Mangalore University
Credit Based III Semester B.B.M Degree Examination,Oct./Nov 2012
(New Syllabus) ( 2006-07 Batch onwards)
BUSINESS MATHEMATICS-Solution
g. Bankers Discount:- For discounting the bill, the bank deduct the interest on the sum due for the
period between the date of exchange and the due date. The interst charged by the bank on face value
of the bill is known as Bankers Discount
Bankers Discount B.D= Anr/100
A= Face value of the bill, r= rate of interest, n=period
h. Simple Interest I =Prt
P=4000 r=9%, t=3years
I= 4000* 9*3/100
= Rs.1080/-
i. x^2 -27x+50=0
(x-25)(x-2)
= x=25, x=2
j.Perpetuity:- An Annuity which continues for ever is perpetuity.
example:- Water tax paid to Muncipality
k. I=Prt
I=480, t=4yrs, r=?, P=1500
480= 1500*4*r/100
r= 480*100/(1500*4)
= 8%
l. Slack Variable:-
In L.P.P, the constraint may be '<=', or '>=' or '= '. To make '<=' constraint to '=', we add a
variable S1, where s1>=0 , to left side of the constraint. This variable is called Slack variable.
eg:-x1+x2<=10 , x1+x2+s1=10
2 Solve using cramer's rule
x+y+z= 3
2x+ y+2z= 5
4x-y+3z=6
A= 1 1 1
2 1 2
4 -1 3
|A|= 1(3+2)-1(6-8)+1(-2-4)
= 5+2-6
= 1
A1= 3 1 1
5 1 2
6 -1 3
|A1|= 3( 3+2)-1(15-12)+1(-5-6)
=15-3-11
= 1
A2= 1 3 1
2 5 2
4 6 3
=1(15-12)-3(6-8)+1(12-20)
= 3+6-8
= 1
A3= 1 1 3
2 1 5
4 -1 6
|A3| = 1(6+5) -1(12-20)+3(-2-4)
= 11+8-18
=1
X=|A1|/|A|= 1/1=1
Y=|A2|/|A|= 1/1=1
Z= |A3|/|A|= 1/1=1
3. Let the three number be a-d,a,a+d
sum of three numbers= a-d+a+a+d= 18
= 3a= 18, a=18/3= 6
product of three numbers= (a-d)a(a+d)= 192
(a-d)(a+d)= 192/6
a^2 - d^2= 32
6^2-d^2= 32
36-32= d^2
d=2
The three numbers are a-d= 6-2=4
a=6
a+d=6+2=8
(working note:- 4+6+8=18,4*6*8=192)
4.Let the current age of wife=x , Current age of Husband=y
Ratio 8 years back = (x-8)/(y-8)= 11/14
= 14(x-8)=11(y-8)
= 14x-112-11y+88=0
=14x-11y=24 ----------1
As per current ratio x/y=5/6
6x-5y=0 -----------2
solving equation 1 and 2
14 x-11y=24-------------1
6 x- 5y = 0 -------------2
multiplying equation 1*5 and equation 2* 11
(14*5)x -(11*5)y= (24*5)
(6*11)x- (5*11)y= (0*11)
70x-55y=120------------------3
66x-55y=0---------------------4
4x=120 eq 3-4
x=120/4
= 30
70*30 -120= 55y
1980=55y
y=36
age of wife=30 years and age of husband=36.
5.
6.
Let 12th march is Zero date
No.of days till 29th March= 17 days
No.of days till 14th April= 17+2+14=34days
No.of days till 6th June = 19+30+31+6=86days
Bill no. face value no.of days sum(nipi)
1 300 0 0
2 200 17 3400
3 400 34 13600
4 600 86 51600
sum pi=1500 sum(nipi)= 68600
equated due date= sum(nipi)/sum pi
= 68600/1500
= 45.7 days
The equated due date is 46 days ahead of march 12th ie, 19+27
= April 27.
7.
F=p[1+ i/2]^2n
P=30000, i=16/100, n=2
= 30000[1+ .16/2]^2*2
= 30000*(1.08)^4
= Rs.40814.66/-
8.
Let x: number of units of Model A to be produced
y: '' B "
Model A Model B Requirement
No.of units x y
Profit 10 12 Maximize
Labour 3x 2x <=500
demand 200 250
Maximize Z= 10x+12y
subject to 3x+ 2x<=500
x>=200,y>=250
9.a)
4 0 1 3 2 1
A= 2 3 1 B= 1 3 2
4 1 3 2 3 4
AB= 12+0+2 8+0+3 4+0+4
6+3+2 4+9+3 2+6+4
12+1+6 8+3+9 4+2+12
= 15 11 8
11 16 12
19 20 18
BA=> 3 2 1 4 0 1
B= 1 3 2 A= 2 3 1
2 3 4 4 1 3
BA= 12+4+4 6+1 3+2+3
4+6+8 9+2 1+3+6
8+6+16 9+4 2+3+12
= 20 7 8
18 11 10
30 3 17
AB not equal to BA
b).
A= 4 6 1
1 3 2
1 4 2
mod A= 4(6-8)-6(2-2)+1(4-3)
= 4(-2)-6(0)+1(1)
= -8+1=-7
4 1 1
A transpose = 6 3 4
1 2 2
a11=6-8=-2; a12=12-4=8; a13=12-3=9
a21=2-2=0; a22=8-1=7; a23=8-1=7
a31=4-3=1;a32=16-6=10;a33=12-6=6
-2 8 9 -2 -8 9
0 7 7 => 0 7 -7
1 10 6 1 -10 6
M^-1= 1/det(M) *adj M
= 1/-7( -2 -8 9
0 7 - 7
1 -10 6)
= 2/7 8/7 -9/7
0 -1 1
-1/7 10/7 -6/7
9.
tn=a+(n-1)d
t5= a+4d=17
t7=a+6d=24
a+4d=17 a+6d=24
a+6d=24 a+4d=17
2d=7 d=7/2
a+4d=17
a+4*7/2=17
a=17-14=3; first term-a=3; common difference- d=7/2
10.a)
50 men=10 days earn Rs.60000/-
x men =10 days earn Rs.100000/-
50/x = 60000/100000 =83.33men=>84 men
b). cash price= Rs.92.20/-
Cash Price= Invoice Price-Discount
92.20= x-x*4/100
92.20=96/100*X
x=96.0416/- = Rs.96.05/=
invoice price=marked price-trade discount
96.05=x-5/100*x
x=101.10= 101.10/-
Marked Price=Rs.101.10/-
c)
I=Prt
P=6000 t=
11.a)
A=500, n=8, i=0.05
F= A((1+i)^n -1)(1+i)
i
= 500((1.05)^8 -1)(1.05)
0.05
= Rs.5013.28/-
b) F= P(1+i)^n
1200= 1000(1+i)^2
1.2=(1+i)^2
1+i=1.0955
i=9.55%
c)F= A((1+i)^n -1)(1+i)
i
6000((1+0.15)^10 -1)(1+.15)
= 0.15
= Rs.140095.66/-
12.L.P.P By Simplex Method
By introducing slack variable S1 and S2 the given L.P.P is written in the standard
form is as follows
The standard form is;-
Maximise Z=4X+5Y+0S1+0S2
s.t 2X+Y+S1=12
3X+2Y+S2=14
and x,y,s1,s2>=0
The matrix form=
Maximise Z= (4 5 0 0)( x
y
s1
s2)
s.t => 2 1 1 0 (x
3 2 0 1 y = 4
s1 5
s2)
=> x 0
y >= 0
s1 0
s2 0