The population of a city is 250000. It is increasing at the rate of 2% every year. The growth in the population after 2 yr is?

Given that,
p = 1058400, R = 5% and n = 2
According the formula,
Population n yr ago = P/(1 + R/100) ⁿ
? Population 2yr ago = 1058400/(1+ 5/100)²
= 1058400 x 20/21 x 20/21 = 960000

Given that,
P = 705600, R = 5% and n = 2

According to the formula,
Population after n yr = p(1 + R/100)ⁿ

? Population after 2 yr.
= 705600 x (1+ 5/100)²
= 705600 x (105/100)²
= 705600 x (21/20 x 21/20) = 777924

Given, R = 15% and R₂ = 20%
Required population
= P(1 + R₁/100) (1 - R₂/100)
= 126800(1 + 15/100) (1- 20/100)
= 126800 (1 + 3/20 ) (1- 1/5)
= 126800 x (23/20) x (4/5) = 116656

Let the maximum marks be N.

According to the question,
(20N/100) + 50 = (40N/100) - 30
? 20N/100 = 80
? N = (80 x 100)/20 = 400

Let maximum marks be N.

According to the question,
40N/100 = 40 + 40
? 40N/100 = 80
? N = 200

Let the number of boys = B
and number of girls = G
Then, 10% of B = 1/4 of G
? B/10 = G/4
? B/G = 10/4 = 5/2
? B : G = 5 : 2

In such case, there is always decrease.
Given that, a = (common increase or decreased) = 7%

According to the formula,
Decreased Percentage = a²/100 %
= 7²/100 %
= 0.49%

Let 1st number be M and 2nd number be M.
According to the question,
48% of M = 60% N
? (M x 48)/100 = (N x 60)/100
? M/N = (60/100) x (100/48) = 5/4
? M : N = 5 : 4

Let the positive number be N.

According to the question
N x (15+ 20)% = 126
? N x 35 = 126 x 100
? N = 12600/35 = 360
So, one-third of the number = 360/3 = 120

B's marks = C's marks + 5% of 400
= 300 + 20 = 320

Now , A's marks = B's marks + 10% of 400
= 320 + 40 = 360