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CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 9 - MCQExams.com
CBSE
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?
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$$750$$
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$$600$$
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$$900$$
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$$150$$
Explanation
The maximum value of b occurs when take equality.
$$\Rightarrow -15x+3000=5x$$
$$\Rightarrow -20x=-3000$$
$$\Rightarrow -20x=-3000$$
$$\Rightarrow x=150$$
Therefore , $$y=5\times150=750$$
Hence maximum possible value of $$b=750$$
Conclude from the following:
$$x - y = 7$$
$$x - z < 7$$
A: $$y$$
B: $$z$$
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The quantity A is greater than B.
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The quantity B is greater than A.
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The two quantities are equal.
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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?
$$y > x$$
$$3y≤-4x+6$$
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$$A$$
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$$B$$
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$$C$$
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$$D$$
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?
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$$(0,3)$$
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$$(1,2)$$
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$$(2,4)$$
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$$(3,3)$$
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.
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$$\displaystyle \frac { 7 }{ 3 } $$
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$$\displaystyle \frac { 5 }{ 3 } $$
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$$\displaystyle \frac { 9 }{ 7 } $$
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$$\displaystyle \frac { 14 }{ 15 } $$
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:
$$9 \geq 7x -5 \geq 5x -11, x\in R$$.
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$$[-4,2]$$
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$$[-3,2]$$
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$$[-3,3]$$
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$$[-3,4]$$
Which coordinate is in the solution set of the system of linear inequalities:
$$3x + 5y \leq 9$$
$$x - 6y \leq 3$$
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$$( 0, 2.5 )$$
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$$( 0 , 2 )$$
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$$( 1 , 0 )$$
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$$( 3.5 , 0 )$$
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?
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0%
0%
0%
0%
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 $$c + d > 0$$
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$$a + b + c > 0$$
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$$ac + bd > 0$$
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$$a^{2} + b^{2} > 0$$
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$$d(a + b) > 0$$
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$$a + b > c + d$$
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
(C) : $$a^{2}+b^{2}> 0$$ 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 $$s + t$$.
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The quantity A is greater than B.
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The quantity B is greater than A.
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The two quantities are equal
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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
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$$18$$
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$$19$$
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$$20$$
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$$21$$
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 $$\sqrt { 3x-8 } <-2$$
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$$\phi$$
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$$(0,1)$$
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$$(0,3)$$
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None of these
Solve: $$\displaystyle\frac{1}{x} < 1$$.
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$$x\epsilon(-\infty, 0)\cup(1, \infty)$$
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$$x\epsilon(-\infty, -2)\cup(1, \infty)$$
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$$x\epsilon(-\infty, 0)\cup(7, \infty)$$
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$$x\epsilon(-\infty, 0)\cup(3, \infty)$$
Which region is described by the shade in the graph given?
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$$2x+3y=3$$
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$$2x+3y>3$$
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$$2x+3y<3$$
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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.
N
ow, 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
$$p > w$$
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True
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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
$$p+2 > w+2$$
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True
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False
Which of the following is the correct graph for $$x\ge 3$$ or $$x\le -2$$?
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Line A
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Line B
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Line C
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Line D
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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.
Which $$x$$ values make the ordered pair $$(x,8)$$ a solution of the following inequality?
$$9x+4y<13$$
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$$x <\dfrac{-19}{9}$$
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$$x < -5$$
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$$x > -5$$
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$$x < 5$$
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
$$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c} + \frac{a^2+c^2}{a+c}\geq a+b+c$$
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True
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False
If $$a,b,c$$ 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
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$$3$$
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$$9$$
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$$27$$
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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?
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$$I$$
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$$II$$
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$$III$$
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$$IV$$
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:
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$$P>10abxy$$
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$$P>16abxy$$
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$$P>4abxy$$
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$$P>2abxy$$
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
$$\implies (ab+xy)(ax+by)\geq (2abxy)$$
If $$P_n$$ denotes the product of first $$n$$ natural numbers, then for all $$n \in N$$ .
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$$P_n\leq n^n$$
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$$P_{n+1}\leq n^n$$
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$$P_n \leq \left(\dfrac {n+1}{2}\right)^n$$
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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,
$$P_n=(1*2*...*n)$$
)
If $$a+b+c=1$$ and $$a,b,c \in R^+$$ such that $$(1-a)(1-b)(1-c)\geq \lambda bac$$, then $$\lambda=$$
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$$2$$
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$$4$$
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$$6$$
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$$8$$
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 $$8$$
If $$a+b+c =15 $$ and $$ ab+bc+ca= 74 $$, then the value of $$ a^3+b^3+c^3-3abc$$ is,
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$$42$$
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$$46$$
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$$45$$
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$$48$$
If $$a^2+b^2+c^2=1$$, then $$ab+bc+ca$$ lies in the interval :
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$$[0,1]$$
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$$[-0.5,1]$$
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$$[0,0.5]$$
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$$[1,2]$$
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
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$$1$$
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$$2$$
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$$4$$
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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
If $$a,b,c$$ are the sides of a triangle, then
$$\dfrac a{b+c-a} +\dfrac b{c+a-b} +\dfrac c{a+b-c}$$
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$$\leq 3$$
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$$\geq 3$$
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$$\leq 2$$
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$$\geq 2$$
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.
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Rs. $$1020$$
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Rs. $$1040$$
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Rs. $$1140$$
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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
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0%
$$2$$
0%
$$1$$
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$$10$$
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 $$1$$.
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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