MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 11

Single Correct Answer Type
A swimmer wishes to cross a

$500$ -m river flowing at

$5$ km h

$^{-1}$ . His speed with respect to water is 3 km h

$^{-1}$ . The shortest possible time to cross the river is:

0%
10 min
0%
20 min
0%
6 min
0%
7.5 min

Explanation

R.E.F image

$V_{s} =$ velocity of swimmer

w.r.t water = 3 km/hr

$V_{w} =$ velocity of water = 5 km/hr

For shortest possible time,

$V_{s}$ should be perpendicular

to river flow

$T_{shortest} = \dfrac{Distance}{V_{S}}$

$= \dfrac{0.5hr}{3\,km/hr}$

$= \dfrac{1}{6}hr$

$= 10\,min$

Option (A)

1163531_985325_ans_9a29c329836e4afab1ff1b4f0132538d.jpg

The symbol

$\left| a \right|$ means

$+a$ if

$a$ is greater than or equal to zero and

$-a$ if

$a$ is less than or equal to zero; the symbol M means "less than"; the symbol > means "greater than"
The set of values

$x$ satisfying the inquality

$\left| 3-x \right| <4$ consists of all

$x$ such that:

0%
${ x }^{ 2 }<49$
0%
${ x }^{ 2 }>1$
0%
$1>{ x }^{ 2 }<49$
0%
$-1< x< 7$
0%
$-7< x< 1$

Explanation

Let

$AB$ be a straight line segment with its midpoint at

$3$ , and its right end at

$B$ . Each if the two intervals from

$A$ to

$3$ and from

$3$ to

$B$ is less than

$4$ . Hence

$B$ is to the left of

$7$ and

$A$ is to the right of

$-1$ or
if

$x\le 3,\left| x-3 \right| <4\quad$ means

$3-x<4.\quad \therefore -1<x;$
if

$x> 3, \left| x-3 \right| <4$ means

$x-3<4\Rightarrow x<7$

$\Rightarrow -1<x<7$

Choose the correct answer from the alternatives given :
A sum of money is divided among A, B, C and D in the ratio of 3 : 7 : 9 : 13 respectively. If the share of B is Rs. 9180 more than the share of A. then what is the total amount of money of A and C together?

0%
Rs. 27540
0%
Rs. 26560
0%
Rs.26680
0%
Rs. 24740

Explanation

Let shares of

$A, B, C\ and\ D$ be

$3x, 7x, 9x\ and\ 13x$ rupees respectively. According to the question,

$(7x - 3x) = 9380$

$\Rightarrow$ 4x = 9180

$\Rightarrow$ x = 2295

Total amount of

$(A + C) = (3x + 9x) = 12x$

= 12

$\times$ 2295 =

$27540$

$A:B=3:4$ and

$B:C=2:3$ , then

$A:B:C$ will be

0%
$3:4:6$
0%
$3:4:12$
0%
$4::6$
0%
$6:4:3$

Explanation

$A:B:C=\cfrac { A }{ B } :1:\cfrac { C }{ B } =\cfrac { 3 }{ 4 } :1:\cfrac { 3 }{ 2 }$

$\Rightarrow A:B:C=3:4:6$

Sameer can row certain distance downstream in 24 h and can come back covering the same distance in 36 h. If the stream flows at the rate of 12 km/h. find the speed of Sameer in still water.

0%
30 km/h
0%
15 km/h
0%
40 km/h
0%
60 km/h

Explanation

Time taken in downstream

$=24h$

Time taken for upstream

$=36h$

Speed of stream

$(v)=12km/h$

Let the distance be

$x\ km$ and speed of sameer in still water be

$u\ km/h$

A/q For downstream

$\cfrac{x}{u+12}=24\\ \implies x=24(u-12)$ ............(1)

For upstream

$\cfrac{x}{u-12}=12\\ \implies x=36(u-12)$ ...........(2)

On comparing(l)&(2)

$24(u + l2) = 36(u - 12)$

$24u + 288 = 36u - 432$

$720 = 12u$

$\Rightarrow$ u = 60 km/hr

we get u=60km/h$$

Hemce speed of Sameer in still water is

$60km/h$

A boat takes

$4$ hours to move

$10\ km$ and back to starting point in still water. If the river flow velocity is

$2\ km/hr$ , then time taken by the boat to move

$10\ km$ upstream and back to starting point is approximately

0%
$4\ hours$
0%
$5\ hours$
0%
$6\ hours$
0%
$7\ hours$

Explanation

Speed of boat in still water

$=\dfrac {10+10}{4}=5\, km/hr$ .

Time taken by boat to cover

$10\, km$ and come back to starting point

$=$ Time taken to go upstream + Time taken to come downstream.

$= \dfrac{{10}}{{5 - 2}} + \dfrac{{10}}{{5 + 2}} = \dfrac{{100}}{{21}} = 4.77 \approx 5\, hr.$

A boat takes two hours to travel 8 km down and 8 km up the river when the water is still. How much time will the boat take to make the same trip when the river starts flowing at 4 kmph?

0%
2 hour
0%
2 hour 40 minute
0%
3 hour
0%
3 hour 40 minute

Explanation

Speed in still water=

$\dfrac{distance}{time}=(8+8)/2=8km/h$

Now speed during going down the stream motion will be

$8+4=12km/h$ so time taken in

$down$ the stream journey

$t_1=\dfrac{distance}{speed}=8/12=2/3 hour$

Speed during up the stream will be

$8-4=4km/h$ so time in going down the stream,

$t_2=\dfrac{distance}{speed}=8/4=2Hour$

Time for the trip,

$t_1+t_2=2Hour40Minutes$ .

Option B is correct.

A boat takes

$2$ hours to go

$8\ km$ and come back in still water lake. The time taken for going

$8\ km$ upstream and coming back with water velocity of

$4\ km/hr$ is:

0%
$140\ min$
0%
$150\ min$
0%
$160\ min$
0%
$170\ min$

Explanation

Velocity of boat in still water

$V_b = \dfrac{8+8}{2} = 8 \ km/h$

Upstream motion :

Velocity of boat

$v_1 = 8-4 = 4 \ km/h$

Time taken

$t_1 = \dfrac{8}{4} = 2 \ hr = 120 \ min$

Downstream motion :

Velocity of boat

$v_2 = 8+4 = 12 \ km/h$

Time taken

$t_2 = \dfrac{8}{12} = \dfrac{2}{3} hr = 40 \ min$

So, total time taken

$t = 120+40 = 160 \ min$

A motor boat going downstream overcomes a float at a point

$A$ .

$60$ minutes later it turns and after some time passes the float at a distance of

$12\ km$ from the point

$A$ . The velocity of the stream is (assuming constant velocity for the boat in still water)

0%
$6\ km\ h^{-1}$
0%
$3\ km\ h^{-1}$
0%
$4\ km\ h^{-1}$
0%
$2\ km\ h^{-1}$

Explanation

Let the speed of stream be v and speed of motorboat be u.
Now distance moved by boat in 1 hour

$= (u+v) \times1 = u+v$
Total time is

$= 6/v$
Speed of boat when it turns back

$= u-v$
=>

$u+v-6 = (u-v)(6/v -1)$
Solving for v we get,

$v = 3 km/hr$

The velocity of a motorboat with respect to still water is

$7 ms^{-1}$ and speed of the stream is

$3 ms^{-1}$ . When the boat starts moving upstream a float was dropped from it. The boat travels

$4.2$ km up stream turned about and caught up with the float. The time is taken by the boat to reach the float is:

0%
25 mins
0%
30 mins
0%
35 mins
0%
48.125 mins

Explanation

Speed of the boat when it travels upstream

$= 7-3 = 4 m/sec$

Speed of the boat when it travels downstream

$= 7+3 = 10 m/sec$

Now,

As soon as the float falls, it will travel downstream at the speed of

$3 m/sec$

And in the meantime, the boat has also traveled upstream a distance of 4.2 km

So,

Time spend on traveling upstream by boat

$= \dfrac{4200}{4} = 1050 sec$

In the meantime, the float would had traveled downstream by =1050×3 = 3150 m

Hence, the total distance between the float and the boat will be =3150 + 4200 = 7350 m

Now, the relative speed between a boat and the float =10-3=4 m/sec

Hence time taken = 7350/4=1837.5 sec

Hence, total time spent = 1837.5+1050=2887.5 sec

This is 2887.5 /60= 48.125 mins

Hence the float and the boat will meet in

$48.125 mins$

Solve inequality and show the graph of the solution,

$7x+3 < 5x+9$

Explanation

$7x + 3 < 5x + 9$

$\Rightarrow 7x - 5x < 9 - 3$

$\Rightarrow 2x < 6$

$\Rightarrow x < 3$

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

$'t_1'$ time, going down stream and

$'t_2'$ time going upstream, what time will be taken by the boat to cover the same distance in still water :-

0%
$\dfrac{t_1 + t_2}{2}$
0%
$\dfrac{t_1 }{2}$ + $\dfrac{3}{4}$ $t_2$
0%
$\dfrac{2t_1t_2}{t_1 + t_2}$
0%
$\dfrac{t_1 + t_2}{2t_1t_2}$

Explanation

Let the velocity of a boat in still water is u and velocity of the river is v and distance is d

down the stream speed will add total velocity is u+v it takes time t_{1}

in upstream speed will be subtracted total velocity is u-v it take time t_{2}

$u+v=\dfrac{d}{t_{1}}$ eq(1)..

$u-v=\dfrac{d}{t_{2}}$ eq(2)..

$2U= \dfrac{d}{t_{1}}+\frac{d}{t_{2}}$

$u = \dfrac{d(t_{1}+t_{2})}{2t_{1}t_{2}}$

time take by the still boat is

$t=\dfrac{d}{u}=\dfrac{2t_{1}t_{2}}{(t_{1}+t_{2})}$

Hence C option is correct

A boat crosses a river of width 200m in the shortest time and is found to experience a drift of 100m is reaching the opposite bank. The time taken now is 't'. If the same boat is to cross the river by shortest path, the time taken to cross will be:

0%
$2t$
0%
$\sqrt 2t$
0%
$3t$
0%
$\dfrac{2t}{\sqrt 3}$

Explanation

Let the velocity of the man and river is

$v_{m}$ and

$v_{r}$

when it cross the river in shortest time t

$v_{m}=\dfrac{200}{t}$

it drift due the river velocity

$v_{r}=\dfrac{100}{t}$

when he cross the river by shortest path than one vector component of v_{m} is cancel with river velocity

$v_{r}$

Let it make a angle as shown in fig

$v_{m}sin\theta=v_{r}$ from this we get

$\dfrac{200}{t}sin\theta=\dfrac{100}{t}$

$\Rightarrow sin\theta=\dfrac{1}{2}$

$\Rightarrow \theta=30^0.$

than man cross the river with velocity

$v_{m}cos\theta=v_{m}cos30^0$

time taken by the man to cross is

$t =\dfrac{200}{v_{m}\dfrac{\sqrt{3}}{2}}=\dfrac{400}{\dfrac{200\sqrt{3}}{t}}=\dfrac{2t}{\sqrt{3}}$

965006_1042518_ans_ac986c34c38a42a58f5e0349f43479ea.jpeg

A man swimming down stream overcome a flot at a point M. After travelling distance D he turned back and passed the float at a distance of D/2 from the point M. then the ratio of speed of swimmer with respect to still water to the speed of the speed of the river will be.

0%
2
0%
3
0%
4
0%
2.5

Explanation

The floor is swimming along with the stream with speed say,r are while the man swimming downstream has a speed of (S + r) where S is the speed of swimming in still water and r the speed of river stream.

Swims upstream till distance

$\dfrac{D}{2}$ with a speed of (s -r)

Total time taken

$\dfrac{D}{S+r} + \dfrac{D}{2(s-r)}$

and by the float

$= \dfrac{Dr}{2}$

$\dfrac{D}{S+r} + \dfrac{D}{2(s-r)}$

$= \dfrac{D}{2}r$

$= 3sr - r^2 = s^2 - r^2$

$\dfrac{S}{r} = 3$

So option (B) is correct

$x+ 10= y + 14$ , then

$x > y$ .

0%
True
0%
False

Explanation

$\textbf{Step 1 : Find the simple relation between x and y}$

$\text{x+10=y+14}$

$\text{x=y+4}$

$\text{Hence , x>y}$

$\textbf{Statement is true}$

Area of the region {

$(x,y):{x}^{2}+{y}^{2}\le 1\le x+y$ } is

0%
$\dfrac{\pi}{4}+\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}-\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}+\dfrac{3}{4}$
0%
$\pi+1$

Explanation

To find area

$x^2+y^2\leqslant |\leqslant x+y$

1) Circle

$x^2+y^2\leqslant |$

$(x-0)^2+(y-0)^2\leqslant |$\Rightarrow $ centere$ (0,0)

$, Radius$ 1$$

$x+y\geqslant |$

Required area is

$A=\int_{1}^{0}\sqrt{1-x^2}-(1-x)$

$A\Rightarrow ^1_0\left[\dfrac{\sin ^{-1}(x)}{2}+\dfrac{x}{2}\sqrt{1-x^2}\right]^1_0\left[x-\dfrac{x^2}{2}\right]$

$\therefore A\Rightarrow \left[\dfrac{\pi}{4}+\dfrac{1}{2}\times 0-0-0\right]-\left[1-0-\dfrac{1}{2}+0\right]$

$\boxed{\therefore A\Rightarrow \dfrac{\pi}{4}-\dfrac{1}{2}}$

A, B and C are three rafts floating in a river such that they always form an equilateral triangle. A swimmer whose swimming speed is constant swims from A to B then from B to C and finally from C to A along straight lines. Time taken by the swimmer is maximum in swimming:

0%
from A to B
0%
from B to C
0%
from C to A
0%
it is same in all the parts.

A man rows upstream a distance of

$9\ km$ or downstream a distance of

$18\ km$ taking

$3$ hours each time. The speed of the boat in still water is

0%
$7\dfrac {1}{2}km/h$
0%
$6\dfrac {1}{2}km/h$
0%
$5\dfrac {1}{2}km/h$
0%
$4\dfrac {1}{2}km/h$

Explanation

Let the speed of a boat in still water be

$x$ and speed of stream is

$y$

Then, speed of boat in downstream is

$x+y$

Speed of boat in upstream

$=x-y$

Now,

$x+y=\frac{18}{3}$

$x+y=6$ ---- (i)

and

$x-y=\frac{9}{3}$

$x-y=3$ ----- (ii)

Add (i) and (ii)

$x=\frac{9}{2}\;Km/hr$

the speed of boat in still water is

$4\frac{1}{2}\;Km/hr$

Number of solutions to

$x+y+z=10$ where

$1\le x,y,z\le 6$ and

$x,y,z\in N$ .

0%
$35$
0%
$36$
0%
$27$
0%
$66$

Explanation

$x+y+z=10$

$0\le x,y,z \le 6$

$(x-1) + (y-1) + (z-1)=7$

$\Rightarrow x_1+y_1+z_1=7$

$0 \le x_1,y_1,z_1, \le 5$

Similar to distributing

$7$ coins in

$3$ beggars

$\therefore$ No. of ways

$= \dfrac{(7+(3-1))!}{7!(3-1)!}$

$=\dfrac{9!}{7!2!} = 36$

Now we need to subtract the cases when

$x_1,y_1,z_1 = 7\ or\ 6$

$\therefore x_1=7\Rightarrow y_1=0,z_1=0$

$\therefore 3$ cases

and

$x_1=6 \Rightarrow y_1=0,z_1=1$

$y_1=1, z_1=0$

$\Rightarrow 3\times 2=6$ cases

$\therefore$

Required answer

$=36-3-6$

$=27$

Number of integral solutions of inequality

$|x + 3| > |2x - 1|$ is

0%
$3$
0%
$4$
0%
$5$
0%
$2$

Explanation

The inequality is

$\left | x+3 \right |>\left | 2x-1 \right |$

$\left | x+3 \right |-\left | 2x-1 \right |>0$

Case (1)

For

$-\infty <x<-3$ , inequality is

$-(x+3)-\left \{ -(2x-1) \right \}>0$

$-x-3+2x-1>0$

$x-4>0$

$x>4$

But we considered

$-\infty <x<-3$

so there is no solution in this case.

Case (2)

For

$-3\leq x<\dfrac{1}{2}$

$(x+3)+(2x-1)>0$

$3x+2>0$

$x>-\dfrac{2}{3}$

so inequality is true for

$-\dfrac{2}{3}<x<\dfrac{1}{2}$

$x=0$ is an integral solution.

Case (3)

For

$\dfrac{1}{2}\leq x<\infty$

$(x+3)-(2x-1)>0$

$x+3-2x+1>0$

$-x+4>0$

$x<4$

so inequality is true for

$\dfrac{1}{2}\leq x<4$

$x=1,2,3$ are the integral solution.

$4$ integral solutions exists for given inequality.

$S$ is the set of all real

$x$ such that

$\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}$ , then

$S$ is equal to

0%
$(-\infty, \dfrac{-1}{2})$
0%
$(-\infty, \dfrac{-2}{3})$
0%
$(-\infty, \dfrac{2}{3})$
0%
$( \dfrac{2}{3}, \infty)$

Explanation

Given,

$\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}$

or,

$2x^2+5x+2\lt 2x^2+2x$

or,

$3x+2<0$

or,

$x<\dfrac{-2}{3}$ .

$S=\left(-\infty,\dfrac{-2}{3}\right)$ ..

A boat travels a distance of

$80 \ km$ in

$4$ hours upstream and same distance down stream in

$2$ hours in a river. Then the velocities of boat and stream are (in Kmph)

0%
$20, 10$
0%
$30, 10$
0%
$30, 20$
0%
None of these

Explanation

Let

$v_b ,v_r b$ e the speed of boat and river respectively

$v_b+v_r=\dfrac{80}{4}=20\ldots(1)$

$v_b-v_r=\dfrac{80}{2}=40\ldots(2)$

solving we get

$v_b=30,v_r=10$

The solution of

$x$ in the inequation,

$\dfrac{1}{x-3} < 0$ if

$x \in R$ is

0%
$x\in (-\infty,3]$
0%
$x\in (-\infty,3)$
0%
$x\in (-\infty,-3)$ $\cup (3,\infty)$
0%
$x\in (-\infty,-3)$

The value of

$x$ , which satisfies the inequation,

$\left (\log _{ (1/2) }{ x } \ge \log _{ (1/3) }{ x }\right )$ is

0%
$(0,1]$
0%
$(0,1)$
0%
$[0,1)$
0%
none

Explanation

$\log_{1/2}x\geq\log_{1/3}x$

We can write this as

$\Rightarrow \dfrac{\log x}{\log {1/2}}\geq\dfrac{\log x}{\log {1/3}}$

$\Rightarrow \dfrac{-\log x}{\log {2}}\geq\dfrac{-\log x}{\log {3}}$

$\Rightarrow \dfrac{\log x}{\log {2}}\leq\dfrac{\log x}{\log {3}}$

$\Rightarrow \dfrac{\log x}{\log {2}}-\dfrac{\log x}{\log {3}}\leq 0$

$\Rightarrow \log x\left(\dfrac{1}{\log {2}}-\dfrac{1}{\log {3}} \right)\leq 0$

$\Rightarrow \log x\left(\dfrac{\log 3-\log 2}{\log {2}.\log {3}} \right)\leq 0$

$\Rightarrow \log x\left(\log{\dfrac{3}{2}} \right)\leq 0$

$\log\dfrac{3}{2}>0$ then

$\log x \leq 0$

This is only possible when

$x$ belongs to

$(0,1]$

The largest interval among the following for which

${x}^{12}-{x}^{9}+{x}^{4}-x+1> 0$ is

0%
$-4< x\le 0$
0%
$0< x< 1$
0%
$-100< x< 100$
0%
$-\infty< x< \infty$

Explanation

$\Rightarrow$ If $x\leq 0$ , then $x^{12}-x^9+x^4-x+1>0$

$\Rightarrow$ If $0<x\leq 1$ , then $x^{12}+x^4(1-x^5)+(1-x)>0$

$\Rightarrow$ If $x>1$ , then $x^9(x^3-1)+x(x^3-1)+1>0$

So the expression $x^{12}-x^9+x^4-x+1>0$ is valid for $-\infty<x<\infty$

$x> \sqrt{(1-x)}$ and

$x<0.5$ have infinitely many solutions

0%
True
0%
False

the set of values of

$'x'$ satisfying the inequation

$|x+5|\geq 10$ are

0%
$x\in (-15, 5]$
0%
$x\in (-5, 5]$
0%
$x\in (-\infty, -15]\cup [5, \infty)$
0%
$x\in [-\infty, -15]\cup [5, \infty)$

A man rows a boat with a speed of 18 km/hr in north-west direction. The shoreline makes an angle of

$15^o$ south of west. Obtain the component of the velocity of the boat along the shoreline.

0%
$9 \ km/hr$
0%
$18 \dfrac{\sqrt{3}}{2} \ km/hr$
0%
$18 cos (15^o) \ km/hr$
0%
$18 cos (75^o) \ km/hr$

Explanation

Given,

The speed of boat

$=18km/hr$ in north west direction. That is velocity makes an angle of

$45^0$ with west.

We have to find the velocity along shoreline or parallel to river.

We can see in figure that the velocity V makes an angle $(45+15) =60^0$ with the river.

So,

$V\,cos\theta = V\, cos60^0$

$= 18\times \dfrac12$

$=9\, km/hr$

1470926_1431588_ans_656e55f356c649849aad451755a8c71e.png

If z satisfies the inequality

$|z-1-2i| \leq 1$ , then

0%
$min (arg (z)) = tan^{-1} (\dfrac{3}{4})$
0%
$max (arg (z)) =\dfrac{\pi}{6}$
0%
$min ( |z|) = \sqrt{5}$
0%
$max ( |z|) = \sqrt{5}$

$|x - 2| + |x + 1| \ge 3$ , then complete solution set of this inequation is

0%
$[1 , \infty)$
0%
$(-\infty , -2]$
0%
$R$
0%
$[-2 , 1]$

Explanation

The inequation is

$\left | x-2 \right |+\left | x+1 \right |\geq 3$

Case (1)

For

$-\infty < x< -1$ ............ (1)

the inequation is

$-(x-2)-(x+1)\geq 3$

$-x+2-x-1\geq 3$

$-2x+1\geq 3$

$-2x\geq 2$

$x\leq -1$ ................ (2)

From (1) and (2), the inequation is true for

$-\infty < x< -1$

Case (2)

For

$-1\leq x< 2$ ,

the inequation is

$-(x-2)+x+1\geq 3$

$-x+2+x+1\geq 3$

$3\geq 3$ , which is true.

So the inequation is true for

$-1\leq x< 2$

Case (3)

For

$2\leq x< \infty$ .............. (3)

the inequation is

$x-2+x+1\geq 3$

$2x-1\geq 3$

$2x\geq 4$

$x\geq 2$ ........... (4)

From (3) and (4),

$2\leq x< \infty$

Therefore, the inequation is true for all real values of

$x$

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 11 - 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 11 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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