MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 7

Solve the following inequality.

$\displaystyle \frac{3x+6}{2x-12}\le 0$

0%
$-2 \le x < 6$
0%
$-2 \le x < -6$
0%
$2 \le x < 6$
0%
None of these

Explanation

The given inequality is,

$\frac { 3x+6 }{ 2x-12 } \le 0$

There are following two possibilies-

Case (i)

$3x+6\ge 0$ and

$2x-12<0$

$\therefore 3x\ge -6$ and

$2x<12$

$\therefore x\ge -2$ and

$x<6$

Thus, we can write domain of the solution as,

$-2\le x<6$

Case (ii)

$3x+6\le 0$ and

$2x-12>0$

$\therefore 3x\le -6$ and

$2x>12$

$\therefore x\le -2$ and

$x>6$

Satisfying these two conditions simultaneously is not possible. Thus solution of this inequality is

$-2\le x<6$

Thus, Answer is option (A)

Solve the following inequality.

$\displaystyle \frac{2}{p-1}\ge \frac{3}{4}$

0%
$1 < p \le \frac{11}{3}$
0%
$1 < p \le \frac{13}{3}$
0%
$2 < p \le \frac{11}{3}$
0%
None of these

Explanation

The given inequality is,

$\frac { 2 }{ p-1 } \ge \frac { 3 }{ 4 }$

Subtract

$\frac { 3 }{ 4 }$ from both the sides, we get

$\frac { 2 }{ p-1 } -\frac { 3 }{ 4 } \ge \frac { 3 }{ 4 } -\frac { 3 }{ 4 }$

$\therefore \frac { 2 }{ p-1 } -\frac { 3 }{ 4 } \ge 0$

Performing cross multiplication, we get,

$\frac { \left( 2\times 4 \right) -3\left( p-1 \right) }{ 4\left( p-1 \right) } \ge 0$

$\therefore \frac { 8-3p+3 }{ 4p-4 } \ge 0$

$\therefore \frac { 11-3p }{ 4p-4 } \ge 0$

There are following two possibilities for this inequality-

Case (i)

$\therefore 11-3p\ge 0$ and

$4p>4$

$\therefore 11\ge 3p$ and

$p>1$

$\therefore p\le \frac { 11 }{ 3 }$ and

$p>1$

Thus, range of values of p is,

$1<p\le \frac { 11 }{ 3 }$

Case (ii)

$11-3p\le 0$ and

$4p-4<0$

$\therefore 11\le 3p$ and

$4p<4$

$\therefore p\ge \frac { 11 }{ 3 }$ and

$p<1$

Satisfying these two conditions simultaneously is not possible. Thus, solution of the given inequality is

$1<p\le \frac { 11 }{ 3 }$

Thus, Answer is option (A)

Solve the following inequality.

$\displaystyle \frac{2}{m+3}\le 1$

0%
$m < -3$ or $m \ge -1$
0%
$m < -3$ or $m \ge -3$
0%
$m < -4$ or $m \ge -1$
0%
None of these

Solve the following inequality.

$\displaystyle \frac{p-5}{3-p}\le 0$

0%
$p < 3$ or $p\ge 5$
0%
$p < -3$ or $p\ge 5$
0%
$p < 3$ or $p\ge -5$
0%
None of these

Solve the following inequality.

$\displaystyle \frac{4}{x+2}>2$

0%
$(-2,0)$
0%
$(2,0)$
0%
$(-6,0)$
0%
None of these

Solve the following inequality.

$\displaystyle \dfrac{2z-5}{z-7}\le 0$

0%
$\dfrac{5}{7}\le z < 7$
0%
$\dfrac{5}{7}\le z < -7$
0%
$\dfrac{-5}{7}\le z < 7$
0%
None of these

Solve the following inequality.

$\displaystyle \frac{2t+7}{t-4}\ge 3$

0%
$4 < t \le 19$
0%
$4 < t \le -19$
0%
$-4 < t \le 19$
0%
None of these

Solve the following inequality.

$\displaystyle \frac{4-x}{x+3}>0$

0%
$-3<x<4$
0%
$3<x<4$
0%
$-3<x<-4$
0%
None of these

Solve the following inequality.

$\displaystyle \frac{3x-1}{x}\le -1$

0%
$0 < x< \dfrac{1}{4}$
0%
$0 < x< \dfrac{2}{4}$
0%
$0 < x< \dfrac{1}{6}$
0%
None of these

The shaded region in the figure shown represents the solution set of

0%
$x-y\ge 0$ , $x+y\le 0$
0%
$x-y< 0$ , $x+y< 0$
0%
$x-y\ge 0$ , $x+y \ge 0$
0%
$x-y\le 0$ , $x+y\ge 0$

Explanation

The shaded region represents

$x-y\ge 0$

$x+y\ge 0$

$\therefore$ (3) is correct.

Solve the following inequality.

$\displaystyle \frac{x^2-16}{(x-1)^2}<0$

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

Solve the following inequality.

$\displaystyle u\le \frac{4}{u-3}$

0%
$u\le -1$ and $3<u\le4$
0%
$u\le -5$ and $3<u\le4$
0%
$u\le -5$ and $3<u\le7$
0%
None of these

From a container of milk,

$5$ litres of milk is replaced with

$5$ litres of water. This process is repeated again. Thus in two attempts the ratio of milk and water became

$81:19.$ The initial amount of milk in the container was

0%
$50$ litres
0%
$45$ litres
0%
$40$ litres
0%
$25$ litres

Explanation

Milk remaining/water

$=81/19$

Milk remaining/milk inital

$=81/100$

$\therefore Remaining \space milk=initial \space milk \space [1-(quantity \space taken \space out /total \space amount)]^n$

$\Rightarrow 81x=100x(1-5/k)^2$

$\Rightarrow \frac{81}{100}=(1-5/k)^2$

$\Rightarrow (\frac{9}{10})^2=(1-5/k)^2$

$\Rightarrow 5/k=1-\frac{9}{10}$

$\Rightarrow 5/k=\frac{1}{10}$

$\Rightarrow k=50 \space litre$

Solve the following inequality.

$\displaystyle \frac {3x+8}{x-1}< -2$

0%
$-\frac{6}{5}<x<1$
0%
$-\frac{5}{5}<x<1$
0%
$-\frac{6}{5}<x<6$
0%
None of these

Solve the following inequality.

$\displaystyle \frac{x+1}{x-5} \le 0$ .

0%
$-1\le x<5$
0%
$-5\le x<5$
0%
$-1\le x<7$
0%
None of these

Zinc and copper are in the ratio of 5 : 3 in 200 gm of alloy. How much grams of copper be added to make the ratio as 3 : 5 ?

0%
400/3
0%
1/200
0%
72
0%
66

Which set of points is in the solution set for the system of inequalities:

$x-y>1$ and

$y<2x-1$ ?

0%
$(-1, -1)$
0%
$(-2, -1)$
0%
$(0, 1)$
0%
$(0, -2)$

Solve the following inequality.

$\displaystyle \frac{3x+1}{x+4}\ge 1$

0%
$(-\infty, -4)$ and $[\dfrac{3}{2}, \infty)$
0%
$(-\infty, 4)$ and $[\dfrac{3}{2}, \infty)$
0%
$(-\infty, -6)$ and $[\dfrac{3}{2}, \infty)$
0%
None of these

Divide

$Rs.370$ into three parts such that second part is

$\dfrac{1}{4}$ of the third part and the ratio between the first and the third part is

$3:5$ . Find each part.

0%
$120,50,200$
0%
$100,50,200$
0%
$240,25,200$
0%
$120,50,100$

A bag contains Rs.

$510$ in the form of

$50p$ ,

$25p$ and

$20p$ coins in the ratio

$2:3:4$ . Find the number of coins of each type

0%
$200,\,300,\,400$
0%
$100,\,150,\,200$
0%
$400,\,600,\,800$
0%
$600,\,900,\,1200$

Explanation

Let the number of

$50p$ ,

$25p$ and

$20p$ coins be

$2x,\,3x,\,and\,4x$ .
Then

$2x\times \dfrac{50}{100}+3x\times\dfrac{25}{100}+4x\times\dfrac{20}{100}=510$

$\Longrightarrow\dfrac{x}{1}+\dfrac{3x}{4}+\dfrac{4x}{5}=510$

$\Longrightarrow \dfrac{20x+15x+16x}{20}=510$

$\Longrightarrow \dfrac{51x}{20}=510$

$\Longrightarrow x=\dfrac{510\times20}{51}$

$x=200$

$2x=2\times200=400$

$3x=3\times200=600$

$4x=4\times200=800$ .
Therefore, number of

$50p$ coins,

$25p$ coins and

$20p$ coins are

$400,\,600,\,800$ respectively.

The ratio of number of boys and girls is

$4:3$ . If there are

$18$ girls in a class, then find the total number of students in the class.

0%
$40$
0%
$41$
0%
$42$
0%
$43$

Explanation

Number of girls in the class

$=18$

Ratio of boys and girls

$=4:3$

$\dfrac {Boys}{Girls} =\dfrac{4}{5}$

$\dfrac{Boys}{18}=\dfrac{4}{5}$

Boys

$=\dfrac{4\times18}{3}=24$

Therefore, total number of students

$=24+18=42$ .

Divide Rs.

$290$ among A, B, C in the ratio

$\dfrac{3}{2},\dfrac{5}{4}$ and

$\dfrac{3}{8}$ .

0%
$120,\,100,\,30$
0%
$120,\,30,\,60$
0%
$30,\,60,\,15$
0%
$60,\,50,\,15$

Explanation

The L.C.M of

$2,\,4,\,8$ is

$8$ .
So we have

$\dfrac{3}{2}\times8:\dfrac{5}{4}\times8:\dfrac{3}{8}\times8=12:10:3$
Therefore, share of

$A=\dfrac{12}{29}\times290=Rs. 120$
Share of B

$=\dfrac{10}{29}\times290=Rs. 100$
Share of C

$=\dfrac{3}{29}\times290=Rs. 30$

Ratio of two numbers is 2 : 3 and their LCM isFind the two numbers.

0%
12, 18
0%
8 , 12
0%
18, 24
0%
16, 24

Explanation

Let the numbers be 2x and 3x

LCM of 2x and 3x=6x

According to the question LCM is 36 then

$\therefore 6x=36$

$\Rightarrow x=\frac{36}{6}=6$

$\therefore$ number are

$2\times 6=12$ and

$3\times 6=18$

The ratio

$4^{3.5}:2^5$ is same as

0%
$2:1$
0%
$4:1$
0%
$7:5$
0%
$7:10$

Explanation

$\dfrac{4^{3.5}}{2^5}=\dfrac{2^{2(3.5)}}{2^5}=\dfrac{2^7}{2^5}=2^2=4:1$

Rs.

$1210$ were divided among A, B, C so that

$A:B=5:4$ ,

$B:C=9:10$ . then C gets

0%
Rs. $340$
0%
Rs. $400$
0%
Rs. $450$
0%
Rs. $475$

Explanation

$B:C=\begin{pmatrix}9\times \dfrac{4}{9}\end{pmatrix}:\begin{pmatrix}10\times \dfrac{4}{9}\end{pmatrix}=4:\dfrac{40}{9}$

$A:B:C=5:4:40/9=45:36:40$
Sum of ratio terms

$=45+36+40=121$
C"s share

$=1210\times \dfrac{40}{121}=Rs. 400$

Production of a company A is 120 % of the production of company B and 80 % of the production of company C. What is the ratio between production of companies A, B and C respectively?

0%
6 : 5 : 9
0%
6 : 5 : 4
0%
12 : 10 : 15
0%
10 : 12 : 15

Explanation

Let the production of company B=100

Then production of company A =120

$\therefore$ Production of company C

$=\frac{120}{80}{100}=150$

Then

$A:B:C=120:100:150\Rightarrow 12:10:15$

A stone is projected vertically up from the bottom of a water tank. Assuming no water resistance it will go up & come down in same time but if water drag is present then the time it takes to go up,

${t}_{up}$ and the time it takes to come down,

${t}_{down}$ are related as

0%
${t}_{up} > {t}_{down}$
0%
${t}_{up} = {t}_{down}$
0%
${t}_{up} < {t}_{down}$
0%
Can not say

Explanation

While moving up & while moving down

$\because \quad \quad { a }_{ up }>{ a }_{ down }$

Deceleration is more while going up than while coming down.

Hence to cover same distance $t_{up} < t_{down}$

Simplify

$\dfrac{5}{2}:\dfrac{3}{8}:\dfrac{4}{9}$

0%
$180:27:32$
0%
$10:6:8$
0%
$15:9:12$
0%
$25:15:20$

Explanation

The L.C.M of

$2,\,8,\,9=2\times2\times2\times3\times3=8\times9=72$
Now, multiplying each fraction by the L.C.M.

$\Longrightarrow\dfrac{5}{2}\times72=160$

$\Longrightarrow \dfrac{3}{8}\times72=27$

$\Longrightarrow \dfrac{4}{9}\times72=32$
So, the ratio becomes

$180:27:32$

$\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{7}$ , then

$\dfrac{a+b+c}{c}$ is equal to

0%
$7$
0%
$2$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{7}$

Explanation

Let

$\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{7}=k$ , then

$a=3k$ ,

$b=4k$ ,

$c=7k$

$\Longrightarrow \dfrac{a+b+c}{c}=\dfrac{3k+4k+7k}{7k}=2$

The ratio of three numbers is

$3:4:5$ and sum of their squares is

$1250$ . The sum of the numbers is

0%
$30$
0%
$50$
0%
$60$
0%
$90$

Explanation

Let the no.s be

$3x,\,4x,\,5x$ then,

$9x^2+16x^2+25x^2=1250$

$\Longrightarrow 50x^2=1250$

$\Longrightarrow x=5$
Sum of numbers

$=3x+4x+5x=12x=12\times5=60$

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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