MBA SOLVED ASSIGNMENTS: BASIC MATHEMATICS

BCA 1^st Semester

BC0033 – 02

BASIC MATHEMATICS

1. How many 3 digit numbers can be formed by using the digits 2, 3, 5, 6, 7, 9 repetitions not being allowed

i) How many of these are less than 400 ?

ii) How many of these are even ?

iii) How many of these are multiples of 5 ?

Ans:

There are 6 digit and we require 3 digit number.

This means we have to fill up 3 places ( units, tenths, and hundredths ) by using 6 digits. Units place can be filled up in 6 ways as we can put any one of the six digit. After this 5 digits are left behind and 10^th place can be filled up in 5 ways and similarly 100^th place can be filled up in 4 ways.

100 10 1

4 5 6

Therefore the number of ways of filling up the 3 places = 6 x 5 x 4 = 120

We can just write that no. of ways of filling up 3 places using 6 digits is

Therefore the number of 3 digits numbers =120

i) Since the numbers should be less than 400, we

can put either 2 or 3 in the 100’s place. So the

100’s place can be filled up in 2 ways.

The other 2 places can filled up with remaining

5 digits means 5 ways.

100	10	1

		2 or 6

ii) In units place we must have either 2 or 6. so the units place can be filled up in 2 ways and other places in remaining 5 digits means 5 ways.

iii) A multiple of 5 ends with 0 of 5. as 0 is not one of the digits here, we must have 5 only in units place. So, units place can be filled up in 1 way, and other two places in 5 ways.

100	10	1

		5

2. Calculate the probabilities of selecting at random

(a) the winning horse in a race in which 10 horses are running

(b) the winning horses in both the first and second races if there are 10 horses in each race

Solutions:

a) the winning horse in a race in which 10 horses are running.

There are 10 horses running in a race. A horse selected at random

may be any one of the 10 horses. Therefore, there are 10 equally likely, mutually exclusive and exhaustive outcomes.

Let event A be selected winning horse.

There is 1 winning horse in a race. Therefore, 1 outcome is favorable to event A.

Therefore P [Horse] = P [ A ]

P [ A ] =

The probabilities of selecting at random the winning horse in a race in which 10 horses are running is =

Ans.

b) the winning horses in both the first and second races if there are 10 horses in each race

There are 10 horses running in each row. A horse selected at random may be any one of the 10 horse.

Let event A and B be -

A. selected winning horse in first race

B. selected winning horse in second race.

Therefore

The probabilities of selecting at random the winning horses in both races in which 10 horses are running in each race is =

Ans

3. Define radian and prove that radian is a constant angle.

Ans:

A radian is the measure of an angle at the centre of a circle subtended by an arc, whose arc length is equal to the radius of the circle.

A radian is the angle formed by a half line that has been rotated about its end point

of a complete rotation.

Let O be the centre of circle of radius r and if the circular arc AB is equal in length to the radius r of the circle, then by definition AB = 1 radian.

A radian is the angle subtended at the centre of a circle by an arc equal to the radius of the circle. O is the centre of a circle. A and B are points on the circle such that are AB = radius OA. Then AOB is called one radian of one circular measure.

We can write

Radian is a constant angle.

Proof:

Let PQR be a circle with centre O and radius r. Consider an arc AB of the circle whose length is equal to radius of the circle. Draw OA and OB and Produce AO to cut the circle at C.