5 Marks Questions

Question 15 Marks

A company produces soft drinks that have a contract which requires that a minimum of 80 units of chemical A and 60 units of the chemical B go into each bottle of the drink. The chemicals are available in prepared mix packets from two different suppliers. Supplier S had a packet of mix of 4 units of A and 2 units of B that costs ₹10. The supplier T has a packet of mix of 1 unit of A and 1 unit of B that costs ₹4. How many packets of mixes from S and T should the company purchase to honour the contract requirement and yet maintain the minimum cost? Make a LPP and solve graphically.

Answer

Let x and y units of packet of mixes are purchased from S and T, respectively. If Z is the total cost, then
Z = 10x + 4y...(i)
is objective function which we have to minimize.
Here, constraints are:
$4 x+y \geq 80$...(ii)
$2 x+y \geq 60$...(iii)
Also, $x \geq 0$...(iv)
$y \geq 0$...(v)
Now, 4x + y = 80

x	0	20
y	80	0

and 2x + y = 60...(2)

x	0	30
y	60	0

On plotting the graph of above constraints or inequalities (ii), (iii), (iv) and (v), we get shaded region having corner points A, P, B as feasible region. For coordinate of P

Point of intersection of
2x + y = 60 ...(vi)
and 4x + y = 80 ...(vii)
From (vi) - (vii),
2x + y - 4x - y = 60 - 80
- 2x = - 20
x = 10
y = 40
$\because$ Co-ordinates of $P=(10,40)$
Now the value of Z is evaluated at corner points in the following table:

Corner Point	Z = 10x + 4y
A(30, 0)	300
P(10,40)	$260 \leftarrow$ Minimum
B(0, 80)	320

Since, feasible region is unbounded. Therefore, we have to draw the graph of the inequality.
10x + 4y < 260
i.e., 5x + 2y < 130 ...(viii)
Since, the graph of inequality (viii) does not have any point common.
So, the minimum value of Z is 260 at (10, 40).
i.e., minimum cost of each bottle is ₹260, if the company purchases 10 packets of mixes from S and 40 packets of mixes from supplier T.

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Question 25 Marks

A dealer in rural area wishes to purchase a number of sewing machines. He has only ₹5760 to invest and has space for at most 20 items of storage. An electronic sewing machine cost him ₹360 and a manually operated sewing machine ₹240. He can sell an electronic sewing machine at a profit of ₹22 and a manually operated sewing machine at a profit of ₹18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit? Make it as a LPP and solve it graphically

Answer

Let x and y be electronic and manually operated sewing machine purchased respectively.
$\therefore$ LPP is
Maximize Z = 22x + 18y
Subject to,
$\begin{aligned} 360 x+240 y & \leq 5760 \\ 3 x+2 y & \leq 48\end{aligned}$
or $x+y \leq 20$
$x \geq 0, y \geq 0$
For equation 3x + 2y = 48

x	16	0
y	0	24

and x + y = 20

x	0	20
y	20	0

Vertices of feasible region are:
A(0, 20) B(8, 12) C(16,0) and O(0, 0) where B(8, 12) is the point of intersection of equation 3x + 2y = 48 and x + y = 20

Value of Z at corner points:

Corner points	Value of objective function Z = 22x + 18y
O(0, 0)	Z = 0
A(0, 20)	Z = 360
B(8, 12)	$Z =392 \rightarrow Max$
C(16,0)	Z = 352

$\therefore$ For maximum Z, Electronic machines = 8 , Manual machines = 12.

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Question 35 Marks

A manufacturer produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of ₹ 30 per trunk on the first type of trunk and ₹ 25 per trunk on the second type. Formulate a Linear Programming Problem to find out how many trunks of each type he must make each day to maximize his profit.

Answer

Let x units of A type and y units of B type be produced.
$\therefore$ Max Z = 30x + 25y
Under the following restrictions
$\begin{aligned} 3 x+3 y & \leq 18 \\ 3 x+2 y & \leq 15 \\ x & \geq 0, y \geq 0\end{aligned}$
Now x + y = 6

x	0	6	3
y	6	0	3

and 3x + 2y = 15

x	0	5	3
y	7.5	0	3

We draw the two constraints on the graph paper and indicate the feasible region. The region OABC represents the set of all feasible solutions.
The co-ordinates of the corner points of the feasible region are O(0, 0) A(5, 0) B(3, 3) and C(0, 6).

Now, Z = 30x + 25y
At O(0, 0), Z = 30(0) + 25(0) = 0
At A(5, 0), Z = 30(5) + 25(0) = 150
At B(3, 3), Z = 30(3) + 25(3) = 165
At C(0,6), Z = 30(0) + 25(6) = 150
Maximum value of Z = 165 when x = 3 and y = 3
Hence, 3 trunks of first type and 3 trunks of second type should be produced each day in order to get the maximum profit.

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Question 45 Marks

A new cereal, formed of a mixture of bran and rice, contains at least 88 grams of protein and at least 36 milligram of iron. Knowing that bran contains 80 gram of protein and 40 milligram of iron per kilogram, and that of rice contains 100 gram of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing a kilogram of this new cereal if bran costs ₹28 per kilogram and rice costs ₹25 per kilogram.

Answer

Let x kg of bran and y kg rice be produced.
Minimize, z = 28x + 25y
Subject to constraints
$\frac{80}{1000} x+\frac{100}{1000} y \geq \frac{88}{1000}$
$20 x+25 y \geq 22$...(1)
and $\frac{0.040}{1000} x+\frac{0.030}{1000} y \geq \frac{0.036}{1000}$
$20 x+15 y \geq 18$...(2)
$x \geq 0, y \geq 0$
First we draw the line AB and CD whose equations are $20 x+25 y \geq 22$ and $20 x+15 y \geq 18$, respectively.
Now, 20x + 25y = 22

x	0	1.1
y	0.88	0

20x + 15y = 18

x	0	0.9
y	1.2	0

The feasible region is OBPCO

The vertices of the feasible region are B(1.1, 0) P (0.6,0.4) and C(0,1.2) P is the point of intersection of the lines 20x + 25y = 22 and 20x + 15y = 18 Solving these we get the point P(0.6,0.4)
The minimum value of Z is:

Corner points	Z = 28x + 25y
At B(1.1, 0)	Z = 28 x 1.1 + 25 x 0 = 30.8
At P(0.6,0.4)	Z = 28 x 0.6 + 25 x 0.4 = 26.8
At C(0,1.2)	Z = 0 + 25 x 1.2 = 30

Hence, the minimum value of Z is 26.8 at the point P(0.6,0.4)
Since, feasible region is unbounded,
$\therefore$ we have to draw the graph 28x + 25y < 26.8
Since the graph of inequality does not have any common part.
So, the minimum value is 26.8.

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Question 55 Marks

A company produces two types of items, P and Q. Manufacturing of both items requires the metals gold and copper. Each unit of item P requires 3 gms of gold and 1 gm of copper while that of item Q requires 1 gm of gold and 2 gm of copper. The company has 9 gm of gold and 8 gm of copper in its store. If each unit of item P makes a profit of ₹50 and each unit of item Q makes a profit of ₹60, determine the number of units of each item that the company should produce to maximize profit. What is the maximum profit?

Answer

Let $x_1$ = number of units of item P
$x_2$ = number of units of item Q
Max.$Z=50 x_1+60 x_2$
Subject to the constraints
$\begin{array}{l}3 x_1+x_2 \leq 9 \\ x_1+2 x_2 \leq 8\end{array}$
such that $x_1, x_2 \geq 0$
Now, $3 x_1+x_2=9$...(i)

$x_1$	0	3
$x_1$	9	0

and $x_1+2 x_2=8$ ...(ii)

$x_1$	0	8
$x_2$	4	0

The corner points of the feasible region are O(0, 0) A(3, 0) B(0, 4) and C(2,3)

Corner points	$Z=50 x_1+60 x_1$
at O(0, 0),	Z = 0
at A(3, 0),	Z = 50 x 3 + 60 x 0 = 150
at C(2,3),	Z = 50 x 2 + 60 x 3 = 280
at B(0, 4),	Z = 50 x 0 + 60 x 4 = 240

$\therefore$ The company should produce 2 items of type P and 3 items of type Q.
$\therefore$ Maximum profit = 280

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Question 65 Marks

A manufacturer manufactures two types of tea-cups, A and B. Three machines are needed for manufacturing the tea cups. The time in minutes required for manufacturing each cup on the machines is given below:

Type of Cup	Time in minutes
Type of Cup	Machine I	Machine II	Machine III
A	12	18	6
B	6	0	9

Each machine is available for a maximum of six hours per day. If the profit on each cup of type A is ₹1.50 and that on each cup of type B is ₹1.00, find the number of cups of each type that should be manufacturing in a day to get maximum profit.

Answer

Let x be the number of A type tea cups and y be the number of B type tea cups.

The problem can be formulated as
Max. Z = 1.5x + 1y
i.e., Max. Z = 1.5x + y
Subject to the constraints:
$\begin{aligned} 12 x+6 y & \leq 360, \text { i.e., } 2 x+y \leq 60 \\ 18 x+0 . y & \leq 360, \text { i.e., } x \leq 20 \\ 6 x+9 y & \leq 360 \text {, i.e., } 2 x+3 y \leq 120\end{aligned}$
and $x \geq 0, y \geq 0$ (Non-negative constraints)
We draw the straight lines:
2x + y = 60 x = 20 2x + 3y = 120
For 2x + y = 60

x	30	0	40
y	0	60	20

For 2x + 3y = 120

x	60	0	30
y	0	40	20

The shaded portion shows the feasible region.
The points B and C are points of intersection of lines
2x + y = 60 with 2x + 3y = 120 and x = 20 with
2x + y = 60 respectively.

Thus, the 5 corner points of the feasible region are O(0, 0) A(0, 40) B(15, 30) C(20,20) and D(20, 0)
z = 1.5x + y

Corner Point	Value of Z1.5x + y
O(0, 0)	0
A(0, 40)	40
B(15, 30)	52.5(Max.)
C(20,20)	50
D(20, 0)	30

Clearly, the maximum profit is at x = 15 and y = 30
Thus, the manufacturer should manufacture 15 cups of type A and 30 cups of type B to get maximum profit in a day.

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Question 75 Marks

Two tailors P and Q earn ₹ 150 and ₹ 200 per day respectively. P can stitch 6 shirts and 4 trousers a day, while Q can stitch 10 shirts and 4 trousers per day. How many days should each work to produce at least 60 shirts and 32 trousers at minimum labour cost?

Answer

Let the tailor P work for x days and the tailor Q work for y days respectively.
Here, the problem can be formulated as an L.P.P. as follows:
Minimize Z = 150x + 200y
Subject to the constraints:
$6 x+10 y \geq 60$
or $3 x+5 y \geq 30$ ....(i)
4x + 4y >= 32
or $x+y \geq 8$....(ii)
and $x \geq 0, y \geq 0$...(iii)
Converting them into equations we obtain the following equations:
3x + 5y = 30, x + y = 8
$y=\frac{30-3 x}{5}$
y = 8 - x
3x + 5y = 30

x	0	10	5
y	6	0	3

x + y = 8

x	0	8	5
y	8	0	3

The shaded region in the diagram represent the feasible region.
The corner points are
A(10, 0) B(5, 3) and C(0,8)
At corner point value of Z = 150x + 200y
At A(10, 0) Z = 1500
At B(5, 3) Z = 150 x 5 + 200 x 3
= 750+600 = 1350
At C(0,8) Z = 1600
As the feasible region is unbounded, we draw the graph of the half-plane.
150x + 200y < 1350
3x + 4y < 27
There is no point common with the feasible region, therefore, Z has minimum value.
Minimum value of Z is 1350 and it occurs at the point B(5, 3).
Hence, the labour cost is 1350 when P works for 5 days and Q works for 3 days.

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Question 85 Marks

A company manufactures two types of toys A and B. A toy of type A requires 5 minutes for cutting and 10 minutes for assembling. A toy of type B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours available for cutting and 4 hours available for assembling the toys in a day. The profit is ₹50 each on a toy of type A and ₹60 each on a toy of type B. How many toys of each type should the company manufacture in a day to maximize the profit? Use linear programming to find the solution.

Answer

	Toy A	Toy B	Time in a day
Cutting time	5 min	8 min	180 min
Assembling time	10 min	8 min	240 min

Objective function
Minimize: Z = 50x + 60y
Subject to: $10 x+8 y \leq 240$
or $5 x+4 y \leq 120$....(i)
$x \geq 0, y \geq 0$....(ii)
$5 x+8 y \leq 180$...(iii)
For, 5x + 8y = 180

	A	B	E
x	0	36	12
y	22.5	0	12

For, 5x + 4y = 120

	C	D	E
x	0	24	12
y	30	0	15

Corner points	Z = 50x + 60y
At O(0, 0)	0
At D(24, 0)	1200
At E(12, 15)	1500(Max.)
At A(0, 22.5)	1350

Hence the maximum profit is ₹1500 at E(12, 15).
Therefore 12 units of type A and 15 units of type B is produced.

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Question 95 Marks

A dietician wishes to mix two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin and 8 units of vitamin C. The vitamin contents of one kg food is given below:

Food	Vitamin A	Vitamin B	Vitamin A
X	1 unit	2 units	3 units
Y	2 units	2 units	1 unit

One kg of food X costs ₹24 and one kg of food Y costs ₹36. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.

Answer

Let x be the number of units of food X and y be the number of units of food Y be mixed to obtain the desired diet. Then LPP of the given problem is:
Minimise, Z = 24x + 36y
Subject to the constraints,
$x+2 y \geq 10,2 x+2 y \geq 12$
i.e., $x+y \geq 6$ and $3 x+y \geq 8$
$x \geq 0, y \geq 0$
We draw the lines x + 2y = 10 x + y = 6, 3x + y = 8 and obtain the feasible region (unbounded and convex) as shown in the figure.

Thus, corner points are A(0, 8) B(1, 5) C(2,4) and D(10, 0)
The values of Z(in ₹) at these points are given in the following table:

Corner Point	Objective Function Z = 24x + 36y
A(0, 8)	Z = 24 x 0 + 36 x 8 = 288
B(1, 5)	Z = 24 x 1 + 36 x 5 = 204
C(2,4)	Z = 24 x 2 + 36 x 4 =192(Max.)
D(10, 0)	Z = 24 x 10 + 36 x 0 = 240

As the feasible region is unbounded, we draw the graph of the half plane 24x + 36y < 192 i.e., 2x + 3y < 16 and note that there is no point common with the feasible region. The minimum value of Z is ₹ 192. It occurs at C(2,4).
i.e., when 2 kg of food X and 4 kg of food Y are mixed to get the desired diet.

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Question 105 Marks

A company manufacturers two types of products A and B. Each unit of A requires 3 gram of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 gram of chromium. The firm can produce 9 gram of nickel and 8 grams of chromium. The profit is ₹40 on each unit of product of type A and ₹50 on each unit of type B. How many units of each type should the company manufactures so as to earn maximum profit? Use linear programming to find the solution.

Answer

Let x = Number of units of type A
y = Number of units of type B
Maximize Z = 40x + 50y
Subject to the constraints,
$\begin{array}{l}3 x+y \leq 9 \\ x+2 y \leq 8\end{array}$
and $x \geq, 0, y \geq 0$
Consider the equation,
3x + y = 9

x	0	3	2
y	9	0	3

and x + 2y = 8

x	0	8	2
y	4	0	3

Shaded region in the diagram represents the feasible solution.
Now, we can determine the maximum value of Z by evaluating the value of Z at the four points (vertices) as shown below.

Vertices	Z = 40x + 50y
(0,0)	Z = 40 x 0 + 50 x 0 = 0
(3,0)	Z = 40 x 3 + 50 x 0 =₹120
(0,4)	Z = 40 x 0 + 50 x 4 =₹200
(2, 3)	Z = 40 x 2 + 50 x 3 =₹230

From graph,
Maximum profit, Z = ₹ 230
$\therefore$ Number of units of type A is 2 and number of units of type B is 3.

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Question 115 Marks

A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After soil test, it is found that atleast 7 kg of nitrogen and same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs ₹5.00 per kg and the type B costs ₹8.00 per kg. Using Linear Programming, find how many kilograms of each type of the fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.

Answer

Let the fertilizer of type A be x kg and type B be y kg.

Fertilizer	Nitrogen	Phosphoric Acid	Quantity	Cost
A B	10% 5%	6% 10%	x kg y kg	₹5 ₹8
Total	7 kg	7 kg

According to Question:

Nitrogen	Phosphoric Acid
Quantity of Nitrogen in A = 10% Quantity of Nitrogen in B = 5% Minimum availability = 7kg	Quantity of Phosphoric Acid in A = 6% Quantity of Phosphoric Acid in B = 10% Minimum availability = 7kg
$\therefore$ 10% of x + 5% of $y \geq 7$ $\frac{10}{100} x+\frac{5}{100} y \geq 7$ $2 x+y \geq 140$ and $x \geq 0, y \geq 0$	$\therefore$ $6 \%$ of $x+10 \%$ of $y \geq 7$ $\frac{6}{100} x+\frac{10}{100} y \geq 7$ $3 x+5 y \geq 350$ and $x \geq 0, y \geq 0$

$\therefore$ Min. Z $=5 x+8 y$
Subject to constraints
$2 x+y \geq 140$
$3 x+5 y \geq 350$
$x \geq 0, y \geq 0$
2x + y = 140 ...(i)

x	70	0	50
y	0	140	40

3x + 5y = 350 .....(ii)

x	0	$\frac{350}{3}$
y	70	0

On solving eq (i) and (ii), we get x = 50 and y = 40

We draw the lines 2x + y = 140 and 6x + y = 70 and obtain the feasible region (unbounded convex) as shown in figure.
Thus, corner points are A( 0 , 140), B( 50 , 40), $C\left(\frac{350}{3}, 0\right)$.
The values of Z at these points are given in following table:

Corner Points	Objective function, z = 5x + 8y
A(0, 140)	Z = 5 x 0 + 8 x 140 = 1120
B(50, 40)	$Z=5 \times 50+8 \times 40=570 \rightarrow$minimum
$C\left(\frac{350}{3}, 0\right)$	$Z=5 \times \frac{350}{3}+8 \times 0=583 \frac{1}{3}$

As the feasible area is unbounded, Hence, 500 may or may not be the minimum value of Z.
For this, we need to graph inequality.
underline 5x + 8y < 570

x	114	0
y	0	71.25

Since, there is no common point between the feasible region and 5x + 8y < 570
Hence, the cost will be minimum,
if Fertilizer of type A used = 50 kg
Fertilizer of type B used = 40 kg
Minimum cost = 570

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Question 125 Marks

A manufacturing company makes two types of teaching aids A and B of Mathematics for class X. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of ₹80 on each piece of type A and ₹120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch.

Answer

Let the number of two types of teaching aids A and B be x and y, respectively.
Write the given data in tabular for mas following:

Item	No. of aids	Time on fabricating (in hrs)	Time on finishing (in hrs)	Profit (in)
A	x	9x	x	80x
B	y	12y	3y	120y
Total	x + y	9x + 12y	x + 3y	80x + 120y
Availability		180	30

The profit on type A is 80 and type B is 120. Thus, the required LPP is
Max. Z = 80x + 120y ...(i)
Subject to constraints
$9 x+12 y \leq 180$
or $3 x+4 y \leq 60$ ...(ii)
$x+3 y \leq 30$ ...(iii)
$x \geq 0, y \geq 0$
Consider the equation,
3x + 4y = 60

x	0	20
y	15	0

and x + 3y = 30

x	0	30
y	10	0

On solving eq, (ii) and (iii), we get
x = 12 and y = 6
The graphical representation of the system of inequations is as shown below

From graph, feasible region is
OABCO, whose corner points are (0, 0), A(20, 0) B(12, 6) and C(0,10)
The values of Z at corner points are as follows:

Corner Points	Value of Z = 80x + 120y
O(0, 0)	Z = 80 x 0 + 120 x 0 = 0
A(20, 0)	Z = 80 x 20 + 120 x 0 = 1600
B(12, 6)	Z = 80 x 12 + 120 x 6 = 1680
C(0,10)	Z = 80 x 0 + 120 x 10 = 1200

From the table, the maximum value of Z is ₹ 1680.
Hence, 12 pieces of type A and 6 pieces of type В should be manufactured per week to get maximum profit of ₹ 1680 per week.

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Applied Maths 5 Marks Questions

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