The price of an article has been reduced by 25%. In order to restore the original price, the new price must be increased by?

Let income = Rs. 100
Then expenditure = Rs. 75 and saving = Rs. 25
New expenditure = 110% of Rs. 75 = Rs. 165 / 2
New saving = Rs. (120 - 165/2) = Rs. 75/2
Increase in saving = Rs. (75/2 - 25) = Rs. 25/2
? Increase% = (25/2) x (1/25) x 100% = 50%

B's salary is less than A's by (30 x 100)/130 % = 23¹/₁₃%

P = 6q. Thus q is less than p by 5q
? q is less than p by = {(6q - q)/6q} x 100%
= ( 5q/6q) x 100 % = 83¹/_{3 %}

Let original wages = Rs. 100
Reduced wages = Rs. 50
Increased wages = 150% of Rs. 50 = (150 x 50)/100 = Rs. 75
? Loss = 25%

Increase in 3 years over 100 = 100 x {1 + 7/(2 x 100)}³
= (100 x 207/200 x 207/200 x 207/200)
= (200 + 7)³ / 80000
= 88869743 / 80000 = 110.8718

Required increase % = (110.8 - 100)% = 10.8%

Let the original fraction be p/q
Now, (120% of p) / (90% of q) = 16 / 21
? 4p / 3q = 16/21
? p / q = (16/21) x (3/4) = 4 / 7

? Milk = 90% 0f 40 = 36 litres and water = 4 litres.
Let L litre water to be added in order to make 20% water in new mixture.
? {(4 + L) / (40 + L)} x 100 = 20
? 20 x (40 + L) = 100 x (4 + L)
? 80 x L = 400
? L = 5 litres .

Suppose boys = 3N and girls = 2N
Not adults = (80% of 3N) + (75% of 2N) = (12N/5 + 3N/2) = 39N/10
? Required percentage = (39N/10 / 1/5N) x 100 % = 78%

Let the total number of students be 100
Then, n(A ? B) = n(A) + n(B) - n(A ? B) = (72 + 44 - 100)% = 16%
Now,16% of N = 40
? (16 x N) / 100 = 40
? N = (100 x 40) / 16 = 250

Let side of square = 100 cm
Area = (100 x 100)cm² = 10000 cm²
New area = (130 x 130)cm² = 16900 cm²
Increase in area = (6900 / 10000) x 100% = 69%