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 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
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
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 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
Total marks scored by Mathew in all subjests = 42 + 51 + 58 + 35 + 48 = 234
? Maximum marks is 60 in any subject.
? Maximum marks = 60 x 5 = 300
? Percentage of Mathew's marks
= [(Marks obtained / Maximum marks)] x 100
= (234 / 300) x 100
= 234/3 = 78%
Net effect on area = -20 + 10 + [(-20 )(10)/100] %
= (-10 - 2) % = -12%
Now, after this mistake new area = (100 - 12)% of 200
= (88/100) x 200
= 176 sq cm
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