Maths M.C.Q (1 Marks)

3. $\Big(\frac{7}{2},\frac{9}{4}\Big)$

4. q = 3p

Solution:

Let z0​ be the maximum value of z in the feasible region.

Since maximum occurs at both (15, 15) and (0, 20)$, the value z0​ is attained at  both (15, 15) and (0, 20).

⟹ z0​ = p(15) + q(15) and z0​ = p(0) + q(20)

⟹ p(15) + q(15) = p(0) + q(20)

⟹ 15p = 5q

⟹ 3p = q

2. 1 : 4

Solution:

Given: d₁ ​= 2cm

d₂​ = 4cm

Since the diameter are 2cm and 4cm.

The replacement ratio of the two pipes are 1cm and 2cmr₁ ​= 1cm

r₂​ = 2cm

Square of the ratio of the pipes are 1 and 4

$\therefore$ The ratio of rates of flow in two pipes $=1:\frac{1}{4}$

$\Rightarrow\frac{1}{4}$

2. P = 5y - 2x

Solution:

Let the number of normal calculators produced in a day be x andthe number of scientific calculators produced in a day be y the minimum of total calculators to be produced per day is 200

$\Rightarrow\text{x}+\text{y}\leq200$

Given, the minimum number of normal calculators to be produced per day is 100

$\Rightarrow\text{x}\geq100$

andthe minimum number of scientific calculators to be produced per day is 80

 $\Rightarrow\text{y}\geq80$

Also given, the maximum number of normal calculators can be produced per day is 200

$\Rightarrow\text{x}\leq200$

 andthe maximum number of scientific calculators can be produced per day is 170

$\Rightarrow\text{x}\leq170$

A normal calculator incurred a loss of Rs. 2

For x normal calculators, the loss is Rs. 2x

A scientific calculator gained a profit of Rs. 5

For xy scientific calculators, the gain is Rs. 5y

Therefore, profit of the manufacturer P = 5y - 2x.

3. {(1, 4)(2, 4)(3, 4)}

Solution:

Given,

A = {1, 2, 3}

B = {3, 4, 5}

C = {4, 6}

Now, $\text{B}\cap\text{C}=\{{4\}}$

$\therefore\text{A}\times(\text{B}\cap\text{C})=\{(1,4),(2,4),(3,4)\}$

2. $\text{p}=\frac{\text{q}}{2}$

1. 5p = 2q

Solution:

Maximum of Z occurs at (25, 20) and at (0, 30).

Hence, equating the vales of Z at these points, we get 25p + 20q = 30q

$\therefore$ 5p = 2q

This is the required relation.

Also as p, q > 0, the value of Z is always positive and hence, is greater at (25, 20) and at (0, 30) than at (0,10) and (5, 5).

1. $(0,0)$

Solution:

$\text{x}\leq0$ and $\text{y}\leq0$ represents third Quadrant $\text{x}+\text{y}\leq2$ represents the region below the line $\text{x}+\text{y}\leq2$ (the region which contains origin)

The common region of given set of equations is third quadrant (including negative x axis and negative y axis)

Since x and y values are $\leq0$ in the third quadrant, the maximum value of 3x + 2y occurs at x = 0 and y = 0 and the maximum value is 0.

2. $2\text{x}+\text{y}\geq10,\text{x}\geq0,\text{y}\geq0$

3. 3

Solution:

Since in the given function Z = 40x + 50y, two variables are used.

So, the two constraints will be $\text{x}\geq0,\text{y}\geq0$ and the third one will be of the type

$\text{ax}+\text{by}\geq\text{c}.$

Hence, at least 3 constraints are required.

2. 7

Solution:

Formula used:

$\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2)$

$=(\sqrt{(\text{x}+\text{y})^{2}-4\text{xy}})[(\text{x}+\text{y})^{2}-\text{xy}]$

$=(\sqrt{(3)^{2}-4(2})[(3)^{2}-2]$

$=(\sqrt{1})(7)=7$

4. $\text{z}>4$

Solution:

$28+7\text{z}>34+5.50\text{z}$

$\rightarrow1.50\text{z}>6$

$\rightarrow\text{z}>\frac{6}{1.5}\text{z}>4$

4. None of the above.

Solution:

We know that $ \text{A}.\text{M}.\geq\text{G}.\text{M}.$

Therefore, $ \frac{{\text{b}^2+\text{c}^2}​}{2}\geq\sqrt{\text{b}^2\text{c}^2}​$

$\Rightarrow\text{b}^{2}+\text{c}^{2}\geq2\text{bc}$

Multiplying a on both sides doesn’t change the inequality.

Since, given that a is positive.

$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(1)$

Similarly, $\text{b}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(2)$

and $\text{c}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(3)$

adding (1), (2) and (3) we get

$\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{c}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq2\text{abc}+2\text{abc}+2\text{abc}$

$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{b}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq6\text{abc}$

Therefore $\lambda$ is 6

3. Which does not form boundary of feasible region.

Solution:

A constraint in an LP model becomes redundant when the feasible region doesnt change by the removing the constraint.

For example, $2\text{x}+\text{y}\geq10$ and $6\text{x}+3\text{y}\geq30$ are constraints.

$6\text{x}+3\text{y}\geq30$

$\Rightarrow3\times(2\text{x}+\text{y})\geq3\times10$

$\Rightarrow2\text{x}+\text{y}\geq10$

 which is same as the first constraint.

Therefore, $6\text{x}+3\text{y}\geq30$  can be removed.

By removing this constraint the feasible region doesnt change.

If the boundary of the feasible region is removed then feasible solution set changes.

Hence, redundant constraint cannot be the boundary of the feasible region.

2. $\text{p}=\frac{\text{q}}{2}$

Solution:

So, condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) is,

p + q = 3p

⇒ 2p = q

$\therefore\text{p}=\frac{\text{q}}{2}$

2. 7

Solution:

The feasible region determined by the system of constraints, $\text{x}+3\text{y}\geq3,\text{x}+\text{y}\geq2,$ and $\text{x},\text{y}\geq0$ is given below

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(3, 0), $ \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big)$ and C(0, 2)

The values of Z at these corner points are given below

7 may or may not be the minimum value of Z because the feasible region is unbounded

For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check the resulting half - plane have common points with the feasible region or not.

Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7.

Thus, the minimum value of Z is 7 at point $ \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big).$

3. 2

Solution:

Better approch is with graphs.Considering graphs in eqaution we get

 $\text{x}^{2}-\text{x}\sin\text{x}-\cos\text{x}=0$

$\text{x}^{2}=\text{x}\sin\text{x}+\cos\text{x}$

Let  $\text{f}(\text{x})=\text{x}^{2},\text{g}(\text{x})=\text{x}\sin\text{x}+\cos\text{x}$

Using graphical methods,we can do the graph of f(x) and g(x)

The graph f(x) and g(x) intersects at two points between $(-\infty,\infty)$

3. 16

Solution:

 The feasible region determined by the constraints, $\text{x}+\text{y}\leq4,\text{x}\geq0,$ $\text{y}\geq0,$ is given below

O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given below:

Hence, the maximum value of Z is 16 at point B (0, 4)