The sum of 15% of a positive number and 20% of the same number is 126. What is one-third of that number

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

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

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

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