MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 1

$5$ mangoes and

$4$ oranges cost as much as

$3$ mangoes and

$7$ oranges. What is the ratio of the cost of one mango to that of one orange?

0%
$\;4\,\colon\,7$
0%
$\;7\,\colon\,4$
0%
$\;3\,\colon\,2$
0%
$\;2\,\colon\,3$

Explanation

Let the cost of one mango be

$x$ paise and that of one orange be

$y$ paise.

$\;\;\;\;\;\;\;\;\;$ Then,

$5x+4y=3x+7y$

$\;\;\;\;\;\;\;\;\;\Rightarrow\,2x=3y\;\;\Rightarrow\;\;\displaystyle\frac{x}{y}=\displaystyle\frac{3}{2}$

$\;\;\;\;\;\;\;\;\;\Rightarrow\,x\,\colon\,y=3\,\colon\,2$ .

State whether True or False

The ratio of the following.

$5$ m to

$10$ km ratio is

$1:2000$

0%
True
0%
False
0%
Ambiguous
0%
Cannot be determined

Explanation

Required Ratio is

$=\displaystyle \frac { 5 m }{ 10 km } =\frac { 5 m }{ 10000 m } =\frac { 1 }{ 2000 }=1:2000$

State whether the given statement is True or False.

The ratio of 50 paise to Rs. 5 ratio is 1: 10

0%
True
0%
False
0%
Ambiguous
0%
Cannot be determined

Explanation

Required Ratio is

$=\displaystyle \frac { 50p }{ Rs5 } =\frac { 50p }{ 500p } =\frac { 1 }{ 10 }=1:10$

Ordered pair that satisfy the equation

$x + y + 1 < 0$ is:

0%
$(0, -1)$
0%
$(-2 , 0)$
0%
$(2, - 4)$
0%
Both (B) and (C)

Explanation

Given inequation is

$x+y+1<0$

From option A,

$0+\left ( -1 \right )+1 <0$

$\Rightarrow 0<0$ which is false
Hence,

$(0,-1)$ is not a solution.

From option B,

$-2+0+1 <0$

$\Rightarrow -1<0$ which is true
Hence,

$(-2,0)$ there is a solution

From option C,

$2-4+1 <0$

$\Rightarrow -1<0$ which is true.
Hence,

$(2,-4)$ is a solution
Hence, Option B and C are the solutions.

What must be subtracted from each term of the ratio

$4 : 7$ so that the ratio becomes

$2 : 5$ ?

0%
$-2$
0%
$-1$
0%
$2$
0%
$1$

Explanation

$\displaystyle \frac{4-x}{7-x}=\frac{2}{5}$

$20 - 5x = 14 - 2x$

$3x = 6$

$x = 2$

The ratio of Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour is 1:2

0%
True
0%
False
0%
Ambiguous
0%
Cannot be determined

Explanation

Required Ratio is

$=\displaystyle \frac { 15 }{ 30 } =\frac { 1 }{ 2 } =1:2$

Find the order relation between the following pairs of ratios. a, b are integers.

$\displaystyle \frac{a-1}{b-1}, \frac{a+1}{b+1}$ where

$a > b$ and

$b \neq \pm 1$ .

0%
$\displaystyle \frac{a-1}{b-1} > \frac{a+1}{b+1}$
0%
$\displaystyle \frac{a-1}{b-1} < \frac{a+1}{b+1}$
0%
Cannot be determined
0%
None of these

Explanation

Now, $a>b$

$=>\dfrac{a}{b}>1$

$\therefore \dfrac{a}{b}>\dfrac{a+1}{b+1}$ and

$\dfrac{a-1}{b-1}>\dfrac{a}{b}$

Thus, $\dfrac{a-1}{b-1}>\dfrac{a+1}{b+1}$

The ratio of the two areas of two triangles with the common base is

$6:5$ . Height of the larger triangle is

$9\, cm$ . Then find the corresponding height of the smaller triangle.

0%
Height of the smaller triangle is $7.5\, \, cm$ .
0%
Height of the smaller triangle is $6.5\, \, cm$ .
0%
Height of the smaller triangle is $5.5\, \, cm$ .
0%
Height of the smaller triangle is $4.5\, \, cm$ .

Explanation

Base of two triangles is the same. Height of first triangle is 9 cm.

$\frac{Area {\triangle}_1}{Area {\triangle}_2} = \frac{\frac{1}{2} b \times h_1}{\frac{1}{2} b \times h_1}$

$\frac{Area {\triangle}_1}{Area {\triangle}_2} = \frac{h_1}{h_2}$

$\frac{6}{5} = \frac{9}{h_2}$

$h_2 = \frac{9 \times 5}{6}$

$h_2 = 7.5$ cm

Solve the inequality:

$3-2x\leq 9$

0%
$x\leq 8$
0%
$x\geq -6$
0%
$x\leq -3$
0%
$x\geq -3$

Explanation

$3-2x\leq 9$

Add -3 both sides we get

$3-2x-3\leq 9-3$

$\Rightarrow -2x\leq 6$

Divided by -2 both sides we get

$x\geq -3$

The ratio of the velocities of the hour hand and minute hand of a clock is:

0%
$\;1\,\colon\,12$
0%
$\;1\,\colon\,1$
0%
$\;5\,\colon\,1$
0%
$\;12\,\colon\,1$

Explanation

The minute hand of a clock covers

$360$ deg in

$60$ min or

$6$ deg/min.

The hour hand of a clock covers

$30$ deg in

$60$ min or

$0.5$ deg/min.

The ratio of the angular speed of the minute to the hour hand

$= \dfrac{0.5}6 = 1:12.$

Solve the inequality:

$\displaystyle 4\left (3-x \right )\leq 16$

0%
$x\ge -7\quad$
0%
$x\ge -1\quad$
0%
$\ x\le -1\quad$
0%
None of these

Explanation

The given inequation is

$\quad 4(3-x)\le 16\\ \Longrightarrow (3-x)\le 4\\ \Longrightarrow -x\le 4-3\\ \Longrightarrow -x\le 1\\ \Longrightarrow x\ge -1\quad$

Ans- option B.

$47.2506$

$\displaystyle =4A+\frac{7}{B}+2C+\frac{5}{D}+6E,$ then the value of

$5A + 3B + 6C + D + 3E$ is:

0%
$63.6003$
0%
$53.603$
0%
$153.6003$
0%
$213.0003$

Explanation

Given

$47.2506=4A+\frac{7}{B}+2C+\frac{5}{D}+6E..............(1)$

Now write

$47.2506=40+7+0.2+0.05+0.0006...........................(2)$

comparing (1) and (2) we get

$4A=10\Rightarrow A=10$

$\frac{7}{B}=7\Rightarrow B=1\ and\ \ 2C=0.2\Rightarrow C=0.1$

$\frac{5}{D}=0.05\Rightarrow D=\frac{5}{0.05}$

$=100$

$6E=0.0006\Rightarrow E=0.0001$

So velue of

$5A+3B+6C+D+3E$ =

$(5\times 10)+(3\times 1)+(6\times 0.1)+100+(3\times 0.0001)$

$=50+3+0.6+100+0.0003=153.6003$

The sum of the squares of

$2$ numbers is

$146$ and the square root of one of them is

$\displaystyle \sqrt {5}$ . The cube of the other number is

0%
$1111$
0%
$1221$
0%
$1331$
0%
$1441$

Explanation

Let the two numbers be x and y

$\displaystyle x^{2}+y^{2}=146\: \: \: and\: \: \: \sqrt{x}=\sqrt{5}\Rightarrow x=5$

$\displaystyle \therefore 5^{2}+y^{2}=146\Rightarrow 25+y^{2}=146\Rightarrow y^{2}=121$

$\displaystyle \Rightarrow y=\sqrt{121}=11$

$\displaystyle \therefore \left ( y \right )^{3}=\left ( 11 \right )^{3}=1331$

Solve the given inequality:

$−5≤3x−2<7$ , where

$x\epsilon R$

0%
$-1 ≤ x <3$
0%
$-1 ≤ x <2$
0%
$-2 ≤ x <3$
0%
None

Explanation

The given inquations are

$-5\le 3x-2<7..........(ii)\quad$

$\quad (i)\longrightarrow -5\le 3x-2\\ \Longrightarrow -5+2\le 3x\\ \Longrightarrow -3\le 3x\\ \Longrightarrow -1\le x.\quad$

Again

$\quad (ii)\longrightarrow 3x-2<7\\ \Longrightarrow 3x<7+2\\ \Longrightarrow 3x<9\\ \Longrightarrow x<3.\quad$ \\

$\quad \therefore \quad -1\le x<3.\quad$

The solution of

${\left | 2x-3 \right |<5}$ is

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

Explanation

$|2x-3|<5$

$\Rightarrow -5<2x-3<5$

$\Rightarrow -5+3<2x<5+3$

$\Rightarrow -2<2x<8$

$\Rightarrow -1<x<4\Rightarrow x\in (-1,4)$

A boat takes

$2\ hours$ to travel

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

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

$8\ km$ and coming back is

0%
$160\ minutes$
0%
$80\ minutes$
0%
$320\ minutes$
0%
$180\ minutes$

Explanation

Speed of boat relative to water

$= 8\ km/h$
Speed downstream

$= 12\ km/h$
Speed upstream

$= 4\ km/h$
Time taken with water flow

$= \dfrac{8}{12} + \dfrac{8}{4} hrs = 2+ \dfrac{2}{3}\ hrs = 160\ min$

In a lottery a total of

$200$ prizes are to be given. A prize is either Rs

$500$ or Rs

$100$ . If the total prize money is Rs

$50000$ , then the number of Rs

$500$ and Rs

$100$ prizes are?

0%
$70, 130$
0%
$75, 125$
0%
$60, 140$
0%
$80, 120$

Explanation

Let

$x$ =Number of Rs

$500$ prizes given

$y$ =number of Rs

$100$ prizes given

Since it is given that in that lottery the total of Rs 200 prizes are to be given.

$\Rightarrow$

$x+y=200$

$\rightarrow$ 1

Since the total prize money 50000

$\Rightarrow 500x+100y=50000$

$\rightarrow$ 2

Solving equation 1 and 2 simultaneously

$\dfrac { x+y=200\\ 5x+y= 500 }{ -4x=-300 }$

$x=75$

$x+y=200\\ \Rightarrow y=125$

$X$ is

$36$ years old and

$Y$ is

$16$ years old. In how many years will

$X$ be twice as old as

$Y$ ?

0%
$1$ year
0%
$2$ years
0%
$3$ years
0%
$4$ years

Explanation

Let

$X$ be twice as old as

$Y$ after

$x$ years

Then after

$x$ years

$X$ 's age

$= (36+ x)$ yrs and

$Y$ 's age

$= (16+x)$ yrs
Given

$36 + x = 2(16 + x)$

$\displaystyle \Rightarrow$

$36 + x = 32 + 2x$

$\displaystyle \Rightarrow x=4$ years

$\dfrac{-3}{5}= \dfrac{-24}{x}$ , then

$x=$

0%
$40$
0%
$-40$
0%
${\pm }$ $40$
0%
None

Explanation

$\cfrac{-3}{5} =\cfrac{-24}{x}$

$\Rightarrow -3 \times x=-24 \times 5$

$\Rightarrow x=\cfrac{-24 \times5}{-3} =40$

$\Rightarrow x=40$ .

If a : b = b : c then

$\displaystyle a^{4}:b^{4}$ is equal to

0%
$\displaystyle ac:b^{2}$
0%
$\displaystyle a^{2}:c^{2}$
0%
$\displaystyle c^{2}:a^{2}$
0%
$\displaystyle b^{2}:ac$

Explanation

$\displaystyle \frac{a}{b}=\frac{b}{c}\Rightarrow b^{2}=ac\therefore \frac{a^{4}}{b^{4}}=\frac{a^{4}}{\left ( ac \right )^{2}}=\frac{a^{4}}{a^{2}c^{2}}=\frac{a^{2}}{c^{2}}$

Solve

$\displaystyle 5\left (2x+1 \right)\leq 35,$ where x is a natural number.

0%
$x\le3$
0%
$x= \{1,3\}$
0%
$x=3$
0%
$x= {3,4,5,6.......\infty}$

Explanation

The given inequation is

$\quad 5\left( 2x+1 \right) \le 35\\ \Rightarrow 2x+1\le 7\\ \Rightarrow 2x\le 6\\ \Rightarrow x\le 3\\ \because \quad x\in N\quad \\ \therefore \quad x=\left\{ 1,2,3 \right\} .\quad$

Solve the inequality:

$\displaystyle \left | 4x-3 \right |\geq 5$

0%
$\displaystyle \left ( \infty ,-2 \right )\cup \left (-1/2 ,\infty \right )$
0%
$\displaystyle \left ( -\infty ,-1/2 \right )\cup \left (2 ,\infty \right )$
0%
$\displaystyle \left ( -\infty ,1/2 \right )\cup \left (-2 ,-\infty \right )$
0%
none of the above

Explanation

Given inequation is

$\ \left| 4x-3 \right| \ge 5\\ \ 4x-3\ge 5.......(i)\ \& \ \\ -(4x-3)\ge 5........(ii)\$

$\ (i)\Rightarrow 4x-3\ge 5\\ \Longrightarrow 4x\ge 5+3\\ \Longrightarrow 4x\ge 8\\ \Longrightarrow x\ge 2\\ \therefore \ x\in \left\{ 2,\propto \right\} .\\ (ii)\Rightarrow -(4x-3)\ge 5\\ \Longrightarrow -4x\ge 5-3\\ \Longrightarrow -4x\ge 2\\ \Longrightarrow x\ge -\dfrac { 1 }{ 2 } .\\ \therefore \ x\in \left\{ -\propto ,-\dfrac { 1 }{ 2 } \right\} .\$
So, the solutions set is

$\ x\in \left\{ -\propto ,-\dfrac { 1 }{ 2 } \right\} \cup \left\{ 2,\propto \right\}$

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

let the number of boys and girls in the college be 7x and 8x

their increased number is (120% of 7x) =

$\Rightarrow \frac{120}{100}\times 7x$

$\Rightarrow \frac{42x}{5}$

And (10% of 8x).= $\Rightarrow \frac{110}{100}\times 8x$

$\Rightarrow \frac{44x}{5}$

$\therefore$ Required Ratio= $\frac{42x}{5}:\frac{44x}{5}$

$21:22$

A number to the 8th power divided by the same number to the 4th power isWhat is the number?

0%
2
0%
6
0%
4
0%
8

Explanation

Let the number is x
then according to the question

$\Rightarrow \frac{x^8}{x^4}=16$

$\Rightarrow x^{8-4}=16$

$\Rightarrow x^4=16$

$\Rightarrow x^4=2^4$

$\Rightarrow x=2$

Solve the inequality:

$\displaystyle \left | 2x+6 \right |\leq 16$

0%
$-10\le x \le -5$
0%
$-16\le x \le 0$
0%
$-11\le x \le 5$
0%
$-16\le x \le 16$

Explanation

The given inequation is

$\quad \left| 2x+6 \right| \le 16\\ \Longrightarrow -16\le 2x+6\le 16\\ \Longrightarrow -22\le 2x\le 10\\ \Longrightarrow -11\le x\le 5\quad$

Ans- Option C.

$a + b + c = abc$ , then how many real number solutions are possible ?

0%
Unique Solution
0%
No Solution
0%
Two Solutions
0%
Infinite Solutions

Explanation

Given that:

$a+b+c$

$=abc$

Since

$a+b+c=abc$

$\Rightarrow$

$abc-c-abc$ .

$c(ab-1)=a+b$

$\dfrac { a+b }{ ab-1 }$

where

$ab-1\ne0$

so we can substitute a and b with b any real number so we can say that we can have infinite number of real solution except 1

$\displaystyle \left | x+3 \right |\leq 3$ , then find minimum value of

$|x+1|$ .

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

Explanation

Given:

$\left| x+3 \right| \le 3$

Solution:The equation

$\left| x+3 \right| \le 3$ represents a circle whose centre is (-3,0) and radius is 3 units

Since x lies on the circle,the value x can take

so that the maximum is (-6,0) and minimum is (-1,0) i.e

Hence the minimum value of

$\left| x+1 \right| \ is 0$

$\displaystyle \left | x+3 \right |> -5$ then find the solution set for the inequality

0%
$R$
0%
$N$
0%
$W$
0%
$I$

Explanation

Given,

$\displaystyle \left | x+3 \right |> -5$
Clearly

$\displaystyle \left | x+3 \right |$ is a positive value for all

$\displaystyle x\epsilon R$ , which is always greater than any negative value i.e.,

$- 5$ .

$\displaystyle \therefore$ Solution set is

$R$ .

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

$15 < 4x + 3 \le 31$ ?

Explanation

$15 < 4x + 3 \le 31$ ........... (Given)

$\implies 15-3 < 4x+3-3 \leq 31-3$

$\implies 12 < 4x \leq 28$

$\implies 3 < x \le 7$ ......... (Divided by

$4$ )

Hence, option C is correct.

State true or flase

If you multiply an inequality by a negative number, inequality reverses

0%
False
0%
True

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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