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 original price be ? 100
Then, reduced price = ? 94
? Increase in consumption
= [(100 - 94)/94] x 100 %
= (6/94) x 100 %
= 618/47%
Let original price be ? 100
Then, increased price = ? 125
? Reduction in consumption
= [(125 - 100)/125 ] x 100%
= (25/125) x 100 %
= 20%
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
Population after 2 yr
= P (1 + R/100)2
= 250000 (1 + 2/100)2
? 250000 x (51/50) x (51/50) = 260100
? Growth = 260100 - 250000 = 10100
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
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