MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 5

A person bought some articles at the rate of

$5$ per rupee and the some number at the rate of

$4$ per rupee. He mixed both the types and sold at the rate of

$9$ for Rs.

$2$ . In this business he suffered a loss of Rs.

$3$ . The total number of articles bought by him was

0%
$1090$
0%
$1080$
0%
$540$
0%
$545$

Explanation

Suppose he purchased x articles of each type.
Then, Total C.P.

$=\dfrac {x}{5}+\dfrac {x}{4}=Rs \dfrac {9x}{20}$
Total S.P.

$=\dfrac {2}{9}\times 2x=Rs\frac {4x}{9}$
Total loss

$=\dfrac {9x}{20}-\dfrac {4x}{9}$

$\Rightarrow \dfrac {81x-80x}{180}=3\Rightarrow x=3\times 180=540$

$\therefore$ Total number of articles

$=2\times 540=1080$ .

$\displaystyle\frac{a}{3}=\displaystyle\frac{b}{4}=\displaystyle\frac{c}{7}$ , then the value of

$\displaystyle\frac{a+b+c}{c}$ is

0%
$\;2$
0%
$\;7$
0%
$\;\displaystyle\frac{1}{2}$
0%
$\;\displaystyle\frac{1}{7}$

Explanation

Let

$a=3x,\;\;b=4x,\;\;c=7x$ . Then

$\;\;\;\;\;\;\;\;\displaystyle\frac{a+b+c}{c}=\displaystyle\frac{3x+4x+7x}{7x}=\displaystyle\frac{14x}{7x}=2$ .

From a number of mangoes, a man sells half the number of existing mangoes plus

$1$ to the first customer, then sells

$\displaystyle \frac{1}{3}$ rd of the remaining number of mangoes plus

$t$ o the second customer, then

$\displaystyle \frac{1}{4}$ of the remaining number of mangoes plus

$1$ to third customer and

$\displaystyle \frac{1}{5}$ th of the remaining number of mangoes plus

$1$ to the fourth customer. He then finds that he does not have any mango left. How many mangoes did he have originally ?

0%
$12$
0%
$14$
0%
$15$
0%
$13$

Explanation

Let the number of mangoes that the man has originally

$= x$
Number of mangoes sold to Balance
1st customer

$\displaystyle =\frac{x}{2}+1$

$\dfrac{x-2}{2}$

2nd customer

$\displaystyle =\frac{x-2}{6}+1$

$\dfrac{x-5}{3}$
3rd customer

$\displaystyle =\frac{x-5}{12}+1$

$\dfrac{x-9}{4}$
4th customer

$\displaystyle =\frac{x-9}{20}+1$

$0$

$\displaystyle \frac{x-9}{20}+1=\frac{x-9}{4}$

$\Rightarrow \dfrac{x+11}{20}=\dfrac{x-9}{4}$

$\displaystyle \Rightarrow 4x+44=20x-180$

$\displaystyle \Rightarrow 16x=224$

$\Rightarrow x = 14$

Rs.

$750$ is divided among

$A, B$ and

$C$ in such a manner that

$A:B = 5:2$ and

$B:C = 7:13$ . What is

$A'$ s share?

0%
$Rs. 350$
0%
$Rs. 260$
0%
$Rs. 140$
0%
$Rs. 250$

Explanation

Since,

$A : B = 5 : 2$ and

$B : C = 7 : 13$

$\therefore \displaystyle \frac AB = \frac 52$ and

$\displaystyle \frac BC = \frac 7{13}$

$\Rightarrow \displaystyle \frac AB = \frac 52\times \frac 77$ and

$\displaystyle \frac BC = \frac 7{13}\times \frac 22$

$\Rightarrow \displaystyle \frac AB = \frac {35}{14}$ and

$\displaystyle \frac BC = \frac {14}{26}$

$\Rightarrow A : B : C = 35 : 14 : 26$
Let

$A's$ share

$= 35x$

$B's$ share

$= 14x$

$C's$ share

$= 26x$

$\therefore 35x+14x+26x = 750$

$\Rightarrow 75x= 750$

$\Rightarrow x=10$

$\Rightarrow A's$ share

$=35x= 35\times 10 = 350$
Option A is correct.

The ratio of first and second class train fares between two station is

$3:1$ and that of the number of passengers travelling between these stations by first and second class is

$1:50.$ If on a particular day

$Rs. 1325$ be collected from the passengers travelling between these stations, then the amount collected from the second class passengers is:

0%
$Rs. 1250$
0%
$Rs. 1000$
0%
$Rs. 850$
0%
$Rs. 750$

Explanation

Let the 1st class fare is Rs.

$3x$ and the

$2$ nd class fare is Rs.

$x$ .
Let the number of passengers travelling in 1st class =

$y$
Then, the number of passengers travelling in 2nd class =

$50y$

$\therefore 3xt + 50xt = 1325 \Rightarrow 53xy = 1325 \Rightarrow xy = Rs. 25$

$\therefore Amount \ collected \ from \ 2nd \ class \ passengers = 50 \times Rs. 25 = Rs. 1250$

The solution to the inequality

$\displaystyle -2x+\left ( 3^{3}-5^{2} \right )\geq 4$ is

0%
$\displaystyle x\geq -1$
0%
$\displaystyle x\leq -1$
0%
$\displaystyle x>-2$
0%
$\displaystyle x<2$

Explanation

$\displaystyle -2x+\left ( 3^{3}-5^{2} \right )\geq 4$

Simplifying, we get

$-2x+27-25\geq 4$

$-2x+2\geq 4$

$-2x\geq 2$

$2x\leq -2$

$x\leq -1$

The inequality

$\displaystyle \left | 3-p \right |-4\leq1$ , then the solution for p is

0%
$-2<p<8$
0%
$-4<p<8$
0%
$2<p<8$
0%
$-2<p<5$

How many integers are there in the solution set of

$\displaystyle \left | 2x+6 \right |< \frac{19}{2}?$

0%
One
0%
Two
0%
Fourteen
0%
Nine

Explanation

Simplifying, we get

$\dfrac{-19}{2}<2x+6<\dfrac{19}{2}$

Hence

$-19<4x+12<19$

$-31<4x<7$

$\dfrac{-31}{4}<x<\dfrac{7}{4}$

$-7.75<x<1.75$
Hence the integers in the above solution set are

$\{-7,-6,-5,-4,-3,-2,-1,0,1\}$
Hence in total 9 integers.

The ratio of the ages of two boys is

$5 : 6.$ After

$2$ years, the ratio of their ages will be

$7 : 8.$ The ratio of the their ages after

$10$ years will be

0%
$15 : 16$
0%
$17 : 18$
0%
$11 : 12$
0%
$22 : 24$

Explanation

Let the ages of the two boys be 5x and 6x.
Given,

$\displaystyle \frac{5x + 2}{6x + 2} = \frac{7}{8} \Rightarrow 40x + 16 = 42x + 14$

$\Rightarrow 2x = 2 \Rightarrow x = 1$

$\therefore$ Required ratio

$= (5x + 10) : (6x + 10) = (5 \times 1 + 10) : (6 \times 1 + 10) = 15 : 16$

$x:y = 7:3,$ then the value of

$\displaystyle \frac{xy+y^2}{x^2-y^2}$ is:

0%
$\displaystyle \frac{3}{4}$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{3}{7}$
0%
$\displaystyle \frac{7}{3}$

Explanation

Let

$x = 7a; y = 3a$

So,

$\displaystyle \frac {xy + {y}^{2}}{{x}^{2} - {y}^{2}} =\frac {(7a)(3a)+(3a)^{2}}{(7a)^{2}-(3a)^{2}}$

$= \displaystyle \frac {21{a}^{2} + 9{a}^{2}}{49{a}^{2} - 9{a}^{2}}$

$=\displaystyle \frac {30{a}^{2}}{40{a}^{2}}$

$=\displaystyle \frac {3}{4}$

Hence, option A.

The inequality

$\displaystyle -1\leq 2x+4<5,$ Find the solution for x

0%
(a) $\displaystyle x=\left \{ -3,-1,0 \right \}$
0%
(b) $\displaystyle x=\left \{ -2,-1,1 \right \}$
0%
(c) $\displaystyle x=\left \{ -2,1,0 \right \}$
0%
(d) $\displaystyle x=\left \{ -2,-1,0 \right \}$

Explanation

$\displaystyle -1\leq 2x+4<5\Rightarrow -1-4\leq 2x<5-4\Rightarrow -5\leq 2x<1\Rightarrow -\frac{5}{2}\leq x<\frac{1}{2}\Rightarrow x$ can take the values between

$\displaystyle -2\frac{1}{2}\:and\:\frac{1}{2}\Rightarrow$ If x is an integer, then

$\displaystyle x=\left \{ -2,-1,0 \right \}$

The solution to the inequality

$\displaystyle \left | 10-2x \right |>6$ is

0%
$\displaystyle x<-2$ and $x<8$
0%
$\displaystyle x<-2$ and $x>8$
0%
$\displaystyle x>2$ and $x<-8$
0%
$\displaystyle x<2$ or $x>8$

Explanation

Given inequality is

$\displaystyle \left | 10-2x \right |>6$

Simplifying, we get

$10-2x>6$ and

$10-2x<-6$

$2x<4$ and

$16<2x$

$x<2$ and

$x>8$
Therefore the required solution set is

$x\epsilon(-\infty,2)\cup(8,\infty)$ .

In given figure, number line represents the solution of inequality ____ .

0%
$\displaystyle 2x-4<16$
0%
$\displaystyle 2x-6<10$
0%
$\displaystyle 2x-6>12$
0%
$\displaystyle 2x-4>16$

Explanation

$\displaystyle 2x-6<10\Rightarrow 2x< 16\Rightarrow x<8$

Which of the following is the solution set of

$\displaystyle \left | \frac{2}{3}x-5 \right |>8?$

0%
$\displaystyle \left \{ x:x<\frac{39}{2}\:or\:x<-\frac{9}{2} \right \}$
0%
$\displaystyle \left \{ x:x>\frac{39}{2}\:or\:x>-\frac{9}{2} \right \}$
0%
$\displaystyle \left \{ x:x>\frac{39}{2}\:or\:x<-\frac{9}{2} \right \}$
0%
$\displaystyle \left \{ x:x>\frac{9}{2}\:or\:x>\frac{-39}{2} \right \}$

Explanation

Simplifying, we get

$|2x-15|>24$
Hence

$2x-15<-24$ and

$2x-15>24$
Hence

$2x<-9$

$x<\dfrac{-9}{2}$ and

$2x>39$

$x>\dfrac{39}{2}$
Hence

$x\epsilon (-\infty,\dfrac{-9}{2})\cup(\dfrac{39}{2},\infty)$ .

The region for which

$\displaystyle x\geq 4$ is a part of the:

0%
first and second quadrants
0%
second and third quadrants
0%
third and fourth quadrants
0%
fourth and first quadrants

Explanation

$x\geq 4$

$x\epsilon [4,\infty)$
This consist right half of the co-ordinate plane excluding

$x\epsilon[0,4)$
Hence, the first and fourth quadrants.

$\displaystyle \left ( 2x-y<7 \right )\:and\:\left ( x+4y<11 \right ),$ then which one of the following is corect?

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

Explanation

$2x-y<7$ and

$x+4y<11$
Adding both we get

$3x+3y<18$

$x+y<6$

You are buying a carpet for a rectangular room. The carpet can be at most 12 m lone and 6 m wide. Which inequality represents the area of the carpet is square metres?

0%
$\displaystyle A\leq 36$
0%
$\displaystyle A\geq 36$
0%
$\displaystyle A\leq 72$
0%
$\displaystyle A\geq 72$

Explanation

Let l be the length of the carpet and b be the breadth.
It is given that

$l\leq 12m$

$b\leq 6m$
Area=A

$=l\times b$
Hence maximum area is achieved when

$l=12cm$ and

$b=6cm$
Hence

$A_{max}=l\times b$

$=72m^{2}$
Hence

$A\leq 72m^{2}$

The solution set of the inequality

$\displaystyle 2\left ( 4x-1 \right )\leq 3\left ( x+4 \right )$ is

0%
$\displaystyle x>\frac{14}{5}$
0%
$\displaystyle x<7$
0%
$\displaystyle x\leq \frac{14}{5}$
0%
$\displaystyle x\geq 7.5$

Explanation

Simplifying we get

$8x-2\leq 3x+12$

$5x\leq 14$

$x\leq \dfrac{14}{5}$
Hence

$x\epsilon(-\infty,\dfrac{14}{5}]$

The graph of which inequality is shown below:

0%
$\displaystyle y-x\leq 0$
0%
$\displaystyle x-y\leq 0$
0%
$\displaystyle y+x\leq 0$
0%
None of the above

Explanation

The equation of the above straight line is

$y=-x$
or

$x+y=0$ .
Now the shading in the above graph is towards the negative part (where x is negative).
Also the line is dark and not dotted.This indicates that the points on the line are part of the inequality.
Hence the required inequality is

$x+y\leq 0$ .

The greatest value of x that satisfies the inequality

$\displaystyle 2x+3<25,$ where x is a prime number is

0%
11
0%
7
0%
10
0%
2

Explanation

$2x+3<25$

$2x<22$

$x<11$
Hence the greatest prime number satisfying the inequality will be the prime number just preceding 11.
Hence

$x=7$ .

The area of the plane region

$\displaystyle \left | x \right |\leq 5;\left | y\right |\leq 3$ is

0%
$15$ sq units
0%
$34$ sq units
0%
$60$ sq units
0%
$120$ sq units

Explanation

$\displaystyle \left | x \right |\leq 5\Rightarrow -5\leq x\leq 5\:\left | y \right |\leq 3\Rightarrow -3\leq y\leq 3\therefore$
Area bounded by the rectangle so formed

$\displaystyle =AB\times\:AD=10\times\:6=60$ sq units.

The solution set of

$\displaystyle x\geq 5,y\geq 0\:and\:x\leq 0$ is

0%
$\displaystyle x\geq -5,y=0$
0%
$\displaystyle x=5,y=0$
0%
$\displaystyle x\geq -5,y\leq 0$
0%
$\displaystyle x\leq /5,y\geq 0$

Explanation

From the above conditions

$x\geq 5$ and

$x\leq 0$
Hence

$x\epsilon(-\infty,0]\cup[5,\infty)$
and

$y\geq 0$ implies

$y\epsilon [0,\infty)$
Hence

$x=5,y=0$ is an obvious solution.

The shaded region is represented by the inequation:

0%
$\displaystyle y\geq x$
0%
$\displaystyle y\geq- x$
0%
$\displaystyle y\geq \left | x \right |$
0%
$\displaystyle y\leq \left | x \right |$

Explanation

The equations of both the lines in the above graph are

$y=-x$ and

$y=x$
Hence if we put them together we get

$y=|x|$
Now
Let us take a point inside the shaded region.
Let it be

$(0,2)$
Now

$2>0$

$y>|x|$
Hence the required inequality is

$y\geq |x|$ .

Solve the inequality:

$\displaystyle \left | 1-x \right |>3.$

0%
$\displaystyle x>4\:or\:x<-1$
0%
$\displaystyle x>2\:or\:x<-2$
0%
$\displaystyle x>5\:or\:x<-2$
0%
$\displaystyle x>4\:or\:x<-2$

Explanation

$|1-x|>3$
Now

$|1-x|=|x-1|$
Hence

$|x-1|>3$

$x-1>3$ and

$x-1<-3$

$x>4$ and

$x<-2$
Hence

$x\epsilon(-\infty -2)\cup(4,\infty)$

The shaded region is represented by the inequality:

0%
$\displaystyle y-2x\leq -1$
0%
$\displaystyle x-2y\leq -1$
0%
$\displaystyle y-2x\geq -1$
0%
$\displaystyle x-2y\geq -1$

Explanation

The equation of the line is given as

$y-2x=-1$
Now the shading is away from the origin.
Hence, at y=0 and x=0 the inequality is not true.
we know

$0>-1$
Hence at origin, the inequality must be of the type

$0<-1$ ....(since inequality is not true at origin).
Hence

$y-2x<-1$
Also the points on the line is a part of the inequality.
Hence

$y-2x\leq -1$

If x : y = 3 : 2 then the ratio

$\displaystyle 2x^{2}+3y^{2}:3x^{2}-2y^{2}$ is

0%
12 : 5
0%
6 : 5
0%
30 : 19
0%
5 : 3

Explanation

$\dfrac{x}{y}$

$= \dfrac{3}{2}$

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

$=\dfrac{9}{4}$

$= \dfrac{2x^{2}+3y^{2}}{3x^2-2y^2}$

$=\dfrac{2\times\dfrac{9}{4}+3}{3\times\dfrac{9}{4}-2}$

$=\dfrac{\dfrac{4}{18+12}}{\dfrac{27-8}{4}}=\dfrac{30}{19}=30:19$

$(25)^x\, =\, (125)^y$ then

$x\, :\, y$ = ...........

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

Explanation

$(5^2)^x\, =\, (5^3)^y$

$\Rightarrow\, 5^{2x}\, =\, 5^{3y}$

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

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

$\Rightarrow\, x\, :\, y\, =\, 3\, :\, 2$

Rs 180 is to be divided among 66 persons (men and women) The ratio of the total amount of money received by men and women is 5 : 4 But the ratio of the money received by each man and woman is 3 : 2 The number of men is

0%
20
0%
24
0%
30
0%
36

Explanation

Let the number of men be x.
Then, number of women

$\displaystyle =66-x$
Money received by x men

$\displaystyle =\frac{5}{\left ( 5+4 \right )}\times\:RS\:180=RS\:100$

$\therefore$ Money received by $$\displaystyle \left ( 66-x \right )women=\left ( 180-100 \right )=Rs\;80\$$

$Given,\dfrac{100}{x}:\dfrac{80}{66-x}=3:2\Rightarrow \dfrac{100}{x}\times\dfrac{66-x}{80}=\dfrac{3}{2}\Rightarrow \dfrac{200}{x}=\dfrac{240}{66-x}\Rightarrow 240x=13200-200x$

$\Rightarrow 440x=13200\Rightarrow x=\dfrac{13200}{440}=30$

Given

$\displaystyle a>0,b>0,a>b\:and\:c\neq 0.$ Which inequality is not always correct?

0%
$\displaystyle a+c>b+c$
0%
$\displaystyle a-c>b-c$
0%
$\displaystyle ac>bc$
0%
$\displaystyle \frac{a}{c^{2}}>\frac{b}{c^{2}}$

Explanation

It is given that both a and b are positive and a>b.
But it is not given whether c is positive or negative.
Now

$a>b$

$a\pm c>b\pm c$
Also

$c^{2}>0$ for any real value of c.
Hence

$\dfrac{a}{c^{2}}>\dfrac{b}{c^{2}}$
However.

$ac>bc$ is only true if

$C>0$
If

$c<0$

$ac<bc$
Hence answer is Option C.

The expenses on rice, fish and oil of a family are in the ratio

$12 : 17 : 13.$ The prices of these articles are increased by

$20\%, 30\%$ and

$50\%$ respectively. The total expenses of the family are increased by:

0%
$\displaystyle 14\frac{1}{3}$ %
0%
$\displaystyle 7\frac{1}{3}$ %
0%
$\displaystyle 56\frac{1}{3}$ %
0%
$\displaystyle 33\frac{1}{3}$ %

Explanation

Let the expenses on rice, fish & oil were

$12x, 17x$ &

$13x$ when x is in Rs

Then the total expenses

$=12x+17x+13x=42x$ .

Now rise in price of rice=

$3x\times \cfrac { 20 }{ 100 } =\cfrac { 3x }{ 5 } ,$

rise in price of fish=

$17x\times \cfrac { 30 }{ 100 } =\dfrac { 51x }{ 10 } ,$ and

rise in price of oil=

$13x\times \cfrac { 50 }{ 100 } =\dfrac { 13x }{ 2 } .$

So the total rise in expenses=

$\cfrac { 12x }{ 5 } +\cfrac { 51x }{ 10 } +\dfrac { 13x }{ 2 } =\dfrac { 140x }{ 10 } =14x.$

$\therefore$ The percentage of rise of total expenses

$\dfrac { 14x }{ 42x } \times 100$ %

$33\dfrac { 1 }{ 3 } %$ %

Ans- Option D.

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 5 - 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 5 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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