Courses

Solution- Linear Programming Test-1 Class 12 Notes | EduRev

Class 12 : Solution- Linear Programming Test-1 Class 12 Notes | EduRev

``` Page 1

CBSE TEST PAPER-01
CLASS - XII MATHEMATICS (Linear Programming)
Ans 01. Let food P consist of x Kg and food Q consists of Y Kg.
Z = 60x + 80y
3x + 4y = 8
5x + 2y = 11
x = 0
y = 0
Hence, Cost is minimum  = Rs 160
When x = 2,   y =
1
2
Ans 02. Let x  be number of cakes of first kind and y the number of cakes of other kind.
Z = x + y
200x + 100y = 5000
? 2x + y = 50
25x + 50y = 1000
? x + 2y = 40
x = 0, y = 0
Maximum number of cakes
Z = 30
When x = 20, y = 10.
Ans 03.  Let the number of cricket and the number of cricket bats to be made in a day be x
and y respectively.
Z = x + y
and also P = 20x + 10y
3
3 42
2
2 28
3 24
0, 0
x y
x y
x y
x y
+ =
? + =
+ =
= =
(i) Maximum Z = 16 at  x = 4  y = 12
(ii) P = 20×4 +10×12
= 200
Page 2

CBSE TEST PAPER-01
CLASS - XII MATHEMATICS (Linear Programming)
Ans 01. Let food P consist of x Kg and food Q consists of Y Kg.
Z = 60x + 80y
3x + 4y = 8
5x + 2y = 11
x = 0
y = 0
Hence, Cost is minimum  = Rs 160
When x = 2,   y =
1
2
Ans 02. Let x  be number of cakes of first kind and y the number of cakes of other kind.
Z = x + y
200x + 100y = 5000
? 2x + y = 50
25x + 50y = 1000
? x + 2y = 40
x = 0, y = 0
Maximum number of cakes
Z = 30
When x = 20, y = 10.
Ans 03.  Let the number of cricket and the number of cricket bats to be made in a day be x
and y respectively.
Z = x + y
and also P = 20x + 10y
3
3 42
2
2 28
3 24
0, 0
x y
x y
x y
x y
+ =
? + =
+ =
= =
(i) Maximum Z = 16 at  x = 4  y = 12
(ii) P = 20×4 +10×12
= 200
Ans 04. Let the manufacture produce x nuts and y bolts
Z = 17.50x + 7y
x + 3y = 12
3x + y = 12
X 1 y = 0
Maximum profit
Z = Rs 73. 50 at
X = 3, y = 3
Ans 05.  Let the manufacturer produce x packages of screw A and y packages Screw B.
Z = 7x +10y
4x + 6y = 240
2 3 120
6 3 240
x y
x y
? + =
+ =
2 80 x y ? + =
x=0,y =0
profit is maximum = 410
When 30 packages of screw A and 20 package
of screw B.
Ans 06.  Let x be pedestal lamps and y wooden shades
Z = 5x + 3y
2x + y = 12
3x + 2y = 20
x = 0, y = 0
profit maximum
when 4 pedestal lamps
Ans 07.  Let x souvenirs of type A and y souvenirs of type B
Z = 5x + 6y
5x + 8y = 200
10x + 8y = 240
? 5x + 4y = 120
x = 0, y = 0.
Maximum profit is Rs 160
When 8 souvenirs of Type A and 20 souvenirs
of type B.
Page 3

CBSE TEST PAPER-01
CLASS - XII MATHEMATICS (Linear Programming)
Ans 01. Let food P consist of x Kg and food Q consists of Y Kg.
Z = 60x + 80y
3x + 4y = 8
5x + 2y = 11
x = 0
y = 0
Hence, Cost is minimum  = Rs 160
When x = 2,   y =
1
2
Ans 02. Let x  be number of cakes of first kind and y the number of cakes of other kind.
Z = x + y
200x + 100y = 5000
? 2x + y = 50
25x + 50y = 1000
? x + 2y = 40
x = 0, y = 0
Maximum number of cakes
Z = 30
When x = 20, y = 10.
Ans 03.  Let the number of cricket and the number of cricket bats to be made in a day be x
and y respectively.
Z = x + y
and also P = 20x + 10y
3
3 42
2
2 28
3 24
0, 0
x y
x y
x y
x y
+ =
? + =
+ =
= =
(i) Maximum Z = 16 at  x = 4  y = 12
(ii) P = 20×4 +10×12
= 200
Ans 04. Let the manufacture produce x nuts and y bolts
Z = 17.50x + 7y
x + 3y = 12
3x + y = 12
X 1 y = 0
Maximum profit
Z = Rs 73. 50 at
X = 3, y = 3
Ans 05.  Let the manufacturer produce x packages of screw A and y packages Screw B.
Z = 7x +10y
4x + 6y = 240
2 3 120
6 3 240
x y
x y
? + =
+ =
2 80 x y ? + =
x=0,y =0
profit is maximum = 410
When 30 packages of screw A and 20 package
of screw B.
Ans 06.  Let x be pedestal lamps and y wooden shades
Z = 5x + 3y
2x + y = 12
3x + 2y = 20
x = 0, y = 0
profit maximum
when 4 pedestal lamps
Ans 07.  Let x souvenirs of type A and y souvenirs of type B
Z = 5x + 6y
5x + 8y = 200
10x + 8y = 240
? 5x + 4y = 120
x = 0, y = 0.
Maximum profit is Rs 160
When 8 souvenirs of Type A and 20 souvenirs
of type B.
Ans 08.  Let the merchant stock x desktop computers and y portable computer.
Z = 4500x + 5000y
x + y =250
25000x + 40000y = 700000
? 5x + 8y = 1400
x = 0,  y = 0.
Profit is maximum = 1150000
When 250 desktop computers and 50
portable computers are stocked.
Ans 09.  Let the diet contain x unit of food F 1 and y units of food F 2.
Z = 4x +6y
3x + 6y = 80
4x + 3y = 100
X = 0, y = 0
Z is minimum when 24 units of food F 1 and
4
3
unit of food F 2 are mixed minimum cost =
104 .
Ans 10.  Let the farmer use x Kg of F 1 and y Kg of F 2.
Z = 6x + 5y
10 5
14
100 100
6 10
14
100 100
0, 0.
x y
x y
x y
+ =
+ =
= =

Minimum cost
Z = 100
at  x = 100
y = 80
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;