A papaya was planted 2 years ago. It increases at the rate of 20% every year. If at present, the height of the tree is 540 cm, what was it when the tree was planted?

Population after 2 years
=8000(1 + 10/100) (1 + 20/100)
= (8000 x 11/10 x 6/5) = 10560

Let price = Rs. 100, sale = 100
Then sale value = Rs. (100 x 100) = Rs. 10000
New sale value = Rs. (70 x 120) = Rs. 8400
Decrease% = (1600/10000) x 100 % = 16%

Alcohol in 15 litres =(20 x 15)/100 litres = 3 litres
Now, alcohol in 18 litres = 3 litres

? strength of alcohal =(3/18) x 100 % = 16²/₃%

Let tax = Rs. 100 and Consumption = 100 units
Original expenditure = Rs. (100 x 100) = 10000
New expenditure = Rs. (120 x 80) = Rs. 9600
? Decreased in expenditure = (400 x 100)/1000 % = 4%

Reduction in consumption = (20/120) x 100% = 50/3%
? Reduction in consumption /Original consumption = (50/3) / 100 = 1/6 = 1 : 6

Value of machine 2 years later = Rs. [ 4000 x {(1 - 10)/10)}²]
= Rs. (4000 x 9/10 x 9/10)
= Rs. 3240

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

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 %}

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