Aptitude - Percentage
11.
(180% of ?) ÷ 2 = 504
Answer
180100 x x 
÷ 2 = 504 ⇒ 180x200 = 504
x = 504 x 200180 = 560
Answer :
Option CExplanation :
180100 x x ÷ 2 = 504 ⇒ 180x200 = 504x =
504 x 200180 = 560 12.
x% of y is y% of ?
Answer
Let x% of y = y% of z. Then, x100 x y = y100 x z
⇒ z = x
∴ x% of y = y% of x
Answer :
Option AExplanation :
Let x% of y = y% of z. Then, x100 x y = y100 x z
⇒ z = x
∴ x% of y = y% of x
13.
When a number is first increased by 10% and then reduced by 10%, the number :
Answer
Let the given number be x
Increased number = (110% of x) = 110100 x x = 11x10
Finally, reduced number = 90% of 11x10 = 99x100
Decrease =x − 99x100 = x100
Decrease % = x100 x 1x x 100 % = 1%
Answer :
Option BExplanation :
Let the given number be x
Increased number = (110% of x) =
110100 x x = 11x10Finally, reduced number =
90% of 11x10 = 99x100Decrease =
x − 99x100 = x100Decrease % =
x100 x 1x x 100 % = 1% 14.
If the side of a square is increased by 30%, then the area is increased by :
Answer
Let side = 10 cm. New side = 10 x 130100 cm = 13 cm
Original area = (10 x 10)cm2 = 100cm2
New area = (13 x 13)cm2 = 169cm2
Increase in area = 69%
Answer :
Option DExplanation :
Let side = 10 cm. New side =
10 x 130100 cm = 13 cmOriginal area = (10 x 10)cm2 = 100cm2
New area = (13 x 13)cm2 = 169cm2
Increase in area = 69%
15.
One litre of water is evaporated from 6 litres of a solution containing 5% salt. The percentage of salt in the remaining solution is :
Answer
Salt in 6 litres = 5% of 6 litre = 5100 x 6 litre = 0.30 litre
Salt in new solution = 0.305 x 100 % = 6%
Answer :
Option CExplanation :
Salt in 6 litres = 5% of 6 litre =
5100 x 6 litre = 0.30 litreSalt in new solution =
0.305 x 100 % = 6% Jump to page number :