MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 3

Which of the following could be the graph of all values of

$x$ that satisfy the inequality

$2-5x\le -\cfrac{6x-5}{3}$

Explanation

$2-5x\le\dfrac{-(6x-5)}{3}$

$\Rightarrow 6-15x\le\, -6x+5$

$\Rightarrow 1\le 9x$

$\Rightarrow \dfrac{1}{9} \le x$

$\Rightarrow x\ge \dfrac{1}{9}$

Solution set of inequality

$4x+3<5x+7$ is

$(-4, \infty)$ .

0%
True
0%
False

Explanation

$4x+3<5x+7$

$4x+3-7<5x+7-7$

$4x-4<5x$

$4x-4-4x<5x-4x$

$-4<x$

Thus, the solution set of the given inequality is

$(-4,\infty)$ .

$A : B: C = 2 : 3 : 4$ , then

$\dfrac {A}{B} :\dfrac {B}{C} : \dfrac {C}{A}$ is equal to

0%
$4 :9 : 16$
0%
$8 : 9 : 12$
0%
$8 : 9 : 16$
0%
$8 : 9 : 24$

Explanation

$A : B : C = 2 : 3 : 4$

$A/B : B/C : C/A$

$Let A = 2x , B = 3x\ and\ C = 4x$

$Then,$

$$ AB=2x3x=23$$AB=2x 3x=23$$$$

$$BC=3x4x=34$$BC=3x4x=34

$$CA=4x2x=21$$CA=4x2x=21

$$A/B : B/C : C/A=2/3:3/4:2/1=8:9:24$$

A man swims to and fro along the bank of a river with a velocity v relative to water. If the velocity of flow is u,the average speed of the man (for to and motion)is

0%
$\frac { v^2-u^2 }{ v }$
0%
$\sqrt { uv }$
0%
$\frac { u+v }{ 2 }$
0%
$\frac { v^2-u^2 }{ 2v }$

Explanation

Speed of river, $u$

Speed of man relative to water, $v$

Downward speed of man w.r.t ground, $v+u$

Upward speed of man w.r.t ground, $v-u$

Downward time, ${{t}_{1}}=\dfrac{s}{v+u}\,\,$ and upward time ${{t}_{2}}=\dfrac{s}{v-u}$

Average speed = total distance/ total time

Average speed, ${{V}_{av}}=\dfrac{s+s}{\dfrac{s}{v+u}\,+\dfrac{s}{v-u}\,}=\dfrac{{{v}^{2}}-{{u}^{2}}}{v}$

Hence, average speed is $\dfrac{{{v}^{2}}-{{u}^{2}}}{v}$

The value of

$x$ for which

$12 x - 6 < 0$ and

$12 - 3 x < 0$

0%
$\phi$
0%
$R$
0%
$R - \{ 0 \}$
0%
$None$

Explanation

$\\12x-6<0\>or\>x<(\frac{1}{2})\\and\>12-3x<0\>or\>x>4\\clearly\>there\>is\>no\>common\>value\>of\>x\\hence\>solution\>set\>will\>be\>a\>null\>set\>\phi.$

1116650_1197659_ans_f4b8378570f04a119ff7fc41b0ce7c03.JPG

The ratio of coins of

$1\ Rs\ 50$ paise and

$25$ pasie in a bag is

$4 : 5 : 6$ . If total number ofcoins is

$120$ then the number of coins of

$25$ paise

0%
$None\ of\ these$
0%
$48$
0%
$40$
0%
$32$

A diamond price is proportional to square of its weight it gets broken onto

$4$ pices with its weight in the ratio

$1:2:3:4$ . The total price of diamond after breaking is

$70000$ less than the diamond in original.What was the price of the original diamond?

0%
$70,000$
0%
$1,00,000$
0%
$1,40,000$
0%
$none\ of\ these$

Explanation

$Let\text{ the weight of the four piece of diamond be x,2x,3x,4x}\,\text{respectively} \\$

$\text{The weight of the original diamond =x+2x+3x+4x=10x} \\$

$\text{Let the price of the original diamond =k}\left( 100{{x}^{2}} \right) \\$

$\cos t\,\text{ of four pieces=k}\left( {{x}^{2}}+4{{x}^{2}}+9{{x}^{2}}+16{{x}^{2}} \right)=k\left( 30{{x}^{2}} \right) \\$

$Loss\text{ in the total value of the diamond=k}\left( \text{100}{{\text{x}}^{2}} \right)-k\left( 30{{x}^{2}} \right)=7000 \\$

$\Rightarrow k\left( 70{{x}^{2}} \right)=70000 \\$

$\Rightarrow k{{x}^{2}}=1000 \\$

$Hence\,\text{ price of the original diamond =k}\left( 100{{x}^{2}} \right) \\$

$\,\,=RS100000 \\$

A body is just floating on the surface of a liquid. The density of the body is same as that of the liquid. The body is slightly pushed down. What will happen to the body

0%
It will slowly come back to its earlier position
0%
It will remain submerged, where it is left.
0%
It will sink
0%
It will come out violently

A motor boat covers the distance between two places on a river and returns in

$14 \ hrs$ . If the velocity of the boat in still water is

$35 \ km/h$ and velocity of water in the river is

$5 \ km/h$ , then the distance between the two places is

0%
$100 \ km$
0%
$240 \ km$
0%
$220\ km$
0%
$180\ km$

Explanation

Speed downstream

$= 35+5 = 40 \ km/hr$
Speed upstream

$= 35-5 = 30 \ km/hr$
Let distance be

$x \ km$ .

Then, total time

$= 14 = \dfrac{x}{40} +\dfrac{x}{30}$
=>

$x =240\ km$

$\left | x-2 \right |=p$ , when x < 2, then

$x-p=$

0%
-2
0%
2
0%
$2-2p$
0%
$2p-2$

Explanation

$|x-2|=p$

$x-2=\pm p$

$x-2=-p$ and

$x-2=p$

$x=2-p$ .....(i) and

$x=p+2$ .....(ii)
Hence

$x-p=2-2p$ and

$x-p=2$
However x < 2
Therefore

$x-p<2$ since

$x<2$
Hence answer is

$2-2p$

Wind is blowing from west to east along two parallel tracks. A train is moving on each track in opposite directions. They have same speed when no wind is blowing. Now, one train has speed double that of the other. The speed of each train is

0%
Equal to that of wind
0%
Double that of wind
0%
Three times that of wind
0%
Half of that of wind

Explanation

Let the speed of train be x and speed of wind be y.
On the track where train and wind are in same direction net speed is x+y.
On the track where train and wind are in opposite direction net speed is x-y.
Now , as per question,

$x+y = 2(x-y) \Rightarrow x = 3y$

The mean of 45, 38.5, 42.4, 44.3, 35.8, 39.1, 41.5, 39.4 is.

0%
44.3
0%
35
0%
40.7
0%
29.6

An aeroplane flies along a straight line from A to B with a speed

$v$

$_{0}$ and back again with the same speed

$v$

$_{0}$ . A steady wind

$v$ is blowing. If

$AB = l$ then

0%
total time for the trip is $\dfrac{2v_{0}l}{v_{0}^{2}-v^{2}}$ if wind blows along the line AB
0%
total time for the trip is $\dfrac{2l}{\sqrt{v_{0}^{2}-v^{2}}}$ , if wind blows perpendicular to the line AB
0%
total time for the trip decrease because of the presence of wind
0%
total time for the trip increase because of the presence of wind

Explanation

Wind increases time. That is always true. Hence option D is correct
Option A.=> time

$= \dfrac{l}{(v_0 + v)} + \dfrac{l}{(v_0 - v)} = \dfrac{2v_0l}{(v_0^2 - v^2)}$
option B => velocity along the line =

$\sqrt{ v_0^2 - v^2}$ for both ways.

Hence time =

$\dfrac{2l}{\sqrt{ v_0^2 - v^2}}$

A motor ship covers the distance of

$300 km$ between two localities on a river in

$10$ hours downstream and in

$12$ hours upstream. Find the flow velocity of the river assuming that these velocities are constant.

0%
2.0 km/h
0%
2.5 km/h
0%
3 km/h
0%
3.5 km/h

Explanation

Let

$v_b$ be the velocity of the boat and

$v_r$ be the velocity of the river
During downstream, time taken is

$10 = \dfrac{300}{v_b + v_r}$

$v_b + v_r = 30\ km/h \rightarrow (1)$

During upstream, the relative velocity of the boat with respect to river is

$=v_b - v_r$

$12 = \dfrac{300}{v_b - v_r}$

$v_b - v_r = 25\ km/h \rightarrow (2)$

Equation (1) - equation (2) gives:

$2v_r = 5$

$v_r = 2.5\ km/h$

A boat takes

$2 h$ to travel

$8 km$ and back in still water. In a lake with water velocity of

$4 km/h$ , the time taken for going upstream of

$8 km$ and coming back is

0%
$120 min$
0%
$160 min$
0%
$200 min$
0%
none of these

Explanation

Time taken by the boat to cover a total distance of

$16\ km$ is

$2\ hr$ .
Velocity of the boat

$= \cfrac{16}{2}$

$= 8\ km/h$
Velocity of the river is,

$V_r = 4\ km/h$
Time taken for upstream

$= \cfrac{d}{V_b - V_r}$

$t_u= \cfrac{8}{8-4} = 2\ hrs$

Time taken for downstream

$= \cfrac{d}{V_b + V_r}$

$t_d = \cfrac{8}{8+4}$

$t_d = \cfrac{2}{3} hrs$
Total time

$= \cfrac{2}{3} + 2 hrs$

$= \cfrac{8}{3} hrs = 160\ mins$

The width of river is 1 km. The velocity of boatis 5 km/hr. The boat covered the width of riverwith shortest will possible path in 15 min. Thenthe velocity of river stream is :

0%
3 km/hr
0%
4 km/hr
0%
$\sqrt { 29 } km/hr$
0%
$\sqrt { 41 } km/hr$

Explanation

$t=\dfrac { d }{ \sqrt { { u }^{ 2 }-{ v }^{ 2 } } }$

$\dfrac { 1 }{ 4 } =\dfrac { 1 }{ \sqrt { { (5) }^{ 2 }-{ v }^{ 2 } } }$

$\dfrac { 1 }{ 16 } =\dfrac { 1 }{ 25-{ v }^{ 2 } }$
v = 3 km/hr.

Solve the linear inequations and represent the solution set on the number line.
(a)

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

(b)

$-2\le \cfrac{1}{2}-\cfrac{2x}{3}\leq 1\cfrac{5}{6},x\in N$

0%
(a) $\left [ -3,1 \right ]$

(b) $\dfrac{15}{4}\geq x\geq -2$
0%
(a) $\left [ 0,2 \right ]$

(b) $\dfrac{15}{4}\geq x\geq -2$
0%
(a) $\left [ -3,2 \right ]$

(b) $\dfrac{15}{4}\geq x\geq -2$
0%
(a) $\left [ 0,5 \right ]$

(b) $\dfrac{15}{4}\geq x\geq -2$

Explanation

Part

$(a)$

$9\ge7x-5\ge5x-11$

Now,

$9\ge7x-5$

$14\ge7x$

$2\ge x$

Now,

$7x-5\ge5x-11$

$2x\ge -6$

$x\ge -3$

Therefore,

$x\in [-3, 2]$

Part

$(b)$

$-2\le \dfrac{1}{2}-\dfrac{2x}{3}\le1\dfrac{5}{6}$

Now,

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

$-\dfrac{15}{6}\le-\dfrac{4x}{6}$

$\dfrac{15}{4}\ge x$

Now,

$\dfrac{1}{2}-\dfrac{2x}{3}\le1\dfrac{5}{6}$

$\dfrac{1}{2}-\dfrac{2x}{3}\le\dfrac{11}{6}$

$-\dfrac{2x}{3}\le\dfrac{8}{6}$

$-\dfrac{2x}{3}\le\dfrac{4}{3}$

$-2x\le 4$

$-x\le 2$

$x\ge -2$

Therefore,

$\dfrac{15}{4}\ge x\ge -2$

Hence, this is the answer.

The range of values of

$x$ satisfying the following inequation,

$\dfrac{1}{2-\left | x \right |}\geq 1$ is

0%
$x\in [-2,-1]\cup [1,2]$
0%
$x\in (-2,-1)\cup (1,2)$
0%
$x\in (-2,-1]\cup (1,2]$
0%
$x\in (-2,-1]\cup [1,2)$

Explanation

Given,

For

$\dfrac{1}{2-|x|}$ to be real

$x \neq \ 2 \ or \ -2........(1)$

Now in the equation,

$\dfrac{1}{2-|x|}\ge 1$

or,

$2-|x|\le 1$

or,

$|x|\ge 1$

$\Rightarrow x\le -1$ or

$x\ge 1$ .

$\Rightarrow x\in(-2,-1]$ or

$x\in [1,2) \ (Considering \ eqn. \ 1)$

$x\in (-2,-1]\cup [1,2)$ .

Identify solution set for

$\left | 4-x \right |+1< 3$

0%
$(2,3)$
0%
$(2,6)$
0%
$(3,6)$
0%
$(4,3)$

Explanation

Therefore

$-2<4-x<2$

$4-x<2$

$x>2$ ...(i)

$-2<4-x$

$x<6$ ...(ii)
From i and ii

$x\: \epsilon(2,6)$
Answer is B

What is the solution set for

$\dfrac{\left | x-2 \right |}{x-2}> 0$ ?

0%
$(2, \infty )$
0%
$(-\infty ,-2)$
0%
$(-\infty ,0)$
0%
(0,2)

Explanation

$\dfrac{\left | x-2 \right |}{x-2}> 0$

When,

$x > 2$ ,

$x -2 > 0$

Thus,

$\dfrac{\left | x-2 \right |}{x-2}> 0$

$\dfrac{x-2 }{x-2}> 0$

$1 > 0$

When,

$x < 2$ ,

$x - 2 < 0$

$\dfrac{\left | x-2 \right |}{x-2}> 0$

$\dfrac{ - x + 2 }{x-2}> 0$

$- 1 \not > 0$

Hence,

$x > 2$ satisfies the equation.

$(2, \infty )$ is the solution set.

What is the solution set for

$\left |\dfrac{2x-1}{x-1} \right |> 2$ ?

0%
$(\frac{-3}{4},1)\cup (1,\infty )$
0%
$(\frac{3}{4},1)\cup (1,\infty )$
0%
$(\frac{1}{4},1)\cup (1,\infty )$
0%
$(\frac{5}{4},1)\cup (1,\infty )$

Explanation

$\left |\dfrac{2x-1}{x-1} \right |> 2$

then,

$x \ne 1$ as denominator will become zero.

$x < 1$

$\dfrac{-2x -1}{-x-1} > 2$

$2x + 1 > 2x + 2$

$1 > 2$ , which is not true.

Now, when

$x > \dfrac{1}{2} \, and \, x < 1$

$\dfrac{2x -1}{x-1} > -2$

$2x -1 > -2x + 2$

$4x > 3$

$x > \dfrac{3}{4}$

Now, when

$x > 1$

$\dfrac{2x -1}{x-1} > 2$

$2x -1 > 2x - 2$

$-1 > -2$ , which is true.

Hence

$x \in (\frac{3}{4},1) \cup (1,\infty)$

7 - 3 (1 - y) = - 5 + 2x

0%
Area of triangle: $18\,\,sq.\,units$ and co-ordinate axes:(1,-3)
0%
Area of triangle: $73\,\,sq.\,units$ and co-ordinate axes:(1,-3)
0%
Data insufficient
0%
None of these

Represent the set of real values of x on the number line satisfying

$\dfrac{1}{2}(2x-1)\leq 2x+\dfrac{1}{2}\leq 5\dfrac{1}{2}+x$ . Also, find the greatest and the values of x satisfying the inequations.

0%
$x\in \left [ 1,5 \right ]$
0%
$x\in \left [ -1,5 \right ]$
0%
$x\in \left [ -4,3 \right ]$
0%
$x\in \left [ 4 ,3 \right ]$

Explanation

$The\ system\ of\ equations\ is\\ \dfrac { 1 }{ 2 } \left( 2x-1 \right) \ge 2x+\dfrac { 1 }{ 2 } \ge 5\frac { 1 }{ 2 } +x.\\ So\quad \dfrac { 1 }{ 2 } \left( 2x-1 \right) \ge 2x+\dfrac { 1 }{ 2 } \\ \Longrightarrow 2x-1\ge 4x+1\\ \Longrightarrow 2x-4x=1+1\\ \Longrightarrow x\ge -1.\\ Again\quad \\ 2x+\dfrac { 1 }{ 2 } \ge 5\frac { 1 }{ 2 } +x\\ \Longrightarrow 4x+1\ge 11+2x\\ \Longrightarrow 4x-2x\ge 11-1\\ \Longrightarrow 2x\ge 10\\ \Longrightarrow x\ge 5.\\ \therefore \ x\in \left[ -1,5 \right] .$

Solve the following inequality:

$\dfrac{3x-1}{5}\leq 2-\dfrac{x+1}{3}$

0%
$x\leq 2$
0%
$x> 2$
0%
$2$
0%
$0$

Explanation

$\dfrac{3x-1}{5}\leq 2-\dfrac{x+1}{3}$

$\implies 9x - 3 \leq 30 - 5x -5$

$\implies 14x \leq 28$

$\implies x \leq 2$

What is the solution set for

$\dfrac{2x+3}{3x-7} >0$

0%
$\left(-\infty ,-\dfrac{1}{2}\right)\cup \left(\dfrac{2}{3},\infty \right)$
0%
$\left(-\infty ,-\dfrac{3}{2}\right)\cup \left(\dfrac{7}{4},\infty \right)$
0%
$\left(-\infty ,-\dfrac{3}{2}\right)\cup \left(\dfrac{7}{3},\infty \right)$
0%
$\left(-\infty ,-\dfrac{1}{2}\right)\cup \left(\dfrac{7}{4},\infty \right)$

Explanation

$\dfrac{2x+3}{3x-7} >0$ - (Given)

$3x-7\neq 0$

$x\neq \dfrac{7}{3}$ .....(1)

Now, multiply by

$(3x-7)^2$ to both sides, we get

$(2x+3)(3x-7)>0$
Either both

$(2x+3)$ and

$(3x-7)$ will be positive or negative.

Case I :

$(2x+3)>0$ ,

$(3x-7)>0$

$x>-\dfrac{3}{2}, x>\dfrac{7}{3}$ .....(2)

Case II:

$(2x+3)<0$ ,

$(3x-7)<0$

$x<-\dfrac{3}{2}, x<\dfrac{7}{3}$ .....(3)

So, from (1),(2) and (3), we get

$\left(-\infty ,-\dfrac{3}{2}\right)\cup \left(\dfrac{7}{3},\infty \right)$

Identify the solution set for

$\dfrac{1}{\left | x \right |-3}< \dfrac{1}{2}$

0%
(-3,2)
0%
$(-\infty ,-5)\cup (5,\infty )$
0%
$(5,\infty )$
0%
$(-\infty ,-5)\cup (-3,3)\cup (5,\infty )$

Explanation

$\dfrac{1}{\left | x \right |-3}< \dfrac{1}{2}$
Solution:
Since,

$\dfrac{1}{\left | x \right |-3}< \frac{1}{2}$

$\therefore |x|-3 > 2$

$\therefore |x|>5$

$\therefore x<-5$ and

$x>5$
Hence, solution set for

$\dfrac{1}{\left | x \right |-3}< \dfrac{1}{2}$ is

$(-\infty ,-5)\cup (5,\infty )$

Solve the following inequalities:

$\dfrac{1}{\left | x \right |-3}< \dfrac{1}{2}$

0%
$(-\infty ,5)\cup (5,\infty )$
0%
$(-\infty ,-5)\cup (5,\infty )$
0%
$(\infty ,-5)\cup (5,\infty )$
0%
None of these

Explanation

Given,

$\frac{1}{\left | x \right |-3}< \frac{1}{2}$

$=> \frac{1}{\left | x \right |-3} - \frac{1}{2} < 0$

$=> \frac {2 - \left | x \right | + 3 }{(\left | x \right |-3)\times 2} < 0$

$=> 5 - \left | x \right | < 0$

Now, when

$x > 0$ , we have

$\left | x \right | = x$
So,

$5 - x < 0$

$=> x > 5$ -- (1)

And when

$x < 0$ , we have

$\left | x \right | = - x$
So,

$5 - (-x) < 0$

$5 + x < 0$

$=> x < -5$ -- (2)

From

$(1), (2)$

$\left( -\infty ,-5 \right) \cup \left( 5,\infty \right)$

Identify the solution set for

$\dfrac{7x-5}{8x+3} >4.$

0%
$\dfrac{-17}{25},\dfrac{-3}{8}$
0%
$\dfrac{-5}{7},\dfrac{-3}{8}$
0%
$\dfrac{-31}{28},\dfrac{-3}{8}$
0%
$\dfrac{-13}{15},\dfrac{-3}{8}$

Explanation

$\dfrac{7x-5}{8x+3} >4$

$\dfrac{7x-5}{8x+3}-4 >0$

$\dfrac{-25x - 17}{8x+3} > 0$

$\dfrac{25x + 17}{8x+3} < 0$

$25x + 17 < 0$

$x < \dfrac{-17}{25}$

$8x + 3 < 0$

$x < \dfrac{-3}{8}$
Therefore, on a number line x is defined in 3 intervals,

$(-\infty, \dfrac{-17}{25})$ ,

$(\dfrac{-17}{25}, \frac{-3}{8})$ and

$(\dfrac{-3}{8}, \infty)$
and out of all these

$x$ is defined only in

$(\dfrac{-17}{25}, \dfrac{-3}{8})$

The solution set for

$\left | x \right |> 7$ is

0%
$(-\infty ,-7)$
0%
$(7, \infty )$
0%
$(-\infty ,-7)\cup (7, \infty )$
0%
$(-\infty ,-7]\cup [7, \infty )$

Explanation

$|x|>7$

$|x|-7>0$

Case-I

when

$x>0$

then

$x-7>0$

$\Rightarrow x>7$

Case-II

when

$x<0$

$-x-7>0$

$\Rightarrow x<-7$

Therefore

$x>7$ or

$x<-7$

$x\:\epsilon\:(7,\infty)$ or

$x\:\epsilon\:(-\infty,-7)$

Hence the solution set for the inequality is

$(-\infty,-7)\cup (7,\infty)$

Solve the following inequalities:

$\dfrac{\left | x+2 \right |-x}{x}< 2$

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

Explanation

$\frac{\left | x+2 \right |-x}{x} < 2$

$\Rightarrow \left | x +2 \right |-x <2x$

$\Rightarrow x + 2 - x < 2x$

$\Rightarrow 2x > 2$

$\Rightarrow x >1$

$\therefore x = \left ( 1 , \infty \right )$

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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