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%
Value of machine 2 years later = Rs. [ 4000 x {(1 - 10)/10)}2]
= Rs. (4000 x 9/10 x 9/10)
= Rs. 3240
? 540 = H(1 + 20/100)2
? H = (540 x 5/6 x 5/6) = 375 cm
Population after 2 years
=8000(1 + 10/100) (1 + 20/100)
= (8000 x 11/10 x 6/5) = 10560
B's salary is less than A's by (30 x 100)/130 % = 231/13%
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%
Let original price = Rs. 100
Reduced price = Rs. 75
Increase on Rs. 75 = Rs. 25
Increase on Rs. 100 = (25 x 100)/75 % = 331/3 %
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 .
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