MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 6

If p : q = r : s = t : u = 2 : 3 then what is (mp + nr + ot) : (mq + ns + ou) equal to?

0%
2 : 3
0%
5 : 6
0%
7 : 9
0%
2 : 5

Explanation

Given

$p : q = r : s = t : u = 2 : 3$

$\implies \dfrac{p}{q}=\dfrac{r}{s}=\dfrac{t}{u}=\dfrac{2}{3}$

Let

$p = r = t = 2x$ and

$q = s = u = 3x$

Then

$\displaystyle \frac{mp+nr+ot}{mq+ns+ou}=\frac{m\times 2x+n\times 2x+o\times 2x}{m\times 3x+n\times 3x+o\times 3x}=\frac{2x(m+n+o)}{3x(m+n+o)}=\frac{2}{3}=2:3$

If x is an integer greater than-10, but less than 10 and

$\displaystyle \left | x-2 \right |<3,$ then the values of x are

0%
$\displaystyle -10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,$
0%
$\displaystyle 0,1,2,3,4,5,6,7,8,9,10,$
0%
$\displaystyle 0,1,2,3,4,$
0%
$\displaystyle -1,0,1$

Explanation

It is given that

$-10<x<10$
That is

$|x|<10$ ...(i)
Now

$|x-2|<3$

$-3<x-2<3$

$-1<x<5$ ...(ii)
Hence
from ii,

$x\epsilon (-1,5)$ and from i

$x\epsilon (-10,10)$
Hence

$x\epsilon (-1,5)$
Now x is an integer.
Hence the set of integers falling in the above solution set is

$\{0,1,2,3,4\}$ .

If x : a = y : b = z : c then

$\displaystyle \frac{ax-by}{\left ( a+b \right )\left ( x-y \right )}+\frac{by-cz}{\left ( b+c \right )\left ( y-z \right )}+\frac{cz-ax}{\left ( c+a \right )\left ( z-x \right )}$ is equal to

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

Explanation

$\displaystyle \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\left ( say \right )\Rightarrow x=ak,ynl.z=ck$

$\therefore \dfrac{ax-by}{\left ( a+b \right )\left ( x-y \right )}+\dfrac{by-cz}{\left ( \left ( b+c \right ) \right )\left ( y-c \right )}+\dfrac{cz-ax}{\left ( c+a \right )\left ( z-x \right )}$

$=\dfrac{a\times\:ak-b\times\:bk}{\left ( \left ( a+b \right ) \right )\left ( ak-bk \right )}+\dfrac{b\times\:bk-c\times\:ck}{\left ( b+c \right )\left ( bk-ck \right )}+\dfrac{c\times\:ck-c\times\:ak}{\left ( c+a \right )\left ( ck-ak \right )}$

$=\dfrac{k\left ( a^{2}-b^{2} \right )}{k\left ( a^{2}-b^{2} \right )}+\dfrac{k\left ( b^{2}-c^{2} \right )}{k\left ( b^{2}-c^{2} \right )}+\dfrac{k\left ( c^{2}-a^{2} \right )}{k\left ( c^{2}-a^{2} \right )}=1+1+1=3$

A leak in the bottom of the tank can empty it in

$6$ hr. A pipe fills the tank at

$4$ ltr/min. When the tank is full the inlet is opened but due to the leak, the tank is emptied in

$8$ hour. What is the capacity of the tank ?

0%
$5,260 L$
0%
$5,760 L$
0%
$5,846 L$
0%
$6,970 L$

Explanation

Solution:Work done by the inlet in 1 hour=

$\dfrac { 1 }{ 6 } -\dfrac { 1 }{ 8 } =\dfrac { 1 }{ 24 }$

Work done by inlet in 1 min

$\dfrac { 1 }{ 24 } \times \dfrac { 1 }{ 60 } =\dfrac { 1 }{ 1440 }$

Value of

$\dfrac { 1 }{ 1440 }$ part=4 litres

Volume of whole=

$(1440\times 4)$ litres

$5760$ litres

The solution set of inequality $$\displaystyle 2 sin^2 x - 5 sin x + 2> 0 if x \varepsilon [0, 2 \pi] is

0%
$\displaystyle [0, \frac {\pi}{6}]$
0%
$\left[\dfrac {5 \pi}{6}, 2 \pi\right]$
0%
$\displaystyle \left[0, \frac {\pi}{6} \right] \cup \left[\frac {5 \pi}{6}, 2 \pi\right]$
0%
None of these

Explanation

Given:

${ 2\sin }^{ 2 }x-5{\ sin }^{ 2 }x+2>0$

$ifx\epsilon \left[ 0,2\pi \right]$

Solution:

$2{ sin }^{ 2 }x-5x-3=0$

Factorise:

$(2\sin { x+1)(\sin { x-3)=0 } } \\ \Rightarrow \sin { x=\dfrac { 1 }{ 2 } }$ or

$ii)\sin { x=3 }$

There are no real solution for

$\sin { x=3 }$

$i)\sin { x } =\dfrac { -1 }{ 2 } \\ x={ sin }^{ -1 }\left( \dfrac { -1 }{ 2 } \right) \\ x=+\dfrac { \pi }{ 6 } \\$

Hence the solution would be in the range of

$\left[ 0,\dfrac { \pi }{ 6 } \right]$ for the internal

$\left[ 0,2\pi \right]$

A boat takes

$2$ hours to go

$8 \;km$ and come back in still water lake. With velocity of

$4\;km/hr$ , the time taken for going upstream of

$8\;km$ and coming back is

0%
140 minutes
0%
150 minutes
0%
160 minutes
0%
170 minutes

Explanation

$\displaystyle \frac{8}{v_{rm}}+\frac{8}{v_{rm}}=2\Rightarrow v_{mr}=8km/h$

$\displaystyle \Rightarrow t=\frac{8}{v_{mr}+v_{r}}+\frac{8}{v_{mr}-v_{r}}$

$\displaystyle \Rightarrow t=\frac{8(v_{mr}-v_{r}+v_{mr}+v_{r})}{v^{2}_{mr}-v^{2}_{r}}=\frac{16v_{mr}}{v^{2}_{mr}-v^{2}_{r}}$

$\displaystyle \frac{16\times 8}{64-16}=\frac{16\times 8}{48}h=\frac{8}{3}\times 60=160\:min$

A boy is now

$a$ years old, and his father is

$5a$ years old. What will the father be when the boy is

$3a$ years old? How old was the father when the boy was born?

0%
$7a, 4a$ years
0%
$4a, 10a$ years
0%
$12a , 3a$ years
0%
$15a, 3a$ years

Explanation

When the boy is

$a$ years old, the father is

$5a$ years old

So, when the boy is

$3a$ years old, its age increased by

$2a$ . This means, fathers age also increased by

$2a$ years and will be

$5a + 2a = 7a$ years old.

And when the boy was born, fathers age will be

$5a - a = 4a$ years old.

The complete solution set of the inequation

$\displaystyle x-\frac{2(k-1)}{5}\leq \frac{2}{3k}(x+1)$ given by

0%
$\displaystyle (-\infty ,2)$ if $\displaystyle k>\frac{2}{3}$
0%
$\displaystyle [-\infty ,2)$ if $\displaystyle k>\frac{2}{3}$
0%
$\displaystyle (-\infty ,2]$ if $k < 0$
0%
All of these

$a, b, c, d$ are positive real numbers, such that

$a + b + c + d = 2$ , then

$M = (a + b) (c + d)$ satisfies the

0%
$\displaystyle 0\leq M\leq 1$
0%
$\displaystyle 1\leq M\leq 2$
0%
$\displaystyle 2\leq M\leq 3$
0%
$\displaystyle 3\leq M\leq 4$

Explanation

Given:

$a,b,c ,d$ are positive real number

$\Rightarrow$

$a+b+c+d=2 \ and \ M=(a+b))c+d)$

Solution:

Since we know for two positive number Arthematic mean

$\ge$ Geometric mean

Let us consider

$(a+b) and (c+d)$ as two numbers

Hence

$A.M\ge G.M$

$\Rightarrow$

$\dfrac { (a+b)+(c+d) }{ 2 } \ge \sqrt { (a+b)(c+d) }$

$\Rightarrow$

$\dfrac { 2 }{ 2 } \ge \sqrt { M }$

$\Rightarrow 0\le M\le 1$

$p,q$ and

$r$ are positive real number, then the quantity

$\dfrac{( p + r )}{ ( q + r)}$ is

0%
$> \dfrac{p}{q}$ if $p > q$
0%
$= \dfrac{p}{q}$ if $p > q$
0%
$> \dfrac{p}{q}$ if $p < q$
0%
$< \dfrac{p}{q}$ if $p >q$

Explanation

Given:

$p,q , r$ are positive real number

Case 1: If

$p>q$

then

$\dfrac { p }{ q } >1$

$\Rightarrow p+r>q+r$

$\dfrac { p+r }{ q+r } >1$

Let us assign some values

Let

$p=2,q=1, r=3$

$\dfrac { p+r }{ q+r } =\dfrac { 5 }{ 4 } =1.25$

ii)

$\dfrac { p }{ q } =\dfrac { 2 }{ 1 } =2$

$\Rightarrow \dfrac { p }{ q } >\dfrac { p+r }{ q+r }$ ......

$(1)$

Case 2: if

$p<q$

$\dfrac { p }{ q } <1$

$\Rightarrow p+r<q+r$

Let us assign some value

$p=1, q=2, r=3$

$\dfrac { p }{ q } =\dfrac { 1 }{ 2 } =0.5$

ii)

$\dfrac { p+r }{ q+r } =\dfrac { 1+3 }{ 2+3 } =\dfrac { 4 }{ 5 } =0.8$

$\Rightarrow \dfrac { p }{ q } <\dfrac { p+r }{ q+r }$ ......

$(2)$

From equation

$(1)$ and

$(2),$ we get

for

$p>q$ ,

$\dfrac { p }{ q } <\dfrac { p+r }{ q+r }$

$\displaystyle l^{2}+m^{2}+n^{2}=5,$ then

$( lm + mn + ln)$ is

0%
$\displaystyle \geq (-5/2)$
0%
$\displaystyle \geq (-1)$
0%
$\displaystyle \geq 5$
0%
Option A & C both

Explanation

We have,

$(l+m+n)^2 \ge 0$ as a square is always positive

$=> l^2 + m^2 + n^2 + 2(lm+mn+ln) \ge 0$

$=>5 + 2(lm+mn+ln) \ge 0$

$=> lm+mn+ln) \ge -\dfrac {-5}{2}$

$\triangle ABC$ , if

$D$ is a point in

$BC$ and divides it in the ratio

$3:5$ i.e., if

$BD:DC=3:5$ then,

$ar(\triangle ADC) : ar(\triangle ABC) =?$

0%
$3:5$
0%
$3:8$
0%
$5:8$
0%
$8:3$

Explanation

Given:

$BD=3x, DC=5x$

$\displaystyle \left ( \frac {Area of \triangle ADC}{Area of \triangle ABC}\right)=\frac {\displaystyle \frac {1}{2}\times DC\times AE}{\displaystyle \frac {1}{2}\times (BD+DC)\times AE}$

$\displaystyle =\frac {5x \times AE}{8x \times AE}=\frac {5}{8}$

Solve for

$\displaystyle x: \left | x + 4 \right |< 12$

0%
$\displaystyle \left [ -16,7 \right ]$
0%
$\displaystyle \left ( -16,4 \right ]$
0%
$\displaystyle \left [ -16,3 \right )$
0%
$\displaystyle \left ( -16,8 \right )$

Explanation

Given,

$\left| x+4 \right| <12$

Case 1: When

$(x+4) > 0 \Rightarrow \left| x+4 \right| = x + 4$
Then,

$x +4 < 12 \Rightarrow x < 8$ --- (1)

Case 2: When

$(x+4) < 0\Rightarrow \left| x+4 \right| = -(x + 4)$
Then,

$-x - 4 < 12 \Rightarrow -x < 16 \Rightarrow x < -16$ --- (2)

From case 1 and 2, we get the solution of

$x$ as

$(-16,8)$

The speed of motor boat in still water

$20\ km/hr$ and river flow is

$5\ km/hr$ . A float is dropped from the boat when it start moving upstream. After moving

$1.5\ km$ the boat returns back. The boat will catch the float (from initial instant) after

0%
$6\ min$
0%
$12\ min$
0%
$10\ min$
0%
$15\ min$

Explanation

Speed of boat in upstream = $15\ km/hr$

Distance = $1.5\ km$

Time = $t_1=\dfrac {1.5}{15}$ = $6\ min$

Distance move by float in this time

$5\times \dfrac {1}{10}=\dfrac {1}{2}\ km$

Speed of boat w.r.t. float,

$20 km/hr , t_2= \dfrac {2\ km}{20}$ = $6\ min$ .

Let

$a, b, c, x, y, z$ be positive real numbers such that

$a+b+c = x+y+z$ and

$abc = xyz.$ Further, suppose that

$a \leq x < y < z \leq c$ and

$a < b < c.$ then

$a = x, b = y,$ and

$c = z$

0%
True
0%
False

A motor ship covers the distance of 300 km between two localities on a river in 10 hrs downstream and in 12 hrs upstream. Find the flow velocity of the river assuming that these velocities are constant.

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

Explanation

Let

$\displaystyle { v }_{ 1 }$ e the flow velocity of the be ship and

$\displaystyle { v }_{ 2 }$ be the velocity of the river flow
In downward stream,

$\displaystyle { t }_{ 1 }=\frac { s }{ { v }_{ 1 }+{ v }_{ 2 } }$

$\displaystyle \therefore \quad 10=\frac { 300 }{ { v }_{ 1 }+{ v }_{ 2 } }$

$\displaystyle or\quad { v }_{ 1 }+{ v }_{ 2 }=30.....(i)$
In upstream,

$\displaystyle 12=\frac { 300 }{ { v }_{ 1 }-{ v }_{ 2 } }$

$\displaystyle { v }_{ 1 }-{ v }_{ 2 }=25$
Solving equation (i) and (ii), we get

$\displaystyle { v }_{ 2 }=2.5\quad km/hr$

Solve for

$x:$

$\displaystyle \left | 3x+2 \right |\geq 7$

0%
$\displaystyle \left ( -\infty ,-3 \right ]\cup \left [ \frac{5}{3},\infty \right )$
0%
$\displaystyle \left ( -\infty ,-\dfrac53 \right ]\cup \left [3,\infty \right )$
0%
$\displaystyle \left ( -3,0 \right ]\cup \left [ \frac{5}{3},\infty \right )$
0%
$\displaystyle \left ( -5 ,-3 \right ]\cup \left [ \frac{5}{3},\infty \right )$

Explanation

Given

$\mid 3x+2 \mid \geq 7$

$\implies -7 \geq 3x+2 \geq 7$

$\implies 3x+2 \leq -7$ (or)

$3x+2 \geq 7$

$\implies 3x \leq -7-2$ (or)

$3x \geq 7-2$

$\implies x \leq \dfrac{-9}{3}$ (or)

$x \geq \dfrac{5}{3}$

$\implies x \leq -3$ (or)

$x \geq \dfrac{5}{3}$

$\implies (-\infty, -3] \cup [\dfrac{5}{3},\infty)$

solve for

$\displaystyle x:\left | x-3 \right |> 4$

0%
x < - 1
0%
x > 7
0%
x < 2
0%
(1) or (2)

Explanation

Given,

$\left| x-3 \right| >4$

Case 1: When

$x-3> 0 \Rightarrow \left| x-3\right| = x-3$

Then,

$x-3>4$

$\Rightarrow x > 7$ -- (1)

Case 2: When

$x-3 < 0 \Rightarrow \left| x-3 \right| = -(x-3)$

Then,

$-x + 3 > 4$

$\Rightarrow -x > 1$

$\Rightarrow x < -1$ --- (2)

From, solutions

$1, 2$ , we get

$\Rightarrow x > 7$ or

$x < -1$

$\displaystyle \left | x + 7 \right |> -8$ then find the solution set

0%
Q
0%
N
0%
W
0%
R

Explanation

Given,

$\left| x+7 \right| >-8$

We know that

$\left| x+7 \right|$ will always be a positive value irrespective of the value of

$x$

So, the solution of the given inequality is

$R$

$\displaystyle 12\frac{3}{10}=$ _________

0%
12.10
0%
10.3
0%
12.3
0%
123.10

Explanation

In the given mixed fraction

$12\dfrac {3}{10}$ , the whole number is

$12$ , that will be written to the left of the decimal point.

The fraction has its denominator as a power of

$10$ , which means it can be written as a decimal with

$1$ decimal place that is

$\dfrac {3}{10}=0.3$

Therefore,

$12\dfrac {3}{10}$ is equal to:

$12\dfrac { 3 }{ 10 } =12\times 0.3=12.3$

Hence,

$12\dfrac { 3 }{ 10 }=12.3$ .

What are the real values of x that satisfy the inequations

$6x + 9 < 3x +5$ and

$4x + 7 > 2x - 5$ ?

0%
$\displaystyle \left ( -6,\frac{-4}{3} \right )$
0%
$\displaystyle \left (\frac{-2}{3},5 \right )$
0%
$\displaystyle \left (\frac{7}{3},10 \right )$
0%
$\displaystyle \left (-5,\frac{2}{3}\right )$

Explanation

Given,

$6x+9 < 3x + 5$

$=> 3x < -4$

$=> x < - \frac {4}{3}$ --- (1)

And

$4x+7 > 2x-5$

$=> 2x > -12$

$=> x > -6$ --- (2)

From, solutions

$1, 2$ , we get

$\left(-6 , -\dfrac {4}{3}\right)$

Which of the following number line represents the solution of the inequality

$-6x + 12 > -7x + 17$ ?

Explanation

Given,

$-6x+12>-7x+17$

$\Rightarrow -6x+7x>17-12$

$\Rightarrow x > 5$

Hence, option A is correct.

Solve the inequality. Write the solution set in interval notation.

$\left|5-4x\right| > 8$

0%
$\left( \displaystyle\frac { 3 }{ 4 } , -\displaystyle\frac { 13 }{ 4 } \right)$
0%
$\left( -\infty ,-\displaystyle\frac { 3 }{ 4 } \right) \cup \left( \displaystyle\frac { 13 }{ 4 } ,\infty \right)$
0%
$\left( -\infty ,\displaystyle\frac { 1 }{ 4 } \right) \cup \left( -\displaystyle\frac { 15 }{ 4 } ,\infty \right)$
0%
$\left( -\displaystyle\frac { 13 }{ 4 } ,-\displaystyle\frac { 3 }{ 4 } \right)$

Explanation

Given inequality is,

$\left| 5-4x \right| >8$

Case i) Value of x is positive-

$\therefore \left| 5-4x \right| =5-4x$

$\therefore 5-4x>8$

Adding 4x to both sides, we get,

$5-4x+4x>8+4x$

$\therefore 5>8+4x$

Subtract 8 from both sides,

$\therefore 5-8>8+4x-8$

$\therefore -3>4x$

$\therefore x<\frac { -3 }{ 4 }$

Thus, an interval is,

$\left( -\infty ,\frac { -3 }{ 4 } \right)$

Case ii) Value of x is negative-

$∴|5−4x|=-(5−4x)$

$∴|5−4x|=4x-5$

$∴4x-5>8$

Adding 5 to both sides, we get,

$∴4x-5+5>8+5$

$∴4x>13$

$∴x>\frac { 13 }{ 4 }$

Thus, an interval is,

$\left( \frac { 13 }{ 4 } ,\infty \right)$

Thus, combining both the results, we get,

$\left( -\infty ,\frac { -3 }{ 4 } \right) \bigcup \left( \frac { 13 }{ 4 } ,\infty \right)$

Answer is Option (B)

Solve the rational inequality. Write the solution set in interval notation.

$\displaystyle\frac{1}{x+10} > 0$

0%
$\left( -\infty ,10 \right)$
0%
$\left( 10,-\infty \right)$
0%
$\left( -10,\infty \right)$
0%
$\left[ 10,\infty \right]$

Explanation

The given inequality is,

$\frac { 1 }{ x+10 } >10$

We know that

$1>0$

Thus, numerator is greater than 1. For LHS to be positive, denominator must be greater than zero.

$\therefore x+10>0$

$\therefore x>-10$

Thus, Interval of the solution is,

$\left( -10,\infty \right)$

Thus, Answer is option (C)

Solve the polynomial inequality. Express the solution set in interval notation.

$13x^2-5x\le 0$

0%
$(-\infty,0)\cup\left(\dfrac{5}{13},\infty\right)$
0%
$\left(0, -\dfrac{5}{13}\right)$
0%
$(-\infty,0)\cup\left[-\dfrac{5}{13},\infty\right]$
0%
$\left[0, \dfrac{5}{13}\right]$

Explanation

Given,

$13x^2-5x\le0$

$\Rightarrow x(13x-5)\le0$ ...(i)

Two cases arises:
Case I:
(i) holds,
if

$x\ge 0$ and

$(13x-5)\le 0$

$x\ge 0$ and

$x\le \dfrac 5{13}$

$0\le x \le\dfrac5{13}$

$x\in\left[0,\dfrac5{13}\right]$

Case II:
(i) holds,
if

$x\le 0$ and

$(13x-5)\ge 0$

$x\le 0$ and

$x\ge \dfrac 5{13}$

no such x exist.
On combining case I and II, we get

$x\in\left[0,\dfrac5{13}\right]$

$\displaystyle\frac { 5 }{ x } >3$

0%
$(-\infty,\dfrac 53)$
0%
$(-\infty,\dfrac 43)$
0%
$(-\infty,\dfrac 57)$
0%
None of these

Explanation

R.E.F image

Given, inequality

$\dfrac{5}{x}>3$

$5>3x$

$0>3x-5$

$3x-5<0$

$x\epsilon \left ( -\infty ,\dfrac{5}{3} \right )$

Option A is correct

1140147_426593_ans_ac602e4593d846248d8bc7ca6d9468d3.png

$\displaystyle\frac { 5x-8 }{ x-5 } \ge 2$

0%
$[-\dfrac 23,5] \cup [5,\infty]$
0%
$(-\infty,-\dfrac 23]\cup(5,\infty)$
0%
$[-\dfrac 23,5)$
0%
None of these

Explanation

R.E.F image

Given inequality

$\dfrac{5x-8}{x-5}\geq 2$

obviously

$x\neq 5$

$5x-8\geq 2(x-5)$

$5x-8\geq 2x-10$

$3x+2\geq 0$

$x\epsilon [\dfrac{-2}{3},5) \cup (5,\infty )$

1140141_426595_ans_bd95a66b9e5a408580c90447675ff9bb.png

Solve the following inequality.

$\displaystyle \frac{x-1}{x+2}<0$

0%
$-2< x < 1$
0%
$-2< x < 7$
0%
$-4< x < 1$
0%
None of these

The point which does not belong to the feasible region of the LPP:
Minimize:

$Z=60x+10y$
subject to

$3x+y \ge 18$

$2x+2y \ge 12$

$x+2y\ge 10$

$x,y \ge 0$ is

0%
$(0,8)$
0%
$(4,2)$
0%
$(6,2)$
0%
$(10,0)$

Explanation

We test whether the inequalitiies are satisfied or not

$(0,8)$ ,

$3(0)+8 \ge 8$

$8\ge 8$ is true.

$2(0)+2(8)=16\ge 12$ is true.

$0+2(8)=16 \ge 10$ is true.

$\therefore$

$(0,8)$ is in the feasible region.

$(4,2)$ ,

$3(4)+2=14 \ge 8$

$2(4)+2(2)=16\ge 12$

$4+2(2)=8\ge 10$ is not true

$\therefore$

$(4,2)$ is not a point in the feasible region

$\therefore$ (2) is correct

Solve the following inequality.

$\displaystyle \frac{a-1}{a}>0$

0%
$a < 0$ or $a > 1$
0%
$a < 0$ or $a > 5$
0%
$a < 0$ or $a > 2$
0%
None of these

Explanation

The given inequality is,

$\frac { a-1 }{ a } >0$

Case (1)- Numerator and Denominator are positive (greater than zero)

$\therefore a-1>0$

and

$a>0$

$\therefore a>1$

and

$a>0$

These both conditions are satisfied when

$a>1$

Case (ii) Numerator and Denominator both are negative (Less than zero)

$\therefore a-1<0$

and

$a<0$

\therefore

$a<1$

and

$a<0$

These both conditions are satisfied when

$a<0$

Thus, the solution is,

$a<0$ or

$a>1$

Thus, Answer is option (A)

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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