Then Value of the machine after n years = P((1-R)/100))n
If the present value of a machine = P and it depreciates at the rate of R% per annum,
Solved Examples
1. If A = x% of y and B = y% of x, then which of the following is true? | |
A. None of these | B. A is smaller than B. |
C. Relationship between A and B cannot be determined. | D. If x is smaller than y, then A is greater than B. |
E. A is greater than B. |
Answer : Option A
Explanation :
A = xy/100 ………….(Equation 1)
B = yx/100.................(Equation 2)
From these equations, it is clear that A = B
2. If 20% of a = b, then b% of 20 is the same as: | |
A. None of these | B. 10% of a |
C. 4% of a | D. 20% of a |
Answer : Option C
Explanation :
20% of a = b
=> b = 20a/100
b% of 20 = 20b/100=(20a/100) × 20/100
=(20×20×a)/(100×100)=4a/100 = 4% of a
3. Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B. | |
A. 2 : 1 | B. 1 : 2 |
C. 1 : 1 | D. 4 : 3 |
Answer : Option D
Explanation :
5% of A + 4% of B = 2/3(6% of A + 8% of B)
5A/100+4B/100=2/3(6A/100+8B/100)
⇒5A+4B=2/3(6A+8B)
⇒15A+12B=12A+16B
⇒3A=4B
⇒AB=43⇒A:B=4:3
4. The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year? | |
A. 4% | B. 6% |
C. 5% | D. 50% |
Answer : Option C
Explanation :
Increase in the population in 10 years = 2,62,500 - 1,75,000 = 87500
% increase in the population in 10 years = (87500/175000)×100=8750/175=50%
Average % increase of population per year = 50%/10=5%
5. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get? | |
A. 57% | B. 50% |
C. 52% | D. 60% |
Answer : Option A
Explanation :
Votes received by the winning candidate = 11628
Total votes = 1136 + 7636 + 11628 = 20400
Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%
6. A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally? | |
A. 420 | B. 700 |
C. 220 | D. 400 |
Answer : Option B
Explanation :
He sells 40% of oranges and still there are 420 oranges remaining
=> 60% of oranges = 420
⇒(60×Total Oranges)/100=420
⇒Total Oranges/100=7
⇒ Total Oranges = 7×100=700
7. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets? | |
A. 499/11 % | B. 45 % |
C. 500/11 % | D. 489/11 % |
Answer : Option C
Explanation :
Total runs scored = 110
Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60
Total runs scored by running between the wickets = 110 - 60 = 50
Required % = (50/110)×100=500/11%
8. What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit? | |
A. 2023% | B. 20% |
C. 21% | D. 2223% |
Answer : Option B
Explanation :
Total numbers = 70
Total numbers in 1 to 70 which has 1 in the unit digit = 7
Total numbers in 1 to 70 which has 9 in the unit digit = 7
Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14
Required percentage = (14/70)×100=140/7=20%
9. 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, what was the number of valid votes that the other candidate got? | |
A. 2800 | B. 2700 |
C. 2100 | D. 2500 |
Answer : Option B
Explanation :
Total number of votes = 7500
Given that 20% of Percentage votes were invalid
=> Valid votes = 80%
Total valid votes = (7500×80)/100
1st candidate got 55% of the total valid votes.
Hence the 2nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = (total valid votes ×45)/100
=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700
10. In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State? | |
A. 8200 | B. 7500 |
C. 7000 | D. 8000 |
Answer : Option D
Explanation :
State A and State B had an equal number of candidates appeared.
In state A, 6% candidates got selected from the total appeared candidates
In state B, 7% candidates got selected from the total appeared candidates
But in State B, 80 more candidates got selected than State A
From these, it is clear that 1% of the total appeared candidates in State B = 80
=> total appeared candidates in State B = 80 x 100 = 8000
=> total appeared candidates in State A = total appeared candidates in State B = 8000
1. What is the role of percentage in the quantitative aptitude section of the Civil Service Examination? |
2. How can I improve my skills in solving percentage-related questions for the Civil Service Examination? |
3. Can you provide some examples of percentage-based questions that may appear in the Civil Service Examination? |
4. Are there any shortcuts or tricks to solve percentage-related questions quickly in the Civil Service Examination? |
5. Are there any recommended books or resources to prepare specifically for the percentage-related questions in the Civil Service Examination? |
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