MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 2

${G}^{2}< G$ , which of the following could be

$G$ ?

0%
$1$
0%
$\cfrac{23}{7}$
0%
$\cfrac{7}{23}$
0%
$-4$
0%
$-2$

Explanation

${G}^{2}< G$ , then

$G$ must be positive (since

${G}^{2}$ will never be negative), and

$G$ must be less than

$1$ , because otherwise

${G}^{2}> G$ . Thus,

$0< G< 1$ . You can eliminate (D) and (E) since they violate the condition that

$G$ be positive. Then test (A):

$1$ is not less than

$1$ , so you can eliminate (A). (B) is larger than

$1$ , so only (C) satisfies the inequality.

$4x-12\ge x+9$ , which of the following must be true?

0%
$x>6$
0%
$x< 7$
0%
$x> 7$
0%
$x>8$
0%
$x< 8$

Explanation

$4x-12\ge x+9$

$3x\ge 21$

$x\ge 7$

You were asked to pick the answer that must be true. If

$x$ greater than or equal to

$7$ , then

$x$ could be

$7, 7.3, 8, 9.2$ and so on. Which of the five answers contains an expression that covers all possible values of

$x$ ? Most people will immediately look answer (C)

$x> 7$ , but be careful! Does

$x$ have to be greater cover some of the possible values for

$x$ , but not all of then. Answer (B) does not share anything in common with

$x> 7$ , so its wrong. You are left with answer (A). Why must it be true that

$x$ is greater than

$6$ . The logic here is very similar to that of Data Sufficiency: if

$x\ge 7$ were a statement, it would be sufficient to establish that

$x> 6$

Which equation has the solution shown on the number line?

0%
$1\leq x < 4$ .
0%
$x < 1$
0%
$x\neq 1$
0%
$x < 0$

Explanation

The red line represents all the numbers that satisfy the equation

$1\leq x < 4$ .

Two numbers are in the ratio

$3:4$ . If the sum of numbers is

$63$ , find the numbers.

0%
$27\;\&\;36$
0%
$27\;\&\;72$
0%
$54\;\&\;36$
0%
$9\;\&\;27$

Explanation

Sum of the terms of the ratio

$=3+4=7$
Sum of numbers

$=63$
Therefore, first number

$=\dfrac{3}{7}\times63=27$
Second number

$=\dfrac{4}{7}\times63=36$
Therefore, the two numbers are

$27\;\&\;36$

$3x \ge 8$ has solution in

0%
Ist and IInd quadrant
0%
IIIrd and IVth quadrant
0%
IInd and IIIrd quadrant
0%
Ist and IVth quadrant

Explanation

From the figure, it is clear that

$3x\ge 8$ has solution in Ist and IVth quadrant.
1046978_460457_ans_60f217bce6324d70b7a9a23ad4513821.jpg

State whether True or False

$5B> 4B+1$ , Then,

${B}^{2}> 1$ ?

0%
True
0%
False

Explanation

$5B> 4B+1$

$B> 1$

The squares of all numbers greater than

$1$ are also greater than

$1$ , so

${B}^{2}> 1$

Which equation has the solution shown on the number line?

0%
$x \geq 1$
0%
$x \geq -6$
0%
$x\neq 1$
0%
$x < 0$

Explanation

The red line represents all the numbers that satisfy the equation

$x \geq -6$ .

Which equation has the solution shown on the number line?

0%
$x \geq 1$
0%
$x \geq -6$
0%
$x\leq 6$
0%
$x < 0$

Explanation

The red line represents all the numbers that satisfy the equation

$x \leq 6$ .

Which equation has the solution shown on the number line?

0%
$-3\leq x < 0$ .
0%
$3\leq x < 6$ .
0%
$x\neq 1$
0%
$x < 0$

Explanation

The red line represents all the numbers that satisfy the equation

$-3\leq x < 0$ .

If a

$:$ b

$=2:5$ , then find

$(3a+4b):(5a+6b)$ .

0%
$\displaystyle\frac{11}{20}$
0%
$\displaystyle\frac{13}{20}$
0%
$\displaystyle\frac{1}{20}$
0%
$\displaystyle\frac{3}{20}$

Explanation

$\dfrac{a}{b} = \dfrac{2}{5}\Rightarrow b = (5/2^{a}) \therefore \dfrac{3a+4b}{5a+6b} = \dfrac{3a+4\times \dfrac{5}{2}b}{5a+6\times \dfrac{5}{2}b}$

$\Rightarrow \dfrac{3a+4b}{5a+6b} = \dfrac{13a}{20a} = \boxed {\dfrac{13}{20}}$

1103271_515469_ans_179475ba4f664722a5d1c9b5c83453e2.jpg

$|x-2|<3$ , then '

$x$ ' lies between

0%
$1$ and $5$
0%
$-1$ and $5$
0%
$-1$ and $1$
0%
none

Explanation

$|x-2|<3$

$\Rightarrow -3<(x-2)<3$

$\Rightarrow \boxed {-1<x<5}$

1146663_515277_ans_cb9de85bdd924fb387ab19bf74ee5f94.jpg

The half planes contains the corresponding point:

$7x\geq 2y+x$ , point

$(1, 0)$ .

0%
True
0%
False

Explanation

$7x\geq 2y+x\Rightarrow 6x-2y\geq 0$

put

$(1,0)\Rightarrow 6\geq 0$ hence True

1106603_515103_ans_7fab1759aa6c4fb7a160b50eac84bade.jpg

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

$2x+1 \ge 9$ ?

Explanation

Given,

$2x+1 \geq 9$

$\Rightarrow 2x \geq 9-1$

$\Rightarrow 2x\geq 8$

$\Rightarrow x \geq 4$

$\therefore$ Solution

$=$

$x \geq 4$

Hence, option

$A$ is the answer.

A and B together have Rs.

$1210$ . If

$\dfrac{4}{15}$ of A's amount is equal to

$\dfrac{2}{5}$ of B's amount, how much amount does B have?

0%
Rs. $460$
0%
Rs. $484$
0%
Rs. $550$
0%
Rs. $664$

Explanation

Given,

$\cfrac{4}{15}A$

$=\cfrac{2}{5}B$

$\Rightarrow A = \left ( \cfrac{2}{5}\times \cfrac{15}{4} \right ) B$

$\Rightarrow A = \cfrac{3}{2} B$

$\Rightarrow \cfrac{A}{B} = \cfrac{3}{2}$

$\Rightarrow A : B = 3 : 2$

$\therefore$ B's share = Rs.

$\left ( 1210\times \cfrac{2}{5} \right )$

$= Rs.\, 484$

If a

$:$ b

$=2:5$ , then find the value of

$(2a+3b):(7a+5b)$ .

0%
$\displaystyle\frac{19}{39}$
0%
$\displaystyle\frac{1}{39}$
0%
$\displaystyle\frac{11}{39}$
0%
$\displaystyle\frac{19}{37}$

Explanation

$a< b$ As

$C$ is

$+ve$

$\therefore \boxed {ac<bc}$ & lly

$\boxed {\dfrac{a}{c}<\dfrac{b}{c}}$

$\dfrac{a}{b} = \dfrac{2}{5}\Rightarrow \boxed {b = \dfrac{1}{2}a}$

than

$\dfrac{2a+3b}{7a+5b} = \dfrac{2a+3\times \dfrac{5}{2}a}{7a+5\times \dfrac{5}{2}a} = \dfrac{19a}{39a} = \boxed {\dfrac{19}{39}}$

1106632_515393_ans_9aef047fda0242b69ed1c816dbb3797f.jpg

The following half planes contains origin

$(0, 0)$

$2x+5y\geq x+12$ .

0%
True
0%
False

Explanation

$2x+5y\geq x+12$

$\Rightarrow x+5y-12\geq 0$

(Put)

$(0,0)$ in the

$eq^{n}$

$\Rightarrow -12 \geq 0 \therefore \boxed {False}$

1106597_515102_ans_a01b7e8363cd4aafa312d86518494b03.jpg

Of the following sets, the one that includes all values of x which will satisfy

$2x - 3 > 7 - x$ is:

0%
$x > 4$
0%
$x < 10/3$
0%
$x = 10/3$
0%
$x > 10/3$
0%
$x < 0$

Explanation

$2x - 3 > 7 - x$ . Adding

$x + 3$ to both sides of the inequality, we have

$3x > 10 or x > \frac{10}{3}$ .

Solve:

$\displaystyle\frac{3x+2}{2x+3}\geq 4$ .

0%
$x\epsilon[-2, -3/2]$
0%
$x\epsilon[-2, -3]$
0%
$x\epsilon[-1, -3/2]$
0%
$x\epsilon[0, -3/2]$

Explanation

$\dfrac{3x+2}{2x+3}\geq 4$

$\Rightarrow 2x+3\neq 0\Rightarrow \boxed {x \neq \dfrac{-3}{2}}$

$(3x+2)\geq 4(2x+3)[2x+3>0]$

$\Rightarrow (3x+2)\geq 8x+12$

$\Rightarrow 5x+10\leq 0$

$\Rightarrow x\leq -2$

$(2x+3)>0\Rightarrow \boxed {x>\dfrac{-3}{2}}$

$\therefore \boxed {x\epsilon [-2,-3/2)}$

1146662_515275_ans_6c9e0d8104914e30a00f59ab9db9a284.jpg

Given

$k:p$ is equal to

$1:3$ and

$p:a$ is equal to

$2:3$ , find the ratio of

$k:a$ ?

0%
$2:9$
0%
$1:3$
0%
$1:2$
0%
$2:3$

Explanation

Given,

$k:p=1:3$

and

$p:a=2:3$

Multiply

$k:p$ by

$2$ and

$p:a$ by

$3$ , we get-

$\Rightarrow k:p=2:6$

$\Rightarrow p:a=6:9$

$\Rightarrow k:p:a=2:6:9$

$\therefore k:a=2:9$

$x=$ ............... is identified as a GOLDEN RATIO

0%
$\cfrac { 1+\sqrt { 5 } }{ 2 }$
0%
$0$
0%
$\cfrac { 1+\sqrt { 2 } }{ 2 }$
0%
$1$

Explanation

Two quantities are said to be in golden ratio if their ratio is same as the ratio of their sum to the larger of the two quantities.

Let us take two numbers

$a$ and

$b$ where

$a > b$

As per the definition,

$\cfrac{a}{b} = \cfrac{a + b}{a}$

$\Rightarrow a \times a = b \times (a + b)$

$\Rightarrow a^2 = ab + b^2$

$\therefore a^2 - ab - b^2 = 0$

Assuming a quadratic equation in

$a$ , we get

$a = \cfrac{b \pm \sqrt{b^2 + 4b^2}}{2}$

$\therefore a = \cfrac{b + \sqrt{5}b}{2}$

$\Rightarrow \cfrac{a}{b} = \cfrac{1 + \sqrt{5}}{2}$

This is the ratio of the numbers we had taken and is defined as the golden ratio.

$(2x - 3y < 7)$ and

$(x + 6y < 11)$ , then which one of the following is correct?

0%
$x + y < 5$
0%
$x + y < 6$
0%
$x + y \leq 5$
0%
$x + y \leq 6$

Explanation

$2x - 3y < 7$ and

$x + 6y < 11$ ; adding

$\Rightarrow 3x + 3y < 18$ or

$x + y < 6$ .

A bag contains one rupee

$50$ paise and

$25$ paise coins in the ratio

$5 : 6 : 8$ . If the total amount is Rs.

$420$ , find the total number of coins.

0%
$798$
0%
$789$
0%
$978$
0%
$987$

Explanation

$\Rightarrow$ The ratio of

$Rs.1$ ,

$50$ paise and

$25$ paise coins is

$5:6:8$

$\therefore$ Number of

$Rs.1$ coins =

$5x$ .

$\Rightarrow$ Number of

$50\,paise$ coins =

$6x$

$\Rightarrow$ Number of

$25\,paise$ coins =

$8x$ .

$\therefore$ Value of

$5x$

$Rs.1$ coins =

$Rs.1\times 5x=Rs.5x$ .

$\Rightarrow$ Value of

$6x$

$50\,paise$ coins =

$6x\times 50\,paise=Rs.\dfrac{6x\times 50}{100}=Rs.3x$

$\Rightarrow$ Value of

$8x$

$25\,paise$ coins =

$8x\times 25\,paise=Rs.\dfrac{8\times 25}{100}=Rs.2x$

$\Rightarrow$ Total amount =

$Rs.420$

$\therefore$

$5x+3x+2x=420$

$\therefore$

$10x=420$

$\therefore$

$x=\dfrac{420}{10}=42$

$\therefore$ Number of

$Rs.1$ coins =

$5\times 42=210$

$\therefore$ Number of

$50\,paise$ coins =

$6\times 42=252$

$\therefore$ Number of

$25\,paise$ coins =

$8\times 42=336$

$\Rightarrow$ Total number of coins =

$210+252+336=798$

The ratio of the number of boys and girls in a college is

$7 : 8$ . If the percentage increase in the number of boys and girls be

$20\%$ and

$10\%$ respectively, what will be the new ratio?

0%
$8 : 9$
0%
$17 : 18$
0%
$21 : 22$
0%
Cannot be determined

Explanation

Originally, let the number of boys and girls in the college be

$7x$ and

$8x$ respectively.

The percentage increase in the number of boys and girls is

$20\%$ and

$10\%$ respectively,

Their increased number is (

$120\%$ of

$7x$ ) and (

$110\%$ of

$8x$ ).

$\left ( \dfrac{120}{100}\times 7x \right )$ and

$\left ( \dfrac{110}{100}\times 8x \right )$

$\dfrac{42x}{5}$ and

$\dfrac {44x} {5}$

$\therefore$ The required ratio =

$\left ( \dfrac{42x}{5}: \dfrac{44x}{5}\right )$

$= 21:22$

$x : y = 2 : 3$ and

$2 : x = 1 : 2$ , then the value of

$y$ is _____.

0%
$\dfrac {1}{3}$
0%
$\dfrac {3}{2}$
0%
$6$
0%
$\dfrac{1}{2}$

Explanation

Given,

$\dfrac{2}{x}=\dfrac{1}{2}$

So,

$x=4$

Also given

$\dfrac{x}{y}=\dfrac{2}{3}$

$\Rightarrow 3\times x=2\times y$

$\Rightarrow 3\times (4)=2\times y$

$\Rightarrow y=6$

$x:0.75::8:5$ .Find

$x$

0%
$1.12$
0%
$1.2$
0%
$1.25$
0%
$1.30$

Explanation

$5\times x = (0.75 \times 8)$ =>

$x=$

$\left ( \dfrac{6}{5} \right )$

$= 1.20$

The price of

$98$ notebooks is Rs.

$5551.70$ . What is the cost of

$1$ such notebook?

0%
Rs. $42.85$
0%
Rs. $39.05$
0%
Rs. $62.08$
0%
Rs. $56.65$

Explanation

Cost of

$98$ notebooks

$=Rs. 5551.70$

$\therefore$ Cost of

$1$ notebook

$=Rs. (5551.70\div 98)$

$=Rs. 56.65$ .

Shifting one term from one side of an equation to the another side with a change of sign in known as _____

0%
Commutativity
0%
Transposition
0%
Distributivity
0%
Associativity

Explanation

$Transposition$ is the process of moving a term from one side of an equation to the another side by changing it's sign without affecting the equation.

eg: The equation

$x=y+5$ can be written as

$x-5=y$ by transposition.

So, the answer is

$[B]$ .

Five times

$A's$ money added to

$B's$ money is more than Rs.

$51.00$ . Three times

$A's$ money minus

$B's$ money is Rs.

$21.00$ . If

$a$ represents

$A's$ money in dollars and

$b$ represents

$B's$ money in rupees, then

0%
$a > 9, b > 6$
0%
$a > 9, b < 6$
0%
$a > 9, b = 6$
0%
$a > 9$ , but we can put no bounds on $b$
0%
$2a = 3b$

Explanation

$5a+b>51$

$3a-b=21$

$b=3a-21$

$5a+b>51$

$5a+3a-21>51$

$8a>72$

$a>9$

$when\quad a>9$

$3a-21>(3\times 9)-21$

$b>6$

$x$ satisfies

0%
$0\le x\le 4$
0%
$x\le 0$ or $x\ge 4$
0%
$x\le -2$ or $x\ge 4$
0%
$x\ge -2$ or $x\le 4$

Explanation

$(i)x>3$

$\quad x-1+x-2+x-3\ge 6$

$\quad 3x\ge 12$

$\quad x\ge 4\quad \quad ,\quad x\quad \varepsilon [4,\infty )\quad -(1)$

$(ii)2<x<3$

$\quad \quad x-1+x-2-x+x\ge 6$

$\quad \quad x\ge 6\quad \quad but\quad (2<x<3),\quad x\quad \varepsilon \{ \quad \} \quad -(2)$

$(iii)1<x<2$

$\quad \quad x-1-x+2-x+3\ge 6$

$\quad \quad 4-x\ge 6$

$\quad \quad x\le -2\quad but\quad (1<x<2),\quad x\quad \varepsilon \{ \quad \} \quad -(3)$

$(iv)x<1$

$\quad \quad -x+1-x+2-x+3\ge 6$

$\quad \quad -3x\ge 0$

$\quad \quad \quad x\le 0\quad \quad ,x\quad \varepsilon (-\infty ,0]\quad -(4)$

$from\quad 1,2,3\quad \& \quad 4$

$x\quad \varepsilon (-\infty ,0]\bigcup [4,\infty )$

$\quad$

Solution set of the inequality

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

0%
$(1, \infty)$
0%
$(0, \log_2(4/3))$
0%
$( -1, \infty)$
0%
$(0, 1)$

Explanation

$\cfrac { 1 }{ { 2 }^{ x }-1 } >\cfrac { 1 }{ 1-\cfrac { { 2 }^{ x } }{ 2 } }$

$\cfrac { 1 }{ { 2 }^{ x }-1 } -\cfrac { 1 }{ 1-\cfrac { 2^{ x } }{ 2 } } >0$

$let\quad { 2 }^{ x }=t$

$\cfrac { 1 }{ t-1 } -\cfrac { 2 }{ 2-t } >0$

$\cfrac { 1 }{ t-1 } +\cfrac { 2 }{ t-2 } >0$

$\cfrac { t-2+2t-2 }{ (t-1)(t-2) } >0$

$\cfrac { (t-\cfrac { 4 }{ 3 } ) }{ (t-1)(t-2) } >0$

$t\quad \varepsilon (1,\cfrac { 4 }{ 3 } )\bigcup (2,\infty )$

${ 2 }^{ x }\quad \varepsilon (1,\cfrac { 4 }{ 3 } )\bigcup (2,\infty )$

$x\log _{ 2 }{ 2 } \quad \varepsilon (0,\log _{ 2 }{ \cfrac { 4 }{ 3 } } )$

$x\quad \varepsilon (0,\log _{ 2 }{ \cfrac { 4 }{ 3 } } )\quad$

996973_1046373_ans_333eecff9e6c4976be4a762a44283652.png

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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