MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 1

Solve the following equations by substitution method.

$2x-3y=14; \, \, 5x+2y=16$

0%
$x = 7, y= 1$
0%
$x = 4, y= -2$
0%
$x = 6, y= -2$
0%
$x = 3, y= -1$

Explanation

$2x-3y=14; \, \, 5x+2y=16$

$x = \dfrac {14+3y}{2}$ Put it in second equation.

$5 \times \dfrac {14+3y}{2} +2y =16$

$70 +15y +4y = 32$

$19y = -38$

$y = -2$

$x = \dfrac {14+3y}{2} = \dfrac {14+3(-2)}{2} = 4$

$x = 4$ and

$y = -2$

$x - 2y = -1$ ,

$y=-\dfrac { 1 }{ 2 }$ , then find the value of

$x$

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

Explanation

Given:

$x-2y=-1$ .....................[1]

$y=-\dfrac12$ ..................[2]

Putting

$y=-\dfrac12$ in equation

$1$ :

$x-2\times -\dfrac12=-1$

$\implies x+1=-1$

$\implies x=-2$

Giri is 20 year older than his son. If the sum of their ages ishow old is his son?

0%
5 years
0%
20 years
0%
25 years
0%
30 years

$\cfrac { a }{ x+y } =\cfrac { b }{ y+z } =\cfrac { c }{ z-x }$ , then which of the following equations is true?

0%
$a=b+c$
0%
$c=a+b$
0%
$b=a\times c$
0%
$b=a+c$

Explanation

Let

$\cfrac { a }{ x+y } =\cfrac { b }{ y+z } =\cfrac { c }{ z-x } =k$

$\therefore$

$a=xk+yk........(1)$

$b=yk+zk...........(2)$

$c=zk-xk...........(3)$

equation

$(1)+(3)$

$a+c=zk-xk+xk+yk$

$=zk+yk$

$=b$

$\therefore$

$a+c=b$

Hence, option D is correct.

Given that

$3x-y=-11$ and

$4x+y=4$ . Calculate the value of

$x$ .

0%
$-7$
0%
$-1$
0%
$8$
0%
$15$

Explanation

$3x-y=-11$ ...........

$(i)$

$4x+y=4$ ...........

$(ii)$

Adding

$(i)$ and

$(ii),$ we get

$(3x-y)+(4x+y)=-11+4$

$\Rightarrow 3x-y+4x+y=-11+4$

$\Rightarrow 3x+4x=-11+4$

$\Rightarrow 7x=-7$

$\Rightarrow x=\dfrac{-7}{7}=-1$

The cost of 9 chairs and 3 tables is Rs. 306, while the cost of 6 chairs and 3 tables is Rs.Then the cost of 6 chairs and 1 table is

0%
Rs. 161
0%
Rs. 162
0%
Rs. 169
0%
Rs. 175

Explanation

$9x+3y=306$ ---- (1)

$6x+3y=246$ -----(2)

Subtracting eq(2) from eq(1)

$3x=60$

$x=20$

Substituting

$x=20$ in eq(1)

$9(20)+3y = 306$

$180+3y=306$

$3y=126$

$y=42$

Hence,

$6x+y =162$ .

If,

$\displaystyle \frac{1}{x}+\frac{1}{y}=k$ and

$\displaystyle \frac{1}{x}-\frac{1}{y}=k$ , then the value of y .................

0%
$3$
0%
$-4$
0%
Does not exist.
0%
None of these

Explanation

Let

$\displaystyle \frac{1}{x}+\frac{1}{y}=k$ ......(1)

$\displaystyle \frac{1}{x}-\frac{1}{y}=k$ ......(2)

$(1)-(2) \Rightarrow \displaystyle \frac{2}{y}=0$

$\therefore$ the value of

$y$ does not exist.

Given that

$4x-3y=6$ and

$x-3y=-3$ , the value of

$y$ is

0%
$-3$
0%
$-2$
0%
$\dfrac { 4 }{ 3 }$
0%
$2$

Explanation

$4x-3y=6$ ...........

$(i)$

$x-3y=-3$ ...........

$(ii)$

Multiply

$(ii)$ by

$4,$ we get

$x\times 4-3y\times 4=-3\times 4$

$\Rightarrow 4x-12y=-12$ ...........

$(iii)$

Subtract

$(iii)$ with

$(i)$ , we get

$(4x-3y)-(4x-12y)=6-(-12)$

$\Rightarrow 4x-3y-4x+12y=6+12$

$\Rightarrow -3y+12y=18$

$\Rightarrow 9y=18$

$\Rightarrow y=\dfrac{18}{9}=2$

The sum of two numbers is equal to

$20$ and their difference is

$2.5$ . Find the ratio of the numbers

0%
$9 : 7$
0%
$7 : 9$
0%
$3 : 5$
0%
$2 : 7$

Explanation

Let the numbers be

$x$ and

$y$ , then

$x+y = 20...(1)$

$x-y = 2.5...(2)$

$(1)+(2)\Rightarrow 2x = 22.5$

$\Rightarrow x = 11.25$

$\Rightarrow y = 8.75$

$\therefore x:y = 11.25:8.75$

$\Rightarrow x:y = 9:7$

Solve the following pair of simultaneous equations:

$a\, =\, b\, +\, 2$

$2a\, -\, b\, =\, 7$

0%
$a = -2, b = 9$
0%
$a = 12, b = 6$
0%
$a = -3, b = 7$
0%
$a = 5, b = 3$

Explanation

$a - b = 2$ ...(1)

$2a - b = 7$ ...(2)

Subtract eq(2) from eq(1) we get

$a = 5\quad and \quad b = 3$

Given that

$2y=-6$ and

$x-3y=13$ , the value of

$x=$ ________

0%
$\dfrac { 13 }{ 9 }$
0%
$4$
0%
$19$
0%
$22$

Explanation

Given

$2y=-6$

$\Rightarrow y=\dfrac{-6}{2}$

$\Rightarrow y=-3$ ............

$(i)$

$x-3y=13$ .............

$(ii)$ (given )

Put the value of

$y$ in

$(ii)$ , we get

$x-3(-3)=13$

$\Rightarrow x+9=13$

$\Rightarrow x=13-9=4$

Then

$x=4$

The ratio of income of two persons is

$9 : 7$ and the ratio of their expenditure is

$4 : 3$ . If each of them manages to save Rs.

$2000$ per month, find their monthly income.

0%
$Rs. 18,000 , Rs. 14,000$
0%
$Rs. 1,000 , Rs. 14,000$
0%
$Rs. 18,000 , Rs. 1,000$
0%
None of these

Explanation

Let their salaries be

$9x$ and

$7x$ .

Let their expenditure be

$4y$ and

$3y$ .

According to the question,

$9x-4y=2000$

$—(1)$

$7x-3y=2000$

$—(2)$

From

$(1)$

$x=\dfrac{2000+4y}{9}$

$—(3)$

On putting

$x$ in

$(2)$ , we get

$7\times \dfrac{(2000+4y)}{9}-3y=2000$

$\dfrac{(14000+28y)}{9}-3y=2000$

$\dfrac{14000+28y-27y}{9}=2000$

$14000+y=18000$

$y=4000$

Now, put

$y$ in (3)

$x=\dfrac{2000+4\times 4000}{9}$

$x=\dfrac{2000+16000}{9}$

$x=\dfrac{18000}{9}=2000$

So,

Salary of first person

$=9\times 2000=Rs.\ 18000$

Salary of second person

$=7\times 2000=Rs.\ 14000$

Hence, this is the answer.

$(a+b,1)=(5,a-b)$ , then the value of

$2a+3b=$ ?

0%
$5$
0%
$12$
0%
$8$
0%
$6$

Explanation

According to question

$a + b = 5$

$1 = a - b$

Adding above equations, we get

$2a = 6 \Rightarrow a = 3$

Put

$a=3$ in

$a+b = 5 \Rightarrow 3+b = 5 \Rightarrow b = 2$

$a=3$ and

$b=2$

So value of

$2a + 3b = 6 + 6 = 12$

$\bar{x}-Z=3,$

$\bar{x}+Z =45$ and

$2\bar{x}+Z=3M$ then find

$M$ .

0%
$26$
0%
$22$
0%
$24$
0%
$23$

Explanation

Given equations are

$\bar{x}-Z=3$ ......

$(1)$

$\bar{x}+Z=45$ ......

$(2)$

$3M=2\bar{x}+Z$ .......

$(3)$

Adding 1 and 2 equations, we get

$2\bar{x}=48$

$\therefore \bar{x}=24$
Now,

$\bar{x}+Z=45$

$\therefore Z=45-\bar{x}$

$=45-24=21$

$3M=2\bar{x}+Z$

$=2(24)+Z$

$=48+21$

$=69$

$\therefore M=23$

Solve the following system of equations by elimination method.

$8x-3y=5xy, 6x-5y=-2xy,x\not=0,y\not=0$

0%
$\left ( \dfrac{11}{23}, \dfrac{22}{31} \right )$
0%
$\left ( \dfrac{1}{23}, \dfrac{22}{31} \right )$
0%
$\left ( \dfrac{11}{23}, \dfrac{2}{31} \right )$
0%
None of these

Explanation

Consider the given equations.

$8x - 3y = 5xy$

$-------( 1 )$

$6x - 5y = -2xy$

$-------( 2 )$

From

$(1)$ and

$(2)$

Multiply by

$(8)$ in

$( 2)$ and by

$(6)$ in

$( 1 )$

$48x - 18y = 30xy$

$48x - 40y = -16xy$

—————————

$22y = 46xy$

$x = \dfrac{11}{23}$

$[$ put in

$( 2 )\ ]$

$6\times \dfrac{11}{23} - 5y = -2\times \dfrac{11}{23}y$

$\dfrac{66-115y}{23} = -\dfrac{22}{23}y$

$66-115y= -22y$

$93y=66$

$y=\dfrac{66}{93}$

$y=\dfrac{22}{31}$

Hence, the value of

$x$ is

$\dfrac{11}{23}$ and the value of

$y$ is

$\dfrac{22}{31}$ .

Solve the following system of equations by elimination method.

$\dfrac{3}{x}+\dfrac{5}{y}=\dfrac{20}{xy}, \dfrac{2}{x}+\dfrac{5}{y}=\dfrac{15}{xy}, x\neq 0, y\neq 0$

0%
$(1, 5)$
0%
$(2, 5)$
0%
$(1, 2)$
0%
None of these

Explanation

Consider the given equations.

$\dfrac{3}{x} + \dfrac{5}{y} = \dfrac{20}{xy}$

$5x+3y=20$

$-------( 1 )$

$\dfrac{2}{x} +\dfrac{5}{y} = \dfrac{15}{xy}$

$5x+2y=15$

$-------(2)$

Subtract

$(1)$ and

$(2)$

$5x + 3y = 20$

$5x + 2y =15$

—————————

$y = 5$

$[$ put in

$( 2 )\ ]$

$5x+2\times 5=15$

$5x+10=15$

$5x=5$

$x=1$

Hence, the value of

$x$ is

$1$ and the value of

$y$ is

$5$ .

To eliminate x from equations

$x + y + 3x = 12 \rightarrow$ ....... (1) &

$8x + 3y = 17 \rightarrow$ ..... (2), equation (2) is to be multiplied by ..............( from the given options)

0%
8
0%
$- \dfrac{1}{2}$
0%
3
0%
None of the above

Explanation

The given equations are

$x + y + 3x = 12$ ........ (1)
and

$8x + 3y = 17$ ....... (2)
Equation (i) can be further simplified as

$4x + y = 12$
So, to eliminate x, equation (2) should be multiplied by

$- \dfrac{1}{2}$ .

Solve the following system of equations using elimination method.

$3x+4y=24, 20x-11y=47$

0%
$(4 , 3)$
0%
$(2 , 3)$
0%
$(4 , 2)$
0%
None of these

Explanation

Consider the given equations.

$3x+4y=24$

$-------( 1 )$

$20x-11y=47$

$-------(2)$

From

$(1)$ and

$(2)$

On multiplying by

$20$ in equation

$(1)$ and multiplying by

$3$ in equation

$(2)$ and subtract, we get

$60x + 80y = 480$

$60x -33y =141$

—————————

$113y = 339$

$y=3$

$[$ put in

$( 2 )\ ]$

$20x-11\times 3=47$

$20x-33=47$

$20x=80$

$x=4$

Hence, the value of

$x$ is

$4$ and the value of

$y$ is

$3$ .

The solution of

$8x+5y=9$ ,

$3x+2y=4$ lie on

$x+y=k$ , then

$k=$

0%
$3$
0%
$4$
0%
$5$
0%
$2$

Explanation

$8x+5y=9 ---(1) \times 2$

$3x+2y=4 ---(2) \times 5$

subtracting both

$(16x+10y) - (15x+10y)=18-20$

$x=-2$

Putting

$x=-2$ in

$(2)$

$3x+2y=4$

$\Rightarrow 3(-2)+2y=4$

$\Rightarrow 2y=10$

$\Rightarrow y=5$

Since point

$(-2,5)$ lies on

$x+y=k$

So,

$-2+5=k$

$\Rightarrow k=3$

The solution of

$\sqrt{2}x+\sqrt{3}y=0$ and

$\sqrt{3}x+\sqrt{8}y=0$ is

0%
$(1, 2)$
0%
$(\sqrt{2}, \sqrt{8})$
0%
$(0, 0)$
0%
$(\sqrt{3}, \sqrt{2})$

Explanation

$\sqrt 2 x + \sqrt 3 y = 0 - - - \,\,\,\,\,\left( 1 \right) \times \sqrt 3$

$\sqrt 3 x + \sqrt 8 y = 0 - - - \,\,\,\,\,\left( 2 \right) \times \sqrt 2$

subtracting both

$\left( {\sqrt 6 x + 3y} \right) - \left( {\sqrt 6 x + 4y} \right) = 0 - 0$

$- y = 0 \Rightarrow y = 0$

From (1)

$\sqrt 2 x + \sqrt 3 y = 0$

$\Rightarrow \sqrt 2 x + 0 = 0$

$\Rightarrow x = 0$

So, point of intersection is

$(0,0)$

$7$ pens and

$3$ books cost

$Rs 1395$ , while

$5$ pens and

$4$ books cost

$Rs 1665$ . Find the cost of a book

0%
$360$
0%
$45360$
0%
$35350$
0%
$none\ of\ these$

Explanation

Let the cost of

$1$ pen be

$x$

and the cost of

$1$ book be

$y$

Then

$7x+3y=1395$ -----

$(1) \times 4$

$5x+4y=1665--(2) \times 3$

$4\times (1)$ --------

$3 \times (2)$

$(28x+12y)-(15x+12y)=5580-4995$

$13x=585$

$x=45$

Putting

$x=45$ in (1)

$7 \times 45+3y=1395$

$\Rightarrow 315+3y=1395$

$\Rightarrow 3y=1395-315$

$\Rightarrow 3y=1080$

$\Rightarrow y=360$

Therefore Cost of one pen is Rs

$45$ and cost of one book is Rs

$360$

The sum of two numbers is

$2$ and their difference is

$1$ . Find the numbers.

0%
$2,0$
0%
$1.5,0.5$
0%
$1,1$
0%
$-0.5,2.5$

Explanation

Let two numbers be

$x$ and

$y$

According to question sum of

$x$ and

$y$ is

$2$ and difference is

$1$

$\therefore x+y=2......(1)$ and

$x-y=1......(2)$

Now adding equation

$(1)$ and

$(2)$

$x+y=2$

$\underline{x-y=1}$

$2x\quad=3$

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

$\Rightarrow x=1.5$

Now substitute the value of

$x$ in equation

$(1),$

$1.5+y=2$

$\Rightarrow y=2-1.5$

$\Rightarrow y=0.5$

Solve:

$4x + \frac{6}{y} = 15$

$6x - \frac{8}{y} = 14$

0%
$x=-3,y=2$
0%
$x=2,y=3$
0%
$x=3,y=-2$
0%
$x=3,y=2$

Explanation

Given:

$4x+\frac{6}{y}=15......(1)$

$6x-\frac{8}{y}=14......(2)$

Multiplying equation (1) by 4 and equation (2) by 3 and then adding them,

$16x+\frac{24}{y}=60$

$\underline{18x-\frac{24}{y}=42}$

$34x\quad=102\Rightarrow x=3$

Now putting the value of x in equation (1)

$4\times3+\frac{6}{y}=15\Rightarrow\frac{6}{y}=15-12\Rightarrow\frac{6}{y}=3\Rightarrow y=\dfrac{6}{3}=2$

Solve for

$x$ and

$y$ by using method of substitution :

$0.2x+0.3y=1.3$ ;

$0.4x+0.5y=2.3$

0%
$-2,-3$
0%
$2,-3$
0%
$2,3$
0%
$-2,3$

Explanation

$0.2x+0.3y=1.3 \Rightarrow 2x+3y=13$

$\Rightarrow x=\dfrac{13-3y}{2}$

and

$0.4x+0.5y=2.3\Rightarrow 4x+5y=23$

$\Rightarrow 4\left ( \dfrac{13-3y}{2} \right )+5y=23$

$\Rightarrow 26-6y+5y=23$

$\Rightarrow 3=y$

$\therefore x=\dfrac{13-3\times 3}{2}=2$

To eliminate x, from

$3x+y=7$ and

$-x+2y=2$ second equation is multiplied by ______

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

Explanation

$by\>multiplying\>second\>equation\>with\>3\>,\>we\>get\\-3x+6y=6\\now\>by\>adding\>it\>with\>equation\\3x+y=7,\>values\>of\>unknowns\>can\>be\>calculated$

Solve

$5x-4y+8=0$

$7x+6y-9=0$ .

0%

$y = -\cfrac{101}{58}$
0%

$y = \cfrac{101}{58}$
0%

$y = \cfrac{101}{58}$
0%
None of these

Explanation

$5x-4y+8 = 0$ ...

$\times 7$

$7x+6y-9 = 0$ ....

$\times 5$

$35x-28y+56 = 0$

$35x+30y-45 = 0$

$58y = 101$

$y = \cfrac{101}{58}$

$x = -\cfrac{6}{29}$

Find the value of

$x$ and

$y$ using substitution method:

$x- y = 2$ and

$2x - y = 9$

0%
$7$ and $7$
0%
$7$ and $5$
0%
$-7$ and $-5$
0%
$5$ and $7$

Explanation

$x- y = 2$ -------

$(1)$

$2x - y = 9$ ------

$(2)$
From equation

$(1)$ :

$y = x - 2$
Substitute the value of

$y$ in equation

$(2)$ :

$2x - y = 9$

$2x -x + 2 = 9$

$x = 9 -2$

$x = 7$
Now, Substitute

$x = 7$ in equation

$(1)$ :

$x- y = 2$

$7- y = 2$

$y = 7 -2$

$y = 5$
Therefore the solution is:

$x = 7$ and

$y = 5$

Solve :

$ax + by = a - b$

$bx - ay = a + b$

0%
$x=a,y=-b$
0%
$x=1,y=-1$
0%
$x=-1,y=1$
0%
$x=\dfrac {1}{a},y=-\dfrac {1}{b}$

Explanation

$ax+by=a-b$ ............ (1)

$bx-ay=a+b$ .............. (2)

Multiply equation (1) with

$a$ and equation (2) with

$b$ ,

$a^{2}x+aby=a^{2}-ab$ ................. (3)

$b^{2}x-aby=ab+b^{2}$ ................. (4)

Adding equation (3) and (4), we get

$a^{2}x+b^{2}x=a^{2}+b^{2}$

$\left ( a^{2}+b^{2} \right )x=a^{2}+b^{2}$

$\Rightarrow x=1$

Substitute

$x=1$ in equation (1),

$a(1)+by=a-b$

$by=-b$

$y=-\cfrac{b}{b}=-1$

Solve

$0.2x+0.3y=1.3$

$0.4x+0.5y=2.3$

0%
$x=-2$ and $y=3$
0%
$x=3$ and $y=3$
0%
$x=2$ and $y=3$
0%
None of these

Explanation

Let

$0.2x+0.3y=1.3$ ....

$\left(1\right)$

$0.4x+0.5y=2.3$ ....

$\left(2\right)$

Multiply eqn

$\left(1\right)$ by

$10$ we get

$2x+3y=13$ ....

$\left(3\right)$

Multiply eqn

$\left(2\right)$ by

$10$ we get

$4x+5y=23$ ....

$\left(4\right)$

Solving equations

$\left(3\right)$ and

$\left(4\right)$

$5\times \left(3\right)\Rightarrow 10x+15y=65$

$3\times \left(4\right)\Rightarrow 12x+15y=69$

Subtracting we get

$12x+15y-10x-15y=69-65=4$

$2x=4$ or

$x=2$

Substituting

$x=2$ in eqn

$\left(3\right)$ we get

$2x+3y=13\Rightarrow 2\times 2+3y=13$ or

$3y=13-4=9$ or

$y=3$

Hence

$x=2,y=3$

Given two straight lines

$x + y - 7 = 0$ and

$x - y + 3 = 0$ . Find point of intersection of them.

0%
$(2,5)$
0%
$(-2,-5)$
0%
$(2,-5)$
0%
$(-2,5)$

Explanation

$x+y-7=0.......eq1\\x-y+3=0......eq2$

Add both the equations

Thus we get,

$2x-4=0\\2x=4\\x=\dfrac{4}{2}\\x=2\\2+y-7=0\\y=7-2=5\\(x,y)=(2,5)$

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations 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 Linear Equations Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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