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Linear Programming — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSLinear Programming4 Marks
Question
Linear programming is a method for finding the optimal values (maximum or minimum) of quantities subject to the constraints when relationship is expressed as linear equations or inequations. Based on the above information, answer the following questions.
  1. The optimal value of the objective function is attained at the points:
  1. On X-axis.
  2. On Y-axis.
  3. Which are comer points of the feasible region.
  4. None of these.
  1. The graph of the inequality 3x + 4y < 12 is:
  1. Half plane that contains the origin.
  2. Half plane that neither contains the origin nor the points of the line 3x + 4y = 12.
  3. Whole XOY-plane excluding the points on line 3x + 4y = 12.
  4. None of these.
  1. The feasible region for an LPP is shown in the figure. Let Z = 2x + 5y be the objective function. Maximum of Z occurs at:
  1. (7, 0)
  2. (6, 3)
  3. (0, 6)
  4. (4, 5)
  1. The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points ( 15, 15) and (0, 20) is:
  1. p = q
  2. p = 2q
  3. q = 2p
  4. q = 3p
  1. The comer points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y.
Compare the quantity in Column A and Column B
Column A
Column B
Maximum of Z
325
  1. The quantity in column A is greater.
  2. The quantity in column Bis greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined on the basis of the information supplied.
✓

Answer

  1. (c) Which are comer points of the feasible region.
Solution:
When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at comer points of the feasible region.
  1. (d) None of these.
Solution:
From the graph of 3x + 4y < 12, it is clear that it contains the origin but not the points on the line 3x + 4y = 12.
  1. (d) (4, 5)
Solution:
Maximum of objective function occurs at corner points.
Corner Points
Value of Z = 2x + 5y
(0, 0)
0
(7, 0)
14
(6, 3)
27
(4, 5)
$33\leftarrow\text{Maximum}$
(0, 6)
30
  1. (d) q = 3p
​​​​​​​​​​​​​​Solution:
Value of Z = px + qy at ( 15, 15)= 15p + 15q and that at (0, 20) = 20q. According to given condition, we have 15p + 15q = 20q ⇒ 15p = Sq ⇒ q = 3p.
  1. (b) The quantity in column Bis greater.
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:
Construct the following table of values of the objective function:
Corner Points
Value of Z = 4x + 3y
(0, 0)
4 × 0 + 3 × 0 = 0
(0, 40)
4 × 0 + 3 × 40 = 120
(20, 40)
4 × 20 + 3 × 40 = 200
(60, 20)
$4\times60+3\times20=300\leftarrow\text{Maximum}$
(60, 0)
4 × 60 + 3 × 0 = 240

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$ii.$ If $\vec{a}=4 \hat{i}+6 \hat{j}+12 \hat{k}$, then find the unit vector in direction of $\vec{a}. (1)$
$iii.$ Find area of $\triangle ABC. (2)$
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Write the triangle law of addition for $\triangle ABC$. Suppose, if the given slogans are to be placed on a straight line, then the value of $|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|. (2)$
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Ginni purchased an air plant holder which is in the shape of a tetrahedron.
Let A, B, C, and Dare the coordinates of the air plant holder where $\text{A}\equiv(1,1,1),\text{B}\equiv(2,1,3),\text{C}\equiv(3,2,2)$ and $\text{D}\equiv(3,3,4).$

Based on the above information, answer the following questions.
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  3. $\hat{\text{i}}+2\hat{\text{k}}$
  4. $-2\hat{\text{i}}-\hat{\text{k}}$
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  1. $2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
  2. $2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
  3. $-2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
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  2. $\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
  3. $3\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
  4. $2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
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  1. $\frac{\sqrt{11}}{2}\text{sq}.\text{units}$
  2. $\frac{\sqrt{14}}{2}\text{sq}.\text{units}$
  3. $\frac{\sqrt{13}}{2}\text{sq}.\text{units}$
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  1. $\frac{1}{\sqrt{17}}(2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
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On the basis of above information, answer the following questions.
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  1. (1, 0)
  2. (0, 1)
  3. (1, 1)
  4. (0, 0)
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  4. None of these.
  1. Value of the integral $\int\limits_{-1}^{0}(\text{x}+1)\text{dx}$ is:
  1. $\frac{1}{2}$
  2. $\frac{2}{3}$
  3. $\frac{3}{4}$
  4. $\frac{1}{3}$
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  1. 0
  2. -1
  3. 2
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  1. $\frac{1}{2}\text{ sq}.\text{units}$
  2. $\frac{3}{2}\text{ sq}.\text{units}$
  3. $\frac{3}{4}\text{ sq}.\text{units}$
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Image

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If $\mathrm{A}$ and $\mathrm{B}$ are two matrices such that $\mathrm{AB}=\mathrm{B}$ and $\mathrm{BA}=\mathrm{A}$, then find $\mathrm{A}^2+\mathrm{B}^2$ in terms of $\mathrm{A}$ and $\mathrm{B}$.

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  1. $\frac{1}{5}$
  2. $\frac{1}{6}$
  3. $\frac{1}{3}$
  4. None of these
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  2. $\frac{1}{3}$
  3. $\frac{1}{6}$
  4. $\frac{3}{10}$
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  2. $\frac{5}{12}$
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  4. $\frac{11}{20}$
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  1. $\frac{11}{100}$
  2. $\frac{1}{3}$
  3. $\frac{29}{100}$
  4. $\frac{1}{2}$
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  1. $\text{x}^\text{x}(1+\log\text{x})$
  2. $\text{x}^\text{x}(1-\log\text{x})$
  3. $-\text{x}^\text{x}(1+\log\text{x})$
  4. $\text{x}^\text{x}\log\text{x}$
  1. Differentiate $x^x + a^{x }+ x^a + a^a \ w.r.t. x.$
  1. $(1+\log\text{x})+(\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1})$
  2. $\text{x}^\text{x}(1+\log\text{x})+\log\text{a}+\text{ax}^{\text{a}-1}$
  3. $\text{x}^\text{x}(1+\log\text{x})+\text{x}^\text{a}\log\text{x}+\text{ax}^{\text{a}-1}$
  4. $\text{x}^\text{x}(1+\log\text{x})+\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$
  1. If $\text{x}=\text{e}^\frac{\text{x}}{\text{y}},$ then find $\frac{\text{dy}}{\text{dx}}.$
  1. $-\frac{(\text{x}+\text{y})}{\text{x}\log\text{x}}$
  2. $-\frac{(\text{x}-\text{y})}{\text{x}\log\text{x}}$
  3. $\frac{(\text{x}+\text{y})}{\text{x}\log\text{x}}$
  4. $\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
  1. If $y = (2 - x)^3(3 + 2x)^5,$ then find $\frac{\text{dy}}{\text{dx}}.$
  1. $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{15}{3+2\text{x}}-\frac{8}{2-\text{x}}\Big]$
  2. $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{15}{3+2\text{x}}+\frac{3}{2-\text{x}}\Big]$
  3. $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{10}{3+2\text{x}}-\frac{3}{2-\text{x}}\Big]$
  4. $(2-\text{x})^3(3+2\text{x})^5\cdot\Big[\frac{10}{3+2\text{x}}+\frac{3}{2-\text{x}}\Big]$
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  1. $\text{x}^\text{x}\text{e}^{2\text{x}+5}$
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  3. $\text{x}^\text{x}\text{e}^{2\text{x}+5}(1-\log\text{x})$
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Two farmers Shyam and Balwan Singh cultivate only three varieties of pulses namely Urad, Masoor and Mung. The sale (in ₹) of these varieties of pulses by both the farmers in the month of September and October are given by the following matrices A and B.

September sales (in ₹)
$\begin{matrix}\ \ \ \ \ \ \ \ \ \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\text{A}=\begin{bmatrix}10000&20000&30000\\50000&30000&10000\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$
October sales (in ₹)
$\begin{matrix}\ \ \ \ \ \ \ \ \ \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\text{B}=\begin{bmatrix}10000&20000&30000\\50000&30000&10000\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$
Using algebra of matrices, answer the following questions.
  1. The combined sales of Masoor in September and October, for farmer Balwan Singh, is:
  1. ₹ 80000
  2. ₹ 90000
  3. ₹ 40000
  4. ₹ 135000
  1. The combined sales of Urad in September and October, for farmer Shyam is:
  1. ₹ 20000
  2. ₹ 30000
  3. ₹ 36000
  4. ₹ 15000
  1. Find the decrease in sales of Mung from September to October, for the farmer Shyam.
  1. ₹ 24000
  2. ₹ 10000
  3. ₹ 30000
  4. No change
  1. If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October.
  1. $\begin{matrix} \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\begin{bmatrix}100&\ \ \ \ \ \ 200&\ \ \ \ \ 220\\400&\ \ \ \ \ \ 300&\ \ \ \ \ 200\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$
  2. $\begin{matrix} \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\begin{bmatrix}100&\ \ \ \ \ \ 200&\ \ \ \ \ 120\\400&\ \ \ \ \ \ 200&\ \ \ \ \ 200\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$
  3. $\begin{matrix} \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\begin{bmatrix}150&\ \ \ \ \ \ 200&\ \ \ \ \ 220\\400&\ \ \ \ \ \ 200&\ \ \ \ \ 280\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$
  4. $\begin{matrix} \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\begin{bmatrix}100&\ \ \ \ \ \ 200&\ \ \ \ \ 120\\250&\ \ \ \ \ \ 200&\ \ \ \ \ 220\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$
  1. Which variety of pulse has the highest selling value in the month of September for the farmer Balwan Singh?
  1. Urad
  2. Masoor
  3. Mung
  4. All of these have the same price
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Area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the determinant:
$\Delta=\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}$
Since, area is a positive quantity,
so we always take the absolute value of the determinant $\Delta.$ Also, the area of the triangle formed by three collinear points is zero.
Based on the above information, answer the following questions.
  1. Find the area of the triangle whose vertices are $(-2, 6), (3, -6)$ and $(1, 5).$
  1. $30\ \text{sq. units}$
  2. $35\ \text{sq. units}$
  3. $40\ \text{sq. units}$
  4. $15.5\ \text{sq. units}$
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  1. $4$
  2. $\frac{7}{140}$
  3. $47$
  4. $\frac{40}{7}$
  1. If the area of a triangle $\text{ABC},$ with vertices $A(1, 3), B(0, 0)$ and $C(k, 0)$ is $3$ sq. units, then a value of $k$ is:
  1. $2$
  2. $3$
  3. $4$
  4. $5$
  1. Using determinants, find the equation of the tine joining the points $A(1, 2)$ and $B(3, 6).$
  1. $y = 2x$
  2. $x = 3y$
  3. $y = x$
  4. $4x - y = 5$
  1. If $\text{A}\equiv(11,7),\text{B}\equiv(5,5)$ and $\text{C}\equiv(-1,3),$ then:
  1. $\Delta\text{ABC}$ is scalene triangle.
  2. $\Delta\text{ABC}$ is equilateral triangle.
  3. $A, B$ and $C$ are collinear.
  4. None of these.
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A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by cab, metro, bike or by other means of transport are respectively $0.3, 0.2, 0.1$ and $0.4.$ 'Tile probabilities that he will be late are $0.25, 0.3, 0.35$ and $0.1$ if he comes by cab, metro, bike and other means of
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  1. $\frac{5}{14}$
  2. $\frac{2}{7}$
  3. $\frac{5}{21}$
  4. $\frac{1}{6}$
  1. When the doctor arrives late, what is the probability that he comes by cab?
  1. $\frac{4}{21}$
  2. $\frac{1}{7}$
  3. $\frac{5}{14}$
  4. $\frac{2}{21}$
  1. When the doctor arrives late, what is the probability that he comes by bike?
  1. $\frac{5}{21}$
  2. $\frac{4}{7}$
  3. $\frac{5}{6}$
  4. $\frac{1}{6}$
  1. When the doctor arrives late, what is the probability that he comes by other means of transport?
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  2. $\frac{5}{14}$
  3. $\frac{4}{21}$
  4. $\frac{2}{7}$
  1. What is the probability that the doctor is late by any means?
  1. $1$
  2. $0$
  3. $\frac{1}{2}$
  4. $\frac{1}{4}$
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  1. The equation of plane passing through the points A, Band C is:
  1. 3x - 4y + z = 0
  2. 3x - 2y + z = 0
  3. 3x - 2y + z = 0
  4. 4x - 3y + 3z = 0
  1. The height of the tower from the ground is:
  1. 6 units
  2. 5 units
  3. $\frac{17}{\sqrt{14}}\text{units}$
  4. $\frac{5}{\sqrt{14}}\text{units}$
  1. The equation of line of perpendicular drawn from the peak of tower to the ground is:
  1. $\frac{\text{x}-6}{3}=\frac{\text{y}-4}{-2}=\frac{\text{z}-9}{1}$
  2. $\frac{\text{x}-6}{3}=\frac{\text{y}-5}{-2}=\frac{\text{z}-9}{1}$
  3. $\frac{\text{x}-6}{3}=\frac{\text{y}-4}{2}=\frac{\text{z}-9}{1}$
  4. $\frac{\text{x}-6}{3}=\frac{\text{y}-5}{2}=\frac{\text{z}-9}{1}$
  1. The coordinates of foot of perpendicular drawn from the peak of tower to the ground are:
  1. $\Big(\frac{33}{14},\frac{104}{14},\frac{109}{14}\Big)$
  2. $\Big(\frac{33}{14},\frac{109}{14},\frac{104}{14}\Big)$
  3. $\Big(\frac{33}{14},\frac{105}{14},\frac{109}{14}\Big)$
  4. None of these
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  1. $\frac{1}{2}\sqrt{14}\text{sq}.\text{units}$
  2. $\frac{3}{2}\sqrt{14}\text{sq}.\text{units}$
  3. $\sqrt{14}\text{sq}.\text{units}$
  4. $2\sqrt{14}\text{sq}.\text{units}$
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