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CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 5 - MCQExams.com
CBSE
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
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$$1090$$
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 $$=2\times 540=1080$$.
If $$\displaystyle\frac{a}{3}=\displaystyle\frac{b}{4}=\displaystyle\frac{c}{7}$$, then the value of $$\displaystyle\frac{a+b+c}{c}$$ is
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$$\;2$$
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$$\;7$$
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$$\;\displaystyle\frac{1}{2}$$
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$$\;\displaystyle\frac{1}{7}$$
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 ?
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$$12$$
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?
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$$Rs. 350$$
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:
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$$Rs. 1250$$
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
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$$\displaystyle x\geq -1$$
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$$\displaystyle x\leq -1$$
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$$\displaystyle x>-2$$
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$$\displaystyle x<2$$
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
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$$ -2<p<8$$
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$$-4<p<8$$
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$$2<p<8$$
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$$-2<p<5$$
How many integers are there in the solution set of $$\displaystyle \left | 2x+6 \right |< \frac{19}{2}?$$
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One
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Two
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Fourteen
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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
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$$15 : 16$$
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$$17 : 18$$
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$$11 : 12$$
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$$22 : 24$$
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 $$= (5x + 10) : (6x + 10) = (5 \times 1 + 10) : (6 \times 1 + 10) = 15 : 16$$
If $$x:y = 7:3,$$ then the value of $$\displaystyle \frac{xy+y^2}{x^2-y^2}$$ is:
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{4}{3}$$
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$$\displaystyle \frac{3}{7}$$
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$$\displaystyle \frac{7}{3}$$
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
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(a)
$$\displaystyle x=\left \{ -3,-1,0 \right \}$$
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(b)
$$\displaystyle x=\left \{ -2,-1,1 \right \}$$
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(c)
$$\displaystyle x=\left \{ -2,1,0 \right \}$$
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(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
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$$\displaystyle x<-2$$ and $$x<8$$
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$$\displaystyle x<-2$$ 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 ____ .
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$$\displaystyle 2x-4<16$$
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$$\displaystyle 2x-6<10$$
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$$\displaystyle 2x-6>12$$
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$$\displaystyle 2x-4>16$$
Explanation
$$\displaystyle 2x-6<10\Rightarrow 2x< 16\Rightarrow x<8$$
Which of the following is the solution set of $$\displaystyle \left | \frac{2}{3}x-5 \right |>8?$$
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$$\displaystyle \left \{ x:x<\frac{39}{2}\:or\:x<-\frac{9}{2} \right \}$$
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$$\displaystyle \left \{ x:x>\frac{39}{2}\:or\:x>-\frac{9}{2} \right \}$$
0%
$$\displaystyle \left \{ x:x>\frac{39}{2}\:or\:x<-\frac{9}{2} \right \}$$
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$$\displaystyle \left \{ x:x>\frac{9}{2}\:or\:x>\frac{-39}{2} \right \}$$
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:
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first and second quadrants
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second and third quadrants
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third and fourth quadrants
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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.
If $$\displaystyle \left ( 2x-y<7 \right )\:and\:\left ( x+4y<11 \right ),$$ then which one of the following is corect?
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$$\displaystyle x+y<5$$
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$$\displaystyle x+y<6$$
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$$\displaystyle x+y\leq 5$$
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$$\displaystyle x+y\geq 6$$
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?
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$$\displaystyle A\leq 36$$
0%
$$\displaystyle A\geq 36$$
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$$\displaystyle A\leq 72$$
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$$\displaystyle A\geq 72$$
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
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$$\displaystyle x>\frac{14}{5}$$
0%
$$\displaystyle x<7$$
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$$\displaystyle x\leq \frac{14}{5}$$
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:
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$$\displaystyle y-x\leq 0$$
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$$\displaystyle x-y\leq 0$$
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$$\displaystyle y+x\leq 0$$
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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
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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
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$$15$$ sq units
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$$34$$ sq units
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$$60$$ sq units
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$$120$$ 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
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$$\displaystyle x\geq -5,y=0$$
0%
$$\displaystyle x=5,y=0$$
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$$\displaystyle x\geq -5,y\leq 0$$
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$$\displaystyle x\leq /5,y\geq 0$$
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:
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$$\displaystyle y\geq x$$
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$$\displaystyle y\geq- x$$
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$$\displaystyle y\geq \left | x \right |$$
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$$\displaystyle y\leq \left | x \right |$$
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:
$$\displaystyle \left | 1-x \right |>3.$$
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$$\displaystyle x>4\:or\:x<-1$$
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$$\displaystyle x>2\:or\:x<-2$$
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$$\displaystyle x>5\:or\:x<-2$$
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:
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$$\displaystyle y-2x\leq -1$$
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
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0%
12 : 5
0%
6 : 5
0%
30 : 19
0%
5 : 3
Explanation
$$ \dfrac{x}{y}$$ $$= \dfrac{3}{2}$$
$$= \dfrac{x^{2}}{y^{2}}$$ $$=\dfrac{9}{4}$$
$$ = \dfrac{2x^{2}+3y^{2}}{3x^2-2y^2}$$
$$=\dfrac{2\times\dfrac{9}{4}+3}{3\times\dfrac{9}{4}-2}$$
$$=\dfrac{\dfrac{4}{18+12}}{\dfrac{27-8}{4}}=\dfrac{30}{19}=30:19$$
If $$(25)^x\, =\, (125)^y$$ then $$x\, :\, y$$ = ...........
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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
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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.$$ Which inequality is not always correct?
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$$\displaystyle a+c>b+c$$
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$$\displaystyle a-c>b-c$$
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$$\displaystyle ac>bc$$
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$$\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\% $$ respectively. The total expenses of the family are increased by:
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0%
$$\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 $$ T
he percentage of rise of total expenses
=
$$\dfrac { 14x }{ 42x } \times 100$$%
=
$$33\dfrac { 1 }{ 3 } %$$%
Ans- Option D.
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