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Solve the following : A person purchases a television by paying ₹20,000 in cash and promising to pay ₹1,000 at end of every month for the next 2 years. If money is worth 12% p. a. converted monthly, - Mathematics and Statistics

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Question

Solve the following :

A person purchases a television by paying ₹20,000 in cash and promising to pay ₹1,000 at end of every month for the next 2 years. If money is worth 12% p. a. converted monthly, find the cash price of the television. [(1.01)–24 = 0.7875]

Sum
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Solution

Person buys the television for ₹20,000 in cash.
∴ First payment = ₹20,000
Remaining value of the television was paid in monthly instalments of ₹1,000.
∴ C= ₹1,000,
The duration of monthly installments is of 2 years.
∴ n = 24
Rate of interest is 12% p.a.

∴ r = `(12)/(12)` = 1% p.m.

∴ i = `"r"/(100) = (1)/(100)` = 0.01
The amount is paid at the end of every month.
∴ It is an immediate annuity.
Now, to find sum of all instalments we have to find present value.

∴ P = `"C"/"i"[1 - (1 + "i")^-"n"]`

∴ P = `(1,000)/(0.01)[1 - (1 + 0.01)^24]`

= 1,00,000 [1 – (1.01)–24]
= 1,00,000 (1 – 0.7875)
= 1,00,000 x 0.2125
∴ P = ₹21,250
∴ Cash price of the television = First Payment + Present Value
= 20,000 + 21,250
= ₹41,250
∴ Cash price of the television is ₹41,250.

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Annuity
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Chapter 2: Insurance and Annuity - Miscellaneous Exercise 2 [Page 31]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 2 Insurance and Annuity
Miscellaneous Exercise 2 | Q 4.18 | Page 31

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[Given (1.1)4 = 1.4641]


The future amount, A = ₹ 10,00,000

Period, n = 20, r = 5%, (1.025)20 = 1.675

A = `"C"/"I" [(1 + "i")^"n" - 1]`

I = `5/200` = `square` as interest is calculated semi-annually

A = 10,00,000 = `"C"/"I" [(1 + "i")^"n" - 1]`

10,00,000 = `"C"/0.025 [(1 + 0.025)^square - 1]`

= `"C"/0.025 [1.675 - 1]`

10,00,000 = `("C" xx 0.675)/0.025`

C = ₹ `square`


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⇒ r = 4%

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⇒ P' = `(C(1 + i))/i [1 - (1 + i)^-n]`

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= `(2000(square))/square [1 - (square)^-4]`

= 50,000`(square)`[1 – 0.8548]

= ₹ 7,550.40


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