Question

$\mathfrak{Question}$

A machine which was purchased 3 year ago at Rs. 6,25,000 depreciates at the rate of 8 p.c.p.a. What is its value today?

Option :-
a) rs. 480000
b) rs. 486680
c) rs. 458000
d) rs. 495670​

Answer 1

Given :-

• A machine was Purchased 3 years ago at Rs. 6,25,000
• It's value depreciates at the rate of 8 p.c.p.a

To Find :-

• Machine's value today

Solution :-

❒ Here , in this question :

➼ Principal ( P ) = Cost price of the machine = 6,25,000

➼ Amount ( A ) = Cost price after 3 years

➼ R = Rate of depreciation = 8 %

➼ N = 3 years

$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$

$\underlined{\boxed{A=P \bigg\{ 1 + \dfrac{R}{100}\bigg\}^{N} }}$

→ Let's solve by putting the values in this formula

$\sf \leadsto 6,25,000 \bigg\{ 1 + \dfrac{-8}{100} \bigg\}^{3}$

( Here , -8 is taken because value of machine is decreasing )

$\sf \leadsto 6,25,000 \bigg\{ \dfrac{100+(-8)}{100} \bigg\}^{3}$

$\sf \leadsto 6,25,000 \bigg\{\dfrac{92}{100}\bigg\}^{3}$

$\sf \leadsto 6,25,000 \bigg\{\dfrac{23}{25} \bigg\}^{3}$

$\sf \leadsto 6,25,000 \times \dfrac{23}{25} \times \dfrac{23}{25} \times \dfrac{23}{25}$

$\sf \leadsto 40 \times 23 \times 23 \times 23$

$\sf \leadsto 4,86,680$

∴ Value of the machine today is Rs. 4,86,680

  ( Option ' b ' )

Answer 2

$\huge{ \sf{ \red{ \underline{❥︎ Answer}}}}$

Given :-

• A machine was Purchased 3 years ago at Rs. 6,25,000
• It's value depreciates at the rate of 8 p.c.p.a

To Find :-

Machine's value today

Solution :-

❒ Here , in this question :

➼ Principal ( P ) = Cost price of the machine = 6,25,000

➼ Amount ( A ) = Cost price after 3 years

➼ R = Rate of depreciation = 8 %

➼ N = 3 years

Main Answer :-

Value of the machine today is Rs. 4,86,680

( Option ' b ' )

How to solve the problem. This is already given!!