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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Solve the following : A man borrowed some money and paid back in 3 equal installments of ₹2,160 each. What amount did he borrow if the rate of interest was 20% per annum compounded annually? - Mathematics and Statistics

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प्रश्न

Solve the following :

A man borrowed some money and paid back in 3 equal installments of ₹2,160 each. What amount did he borrow if the rate of interest was 20% per annum compounded annually? Also find the total interest charged. [(1.2)3 = 0.5787]

बेरीज

उत्तर

Given, C = ₹2,160, n = 3 years, r = 20% p.a.

∴ i  = `"r"/(100) = (2)/(100)` = 0.2

Here, we have to find present value of annuity.

∴ P = `"C"/"i"[1 - (1 + "i")^-"n"]`

= `(2,160)/(0.2)[1 - (1 + 0.2)^-3]`

= 10,800[1 – (1.2)–3]
= 10,800[1 – 0.5787]
= 10,800[0.4213]
∴ P = ₹4,550
The man has paid 3 equal instalments of ₹2,160 each.
∴ Total paid value of instalments
= 3 x 2,160
= ₹6,480
Interest = Total paid value of instalments – Present Value
= 6,480 – 4,550
= ₹1,930.

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Annuity
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Insurance and Annuity - Miscellaneous Exercise 2 [पृष्ठ ३१]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 2 Insurance and Annuity
Miscellaneous Exercise 2 | Q 4.2 | पृष्ठ ३१

संबंधित प्रश्‍न

Find accumulated value after 1 year of an annuity immediate in which ₹ 10,000 is invested every quarter at 16% p.a. compounded quarterly. [Given (1.04)4 = 1.1699]


A lady plans to save for her daughter’s marriage. She wishes to accumulate a sum of ₹4,64,100 at the end of 4 years. What amount should she invest every year if she gets an interest of 10% p.a. compounded annually? [Given (1.1)4 = 1.4641]


Find the present value of an annuity due of ₹ 600 to be paid quarterly at 32% p.a. compounded quarterly. [Given (1.08)−4 = 0.7350]


Fill in the blank :

The intervening time between payment of two successive installments is called as ___________.


Fill in the blank :

An annuity where payments continue forever is called __________.


Fill in the blank :

If payments of an annuity fall due at the end of every period, the series is called annuity __________.


State whether the following is True or False :

Payment of every annuity is called an installment.


State whether the following is True or False :

The present value of an annuity is the sum of the present value of all installments.


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]


Solve the following :

Some machinery is expected to cost 25% more over its present cost of ₹6,96,000 after 20 years. The scrap value of the machinery will realize ₹1,50,000. What amount should be set aside at the end of every year at 5% p.a. compound interest for 20 years to replace the machinery? [Given (1.05)20= 2.653]


Multiple choice questions:

In an ordinary annuity, payments or receipts occur at ______


Multiple choice questions:

The present value of an immediate annuity of ₹ 10,000 paid each quarter for four quarters at 16% p.a. compounded quarterly is ______


State whether the following statement is True or False:

Annuity contingent begins and ends on certain fixed dates


The present value of an immediate annuity for 4 years at 10% p.a. compounded annually is ₹ 23,400. It’s accumulated value after 4 years would be ₹ ______


If payments of an annuity fall due at the beginning of every period, the series is called annuity ______


A company decides to set aside a certain sum at the end of each year to create a sinking fund, which should amount to ₹ 4 lakhs in 4 years at 10% p.a. Find the amount to be set aside each year?
[Given (1.1)4 = 1.4641]


For an annuity due, C = ₹ 2000, rate = 16% p.a. compounded quarterly for 1 year

∴ Rate of interest per quarter = `square/4` = 4

⇒ r = 4%

⇒ i = `square/100 = 4/100` = 0.04

n = Number of quarters

= 4 × 1

= `square`

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

⇒ P' = `(square(1 + square))/0.04 [1 - (square + 0.04)^-square]`

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

= 50,000`(square)`[1 – 0.8548]

= ₹ 7,550.40


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