Students can download 11th Business Maths Chapter 2 Algebra Ex 2.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.2

### Samacheer Kalvi 11th Business Maths Algebra Ex 2.2 Text Book Back Questions and Answers

Question 1.

Find x if \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\)

Solution:

Given that \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\)

\(\frac{1}{6 !}+\frac{1}{7 \cdot 6 !}=\frac{x}{8 \cdot 7 \cdot 6 !}\)

Cancelling all 6! we get

\(\frac{1}{1}+\frac{1}{7}=\frac{x}{8 \times 7}\)

\(\frac{7+1}{7}=\frac{x}{8 \times 7}\)

\(\frac{8}{7}=\frac{x}{8 \times 7}\)

x = \(\frac{8}{7}\) × 7 × 8 = 64

Question 2.

Evaluate \(\frac{n !}{r !(n-r) !}\) when n = 5 and r = 2.

Solution:

\(\frac{n !}{r !(n-r) !}=\frac{5 !}{2 !(5-2) !}=\frac{5 !}{2 ! \times 3 !}=\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1}=10\)

Question 3.

If (n + 2)! = 60[(n – 1)!], find n.

Solution:

Given that (n + 2)! = 60(n – 1)!

(n + 2) (n + 1) n (n – 1)! = 60(n – 1)!

Cancelling (n – 1)! we get,

(n + 2)(n + 1)n = 60

(n + 2)(n + 1)n = 5 × 4 × 3

Both sides we consecutive product of integers

∴ n = 3

Question 4.

How many five digits telephone numbers can be constructed using the digits 0 to 9 If each number starts with 67 with no digit appears more than once?

Solution:

Given that each number starts at 67, we need a five-digit number. So we have to fill only one’s place, 10’s place, and 100th place. From 0 to 9 there are 10 digits. In these digits, 6 and 7 should not be used as a repetition of digits is not allowed. Except for these two digits, we have 8 digits. Therefore one’s place can be filled by any of the 8 digits in 8 different ways. Now there are 7 digits are left.

Therefore 10’s place can be filled by any of the 7 digits in 7 different ways. Similarly, 100th place can be filled in 6 different ways. By multiplication principle, the number of telephone numbers constructed is 8 × 7 × 6 = 336.

Question 5.

How many numbers lesser than 1000 can be formed using the digits 5, 6, 7, 8, and 9 if no digit is repeated?

Solution:

The required numbers are lesser than 1000.

They are one digit, two-digit or three-digit numbers.

There are five numbers to be used without repetition.

One digit number: One-digit numbers are 5.

Two-digit number: 10th place can be filled by anyone of the digits by 5 ways and 1’s place can be 4 filled by any of the remaining four digits in 4 ways.

∴ Two-digit number are 5 × 4 = 20.

Three-digit number: 100th place can be filled by any of the 5 digits, 10th place can be filled by 4 digits and one’s place can be filled by 3 digits.

∴ Three digit numbers are = 5 × 4 × 3 = 60

∴ Total numbers = 5 + 20 + 60 = 85.