Conclude from the following: $$x - y = 7$$ $$x - z < 7$$ A: $$y$$ B: $$z$$

The quantity B is greater than A.

The quantity A is greater than B.

The relationship cannot be determined from the information given.

MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 9

$y\leq -15x + 3000$

$y\leq 5x$
In the xy-plane, if a point with coordinates

$(a, b)$ lies in the solution set of the system of inequalities above, what is the maximum possible value of b?

0%
$750$
0%
$600$
0%
$900$
0%
$150$

Explanation

The maximum value of b occurs when take equality.

$\Rightarrow -15x+3000=5x$

$\Rightarrow -20x=-3000$

$\Rightarrow x=150$
Therefore ,

$y=5\times150=750$
Hence maximum possible value of

$b=750$

Conclude from the following:

$x - y = 7$

$x - z < 7$

$y$

$z$

0%
The quantity A is greater than B.
0%
The quantity B is greater than A.
0%
The two quantities are equal.
0%
The relationship cannot be determined from the information given.

Explanation

given,

$x-y=7 \implies x=y+7$

substituting

$x=y+7$ in

$x-z<7$ we get

$\implies y+7-z<7$

$\implies y-z<7-7$

$\implies y-z<0$

$\implies y<z \implies z>y$

Therefore, quantity

$B:z$ is greater than

$A:y$

A graph and the system of inequalities are shown above. Which region of the graph could represent the solution for the system of in equations?

$y > x$

$3y≤-4x+6$

0%
$A$
0%
$B$
0%
$C$
0%
$D$

Explanation

given

$y>x$

and

$3y\leq -4x+6$

first, draw the graph for equations

$y =x$ and

$3y= -4x+6$

$y=x$ is the line which passes through the origin as shown in the above fig

Hence,

$y>x$ includes the above region of the line.

similarly, for

$3y= -4x+6$

substitute y=0 we get,

$-4x+6=0 \implies x=1.5$

substitute x=0 we get,

$3y=6 \implies y=2$

therefore,

$3y= -4x+6$ line passes through (1.5,0) and (0,2) as shown in fig.

Hence,

$3y\leq -4x+6$ includes the region below the line.

the intersection region is the D region as shown in above figure.

If the solution set for the system

$\begin{cases} y>x \\ y\le -\dfrac { 3 }{ 7 } x+5 \end{cases}$ is given by the above figure, then which of the following is NOT a solution to the system?

0%
$(0,3)$
0%
$(1,2)$
0%
$(2,4)$
0%
$(3,3)$

Explanation

As per the given figure,

$(0,3)$ and

$(1,2)$ lie in the desired region, hence these are solutions of the system.

Whereas

$(2,4)$ and

$(3,3)$ lie on the lines but

$(3,3)$ does not satisfy the equation

$y>x$ .

Hence, it is not a solution to the system.

At Anthony's High School, the ratio of juniors to seniors is 4 to 3, the ratio of seniors to sophomores is 5 to 4, and the ratio of freshmen to sophomores is 7 toFind the ratio of freshmen to seniors.

0%
$\displaystyle \frac { 7 }{ 3 }$
0%
$\displaystyle \frac { 5 }{ 3 }$
0%
$\displaystyle \frac { 9 }{ 7 }$
0%
$\displaystyle \frac { 14 }{ 15 }$

Explanation

Ratio of juniors to seniors

$= 4:3$

Ratio of senior to sophomores

$=5:4$

Ratio of freshmen to sophomores

$= 7:6$

Ratio of juniors to seniors to sophomores

$=20:15:12$

Equivalent ratio of freshmen to sophomores

$=14:12$

Ratio of juniors to seniors to sophomores to freshmen

$=20:15:12:14$

So, ratio of freshmen to seniors

$=14:15$

Solve the linear inequation:

$9 \geq 7x -5 \geq 5x -11, x\in R$ .

0%
$[-4,2]$
0%
$[-3,2]$
0%
$[-3,3]$
0%
$[-3,4]$

Which coordinate is in the solution set of the system of linear inequalities:
$3x + 5y \leq 9$

$x - 6y \leq 3$

0%
$( 0, 2.5 )$
0%
$( 0 , 2 )$
0%
$( 1 , 0 )$
0%
$( 3.5 , 0 )$

Explanation

By option verification,

i.e., substituting the options in given equation and verifying

Given,

$3x+5y\leq 9$ and

$x-6y\leq 3$

substituting option A i.e.,

$(x,y)=(0,2.5)$

$3x+5y\leq 9\implies 3\times 0+5\times 2.5\leq 9 \implies 12.5\leq 9$ False

substituting option B i.e.,

$(x,y)=(0,2)$

$3x+5y\leq 9\implies 3\times 0+5\times 2\leq 9 \implies 10\leq 9$ False

substituting option C i.e.,

$(x,y)=(1,0)$

$3x+5y\leq 9\implies 3\times 1+5\times 0\leq 9 \implies 3\leq 9$ True

$x-6y\leq 3\implies 1-6\times 0\leq 3 \implies 1\leq 3$ True

substituting option D i.e.,

$(x,y)=(3.5,0)$

$3x+5y\leq 9\implies 3\times 3.5+5\times 0\leq 9 \implies 17.5\leq 9$ False

Therefore,

$(x,y)=(1,0)$ is the solution set of

$3x+5y\leq 9$ and

$x-6y\leq 3$

Which of the following is the graph of the region

$1 < x+y <2$ in the standard

$\left(x, y\right)$ coordinate plane?

Explanation

given,

$1<x+y<2$

$\implies x+y>1$ and

$x+y<2$

graph for equations

$x+y=1$ and

$x+y=2$

for

$x+y=1$

substitute

$y=0$ , we get

$x=1$

substitute

$x=0$ , we get

$y=1$

therefore,

$x+y=1$ line passes through

$(0,1)$ and

$(1,0)$

Hence,

$x+y>1$ includes the region above the line.

for

$x+y=2$

substitute

$y=0$ , we get

$x=2$

substitute

$x=0$ , we get

$y=2$

therefore,

$x+y=2$ line passes through

$(0,2)$ and

$(2,0)$

Hence,

$x+y<2$ includes the region below the line.

Option D has the lines passing through

$(0,1)$ and

$(1,0)$ and

$(0,2)$ and

$(2,0)$

Find the correct inequality if

$a + b > 0$ and

$c + d > 0$

0%
$a + b + c > 0$
0%
$ac + bd > 0$
0%
$a^{2} + b^{2} > 0$
0%
$d(a + b) > 0$
0%
$a + b > c + d$

Explanation

Given

$a+b > 0$ and

$c+d > 0$ then correct option is

(A) :

$a+b+c> 0$ is wrong because we can no t comment od value of

$c$ , it is wrong if

$a=1,b=1$ and

$c=-5, d= 6$ ,

$c+d=1>0$ ,

$a+b+c <0$

(B) :

$ac+bd > 0$ is wrong if

$a=3,b=1,c=-5, d=6$ even in this case above given condition are true

$a^{2}+b^{2}> 0$ is correct, because square of any number is always positive

(D) :

$d(a+b)> 0$ is wrong, if

$c>d$ and

$d$ is negative and

$a,b$ are positive

(E) :

$a+b> c+d$ is wrong because we can not comment on which number is big

Conclude from the given information:

$x, y, s$ , and

$t$ are positive integers.

$x < y < 10$

$s < t < 10$

A: The maximum possible value of

$y - x$ .

B: The minimum possible value of

$s + t$ .

0%
The quantity A is greater than B.
0%
The quantity B is greater than A.
0%
The two quantities are equal
0%
The relationship cannot be determined from the information given.

Explanation

Given

$x<y<10$ and

$x,y$ are positive integers.

So the maximum value of

$y-x$ occurs when

$y$ is maximum and

$x$ is minimum.

Therefore maximum value of

$y-x$ is

$9-1 = 8$ .

Given

$s<t<10$ and

$s,t$ are positive integers.

So the minimum value of

$s+t$ occurs when

$s$ is minimum and

$t$ is minimum.

Therefore minimum value of

$s+t$ is

$1+1 = 2$ .

Therefore the quantity in

$1$ st column is greater.

So, option

$A$ is correct.

A stick of length

$20$ units is to be divided into

$n$ parts so that the product of the lengths of the parts is greater than unity. The maximum possible value of

$n$ is

0%
$18$
0%
$19$
0%
$20$
0%
$21$

Explanation

Given, a stick is divided into

$n$ parts which is of length

$20$ units

Let the

$n$ parts be

$x_1, x_2, ...,x_n$

$x_1+x_2+x_3+...+x_n=20$

Given,

$x_1 .x_2. ....x_n >1$ -------(1)

We know that

$A.M. \geq G.M.$

$A.M.$ of

$x_1, x_2, ...,x_n = \dfrac{x_1+x_2+...+x_n}{n}$

$G.M.$ of

$x_1, x_2, ...,x_n = \sqrt[n]{x_1.x_2.....x_n}$

Therefore,

$\dfrac{x_1+x_2+...+x_n}{n}\geq \sqrt[n]{x_1.x_2.....x_n}$

$\dfrac{20}{n}\geq \sqrt[n]{x_1.x_2.....x_n}$ ...(since,

$x_1+x_2+x_3+...+x_n=20$ )

taking

$n^{th}$ power on both sides.

$(\dfrac{20}{n})^n\geq {x_1.x_2.....x_n}$ -------(2)

From (1) and (2) we get

$(\dfrac{20}{n})^n> 1$

$\implies \dfrac{20}{n}>1$ (since,

$\sqrt[n]1=1$ )

$n<20$

Therefore, the maximum possible value of

$n = 19$

Solve the inequation

$\sqrt { 3x-8 } <-2$

0%
$\phi$
0%
$(0,1)$
0%
$(0,3)$
0%
None of these

Solve:

$\displaystyle\frac{1}{x} < 1$ .

0%
$x\epsilon(-\infty, 0)\cup(1, \infty)$
0%
$x\epsilon(-\infty, -2)\cup(1, \infty)$
0%
$x\epsilon(-\infty, 0)\cup(7, \infty)$
0%
$x\epsilon(-\infty, 0)\cup(3, \infty)$

Which region is described by the shade in the graph given?

0%
$2x+3y=3$
0%
$2x+3y>3$
0%
$2x+3y<3$
0%
None of these

Explanation

The graph of the function

$2x+3y=3$ is shown in the figure.

It is clear from the graph,

$(0,0)$ does not lie on the line.

Now, put

$(0,0)$ in the equation of line.

$2(0)+3(0)=3$

$0\neq 3$ hence,

$0<3$ is true

Hence, region which lies below the graph is denoted by

$2x+3y<3$

It is clear from the graph,

$(0,2)$ does not lie on the line.

Now, put

$(0,2)$ in the equation of line,

$2(0)+3(2)=3$

$0\neq 6$ hence,

$6>3$ is true.

Hence, the region which lies above the graph is denoted by

$2x+3y>3$ .

State true or false.

The diagram shows a scale drawing of a garden which is pm long and wm wide.

Inequality between p and w is

$p > w$

0%
True
0%
False

State true or false.

If a path of

$2$ m width is added to two sides as shown in the diagram,then a new inequality between the length and the width is

$p+2 > w+2$

0%
True
0%
False

Which of the following is the correct graph for

$x\ge 3$ or

$x\le -2$ ?

0%
Line A
0%
Line B
0%
Line C
0%
Line D
0%
Line E

Explanation

It is stated

$x \ge 3$ that means including

$3$ and till infinity (positive)

$= \{3, 4, 5, 6, ....\}$

And, it also stated

$x \le -2$ that means including

$-2$ and till infinity (negative)

$= \{3, 4, 5, 6, ... \}$

So, the line D perfectly defines the two stated i.e.

$x \ge 3$ and

$x \le -2$

$\therefore$ Option D is correct.

2106130_537737_ans_0c2aad2c61b04aab87ab1316046e7ef4.png

Which $x$ values make the ordered pair $(x,8)$ a solution of the following inequality?

$9x+4y<13$

0%
$x <\dfrac{-19}{9}$
0%
$x < -5$
0%
$x > -5$
0%
$x < 5$

Explanation

Given inequality is

$9x+4y<13$

Now substitute the point

$(x,8)$ in the given inequality

we get

$9x+32<13$

$\Rightarrow 9x < -19$

$\Rightarrow x < -\dfrac{19}{9}$

Let a, b, c be positive real numbers, then

$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c} + \frac{a^2+c^2}{a+c}\geq a+b+c$

0%
True
0%
False

$a,b,c$ are three distinct positive real number , then the least value of

$\dfrac{(1+a+a^2)(1+b+b^2)(1+c+c^2)}{abc}$ , is

0%
$3$
0%
$9$
0%
$27$
0%
None of the above.

Explanation

since,

$A.M. \geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

let

$a=1, b=a$ and

$c=a^2$

$\implies \dfrac{1+a+a^2}{3}\geq \sqrt[3]{1*a*a^2}$

$\implies \dfrac{1+a+a^2}{3}\geq a$

$\implies \dfrac{1+a+a^2}{a}\geq 3$ ------(1)

similarly,

$\implies \dfrac{1+b+b^2}{b}\geq 3$ ------(2)

and

$\implies \dfrac{1+c+c^2}{c}\geq 3$ ------(3)

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

$\implies (\dfrac{1+a+a^2}{a})(\dfrac{1+b+b^2}{b})(\dfrac{1+c+c^2}{c})\geq 3*3*3$

$\implies \dfrac{(1+a+a^2)(1+b+b^2)(1+c+c^2)}{abc}\geq 27$

Which quadrant does the solution lie in?

0%
$I$
0%
$II$
0%
$III$
0%
$IV$

Explanation

Given,

$2x+3y>2$

$x-y<0 \implies y>x$

$y\geq 0$

$x\leq 3$

first, draw the graph for equations

$2x+3y=2$

$x-y=0 \implies x=y$

$y= 0 \implies x-axis$

$x= 3$

$x=y$ is the line which passes through the origin as shown in the above fig

Hence,

$y>x$ includes the above region of the line.

$y= 0 \implies x-axis$ . Hence,

$x\leq 3$ includes the left region of the line.

$x= 3 \implies$ line parallel to x-axis. Hence,

$x\leq 3$ includes the left region of the line.

for

$2x+3y=2$

substituting y=0, we get

$2x=2\implies x=1$

substituting x=0, we get

$3y=2\implies y=\dfrac{2}{3}$

Therefore, line

$2x+3y=2$ passes through (1,0) and (0,2/3) as shown in the above figure

Hence,

$2x+3y>2$ includes the above region of the line.

Therefore, the blue shaded region is the feasible region which is present in I quadrant as shown in the figure.

$P=(ab+xy)(ax+by)$ , then:

0%
$P>10abxy$
0%
$P>16abxy$
0%
$P>4abxy$
0%
$P>2abxy$

Explanation

$A.M. = \dfrac{a+b}{2}$ and

$G.M. = \sqrt{ab}$

Subtracting above equations we get

$A.M.- G.M. = \dfrac{a+b}{2}-\sqrt{ab}$

$= \dfrac{a+b-2\sqrt{ab}}{2}$ ------- (1)

$=\dfrac{(\sqrt{a}-\sqrt{b})^2}{2}\geq 0$ (since, square term is always

$\geq 0$ )

$\implies A.M.-G.M.\geq 0$

Therefore,

$A.M. \geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

squaring on both sides.

$\implies (\dfrac{a+b}{2})^2\geq {ab}$

$\implies(a+b)^2\geq {4ab}$

Let

$a=ab$ and

$b=xy$

$\implies (ab+xy)^2\geq 4abxy$ -------(1)

Let

$a=ax$ and

$b=by$

$\implies (ax+by)^2\geq 4abxy$ -------(2)

multiplying (1) and (2)

$\implies (ab+xy)^2(ax+by)^2\geq (4abxy)^2$

applying square root on both sides

$\implies (ab+xy)(ax+by)\geq (4abxy)$

Let

$a=2ab$ and

$b=2xy$

$\implies (2ab+2xy)^2\geq 4abxy$

$\implies (ab+xy)^2\geq abxy$ -------(3)

multiplying (3) and (2)

$\implies (ab+xy)^2(ax+by)^2\geq 4(abxy)^2$

applying square root on both sides

$\implies (ab+xy)(ax+by)\geq (2abxy)$

$P_n$ denotes the product of first

$n$ natural numbers, then for all

$n \in N$ .

0%
$P_n\leq n^n$
0%
$P_{n+1}\leq n^n$
0%
$P_n \leq \left(\dfrac {n+1}{2}\right)^n$
0%
None of the above.

Explanation

Given,

$n\in N$

Natural numbers are counting numbers whose set is

$N=\{1,2,...\}$

Given,

$P_n$ is the product of first

$n$ natural numbers.

Therefore,

$P_n=(1*2*...*n)$

We know that,

$A.M. \geq G.M.$

$A.M.$ of first

$n$ natural numbers =

$\dfrac{1+2+...+n}{n}$

$G.M.$ of first

$n$ natural numbers =

$\sqrt[n]{1*2*...*n}$

Therefore,

$\dfrac{1+2+...+n}{n}\geq \sqrt[n]{1*2*...*n}$

$\implies \dfrac{n(n+1)}{2n}\geq \sqrt[n]{1*2*...*n}$

since, the sum of first

$n$ natural numbers is

$\dfrac{n(n+1)}{2}$

$\implies (\dfrac{(n+1)}{2})^n\geq {1*2*...*n}$

$\implies P_n\leq (\dfrac{n+1}{2})^n$ (since,

$P_n=(1*2*...*n)$ )

$a+b+c=1$ and

$a,b,c \in R^+$ such that

$(1-a)(1-b)(1-c)\geq \lambda bac$ , then

$\lambda=$

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Given,

$a+b+c=1$

we know that,

$A.M.\geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

$\implies \dfrac{1}{3}\geq \sqrt[3]{abc}$

$\implies \sqrt[3]{abc}\leq \dfrac{1}{3}$

cubing on both sides

$\implies {abc}\leq \dfrac{1}{27}$ --------------(1)

Similarly

$\implies \dfrac{1-a+1-b+1-c}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

$\implies \dfrac{3-(a+b+c)}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

$\implies \dfrac{3-1}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

$\implies \dfrac{2}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

cubing on both sides

$\implies \dfrac{1}{(1-a)(1-b)(1-c)}\leq \dfrac{27}{8}$ ---------------(2)

multiplying (1) and (2) we get

$\implies \dfrac{abc}{(1-a)(1-b)(1-c)}\leq \dfrac{1}{27}*\dfrac{27}{8}$

$\implies \dfrac{abc}{(1-a)(1-b)(1-c)}\leq \dfrac{1}{8}$

$\implies \dfrac{1}{(1-a)(1-b)(1-c)}\leq \dfrac{1}{8abc}$

$\implies {(1-a)(1-b)(1-c)}\geq {8abc}$

Therefore

$\lambda$ is

$8$

$a+b+c =15$ and

$ab+bc+ca= 74$ , then the value of

$a^3+b^3+c^3-3abc$ is,

0%
$42$
0%
$46$
0%
$45$
0%
$48$

$a^2+b^2+c^2=1$ , then

$ab+bc+ca$ lies in the interval :

0%
$[0,1]$
0%
$[-0.5,1]$
0%
$[0,0.5]$
0%
$[1,2]$

Explanation

we know that

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$ and

$(a+b+c)^2\geq 0$ for any real a,b,c

Given,

$a^2+b^2+c^2=1$

Therefore,

$1+2(ab+bc+ca)\geq 0$

$(ab+bc+ca)\geq -\dfrac{1}{2}$

Since,

$A.M. \geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

$\implies {a+b}\geq 2\sqrt{ab}$

Assume

$a=a^2$ and

$b=b^2$

$\implies {a^2+b^2}\geq 2ab$ ------(1)

similarly,

${b^2+c^2}\geq 2bc$ ------(2)

${c^2+a^2}\geq 2ac$ ------(3)

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

$a^2+b^2+c^2\geq ab+bc+ca$

Since,

$a^2+b^2+c^2=1$

$(ab+bc+ca)\leq 1$

Therefore,

$ab+bc+ca$ lies in the interval

$[-\dfrac{1}{2},1]$

The minimum value of

$4^a+4^{1-a}, a\in R$ , is

0%
$1$
0%
$2$
0%
$4$
0%
None of the above.

Explanation

We know that

$A.M.\geq G.M.$

$\implies \dfrac{4^a+4^{1-a}}{2}\geq \sqrt{4^a*4^{1-a}}$

$\implies \dfrac{4^a+4^{1-a}}{2}\geq \sqrt{4^{a+1-a}}$

$\implies 4^a+4^{1-a}\geq 2\sqrt{4}$

$\implies 4^a+4^{1-a}\geq 2*2$

$\implies 4^a+4^{1-a}\geq 4$

Therefore, the minimum value of

$4^a+4^{1-a}$ is 4

$a,b,c$ are the sides of a triangle, then

$\dfrac a{b+c-a} +\dfrac b{c+a-b} +\dfrac c{a+b-c}$

0%
$\leq 3$
0%
$\geq 3$
0%
$\leq 2$
0%
$\geq 2$

Explanation

Given,

$\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$

let

$b+c-a=x$

$c+a-b=y$ and

$a+b-c=z$

Therefore,

$x+y=b+c-a+c+a-b\implies x+y=2c \implies c=\dfrac{x+y}{2}$

$y+z=c+a-b+a+b-c\implies y+z=2a \implies a=\dfrac{y+z}{2}$ and

$x+z=b+c-a+a+b-c\implies x+z=2b \implies b=\dfrac{x+z}{2}$

substituting the above expressions in the given equation we get

$\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}=\dfrac{y+z}{2x}+\dfrac{x+z}{2y}+\dfrac{x+y}{2z}$

multiplying 2 on both sides

$2(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c})=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}$

$2(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c})\geq 6$

since,

$\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{y}{z}\geq 6$

Therefore,

$(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c})\geq 3$

A manufacturer produces two products

$A$ and

$B$ . Product

$A$ fetches him a profit of Rs.

$24$ and product

$B$ fetches him a profit of Rs.

$14$ . It takes

$15$ minutes to manufacture one unit of product

$A$ and

$5$ minutes to manufacture one unit of product

$B$ . There is a limit of

$30$ units to the quantity of

$B$ of manufactured due to a constraint on the availability of materials. The minimum profit that the company decides to make is Rs.

$1000$ . The workers work for a maximum of

$10$ hours a day. Find the maximum daily profit.

0%
Rs. $1020$
0%
Rs. $1040$
0%
Rs. $1140$
0%
Rs. $1240$

Explanation

Given, The profit from the product

$A$ is

$Rs. 24$ and

The profit from the product

$B$ is

$Rs. 14$

The maximum number of units of

$B$ manufactured in a day is

$30$

i.e.,

$y\leq 30$ --------- (1)

Let the number of units of

$A$ manufactured in a day be

$x$

Therefore the profit for a day from the product

$A$ is

$24x$

Let the number of units of

$B$ manufactured in a day be

$y$

Therefore the profit for a day from the product

$B$ is

$14y$

The minimum profit for a day is

$Rs.1000$

Therefore, the total profit from products

$A$ and

$B$ should be more than

$Rs.1000$

i.e.,

$24x+14y\geq 1000$ --------- (2)

Given, time taken to manufacture one product of

$A$ is

$15\space min$ and

time taken to manufacture one product of

$B$ is

$5\space min$

Therefore, time taken to manufacture

$x$ products of

$A$ is

$15x\space min$ and

time taken to manufacture

$y$ products of

$B$ is

$5y\space min$

In a day, a worker works for a maximum of

$10\space hrs = 600\space min$

Therefore, the time taken to manufacture products

$A$ and

$B$ should be less than

$600\space min$

I.e.,

$15x+5y\leq 600$ --------- (3)

The total profit from products

$A$ and

$B$ is

$P=24x+14y$

In the above figure, the blue shaded region is the feasible region with three corner points.

$(\dfrac{290}{12},30), (30,30), (\dfrac{340}{9},\dfrac{20}{3})$

$(\dfrac{290}{12},30)$ is the point where

$24x+14y= 1000$ intersects

$y=30$

I.e., substituting

$y=30 \implies 24x+14*30 =1000 \implies x=\dfrac{1000-420}{24}\implies x=\dfrac{290}{12}$

$(30,30)$ is the point where

$15x+5y= 600$ intersects

$y=30$

I.e., substituting

$y=30 \implies 15x+5*30 =600 \implies x=\dfrac{600-150}{15}\implies x=30$

$(\dfrac{340}{9},\dfrac{20}{3})$ is the point where

$24x+14y= 1000$ intersects

$15x+5y= 600$

I.e., solving the two equations, we get

$x=\dfrac{340}{9}$ and

$y=\dfrac{20}{3}$

Now substituting the corner points the profit equation,

substituting

$(\dfrac{290}{12},30) \implies P=24*\dfrac{290}{12}+14*30=1000$

substituting

$(30,30) \implies P=24*30+14*30=1140$

substituting

$(\dfrac{340}{9},\dfrac{20}{3}) \implies P=24*\dfrac{340}{9}+14*\dfrac{20}{3}=1000$

$1140$ is the maximum profit

Roots of the equation

$f(x)=x^6-12x^5+bx^4+cx^3+dx^2+ex+64=0$ are positive. Remainder when

$f(x)$ is divided by

$x-1$ is

0%
$2$
0%
$1$
0%
$3$
0%
$10$

Explanation

Given

${ x }^{ 6 }-12{ x }^{ 5 }+b{ x }^{ 5 }+c{ x }^{ 3 }+d{ x }^{ 2 }+ex+64=0$

Since

${ \alpha }_{ 1, }{ \alpha }_{ 2 },{ \alpha }_{ 3 },{ \alpha }_{ 4 },{ \alpha }_{ 5 },{ \alpha }_{ 6 }$ are the roots of the equation

Let us consider the sum of the roots

${ \alpha }_{ 1 },{ \alpha }_{ 1 },+.......+{ \alpha }_{ 6 }=12\quad \quad \quad \quad \rightarrow 1$

The product of roots would be

${ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ \alpha }_{ 6 }=64\quad \quad \quad \quad \rightarrow 2$

If we apply arithmetic and geometric mean to equation roots

$A.M >G.M$

${ \alpha }_{ 1 }+{ \alpha }_{ 2 }+.......+{ \alpha }_{ 6 }\ge \quad ({ \alpha }_{ 1 },{ \alpha }_{ 2 }.......+{ \alpha }_{ 6 }{ ) }^{ { 1/6 }_{ }^{ } }$

$\dfrac { 12 }{ 6 } \ge ({ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ \alpha ) }^{ 1/6 }$

${ 2 }^{ 6 }\ge \dfrac { 12 }{ 6 } \ge ({ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ { \alpha }_{ 6 }) }$

${ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ { \alpha }_{ 6 }) }\le 64$

$({ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ { \alpha }_{ 6 }) }=64$

The roots product to equality if all the roots are equal

$\Rightarrow ({ \alpha }_{ 1 }={ \alpha }_{ 2 }=.......={ { \alpha }_{ 6 }) }$

Since the root equation is 2

$(x-2)=0$ defines the root of the equation.

If we divide

$(x-2)$ by

$(x-1)$ the remainder is

$1$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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