Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.35

− − − − −

= (^)
(^132545434) 2 4 13 4
2 .2 .2 .3 .3 .3 .(2 )
( ) 2 ( ) 3 51 4 3^4 5 34 4( ) 2^4
2 (3 )

 + − − − − + −

= 

14 0 4

2 .3

 − −

= 

14 4

2

 − −

=   = 2

Example 45: Simplify

2n 2n 2 n n 2

n 4 n 3 2n

(3 5 3 )(5 3 5 )

5 [9 3 ]

− −

− +

− × − ×

−

Solution : The Given Expression
2n 2n 2 n n 2
n 4 2n 6 2n

(3 5 3 x 3 )(5 3 5 5 )

5 5 [3 x 3 3 ]

− −

−

= − × − × ×

× −

2 n( ) n( )

n 2n 4 6

3 1^5 5 1^3

9 25

5 x3 x [3 1]^1

5

− −

=

−^

4 22 625 275x x

=9 25 728 819=

Example 46 : Simplify

n m 1 m 1 m n 2

m 2m n m 1

4 20 12 15

16 5 9

− − + −

+ −

× × ×

× ×

Solution : The Given Expression
n m 1 m n m n 2
2m 2m n 2m 2

4 (4 5) (4 3) (3 5)

4 5 3

− − + −

+ −

= × × × × × ×

× ×

= 4 n m 1 m n 2m+ − + − − × 5 m 1 m n 2 2m n− + + − − − × 3 m n m n 2 2m 2− + + − − +

4 5^13 1 1^1

4 125 500

= −× − = × =

Example 47 : if a b c and b acx= y= z^2 = prove that.

1 1 2+ =
x z y
Solution : Let a b c kx= y= z=

(^11) y 1
∴ a (k) ,b (k) ,c (k)= x = = z
As per the equation,
2 2 1 1 1 1
b ac (k) k .k k= ⇒ y= x z= x z+
Equating powers on the same base
2 1 1= +
y x z