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BITSAT MATH: ALGEBRA MARATHON BITSAT MATH: ALGEBRA MARATHON
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t.me/mathematicallyinclined BITSAT MATH: ALGEBRA MARATHON
ALGEBRA CHAPTERS 2022 2021 2020 WEIGHTAGE
%
Qs Qs Qs
SEQUENCE & SERIES 2 3 4 8%
MATRICES & DETERMINANTS 3 3 3 8%
BINOMIAL THEOREM 2 2 4 7%
PROBABILITY 1 3 2 4%
QUADRATIC EQUATIONS 2 1 2 4%
P & C 2 2 1 4%
COMPLEX NUMBERS 2 1 1 3%
STATISTICS 2 1 1 3%
LINEAR PROGRAMMING PROBLEM 2 1 0 2%
MATHEMATICAL REASONING 1 0 0 1% BITSAT MATH: ALGEBRA MARATHON
> 2
> 1
> 0
> 3
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 2
> 1
> 0
> 3
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
Both positive
Both negative
Of opposite sign
None of these
Q. Roots of the equation x 2+bx -c=0 (b,c>0) are
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Both positive
Both negative
Of opposite sign
None of these
Q. Roots of the equation x 2+bx -c=0 (b,c>0) are
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
z is purely real
z is purely imaginary
either z is purely real
to purely imaginary
None of these
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
z is purely real
z is purely imaginary
either z is purely real
to purely imaginary
None of these
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
25
-25
-35
35
Q. A Student read common difference of an A.P.
as -2 instead 2 and got the sum of first 5 terms
as -5. Actual sum of first five terms is.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
25
-25
-35
35
Q. A Student read common difference of an A.P.
as -2 instead 2 and got the sum of first 5 terms
as -5. Actual sum of first five terms is.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> 1
> 6
> 9
> None
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 1
> 6
> 9
> None
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
Q. The number of possible outcomes in a throw
of n ordinary dice in which at least one of the
dice shows an odd number is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Q. The number of possible outcomes in a throw
of n ordinary dice in which at least one of the
dice shows an odd number is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> 14
> 21
> 28
> 35
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 14
> 21
> 28
> 35
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
1/4
1/32
1/16
1/8
Q. The mean and variance of a random variable
X having binomial distribution are 4 and 2
respectively, then P(X=1) is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
1/4
1/32
1/16
1/8
Q. The mean and variance of a random variable
X having binomial distribution are 4 and 2
respectively, then P(X=1) is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
Q. Consider an infinite geometric series with
first term a and common ratio r. If its sum is 4
and the second term is 3/4, then:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Q. Consider an infinite geometric series with
first term a and common ratio r. If its sum is 4
and the second term is 3/4, then:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
Q. The negation of the expression q ((~q)p) is
equivalent to
A
B
D
C
(~p)(~q)
p(~q)
(~p)(~q)
(~p)q BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Q. The negation of the expression q ((~q)p) is
equivalent to
A
B
D
C
(~p)(~q)
p(~q)
(~p)(~q)
(~p)q BITSAT MATH: ALGEBRA MARATHON
83667
90000
83660
None of these
Q. The sum of all odd numbers between 1 and
1000 which are divisible by 3 is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
83667
90000
83660
None of these
Q. The sum of all odd numbers between 1 and
1000 which are divisible by 3 is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
Exactly 3 solutions
A unique solution
No solution
Infinite number of
solutions
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Exactly 3 solutions
A unique solution
No solution
Infinite number of
solutions
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> 6
> 3
> 2
> 5
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 6
> 3
> 2
> 5
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> 0
> 1
> 2
> 4pqr
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 0
> 1
> 2
> 4pqr
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
Q. A person invites a party of 10 friends at dinner and
place them so that 4 are on one round table and 6 on
the other round table. The number of ways in which he
can arrange the guests is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Q. A person invites a party of 10 friends at dinner and
place them so that 4 are on one round table and 6 on
the other round table. The number of ways in which he
can arrange the guests is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
-2, 4
-4, 2
2, 6
None
Q. There are four numbers of which the first three are in
G.P. and the last three are in A. P. whose common difference
is 6. If the first and the last numbers are equal then two
other numbers are.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
-2, 4
-4, 2
2, 6
None
Q. There are four numbers of which the first three are in
G.P. and the last three are in A. P. whose common difference
is 6. If the first and the last numbers are equal then two
other numbers are.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
1/6
1/8
1/9
1/5
Q. The probability of getting 10 in a single
throw of three fair dice is:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
1/6
1/8
1/9
1/5
Q. The probability of getting 10 in a single
throw of three fair dice is:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> 4
> 2
> 1
> 3
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 4
> 2
> 1
> 3
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
The problem is to be re -evaluated.
Solution is not defined.
The objective function has to be
modified.
The change in constraints is
ignored.
Q. If the constraints in a linear programming
problem are changed then
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
Solution:
> Correct option is A)
BITSAT MATH: ALGEBRA MARATHON
The problem is to be re -evaluated.
Solution is not defined.
The objective function has to be
modified.
The change in constraints is
ignored.
Q. If the constraints in a linear programming
problem are changed then
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
3/17
14/17
3/10
7/10
Q. Girl students constitute 10% of I year and 5% of II year at Roorkee
University. During summer holidays 70% of the I year and 30% of II -
year students are given a project. The girls take turns on duty
in canteen. The chance that I year girl student is on duty in a randomly
selected day is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
3/17
14/17
3/10
7/10
Q. Girl students constitute 10% of I year and 5% of II year at Roorkee
University. During summer holidays 70% of the I year and 30% of II -
year students are given a project. The girls take turns on duty
in canteen. The chance that I year girl student is on duty in a randomly
selected day is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
5
6
4
None of these
Q. In a binomial distribution, the mean is 4 and
variance is 3. then its mode is:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
5
6
4
None of these
Q. In a binomial distribution, the mean is 4 and
variance is 3. then its mode is:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
1
2
3/2
5/2
Q. 21/4 .2 2/8 .2 3/16 .2 4/32 . is equal to:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
1
2
3/2
5/2
Q. 21/4 .2 2/8 .2 3/16 .2 4/32 . is equal to:
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
9!
10!
3!10!
3!9!
Q. In how many ways can 12 gentlemen sit
around a round table so that three specified
gentlemen are always together?
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
9!
10!
3!10!
3!9!
Q. In how many ways can 12 gentlemen sit
around a round table so that three specified
gentlemen are always together?
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> 90
> 120
> 96
> 240
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 90
> 120
> 96
> 240
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
10
60
15
125
Q. The number of ways in which first, second
and third prizes can be given to 5 competitors is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
10
60
15
125
Q. The number of ways in which first, second
and third prizes can be given to 5 competitors is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
0.99
0.871
0.891
0.762
Q. The probability that certain electronic component fails when first
used is 0.10. If it does not fail immediately , the probability that it
lasts for one year is 0.99. The probability that a new component
will last for one year is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
0.99
0.871
0.891
0.762
Q. The probability that certain electronic component fails
when first used is 0.10. If it does not fail
immediately, the probability that it lasts for one
year is 0.99. The probability that a new component will
last for one year is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> 0
> 1
> -1
> None of these
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
Solution:
[since, |z 1 + z 2| = |]
So, in general BITSAT MATH: ALGEBRA MARATHON
> 0
> 1
> -1
> None of these
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
Zero
Unique
Infinite
None of these
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON BITSAT MATH: ALGEBRA MARATHON
Zero
Unique
Infinite
None of these
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
Q. The statement (p(~q))(p(~q)) is
A
B
D
C
Equivalent to (~p) (~q)
A tautology
Equivalent to p q
A contradiction BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Q. The statement (p(~q))(p(~q)) is
A
B
D
C
Equivalent to (~p) (~q)
A tautology
Equivalent to p q
A contradiction BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> 0
> 1
> 2
> -1
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 0
> 1
> 2
> -1
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> 4
> 6
> 32
> 5
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> 4
> 6
> 32
> 5
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> p
> q
> ~q
> ~p
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> p
> q
> ~q
> ~p
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
Q. Sum of n terms of the series 8 + 88 + 888 +
equals
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Q. Sum of n terms of the series 8 + 88 + 888 +
equals
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
Vertices of an
equilateral triangle
Vertices of an
isosceles triangle
Collinear
None of these
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
Vertices of an
equilateral triangle
Vertices of an
isosceles triangle
Collinear
None of these
Q.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
100
200
300
400
Q. How many numbers lying between 999 and
10000 can be formed with the help of the digits
0, 2, 3, 6, 7, 8, when the digits are not repeated?
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
100
200
300
400
Q. How many numbers lying between 999 and
10000 can be formed with the help of the digits
0, 2, 3, 6, 7, 8, when the digits are not repeated?
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
> Q.
> A
> B
> D
> C
BITSAT MATH: ALGEBRA MARATHON
25.5
24.0
22.0
20.5
Q. If in a frequency distribution, the mean and
median are 21 and 22 respectively, then its
mode is approximately .
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
25.5
24.0
22.0
20.5
Q. If in a frequency distribution, the mean and
median are 21 and 22 respectively, then its
mode is approximately.
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
0.50
0.77
0.35
0.87
Q. A random variable X has the probability distribution
For the events E = {X is a prime number} and F = {X<4} then
P(EF)is
A
B
D
CBITSAT MATH: ALGEBRA MARATHON
> Solution:
BITSAT MATH: ALGEBRA MARATHON
0.50
0.77
0.35
0.87
Q. A random variable X has the probability distribution
For the events E = {X is a prime number} and F = {X<4} then
P(EF)is
A
B
D
C