Problems on Percentages

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1.  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.	45 5 % 11
C.	54 6 % 11	D.	55%

Solution : Number of runs made by running = 110 - (3 x 4 + 8 x 6)

= 110 - (60)

= 50.

Required percentage =		50	x 100	% = 45	5	%
		110			11
2. 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 Solution : Let the amount of taxable purchases be Rs.x. Then, 6% of x = 30/100 x ‹=› (30/100×100/6) = 5. Cost of tax free items= Rs.[25 - (5 + 0.30)] = Rs. 19.70 3. A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction ?   A.  8% B.  9%   C.  10% D.  11% Solution : Actual price= Rs. (25 + 2.50) = Rs. 27.50 Therefore, saving = (2.50 / 27.50 ×100)% = 100 / 11% = 9×1/11% = 9%. 4. 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 Solution : Let their marks be (x+9) and x. Then, x+9 = 56/100(x + 9 +x) ‹=› 25(x+9) ‹=› 14 (2x + 9) ‹=›3x = 99 ‹=›x = 33. SO, their marks are 42 and 33.

5. The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?  

 A.  Rs. 32.500 B.  Rs. 48,750   C.  Rs. 76,375 D.  Rs. 81,250

Solution :

 Amount paid to car owner= 90% of 85% of Rs. 3,25,000.

                                           = Rs. (90/100 ×85/100 ×325000)

                                           = Rs. 2,48,625.

Required differene= Rs. (325000 - 248625) = Rs. 76,375.

6. What percent of a day is 3 hours?  

A.  12×1/2% B.  16×1/2%   C.  18×2/3% D.  22×1/2%

Solution :

Required percentage= (3/24×100)% = (25 / 2)%. = 12×1/2%.

7. If number x is 10% less than another number y and y is 10% more than 125, then x is equal to   A.  123.75 B.  140.55   C.  143 D.  150

Solution :

 y = 125 + 10% of 125

                     ‹=› 125 + 12.50 = 137.50.

x = 137.50 - 10% of 137.50‹=› 137.50 - 13.75 = 123.75.

8. 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%

Solution : Increase in 10 years= (262500 - 175000) = 87500.

Increase%= (87500/175000×100)% = 50%.

Required average= (50/10)% = 5%.

9. A student has to obtain 33% of the total marks to pass. He got 125 marks and failed by 40 marks. The maximum marks are

  A.  300 B.  500   C.  800 D.  1000

Solution :

Let their maximum marks be
Then, 33% of x = 125 + 40
	‹=› 33/100×x= 165
	x‹=› (165×100/33)
	‹=› 500.

10. In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination ?  

 A.  30,000 B.  35,000   C.  37,000 D.  None of these

Solution :

Let the number of applicants be x.Number of eligible candidates	= 95% of x.
Eligible candidates of each other categories	= 15% of (95% of x).
	=(15/100×95/100×x)
	= 57/400×x.
Therefore, 57/400×x	= 4275
	‹=› x =(4275×400 / 57)
	‹=› 30000.

11. How many litres of pure acid are there in 8 litres of a 20% solution?  

A.  1.4 B.  1.5   C.  1.6 D.  2.4

Solution : Quantity of pure acid= 20% of 8 litres = (20/100×8)litres = 1.6 litres.

12. 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

Solution :

20% of a = b	20	a = b.
	100

b% of 20 =		b	x 20		=		20	a x	1	x 20		=	4	a = 4% of a.
		100					100		100				100

13. 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

Solution : 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 +	2	of 48
	3

	80	x = 80
	100

                        x = 100.

14 . 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

Solution :

5% of A + 4% of B =	2	(6% of A + 8% of B)
	3

	5	A +	4	B	=	2	(	6	A +	8	B	)
	100		100			3		100		100

	1	A +	1	B	=	1	A +	4	B
	20		25			25		75

		1	-	1	A =		4	-	1	B
		20		25			75		25

	1	A =	1	B
	100		75

A	=	100	=	4	.
B		75		3

  Required ratio = 4 : 3

15 ) A person spends 20% of his income on rent, 20% of the rest on food, 10% of the remaining on clothes and 10% on groceries. If he is left with Rs. 9520/- find his income.

a. 10000/-

b. 15000/-

c. 20000/-

d. None

Solution :Three successive decreases of 20%, 20% and 10% => 0.8 X 0.8 X 0.9 = 0.576

Again 10% decrease => 0.576 – 0.1 = 0.476.

So, 0.476 x = 9520 => x = 20000

16) From a 20lt solution of alt and water with 20% salt, 2lt of water is evaporated. Find the new % concentration of salt

a. 20%

b. 23%

c. 25%

d. None

Solution:In 20lt, salt = 20% => 4 lt.

New volume = 18 lt (2 lt evaporated)

So, new % = 4/18 X 100 = 22.22%