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2 years ago

1.Find the compound interest on Rs25000 for 3 years at 10% per annum ,Compounded annually.

Mathematics

1 answer:

LekaFEV [45]2 years ago

6 0

Answer:

The compound interest is Rs8275

Step-by-step explanation:

The rule of the compound interest is A = P $(1+\frac{r}{n})^{nr}$ , where

A is the new amount

P is the initial amount

r is the interest rate in decimal

n is the number of periods

t is the time

The interest I = A - P

∵ The amount of investment is Rs25000 for 3 years

∴ P = 25000

∴ t = 3

∵ The rate of interest is 10% per annum, compounded annually

∴ r = 10% = 10 ÷ 100 = 0.1

∴ n = 1 ⇒ compounded annually

→ Substitute these value in the 1st rule above to find the new amount

∵ A = 25000 $(1+\frac{0.1}{1})^{1(3)}$

∴ A = 25000 $(1.1)^{3}$

∴ A = Rs33275

→ Use the 2nd rule above to find the interest

∵ I = 33275 - 25000

∴ I = 8275

∴ The compound interest is Rs8275

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Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

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Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

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This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

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5% during the two years.

10% during the current year. 25% of employees who had lost-time accidents last year will experience a lost-time accident during the current year.

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