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11th Standard English Medium Business Maths Subject Financial Mathematics Book Back 2 Mark Questions with Solution Part - II

Published on: 10/06/2021

QB365 provides detailed and simple solution for every Book back Questions in class 11 Business Maths Subject. It will helps to get more idea about question pattern in every book back questions with solution.

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Std: 11th Standard - Business Maths and Statistics

chapter: - Book back Questions

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Questions:

Generic 2

5 x 2 = 10

If the dividend received from 10% of Rs. 25 shares is Rs. 2000. Find the number of shares.

What is the amount of perpetual annuity of Rs. 50 at 5% compound interest per year?

Find the present value of Rs. 2,000 per annum for 14 years at the rate of interest of 10% per annum. If the payments are made at the end of each payment period. [ (1.1)^–14 = 0.2632]

A limited company wants to create a fund to help their employees in critical circumstances. The estimated expenses per month is Rs. 18,000. Find the amount to be deposited by the company if the rate of compound interest is 15%.

A person pays Rs 64,000 per annum for 12 years at the rate of 10% per year. Find the annuity [(1.1)¹²= 3.3184]

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Answers Key:

Generic 2

5 x 2 = 10

Let x be the number of shares.
Face value of ‘x’ shares = Rs. 25 x
Now \(\frac{10}{100}\times 25x=Rs. 2,000\)
\(\Rightarrow x=\frac{2000\times 100}{25\times 10}=800\)
Hence the number of shares = 800

a = 50, i = \(\frac{5}{100}\) = .05
A = \(\cfrac { a }{ i } =\cfrac { 50 }{ 0.05 } \) Rs.1000

a = Rs. 2000, i = \(\frac{10}{100}\) = 0.1 n = 14
P = \(\cfrac { a }{ i } \left[ 1-\left( 1+i \right) ^{ -n } \right] \)
\(=\cfrac { 2000 }{ 0.1 } \left[ 1-\left( 1.1 \right) ^{ -14 } \right] \)
= 20000[1-0.2632]
= Rs. 14,736

Here a = 18,000 ; i = 0.15 and k = 12.
P = \({\frac{a}{{i}/k}} = {\frac{18,000}{0.15/12}}\)
= \({\frac{18,000}{0.15}} \times {12}\)
\(=\frac{18,00,000\times 12}{15}=14,40,000\)
Therefore the amount to be deposited is Rs. 14,40,000.

Here a = 64,000, n = 12 and i = \(\frac{10}{100}=0.1\)
Amount of Ordinary annuity (A) = \(\frac{a}{i}[(1+i)^n-1]\)
= \(\frac{64000}{0.1}[(1+0.1)^{12}-1]\)
= 6,40,000 [(1.1)¹² – 1]
= 6,40,000[3.3184 – 1]
= 6,40,000 [2.3184]
= 64 x 23184
∴ A = Rs. 14,83,776.