In a test, A scored 10% more than B and B scored 5% more than C. If C scored 300 marks out of 400, then A's marks are

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

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

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

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

After removing 25% of blue balls,
Total blue balls left = 75% of 100
= (75/100) x 100 = 75

After removing 50% of red balls, total red balls left = 50% 0f 50 = (50 x 50) / 100 = 25
? Required percentage = [50/(75 + 25 + 50)] x 100
= (50/150) x 100
= 33¹/₃ %

Let cost prices of two article be 3N and 4N respectively then
(110% of 3N) / (4N + 4) = 3/4
? 1.1N / (N + 1) = 1
? 1.1N = N + 1
? 0.1 N = 1
? N = 10

Thus, cost price of the second article is 4 x 10 = ? 40

Let sale in 2008 = 100
Sale in 2009 = 20
Sale in 2010 = 100
? Required increases = [(100 - 20)/20] x100
= (80/20) x 100 = 400%