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VARVARA [1.3K]
2 years ago
13

Rachit borrowed 15000 from his friend . He gave 3000 at the rate of 15% per annum and the remaining amount at the rate of 18% pe

r annum. How much interest did he pay in 4 years
Mathematics
1 answer:
White raven [17]2 years ago
7 0

Answer:

Therefore he paid $10440 in 4 year.

Step-by-step explanation:

Given, Rachit borrowed $15000 from his friend. He gave $3000 at the rate 15% for 4 year.

Interest(I₁)= Prt                          here P= $3000, r=15% and t = 4 year

                =$(3000×0.15×4)

               =$1800

He gave $(15000-3000)= $12000 at the rate of 18%.

Interest(I₂)= Prt                          here P= $12000, r=18% and t = 4 year

                =$(12000×0.18×4)

                =$8640

Therefore he paid =$(1800+8640)=$10440 in 4 year.

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Step-by-step explanation:

First of all we need to know when does two events become independent:

For the two events to be independent, P(A|B)=P(A) that is if condition on one does not effect the probability of other event.

Here, in our case the only option that satisfies the condition for the events to be independent is P(A|B)=P(A)=0.16. Rest are not in accordance with the definition of independent events.

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50 people consume an average of 110 kg of rice in a week. If 20 more people join this group, how many kilograms of rice will be
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What is the unit rate for 500.48 meters in 27.2 hours?
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2 years ago
A tank contains 500 gallons of salt-free water. A brine containing 0.25 lb of salt per gallon runs into the tank at the rate of
Neporo4naja [7]

Answer:

0.0198 lbs per gallon

Step-by-step explanation:

amount of salt free water = 500 gallons

salt rate in = 1 gal/min

salt rate out = 1 gal/min

amount o salt in brine = 0.25 lb per gallon

Let the amount of salt in the tank be A(t) at any time t.

\frac{dA(t)}{dt} =salt rate in - salt rate out

salt rate in = 0.25 x 1 = 0.25

salt rate out = \frac{A(t)\times 1}{500}

The differential equation is given by

\frac{dA(t)}{dt} =1 - \frac{A(t)\times 1}{500}

where, A(0) = 0

So, the equation becomes

\frac{dA(t)}{dt} + \frac{A(t)}{500} = 1

Here the integrating factor is e^{\frac{dt}{500}}=e^{\frac{t}{500}}

The solution of the above differential equation is given by

A(t)\times e^{\frac{t}{500}} = \int e^{\frac{t}{500}}dt

A(t)\times e^{\frac{t}{500}} = 500\times e^{\frac{t}{500}}+C

where, C is the integrating constant.

A(t)=500+Ce^{-\frac{t}{500}}

Put, A(0) = 0

C = - 500

A(t)=500\left ( 1-e^{-\frac{t}{500} \right )

As concentration is defined as

Concentration = Quantity / Volume

C(t)=\frac{A(t)}{500}

C(t)=1-e^{\frac{-t}{500}}

Now concentration at t = 10 min

Put, t = 10 min

C(10)=1-e^{\frac{-10}{500}}

C (10) = 0.0198 lbs per gallon

Thus, teh concentration after 10 min is 0.0198 lbs per gallon.

7 0
2 years ago
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