Percentage Aptitude Questions PDF

Percentage Aptitude Questions PDF

The word “percent” is derived from the latin words “per centum”, which means “per hundred”.

A percentage is a fraction with a denominator of hundred, It is denoted by the symbol %.
Numerator of the fraction is called the rate per cent.
VALUE OF PERCENTAGE:

Value of percentage always depends on the quantity to which it refers: Consider the statement, “65% of the students in this class are boys”. From the context, it is understood that boys from 65% of the total number of students in the class. To know the value of 65%, the value of the total number of student should be known. If the total number of student is 200, then,

The number of boys =130;

It can also be written as (200) × (0.65) =130.

Note that the expressions 6%, 63%, 72%, 155% etc. Do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.

To express the fraction equivalent to %:

Express the fraction with the denominator 100, then the numerator is the answer.

Example 1:

Express the fraction 11/12 into the per cent.

Solution:

11/12=(11/12×100)/100=(91 2/3)/100=912/3%
To express % equivalent to fraction:
a% =a/100

Example 2:

Express 45 5/6% into fraction.

Solution:

45 5/6% = (45 5/6)/100=275/(6×100)=11/24.

Example 3:

Rent of the house is increased from ` 7000 to `7700. Express the increase in price as a percentage of the original rent.

Solution:

Increase value = Rs 7700 – Rs 7000 = Rs 700
Increase % = (Increas value)/(Original value)×100= 700/7000×100=10
∴ Percentage rise = 10%

Example 4:

The cost of a bike last year was Rs19000. Its cost this year is Rs 17000. Find the per cent decrease in its cost.
Decrease % = (Decreas value)/(Original value) × 100

% decrease = (19000-17000)/19000×100
=2000/19000×100= 10.5%.
∴ Percentage decrease = 10.5%.
If A is x % if C and B is y % of C, then A is x/y × 100% of B.

Example 5:

A positive number is divided by 5 instead of being multiplied by 5. By what per cent is the result of the required correct value?

Solution:

Let the number be 1, then the correct answer = 5

The incorrect answer that was obtained =.

The required % =  = 4%

If two numbers are respectively x% and y% more than a third number, then the first number is % of the second and the second is % of the first.

If two numbers are respectively x% and y% less than a third number, then the first number if % of the second and the second is % of the first.

x% of a quantity is taken by the first, y% of the remaining is taken by the second and    z% of the remaining is taken by third person. Now, if A is left in the fund, then the  initial amount

=(A×100×100×100)/((100-x)(100-y)(100-z)) in the beginning.

x% of a quantity is added. Again, y% of the increased quantity is added. Again z% of the increased quantity is added. Now it becomes A, then the initial amount

=(A×100×100×100)/((100+x)(100+y)(100+z))

Example 6:

3.5% income is taken as tax and 12.5% of the remaining is saved. This leaves Rs. 4,053 to spend. What is the income?

Solution:

By direct method,

Income = (4053×100×100)/((100-3.5)(100-12.5)) = Rs 4800.

If the price of a commodity increases by r%, then reduction in consumption, so as not to increase the expenditure is (r/(100+r)×100)%.

If the price of a commodity decreases by r%, then the increase in consumption, so as not to decrease the expenditure is (r/(100-r)×100)%.

Example 7:

If the price of coal be raised by 20%, then find by how much a householder must reduce his consumption of this commodity so as not to increase his expenditure?

Solution:

Reduction in consumption = (20/(100+20)×100)%

= (20/(100+20)×100)% = 16.67%

POPULATION FORMULA

If the original population of a town is P, and the annual increase is r%, then the population after n years is P(1+r/100)^n and population before n years = P/(1+r/100)^n

If the annual decrease be r%, then the population after n years is P(1-r/100)^n and population before n years = P/(1+r/100)^n

Example 8:

The population of a certain town increased at a certain rate per cent annum. Now it is 456976. Four years ago, it was 390625. What will it be 2 years hence?

Solution:

Suppose the population increases at r% per annum. Then, 390625 (1+r/100)^4 = 456976

(1+r/100)^2 = √(456976/390625)= 676/625

Population 2 years hence = 456976 (1+r/100)^2

= 456976 × 676/625 = 494265 approximately.

Example 9:

The population of a city increase at the rate of 4% per annum. There is an additional annual increase of 1% in the population due to the influx of job seekers. Find percentage increase in the population after 2 years.

Solution:

The net annual increase = 5%

Let the initial population be 100.

Then, population after 2 years = 100×1.05×1.05 = 110.25

Therefore, % increase in population = (110.25-100) = 10.25%

If a number A is increased successively by x% followed by y% and then z%, then the final value of A will be A(1+x/100)(1+y/100)(1+z/100)

In case a given value decreases by an percentage then we will use negative sign before that.

First Increase and then decrease:

If the value is first increased by x% and then decreased by y% then there is (x-y-xy/100)% increase or decrease, according to the +ve or –ve sign respectively.

If the value is first increased by x% and then decreased by x% then there is only decrease which is equal to (x^2/100).

Example 10:

A number is increased by 10% and then it is decreased by 10%. Find the net increase or decrease per cent.

Solution:

% change = (10×10)/100=1%

i.e. 1% decrease.

Average percentage rate of change over a period.

=((New Value-Old Value))/(Old Value)×100/n% where n = period.

The percentage error = (The Error)/(True Value)×100%

SUCCESSIVE INCREASE OR DECREASE

In the value is increased successively by x% and y% then the final increase is given     by (x+y+xy/100)%

In the value is decreased successively by x% and y% then the final decrease is given     by (-x-y-xy/100)%

Example 11:

The price of a car is decreased by 10% and    20% in two successive years. What per cent    of price of a car is decreased after two      years?

Solution:

Put x = -10 and y = -20, then

-10-20+

The price of the car decreases by 28%.

STUDENT AND MARKS

The percentage of passing marks in an examination is x%. If a candidate who scores y marks fails by z marks, then the maximum marks M = 100(y+z)/x

A candidate scoring x% in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more then the minimum required passing marks. Then the maximum marks M = 100(a+b)/x

In an examination x% and y% students respectively fail in two different subjects while z% students fail in both subjects then the % age of student who pass in both the subjects will be {100-(x + y – z)}%

Example 12:

Vishal requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum he could have got?

Solution:

If Vishal has 15 marks more, he could have scored 40% marks.

Now, 15 marks more then 185 is 185+15 =     200

Let the maximum marks be x, then 40% of x = 200

⇒ × x = 200  ⇒ x =500

Thus, maximum marks = 500

Alternate method:

Maximum marks =  (100(185+15))/40=(100×200)/40 = 500

Example 13:

A candidate scores 15% and fails by 30 marks, while another candidate who scores 40% marks, gets 20 marks more then the minimum required marks to pass the pass the examination. Find the maximum marks of the examination.

Solution:

By short cut method:

Maximum marks = (100(30+20))/(40-15) =200

2-DIMENSIONAL FIGURE AND AREA

If the sides of a triangle, square, rectangle, rhombus or radius of a circle are increased by a%, its area is increased by(a(a+200))/100 %

If the sides of a triangle, square, rectangle, rhombus or radius of a circle are decreased by a %

Then its area is decreased by (a(200-a))/100%.

Example 14:

If the radius of a circle is increased by 10%, what is the percentage increase in its area?

Solution:

Let R be the radius of circle.

Area of Circle, A =πR^2

Now, radius is increased by 10%

New radius, R’ = R + 10% of R = 1.1 R

New Area, A’ = π(1.1R)^2= 1.21 πR^2%

increase in area =(1.21πR^2-πR^2)/(πR^2 )×100=21%

Shortcut Method:

So, Area is increased by (10(10+200))/100 = 21%

If the both sides of rectangle are changed by x% and y% respectively, then % effect on area = x + y+xy/100 (+/- according to increase or decrease.

Example 15:

If the length and width of a rectangular garden were each increased by 20%, then what would be the per cent increase in the area of the garden?

Solution:

By direct formula

% increase in area =(20 (20+200))/100=44%

If A’s income is r% more than that of B, then B’s income is less than that of A by (r/(100+r)×100)%

If A’s income is r% less than that of B, then B’s income is more than that of A by (r/(100-r)×100)%

Example 16:

If A’s salary is 50% more than B’s, then by what percent B’s salary is less than A’s salary?

Soluti

Let B’s salary be Rs x

Then, A’s salary = x + 50% of x = 1.5x

B’s salary is less than A’s salary by ((1.5x-x)/1.5x×100)% = 100/3 = 33.33%

Shortcut method,

B’s salary is less than A’s salary by (50/(100+50)×100)%

=50/150×100% = 33.33%

Example 17:

Ravi’s weight is 25% that of Meena’s and 40% that of Tara’s. What percentage of Tara’s weight is Meena’s weight.

Solution:

Let Meena’s weight be x kg and Tara’s weight be y kg. Then Ravi’s weight = 25% of Meena’s weight

= 25/100×x …..(i)

Also, Ravi’s weight = 40% of Tara’s weight

= 40/100×y …..(ii)

From (i) and (ii), we get

25/100×x=40/100×y

⇒ 25x = 40y

⇒ 5x = 8y ⇒ x = 8/5 y

Meena’s weight as the percentage of Tara’s weight

= x/y×100= ( 8/5 y)/y×100

= 8/5×100=160

Hence, Meena’s weight is 160% of Tara’s weight.

Example 18:

The monthly salaries of A and B together amount to `50,000. A spends 80% of his salary and B spends 70% of his salary. If now their saving are the same, then find the salaries of A and B.

Solution:

Let A’s salary by x, then B’s salary (50,000-x)

A spends 80% of his salary and saves 20%

B spends 70% of his salary and saves 30%

Given that

20% of x = 30% of (50,000-x)

20/100×x=30/100×(50,000-x)

50x/100=(30×50,000)/100

⇒ x = (30×50,000×100)/(100×50)=30,000

A’s salary Rs 30,000

B’s salary = Rs 50,000 – Rs 30,000 = Rs20,000

Clock Tricks Questions in Hindi

Formulas and Quick Tricks for Percentage Problems

• a % of b = a * b/100
• If A is x% more than B, then B is less than A by: [x(100) / 100+x]%
• If A is x% less than B, then B is more than A by: [x(100) / 100-x]%
• If A is x% of C and B is y% of C, then A = x/y * B
• If the price of a commodity decreases by P %, then the increase in consumption so that the expenditure remains same, which is: [P(100) / 100-P]%
• If the price of a commodity increases by P%, then the reduction in consumption so that the expenditure remains same, which is: [P(100) / 100+P]%
• If a number is changed (increased/decreased) successively by x% and y%, then net% change is given by [x+y+(xy/100)]%, which represents increase or decrease in value according as the sign is positive or negative
• If two parameters A and B are multiplied to get a product and if A is changed by x% and another parameter B is changed by y%, then the net% change in the product (A * B) is given [x+y+(xy/100)]%
• In an examination, the minimum pass percentage is x%. If a student secures y marks and fails by z marks, then the maximum marks in the examination is 100(y+z)/x
• If a number A is increased successively by x% followed by y% and then by z%, then the final value of A will be: A(100+x / 100) (100+y / 100) (100+z / 100)

Questions and Solved Examples on Percentage Problems

For a candidate to clear an examination, he/she must score 55% marks. If he/she gets 120 and fails by 78 marks, the total marks for the examination is:
A. 300
B. 320
C. 360
D. 400

Explanation:

• Here the mark obtained by the candidate is 120 and the candidate fails by 78 marks
• Therefore the passing marks is (120+78) = 198
• Let the total marks be x. Then,
• => 55/100 * x = 198
• => x = 360

A property decreases in value every year at the rate of 6 1/4% of its value at the beginning of the year. It’s value at the end of 3 years was Rs. 21,093. Find its value at the beginning of the 1st year?
A. Rs. 18,060.36
B. Rs. 18,600
C. Rs. 25,600.24
D. Rs. 32,000.50

Explanation:

• 6 1/4% = 1/16
• x * 15/16 * 15/16 * 15/16 = 21093
• x = 25600.24

80 is what percent of 64?
A. 75%
B. 85%
C. 120%
D. 125%

Explanation: Let x percent of 64 be 80

• 64 * x/100 = 80 => x = (80 * 100)/64 => x = 125
• Hence, 80 is 125% of 64

How much is 80% of 40 is greater than 4/5 of 25?
A. 4
B. 6
C. 9
D. 12

Explanation: (80/100) * 40 – (4/5) * 25

32 – 20 = 12

40% of a number is more than 20% of 650 by 190. Find the number?
A. 600
B. 700
C. 800
D. 900

Explanation: (40/100) * X – (20/100) * 650 = 190

• 2/5 X = 320
• X = 800

40 is subtracted from 60% of a number, the result is 50. Find the number?
A. 110
B. 130
C. 140
D. 150

Explanation:

• (60/100) * X – 40 = 50
• 6X = 900
• X = 150

If A got 80 marks and B got 60 marks, then what percent of A’s mark is B’s mark?
A. 60%
B. 65%
C. 75%
D. 80%

Explanation: A’s marks = 80; B’s marks = 60.

• Let x% of A = B => x/100 * 80 = 60
• => x = (60 * 100)/80 = 75
• B’s marks is 75% of A’s marks

A and B’s salaries together amount to Rs. 2,000. A spends 95% of his salary and B spends 85% of his. If now their savings are the same, what is A’s salary?
A. Rs.500
B. Rs.750
C. Rs.1250
D. Rs.1500

Explanation: (5/100) A = (15/100) B

• A = 3B
• A + B = 2000
• 4B = 2000 => B = 500
• A = 1500

5% people of a village in Sri Lanka died by bombardment, 15% of the remainder left the village on account of fear. If now the population is reduced to 3553, how much was it in the beginning?
A. 3800
B. 4200
C. 4400
D. 5500

Explanation: X * (95/100) * (85/100) = 3553

• X = 4400

In a factory, there are 40% technicians and 60% non-technicians. If the 60% of the technicians and 40% of non-technicians are permanent employees, then the percentage of workers who are temporary is?
A. 32%
B. 42%
C. 52%
D. 62%

Explanation: Total = 100

• T = 40, NT = 60
• 40*(60/100) = 24, 60*(40/100) = 24
• 24 + 24 = 48 => 100 – 48 = 52%

Clocks Practice Questions

मिनट की सुई 6 मिनट मे कितने डिग्री का कोण बनाएगी ?

• 6 डिग्री
• 12 डिग्री
• 24 डिग्री
• 36 डिग्री

12 घंटे मे संपाती कितनी बार बनता है ?

• 12
• 22
• 11
• 24

03:40 बजे मिनट और घंटे की सुई के बीच कितने डिग्री का कोण बनेगा ?

• 320 डिग्री
• 340 या 20 डिग्री
• 360 डिग्री
• कोई नहीं

गाड़ी की सुईयां सायं 2 बजे से 7 बजे तक कितनी बार समकोण बनाएगी ?

• 10 बार
• 9 बार
• 8 बार
• इनमे से कोई नहीं

4 बजकर 15 मिनट पर घंटे और मिनट की सुई के मध्य कितने डिग्री का कोण बनेगा ?

• 30 डिग्री
• 60 डिग्री
• 37.5 डिग्री
• 45 डिग्री

सुबह 4 बजे से 10 बजे के बीच घड़ी की सुईयां कितनी बार ऊपर नीचे होंगी ?

• 5 बार
• 6 बार
• 11 बार
• 23 बार

यदि कोई घड़ी किसी दर्पण मे पौने तीन बजे का समय बता रही है तो वास्तविक समय क्या होगा ?

• 07:30
• 09:15
• 08:45
• 06:45

एक कक्षा मे शिक्षक 09:55 बजे पहुँचा और पप्पू 45 मिनट बाद आया उसे 10 मिनट की देरी हुई टी शिक्षक नियमित समय से कितना समय पहले पहुँचा ?

• 10 मिनट
• 15 मिनट
• 45 मिनट
• 35 मिनट

दो घड़ियों मे से एक 5 मिनट तेज हो जाती है तथा दूसरी घड़ी 5 मिनट सुस्त हो जाती है यदि दोनों घड़िया दोपहर 12 बजे एक समान समय बताती है तो वे पुनः एक साथ कब एक समान समय बताएंगी ?

• 60
• 62
• 76
• 72

जयपुर से हर 30 मिनट मे एक बस दिल्ली के लिए रवाना होती है। एक पुछताछ क्लर्क ने बताया की एक बस अभी 10 मिनट पहले छूटी है और अगली बस 09:35 पर आएगी तो ये सूचना किस वक्त पर दी गई ?

• 09:15 बजे
• 08:55 बजे
• 09:08 बजे
• 09:10 बजे

एक बस पड़ाव मे निम्नलिखित सूचना प्रसारित की जा रही थी, सूरत के लिए 15 मिनट पहले एक बस खुल चुकी है, नियमानुसार प्रत्येक 45 मिनट के बाद सूरत के लिए एक बस है। अगली बस का निर्धारित समय 08:30 AM है बताइए की कितने बजे यह जानकारी प्रसारित की गई थी ?

• 08:00 AM
• 09:00 AM
• 10:45 AM
• 08:15 AM

दिल्ली से आगरा के लिए हर 40 मिनट मे एक बस चलती है। पूछताछ ऑफिसर ने बताया की बस 10 मिनट पहले निकाल चुकी है। दूसरी बस सुबह 10:45 मिनट पर निकलेगी। पूछताछ ऑफिसर ने यात्री को किस वक्त सूचना दी ?

• 09:55 AM
• 10:05 AM
• 10:35 AM
• 10:15 AM

मयंक अपने ऑफिस पर 08:30 पर पहुँच गया जो 15 मिनट समयपूर्व था। बताइए ऑफिस का निर्धारित समय क्या था ?

• 08:30
• 08:15
• 08:45
• 08:05

यदि जल प्रतिबिम्बित समय 10 बजकर 15 मिनट हो रहे हो तो इस स्थिति मे वास्तविक समय क्या होगा ?

• 08:11
• 11:15
• 03:50
• 02:45

यदि किसी घड़ी मे वास्तविक समय 3 बजकर 15 मिनट हो रहा हो तो इस स्थिति मे दर्पण मे समय क्या होगा ?

• 08:45
• 09:15
• 08:15
• 09:45

यदि किसी लम्बवत दर्पण मे प्रतिबिम्बित समय 1 बजकर 40 मिनट हो रहा हो तो वास्तविक समय क्या होगा ?

• 11:40
• 05:45
• 10:20
• 11:20

एक घड़ी 4 बजे का समय दर्शा रही है, घंटे कइ सुई के 90 डिग्री घूमने के बाद क्या समय होगा?

• 6 बजे
• 7 बजे
• 8 बजे
• 9 बजे

यदि किसी घड़ी मे 10 बजकर 10 मिनट हो रहे हो तो इस स्थिति में घंटे और मिनट की सुई के कोण का माप क्या होगा ?

• 105 डिग्री
• 115 डिग्री
• 125 डिग्री
• 75 डिग्री

शाम के चार बजे से शाम के 10 बजे तक घड़ी की सुइयाँ कितनी बार एक-दूसरे के साथ 90 डिग्री का कोण बनती है

• 9
• 12
• 6
• 11

7:20 मिनट पर घड़ी की बड़ी और छोटी सुई के बीच कितने अंश का कोण बनेगा ?

• 160 डिग्री
• 100 डिग्री
• 260 डिग्री
• 120 डिग्री

Maths Questions

Q.1. A sum of Rs. 25000 amounts to Rs. 31000 in 4 years at the rate of simple interest what is the rate of interest?

(a) 3%

(b) 4%

(c) 5 %

(d) 6 %

(e) None of these

Q.2. Kamla took a loan of Rs. 2400 with simple interest for as many years as the rate of interest. If she paid Rs. 864 as interest at the end of the loan period, what was the rate of interest?

(a) 3.6

(b) 6

(c) 18

(d) cannot determined

(e) None of these

Q.3. What is the present worth of Rs. 264 dues in 4 years at 10% simple interest per annum?

(a) 170.20

(b) 166

(c) 188.57

(d) 175.28

Q.4. A sum fetched a total simple interest of Rs. 8016.25 at the rate of 6 p.c.p.a in 5 years what is the sum?

(a) 24720.83

(b) 26730.33

(c) 26720.83

(d) 26710.63

(e) None of these

Q.5.  8. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs 800 become in 3 years?

(a) 1020.80

(b) 1025

(c) 1052

Q. 6.  The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is

a.   74

b.   78

c.   80

d.   90

Q.7.   The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a.   3

b.   6

c.   9

d.   8.2

Q.8.  √6+√6+√6+… is equal to –

a. 2

b. 5
c. 4
d. 3

Q.9. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Q.10. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67