Question

1. A machinery worth ₹10,500 depreciated by 5%. Find its value after one year.


2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.​

Answer 1

Answer:

1) 9975. 2)

Step-by-step explanation:

1) 5% of 10,500=(

100

5

)×10,500

=525

=10,500−525

=9,975

2) Answered 2 years, 4 months ago

Current population of city

= 12 lakh

1200000

Population after two years =

Thus, the population after two years is 1297920.

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Answer 2

Answer:

Question :-

1. A machinery worth ₹10,500 depreciated by 5%. Find its value after one year.

$\mathbb \pink{\underline{\underline{ANSWER}}}$

Given,

Principal = 10500, R = 5%, Time n = 1 year.

We know that A = P(1 + r/100)^n

⇒ 10500(1 + 5/100)^1

⇒ 10500(1 + 1/20)

⇒ 10500(21/20)

⇒ 11025.

Therefore, value after 1 year = 11025.

_______________________

Question :-

2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.

$\mathbb \pink{\underline{\underline{ANSWER}}}$

Formula is a =  p(1+r/100)ⁿ           ------------------ (i)

Given p = 1200000, r = 4, n = 2

On submitting values in (i) we get, 

 a = 1200000(1+4/100)²

    =  1200000(1 + 1/25)²

    = 1200000(26/25)²

    = 1200000 × 26/25 × 26/25

    = 1248000 × 26/25

    = 1297920

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