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Markdown Content: 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