Wednesday, July 4, 2012

Aptitude Questions part 1


1.If 2x-y=4 then 6x-3y=?
(a)15
(b)12
(c)18
(d)10

Ans. (b)
2.If x=y=2z and xyz=256 then what is the value of x?

(a)12
(b)8
(c)16
(d)6

Ans. (b)
3. (1/10)18 - (1/10)20 = ?
(a) 99/1020
(b) 99/10
(c) 0.9
(d) none of these
Ans. (a)
4.Pipe A can fill in 20 minutes and Pipe B in 30 mins
and Pipe C can
empty the same in 40 mins.If all of them work
together, find the time
taken to fill the tank
(a) 17 1/7 mins
(b) 20 mins
(c) 8 mins
(d) none of these
Ans. (a)
5. Thirty men take 20 days to complete a job working 9
hours a day.How
many hour a day should 40 men work to complete the
job?
(a) 8 hrs
(b) 7 1/2 hrs
(c) 7 hrs
(d) 9 hrs
Ans. (b)
6. Find the smallest number in a GP whose sum is 38
and product 1728
(a) 12
(b) 20
(c) 8
(d) none of these
Ans. (c)
7. A boat travels 20 kms upstream in 6 hrs and 18 kms
downstream in 4
hrs.Find the speed of the boat in still water and the
speed of the
water current?
(a) 1/2 kmph
(b) 7/12 kmph
(c) 5 kmph
(d) none of these
Ans. (b)
8. A goat is tied to one corner of a square plot of
side 12m by a rope
7m long.Find the area it can graze?
(a) 38.5 sq.m
(b) 155 sq.m
(c) 144 sq.m
(d) 19.25 sq.m
Ans. (a)
9. Mr. Shah decided to walk down the escalator of a
tube station. He
found that if he walks down 26 steps, he requires 30
seconds to
reach the bottom. However, if he steps down 34 stairs
he would only
require 18 seconds to get to the bottom. If the time
is measured from
the moment the top step begins to descend to the
time he steps off
the last step at the bottom, find out the height of
the stair way in
steps?
Ans.46 steps.
10. The average age of 10 members of a committee is
the same as it was
4 years ago, because an old member has been replaced
by a young
member. Find how much younger is the new member ?
Ans.40 years.
11. Three containers A, B and C have volumes a, b, and
c respectively;
and container A is full of water while the other two
are empty. If
from container A water is poured into container B
which becomes 1/3
full, and into container C which becomes 1/2 full, how
much water is
left in container A?
12. ABCE is an isosceles trapezoid and ACDE is a
rectangle. AB = 10
and EC = 20. What is the length of AE?
Ans. AE = 10.
13. In the given figure, PA and PB are tangents to the
circle at A and
B respectively and the chord BC is parallel to
tangent PA. If AC = 6
cm, and length of the tangent AP is 9 cm, then what
is the length of
the chord BC?
Ans. BC = 4 cm.
15 Three cards are drawn at random from an ordinary
pack of cards.
Find the probability that they will consist of a king,
a queen and an ace.
Ans. 64/2210.
16. A number of cats got together and decided to kill
between them
999919 mice. Every cat killed an equal number of
mice. Each cat
killed more mice than there were cats. How many cats
do you think
there were ?
Ans. 991.
17. If Log2 x - 5 Log x + 6 = 0, then what would the
value / values of
x be?
Ans. x = e2 or e3.
18. The square of a two digit number is divided by
half the number.
After 36 is added to the quotient, this sum is then
divided by 2.
The digits of the resulting number are the same as
those in the
original number, but they are in reverse order. The
ten's place of
the original number is equal to twice the difference
between its
digits. What is the number?
Ans. 46
19.Can you tender a one rupee note in such a manner
that there shall
be total 50 coins but none of them would be 2 paise
coins.?
Ans. 45 one paisa coins, 2 five paise coins, 2 ten
paise coins, and 1
twenty-five paise coins.
20.A monkey starts climbing up a tree 20ft. tall. Each
hour, it hops
3ft. and slips back 2ft. How much time would it take
the monkey to
reach the top?
Ans.18 hours.
21. What is the missing number in this series? 8 2
14 6 11 ? 14 6 18 12
Ans. 9
22. A certain type of mixture is prepared by mixing
brand A at Rs.9 a
kg. with brand B at Rs.4 a kg. If the mixture is worth
Rs.7 a kg., how
many kgs. of brand A are needed to make 40kgs. of
the mixture?
Ans. Brand A needed is 24kgs.
23. A wizard named Nepo says "I am only three times my
son's age. My
father is 40 years more than twice my age. Together
the three of us
are a mere 1240 years old." How old is Nepo?
Ans. 360 years old.
24. One dog tells the other that there are two dogs in
front of me.
The other one also shouts that he too had two behind
him. How many are
they?
Ans. Three.
25. A man ate 100 bananas in five days, each day
eating 6 more than
the previous day. How many bananas did he eat on the
first day?
Ans. Eight.
26. If it takes five minutes to boil one egg, how long
will it take to
boil four eggs?
Ans. Five minutes.
27. The minute hand of a clock overtakes the hour hand
at intervals of
64 minutes of correct time. How much a day does the
clock gain or lose?
Ans. 32 8/11 minutes.
28. Solve for x and y: 1/x - 1/y = 1/3, 1/x2 + 1/y2
= 5/9.
Ans. x = 3/2 or -3 and y = 3 or -3/2.
29. Daal is now being sold at Rs. 20 a kg. During last
month its rate
was Rs. 16 per kg. By how much percent should a family
reduce its
consumption so as to keep the expenditure fixed?
Ans. 20 %.
30. Find the least value of 3x + 4y if x2y3 = 6.
Ans. 10.
31. Can you find out what day of the week was January
12, 1979?
Ans. Friday.
32. A garrison of 3300 men has provisions for 32 days,
when given at a
rate of 850 grams per head. At the end of 7 days a
reinforcement
arrives and it was found that now the provisions will
last 8 days
less, when given at the rate of 825 grams per head.
How, many more men
can it feed?
Ans. 1700 men.
33. From 5 different green balls, four different blue
balls and three
different red balls, how many combinations of balls
can be chosen
taking at least one green and one blue ball?
Ans. 3720.
34. Three pipes, A, B, & C are attached to a tank. A &
B can fill it
in 20 & 30 minutes respectively while C can empty it
in 15 minutes.
If A, B & C are kept open successively for 1 minute
each, how soon
will the tank be filled?
Ans. 167 minutes.
35. A person walking 5/6 of his usual rate is 40
minutes late. What is
his usual time? Ans. 3 hours 20 minutes.

36.For a motorist there are three ways going from City
A to City C. By
way of bridge the distance is 20 miles and toll is
$0.75. A tunnel
between the two cities is a distance of 10 miles and
toll is $1.00 for
the vehicle and driver and $0.10 for each passenger. A
two-lane
highway without toll goes east for 30 miles to city B
and then 20
miles in a northwest direction to City C.


1. Which is the shortest route from B to C

(a) Directly on toll free highway to City C
(b) The bridge
(c) The Tunnel
(d) The bridge or the tunnel
(e) The bridge only if traffic is heavy on the toll
free highway

Ans. (a)


2. The most economical way of going from City A to
City B, in terms of
toll and distance is to use the

(a) tunnel
(b) bridge
(c) bridge or tunnel
(d) toll free highway
(e) bridge and highway

Ans. (a)


3. Jim usually drives alone from City C to City A
every working day.
His firm deducts a percentage of employee pay for
lateness. Which
factor would most influence his choice of the bridge
or the tunnel ?

(a) Whether his wife goes with him
(b) scenic beauty on the route
(c) Traffic conditions on the road, bridge and tunnel
(d) saving $0.25 in tolls
(e) price of gasoline consumed in covering additional
10 miles on the
bridge

Ans. (a)


4. In choosing between the use of the bridge and the
tunnel the chief
factor(s) would be:
I. Traffic and road conditions
II. Number of passengers in the car
III. Location of one's homes in the center or
outskirts of one of the
cities
IV. Desire to save $0.25

(a) I only
(b) II only
(c) II and III only
(d) III and IV only
(e) I and II only

Ans. (a)


37.The letters A, B, C, D, E, F and G, not necessarily
in that order,
stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F

1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F

Ans. (a)


2. A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E

Ans. (a)


3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16

Ans. (a)

4. A - F = ?
(a) 1
(b) 2
(c) 3
(d) 4
(e) Cannot be determined

Ans. (a)


5. An integer T is as much greater than C as C is
greater than E. T
can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined

Ans. (a)


6. The greatest possible value of C is how much
greater than the
smallest possible value of D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6

Ans. (a)



38.
1. All G's are H's
2. All G's are J's or K's
3. All J's and K's are G's
4. All L's are K's
5. All N's are M's
6. No M's are G's


1. If no P's are K's, which of the following must be
true?

(a) All P's are J's
(b) No P is a G
(c) No P is an H
(d) If any P is an H it is a G
(e) If any P is a G it is a J

Ans. (a)


2. Which of the following can be logically deduced
from the conditions
stated?

(a) No M's are H's
(b) No M's that are not N's are H's
(c) No H's are M's
(d) Some M's are H's
(e) All M's are H's

Ans. (a)


3. Which of the following is inconsistent with one or
more of the
conditions?

(a) All H's are G's
(b) All H's that are not G's are M's
(c) Some H's are both M's and G's
(d) No M's are H's
(e) All M's are H's

Ans. (a)


4. The statement "No L's are J's" is
I. Logically deducible from the conditions stated
II. Consistent with but not deducible from the
conditions stated
III. Deducible from the stated conditions together
with the additional
statement "No J's are K's"

(a) I only
(b) II only
(c) III only
(d) II and III only
(e) Neither I, II nor III

Ans. (a)



39.In country X, democratic, conservative and justice
parties have
fought three civil wars in twenty years. TO restore
stability an
agreement is reached to rotate the top offices
President, Prime
Minister and Army Chief among the parties so that each
party controls
one and only one office at all times. The three top
office holders
must each have two deputies, one from each of the
other parties. Each
deputy must choose a staff composed of equally members
of his or her
chiefs party and member of the third party.

1. When Justice party holds one of the top offices,
which of the
following cannot be true

(a) Some of the staff members within that office are
justice party members
(b) Some of the staff members within that office are
democratic party
members
(c) Two of the deputies within the other offices are
justice party members
(d) Two of the deputies within the other offices are
conservative
party members
(e) Some of the staff members within the other offices
are justice
party members.

Ans. (a)


2. When the democratic party holds presidency, the
staff of the prime
minister's deputies are composed
I. One-fourth of democratic party members
II. One-half of justice party members and one-fourth
of conservative
party members
III. One-half of conservative party members and
one-fourth of justice
party members.

(a) I only
(b) I and II only
(c) II or III but not both
(d) I and II or I and III
(e) None of these

Ans. (a)


3. Which of the following is allowable under the rules
as stated:

(a) More than half of the staff within a given office
belonging to a
single party
(b) Half of the staff within a given office belonging
to a single party
(c) Any person having a member of the same party as
his or her
immediate superior
(d) Half the total number of staff members in all
three offices
belonging to a single party
(e) Half the staff members in a given office belonging
to parties
different from the party of the top office holder in
that office.

Ans. (a)


4. The office of the Army Chief passes from
Conservative to Justice
party. Which of the following must be fired.

(a) The democratic deputy and all staff members
belonging to Justice party
(b) Justice party deputy and all his or hers staff
members
(c) Justice party deputy and half of his Conservative
staff members in
the chief of staff office
(d) The Conservative deputy and all of his or her
staff members
belonging to Conservative party
(e) No deputies and all staff members belonging to
conservative parties.

Ans. (a)



40.In recommendations to the board of trustees a
tuition increase of
$500 per year, the president of the university said
"There were no
student demonstrations over the previous increases of
$300 last year
and $200 the year before". If the president's
statement is accurate
then which of the following can be validly inferred
from the
information given:
I. Most students in previous years felt that the
increases were
justified because of increased operating costs.
II. Student apathy was responsible for the failure of
students to
protest the previous tuition increases.
III. Students are not likely to demonstrate over new
tuition increases.

(a) I only
(b) II only
(c) I or II but not both
(d) I, II and III
(e) None

Ans. (a)

41. The office staff of XYZ corporation presently
consists of three
bookeepers--A, B, C and 5 secretaries D, E, F, G, H.
The management is
planning to open a new office in another city using 2
bookeepers and 3
secretaries of the present staff . To do so they plan
to seperate
certain individuals who don't function well together.
The following
guidelines were established to set up the new office
I. Bookeepers A and C are constantly finding fault
with one another
and should not be sent together to the new office as a
team
II. C and E function well alone but not as a team ,
they should be
seperated
III. D and G have not been on speaking terms and
shouldn't go together
IV Since D and F have been competing for promotion
they shouldn't be a
team
1.If A is to be moved as one of the bookeepers,which
of the following
cannot be a possible working unit.

A.ABDEH
B.ABDGH
C.ABEFH
D.ABEGH

Ans.B


2.If C and F are moved to the new office,how many
combinations are
possible

A.1
B.2
C.3
D.4

Ans.A


3.If C is sent to the new office,which member of the
staff cannot go
with C

A.B
B.D
C.F
D.G

Ans.B


4.Under the guidelines developed,which of the
following must go to the
new office

A.B
B.D
C.E
D.G

Ans.A


5.If D goes to the new office,which of the following
is/are true

I.C cannot go
II.A cannot go
III.H must also go

A.I only
B.II only
C.I and II only
D.I and III only

Ans.D


42.After months of talent searching for an
administrative assistant to
the president of the college the field of applicants
has been narrowed
down to 5--A, B, C, D, E .It was announced that the
finalist would be
chosen after a series of all-day group personal
interviews were
held.The examining committee agreed upon the following
procedure

I.The interviews will be held once a week
II.3 candidates will appear at any all-day interview
session
III.Each candidate will appear at least once
IV.If it becomes necessary to call applicants for
additonal
interviews, no more 1 such applicant should be asked
to appear the
next week
V.Because of a detail in the written applications,it
was agreed that
whenever candidate B appears, A should also be
present.
VI.Because of travel difficulties it was agreed that C
will appear for
only 1 interview.
1.At the first interview the following candidates
appear A,B,D.Which
of the follwing combinations can be called for the
interview to be
held next week.

A.BCD
B.CDE
C.ABE
D.ABC

Ans.B


2.Which of the following is a possible sequence of
combinations for
interviews in 2 successive weeks

A.ABC;BDE
B.ABD;ABE
C.ADE;ABC
D.BDE;ACD

Ans.C


3.If A ,B and D appear for the interview and D is
called for
additional interview the following week,which 2
candidates may be
asked to appear with D?

I. A
II B
III.C
IV.E
A.I and II
B.I and III only
C.II and III only
D.III and IV only

Ans.D


4.Which of the following correctly state(s) the
procedure followed by
the search committee

I.After the second interview all applicants have
appeared at least once
II.The committee sees each applicant a second time
III.If a third session,it is possible for all
applicants to appear at
least twice

A.I only
B.II only
C.III only
D.Both I and II

Ans.A


43. A certain city is served by subway lines A,B and C
and numbers 1 2
and 3
When it snows , morning service on B is delayed
When it rains or snows , service on A, 2 and 3 are
delayed both in the
morning and afternoon
When temp. falls below 30 degrees farenheit afternoon
service is
cancelled in either the A line or the 3 line,
but not both.
When the temperature rises over 90 degrees farenheit,
the afternoon
service is cancelled in either the line C or the
3 line but not both.
When the service on the A line is delayed or
cancelled, service on the
C line which connects the A line, is delayed.
When service on the 3 line is cancelled, service on
the B line which
connects the 3 line is delayed.
Q1. On Jan 10th, with the temperature at 15 degree
farenheit, it
snows all day. On how many lines will service be
affected, including both morning and afternoon.
(A) 2
(B) 3
(C) 4
(D) 5
Ans. D

Q2. On Aug 15th with the temperature at 97 degrees
farenheit it begins
to rain at 1 PM. What is the minimum number
of lines on which service will be affected?
(A) 2
(B) 3
(C) 4
(D) 5
Ans. C

Q3. On which of the following occasions would service
be on the
greatest number of lines disrupted.
(A) A snowy afternoon with the temperature at 45
degree farenheit
(B) A snowy morning with the temperature at 45 degree
farenheit
(C) A rainy afternoon with the temperature at 45
degree farenheit
(D) A rainy afternoon with the temperature at 95
degree farenheit
Ans. B

44. In a certain society, there are two marriage
groups, red and
brown. No marriage is permitted within a group. On
marriage, males
become part of their wives groups; women remain in
their own group.
Children belong to the same group as their parents.
Widowers and
divorced males revert to the group of their birth.
Marriage to more
than one person at the same time and marriage to a
direct descendant
are forbidden
Q1. A brown female could have had
I. A grandfather born Red
II. A grandmother born Red
III Two grandfathers born Brown
(A) I only
(B) III only
(C) I, II and III
(D) I and II only
Ans. D

Q2. A male born into the brown group may have
(A) An uncle in either group
(B) A brown daughter
(C) A brown son
(D) A son-in-law born into red group
Ans. A

Q3. Which of the following is not permitted under the
rules as stated.
(A) A brown male marrying his father's sister
(B) A red female marrying her mother's brother
(C) A widower marrying his wife's sister
(D) A widow marrying her divorced daughter's
ex-husband
Ans. B

Q4. If widowers and divorced males retained their
group they had upon
marrying which of the following would be permissible (
Assume that no
previous marriage occurred)
(A) A woman marrying her dead sister's husband
(B) A woman marrying her divorced daughter's
ex-husband
(C) A widower marrying his brother's daughter
(D) A woman marrying her mother's brother who is a
widower.
Ans. D

Q5. I. All G's are H's
II. All G's are J's or K's
III All J's and K's are G's
IV All L's are K's
V All N's are M's
VI No M's are G's
45. There are six steps that lead from the first to
the second floor.
No two people can be on the same step
Mr. A is two steps below Mr. C
Mr. B is a step next to Mr. D
Only one step is vacant ( No one standing on that step
)
Denote the first step by step 1 and second step by
step 2 etc.
1. If Mr. A is on the first step, Which of the
following is true?
(a) Mr. B is on the second step
(b) Mr. C is on the fourth step.
(c) A person Mr. E, could be on the third step
(d) Mr. D is on higher step than Mr. C.
Ans: (d)
2. If Mr. E was on the third step & Mr. B was on a
higher step than
Mr. E which step must be vacant
(a) step 1
(b) step 2
(c) step 4
(d) step 5
(e) step 6
Ans: (a)
3. If Mr. B was on step 1, which step could A be on?
(a) 2&e only
(b) 3&5 only
(c) 3&4 only
(d) 4&5 only
(e) 2&4 only
Ans: (c)
4. If there were two steps between the step that A was
standing and
the step that B was standing on, and A was on a higher
step than D , A
must be on step
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
Ans: (c)

5. Which of the following is false

i. B&D can be both on odd-numbered steps in one
configuration
ii. In a particular configuration A and C must either
both an odd
numbered steps or both an even-numbered steps
iii. A person E can be on a step next to the vacant
step.
(a) i only
(b) ii only
(c) iii only
(d) both i and iii
Ans: (c)

46. Six swimmers A, B, C, D, E, F compete in a race.
The outcome is as
follows.
i. B does not win.
ii. Only two swimmers separate E & D
iii. A is behind D & E
iv. B is ahead of E , with one swimmer intervening
v. F is a head of D
1. Who stood fifth in the race ?
(a) A
(b) B
(c) C
(d) D
(e) E
Ans: (e)
2. How many swimmers seperate A and F ?
(a) 1
(b) 2
(c) 3
(d) 4
(e) cannot be determined
Ans: (d)
3. The swimmer between C & E is
(a) none
(b) F
(c) D
(d) B
(e) A
Ans: (a)


4. If the end of the race, swimmer D is disqualified
by the Judges
then swimmer B finishes in which place
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Ans: (b)
47. Five houses lettered A,B,C,D, & E are built in a
row next to each
other. The houses are lined up in the order A,B,C,D, &
E. Each of the
five houses has a colored chimney. The roof and
chimney of each
housemust be painted as follows.
i. The roof must be painted either green,red ,or
yellow.
ii. The chimney must be painted either white, black,
or red.
iii. No house may have the same color chimney as the
color of roof.
iv. No house may use any of the same colors that the
every next house
uses.
v. House E has a green roof.
vi. House B has a red roof and a black chimney
1. Which of the following is true ?
(a) At least two houses have black chimney.
(b) At least two houses have red roofs.
(c) At least two houses have white chimneys
(d) At least two houses have green roofs
(e) At least two houses have yellow roofs
Ans: (c)
2. Which must be false ?
(a) House A has a yellow roof
(b) House A & C have different color chimney
(c) House D has a black chimney
(d) House E has a white chimney
(e) House B&D have the same color roof.
Ans: (b)
3. If house C has a yellow roof. Which must be true.
(a) House E has a white chimney
(b) House E has a black chimney
(c) House E has a red chimney
(d) House D has a red chimney
(e) House C has a black chimney
Ans: (a)
4. Which possible combinations of roof & chimney can
house
I. A red roof 7 a black chimney
II. A yellow roof & a red chimney
III. A yellow roof & a black chimney

(a) I only
(b) II only
(c) III only
(d) I & II only
(e) I&II&III
Ans: (e)
48. Find x+2y
(i). x+y=10
(ii). 2x+4y=20
Ans: (b)


49. Is angle BAC is a right angle
(i) AB=2BC
(2) BC=1.5AC
Ans: (e)
50. Is x greater than y
(i) x=2k
(ii) k=2y
Ans: (e)
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Mathematical Formulae

Numbers
-----------
1. A number is divisible by 2, if its unit’s place digit is 0, 2, 4, or 8

2. A number is divisible by 3, if the sum of its digits is divisible by 3

3. A number is divisible by 4, if the number formed by its last two digits is divisible by 4

4. A number is divisible by 8, if the number formed by its last three digits is divisible by 8

5. A number is divisible by 9, if the sum of its digits is divisible by 9

6. A number is divisible by 11, if, starting from the RHS,
(Sum of its digits at the odd place) – (Sum of its digits at even place) is equal to 0 or 11x

7. Product of two numbers = Their H. C. F. × Their L. C. M.

Simple Interest
-----------------
1. Let Principle = P, Rate = R% per annum and Time = T years. Then,

a. S.I. = ( P × R × T ) / 100

b. P = ( 100 × S.I. ) / ( R × T ),

c. R = ( 100 × S.I. ) / ( P × T ),

d. T = ( 100 × S.I. ) / ( P × R ).

Compound Interest
---------------------
1. Let Principle = P, Rate = R% per annum and Time = T years. Then,

I. When interest is compounded Annually,
Amount = P (1 + R/100)N

II. When interest is compounded Half-yearly:
Amount = P (1 + R/2/100)2N

III. When interest is compounded Quarterly:
Amount = P (1 + R/4/100)4N

2. When interest is compounded Annually, but the time is in fraction, say 3⅞ years.
Then, Amount = P (1 + R/100)3 × (1 + ⅞R/100)

3. When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd, and 3rd year
respectively,
Then, Amount = P (1 + R1/100) (1 + R2/100) (1 + R3/100)

4. Present worth of Rs. x due n years hence is given by:
Present Worth = x / (1 + R/100)n

Logarithms
-------------
1. Logarithm: If a is a positive real number, other than 1 and am = x, then we write m = loga x
and say that the value of log x to the base a is m.

2. Properties of Logarithms:

a. loga (xy) = loga x + loga y

b. loga (x/y) = loga x - loga y

c. logx x = 1 (i.e. Log of any number to its own base is 1)

d. loga 1 = 0 (i.e. Log of 1 to any base is 0)

e. loga (xp) = p loga x

f. loga x = 1 / logx a

g. loga x = logb x / logb a
= log x / log a (Change of base rule)

h. When base is not mentioned, it is taken as 10

i. Logarithms to the base 10 are known as common logarithms

Algebra
-----------
1. (a + b)2 = a2 + 2ab + b2

2. (a - b)2 = a2 - 2ab + b2

3. (a + b)2 - (a - b)2 = 4ab

4. (a + b)2 + (a - b)2 = 2(a2 + b2)

5. (a2 – b2) = (a + b)(a - b)

6. (a3 + b3) = (a + b)(a2 - ab + b2)

7. (a3 – b3) = (a - b)(a2 + ab + b2)

8. Results on Division:
Dividend = Quotient × Divisor + Remainder

9. An Arithmetic Progression (A. P.) with first term ‘a’ and Common Difference ‘d’ is given
by:
[a], [(a + d)], [(a + 2d)], … … …, [a + (n - 1)d]
nth term, Tn = a + (n - 1)d
Sum of first ‘n’ terms, Sn
= n/2 (First Term + Last Term)

10. A Geometric Progression (G. P.) with first term ‘a’ and Common Ratio ‘r’ is given by:
a, ar, ar2, ar3, … … …, arn-1
nth term, Tn = arn-1
Sum of first ‘n’ terms Sn = [a(1 - rn)] / [1 - r]

11. (1 + 2 + 3 + … … … + n) = [n(n + 1)] / 2

12. (12 + 22 + 32 + … … … + n2) = [n(n + 1)(2n + 1)] / 6

13. (13 + 23 + 33 + … … … + n3) = [n2(n + 1)2] / 4

Percentage
------------------
1. To express x% as a fraction, we have x% = x / 100

2. To express a / b as a percent, we have a / b = (a / b × 100) %

3. If ‘A’ is R% more than ‘B’, then ‘B’ is less than ‘A’ by
OR
If the price of a commodity increases by R%, then the reduction in consumption, not
to increase the expenditure is
{100R / [100 + R] } %

4. If ‘A’ is R% less than ‘B’, then ‘B’ is more than ‘A’ by
OR
If the price of a commodity decreases by R%, then the increase in consumption, not to
increase the expenditure is
{100R / [100 - R] } %

5. If the population of a town is ‘P’ in a year, then its population after ‘N’ years is
P (1 + R/100)N

6. If the population of a town is ‘P’ in a year, then its population ‘N’ years ago is
P / [(1 + R/100)N]

Profit & Loss
---------------

7. If the value of a machine is ‘P’ in a year, then its value after ‘N’ years at a depreciation of
‘R’ p.c.p.a is
P (1 - R/100)N

8. If the value of a machine is ‘P’ in a year, then its value ‘N’ years ago at a depreciation of
‘R’ p.c.p.a is
P / [(1 - R/100)N]

9. Selling Price = [(100 + Gain%) × Cost Price] / 100
= [(100 - Loss%) × Cost Price] / 100

Ratio & Proportion
-----------------------
1. The equality of two ratios is called a proportion. If a : b = c : d, we write a : b :: c : d and
we say that a, b, c, d are in proportion.
In a proportion, the first and fourth terms are known as extremes, while the second and
third are known as means.

2. Product of extremes = Product of means

3. Mean proportion between a and b is=(ab)^(1/2)

4. The compounded ratio of the ratios (a : b), (c : d), (e : f) is (ace : bdf)

5. a2 : b2 is a duplicate ratio of a : b

6. a^(1/2).b^(1/2) : is a sub-duplicate ration of a : b

7. a3 : b3 is a triplicate ratio of a : b

8. a1/3 : b1/3 is a sub-triplicate ratio of a : b

9. If a / b = c / d, then, (a + b) / b = (c + d) / d, which is called the componendo.

10. If a / b = c / d, then, (a - b) / b = (c - d) / d, which is called the dividendo.

11. If a / b = c / d, then, (a + b) / (a - b) = (c + d) / (c - d), which is called the componendo &
dividendo.

12. Variation: We say that x is directly proportional to y if x = ky for some constant k and we
write, x α y.

13. Also, we say that x is inversely proportional to y if x = k / y for some constant k and we
write x α 1 / y.

Area
------
1. Rectangle:

a. Area of a rectangle = (length × breadth)

b. Perimeter of a rectangle = 2 (length + breadth)

2. Square:

a. Area of square = (side)2

b. Area of a square = ½ (diagonal)2

3. Area of 4 walls of a room
= 2 (length + breadth) × height

4. Triangle:

a. Area of a triangle = ½ × base × height

b. Area of a triangle = , where
s = ½ (a + b + c), and a, b, c are the sides of the triangle

c. Area of an equilateral triangle = / 4 × (side)2

d. Radius of incircle of an equilateral triangle of side a = a / 2

e. Radius of circumcircle of an equilateral triangle of side a = a /

5. Parallelogram/Rhombus/Trapezium:

a. Area of a parallelogram = Base × Height

b. Area of a rhombus = ½ × (Product of diagonals)

c. The halves of diagonals and a side of a rhombus form a right angled triangle with
side as the hypotenuse.

d. Area of trapezium = ½ × (sum of parallel sides) × (distance between them)

6. Circle/Arc/Sector, where R is the radius of the circle:

a. Area of a circle = πR2

b. Circumference of a circle = 2πR

c. Length of an arc = Ө/360 × 2πR

d. Area of a sector = ½ (arc × R)
= Ө/360 × πR2

Volume & Surface Area
-------------------------
1. Cuboid:
Let length = l, breadth = b & height = h units Then,

a. Volume = (l × b × h) cu units

b. Surface Area = 2 (lb + bh + hl) sq. units

c. Diagonal = units

2. Cube:
Let each edge of a cube be of length a. Then,

a. Volume = a3 cu units

b. Surface Area = 6a2 sq. units

c. Diagonal = ( l^2 + b^2 +h^2)^1/2 units

3. Cylinder:
Let radius of base = r & height (or length) = h. Then,

a. Volume = (πr2h) cu. units

b. Curved Surface Area = (2πrh) sq. units
c. Total Surface Area = 2πr(r + h) sq. units

4. Cone:
Let radius of base = r & height = h. Then,

a. Slant height, l =(h^2 + R^2)^1/2 units

b. Volume = (⅓ πr2h) cu. units

c. Curved Surface Area = (πrl) sq. units

d. Total Surface Area = πr(r + l) sq. units

5. Sphere:
Let the radius of the sphere be r. Then,

a. Volume = (4/3 πr3) cu. units

b. Surface Area = (4πr2) sq. units

6. Hemi-sphere:
Let the radius of the sphere be r. Then,

a. Volume = (2/3 πr3) cu. units

b. Curved Surface Area = (2πr2) sq. units

c. Total Surface Area = (3πr2) sq. units

Stocks & Shares
-----------------
1. Brokerage: The broker’s charge is called brokerage.

2. When stock is purchased, brokerage is added to the cost price.

3. When the stock is sold, brokerage is subtracted from the selling price.

4. The selling price of a Rs. 100 stock is said to be:

a. at par, if S.P. is Rs. 100 exactly;

b. above par (or at premium), if S.P. is more than Rs. 100;

c. below par (or at discount), if S.P. is less than Rs. 100.

5. By ‘a Rs. 800, 9% stock at 95’, we mean a stock whose face value is Rs. 800, annual
interest is 9% of the face value and the market price of a Rs. 100 stock is Rs. 95.

True Discount
----------------
1. Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per
annum. Clearly, Rs. 100 at 14% will amount to Rs. 156 in 4 years. So, the payment of
Rs. 100 now will clear off the debt of Rs. 156 due 4 years hence. We say that:
Sum due = Rs. 156 due 4 years hence;
Present Worth (P.W.) = Rs. 100;
True Discount (T.D.) = Rs. (156 - 100)
= (Sum due) – (P.W.)

2. T.D. = Interest on P.W.

3. Amount = (P.W.) + (T.D.)

4. Interest is reckoned on R.W. and true discount is reckoned on the amount

5. Let rate = R% per annum & time = T years. Then,

a. P.W. = (100 × Amount) / (100 + [R × T])
= (100 × T.D.) / (R × T)

b. T.D. = (P.W.) × R × T / 100
= ([Amount] × R × T) / (100 + [R × T])

c. Sum = ([S.I.] × [T.D.]) / ([S.I.] – [T.D.])

d. (S.I.) – (T.D.) = S.I. on T.D.

e. When the sum is put at compound interest, then
P.W. = Amount / (1 + R/100)T

Banker’s Discount
---------------------
1. Banker’s Discount (B.D.) is the S.I. on the face value for the period from the date on
which the bill was discounted and the legally due date.

2. Banker’s Gain (B.G.) = (B.D.) – (T.D.) for the unexpired time

3. When the date of the bill is not given, grace days are not to be added

4. B.D. = S.I. on bill for unexpired time

5. B.G. = (B.D.) – (T.D.)
= S.I. on T.D.
= (T.D.)2 / P.W.

6. T.D. =(P.W x B.G)^1/2

7. B.D. = (Amount × Rate × Time) / 100

8. T.D. = (Amount × Rate × Time) / (100 + [Rate × Time])

9. Amount = (B.D. × T.D.) / (B.D. – T.D.)

10. T.D. = (B.G. × 100) / (Rate × Time)

Partnership
--------------
1. If a number of partners have invested in a business and it has a profit, then
Share Of Partner = (Total_Profit × Part_Share / Total_Share)
Chain Rule

2. The cost of articles is directly proportional to the number of articles.

3. The work done is directly proportional to the number of men working at it.

4. The time (number of days) required to complete a job is inversely proportional to the
number of hours per day allocated to the job.

5. Time taken to cover a distance is inversely proportional to the speed of the car.
Time & Work

6. If A can do a piece of work in n days, then A’s 1 day’s work = 1/n.

7. If A’s 1 day’s work = 1/n, then A can finish the work in n days.

8. If A is thrice as good a workman as B, then:
Ratio of work done by A and B = 3 : 1,
Ratio of times taken by A & B to finish a work = 1 : 3

Pipes & Cisterns
-----------------
1. If a pipe can fill a tank in ‘x’ hours and another pipe can empty the full tank in ‘y’ hours
(where y > x), then on opening both the pipes, the net part of the tank filled in 1 hour is
(1/x – 1/y)

Time And Distance
----------------------
1. Suppose a man covers a distance at ‘x’ kmph and an equal distance at ‘y’ kmph, then
average speed during his whole journey is
[2xy / (x + y)] kmph

Trains
--------
1. Lengths of trains are ‘x’ km and ‘y’ km, moving at ‘u’ kmph and ‘v’ kmph (where, u > v) in
the same direction, then the time taken y the over-taker train to cross the slower train is
[(x + y) / (u - v)] hrs

2. Time taken to cross each other is
[(x + y) / (u + v)] hrs

3. If two trains start at the same time from two points A and B towards each other and after
crossing they take a and b hours in reaching B and A respectively.
Then, A’s speed : B’s speed = ( b^(1/2) : a^(1/2) ).

4. x kmph = (x × 5/18) m/sec.

5. y metres/sec = (y × 18/5) km/hr.

Boats & Streams
-------------------
1. If the speed of a boat in still water is u km/hr and the speed of the stream is v hm/hr,
then:
Speed downstream = (u + v) km/hr.
Speed upstream = (u - v) km/hr.

2. If the speed downstream is a km/hr and the speed upstream is b km/hr, then:
Speed in still water = ½ (a + b) km/hr.
Rate of stream = ½ (a - b) km/hr.

Alligation or Mixture
----------------------
1. Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at
the given price must be mixed to produce a mixture at a given price.

2. Mean Price: The cost price of a quantity of the mixture is called the mean price.


Numerical solution of nonlinear equations
------------------------------------------
We want to develop numerical methods for solving nonlinear
equations of the form
f(x) = 0. (1)
Examples:
(i) f(x) = x2 − A = 0, A > 0 (⇒ x = ±√A).
(ii) f(x) = x3 − 3x2 + 2 = 0.
(iii) f(x) = 1
x − A = 0 (⇒ x = 1
A).
(iv) f(x) = sin x − 1
x = 0.
Numerical solution of nonlinear equations
Idea: Approximate Eqn. (1) by an iterative process (recursion)
xn+1 = F(xn), x1 is given, (2)
such that
lim
n→∞
xn = X, with f(X) = 0.
Easy to implement (1) on a computer.
Goal: find a systematic way for obtaining iterative process of the
form (2) for solving equations of the form (1).
Numerical solution of nonlinear equations
Consider Example (i), f(x) = x2 − A = 0, A > 0.
x2 − A = 0 ⇒ 2x2 − x2 = A
⇒ 2x2 = x2 + A ⇒ x =
1
2 x +
A
x .
Use the following iterative scheme:
xn+1 =
1
2 xn +
A
xn . (3)
Choose x1 (initial guess), use (3) to calculate approximately the
square root of A. Let A = 25 ⇒ X = ±5.
Numerical solution of nonlinear equations
Choose x1 = 1:
x1 = 1 ⇒ x2 = 13.0 ⇒ x3 = 7.4615
⇒ x4 = 5.406 ⇒ x5 = 5.0152 . . .
i.e. the iterative process converges to X = 5. Take now x1 = −1.
x1 = −1 ⇒ x2 = −13.0 ⇒ x3 = −7.4615
⇒ x4 = −5.406 ⇒ x5 = −5.0152 . . .
i.e. the iterative process converges to X = −5. Choose x1 = 8:
x1 = 8 ⇒ x2 = 5.5625 ⇒ x3 = 5.0284
⇒ x4 = 5.0001 . . .
The iteration process converges faster to the root of the equation.
Numerical solution of nonlinear equations
When we have an iterative procedure of the form (2) which we use
to calculate the roots of (1) we want to address the following issues.
1. Does the iterative process converge, i.e. is there a limit of xn as
n → ∞?
2. Does the iterative process converge to the right value (i.e. one
of the roots of the equation F(x) = 0)?
3. How does its accuracy increase with n (i.e. is it practical)?
4. How does the choice of x1 affect the convergence of the method
and the value to which it converges (in the case when the
equation has more than one roots)?
A iterative process of the form (2) for solving an equation of the
form (1) is called a fixed point method.
Numerical solution of nonlinear equations
Consider example (ii) x3 − 3x2 + 2 = 0:
x3 − 3x2 + 2 = 0 ⇒ x3 = 3x2 − 2
⇒ x = 3 −
2
x2 .
• Use the iterative process
xn+1 = 3 −
2
x2
n
.
• If this method converges, to which of the 3 roots of the
equation does it converge?
Numerical solution of nonlinear equations
If an iterative process is to be convergent, then we have
lim
n→∞xn+1 − xn = 0,
i.e. the distance between two subsequent elements of the sequence
{xn}∞
n=1 generated by (2) becomes smaller as n increases.
We expect that the starting value x1 affects the eventual
convergence of the iterative process to a limit. It might be that the
iterative process that we have chosen converges only for certain
values of x1.
If we sketch the curves y = x, y = F(x), their intersection satisfies
x = F(x). Start with x1. The iterates generated by (2) should give
x = X.
Numerical solution of nonlinear equations
Calculation of √A.
• Consider the iterative process
xn+1 =
1
2 xn +
A
xn ,
• which approximates √A.
• Assume 0 < x2 =" 1">
1
2 √A +
A
√A
= √A ⇒ x2 > √A.
Numerical solution of nonlinear equations
Now:
x3 =
1
2 x2 +
A
x2
< n =" 2,"> √A. Then xn+1 < n =" 1," a =" 1" a =" 1" a =" 1" 1 =" xn(2" x =" 0" 1 =" Ax" 1 =" 2" x =" x(2" xn =" 1" x1 =" 0" x1 =" 2" xn =" 0," n =" 1,">
2
A ⇒ lim
n→∞
xn = −∞.
Numerical solution of nonlinear equations
Summary
• An iterative process can converge or diverge, depending on the
initial choice x1.
• The root to which the iterative process converges depends on
the initial choice x1.
• The convergence can be fast (quadratic).
How to choose the iterative process for solving f(x) = 0.
A given equation f(x) = 0 may often be written in several different
ways as x = F(x).
Example: f(x) = x2 − 6x + 2 = 0 with roots X1,2 = 3 ± √7. It
can be written as x = F(x) in many different ways:
1. x = x
2
−2
2x−6 ⇒ the iterative process is xn+1 = x
2
−2
2xn−6 :
x2−6x+2 = 0 ⇒ 2x2−6x+2 = x2 ⇒ 2x2−6x = x2−2 ⇒ x =
x2 − 2
2x − 6
.
2. x = 6 − 2
x ⇒ the iterative process is xn+1 = 6 − 2
xn
:
x2 − 6x + 2 = 0 ⇒ x2 = 6x − 2 ⇒ x = 6 −
2
x
.
3. x = 1
6x2 + 1
3 ⇒ the iterative process is xn+1 = 1
6x2
n + 1
3 :
6x = x2 + 2 ⇒ x =
1
6
x2 +
1
3
.
How to choose the iterative process for solving f(x) = 0.
The resulting different iterative processes xn+1 = F(xn) have, in
general, different convergence properties. We want to choose the
iterative process that has the best convergence properties.
This means that not only does it converge to the roots of the
equation f(x) = 0, but also that it converges sufficiently fast. We
want to develop criteria that enable us to choose the best iterative
process for solving a nonlinear equation of the form f(x) = 0.
How to choose the iterative process for solving f(x) = 0.
Suppose that xn+1 = F(xn) with limn→∞ xn = X and
f(X) = 0 ⇒ X = F(X). Let us write
X = xn + ǫn, n = 1, 2, . . .
i.e.
X = approximation + error.
We want to calculate ǫn+1 as a function of ǫn. Assume that ǫn is
small so that we can use the Taylor series expansion:
xn+1 = F(xn) ≈ F(X) − ǫnF′(X) +
1
2
ǫ2
nF′′(X).
But xn+1 = X + ǫn+1 ⇒
ǫn+1 = X − xn+1 ≈ X − F(X) + ǫnF′(X) −
1
2
ǫ2
nF′′(X)
= ǫnF′(X) −
1
2
ǫ2
nF′′(X).
⇒ The size of ǫn+1, relative to the size of ǫn, depends on F′(X).
How to choose the iterative process for solving f(x) = 0.
Indeed:
ǫn+1
ǫn
= F′(X) + O(ǫn). (4)
DEFINITION 1 Assume that F′(X) 6= 0. Then the iterative
process xn+1 = F(xn) is called a first order process.
The convergence of a first order process is guranteed when
F′(X) <> 1. The case F′(X) = 1 requires further study.
How to choose the iterative process for solving f(x) = 0.
We can also define second order processes:
DEFINITION 2 Assume that F′(X) = 0 and that F′′(X) 6= 0.
Then the iterative process xn+1 = F(xn) is called a second order
process.
For a second order process we have that
ǫn+1 = −
1
2
ǫ2
nF′′(X) + O(ǫ3
n) ⇒ ǫn+1 = Cǫ2
n
and the convergence is much faster (quadratic as opposed to
linear). It is to our advantage to use a second order iterative
process to solve the equation f(x) = 0.
How to choose the iterative process for solving f(x) = 0.
Examples
1. The square root formula F(x) = 1
2 􀀀x + A
x .
F′(x) =
1
2 1 −
A
x2 ⇒ F′(√A) =
1
2 1 −
A
A = 0
⇒ We have to calculate the second derivative
at the root X = √A.
F′′(x) =
A
x3 ⇒ F′′(√A) =
1
√A
.
⇒ Second Order Process.
⇒ The convergence is quadratic.
How to choose the iterative process for solving f(x) = 0.
Examples
2. The reciprocal formula F(x) = x (2 − Ax).
F′(x) = (2 − Ax) + x(−A) = 2 − 2Ax
⇒ F′ 1
A = 2 − 2A 1
A = 0
⇒ We have to calculate the second derivative
at the root X =
1
A
.
F′′(x) = −2A ⇒ F′′ 1
A = −2A.
⇒ Second Order Process.
⇒ The convergence is quadratic.
How to choose the iterative process for solving f(x) = 0.
Examples
3. The quadratic equation f(x) = x2 − 6x + 2 = 0. The two roots
of this equation are R1 = 3 + √7, R2 = 3 − √7. Consider the
following 4 different iterative processes.
(a) xn+1 = 6 − 2
xn
.
(b) xn+1 = 1
6x2
n + 1
3 .
(c) xn+1 = √6xn − 2.
(d) xn+1 = x2
−2
2xn−6 .
How to choose the iterative process for solving f(x) = 0.
The performance of the the four different iterative processes is
summarized in the following table.
Iterative process R1 attainable R2 attainable order of process
xn+1 = 6 − 2
xn
. NO YES 1
xn+1 = 1
6x2
n + 1
3 . YES NO 1
xn+1 = √6xn − 2. NO YES 1
xn+1 = x2
−6
2xn−6 . YES YES 2
⇒ The fourth iterative process performs better: both roots are
attainable and it converges faster.
Q: Is there a systematic method for finding the best method for a
given nonlinear equation f(x) = 0? A: Yes, the
Newton–Raphson Method.
The Newton–Raphson Method
Consider the equation f(x) = 0 and let X be the root of this
equation: f(X) = 0. Let x1 be our initial approximation and write
X = x1 + ǫ1.
Use the Taylor series expansion to obtain
0 = f(X) = f(x1 + ǫ1) = f(x1) + ǫ1f′(x1) + . . .
Assuming that f′(x1) 6= 0, we solve this equation for ǫ1 to obtain
ǫ1 ≈ −
f(x1)
f′(x1)
.
Hence, a better approximation to X than x1 would appear to be
x2 = x1 −
f(x1)
f′(x1)
.
The Newton–Raphson Method
We can repeat this procedure to obtain the iterative process
xn+1 = xn −
f(xn)
f′(xn)
=: F(xn). (6)
This is the Newton–Raphson iterative process. The
Newton–Raphson process converges to a particular root of f(x) = 0
if x1 is suitably chosen, assuming that all the roots are attainable.
The process is usually second order convergent if X is a
simple root:
The Newton–Raphson Method
F(x) = x −
f(x)
f′(x) ⇒
F′(x) = 1 −
(f′(x))2 − f(x)f′′(x)
(f′(x))2 =
(f′(x))2 − (f′(x))2 + f(x)f′′(x)
(f′(x))2
=
f(x)f′′(x)
(f′(x))2 ⇒
F′(X) =
f(X)f′′(X)
(f′(X))2 = 0, since f(X) = 0.
If X is not a simple root then the Newton–Raphson process is
usually first order, but it still converges.
The Newton–Raphson Method: Examples
1. The square root formula:
f(x) = x2 − A = 0 ⇒ f′(x) = 2x ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
x2
n − A
2xn
= xn −
xn
2
+
A
2xn
=
1
2 xn +
A
xn .
The Newton–Raphson Method: Examples
2. The reciprocal formula:
f(x) =
1
x − A = 0 ⇒ f′(x) = −
1
x2 ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
1
xn − A
− 1
x2
n
= xn + x2
n 1
xn − A
= xn + xn − Ax2
n
= xn(2 − Axn).
The Newton–Raphson Method: Examples
3. A quadratic equation:
f(x) = x2 − 6x + 2 = 0 ⇒ f′(x) = 2x − 6 ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
x2
n − 6xn + 2
2xn − 6
=
xn(2xn − 6) − x2
n + 6xn − 2
2xn − 6
=
2x2
n − 6xn − x2
n + 6xn − 2
2xn − 6
=
x2
n − 2
2xn − 6
.
The Newton–Raphson process converges to R1 or R2 depending
on the initial guess:
• x1 < r1 =" 3"> 3 ⇒ xn → R2, R2 = 3 + √7.
• x1 = 3 ⇒ x2 = +∞, the method diverges.
The Newton–Raphson Method: Examples
4. The equation sin x − 1
x = 0.
f(x) = sin x −
1
x
= 0 ⇒ f′(x) = cos x +
1
x2 ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
sin xn − 1
xn
cos xn + 1
x2
n
= xn − x2
n
sin xn − 1
xn
x2
n cos xn + 1
= xn −
x2
n sin xn − xn
x2
n cos xn + 1
.
The equation sin x − 1/x = 0 has infinitely many solutions. Let
us find the first positive solution R = 1.11416. Start with
x1 = 0.5.
x1 = 0.5 ⇒ x2 = 0.8117 ⇒ x3 = 1.0413
⇒ x4 = 1.1095 ⇒ x5 = 1.1141.
The Newton–Raphson Method: Examples
0.5 1 1.5 2 2.5 3 3.5 4
−1
−0.5
0
0.5
1
1.5
2
sin(x)
1/x
0.5 1 1.5 2 2.5 3 3.5 4
−1.5
−1
−0.5
0
0.5
a. f1(x) = sin(x), f2(x) = 1/x. b. f(x) = sin(x) − 1/x.
The Newton–Raphson Method: Examples
0 5 10 15 20 25 30
−1
−0.5
0
0.5
1
1.5
2
sin(x)
1/x
f1(x) = sin(x), f2(x) = 1/x.
The Newton–Raphson Method: Examples
We want to show that a solution to the equation
f(x) = sin x − 1
x = 0 exists in the interval (
4 ,
2 ). We have that
f′(x) = cos x +
1
x2 > 0 for x ∈ hπ
4
,
π
2 i.
Consequently, the function f(x) is strictly increasing in the interval
[
4 ,
2 ]. Furthermore,
f(π/4) = sin(π/4) −
1
π/4
=
√2
2 −
4
π
< 2 =" 1"> 0.
Since f(x) is continuous and strictly increasing, there exists an
R ∈ (
4 ,
2 ) such that f(R) = 0. This shows that there exists a root
of the equation f(x) = 0 in the interval (
4 ,
2 ).
The Newton–Raphson Method: Examples
We want to show that the equation
f(θ) = sin(θ) − θ cos(θ) −
1
2
π = 0
in the interval (
2 , 2
3 ).
f′(θ) = cos θ − cos θ + θ sin θ = θ sin θ > 0 for θ ∈ π
2
,

3 .
Consequently, the function f(x) is strictly increasing in the interval
[
2 , 3
2 ]. Furthermore,
f(
π
2
) = sin(
π
2
) −
π
2
cos(
π
2
) −
π
2
= 1 −
π
2
< 3 =" 0.3424"> 0.
Since f(x) is continuous and strictly increasing, there exists an
R ∈ (
2 , 2
3 ) such that f(R) = 0. This shows that there exists a
root of the equation f(x) = 0 in the interval (
2 , 2
3 ).
The Newton–Raphson Method: Examples
Now we find the Newton–Raphson process for this equation. We
have that
f′(θ) = θ sin(θ).
The Newton–Raphson process is
θn+1 = θn −
sin θn − θn cos θn −
2
θn sin θn
.
The root of the equation f(θ) = sin(θ) − θ cos(θ) − 1
2π = 0 in the
interval (
2 , 2
3 ) is R = 1.9057. We calculate it using the
Newton-Raphson process with x1 =
2 :
x1 =
π
2 ⇒ x2 = 1.9342 ⇒ x3 = 1.9058 ⇒ x4 = 1.9057.
The Newton–Raphson Method: Examples
1.5 1.6 1.7 1.8 1.9 2 2.1 2.2
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0 2 4 6 8 10
−8
−6
−4
−2
0
2
4
6
8
a. θ ∈
2 , 2
3 . b. θ ∈ [0, 10].
f(θ) = sin(θ) − θ cos(θ) − 1
2π.
Numerical solution of nonlinear equations
Final Remarks
• One can also define higher order processes:
F′(X) = 0, F′′(X) = 0, . . . F(n)(X) 6= 0.
• This is an nth order process.
• There are many other methods for solving nonlinear equations
other than the Newton–Raphson method: Bisection method,
secant method etc.
• The Newton–Raphson methods works also for equations on the
complex plane:
f(z) = 0, z ∈ C.