UGC NET Paper 1 Mathematical Reasoning & Aptitude.pdf
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About This Presentation
This Presentation is about the Unit 5 Mathematical Reasoning of UGC NET Paper 1 General Studies where we have included Types of Reasoning, Mathematical reasoning like number series, letter series etc. and mathematical aptitude like Fraction, Time and Distance, Average etc. with their solved question...
Slide Content
Mathematical
Reasoning & Aptitude
UGC NET PAPER 1 UNIT 5
Reasoning & Aptitude
UGC NET PAPER 1 UNIT 5
Types of
reasoning.
Number series, Letter
series, Codes, and
Relationships.
Mathematical Aptitude (Fraction,
Time & Distance, Ratio,
Proportion and Percentage, Profit
and Loss, Interest and
Discounting, Averages etc.).
Unit-V
Mathematical
Reasoning
and Aptitude
reasoning.
Number series, Letter
series, Codes, and
Relationships.
Mathematical Aptitude (Fraction,
Time & Distance, Ratio,
Proportion and Percentage, Profit
and Loss, Interest and
Discounting, Averages etc.).
Unit-V
Mathematical
Reasoning
and Aptitude
What is
Reasoning ?
Reasoning ?
When making a choice or addressing an issue,
reasoning is the ability to appraise things rationally
by using logic based on new or existing information.
Reasoning allows you to balance the advantages and
cons of two or more options before deciding on the
best option or the one that best meets your goals. It
also assists you in solving difficulties, dealing with
uncertainty, verifying claims, and carefully
assessing situations to ensure that the decision you
make is in your best interests.
reasoning is the ability to appraise things rationally
by using logic based on new or existing information.
Reasoning allows you to balance the advantages and
cons of two or more options before deciding on the
best option or the one that best meets your goals. It
also assists you in solving difficulties, dealing with
uncertainty, verifying claims, and carefully
assessing situations to ensure that the decision you
make is in your best interests.
Types of
Reasoning
Reasoning
1. Deductive reasoning:
Deductive reasoning is a
method of proving a theory or
hypothesis using formal logic
and observations. Deductive
reasoning starts with a
hypothesis that is then
supported or disproved
through observations or
rational thought.
A marketing division, for
example, analyses data and
confirms that their company’s
most important demographic is
young parents. They opt to give
more of the marketing money to
social media channels that
target that category based on
this information.
Deductive reasoning is a
method of proving a theory or
hypothesis using formal logic
and observations. Deductive
reasoning starts with a
hypothesis that is then
supported or disproved
through observations or
rational thought.
A marketing division, for
example, analyses data and
confirms that their company’s
most important demographic is
young parents. They opt to give
more of the marketing money to
social media channels that
target that category based on
this information.
2. Inductive
reasoning:
To validate observations,
inductive reasoning employs
theories and assumptions. It’s
the polar opposite of
deductive reasoning in that it
requires deducing a general
rule from a specific case or
cases.
Because it employs conclusions
from observations to make
generalisations, the outcomes of
inductive reasoning are not always
certain. Extrapolation, forecasts,
and part-to-whole arguments all
benefit from inductive reasoning.
reasoning:
To validate observations,
inductive reasoning employs
theories and assumptions. It’s
the polar opposite of
deductive reasoning in that it
requires deducing a general
rule from a specific case or
cases.
Because it employs conclusions
from observations to make
generalisations, the outcomes of
inductive reasoning are not always
certain. Extrapolation, forecasts,
and part-to-whole arguments all
benefit from inductive reasoning.
3. Analogical reasoning:
Analogical reasoning is a
style of reasoning that
looks for similarities
between two or more
objects and then uses
those similarities to find
other properties they
share. It is based on the
brain’s tendency to notice
patterns and make
connections.
Analogical reasoning is a
style of reasoning that
looks for similarities
between two or more
objects and then uses
those similarities to find
other properties they
share. It is based on the
brain’s tendency to notice
patterns and make
connections.
4. Abductive reasoning:
Abductive reasoning is a style of reasoning that
reaches a logical conclusion based on an
observation or group of observations. Abductive
reasoning is similar to inductive reasoning, but it
allows you to make the greatest estimates to get
the simplest conclusions. Abduction can help with
both troubleshooting and decision-making,
especially when dealing with uncertainties.
Abductive reasoning is a style of reasoning that
reaches a logical conclusion based on an
observation or group of observations. Abductive
reasoning is similar to inductive reasoning, but it
allows you to make the greatest estimates to get
the simplest conclusions. Abduction can help with
both troubleshooting and decision-making,
especially when dealing with uncertainties.
5. Cause-and-effect
reasoning:
Cause-and-effect reasoning is a style of
reasoning in which the relationship
between two events is demonstrated.
This logic is used to describe what
might happen if a certain action is
taken or why certain things happen
when certain circumstances are met.
When people draw on personal
experience and a drive to improve, this
form of reasoning is frequently used to
guide daily decision-making.
reasoning:
Cause-and-effect reasoning is a style of
reasoning in which the relationship
between two events is demonstrated.
This logic is used to describe what
might happen if a certain action is
taken or why certain things happen
when certain circumstances are met.
When people draw on personal
experience and a drive to improve, this
form of reasoning is frequently used to
guide daily decision-making.
Critical thinking entails delving
deeply into a topic’s rationale in
order to reach a definitive
conclusion. Computing, engineering,
social sciences, and logic all benefit
from it. When it comes to problem-
solving, critical thinking is especially
important when dealing with
technological challenges. It’s used to
determine the validity of artworks,
literature, films, and other forms of
art.
6. Critical
thinking:
deeply into a topic’s rationale in
order to reach a definitive
conclusion. Computing, engineering,
social sciences, and logic all benefit
from it. When it comes to problem-
solving, critical thinking is especially
important when dealing with
technological challenges. It’s used to
determine the validity of artworks,
literature, films, and other forms of
art.
6. Critical
thinking:
7. Decompositional
reasoning:
Decompositional reasoning is the
process of breaking things down into
their basic components in order to
comprehend how each component
contributes to the overall
functionality of the object.
Decompositional reasoning allows an
observer to derive powerful
conclusions about the total by
evaluating each portion separately.
reasoning:
Decompositional reasoning is the
process of breaking things down into
their basic components in order to
comprehend how each component
contributes to the overall
functionality of the object.
Decompositional reasoning allows an
observer to derive powerful
conclusions about the total by
evaluating each portion separately.
Syllogism
A form of arguing in which two
statements are used to prove
that a third statement is true,
for example, ‘All humans are
mortal; I am a human,
therefore I am mortal.’
What is Syllogism ?
statements are used to prove
that a third statement is true,
for example, ‘All humans are
mortal; I am a human,
therefore I am mortal.’
What is Syllogism ?
Modus Ponens and
Modus Tollens
Modus Tollens
In propositional logic, the
modus ponens and
modus tollens are two
types of inference that
can be derived from a
hypothetical proposition
—that is, from a
statement of the form “If
A, then B”
E.g. When an angle is
inscribed in a semicircle,
it is a right angle;
however, this angle is not
a right angle, hence it is
not inscribed in a
semicircle.
modus ponens and
modus tollens are two
types of inference that
can be derived from a
hypothetical proposition
—that is, from a
statement of the form “If
A, then B”
E.g. When an angle is
inscribed in a semicircle,
it is a right angle;
however, this angle is not
a right angle, hence it is
not inscribed in a
semicircle.
Number Series
Number series is an order of numbers which are
not arranged randomly but follow a pattern. Here
in these notes, we will understand how to identify
which kind of pattern is following because
without this it is next to impossible to have a
command on number series. Number series is a
form
of numbers in a certain sequence, where some
numbers are mistakenly put into the series of
numbers and some number is missing in that
series, we need to observe first and then find the
accurate number to that series of numbers.
not arranged randomly but follow a pattern. Here
in these notes, we will understand how to identify
which kind of pattern is following because
without this it is next to impossible to have a
command on number series. Number series is a
form
of numbers in a certain sequence, where some
numbers are mistakenly put into the series of
numbers and some number is missing in that
series, we need to observe first and then find the
accurate number to that series of numbers.
In this type of series pattern can be found by
using difference of terms. If any
pattern is found after the first difference like
square, cube or multiplication then this is called
one
tier series otherwise we need to proceed further
and then it is called two tier series.
Difference series
3--6--9--12--15--18
+3 +3 +3 +3 +3
20 28 37 47 58 70
+8 +9 +10 +11 +12
using difference of terms. If any
pattern is found after the first difference like
square, cube or multiplication then this is called
one
tier series otherwise we need to proceed further
and then it is called two tier series.
Difference series
3--6--9--12--15--18
+3 +3 +3 +3 +3
20 28 37 47 58 70
+8 +9 +10 +11 +12
In this kind of series
the next term is
found by adding the
previous number.
1 1 2 3 5 8 13 21
Fibonacci
Series:
the next term is
found by adding the
previous number.
1 1 2 3 5 8 13 21
Fibonacci
Series:
Prime Number Series
In this kind of series, the next term is found
by adding, multiplying, or dividing
by prime numbers.
2 4 12 60 420 4620
In this kind of series, the next term is found
by adding, multiplying, or dividing
by prime numbers.
2 4 12 60 420 4620
Perfect
Square
Series:
These Types of Series are
based on the square of a
number which is in the
same order and one square
number is missing in that
given series.
Square
Series:
These Types of Series are
based on the square of a
number which is in the
same order and one square
number is missing in that
given series.
Perfect
Cube
Series:
This Types of Series are
based on a cube of a number
which is in the same order
and one cube number is
missing in that given series.
Cube
Series:
This Types of Series are
based on a cube of a number
which is in the same order
and one cube number is
missing in that given series.
Ratio Series
This type of series is based on ration series, where
sequence is in form of ratio in
difference between the numbers. All numbers are
arranged in ratio sequence order.
4096 5120 6400 8000 10000
Here the ratio between each consecutive term is
4:5 or we can say the next term can be found by
multiplying 5/4 to the previous term.
This type of series is based on ration series, where
sequence is in form of ratio in
difference between the numbers. All numbers are
arranged in ratio sequence order.
4096 5120 6400 8000 10000
Here the ratio between each consecutive term is
4:5 or we can say the next term can be found by
multiplying 5/4 to the previous term.
A few things
to notice
while solving
Number
Series
Questions
to notice
while solving
Number
Series
Questions
01
If the number
series gradually
decreases, then
the arithmetic
operation is
subtraction.
02
If the number
series is
gradually
increasing, then
the arithmetic
operation is
addition.
03
If the ratio
between any two
consecutive
numbers on the
sequence is
identical, the line
is a Geometric
Progression (GP)
series.
If the number
series gradually
decreases, then
the arithmetic
operation is
subtraction.
02
If the number
series is
gradually
increasing, then
the arithmetic
operation is
addition.
03
If the ratio
between any two
consecutive
numbers on the
sequence is
identical, the line
is a Geometric
Progression (GP)
series.
04
If the difference
between any two
consecutive
numbers on the
sequence is the
same, then the
sequence is an
Arithmetic
Progression (AP)
series.
06
If there is no direct
relation between two
consecutive numbers,
but there is a relation
between alternate
numbers, then the
series is a hybrid of
two sequences. Solve
the two sequences
separately to obtain
the result.
07
If the sequence
numbers are
increasing in a
multiplicative
manner, then
the arithmetic
operator used
is
multiplication.
05
Have a strong
knowledge of the
cubes and squares
of different
numbers so you can
recognise the
pattern at once.
If the difference
between any two
consecutive
numbers on the
sequence is the
same, then the
sequence is an
Arithmetic
Progression (AP)
series.
06
If there is no direct
relation between two
consecutive numbers,
but there is a relation
between alternate
numbers, then the
series is a hybrid of
two sequences. Solve
the two sequences
separately to obtain
the result.
07
If the sequence
numbers are
increasing in a
multiplicative
manner, then
the arithmetic
operator used
is
multiplication.
05
Have a strong
knowledge of the
cubes and squares
of different
numbers so you can
recognise the
pattern at once.
Questions on
Number Series
Number Series
1. Find out the next number on the sequence
12,7,2,….
Ans: As this series is gradually decreasing, it is a
subtraction series with a difference of 5 between
two numbers.
Therefore, next number = 2-5
=-3
So, the final sequence is 12,7,2,-3.
12,7,2,….
Ans: As this series is gradually decreasing, it is a
subtraction series with a difference of 5 between
two numbers.
Therefore, next number = 2-5
=-3
So, the final sequence is 12,7,2,-3.
2. Find the missing number from the
sequence 3,9,27,…,243.
Ans: On close observation, we find that
this series is a sequence of 3^1,3^2,3^3…
and so on.
Therefore, the missing number is 3^4 = 81.
So, the final sequence is 3, 9, 27, 81, 243.
sequence 3,9,27,…,243.
Ans: On close observation, we find that
this series is a sequence of 3^1,3^2,3^3…
and so on.
Therefore, the missing number is 3^4 = 81.
So, the final sequence is 3, 9, 27, 81, 243.
3. Find out the missing number from the sequences
2,7,17,…,52.
Ans: As this series is gradually increasing, it is an additional
series with a difference of 5 between the two numbers.
With each succession, the addition also increases by 5.
Therefore,
2 + 5 = 7; 7 + 10 = 17; 17 + 15 = 32; 32 + 20 = 52.
Therefore, the missing number of the sequence is 32.
Final sequence = 2, 7, 17, 32, 52.
2,7,17,…,52.
Ans: As this series is gradually increasing, it is an additional
series with a difference of 5 between the two numbers.
With each succession, the addition also increases by 5.
Therefore,
2 + 5 = 7; 7 + 10 = 17; 17 + 15 = 32; 32 + 20 = 52.
Therefore, the missing number of the sequence is 32.
Final sequence = 2, 7, 17, 32, 52.
4. Find the missing number 25: 37:: 49:?
Solution:
Here, 52 = 25 and (5+1)2 + 1 = 37
Similarly, 72 = 49 and (7+1)2 + 1 = 65.
Hence, the missing number is 65.
Solution:
Here, 52 = 25 and (5+1)2 + 1 = 37
Similarly, 72 = 49 and (7+1)2 + 1 = 65.
Hence, the missing number is 65.
5. Find out the missing term in the given series: 11, 17, 39, 85, ?.
Solution:
Given series: 11, 17, 39, 85, ?.
11 + (32 – 3) = 11 + 6 = 17
17 + (52 – 3) = 17 + 22 = 39
39 + (72 – 3) = 39 + 46 = 85
85 + (92 – 3) = 85 + 78 = 163
Therefore, the missing number in the given series is 163.
I.e., 11, 17, 39, 85, 163.
Solution:
Given series: 11, 17, 39, 85, ?.
11 + (32 – 3) = 11 + 6 = 17
17 + (52 – 3) = 17 + 22 = 39
39 + (72 – 3) = 39 + 46 = 85
85 + (92 – 3) = 85 + 78 = 163
Therefore, the missing number in the given series is 163.
I.e., 11, 17, 39, 85, 163.
6. Determine the missing numbers in the series 5, 6, 9, 14, 21, ?.
Solution:
Given the number series: 5, 6, 9, 14, 21, ?.
5 + 1 = 6
6 + 3 = 9
9 + 5 = 14
14 + 7 = 21
21 + 9 = 30.
Hence, the missing number here is 30.
Here, the series follows a pattern of the sum of consecutive odd
numbers.
So, the complete number series is 5, 6, 9, 14, 21, 30.
Solution:
Given the number series: 5, 6, 9, 14, 21, ?.
5 + 1 = 6
6 + 3 = 9
9 + 5 = 14
14 + 7 = 21
21 + 9 = 30.
Hence, the missing number here is 30.
Here, the series follows a pattern of the sum of consecutive odd
numbers.
So, the complete number series is 5, 6, 9, 14, 21, 30.
Letter Series
In this type of problem, a series of letters of English
alphabet will be given, which follow a pattern or a
sequence. The letter series mainly consists of skipping
the letters.
To solve these types of problems, assign numbers 1 to
26 to the letters of English alphabet as shown below. In
some cases, it is useful to assign the numbers in a
reverse order.
alphabet will be given, which follow a pattern or a
sequence. The letter series mainly consists of skipping
the letters.
To solve these types of problems, assign numbers 1 to
26 to the letters of English alphabet as shown below. In
some cases, it is useful to assign the numbers in a
reverse order.
English Alphabet from
Left to Right
Left to Right
English Alphabet
from Right to Left
from Right to Left
One Letter Series
Example 1
A, C, E, G, …, K
(a) I (b) H (c) J (d) M
Solution
The series is A + 2 = C, C + 2 = E, E + 2
= G; G + 2 = I, I +2 = K. The missing
letter is I.
A, C, E, G, …, K
(a) I (b) H (c) J (d) M
Solution
The series is A + 2 = C, C + 2 = E, E + 2
= G; G + 2 = I, I +2 = K. The missing
letter is I.
Example 2
B, E, H, K, N, …
(a) P (b) O (c) Q (d) R
Solution
The series is +3. The missing letter is
N + 3 = Q.
B, E, H, K, N, …
(a) P (b) O (c) Q (d) R
Solution
The series is +3. The missing letter is
N + 3 = Q.
Example 3
D, F, H, I, J, L, …
(a) K (b) O (c) M (d) P
Solution
If the numbers are assigned, the series
becomes 4, 6,
8, 9, 10, 12, and so on, i.e., composite
number series.
The next composite number is 14 and
the corresponding letter is N.
D, F, H, I, J, L, …
(a) K (b) O (c) M (d) P
Solution
If the numbers are assigned, the series
becomes 4, 6,
8, 9, 10, 12, and so on, i.e., composite
number series.
The next composite number is 14 and
the corresponding letter is N.
Combined two
letter series
The first letters of the series follow one logic
and the second letters follow another logic,
and then they pair with each other.
letter series
The first letters of the series follow one logic
and the second letters follow another logic,
and then they pair with each other.
Example 1
AM, BN, CO, DP, …, FR
(a) EQ (b) FT (c) GR (d) ER
Solution
The first letters are A, B, C, D, E, and
F, and the second
letters are M, N, O, P, Q and R.
AM, BN, CO, DP, …, FR
(a) EQ (b) FT (c) GR (d) ER
Solution
The first letters are A, B, C, D, E, and
F, and the second
letters are M, N, O, P, Q and R.
Example 2
AA, CE, EI, GO, …
(a) IU (b) IQ (c) IR (d) IT
Solution
The first letters of all pairs given in the
question follow a sequence of A + 2 = C, C +
2 = E and so on. The second letters are
vowels
AA, CE, EI, GO, …
(a) IU (b) IQ (c) IR (d) IT
Solution
The first letters of all pairs given in the
question follow a sequence of A + 2 = C, C +
2 = E and so on. The second letters are
vowels
Three letter series
This sequence consists of three letters in each term.
The first letters follow one logic, Where the second
letters follow another logic and the third letters
follow some other logic (or the same logic in all the
three cases).
This sequence consists of three letters in each term.
The first letters follow one logic, Where the second
letters follow another logic and the third letters
follow some other logic (or the same logic in all the
three cases).
Example 1
NAB, OEC, PIE, QOG, …
(a) QPH (b) QUH (c) QUI (d) RUK
Solution
The first letters form a series of N, O, P, Q, R and
so on. The second letters form a vowel series and
the third letters form a prime number series
according to their number position.
NAB, OEC, PIE, QOG, …
(a) QPH (b) QUH (c) QUI (d) RUK
Solution
The first letters form a series of N, O, P, Q, R and
so on. The second letters form a vowel series and
the third letters form a prime number series
according to their number position.
Coding &
Decoding
Decoding
In these types of questions,
the letters of the alphabet
are exclusively used. These
letters do not stand for
themselves but are allotted
some artificial values based
on logical patterns or
analogies. By applying
those principles or
observing the pattern
involved, the candidates
are required to decode a
coded word or encode a
word.
Alphabetical Coding
the letters of the alphabet
are exclusively used. These
letters do not stand for
themselves but are allotted
some artificial values based
on logical patterns or
analogies. By applying
those principles or
observing the pattern
involved, the candidates
are required to decode a
coded word or encode a
word.
Alphabetical Coding
Example
If ‘BELONGINGS’ is coded as ‘TABLESTESF’,
then how will you code ‘LINEN’?
(a) BTEAE
(b) BTAEA
(c) BATEA
(d) None of the above
If ‘BELONGINGS’ is coded as ‘TABLESTESF’,
then how will you code ‘LINEN’?
(a) BTEAE
(b) BTAEA
(c) BATEA
(d) None of the above
Example
If ‘POSTED’ is coded as ‘DETSOP’, then how will you code
‘SPEED’?
(a) DEEPS (b) DEESP
(c) DESEP (d) SPEDE
Solution
By careful observation, we can say that the letters have
been written in the reverse order. Hence, SPEED will be
written DEEPS and therefore, (a) is the answer
If ‘POSTED’ is coded as ‘DETSOP’, then how will you code
‘SPEED’?
(a) DEEPS (b) DEESP
(c) DESEP (d) SPEDE
Solution
By careful observation, we can say that the letters have
been written in the reverse order. Hence, SPEED will be
written DEEPS and therefore, (a) is the answer
Example
If ‘TRAIN’ is coded as 23456, then
how will you code ‘RAIN’?
(a) 3456
(b) 3546
(c) 2345
(d) 2456
If ‘TRAIN’ is coded as 23456, then
how will you code ‘RAIN’?
(a) 3456
(b) 3546
(c) 2345
(d) 2456
Solution
Word T R A I N
Code 2 3 4 5 6
These values have been assigned
arbitrarily. The question can be solved
based on the relationship established.
For RAIN, the code is 3456, so (a) is the
answer.
Word T R A I N
Code 2 3 4 5 6
These values have been assigned
arbitrarily. The question can be solved
based on the relationship established.
For RAIN, the code is 3456, so (a) is the
answer.
If rain is water, water is road, road is cloud,
cloud is sky, sky is sea and sea is path, where do
aeroplanes fly?
(a) Road (b) Sea
(c) Cloud (d) Water
Explanation
The aeroplanes fly in the ‘sky’ and the ‘sky’ is
called ‘sea’. Hence, the aeroplanes fly in the
‘sea’.
cloud is sky, sky is sea and sea is path, where do
aeroplanes fly?
(a) Road (b) Sea
(c) Cloud (d) Water
Explanation
The aeroplanes fly in the ‘sky’ and the ‘sky’ is
called ‘sea’. Hence, the aeroplanes fly in the
‘sea’.
Further
Questions
Choosing the Odd Word
Choosing the Odd Number
Choosing the Odd Letter
Choosing the Group of Odd
Words/Letters/Numbers
Questions
Choosing the Odd Word
Choosing the Odd Number
Choosing the Odd Letter
Choosing the Group of Odd
Words/Letters/Numbers
ANALOGY
The meaning of analogy is correspondence. In the
questions based on analogy, a particular relationship is
given and another similar relationship has to be identified
from the alternatives provided to us. Therefore, analogy
tests are meant to test a candidate for the overall
knowledge, power of reasoning, and the ability to
think concisely and accurately.
The meaning of analogy is correspondence. In the
questions based on analogy, a particular relationship is
given and another similar relationship has to be identified
from the alternatives provided to us. Therefore, analogy
tests are meant to test a candidate for the overall
knowledge, power of reasoning, and the ability to
think concisely and accurately.
Example 1
As delicious is related to taste, melodious is
related to
(a) Voice (b) Speak
(c) Music (d) Highness
Answer: (a)
Explanation
Delicious represents good taste. Similarly,
melody describes pleasant voice.
As delicious is related to taste, melodious is
related to
(a) Voice (b) Speak
(c) Music (d) Highness
Answer: (a)
Explanation
Delicious represents good taste. Similarly,
melody describes pleasant voice.
Example 2
Giant: Dwarf: : Genius:?
(a) Wicked (b) Gentle
(c) Idiot (d) Cunning
Answer: (c)
Explanation
As dwarf is the antonym of giant, idiot is
the antonym of genius.
Giant: Dwarf: : Genius:?
(a) Wicked (b) Gentle
(c) Idiot (d) Cunning
Answer: (c)
Explanation
As dwarf is the antonym of giant, idiot is
the antonym of genius.
Example 3
Lamp : Darkness
(a) Fatigue : Exercise
(b) Water : Thirst
(c) Medicine : Illness
(d) Study : Classroom
Answer: (b)
Just as a lamp eliminates darkness, water quenches
thirst
Lamp : Darkness
(a) Fatigue : Exercise
(b) Water : Thirst
(c) Medicine : Illness
(d) Study : Classroom
Answer: (b)
Just as a lamp eliminates darkness, water quenches
thirst
Example 4
Potato : Carrot : Radish
(a) Tomato (b) Spinach
(c) Sesame (d) Groundnut
Solution
All of these crops/vegetables grow
underground.
Answer: (d)
Potato : Carrot : Radish
(a) Tomato (b) Spinach
(c) Sesame (d) Groundnut
Solution
All of these crops/vegetables grow
underground.
Answer: (d)
Example 5
9 : 14 : : 26 : ?
(a) 2 (b) 13 (c) 15 (d) 31
Solution
The relationship is x : (x + 5).
Answer: (d
9 : 14 : : 26 : ?
(a) 2 (b) 13 (c) 15 (d) 31
Solution
The relationship is x : (x + 5).
Answer: (d
Example 6
42 : 56 : : 72 : ?
(a) 81 (b) 90 (c) 96 (d) 100
Solution
The ratio between 42 and 56 is 3 : 4.
Similarly, 72 : 96 depicts the ratio 3 : 4.
Answer: (c)
42 : 56 : : 72 : ?
(a) 81 (b) 90 (c) 96 (d) 100
Solution
The ratio between 42 and 56 is 3 : 4.
Similarly, 72 : 96 depicts the ratio 3 : 4.
Answer: (c)
Blood Relations
Example 1
X and Y are brothers. C and D are sisters. X’s son is D’s
brother. How is Y related to C?
(a) Uncle (b) Grandfather
(c) Father (d) None of the above
Explanation
Y is the brother of X and X’s son is D’s brother. This
implies that D is the daughter of X. As C and D are sisters, C is
also the daughter of X. Hence, Y is the uncle of C.
X and Y are brothers. C and D are sisters. X’s son is D’s
brother. How is Y related to C?
(a) Uncle (b) Grandfather
(c) Father (d) None of the above
Explanation
Y is the brother of X and X’s son is D’s brother. This
implies that D is the daughter of X. As C and D are sisters, C is
also the daughter of X. Hence, Y is the uncle of C.
Direction Sense
Seating
Arrangement
Arrangement
Mathematical
Aptitude
Aptitude
Mathematical
Aptitude
Fraction
Time & Distance
Ratio, Proportion, and Percentage
Profit and Loss
Interest and Discounting
Averages
Aptitude
Fraction
Time & Distance
Ratio, Proportion, and Percentage
Profit and Loss
Interest and Discounting
Averages
If there are 5 apples in a cartoon of 12 apples,
then the fraction of apples for the whole
would be represented as = 5/12
Fraction = Part/Whole = Numerator/Denominator
FRACTIONS
then the fraction of apples for the whole
would be represented as = 5/12
Fraction = Part/Whole = Numerator/Denominator
FRACTIONS
1. Common fraction:
A common fraction is a number written with
a numerator and a denominator, in which
both are natural numbers. For example, 5/12,
17/12, , etc.
A common fraction is a number written with
a numerator and a denominator, in which
both are natural numbers. For example, 5/12,
17/12, , etc.
2. Proper fraction:
A proper fraction that is less than 1 is
known as proper fraction, such as 1/2,
3/4
A proper fraction has the same name as that
ratio.
A proper fraction that is less than 1 is
known as proper fraction, such as 1/2,
3/4
A proper fraction has the same name as that
ratio.
3. Mixed number fraction:
It is basically a whole number plus a proper
fraction. For example.
It is basically a whole number plus a proper
fraction. For example.
4. Improper
fractions:
If we divide each whole unit
into thirds, say, and keep
counting them, then we
will come to 3/3, 4/3, 5/3
and so on. That is, we will
come to fractions that are
equal to or greater than -1.
We call those improper
fractions.
fractions:
If we divide each whole unit
into thirds, say, and keep
counting them, then we
will come to 3/3, 4/3, 5/3
and so on. That is, we will
come to fractions that are
equal to or greater than -1.
We call those improper
fractions.
Questions on
Fractions
Fractions
2. 3/5 of a group of children were
girls. If there were 24 girls, then how
many children were there in the
group?
(a) 32
(b) 36
(c) 40
(d) 42
girls. If there were 24 girls, then how
many children were there in the
group?
(a) 32
(b) 36
(c) 40
(d) 42
Time & Distance
If speed, time and distance are denoted
by S, T, and D,
respectively, then S = D/T; D = S × T and T =
D/S.
To convert from km/h to metre/second
(m/s),
multiply by 5/18.
To convert m/s to km/h, multiply by 18/5.
E.g. Raj covers 60 km in 4 hours then his
speed would be 60/4 = 15 km/hr
by S, T, and D,
respectively, then S = D/T; D = S × T and T =
D/S.
To convert from km/h to metre/second
(m/s),
multiply by 5/18.
To convert m/s to km/h, multiply by 18/5.
E.g. Raj covers 60 km in 4 hours then his
speed would be 60/4 = 15 km/hr
Example 1
A man covers 20 km in 2½ hours.
Find the distance covered in 9 hours.
Solution
Speed = D/T = 20 km/2½ hours = 8
kmph
Distance covered in 9 hours = S × T =
8 × 9 = 72 km
A man covers 20 km in 2½ hours.
Find the distance covered in 9 hours.
Solution
Speed = D/T = 20 km/2½ hours = 8
kmph
Distance covered in 9 hours = S × T =
8 × 9 = 72 km
Example 2
A car completes a journey in 4 hours, the first half
at a speed of 40 kmph and second at 60 kmph.
Find the total distance covered.
Solution
As the total journey is divided into equal parts, the
average speed can be calculated by the formula
2xy/ (x + y) = 2 × 40 × 60/(40 + 60) = 48 kmph.
Distance = S × T = 48 × 4 = 192 km.
A car completes a journey in 4 hours, the first half
at a speed of 40 kmph and second at 60 kmph.
Find the total distance covered.
Solution
As the total journey is divided into equal parts, the
average speed can be calculated by the formula
2xy/ (x + y) = 2 × 40 × 60/(40 + 60) = 48 kmph.
Distance = S × T = 48 × 4 = 192 km.
Example 3
A student walks from his house at a
speed of 3 kmph and reaches the
school 10 minutes late. If he walks
at a speed of 4 kmph, then he
reaches the school 10 minutes
earlier. What is the distance
between his school and his house?
A student walks from his house at a
speed of 3 kmph and reaches the
school 10 minutes late. If he walks
at a speed of 4 kmph, then he
reaches the school 10 minutes
earlier. What is the distance
between his school and his house?
Let the distance = x km
Difference between timings of reaching the
school
at different speeds = 10 + 10 = 20 minutes or
20/60 or
1/3 hours.
Now the difference between timings = x/3 - x/4
= 1/3
= x/3 - x/4 = 1/3
= 4x-3x/12 =
X=4 km
Difference between timings of reaching the
school
at different speeds = 10 + 10 = 20 minutes or
20/60 or
1/3 hours.
Now the difference between timings = x/3 - x/4
= 1/3
= x/3 - x/4 = 1/3
= 4x-3x/12 =
X=4 km
Ratio, Proportion and
Percentage
Percentage
Ratios can also be expressed as fractions. They
represent the basic relationship between two
quantities.
Proportions are in comparison to the whole.
In a mixture of 20 l of milk and 30 l of water, the
ratio of milk and water is 2 : 3. This can be
converted to the fraction of milk in the solution
as 2 : 5 or 2/5th.
As seen, 2/5 is nothing but 2/5 × 100 = 40%.
represent the basic relationship between two
quantities.
Proportions are in comparison to the whole.
In a mixture of 20 l of milk and 30 l of water, the
ratio of milk and water is 2 : 3. This can be
converted to the fraction of milk in the solution
as 2 : 5 or 2/5th.
As seen, 2/5 is nothing but 2/5 × 100 = 40%.
What percentage of 180.50 is
36.1?
1.
(a) 20 (b) 25
(c) 20.50 (d) None of the above
Solution
Lets x% of 180.5 = 36.1
x% = 36.1/180.5
x = 36.1/180.5 × 100 = 20%
36.1?
1.
(a) 20 (b) 25
(c) 20.50 (d) None of the above
Solution
Lets x% of 180.5 = 36.1
x% = 36.1/180.5
x = 36.1/180.5 × 100 = 20%
2. The price of a commodity increases
first by 20% and then by 10%. What is
the net increase in the price?
Solution
Let original price = 100
Price after 1st increase = 100 + 20 = 120
Price after 2nd increase = 120 + (10% of
120) = 132.
Net increase = 132 - 100 = 32%
first by 20% and then by 10%. What is
the net increase in the price?
Solution
Let original price = 100
Price after 1st increase = 100 + 20 = 120
Price after 2nd increase = 120 + (10% of
120) = 132.
Net increase = 132 - 100 = 32%
3. In an exam, a student scored 50%
of the maximum marks and yet
failed by 15 marks. If he had scored
10% more than what he scored,
then he would have just managed to
get the pass percentage. What are
the maximum marks of the paper?
of the maximum marks and yet
failed by 15 marks. If he had scored
10% more than what he scored,
then he would have just managed to
get the pass percentage. What are
the maximum marks of the paper?
Solution
Let maximum passing marks = 100
Actual marks obtained = 50
Had he scored 50 + 10% of 50, i.e., 55 marks, then
he would have scored passing marks. In this
situation, the difference between actual and
passing marks is 5.
Actual difference = 15
5% of maximum marks = 15
Maximum marks = 15 × 100/5 = 300
Let maximum passing marks = 100
Actual marks obtained = 50
Had he scored 50 + 10% of 50, i.e., 55 marks, then
he would have scored passing marks. In this
situation, the difference between actual and
passing marks is 5.
Actual difference = 15
5% of maximum marks = 15
Maximum marks = 15 × 100/5 = 300
4. What is 20% of 30% of 40%
Solution
20/100 x 30/100 x 40/100 =
2.4%
Solution
20/100 x 30/100 x 40/100 =
2.4%
Profit & Loss
Cost Price (C.P.): The price at which an article is
purchased by the seller.
Selling Price (S.P.): The price at which an article is
sold.
Profit or Gain (P): If the difference between S.P. and
C.P. is positive, then the amount is called profit or
gain.
Loss: If the difference between C.P. and S.P. is
positive, then the amount is called loss.
purchased by the seller.
Selling Price (S.P.): The price at which an article is
sold.
Profit or Gain (P): If the difference between S.P. and
C.P. is positive, then the amount is called profit or
gain.
Loss: If the difference between C.P. and S.P. is
positive, then the amount is called loss.
Profit
Percentage
(Profit/Cost
Price) x 100
Loss
Percetange
(Loss/Cost
Price) x 100
Selling Price
Cost Price +
Profit
Cost Price
Selling Price
- Profit
Important Formulae
Percentage
(Profit/Cost
Price) x 100
Loss
Percetange
(Loss/Cost
Price) x 100
Selling Price
Cost Price +
Profit
Cost Price
Selling Price
- Profit
Important Formulae
1. An umbrella was sold at a profit of 20%. What is the
selling price of the umbrella if the shopkeeper procured
it for Rs 180?
(a) 210 (b) 216 (c) 230 (d) 236
Solution
Substituting values in the formula above, we get:
Selling price = Cost Price + Profit
= 180 + 20% of 180
= 180 + 36
=216
selling price of the umbrella if the shopkeeper procured
it for Rs 180?
(a) 210 (b) 216 (c) 230 (d) 236
Solution
Substituting values in the formula above, we get:
Selling price = Cost Price + Profit
= 180 + 20% of 180
= 180 + 36
=216
2. An article is sold for Rs 2400 at a profit of
25%. What would have been the actual profit
or loss if it had been sold at Rs 1800?
Solution
Cost Price = S.P. - Profit
Profit is 25% of Cost meaning 20% of Sales
Therefore, C.P. = 2400 - 20% of 2400 = 1920
Selling at 1800 means a loss of 1800 - 1920 =
120 which is (120/1920)x100 = 6.25%
25%. What would have been the actual profit
or loss if it had been sold at Rs 1800?
Solution
Cost Price = S.P. - Profit
Profit is 25% of Cost meaning 20% of Sales
Therefore, C.P. = 2400 - 20% of 2400 = 1920
Selling at 1800 means a loss of 1800 - 1920 =
120 which is (120/1920)x100 = 6.25%
3. Romit sold his old TV and earned a profit of 10%. If he could have
managed to sell it for Rs 8100 more, then his profit would have been
be 37%. Find the price at which he bought the TV.
(a) Rs 30000 (b) Rs 41000
(c) Rs 44500 (d) Rs 55000
Solution:
We will go answer to question for each option.
Let’s say the first option is the cost price of Rs 30000.
Therefore the Selling Price = 30000 + 10% = 33000
Adding 8100 more to it = 33000 + 8100 = 41100
This means a profit of 11100 which is 11100/30000 = 37%
Therefore 30,000 is the answer
managed to sell it for Rs 8100 more, then his profit would have been
be 37%. Find the price at which he bought the TV.
(a) Rs 30000 (b) Rs 41000
(c) Rs 44500 (d) Rs 55000
Solution:
We will go answer to question for each option.
Let’s say the first option is the cost price of Rs 30000.
Therefore the Selling Price = 30000 + 10% = 33000
Adding 8100 more to it = 33000 + 8100 = 41100
This means a profit of 11100 which is 11100/30000 = 37%
Therefore 30,000 is the answer
4. A man sells an article at a profit of 8 percent. If the cost price
were 10 percent less and the selling price Rs 18 less, then his
profit would have been 15 percent. Find the cost price of the
article.
(a) Rs 430 (b) Rs 450
(c) Rs 220 (d) Rs 400
Solution
Let's go answer to question.
If 400 is the CP then SP would be 400 + 8% = 432
Now new CP is 10% less = 400 - 10% = 360
And, new SP is Rs 18 less = 432 - 18 = 414
Therefore New Profit is SP - CP = 414 - 360 = 54 which is 15% of
360
So, the answer is 400
were 10 percent less and the selling price Rs 18 less, then his
profit would have been 15 percent. Find the cost price of the
article.
(a) Rs 430 (b) Rs 450
(c) Rs 220 (d) Rs 400
Solution
Let's go answer to question.
If 400 is the CP then SP would be 400 + 8% = 432
Now new CP is 10% less = 400 - 10% = 360
And, new SP is Rs 18 less = 432 - 18 = 414
Therefore New Profit is SP - CP = 414 - 360 = 54 which is 15% of
360
So, the answer is 400
Interest &
Discounting
Discounting
Compound InterestSimple Interest
Principal -
The money borrowed or lent out for a certain period is called
the principal or the sum. It is denoted by P.
Simple interest calculations -
Interest is the extra money that the borrower pays for using
the lender’s money. Simple interest is denoted by S.I.
SI = P X N X R/100
P = Principal
N - Number of Years
R = Rate of Interest Per Annum
Simple Interest
The money borrowed or lent out for a certain period is called
the principal or the sum. It is denoted by P.
Simple interest calculations -
Interest is the extra money that the borrower pays for using
the lender’s money. Simple interest is denoted by S.I.
SI = P X N X R/100
P = Principal
N - Number of Years
R = Rate of Interest Per Annum
Simple Interest
The formula for compound interest
is given below:
Compound interests = A – P, where A
is the amount to be available at the
end of
is given below:
Compound interests = A – P, where A
is the amount to be available at the
end of
How much time will it take for an amount of
Rs 900 to yield Rs 81 as interest at 4.5% per
annum of simple interest?
1.
(a) 2 years (b) 3 years
(c) 1 year (d) 4 years
Solution
For one year the Interest @4.5% on 900 would be
40.5
In two years it would be 40.5 x 2 = 81
Therefore it would take two years
Rs 900 to yield Rs 81 as interest at 4.5% per
annum of simple interest?
1.
(a) 2 years (b) 3 years
(c) 1 year (d) 4 years
Solution
For one year the Interest @4.5% on 900 would be
40.5
In two years it would be 40.5 x 2 = 81
Therefore it would take two years
2. Find S.I. on Rs 6250 at 14% per annum for
146 days.
(a) Rs 350 (b) Rs 450
(c) Rs 550 (d) Rs 650
Solution:
SI = PXNXR/100
= 6250 X 146/365 X 14/100
= 350
146 days.
(a) Rs 350 (b) Rs 450
(c) Rs 550 (d) Rs 650
Solution:
SI = PXNXR/100
= 6250 X 146/365 X 14/100
= 350
3. Simple interest on a certain sum of money for
3 years at 8% per annum is half the compound
interest on 4000 for 2 years at 10% per annum.
The sum placed on simple interest is
(a) 1550 (b) 1650
(c) 1750 (d) 2000
Solution:
THE Ci on 4000 for 2 years at 10% p.a. is 840
The SI of 420 at 8% p.a. for 3 years would be on
1750
3 years at 8% per annum is half the compound
interest on 4000 for 2 years at 10% per annum.
The sum placed on simple interest is
(a) 1550 (b) 1650
(c) 1750 (d) 2000
Solution:
THE Ci on 4000 for 2 years at 10% p.a. is 840
The SI of 420 at 8% p.a. for 3 years would be on
1750
4. A certain sum of money amounts to 2500 in a span of 5
years and further to 3000 in a span of 7 years at simple
interest. The sum is
(a) 1800 (b) 2000
(c) 1400 (d) 1250
Explanation
2500 in 5th year and 3000 in 7th year.
So in between 2 years, 500 is increased.
⇒ For a year 500/2 = 250
So, per year it is increasing 250, then in 5 years
⇒ 250 × 5 = 1250
Hence, the initial amount must be 2500 – 1250
= 1250.
years and further to 3000 in a span of 7 years at simple
interest. The sum is
(a) 1800 (b) 2000
(c) 1400 (d) 1250
Explanation
2500 in 5th year and 3000 in 7th year.
So in between 2 years, 500 is increased.
⇒ For a year 500/2 = 250
So, per year it is increasing 250, then in 5 years
⇒ 250 × 5 = 1250
Hence, the initial amount must be 2500 – 1250
= 1250.
5. An amount doubles itself in 15 years, what is
the rate of interest?
(a) 7,85% (b) 9.41%
(c) 6.66% (d) 4.21%
Explanation
Let the principle be P.
As the amount doubles itself the interest is P too.
So P = P × r × 15/100
⇒ r = 100/15 = 20/3% = 6.66%
the rate of interest?
(a) 7,85% (b) 9.41%
(c) 6.66% (d) 4.21%
Explanation
Let the principle be P.
As the amount doubles itself the interest is P too.
So P = P × r × 15/100
⇒ r = 100/15 = 20/3% = 6.66%
6. An amount doubles itself in 6 years, what is
the rate of interest under Compound Interest?
(a) 11% (b) 12%
(c) 14% (d) 10%
Explanation
Use Rule of 72
Time taken to double the amount = 72/Rate of
Interest
= 72/12 = 6 years
Therefore answer is 12%
the rate of interest under Compound Interest?
(a) 11% (b) 12%
(c) 14% (d) 10%
Explanation
Use Rule of 72
Time taken to double the amount = 72/Rate of
Interest
= 72/12 = 6 years
Therefore answer is 12%
Averages
The arithmetic mean or
simply ‘average’ or
‘mean’ of a group of
values is the sum of the
values divided by the
total number of values.
It is one of the measures
of central tendencies.
What is Average ?
simply ‘average’ or
‘mean’ of a group of
values is the sum of the
values divided by the
total number of values.
It is one of the measures
of central tendencies.
What is Average ?
Example 1
If a candidate scores 5, 15, 25, 10, and 15
marks in different subjects, then calculate
the mean marks scored
Solution:
Average /Mean Marks
= (5+15+25+10+15)/ 5 = 70/5 = 14
If a candidate scores 5, 15, 25, 10, and 15
marks in different subjects, then calculate
the mean marks scored
Solution:
Average /Mean Marks
= (5+15+25+10+15)/ 5 = 70/5 = 14
Example 2
The average age of 30 boys in a class is 15
years. If the age of teacher is also included,
then the new average age becomes 16 years.
What is the age of teacher?
Solution:
The Average age of 30 Boys is 15 which means the total age
is 30 boys x 15 years = 350 years
Whereas the Average age of 31 people (30 boys+1 teacher) is
16 years meaning the total age is 31 x 16 = 496 years
Therefore the age of teacher is 496-450 = 46 years
The average age of 30 boys in a class is 15
years. If the age of teacher is also included,
then the new average age becomes 16 years.
What is the age of teacher?
Solution:
The Average age of 30 Boys is 15 which means the total age
is 30 boys x 15 years = 350 years
Whereas the Average age of 31 people (30 boys+1 teacher) is
16 years meaning the total age is 31 x 16 = 496 years
Therefore the age of teacher is 496-450 = 46 years
Example 3
The average monthly income of A and B is 5050. The average
monthly income of B and C is 6250 and the average monthly income
of A and C is 5200. The monthly income of A is
(a) 3500 (b) 4000
(c) 4050 (d) 5000
Solution
Total of A and B = (5,050 × 2) = 10,100 --------- (i)
Similarly, total income of B and C = (6250 × 2) =12,500 -----(ii)
And total of A and C = (5200 × 2) = 10,400 ---------- (iii)
Adding (i), (ii), and (iii), we get, 2(A + B + C) =33,000
Or A + B + C = 16,500
But B + C = 12,500 from Equation (ii)
Hence, A = 16,500 - 12,500 = 4000
The average monthly income of A and B is 5050. The average
monthly income of B and C is 6250 and the average monthly income
of A and C is 5200. The monthly income of A is
(a) 3500 (b) 4000
(c) 4050 (d) 5000
Solution
Total of A and B = (5,050 × 2) = 10,100 --------- (i)
Similarly, total income of B and C = (6250 × 2) =12,500 -----(ii)
And total of A and C = (5200 × 2) = 10,400 ---------- (iii)
Adding (i), (ii), and (iii), we get, 2(A + B + C) =33,000
Or A + B + C = 16,500
But B + C = 12,500 from Equation (ii)
Hence, A = 16,500 - 12,500 = 4000
Calendar
Questions
Questions
Odd days: In a
given period,
the number of
days more than
the complete
weeks is called
odd days.
Leap year: A leap year
has 366 days. Every
year divisible by 4 is
a leap year, if it is not
a century. Only every
4th century is a leap
year and no other
century is a leap year.
Ordinary year: The
year that is not a
leap year is called
an ordinary year.
An ordinary year
has 365 days.
given period,
the number of
days more than
the complete
weeks is called
odd days.
Leap year: A leap year
has 366 days. Every
year divisible by 4 is
a leap year, if it is not
a century. Only every
4th century is a leap
year and no other
century is a leap year.
Ordinary year: The
year that is not a
leap year is called
an ordinary year.
An ordinary year
has 365 days.
01 January 0001 is assumed to be Monday.
1 ordinary year = 365 days = 52 weeks + ‘1
day’.
That one extra day is counted as odd day.
1 leap year = 366 days = (52 weeks + 2 days)
1 leap year has 2 odd days.
100 years = 76 ordinary years + 24 leap years
= (76 × 1 + 24 × 2) odd days = 124 odd days
= (17 weeks + 5 days). It means 5 odd days.
1 ordinary year = 365 days = 52 weeks + ‘1
day’.
That one extra day is counted as odd day.
1 leap year = 366 days = (52 weeks + 2 days)
1 leap year has 2 odd days.
100 years = 76 ordinary years + 24 leap years
= (76 × 1 + 24 × 2) odd days = 124 odd days
= (17 weeks + 5 days). It means 5 odd days.
Example
What was the day of the week on January 1, 2001?
[June 2009]
(a) Friday (b) Tuesday
(c) Sunday (d) Wednesday
Answer: (b)
Explanation
By the end of centuries 400, 800, 1200, 1600, 2000, 2400,
and so on, there are no extra days left. It means that the
last year 2000 was Sunday. As per convention, the week
starts with Monday and hence, the next day after last
day of year 2000, i.e., January 1, 2001, is Monday.
What was the day of the week on January 1, 2001?
[June 2009]
(a) Friday (b) Tuesday
(c) Sunday (d) Wednesday
Answer: (b)
Explanation
By the end of centuries 400, 800, 1200, 1600, 2000, 2400,
and so on, there are no extra days left. It means that the
last year 2000 was Sunday. As per convention, the week
starts with Monday and hence, the next day after last
day of year 2000, i.e., January 1, 2001, is Monday.
Example
January 1, 1995 was a Sunday. What would be the day of
the week on January 1, 1996? [December 2009]
(a) Sunday (b) Monday
(c) Wednesday (d) Saturday
Answer: (b)
Explanation
There is increase of 1 day (odd day) in the subsequent
year. In case of leap year (if the day is after February),
then there will be increase of two days.
Although 1996 is a leap year, the day in question is in
January month, so there will be increase of one day. Thus,
January 1, 1996 is Monday.
January 1, 1995 was a Sunday. What would be the day of
the week on January 1, 1996? [December 2009]
(a) Sunday (b) Monday
(c) Wednesday (d) Saturday
Answer: (b)
Explanation
There is increase of 1 day (odd day) in the subsequent
year. In case of leap year (if the day is after February),
then there will be increase of two days.
Although 1996 is a leap year, the day in question is in
January month, so there will be increase of one day. Thus,
January 1, 1996 is Monday.
VENN DIAGRAMS
THANK YOU
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