P.

Rate of interest = 4% PER annum

Total time = 2 years

As per given data-

Difference between the compound interest and simple interest is Rs. 1

$(\begin{array}{l} \Rightarrow \;\left[ {P \times {{\left( {1 + \frac{4}{{100}}} \right)}^2}} \right] - P\; - \;\frac{{P \times \;4 \times \;2}}{{100}}\; = \;1\\ \Rightarrow \frac{{P \times 104 \times 104}}{{100 \times 100}} - P - \frac{{8P}}{{100}} = 1\\ \Rightarrow \frac{{{{104}^2}P - {{100}^2}P - 800P}}{{100 \times 100}} = 1\;\end{array})$

⇒ 16P = 10000

⇒ P = 10000/16 = 625

<P>Let us assume MONEY lent out to the first person be = Rs. X

So money lent out to second person will be = Rs. (5075 – X)

We know the formula for simple Interest,

SI = (P × R × T)/100

Where,

SI = Simple Interest

P = Principle

R = Rate of Interest

t = Time period

∴ Interest paid by first person = (X × 12 × 5)/100

? = 60X/100

And interest paid by second person = [(5075 – X) × 8 × 7]/100

⇒ ? = [5075 × 56]/100 – [X × 56]/100

Given that,

Interest paid by first person = Interest paid by second person

60X/100 = [5075 × 56]/100 – [X × 56]/100

⇒ 60X = 284200 – 56X

⇒ 60X + 56X = 284200

⇒ 116X = 284200

⇒ X = 2450

Hence the sum lent out at 12% is Rs. 2450

<P>C = P[(1 + R/100)ⁿ - 1]

S = P + C

Where

C = COMPOUND interest

P = Principal

r = rate

n = number of period

S = sum after n periods

In 2 YEARS,

P = 1000

S = 1210

⇒ P + C = 1210

⇒ 1000 + C = 1210

⇒ P[(1 + r/100)ⁿ - 1] = 210

⇒ 1000 × [(1 + r/100)² – 1] = 210

⇒ (1 + r/100)² - 1= 0.210

⇒ (1 + r/100)² = 1.21

⇒ r/100 = (1.21)^1/2 - 1 = 1.1 - 1 = 0.1

⇒ r = 10%

Amount after t years, when interest is compounded ANNUALLY = $({\rm{P}}{\left( {1 + {\rm{R}}/100} \right)^{\rm{t}}})$

Amount $(= 20480{\left( {1 + \frac{{25}}{{4{\rm{\;}} \times 100}}} \right)^2}{\rm{\;}}\left( {1 + \frac{{73}}{{365}} \times \frac{{25}}{{4{\rm{\;}} \times 100}}} \right) = {\rm{Rs}}.{\rm{\;}}23409)$

T = $(1 \FRAC{1}{{2}})$= 3/2 YEARLY = 3 half yearly 

R = 12% per annum = 6% per half yearly

P = Rs. 5000

$(C.I = P [(1+\frac{R}{{100}})^T - 1] \\\Rightarrow C.I = 5000 [(1+\frac{6}{{100}})^3 - 1] \\ \Rightarrow C.I = 5000 \TIMES [\frac{23877}{{125000}}])$

⇒ C.I = 23877/25

∴ C.I = 955.08

LET the AMOUNT be Rs. p

From the problem’s statement:

S.I = (p × r × t)/100

⇒ 25920 = (p × 12 × 6)/100

⇒ p = (25920 × 100)/(72)

⇒ p = 36000

Compound interest can be given as

⇒ CI = p((1 + 8/100)² - 1)

⇒ CI = 36000 × (1.08² - 1)

⇒ CI = 5990.4

Rs. 10000 lent at the rate of INTEREST 8% compounded half yearly,

<P>$(\Rightarrow CI = P\left\{ {\left( {1 + \FRAC{R}{{100}}} \right)t} \right\} - P)$

Let the TWO parts be Rs. ‘X’ and Rs. (9500 - x)

? Simple INTEREST = (Principal × Rate × Time)/100

⇒ Simple interest at 15% p.a. on Rs. x = (x × 15 × 3/2)/100 = 0.225x

⇒ Simple interest at 20% p.a. on Rs. (9500 - x) = [(9500 - x) × 20 × 3/2]/100 = 2850 - 0.3x

Now, total interest = Rs. 2565

⇒ 0.225x + 2850 - 0.3x = 2565

⇒ 0.3x - 0.225x = 2850 - 2565

⇒ 0.075x = 285

⇒ x = 285/0.075 = Rs. 3800

⇒ 9500 - x = 9500 - 3800 = Rs. 5700

For calculating compound INTEREST, Principal = Rs. 4000, Time period = 2 years, Let the rate = 10 %

We know, AMOUNT = Principal (1 + Rate/100)ⁿ

And compound interest = Amount - Principal = 4000 (1 + 10/100)² - 4000 = Rs. 840

It is GIVEN that simple interest is half the compound interest

We know, Simple interest = (P × R × T)/100

Simple interest = Rs. 420, rate = 8%, time = 3 years

⇒ 420 = (P × 8 × 3)/100

⇒ P = Rs. 1750

We KNOW the formula for calculating Simple Interest.

SI = (P × r × t)/100

Where,

SI = Simple Interest

P = Principle

r = RATE of interest (in percentage)

t = Time period

It is given that,

Rate at which Anil lent to SUNIL, r₁ = 8.25%

Rate at which Sunil lent to Devi, r₂ = 11.5%

Earnings by Sunil = Rs. 325

Let the sum lent by Anil to Sunil be P.

Interest earned by Anil, SI₁ = (P × 8.25 × 1)/100

Interest earned by Sunil, SI₂ = (P × 11.5 × 1)/100

Net earnings of Sunil = SI₁ – SI₂ = Rs. 325

⇒ [(P × 11.5 × 1)/100] – [(P × 8.25 × 1)/100] = 325

⇒ 11.5P – 8.25P = 32500

⇒ 3.25P = 32500

⇒ P = Rs. 10000

<P>Let the rate of interest be‘r %’

Money grows to 1331/1000 in three years at compounded annual basis

⇒ (1 + r %)³ = 1331/1000

⇒ (1 + r %)³ = (11/10)³

⇒ (1 + r%) = (11/10)

⇒ 1 + r% = 1.1

⇒ r = 10 %

Let it take ‘t’ years for amount ‘p’ to BECOME ‘3p’ at 10% interest rate

⇒ 3p = p + (p × r × t) /100

⇒ 2p = (p × 10 × t) /100

⇒ t = 20

Given, Principal = Rs. 2500, Rate = 30%, Time = 3 years

When the interest is compounded annually,

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

⇒ Compound interest = 2500 × [(1 + 30/100)³ - 1] = 2500 × 1.197 = Rs. 2992.5

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

⇒ Simple interest = (2500 × 30 × 3)/100 = Rs. 2250

∴ REQUIRED difference = 2992.5 - 2250 = Rs. 742.5

We have,

AMOUNT = PRINCIPAL × (1 + Rate/100)^time

⇒ 10648 = Principal × (1 + 10/100)³

⇒ 10648 = 1.331 × Principal

simple interest on a SUM of money is 16/25 of the principal

let SI = 16 and Principal = 25

⇒ r = t (given)

⇒ SI = (P × r × t)/100

⇒ 16 = (25 × r × r)/100

Let the principal AMOUNT be P.

Now the amount A after 16?years is doubled, hence amount is 2P.

I = P × R × T/100

Where,

P = principal amount

R = RATE of interest

T = time in years = 16²/₃ = 50/3

I = simple interest

Amount A = I + P

According to question

A = P + P × R × T/100

2P = P + P × R × T/100

P = P × R × T/100

R = 100/T

R = 100 × 3/50

R = 6%

We KNOW that,

AMOUNT = Principal + SI

And SI = PRT/100

⇒ 10200 = 8500 + SI

⇒ SI = 1700

⇒ Time = (1700 × 100)/(8500 × 10)

<P>Amount at the end of first year = P + PRT/100 = 1000 + (1000 × 1 × 4)/100 = Rs. 1040

The formula for CI at annual compound interest, including principal sum, is: $(CI\; = \;P{\left( {1\; + \;\FRAC{r}{{100\; \times \;n}}} \right)^{nt}} - P)$

Where:

CI = Compound Interest

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

CI = 1040(1 + 4/100)² - 1040

CI = Rs. 85

We know the formula for compound interest -

$(\Rightarrow {\RM{CI}} = {\rm{}}\left[ {{\rm{P}}\left\{ {{{\left( {1{\rm{}} + {\rm{}}\frac{{\rm{R}}}{{100}}} \right)}^t} - 1} \right\}} \right])$

Where,

CI = Compound interest

P = PRINCIPAL

R = Rate of interest

T = Time period

Let the principal be P

CI = P (1 + 8/100) ² - P = 104P/625

SI = P × 8 × 2/100 = 4P/25

CI - SI = 40

104P/625 - 4P/25 = 40

⇒ Simple interest = (principle × time × RATE)/100 = (275000 × 3 × 8)/100 = 66000

⇒ Total amount to be paid = 275000 + 66000 = RS. 341000

⇒ Compounded Amount = Principal {1 + (rate/100)}^TIME = 7400{1 + (8.5/100)}² = 7400 × (1.085)² = 8711.465

∴ INTEREST = Amount – principal = 8711.465 – 7400 = Rs. 1311.465

Let the amount lends by Ram to Shyam be P.

TOTAL Interest at the end of two YEARS Shyam has to pay = (P × 25/2 × 2)/100 = P/4

Now, the total interest at the end of 2 yrs for MOHAN that is to be paid to Shyam will be = (P × 25 × 2)/100 = P/2

Net interest earned by Shyam = P/2 – P/4 = P/4

∴ Amount earned in percentage by Shyam = P/4P = 25%

SI for 1^st 2 years = 20% of P

SI for next 3 years = 15% of P

SI for next 1 year = 6% of P

TOTAL SI for 6 years = (20% + 15% + 6%) of P

LET the first part be Rs. X

Second part = 12000 – x

SI on x at 6% per annum for 2 years = (x x 6 x 2)/100 = Rs. 3x/25

SI on 12000 – x at 8% per annum for 3 years = {(12000 - x) x 8 x 3}/100

⇒ (72000 – 6X)/25

⇒ 3x/25 = (72000 – 6x)/25

⇒ 3x = 72000 – 6x

⇒ 9x = 72000

⇒ x = 8000

First part = x = 8000

Second Part = 12000 – 8000 = 4000

Let P be the PRINCIPAL amount :

$( \Rightarrow {\RM{\;}}29160 = P \TIMES {\left( {1 + \frac{8}{{100}}} \right)^2} = P \times 1.1664)$

∴ P = 29160/1.1664 = Rs 25000

Given that SIMPLE interest = Rs.y

Rate of interest R = y%

Time PERIOD T = y

We KNOW that simple interest S.I = P × T × R/100

⇒ y = P × y × y /100

⇒ 100/y = P

∴ PRINCIPLE P = 100/y

<P>SI = (P × R × t)/100

SI = Simple interest

P = Principal

R = rate %

t = time in years

Given, Simple interest on a certain SUM for 6 years is $(\frac{9}{{25}})$ of the sum

∴ SI = 9P/25

t = 6 years

R =?

$(\Rightarrow \frac{{9P}}{{25}} = \frac{{P \times 6 \times R}}{{100}})$

Let the rate of INTEREST be r%

Simple interest = (PRINCIPAL × rate × time)/100 = (principal × r × 6)/100

Now,

Amount = principal + simple interest

Principal + 60% of principal = principal + {(principal × r × 6)/100}

⇒ 1.6 = 1 + 0.06r

⇒ 0.6 = 0.06r

⇒ r = 10%

Now,

⇒ COMPOUNDED Amount = principal {1 + (rate/100)}^time = 12000{1 + (10/100)}³ = 12000 × 1.1 × 1.1 × 1.1 = 15972

∴ Interest = Amount – principal = 15972 – 12000 = Rs. 3972

<P>SI = P × R × T/100

⇒ 7056 = P × 8 × 7/100

⇒ P = 7056 × 100/ 56