Arithmatical  Percentage
DIRECTIONS : Problems based on Percentage.
1. 
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was 

A. 2500 

B. 2700 

C. 3000 

D. 3100 
Solution 
Number of valid votes = 80% of 7500 
= 6000. 
Valid votes polled by other candidates  = 45% of 6000 
(45/100×6000) 
= 2700. 

2. 
If a number x is 10% less than another number y and y is 10% more than 125, then x is equal to 

A. 143 

B. 150 

C. 123.75 

D. 140.55 
Solution 
y = 125 + 10% of 125 
= 125 + 12.50 = 137.50. 
x = 137.50  10% of 137.50 
= 137.50  13.75 
= 123.75. 

3. 
In an examination, 35% of the students passed and 455 failed. How many students appeared for the examiantion? 

A. 490 

B. 620 

C. 700 

D. 845 
Solution 
Let the number of students appeared be x. 
Then, 65% of x = 455 
‹=›65 / 100 x = 455 
‹=› x= [455 x 100 / 65] 
= 700. 

4. 
If the price of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increase in his expenditure on sugar, will have to reduce his consumption of sugar by 

A. 10% 

B. 20% 

C. 25% 

D. 30% 
Solution 
Let the original consumption = 100 kg and new consumption = x kg. 
So, 100 x 6 = x × 7.50 
x = 80 kg. 
Reduction in consumption = 20%. 

5. 
From a container having pure milk, 20% is replaced by water and the process is repeated thrice. At the end of the third operation, the milk is 

A. 40% pure 

B. 50% pure 

C. 51.2% pure 

D. 58.8% pure 
Solution 
Let total quantity of original milk = 100 gm. 
Milk after first operation = 80% of 1000 
‹=› 800gm. 
Milk after second operation = 80% of 800 
‹=› 640gm. 
Milk after third operation = 80% of 640 
‹=› 512gm. 
Strength of final mixtures = 51.2%. 
