**Percentage Problems**

In any competitive exam, candidates find difficulty in solving **Percentage Problems, **as it consumes a lot of time. Aspirants can check the basic formulas to solve aptitude problems on percentage. It will become easier for you to manage the time and solve problems on percentage in a short period of time. Candidates you can also acquire important questions based on percentage along with their solutions PDF for this page. The problems we have provided on this page are mostly asked in the quantitative aptitude which is the most important part of the banking and other competitive exam.

It is an advice to all the candidates to practice more and more if you want to maintain you speed. One can easily solve Percentage word problems with the short formulas in minimum time but for the same you need to practice a lot. Candidates you can get Percentage Problems along with their solutions on the below section of the page which is well planned and structured by the team of www.recruitmentinboxx.com

**Percentage Problems**

__Important Formulas for Percentage Calculation__

__Formulas for Percentage of Population__

If the current population is P and it increases at a rate of R% per annum, then

Population after n years = P*(1 + R/100) n

Population n years ago = P/ (1 + R/100) n

__Formulas for Percentage Increase/Decrease__

If the price of an object increases by R%, the reduction in consumption so as not to increase the expenditure = [R/(100 + R)] * 100 %

If the price of an object decreases by R%, then the increase in consumption so as not to decrease the expenditure is = [R/(100 – R)] * 100 %

__Formulas for Depreciation__

If the current value of a car is P and it depreciates at a rate of R% per annum, then

Value of car after n years = P*(1 – R/100) n

Value of car n years ago = P/(1 – R/100) n

Check Out: __Problems on Compound Interest__

__Percentage Problems and Solutions__

Question 1) An increase of 30% in the price of oranges enables a man to buy 6 kg less for Rs. 300. Find the increased price per kg.

Solution: The statement implies that the man has 30% of Rs. 300 available to spend on 6 kgs now. 30% of 300 = 30/100 x 300 = 90

So the increased price per kg is 90/6 = Rs. 15 per kg

Question 2) In an election, candidate A got 75% of the total valid votes. If 15% of the total votes were declared invalid and the total numbers of votes is 560000, find the number of valid vote polled in favour of candidate.

Solution: Total number of invalid votes = 15 % of 560000

= 15/100 × 560000

= 8400000/100

= 84000

Total number of valid votes 560000 – 84000 = 476000

Percentage of votes polled in favour of candidate A = 75 %

Therefore, the number of valid votes polled in favour of candidate A = 75 % of 476000

= 75/100 × 476000

= 35700000/100

= 357000

Question 3) Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

Solution: Let their marks be (x + 9) and x.

Then, x + 9 = 56 (x + 9 + x)

100

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

Get Here: __Maths Formulas PDF__

Question 4) Aaron had Rs 2100 left after spending 30 % of the money he took for shopping. How much money did he take along with him?

Solution:

Let the money he took for shopping be m.

Money he spent = 30 % of m

= 30/100 × m

= 3/10 m

Money left with him = m – 3/10 m = (10m – 3m)/10 = 7m/10

But money left with him = Rs 2100

Therefore 7m/10 = Rs 2100

m = Rs 2100× 10/7

m = Rs 21000/7

m = Rs 3000

Therefore, the money he took for shopping is Rs 3000.

Question 5) An alloy contains 26 % of copper. What quantity of alloy is required to get 260 g of copper?

Solution: Let the quantity of alloy required = m g

Then 26 % of m =260 g

⇒ 26/100 × m = 260 g

⇒ m = (260 × 100)/26 g

⇒ m = 26000/26 g

⇒ m = 1000 g

Question 6) There are 50 students in a class. If 14% are absent on a particular day, find the number of students present in the class.

Solution: Number of students absent on a particular day = 14 % of 50

i.e., 14/100 × 50 = 7

Therefore, the number of students present = 50 – 7 = 43 students.

Check Out: __Compound Interest Formula –With Example__

Question 7) In an examination, 300 students appeared. Out of these students; 28 % got first division, 54 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed.

Solution: The number of students with first division = 28 % of 300 = 28/100 × 300 = 8400/100 = 84

And, the number of students with second division = 54 % of 300

= 54/100 × 300

=16200/100

= 162

Therefore, the number of students who just passed = 300 – (84 + 162) = 54

Question 8) The cost of an article is decreased by 15%. If the original cost is Rs 80, find the decrease cost.

Solution: Original cost = Rs 80

Decrease in it = 15% of Rs 80

= 15/100 × 80

= 1200/100

= Rs 12

Therefore, decrease cost = Rs 80 – Rs 12 = Rs 68

Question 9) Find the number which when decreased by 12 % becomes 198.

Solution: Let the number be m.

Decrease = 12 % of m

= 12/100 × m = 3m /25

Therefore, decrease number = m – 3m/25 = (25m – 3m)/25 = 22m/25

According to the question 22m/25 = 198

22m = 198 × 25

m = 4950/22

m = 225

Do You Know: __How to Prepare For Maths__

Question 10) A number 42 was misread as 24. Find the reading error in per cent.

Solution: Error = 42 – 24 = 18

Therefore, % error = 18/42 = 100%;

[Since, we know decrease% = decrease in value/original value × 100 %]

= 300/7 %

= 42.8 %

__Last Words__:

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