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

$a:b:c=3:4:7$ , then the ratio

$(a+b+c):c$ is equal to

0%
$14:3$
0%
$7:2$
0%
$1:2$
0%
$2:1$

Explanation

Let

$a=3K,b=4K,c=7K$

then,

$\dfrac{a+b+c}{c}=\dfrac{3K+4K+7K}{7K}=\dfrac{14K}{7K}=2$

The Vertices of a triangle are

$A(1,0,1) \quad B(2,-1,4)$ and

$C(3,-4,-1)$ .Then The triangle is

0%
Acute Angled
0%
Obtuse Angled
0%
Equialteral
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Right angled

Explanation

The Vertices are

$A(1,0,1) \quad B(2,-1,4)$ and

$C(3,-4,-1)$

$AB=\sqrt{(1-2)^2+(0+1)^2+(1-4)^2}=\sqrt{1+1+9}=\sqrt {11}$

$BC=\sqrt{(3-2)^2+(-4+1)^2+(-1-4)^2}=\sqrt{1+9+25}=\sqrt {35}$

$AC=\sqrt{(1-3)^2+(0+4)^2+(1+1)^2}=\sqrt{4+16+4}=\sqrt {24}$

$\implies AB^2+BC^2=AC^2$

So it is a right angled triangle.

A person can swim in still water at

$5 \,m/s$ . He moves in river of velocity

$3 \,m/s$ , First down the stream next same distance up the steam the ratio of times takes are

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$1:1$
0%
$1:2$
0%
$1:4$
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$4:1$

Explanation

Distance be

$x$

$t_1 = \dfrac{x}{V_m + V_r}$

$= \dfrac{x}{5+3} = \dfrac{x}{8}$

$t_2 = \dfrac{x}{V_p - V_r}$

$= \dfrac{x}{5-3} = \dfrac{x}{2}$

$\dfrac{t_1}{t_2} = \dfrac{x/8}{x/2} = \dfrac{1}{4}$

$1:4$

Whether the given equation

$x>\sin x > x -\dfrac{x^3}{6}$ for all

$x>0$ . is ?

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True
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False

Solution set of the inequality,

$49^x + 7^{x + 1} - 98 < 0$ , is:

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$\left(-\infty, \dfrac{1}{2}\right)$
0%
$\left(-1, \infty\right)$
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$\left(-\infty, 1\right)$
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$\left(-1, \dfrac{1}{2}\right)$

Explanation

$49^{x}+7^{x+1}-98<0$

$7^{2x}+7.7^{x}-98<0$

$7^{2x}+14.7^{x}-7.7^{x}-98<0$

$\left(7^{x}+14\right)\left(7^{x}-7\right)<0$

$\Rightarrow 7^{x}<7$ and

$7^{x}<-14$

But

$7^{x}$ is always positive

$7^{x}<7$

$x\log 7<\log 7$

i.e,

$x<1$

$x\in \left(-\infty, 1\right)$

The minimum value of

$P = 6 x + 16 y$ subject to constraints

$x \leq 40 , y \geq 20$ and

$x , y \geq 0$ is

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$240$
0%
$320$
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$0$
0%
$280$

Explanation

The minimum value of

$P=6x+16y. x \le 40, y \ge 20$

min. value of

$P=6x+16y$ .

Let

$x=0 \because 0 \le x \le 40.$

and

$y=20. \because y \ge 20.$

$P=6.0+16 \times 20$

$P=320.$

The solution set of the in equation

$|x-1|\geq |x-3|$ , is

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$[-\alpha,2]$
0%
$[2,\alpha]$
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$[1,3]$
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none of these

Explanation

Given :

$\left | x-1 \right |\geq \left | x-3 \right |$

case I :

$x\geq 3 , \left | x-1 \right |\geq \left | x-3 \right |$

$\Rightarrow x-1 \geq x-3$

$2\geq 0$ (true)

case II :

$x< 3, x\geq 1 \left | x-1 \right |\geq \left | x-3 \right |$

$\Rightarrow x-1 \geq 3-x$

$2x\geq 4$

$x \geq 2$

$\Rightarrow x\in [2,3]$

case III :

$x< 1, \left | x-1 \right | \geq \left | x-3 \right |$

$1-x\geq 3-x$

$0\geq 2$ (false)

combining I,II,III

$x\in [2, \infty ]$

$\therefore$ option B is correct.

If the sum of the upstream and downstream speed of a fish is 82 mph , and it travels 105 m upstream in 3hr. Find the time taken by boat to cover 126 m downstream

0%
$2.4$ hrs
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$2.8$ hrs
0%
$2.5$ hrs
0%
$2.68$ hrs

If the sum of the upstream and downstream speed of a fish is

$82$ mph, and it travels

$105$ m. upstrem in

$3$ hr. Find the time taken by a boat to cover

$126$ m downstream.

0%
$2.7$ hrs
0%
$2.8$ hrs
0%
$2.5$ hrs
0%
$2.68$ hrs

Explanation

Let

Speed of boat in still water

$=b$

Speed of still water

$=w$

Then, we know that,

Speed of upstream,

$U=boat-water$

Speed of downstream,

$D=boat+water$

Given,

$U+D=82$

So,

$b-w+b+w=82$

$\implies 2b=82$

and

$b=41\,kmph$

From the given data,

$41-w=\dfrac{105}{3}=35$

$w=6\,kmph$

Now,

$b+w=\dfrac{126}{t}$

$\implies 41+6=\dfrac{126}{t}$

$\implies t=\dfrac{126}{47}=2.68\,hrs$

Area of the region

$\left\{(x, y)/x^{2}+y^{2}\le 1\le x+y\right\}$ is

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$\dfrac{\pi}{4}+\dfrac{1}{2}$
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$\dfrac{\pi}{4}-\dfrac{1}{2}$
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$\dfrac{\pi}{4}+\dfrac{3}{4}$
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$\pi +1$

Find the set of values of

$x$ which satisfies the inequality

${| x+2|}\leq 2x+1.$

0%
$(-1,\infty)$
0%
$[1,\infty)$
0%
$(-2,\infty)$
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None of these

Mark the correct alternative of the following.
If

$a=2b$ , then

$a : b=$ ?

0%
$2 : 1$
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$1 : 2$
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$3 : 4$
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$4 : 3$

Explanation

Given

$a=2b$ .

Then,

$\dfrac{a}{b}=\dfrac{2}{1}$

or,

$a:b=2:1$ .

Mark the correct alternative of the following.
If

$57 : x=51 : 85$ , then the value of x is?

0%
$95$
0%
$76$
0%
$114$
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None of these

Explanation

Now,

$57:x=51:85$

or,

$\dfrac{57}{x}=\dfrac{51}{85}$

or,

$x=57\times \dfrac{85}{51}$

or,

$x=19\times 5$

or,

$x=95$

A boat takes

$4\ hr$ upstream and

$2\ hr$ down the stream for covering the same distance. The ratio of the velocity of boat to velocity of water in the river is

0%
$1 : 3$
0%
$3 : 1$
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$1 : \sqrt{3}$
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$\sqrt{3} : 1$

Explanation

Correct Answer: Option B

Hint: Net velocity of boat for downstream and upstream will be equal to sum and difference of velocity of boat and river respectively.

Explanation of Correct Answer:

Step 1: Finding the velocity of boat and water.

Let the speed of boat in still water be

$x$

$km/h$ and speed of water be

$y$

$km/h$

Then,

speed upstream =

$x-y$

$km/h$

speed downtream =

$x+y$

$km/h$

So,

$x+y=4\rightarrow(i)$

$x-y=2\rightarrow(ii)$

From the above

$2$ equations,

$(i)+(ii)\rightarrow x=3$

$(i)-(ii)\rightarrow y=1$

Step2: Finding the ratio

Ratio of velocity of boat to velocity of water in the river

$=\dfrac{x}{y}$

$=\dfrac{3}{1}$

$\therefore$ Ratio of velocity of boat to velocity of water in the river is $3: 1$

In the given figure D is a point on BC such that BD : DC =

$1 : 3$ . If O is the midpoint of AD, then the ratio of areas of

$\Delta$ AOB and

$\Delta$ ABC will be

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$1 :8$
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$1 :3$
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$1 :4$
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$1 :6$

Explanation

Let

$A$ be origin and

$B(\vec{b})$ and

$C(\vec{C})$ By section formula,

$D=\dfrac{3\vec{b}+\vec{c}}{3+1}=\dfrac{3\vec{b}+\vec{c}}{4}$

$(AOB)=\dfrac{1}{2}(\vec{AB}\times \vec{AO})=\dfrac{1}{2}(\dfrac{3\vec{b}+\vec{c}}{8})$

$=\dfrac{1}{16}(\vec{b}\times \vec{c})$

$\{\vec{b}\times \vec{b}\times \vec{O}\}$

$(\Delta ABC)=\dfrac{1}{2}(\vec{AB}\times \vec{AC})=\dfrac{1}{2}(\vec{b}\times \vec{c})$

$\dfrac{ar(AOB)}{ar(ABC)}=\dfrac{1}{16}\times 2=\dfrac{1}{8}\Rightarrow (A)$

1165149_1225118_ans_3ebf0ec439534c68be2608192cd9d168.png

If a point (h,k) satisfies an inequation ax + by > 4, then the half plane represented by the inequation is

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The half plane contacting the point (h,k) but excluding the point on ax + by =4
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The half plane contacting the point (h , k) and the points on ax +by =4
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Whole xy-plane
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None of these

Explanation

$ax+by\geq 4$ would have the half plane above the line

$ax+by=4$ and also the line

$ax+by=4$

$\Rightarrow (B)$ .

1165339_1242492_ans_06250dc2210a437ba99c60490f96a6d1.jpg

The number of real values of

$\lambda$ for which the system of linear equations

$2x + 4y - \lambda z =0$

$4x +\lambda y + 2z =0$

$\lambda x + 2y +2z =0$
has infinitely many solution is:

0%
0
0%
1
0%
2
0%
3

$\sin ^ { - 1 } ( x ) > \cos ^ { - 1 } ( x )$ holds for

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every $x$
0%
$x \in \left( 0 , \dfrac { 1 } { \sqrt { 2 } } \right)$
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$x \in \left( \dfrac { 1 } { \sqrt { 2 } } , 1 \right]$
0%
not possible

Explanation

$\\sin^{-1}x\>>cos^{-1}x\\sin^{-1}x\>>(\frac{\pi}{2})-sin^{-1}x\\2sin^{-1}x\>>(\frac{\pi}{2})\\sin^{-1}x\>>(\frac{\pi}{4})\\x\>>sin(\frac{\pi}{4})\\x\>>(\frac{1}{\sqrt{2}})\\\therefore\>x\in\left(\frac{1}{\sqrt2 },1\right]$

Three numbers are in the ratio 4 : 5 : 6 and their average is 25.The largest number is :

0%
$30$
0%
$32$
0%
$36$
0%
$42$

Explanation

Let the numbers be

$4x, 5x, 6x$ .

$average =25$

$\dfrac{4x+5x+6x}{3}=25$

$\dfrac{15x}{3}=25$

$x=5$

Hence, the largest number is

$6\times 5=30$ .

A value of 't' satisfying the equation

$\dfrac { 7t+5 }{ 6 } \ge \dfrac { t-1 }{ 2 }$

0%
-1
0%
-3
0%
-4
0%
-6

Explanation

$t$ is greater than

$-2$ Option A is correct.
1820510_1246074_ans_a2445527fbdc4b7e8a576eacb4a72744.jpg

State true or false.

For any real value of "a"

$4 a ^ { 4 } - 4 a ^ { 3 } + 5 a ^ { 2 } - 4 a + 1 \geq 0$

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True
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False

$A$ and

$B$ are in the ratio

$4 : 5$ and the difference of their squares is

$81$ , what us the value of

$B$ ?

0%
$15$
0%
$12$
0%
$36$
0%
$45$

Explanation

$\begin{array}{l} \dfrac { A }{ B } =\dfrac { 4 }{ 5 } ,\, \, \, \, \, \, \, \, \left[ \begin{array}{l} \therefore A=4x \\ \, \, \, \, \, B=5x \end{array} \right] \\ { B^{ 2 } }-{ A^{ 2 } }=81 \\ 25{ x^{ 2 } }-16{ x^{ 2 } }=81 \\ 9{ x^{ 2 } }=81 \\ { x^{ 2 } }=9 \\ x=3 \\ \therefore A=12 \\ \, \, \, \, B=15 \end{array}$

Let

$\overline {AB}$ be a diameter of a circle and

$C$ be a point on

$\overline {AB}$ with

$2AC = BC$ . Let

$D$ and

$E$ points on the circle, such that

$\overline {DC} \bot \overline {AB} \,\,{\rm{and}}\,\,\overline {DE}$ is second diameter, then the ratio of the area of

$\Delta DEC$ to the area of

$\Delta ABD$ is

0%
1 : 3
0%
2 : 3
0%
3 : 1
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none of these

A steamer is going downstream overcome a raft point

$A$ .

$2\ hr$ later it turned back and after some time passed the raft at a distance

$4\ km$ from point

$A$ . The speed of the river is

0%
$1\ km/h$
0%
$2\ km/h$
0%
$3\ km/h$
0%
$4\ km/h$

Explanation

speed of river = distance covered / time elapsed

$=\dfrac{4km} {2hr}$

$=2km/hr$

In what ratio is the line joining A(8,9) and B(-7,4) is divided by

0%
the point $(2,7)$
0%
the x-axis
0%
the y-axis
0%
at the ratio $2:3$

Explanation

R.E.F.Image.

Applying section formula.

$2 = \frac{k(-7)+8}{k+1}, 7=\frac{k(4)+9}{k+1}$

$\Rightarrow 2= \frac{-7k+8}{k+1}, 7 = \frac{4k+9}{k+1}$

$\therefore 7k+7 = 4k+9$

$\Rightarrow 3k=2$

$\Rightarrow k = \frac{2}{3}.$

$\therefore \frac{k}{1} = \frac{2}{3}$

$\therefore Ratio : 2 : 3$

1170810_1271496_ans_ef26e3386f884d019261245bf4418914.jpg

A man can swim in still water at

$4m/s$ . River is flowing at

$2m/s$ . The angle with downstream at which he should swim wo cross the river with minimum drift is:

0%
$120^o$
0%
$150^o$
0%
$30^o$
0%
$60^o$

Explanation

$\vec{V}_{mr} = \vec{V}_m - \vec{V}_r$

$\vec{V}_m = \vec{V}_{mr} + \vec{V}_r$

$\sin \theta = \dfrac{|\vec{V}_r}|{|\vec{V}_{mr}|} = \dfrac{2}{4} = \dfrac{1}{2}$

$\theta = 30^o$
Angle with downstream is

$120^o$

1243415_1611908_ans_3f35aad8ea9a4905be31d9765430bcc7.PNG

The sum

$1\cdot 3+3\cdot 5+5\cdot 7+\dots \dots +(2 n-1)(2 n+1)$ is equal to

0%
$\dfrac{n(4n^{2}+3n-2)}{3}$
0%
$\dfrac{n(4n^{2}+6n-1)}{3}$
0%
$\dfrac{(4n^{2}-6n-1)}{2}$
0%
None of these

Mark the correct alternative of the following.
The sides of a triangle are in the ratio

$2 : 3 : 5$ . If its perimeter is

$100$ cm, the length of its smallest side is?

0%
$2$ cm
0%
$20$ cm
0%
$3$ cm
0%
$5$ cm

Explanation

Given the sides of a triangle are in the ratio

$2:3:5$ .

Let the sides of the triangle be

$2x,3x,5x$ .

Then the perimeter of the triangle is

$2x+3x+5x=10x$ .

Now according to the problem,

$10x=100$

or,

$x=10$ .

So the length of the smallest side of the triangle is

$2\times 10=20$ cm.

Mark the correct alternative of the following.
The angles of a triangle are in the ratio

$1 : 2 : 3$ . The measure of the largest angle is?

0%
$30^o$
0%
$60^o$
0%
$90^o$
0%
$120^o$

Explanation

Given the angles of a triangle are in the ratio is

$1:2:3$ .

Let the angles be

$x,2x,3x$ .

Then we've,

$x+2x+3x=180^o$ [ Since the sum of the angles of a triangles is

$180^o$ ]

or,

$6x=180^o$

or,

$x=30^o$ .

So the largest angle is

$3\times 30^o=90^o$ .

$a:b=2:5$ , then

$\cfrac{3a+2b}{4a+b}=$

0%
$\cfrac{16}{13}$
0%
$\cfrac{13}{16}$
0%
$\cfrac{25}{22}$
0%
$\cfrac{20}{21}$

Explanation

Given,

$a : b = 2 : 5$

Could be written as

$\dfrac{a}{b} = \dfrac{2}{5}$

So,

$5 a = 2 b$

$a = \dfrac{2b}{5}$

Put the value of a in

$\dfrac{3a + 2b}{4a + b}$

$\dfrac{3 \left(\dfrac{2b}{5}\right) + 2b}{4 \left(\dfrac{2b}{5}\right) + b}$

$\dfrac{\dfrac{6b + 10b}{5}}{\dfrac{8b + 5b}{5}}$

$\dfrac{\dfrac{16b}{5}}{\dfrac{13b}{5}}$

$\dfrac{16b}{5}\times \dfrac{5}{13b}$

$\dfrac{16b}{13b}$

$\dfrac{16}{13}$

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

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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