Math, asked by punam0797, 10 months ago

A building is made of cost *2,19,700 at a plot of
cost₹51200. If the rate of plot is increased by
30% annualy and the rate of building is
decreased by 20% annualy, then the cost of plot
and building will be equal after that years-
₹ 51200 में भूखण्ड खरीद कर उस पर ₹2,19,700 की लागत से
बिल्डिंग बनाई गई। यदि भूखण्ड के मूल्य में 30% की वृद्धि तथा
बिल्डिंग के मूल्य में 20% वार्षिक दर से कमी हो, तो कितने वर्ष बाद
दोनों का मूल्य बराबर होगा-
(A)7 yr
(B) 4 yr
C) 3 yr
(D) 2 yr​

Answers

Answered by bhagyashreechowdhury
0

Given:

The cost of the building = Rs. 2,19,700

The cost of the plot = RS. 51200

The rate of annual increase of the plot is 30% and the rate of annual decrease of the building is 20%

To find:

The no. of years after which the cost of the plot and the building will be equal

Formula to be Used:

To solve the above given problems we will use the formula of the appreciation and depreciation of the cost which as follows:

\boxed{Appreciation = P * (1\:+\:\frac{R}{100})^n }\\\\and\\\\\boxed{Depreciation = P * (1 \:-\: \frac{R}{100} )^n}

where

P = present value of the item

R = rate of increase or decrease

n = no. of years

Solution:

Since the no. of years i.e., "n" has to be equal so that the cost of the building and the plot becomes equal so, we will equate the two formulas and substitute the given values:

Present\: value\: of\: the\: building * [1\:-\:\frac{R}{100}]^n \:=\: Present\: value\: of\: the\: plot * [1\:+\:\frac{R}{100}]^n

⇒ 219700 × [1\:-\:\frac{20}{100} ]^n = 51200 × [1\:+\:\frac{30}{100} ]^n

⇒ 2197 × [1\:-\:\frac{1}{5} ]^n = 512 × [1\:+\:\frac{3}{10} ]^n

⇒ 2197 × [\frac{5\:-\:1}{5} ]^n = 512 × [\frac{10\:+\:3}{10} ]^n

⇒ 2197 × [\frac{4}{5} ]^n = 512 × [\frac{13}{10} ]^n ...... (i)

Now,

Let's put each of the values of n from the options given in the question in eq. (i), to find the exact value of it:

Option (A): 7 yr

2197 × [\frac{4}{5} ]^7 = 512 × [\frac{13}{10} ]^7

⇒ 2197 × [0.8]^7 = 512 × [1.3 ]^7

⇒ 2197 × 0.209 = 512 × 6.274

⇒ 459.173 ≠ 3212.28

option A is incorrect

Option (B): 4 yr

2197 × [\frac{4}{5} ]^4 = 512 × [\frac{13}{10} ]^4

⇒ 2197 × [0.8]^4 = 512 × [1.3 ]^4

⇒ 2197 × 0.4096 = 512 × 2.856

⇒ 899.89 ≠ 1462.27

option B is incorrect

Option (C): 3 yr

2197 × [\frac{4}{5} ]^3 = 512 × [\frac{13}{10} ]^3

⇒ 2197 × [0.8]^3 = 512 × [1.3 ]^3

⇒ 2197 × 0.512 = 512 × 2.197

⇒ 1124.86 = 1124.86

option C is correct

Option (D): 2 yr

2197 × [\frac{4}{5} ]^2 = 512 × [\frac{13}{10} ]^2

⇒ 2197 × [0.8]^2 = 512 × [1.3 ]^2

⇒ 2197 × 0.64 = 512 × 1.69

⇒ 1406.08 ≠ 865.28

option D is incorrect

Thus, the cost of the plot and the building will be equal after 3 years.

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