If for any real x, we have $$-1\leq\displaystyle \frac{x^2+nx-2}{x^2-3x+4}\leq 2$$, then $$n$$ belongs to.

$$\left[-1, \sqrt{40}-6\right]$$

$$\left[-\sqrt{40}+6, -1\right]$$

$$\left[-\sqrt{40}+6, \sqrt{40}-6\right]$$

The acceleration experienced by a moving boat after its engine is cut-off, of given by: a = -kv$$^3$$ where k is a constant. If $$v_0$$ is the magnitude of velocity at cut-off, then the magnitude of the velocity at time t after the cut-off is :-

$$\cfrac { v_0 }{\sqrt {1 + 2ktv_0^2 }}$$

$$\cfrac { v_0 }{ 1 + 2ktv_0^2 }$$

$$\cfrac { v_0 }{\sqrt { 1 - 2ktv_0^2 }}$$

The set of values satisfying the inequation $$\left(x+3\right){\left(3x-2\right)}^{5}{\left(7-x\right)}^{5}{\left(5x+8\right)}^{2}\ge0$$

$$\left(-\infty,-3\right]\cup\left[\frac{2}{3},7\right]\cup\left\{-\frac{8}{5} \right\} $$

$$\left(-\infty,-3\right)\cup\left(\frac{2}{3},7\right)\cup\left\{-\frac{8}{5} \right\} $$

$$\left[-\infty,-3\right]\cup\left[\frac{2}{3},7\right]\cup\left\{-\frac{8}{5} \right\} $$

$$\left[-\infty,-3\right)\cup\left[\frac{2}{3},7\right)\cup\left\{-\frac{8}{5} \right\} $$

A man can swim in still water with a speed of $$2\ ms^{-1}$$. If he wants to cross a river of water current speed $$\sqrt {3}\ ms^{-1}$$ along shortest possible path, then in which direction should he swim?

At an angle $$120^{\circ}$$ to the water current

At an angle $$90^{\circ}$$ to the water current

MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 15

If for any real x, we have

- 1 \leq \frac{x^{2} + n x - 2}{x^{2} - 3 x + 4} \leq 2

$-1\leq\displaystyle \frac{x^2+nx-2}{x^2-3x+4}\leq 2$ , then

n

$n$ belongs to.

0%
$\left[-\sqrt{40}+6, -1\right]$
0%
$\left[-\sqrt{40}+6, \sqrt{40}-6\right]$
0%
$\left[-1, \sqrt{40}-6\right]$
0%
None of these

The acceleration experienced by a moving boat after its engine is cut-off, of given by: a = -kv

^{3}

$^3$ where k is a constant. If

v_{0}

$v_0$ is the magnitude of velocity at cut-off, then the magnitude of the velocity at time t after the cut-off is :-

0%
$\cfrac { v_0 }{ 2ktv_0^2 }$
0%
$\cfrac { v_0 }{ 1 + 2ktv_0^2 }$
0%
$\cfrac { v_0 }{\sqrt { 1 - 2ktv_0^2 }}$
0%
$\cfrac { v_0 }{\sqrt {1 + 2ktv_0^2 }}$

Explanation

The regarding acceleration

a = - k v^{3}

$a = -kv^3$

which can be written as

\frac{d v}{d t} = - K v^{3}

$\dfrac{dv}{dt} = -Kv^3$

\frac{d v}{v^{3}} = - K a t

$\dfrac{dv}{v^3} = - Kat$

integrating both side

\frac{- 1}{2 v_{0}^{2}} = - K t + C

$\dfrac{-1}{2v_0^2} = -Kt +C$

C = \frac{- 1}{2 v_{0}^{2}}

$C = \dfrac{-1}{2v_0^2}$

so the equation becomes

\frac{- 1}{2 v^{2}} = - K t - \frac{1}{2 v_{0}^{2}}

$\dfrac{-1}{2v^2} = - Kt -\dfrac{1}{2v_0^2}$

\frac{1}{2 v^{2}} = K t + \frac{1}{2 v_{0}^{2}}

$\dfrac{1}{2v^2} = Kt+ \dfrac{1}{2v_0^2}$

2 v^{2} = \frac{1}{K t + 1 / 2 v_{0}^{2}} = \frac{2 v_{0}^{2}}{2 k t v_{0}^{2} + 1}

$2v^2 = \dfrac{1}{Kt +1/2v_0^2} = \dfrac{2v_0^2}{2ktv_0^2 + 1}$

v = \frac{v_{0}}{\sqrt{1 + 2 k t v_{0}^{2}}}

$v = \dfrac{v_0}{\sqrt {1+2ktv_0^2}}$

Hence(D) is the correct answer

Rs.

3200

$3200$ is divided among

A, B

$A, B$ and

C

$C$ in the ratio of

3 : 5 : 8

$3:5:8$ respectively. What is the difference (in Rs.) between the share of

B

$B$ and

C

$C$ ?

0%
$400$
0%
$600$
0%
$800$
0%
$900$

Explanation

Difference of share of

B

$B$ and

C

$C$

= \frac{3}{16} \times 3200 = R s .600

$=\cfrac { 3 }{ 16 } \times 3200=Rs.600$

The set of values satisfying the inequation

(x + 3) {(3 x - 2)}^{5} {(7 - x)}^{5} {(5 x + 8)}^{2} \geq 0

$\left(x+3\right){\left(3x-2\right)}^{5}{\left(7-x\right)}^{5}{\left(5x+8\right)}^{2}\ge0$

0%
$\left(-\infty,-3\right)\cup\left(\frac{2}{3},7\right)\cup\left\{-\frac{8}{5} \right\}$
0%
$\left(-\infty,-3\right]\cup\left[\frac{2}{3},7\right]\cup\left\{-\frac{8}{5} \right\}$
0%
$\left[-\infty,-3\right]\cup\left[\frac{2}{3},7\right]\cup\left\{-\frac{8}{5} \right\}$
0%
$\left[-\infty,-3\right)\cup\left[\frac{2}{3},7\right)\cup\left\{-\frac{8}{5} \right\}$

A man swimming downstream overcomes a float at a point

M

$M$ . After travelling distance

D

$D$ , he turned back and passed the float at a distance of

D / 2

$D/2$ from the point

M

$M$ . Then the ratio of speed of swimmer with respect to still water to the speed of the river will be:

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

The ratio of the distance carried away by the water current, downstream, in crossing a river, by a person, making same angle with downstream and upstream is

2 : 1

$2 : 1$ . The ratio of the speed of person to the water current cannot be less than

0%
$1/3$
0%
$4/5$
0%
$2/5$
0%
$4/3$

Explanation

Motion of the person making an angle (

a l p h a

$alpha$ ) with the downstream.

The time taken to cross the river

= \frac{d}{v sin α}

$=\dfrac{d}{v\sin\alpha}$

The distance caries away downstream in the same time= speed

\times

$\times$ time

x_{1} = (u + v cos α) \frac{d}{sin α}

$x_1=(u+v\cos \alpha)\dfrac{d}{\sin \alpha}$ ................(i)

Motion of the person making

α

$\alpha$ angle with upstream.

The time taken to cross the river is equal to

\frac{d}{v sin α}

$\dfrac{d}{v\sin\alpha}$

Distance carried away downstream in the same time

x_{2} = [u + v cos (180^{o} - α)] \frac{d}{v sin α}

$x_2=[u+v\cos(180^o-\alpha)]\dfrac{d}{v \sin \alpha}$ ...............(ii)

\Rightarrow x_{2} = (u - v cos α) \frac{d}{v sin α}

$\Rightarrow x_2=(u-v\cos \alpha) \dfrac{d}{v\sin \alpha}$

Given

\frac{(u + v cos α) \frac{d}{v sin α}}{(u - v cos α) \frac{d}{v sin α}} = \frac{2}{1}

$\dfrac{(u+v\cos \alpha) \dfrac{d}{v\sin \alpha}}{(u-v\cos\alpha)\dfrac{d}{v\sin\alpha}}=\dfrac{2}{1}$

\frac{(u + v cos α)}{(u - v cos α)} = \frac{2}{1} \Rightarrow 3 v cos α = u

$\dfrac{(u+v\cos \alpha)}{(u-v\cos \alpha)}=\dfrac{2}{1}\Rightarrow 3v\cos \alpha =u$

\Rightarrow \frac{v}{u} = \frac{sec α}{3}

$\Rightarrow \dfrac{v}{u}=\dfrac{\sec \alpha}{3}$ .............(3)

sec α \geq 1 \Rightarrow \frac{s e c α}{3} \geq \frac{1}{3}

$\sec \alpha \ge 1 \Rightarrow \dfrac{sec \alpha}{3}\ge\dfrac{1}{3}$

From eq (iii)

\frac{V}{u} \geq \frac{1}{3}

$\dfrac{V}{u}\ge\dfrac{1}{3}$

So,vlu cannot be less than

1 / 3

$1/3$ .

A man can swim in still water with a speed of

2 m s^{- 1}

$2\ ms^{-1}$ . If he wants to cross a river of water current speed

\sqrt{3} m s^{- 1}

$\sqrt {3}\ ms^{-1}$ along shortest possible path, then in which direction should he swim?

0%
At an angle $120^{\circ}$ to the water current
0%
At an angle $150^{\circ}$ to the water current
0%
At an angle $90^{\circ}$ to the water current
0%
None of these

Explanation

The speed of the man is $2\;{\rm{m/s}}$ and the speed of the water current is $\sqrt 3 {\rm{m/s}}$ .

The velocity component acting opposite to that of the velocity of water current is given as,

${v_m}\sin \theta = {v_r}$

$2 \times \sin \theta = \sqrt 3$

$\theta = 60^\circ$

The angle between the speed of man and the water current is given as,

${\theta _1} = 60^\circ + 90^\circ$

$= 150^\circ$

Thus, the direction of swimming is at an angle $150^\circ$ to the water current.

A boat covers a certain distance between two spots in a river taking

t_{1}

$t_1$ hrs going downstream and

t_{2}

$t_2$ hrs going upstream. What time will be taken by boat to cover same distance in still water ?

0%
$\dfrac{t_1+t_2}{2}$
0%
$2(t_2=t_1)$
0%
$\dfrac{2 t_1 t_2}{t_1+t_2}$
0%
$\sqrt{t_1 t_2}$

A swimmer crosses a flowing stream of width

$'\omega'$ to and fro in time

$t_1$ . The time taken to cover the same distance up & down the stream is

$t_2$ . if

$t_3$ is the time swimmer would take to swim a distance

$2\omega$ in still water, then:

0%
$t^2_1 = t_2t_3$
0%
$t^2_2 = t_1t_3$
0%
$t^2_3 = t_1t_3$
0%
$t_3 = t_1 + t_2$

Explanation

Let 'u' be the velocity of swimmer in still water and 'v' be velocity of river.

in case 1

Since v and u are perpendicular to each other.

$t_1 = 2 (\dfrac{w}{\sqrt{u^2 - v^2}} )$

In case 2nd

The swimmer goes upstream and downstream.

$t_2 = \dfrac{w}{u+t} + \dfrac{w}{u-t} = \dfrac{2uw}{u^2 - v^2}$

In case 3rd

Still water

$t_3 = \dfrac{2w}{4}$

We see

$t_1^2 = t_2 t_3$

Hence option (A) is correct

The boys and girls in a college are in the ratio

$3:2$ . If

$20\%$ of the boys and

$25\%$ of the girls are adults, the percentage of students who are not adults is?

0%
$58\%$
0%
$67.5\%$
0%
$78\%$
0%
$82.5\%$

Explanation

Lets say there are

$100$ students in the class

The boys and girls are in the ratio

$3:2=3x:2x$ with unknown multiplier of

$'x'$

Total number of student is

$3x+2x=100$

$5x=100$

$x=20$

Number of boys in the class

$= 3x = 60$
Number of girls in the class

$= 2x = 40$

Number of boys who are adults =

$20\%$ of number of boys =

$\dfrac{20}{100}\times60=12$

Number of girls who are adults =

$25\%$ of number of girls =

$\dfrac{25}{100}\times40=10$

Total number of boys and girls who are adults =

$12 + 10 = 22$

Total number of boys and girls who are not adults =

$100 - 22 = 78$

Percentage of boys and girls who are not adults =

$\dfrac{78}{100}\times100=78\%$ .

so option

$C$ is correct.

The ratio of the incomes of

$A, B$ and

$C$ is

$3 : 7 : 4$ and the ratio of their expenditure is

$4 : 3 : 5$ . If in the income of

$Rs. 2400$ , A saves

$Rs. 300$ then the savings of

$B$ and

$C$ is respectively are

0%
$Rs. 4025$ and $Rs. 575$
0%
$Rs. 1575$ and $2625$
0%
$Rs. 2750$ and $Rs. 1025$
0%
$Rs. 3725$ and $Rs. 1525$

Explanation

The ratio of income of

$A,B, C$ is

$3:7:4$ respectively.

Let the income be

$3x, 7x, 4x$ respectively.

The ratio of expenses of

$A,B, C$ is

$4:3:5$ respectively.

Let the expenditure be

$4y, 3y, 5y$ respectively.

It is given that the income of

$A$ is

$2400$ .

$\implies 3x=2400\\\implies x=800$

The value of

$x$ is

$800$ , therefore the income of

$A, B, C$ is

$2400,5600, 3200$ respectively.

Expenditure =income-saving

Expense=

$2400-300=2100$

The expenses of

$A$ is

$2100$ .

$\implies 4y=2100\\\implies y=525$

The value of

$y$ is

$525$ , therefore the expenditure of

$A, B, C$ is

$2100,1575, 2625$ respectively.

Saving of

$b$ is

$5600-1575=4025$

Saving of

$C$ is

$3200-2625=575$

Therefore the saving of

$B$ is

$Rs.4025$ and the saving of

$C$ is

$Rs.575$ .

A man can swim with a speed of

$3m/s$ in still water and river flows at the rate of

$5m/s$ . River width is

$500m$

0%
For shortest path journey $\theta =\tan^{-1}\dfrac{4}{3}$
0%
For shortest path journey $\theta =\tan^{-1}\dfrac{3}{4}$
0%
time taken for shortest path journey is $3\dfrac{17}{36}\ min$
0%
time taken to cross river in shortest time is $2\dfrac{28}{36}\ min$

A boat moves along the flow of of river between two fixed points P and Q. It takes 40 minutes when going downstream and takes 120 minutes when going upstream between these two points. What time it will take in still water to cover the distance equal to PQ?

0%
50 minute
0%
100 minute
0%
60 minute
0%
90 minute

A motor boat travels a distance of 7.5km in a river in 3hrs down stream & 5hrs upstream the velocity of boat in kmph is

0%
$2.0$
0%
$0.5$
0%
$2.5$
0%
$1.5$

A swimmer can swim with velocity a in still water. He jumps in a river flowing with speed b. The swimmer goes a distance d downstream and then returns to original position. The time consumed in the process will be

0%
$\dfrac{3ad}{a^{2}-b^{2}}$
0%
$\dfrac{2ad}{a^{2}-b^{2}}$
0%
$\dfrac{ad}{a^{2}-b^{2}}$
0%
$\dfrac{4ad}{3a^{2}-b^{2}}$

The sum of digits in

${ \left( { 10 }^{ { 2n }^{ 2 }+5n+1 }+1 \right) }^{ 2 }$ , (where n is a positive integer), is

0%
$4$
0%
$6n$
0%
$6+n$
0%
$24n$

A boat of mass

$80\,kg$ is floating on still water. A dog of mass

$20 \,kg$ on the boat at a distance of

$10 \,m$ from the shore. The dog moves on the boat by a distance of

$2 \,m$ towards the shore. The distance of the dog from the shore is:

0%
$11.6 \,m$
0%
$8.4 \,m$
0%
$9.6 \,m$
0%
$10.4 \,m$

A river is following at

$3 m/s$ towards East, a boat is steered toward North at

$4 m/s$ and wind is blowing at

$5 m/s 53^{o}$ North of East. Flag mounted on boat will flutter along

0%
North-East
0%
South-West
0%
West
0%
Stand still

Max. value of

$z = 10 x + 7 y$ subject to

$x \leq 4 , y \leq 8 , x + y \leq 8 , x \geq 0 , y \geq 0$ is

0%
$82$
0%
$65$
0%
$68$
0%
$56$

Explanation

Max value of

$z=10x+7y$ subject to

$x \leq 4,$

$y \leq 8,x+y \leq 8,x\geq 0,y\geq 0$ is

Given constraints are

$x \leq 4$

$y \leq 8$

$x+y \leq 8, x \geq 0,y \geq 0$

Using graphical method :

Let

$x=4$

$y=8$

$x+y=8$

Common region

Let

$A(0,0)$

$B(4,0)$

$C(4,4)$

$D(0,8)$

1186060_1176001_ans_c2bb6c78357c49f785dce6f43485b961.jpeg

$x$ satisfies

$|x-1|+|x-2|+|x-3|\le 6$ , then

0%
$0\le x\le4$
0%
$x\le-2$ or $x\ge4$
0%
$x\le 0$ or $x\ge 4$
0%
$None\ of\ these$

$p, q, r$ and

$s$ are in proportion then

$p$ and

$s$ are

0%
Middle terms
0%
Extreme terms
0%
In ratio
0%
None of these

Explanation

$\textbf{Step 1 : Find relation between p and s}$

$\text{Given , p , q , r and s are in proportion }$

$\text{As p, q , r and s are in proportion }$

$\Rightarrow p : q : : r : s$

$\text{We know that the first term and the fourth term of a proportion are called extreme and the}$

$\text{ second term and the third term are called middle terms or means. }$

$\textbf{Hence, p and s are Extreme terms}$

Find the range of values of

$x$ which satisfies equation

0%
$\left (- \infty, 1 \right] \cup \left[\dfrac {3}{2}, \infty \right)$
0%
$\left (- \infty, 1 \right) \cup \left(\dfrac {3}{2}, \infty \right)$
0%
$\left[1,\dfrac {3}{2}\right]$
0%
None of these

A boat covers the distance between two points in a river in

$6$ hours downstream and

$8$ hrs upstream. A floating body in the lakes crosses these two points in

0%
48 hrs
0%
16 hrs
0%
18 hrs
0%
2 hrs

A vertex of common graph of inequalities

$zx+y\ge 2$ and

$x-y\le 3$ is

0%
$(0,0)$
0%
$(5/3,\ -4/3)$
0%
$(5/3,\ 4/3)$
0%
$(-4/3,\ 5/3)$

${ 2 }^{ \sin x }+{ 2 }^{ \cos x }\ge 2^{ 1-y }$ , then the value of

$y$ is

0%
1
0%
$\sqrt { 2 }$
0%
$\frac{1}{\sqrt2}$
0%
$0$

A boat travels a distance in 4 hrs upstream and same distance downstream in 2 hrs.Then ratio of velocity of the boat to that of water is

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

A steamer takes 10 minute to travel a distance d downstream and 15 minute in upstream for same distance. The steamer can travel the same distance in still water is

0%
12 minute
0%
10 minute
0%
6 minute
0%
15 minute

A boat is travelling with a speed of 27 kmph due east. an observer is situated at 30 m south of the line of travel. the angular velocity of boat relative to the observer in the position shown will be :-

0%
0.125 rad /sec
0%
zero
0%
0.250 rad/sec
0%
0.67 rad/sec

Solve

$\dfrac {x}{x-9} \le \dfrac {1}{1-x}$
Write your solution set in interval notation.

0%
$(-3 < x < 1)\cap (3 < x < 9)$
0%
$(-3 \le x < 1)\cap (3 < x < 9)$
0%
$(-3 \le x < 1)\cap (3 < x < 9)$
0%
$(-3 \le x \le 1)\cap (3 \le x \le 9)$

Explanation

$\text { solution: } \frac{x}{x-9} \leqslant \frac{1}{1-x} \\$

$\Rightarrow \frac{x}{x-9}-\frac{1}{1-x} \leqslant 0 \\$

$\Rightarrow \frac{x-x^{2}-x+9}{(x-9)(1-x)} \leqslant 0$

$\frac{\left(9-x^{2}\right)}{(x-9)(1-x)} \leqslant 0$

$\Rightarrow \frac{\left(x^{2}-9\right)}{(x-9)(x-1)} \leqslant 0$

$\Rightarrow \quad \frac{(x-3)(x+3)}{(x-9)(x-1)} \leqslant 0$

$x \in\left[\begin{array}{ll}-3 & 1\end{array}\right) \cup\left[\begin{array}{ll}3 9\end{array}\right)$

$(-3 \leqslant x<1) \cup(3 \leqslant x<9)$

Graphical solution of

$2 x+y-2>0$

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 15 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 15 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.