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CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Quantification And Numerical Applications
Quiz 8
Assume that neither
x
x
nor
y
y
is equal to 0, to permit division by
x
x
and by
y
y
.
What is the ratio x : y : z ?
(1) x+ y = 2z
(2) 2x+ 3y = z
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0%
$$5:2:1$$
0%
$$3:-2:1$$
0%
$$4:2:1$$
0%
$$5:-3:1$$
Explanation
For this problem, you do not necessarily need to know the value of $$x,y$$ or $$z$$. You simply need to know the ratio $$x: y: z$$ (in other words, the value of $$\dfrac{x}{y}$$ AND the value of $$\dfrac{y}{z}$$). You need to manipulate the information given to see whether you can determine this ratio.
(1) INSUFFICIENT: There is no way to manipulate this equation to solve for a ratio. If you simply solve for $$\dfrac{x}{y}$$, for example, you get a variable expression on the other side of the equation:
$$x+y=2x$$
$$x=2z-y$$
$$\dfrac{x}{y}=\dfrac{2z-y}{y}=\dfrac{2x}{y}=-1$$
2) INSUFFICIENT: As in the previous example, there is no way to manipulate this equation to solve for a ratio. If you simply solve for $$\dfrac{x}{y}$$, for example, you get a variable expression on the other side of they equation:
$$2x+3y=z$$
$$2x=z-3y$$
$$\dfrac{x}{y}=\dfrac{z-3y}{2y}=\dfrac{z}{2y}-\dfrac{3}{2}$$
(1) AND (2) SUFFICIENT: Use substitution to combine the equations:
$$x+y= 2z$$
$$2x+ 3y=z$$
Since $$z = 2x + 3y$$, you can substitute:
$$x +y = 2(2x + 3y)$$
$$x+y= 4x+ 6y$$
Therefore, you can arrive at a value for the ratio $$x:y$$:
$$-3x= 5y$$
$$\dfrac{-3x}{y}=\frac{5y}{y}$$ Divide by y.
$$\dfrac{-3x}{-3y}=\frac{5}{-3}$$
Divide by -3
$$\dfrac{x}{y}=\frac{5}{-3}$$
You can also substitute for $$x$$ to get a value for the ratio $$y:z$$:
$$x+y= 2z$$
$$x= 2z-Y$$
$$2x+ 3y= z$$
$$2(2z -y) + 3y = z$$
$$4z-2y+ 3y= z$$
$$y=-3z$$
$$\dfrac{y}{z}=-3$$
This tells you that $$x: y = -5/3$$, and $$y: z = -3/1$$. Both ratios contain a 3 for the y variable and both also contain a negative sign, so assign the value -3 to $$y$$. This means that x must be 5 and z must be 1. Therefore, the ratio $$x: y: z = 5 :-3 : 1$$.
You can test the result by choosing $$x = 5, y = -3$$, and $$z = 1$$,or $$x = 10,y = -6$$, and $$z = 2$$. In either case, the original equations hold up.
The correct answer is (C).
The ratio of three numbers is $$3:4:7$$ and their product is $$18144$$ the numbers are
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0%
$$9,\,12,\,21$$
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$$18,\,24,\,42$$
0%
$$12,\,16,\,28$$
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$$15,\,20,\,35$$
Explanation
Let the no.s be $$3x,\,4x,\,7x$$. Then,
$$3x\times4x\times7x=18144$$
$$\Longrightarrow\,x^3=216$$
$$\Longrightarrow x=6$$
The numbers are $$18,\,24,\,42$$
$$60kg$$ of alloy A is mixed with $$100kg$$ of alloy B. If alloy A has lead and tin in the ratio $$3:2$$, and alloy B has tin and copper in the ratio $$1:4$$, then amount of tin in the new alloy is
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$$36kg$$
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$$44kg$$
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$$53kg$$
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$$80kg$$
Explanation
Quantity of tin in $$60kg$$ of A $$=60\times \dfrac{2}{5}=24kg$$
Quantity of tin in $$100kg$$ of B $$=100\times \dfrac{1}{5}=20kg$$
Quantity of tin in new alloy $$=\begin{pmatrix}20+24\end{pmatrix}=44kg$$
$$Rs.432$$ is divided amongst three workers A, B and C such that $$8$$ times A"s share is equal to $$12$$ times B"s share which is equal to $$6$$ times C"s share. How much did A get?
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Rs. $$192$$
0%
Rs. $$138$$
0%
Rs. $$144$$
0%
Rs. $$196$$
Explanation
Times A"s share $$=12$$ times B"s share $$=6$$ times C"s share.
Note that this is not the same as the ratio of their wages being $$8:12:6$$
In this case, find out the L.C.M of $$8,\,12\;and\;6$$ and divide the L.C.M by each of the above numbers to get the ratio of their respective shares.
The L.C.M of $$8,\,12\;and\;6$$ and $$6$$ is $$24$$.
Therefore, the ratio $$A:B:C : : \dfrac{24}{8}:\dfrac{24}{12}:\dfrac{24}{6}$$
$$\Rightarrow A:B:C : : 3:2:4$$
The sum of the total wages $$=3x+2x+4x=432$$
$$\Longrightarrow 9x=432$$
or $$\Longrightarrow x=48$$.
Hence, A who gets $$3x$$ will get $$3\times48=Rs.144$$
Three friends Alice, Bond and Charlie divide Rs.$$1105$$ among them in such a way that if Rs.$$10$$, Rs.$$20$$ and Rs.$$15$$ are removed from the sums that Alice, Bond and Charlie received respectively, then the share of the sums that they got will be in the ratio of $$11:18:24$$. How much did charlie receive?
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Rs. $$495$$
0%
Rs. $$500$$
0%
Rs. $$425$$
0%
Rs. $$450$$
Explanation
Let the sums of money received by A, B and C be $$x$$, $$y$$ and $$z$$ respectively.
Then $$x-10:y-20:z-15$$ is $$11a:18a:24a$$
When Rs.$$10$$, Rs.$$20$$ and Rs. $$15$$ are removed, we are removing a total of Rs.$$45$$ from Rs.$$1105$$.
Therefore, $$11a+18a+24a=1105-45=1060$$
$$\Longrightarrow $$$$53a=1060$$
$$\Longrightarrow$$ $$a=20$$.
We know that $$z-15=24a=24\times20=480$$
Therefore, $$z=480+15=Rs.495$$
The ratio of two number is $$3:4$$ and their LCM is $$180$$. The first number is
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0%
$$60$$
0%
$$45$$
0%
$$20$$
0%
$$15$$
Explanation
Let the required no. be $$3x$$ and $$4x$$. Then their LCM is $$12x$$.
$$\Longrightarrow12x=180,$$
$$\Longrightarrow x=15$$. Hence the first no. is $$45$$
The proportion of milk and water in $$3$$ samples is $$2:1,\,3:2$$ and $$5:3$$. A mixture comprising of equal quantities of all $$3$$ samples is made. The proportion of milk and water in the mixture is
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$$227:133$$
0%
$$133:227$$
0%
$$99:61$$
0%
$$61:99$$
Explanation
Proportion of milk in $$3$$ samples is $$\dfrac{2}{3},\dfrac{3}{5},\,\dfrac{5}{8}$$
Proportion of water in $$3$$ samples is $$\dfrac{1}{3},\,\dfrac{2}{5},\,\dfrac{3}{8}$$
Total proportion in milk $$=\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{5}{8}=\dfrac{227}{120}$$
Total proportion in water $$=\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{3}{8}=\dfrac{133}{120}$$
Required proportion $$=227:133$$
If $$d> a$$ and $$L< a$$, which of the following cannot be true?
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$$d+L=14$$
0%
$$d-L=7$$
0%
$$d-L=1$$
0%
$$a-d=9$$
0%
$$a+d=9$$
Explanation
Simplify the inequalities, so that all the inequality symbols point in the same direction. Then line up the inequalities as shown. Finally, combine the inequalities:
$$L< a$$
$$a< d$$ $$\rightarrow$$ $$L< a< d$$
Since $$d$$ is a larger number than $$a, a-d$$ cannot be positive. therefore (D) cannot be true.
If $$\cfrac{AB}{7}> \cfrac{1}{14}$$ and $$A=B$$, which of the following must be greater than $$1$$?
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0%
$$A-2B$$
0%
$$1-A$$
0%
$$2{A}^{2}$$
0%
$${A}^{2}-\cfrac{1}{2}$$
0%
$$A$$
Explanation
$$\cfrac{AB}{7}> \cfrac{1}{14}$$
$$14AB> 7$$ Cross-multiply across the inequality
$$2AB> 1$$ Divide both sides by $$7$$
$$2{A}^{2}> 1$$ Since you know that $$A=B$$, then $$2AB=2{A}^{2}$$
Note that you need to get the expression $$> 1$$ on the right because the question asked what must be greater than $$1$$.
Is $$\cfrac{4}{x}< -\cfrac{1}{3}$$, what is the possible range of values for $$x$$?
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$$-10< x< 0$$.
0%
$$-12< x< 0$$.
0%
$$2< x< 0$$.
0%
$$0< x< 12$$.
Explanation
For this type problem, you have to consider two possibilities: $$x$$ could be positive or negative. When you multiply the inequality by $$x$$, you will need to flip the sign when $$x$$ is negative, but not flip the sign when $$x$$ is positive. However, notice that $$x$$ cannot be positive the left hand side of the inequality is less than $$-\cfrac{1}{3}$$, which means $$\cfrac{4}{3}$$ must be negative. There, $$x$$ must be negative.
Case 1: $$x> 0$$
Not possible
Case 2: $$x< 0$$
$$\cfrac{4}{3}< -\cfrac{1}{3}$$
$$12> -x$$ (flip the sign because $$x$$ is negative)
$$-12< x$$ (divide by $$-1$$, flip it again)
Case 1 is not possible
For Case 2, $$x$$ must be negative AND greater than $$-12$$. Thus, $$-12< x< 0$$.
Which of the following is equivalent to $$-3x+7\le 2x+32$$?
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$$x\ge -5$$
0%
$$x\ge 5$$
0%
$$x\le 5$$
0%
$$x\le -5$$
Explanation
$$-3x+\le 2x+32$$
$$-5x\le 25$$
$$x\ge -5$$
If $$a> 7, a+4> 13$$ and $$2a< 30$$, which of the following must be true?
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$$9< a< 15$$
0%
$$11< a< a15$$
0%
$$15< a< 20$$
0%
$$12< a< 15$$
Explanation
First solve the second and third inequalities. Simplify the inequalities, so that all the inequality symbols point in the same direction. Then, line up the inequalities as shown. finally combine the inequalities:
$$9< a$$
$$a< 15$$ $$\rightarrow$$ $$9< a< 15$$
$$7< a$$
Notice that all the wrong answers are more constrained: the low end is too high. the right answer will both keep out all impossible values of $$a$$ and let in all the possible values of $$a$$
A bookshelf holds both paperback and hardcover books. The ratio of paper back books to hardcover books is $$22$$ to $$3$$. How many paperback books are on the shelf ?
The number of books on the shelf is between $$202$$ and $$247$$, inclusive
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0%
$$205$$
0%
$$2305$$
0%
$$225$$
0%
$$198$$
Explanation
The question stem states that the ratio of paperback books to hardcover books is $$22$$ to $$3$$.Let the number of paperback and hardcover books be $$22x$$ and $$3x$$ respectively.
So, total number of books =$$22x + 3x = 25x$$
So, $$25x$$ lies between $$202$$ and $$247$$
So, the multiple of $$25$$ in between $$202$$ and $$247$$ is $$225$$.
So, $$25x = 225$$ $$\rightarrow$$ $$x = 9 $$
So, no. of paperback books = $$22x = 22 \times 9 = 198$$
Mark the correct alternative of the following.
Two numbers are in the ratio $$7 : 9$$. If the sum of the numbers is $$112$$, then the larger number is?
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0%
$$63$$
0%
$$42$$
0%
$$49$$
0%
$$72$$
Explanation
Given two numbers are in the ratio $$7:9$$.
Let the numbers be $$7x$$ and $$9x$$.
Then according to the problem we get,
$$7x+9x=112$$
or, $$16x=112$$
or, $$x=7$$.
So the largest number is $$7\times 9=63$$.
A bookshelf holds both paperback and hardcover books. The ratio of paper back books to hardcover books is $$22$$ to $$3$$. How many paperback books are on the shelf ?
If $$18$$ paperback books were removed from the shelf and replaced with $$18$$ hardcover books, the resulting ratio of paperback books to hardcover books on the shelf
would be $$4$$ to $$1$$
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0%
$$158$$
0%
$$200$$
0%
$$198$$
0%
$$164$$
Explanation
The question stem states that the ratio of paperback books to hardcover books initially is $$22$$ to $$3$$.
So, let us assume the no. of paperback books and hardcover books be $$22x$$ and $$3x$$ respectively.
Now, according to given conditions,
$$\dfrac{22x - 18}{ 3x+18 }= \dfrac{4}{1} $$
Cross-multiply both the sides,
$$22x -18= 4(3x+18)$$
$$22x -18= 12x+72$$
$$22x -12x= 72+18)$$
$$10x= 90$$
$$x= 9$$
So, the number of paperback books = $$22x$$ = $$22 \times 9$$ = $$198$$
Option C
Two numbers are in the ratio 4:If 10 is added to both the numbers, ratio becomes 2:Find the numbers.
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0%
$$\dfrac{-20}{3}$$ and $$\dfrac{-25}{3}$$
0%
$$\dfrac{-21}{3}$$ and $$\dfrac{-25}{3}$$
0%
$$\dfrac{-22}{3}$$ and $$\dfrac{-25}{3}$$
0%
$$\dfrac{-23}{3}$$ and $$\dfrac{-25}{3}$$
Explanation
Let us assume numbers are $$4x$$ and $$5x$$.
When 10 is added to both the numbers,
The number becomes $$(4x + 10)$$ and $$(5x + 10)$$
Since the new ratio is $$2:1$$
$$\dfrac{(4x+10)}{(5x+10)}=\dfrac{2}{1}$$
By cross multiplying,
$$4x + 10 = 2(5x + 10)$$
$$4x + 10 = 10x + 20$$
$$10x- 4x = 10- 20$$
$$6x = -10$$
$$x = \dfrac{-5}{3}$$
Therefore numbers are,
$$4x = 4 \times (\dfrac{-5}{3}) = \dfrac{- 20}{3}$$
$$5x = 5 \times (\dfrac{-5}{3}) = \dfrac{-25}{3}$$
The shaded region for the inequality $$x+5y\le 6$$ is
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0%
To the non-origin side of $$x+5y=6$$
0%
To the either side of $$x+5y=6$$
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To the origin side of $$x+5y=6$$
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To the neither side of $$x+5y=6$$
Explanation
R.E.F image
for,
$$ x+5y = 6 $$
$$ (x = 0, y = 6/5) $$
$$ (y = 0, x = 6) $$
but, $$ x+5y\leq 6 $$
$$ x+5y-6\leq 0 $$
means, -ve side of coordinate graph
$$ \therefore $$ to the origin side
Which equation has the solution shown on the number line?
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$$-3 > x > -1$$
0%
$$-5 < x < -2$$
0%
$$-2 > x > 0$$
0%
$$-1 > x > -6$$
Explanation
Since shaded region on the graph is from $$-5$$ to $$-2$$
So solution is $$-5<x<-2$$
Hence, option B is correct.
Given that $$x+5>1$$ and $$1-3x>-5$$ , $$x$$ cannot take the value given in one of the following options,which is :
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$$-6$$
0%
$$-3$$
0%
$$0$$
0%
$$\dfrac{1}{2}$$
Explanation
Given, $$x + 5 > 1 $$
$$\Rightarrow x > -4$$
Also given, $$1 - 3x > -5$$
$$\Rightarrow 3x < 6$$ or $$x < 2$$
$$\therefore x \in (-4,2)$$
Thus it cannot take the value of $$-6$$.
Eggs are sold in $$8$$ ounce and $$20$$ ounce boxes. If total weight of eggs sold was less than $$600$$ ounces and total of $$50$$ boxes were sold. Whats is the possible number of boxes sold that contain $$20$$ ounce of eggs?
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0%
$$33$$
0%
$$25$$
0%
$$17$$
0%
$$16$$
Explanation
Eggs are sold in $$8$$ ounce and
$$20$$
ounce boxes.
And total weight of egg sold is was $$600$$ ounces
And total box sold is $$50$$
Let $$x$$ boxes of $$8$$ ounces and $$y$$ boxes of $$20$$ onunces
$$\therefore x+y=50$$....(1)
$$8x+20y< 600$$.....(2)
Multiply (1) with $$8$$, we get
$$8x+8y =400$$......(3)
Subtract (2) by (3), we get
$$12y< 200$$
$$\Rightarrow y< 16.66$$
Says $$y=16$$
What is the closest ratio of $$x$$ to $$y$$ if $$x=0.63a$$ and $$y$$ is $$\dfrac{3}{8}$$ of $$a$$?
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$$0.236$$
0%
$$0.381$$
0%
$$0.595$$
0%
$$1.680$$
Explanation
Given $$x=0.63a$$ and $$y=\dfrac{3}{8}a=0.375a$$
Then ratio $$\dfrac{x}{y}=\dfrac{0.63}{0.375}=1.680$$
Which of the following statement is correct?
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Every LPP admits an optimal solution.
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A LPP admits a unique solution.
0%
If a LLP admits two optimal solutions, then it has an infinite number of optimal solution.
0%
A LPP admits two optimal solutions.
Explanation
If a LPP admits two optimal solutions, then it has an infinite number of optimal solutions.
If $$-1 < x < 0$$ and $$a$$ is a negative integer, then which of the following is NOT true
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$$\sqrt {ax} = ax$$
0%
$$\sqrt {ax} < ax$$
0%
$$|x^{3}| > |x|$$
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$$x^{2}$$ and $$x$$ have opposite signs
Explanation
$$-1<x<0$$. Hence x is negative. Thus $$\sqrt{ax}=ax$$ is true only if $$ax=1$$ or either $$x,a=0$$. Now $$ax\geq \sqrt{ax}$$ will always be true since both $$a$$ and $$x$$ are negative numbers. Now we know that $$|x|<1$$. Hence $$|x^{3}|<|x|$$. Therefore, option C is not true.
Given that $$-7<-3x+5 \le 20$$, the
least possible value which the expression $$6x+1$$ can take is:
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0%
$$-28$$
0%
$$29$$
0%
$$25$$
0%
$$-29$$
Explanation
Given that $$-7 < -3x + 5 \leq 20$$
$$\Rightarrow -7 - 5 < -3x \leq 20 - 5$$
$$\Rightarrow -12 < -3x \leq 15$$
$$\Rightarrow 12 > 3x \geq -15$$
$$\Rightarrow 4 > x \geq -5$$
Since we need the least possible value of $$6x + 1, x$$ must be substituted as $$-5$$, thus giving the value to be $$-30 + 1 = -29$$
If $$xy > 0$$, which of the following has to be true?
I. $$x+y=0$$
II. $$2y-2x < 0$$
III. $${x}^{2}+{y}^{2} > 0$$
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0%
I only
0%
III only
0%
I and III
0%
II and III
Explanation
Given, $$xy \gt 0 \implies x\gt0$$ and $$y\gt 0$$ or $$x< 0$$ and $$y< 0$$
Adding both
$$x\gt 0$$ and
$$y\gt 0$$,
we get
$$x+y >0$$ or $$x+y<0$$
$$\mathrm I.$$
Hence, $$x+y = 0$$ is false
Subtracting $$x$$ from $$y$$ we get,
$$y-x>0$$ or
$$y-x=0$$ or
$$y-x<0$$
$$\mathrm{II}.$$
Hence, $$2y-2x < 0$$ is not always true.
Squaring both, we get
$$x^{2}\gt 0$$ and
$$y^{2}\gt 0$$
Adding both
$$x^2 \gt 0$$ and
$$y^2 \gt 0$$
we get
$$x^2+y^2 >0$$
$$\mathrm{III}.$$
Hence,
$$x^2+y^2 >0$$ is true
The product of $$n$$ positive numbers is $$n^n$$, then their sum is:
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0%
A positive integer
0%
Divisible by $$n$$
0%
Equal to $$n+\dfrac 1n$$
0%
Never less than $$n^2$$
Explanation
We know that $$A.M. \geq G.M.$$
A.M. of $$n$$ positive numbers is $$\dfrac{1+2+3+...+n}{n}$$
G.M. of $$n$$ positive numbers is $$\sqrt[n]{1*2*3*...*n}$$
Given that product of $$n$$ positive numbers is $$n^n$$
Therefore, G.M. is $$\sqrt[n]{n^n}=n$$
Hence, $$\dfrac{1+2+3+...+n}{n}\geq n$$
$$\implies 1+2+3+...+n\geq n^2$$
Therefore, the sum of n positive numbers is never less than $$n^2$$
A burger shop's sells two type of burger, A and B. The selling price of burger A is $$\$ 17,$$ and of burger B is $$\$ 13$$. Ingredient costs for burger A are $$\$450$$ per week, and for burger B are $$\$ 310$$ per week. Assuming the shop sells an equal number of both burgers in one week, at what point will profits for one burger overtake the other?
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0%
After $$35$$ burgers each, burger B profits will overtake burger A profits
0%
After $$145$$
burgers each, burger B profits will overtake burger A profits
0%
After $$35$$
burgers each, burger A profits will overtake burger B profits
0%
After $$145$$
burgers each, burger A profits will overtake burger B profits
Explanation
Let the number of burgers sold in a week be $$2x$$.
Hence, the number '$$A$$" burgers sold is equal to the number of '$$B$$' burgers sold which is in turn equal to $$x$$.
Now the profit for $$A$$ will be $$17x-450$$ and the profit for $$B$$ will be $$13x-310$$.
Assuming burger $$A$$'s profit overtakes $$B$$'s then $$17x-450>13x-310$$ or $$4x>140$$ or $$x>35$$.
Hence, after $$35$$ burgers each, burger $$A$$ profits will overtake burger $$B$$ profits.
Find the condition on $$x$$ given that $$-5x+15 \geq 35$$.
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0%
$$x \leq -4$$
0%
$$x \leq 8$$
0%
$$x \geq -4$$
0%
$$x \geq -8$$
Explanation
Given, $$-5x+15\geq 35$$
$$\Rightarrow- 5x\geq 35-15$$
$$\Rightarrow-5x\geq 20$$
$$\Rightarrow x\leq -4$$
$$ \begin{cases} x+3y\le 18\\ 2x-3y \le 9 \end{cases}$$
If $$(a,b)$$ is the solution of the above system of linear equation and $$a=6$$, find the minimum possible value of $$b$$.
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0%
$$1$$
0%
$$-1$$
0%
$$4$$
0%
$$-5$$
Explanation
$$(6,b)$$ is the solution to the given inequalities.
$$x + 3y \leq 18$$
$$2x - 3y \leq 9$$
When substituted $$x = 6$$, we have
$$6 + 3y \leq 18$$ or $$y \leq 4$$
The other becomes $$12 - 3y \leq 9$$ or $$y \geq 1$$
Thus, the range of $$y$$ becomes $$[1,4]$$
Minimum possible $$y$$ therefore becomes $$1$$.
A dietician allocates Seema $$1600$$ calories per day. Half of the calories come from carbohydates and $$1/4$$th of these calories come from fats. Calculate the difference in number of carbohydrate calories to fat calories, Seema is allocated per day.
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0%
$$150$$
0%
$$250$$
0%
$$400$$
0%
$$800$$
Explanation
Total calories allocated to Seema per day $$=1600$$.
Fats allocated per day $$=\dfrac{1}{4} \times 1600 = 400$$
Carbohydrates allocated per day $$=\dfrac{1}{2} \times 1600 = 800$$
Therefore, the difference is $$400$$ calories.
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