Friday, 21 October 2016

Problems on Percentages

  1. Concept of Percentage:
    By a certain percent, we mean that many hundredths.
    Thus, x percent means x hundredths, written as x%.
    To express x% as a fraction: We have, x% =x.
    100
        Thus, 20% =20=1.
    1005
    To expressaas a percent: We have,a=ax 100%.
    bbb
        Thus,1=1x 100%= 25%.
    44
  2. Percentage Increase/Decrease:
    If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:
    Rx 100%
    (100 + R)
    If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:
    Rx 100%
    (100 - R)
  3. Results on Population:
    Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
    1. Population after n years = P1 +Rn
    100
    2. Population n years ago =P
    1 +Rn
    100
  4. Results on Depreciation:
    Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
    1. Value of the machine after n years = P1 -Rn
    100
    2. Value of the machine n years ago =P
    1 -Rn
    100
    3. If A is R% more than B, then B is less than A byRx 100%.
    (100 + R)
    4. If A is R% less than B, then B is more than A byRx 100%.
    (100 - R)


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.45%
B.
455%
11
C.
546%
11
D.55%
Answer: Option B
Explanation:
Number of runs made by running = 110 - (3 x 4 + 8 x 6)
= 110 - (60)
= 50.
 Required percentage =50x 100% = 455%
11011
2. 
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:
A.39, 30
B.41, 32
C.42, 33
D.43, 34
Answer: Option C
Explanation:
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.
3. 
A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
A.588 apples
B.600 apples
C.672 apples
D.700 apples
Answer: Option D
Explanation:
Suppose originally he had x apples.
Then, (100 - 40)% of x = 420.
60x = 420
100
 x =420 x 100  = 700.
60
4. 
What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?
A.1
B.14
C.20
D.21
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
 Required percentage =14x 100% = 20%.
70
5. 
If A = x% of y and B = y% of x, then which of the following is true?
A.A is smaller than B.
B.A is greater 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.None of these
Answer: Option E
Explanation:
x% of y =xy=yxy% of x
100100
 A = B.
6. 
If 20% of a = b, then b% of 20 is the same as:
A.4% of a
B.5% of a
C.20% of a
D.None of these
Answer: Option A
Explanation:
20% of a = b      20a = b.
100
 b% of 20 =bx 20=20a x1x 20=4a = 4% of a.
100100100100
7. 
In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A.72
B.80
C.120
D.150
E.100
Answer: Option E
Explanation:
Let the number of students be x. Then,
Number of students above 8 years of age = (100 - 20)% of x = 80% of x.
 80% of x = 48 +2of 48
3
80x = 80
100
 x = 100.
8. 
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 : 3
B.1 : 1
C.3 : 4
D.4 : 3
Answer: Option D
Explanation:
5% of A + 4% of B =2 (6% of A + 8% of B)
3
5 A +4 B=26 A +8 B
1001003100100
1 A +1 B=1 A +4 B
20252575
1-1 A = 4-1 B
20257525
1 A =1 B
10075
A=100=4.
B753
 Required ratio = 4 : 3
9. 
A student multiplied a number by3instead of5.
53
What is the percentage error in the calculation?
A.34%
B.44%
C.54%
D.64%
Answer: Option D
Explanation:
Let the number be x.
Then, error =5x -3x =16x.
3515
Error% =16xx3x 100% = 64%.
155x
10. 
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.2700
B.2900
C.3000
D.3100
Answer: Option A
Explanation:
Number of valid votes = 80% of 7500 = 6000.
 Valid votes polled by other candidate = 45% of 6000
=45x 6000= 2700.
100
11. 
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.60%
C.65%
D.90%
Answer: Option A
Explanation:
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
 Required percentage =11628x 100% = 57%.
20400
12. 
Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
A.Rs. 200
B.Rs. 250
C.Rs. 300
D.None of these
Answer: Option B
Explanation:
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
 z +120z = 550
100
11z = 550
5
 z =550 x 5  = 250.
11
13. 
Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
A.Rs. 15
B.Rs. 15.70
C.Rs. 19.70
D.Rs. 20
Answer: Option C
Explanation:
Let the amount taxable purchases be Rs. x.
Then, 6% of x =30
100
 x =30x100 = 5.
1006
 Cost of tax free items = Rs. [25 - (5 + 0.30)] = Rs. 19.70
14. 
Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.
A.Rs. 6876.10
B.Rs. 6999.20
C.Rs. 6654
D.Rs. 7000
Answer: Option A
Explanation:
Rebate = 6% of Rs. 6650 = Rs.6x 6650= Rs. 399.
100
Sales tax = 10% of Rs. (6650 - 399) = Rs.10x 6251= Rs. 625.10
100
 Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
15. 
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
A.4.37%
B.5%
C.6%
D.8.75%
Answer: Option B
Explanation:
Increase in 10 years = (262500 - 175000) = 87500.
Increase% =87500x 100% = 50%.
175000
 Required average =50% = 5%.
10

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