MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 2

The cost of

$4$ pens and

$4$ pencil boxes is Rs

$100$ . Three times the cost of a pen is Rs

$15$ more than the cost of a pencil box. The cost of a pen and a pencil box respectively are

0%
Rs $15$ and Rs $8$
0%
Rs $10$ and Rs $15$
0%
Rs $12$ and Rs $10$
0%
Rs $16$ and Rs $12$

Explanation

Let the cost of a pen and a pencil be

$x$ and

$y$ respectively.

$4x+4y=100$

$x+y=25$ ...(1)

$3x=y+15$

$3x-y=15$ ...(2)
Adding (1) and (2), we get

$4x=40$

$x=10$
Substitute

$x=10$ in equation (1) to get

$y=15$ .

Solve the following pair of simultaneous equations:

$2x+3y = 12; \, \, 5x-3y=9$

0%
$x=1;\, \, y=0$
0%
$x=3;\, \, y=2$
0%
$x=7;\, \, y=3$
0%
$x=2;\, \, y=5$

Explanation

Given equations are

$2x+3y=12\quad\quad\dots(i)$

$5x-3y=9\quad\quad\dots(ii)$

Add equations

$(i)$ and

$(ii)$ to eliminate

$y$ ,

$\begin{array}{l}\underline {\begin{array}{ccccccccccccccc}{2x}& + &{3y}& = &{12}\\{5x}& - &{3y}& = &9\end{array}} \\\begin{array}{ccccccccccccccc}{7x}&{}&{\;\;\;\;\;\;\;}& = &{\;21}\end{array}\\\quad\;\quad\quad\quad \quad x = 3\end{array}$

Substitute this value in equation

$(i)$ ,

$2(3)+3y=12$

$\Rightarrow 6+3y=12$

$\Rightarrow 3y=6$

$\Rightarrow y=2$

Therefore, the solution is

$x=3, y=2$ .

The ages of Hari and Harry are in the ratio

$5:7$ . If four years from now, the ratio of their ages will be

$3:4$ , then the present age of

0%
Hari is $20$ years and Harry is $28$ years.
0%
Hari is $28$ years and Harry is $20$ years.
0%
Hari is $25$ years and Harry is $35$ years.
0%
Hari is $35$ years and Harry is $25$ years.

Explanation

Suppose Hari's present age and Harry's present age are

$x$ and

$y$ years respectively.
Therefore,

$\dfrac{x}{y}=\dfrac{5}{7}$

$7x-5y=0$

$\Rightarrow x=\dfrac{5y}{7}$ ...(1)

After four years, their ages will be

$x+4$ and

$y+4$ years respectively.
Therefore,

$\dfrac{x+4}{y+4}=\dfrac{3}{4}$

$4x+16=3y+12$

$4x-3y=-4$ ...(2)

Putting the value of

$x$ from (1) in equation (2), we get

$4\left(\dfrac{5y}{7}\right)-3y=-4$

$20y-21y=-28$

$y=28$ years

Putting the above value of

$y$ in equation (1), we get

$x=\dfrac{140}{7} = 20$ years.

Hence, option A.

Determine the vertices of the triangle formed by the lines

$y = x, 3y = x$ and

$x + y = 8$ .

0%
$(0, 0), (0, 4), (6, 2)$
0%
$(0, 0), (4, 4), (0, 2)$
0%
$(0, 0), (7, 7), (5, 2)$
0%
$(0, 0), (4, 4), (6, 2)$

Explanation

The vertices of the triangle will be the point of intersection of given lines taken two at a time. So, we need to find the points of intersection of the lines:

A. Taking

$y = x$

$\dots (1)$

$3y = x$

$\dots (2)$

Let, the point of intersection of above two lines is

$A$ .

Subtract

$(1)$ from

$(2)$ :

$2y = 0$

$y = 0$

Put this value of

$y$ in equation

$(1)$ :

$y=x=0$

Hence, the coordinates of A are

$(0,0)$ .

B. Taking

$y = x$

$\dots (2)$

$x+y = 8$

$\dots (3)$

Let, the point of intersection of above two lines is

$B$ .

Subtract

$(2)$ from

$(3)$ :

$x+y-y = 8-x$

$2x = 8$

$x= 4$

Put this value of

$x$ in equation

$(3)$ :

$y=x=4$

Hence, the coordinates of B are

$(4,4)$ .

C. Taking

$x+y = 8$

$\dots (3)$

$x = 3y$

$\dots (1)$

Let, the point of intersection of above two lines is

$C$ .

Subtract equation

$(1)$ from

$(3)$ :

$x+y-x = 8-3y$

$y=8-3y$

$4y = 8$

$y=2$

Put this value of

$y$ in equation

$(1)$ :

$x=3(2) = 6$

Hence, the coordinates of C are

$(6,2)$ .

The coordinates of the triangle are

$(0,0), (4,4)$ and

$(6,2).$

Solve the following pairs of equations:

$\dfrac{x}{a}+\dfrac{y}{b}=a+b;\ \dfrac{x}{a^2}+\dfrac{y}{b^2}=2;\ a,b \neq 0$

0%
$x = 2a^2 , y = 2b^2$
0%
$x = a^3 , y = 2b^2$
0%
$x = a^3 , y =3 b^2$
0%
$x = a^2 , y = b^2$

Explanation

Solution:-

$\cfrac{x}{a} + \cfrac{y}{b} = a + b \; \longrightarrow eq. \left( 1 \right)$

$\cfrac{x}{{a}^{2}} + \cfrac{y}{{b}^{2}} = 2 \; \longrightarrow eq. \left( 2 \right)$

$\text{By multiplying eq.} \left( 1 \right) \; by \; \cfrac{1}{a}, \text{we get}$

$\cfrac{x}{{a}^{2}} + \cfrac{y}{ab} = 1 + \cfrac{b}{a} \; \longrightarrow eq. (3)$

On subtracting eq.

$\left( 3 \right) \; from \; \left( 2 \right)$ , we get

$\cfrac{x}{{a}^{2}} + \cfrac{y}{{b}^{2}} - \cfrac{x}{{a}^{2}} - \cfrac{y}{ab} = 2 - 1 - \cfrac{b}{a}$

$\Rightarrow \; \cfrac{y}{b} \left( \cfrac{1}{b} - \cfrac{1}{a} \right) = 1 - \cfrac{b}{a} \Rightarrow \cfrac{y}{b} = b$

$\Rightarrow \; y = {b}^{2}$

On putting the value of y in eq.

$\left( 1 \right)$ , we get

$\cfrac{x}{a} + \cfrac{{b}^{2}}{b} = a + b \Rightarrow x = {a}^{2}$

Hence,

$x = {a}^{2} \; \& \; y = {b}^{2}$ .

$2x + y = 23$ and

$4x - y = 19$ , find the values of

$5y - 2x$

0%
$31$
0%
$22$
0%
$65$
0%
$10$

Explanation

The given equations are

$2x+y=23$ ...(i)

$4x-y=19$ ...(ii)

On adding (i) and (ii), we get

$6x=42\Rightarrow x=7$

On putting

$x=7$ in (i), we get

$14+y=23\Rightarrow y=9$

Now,

$5y-2x=5\times9-2\times7=31$

Use the method of substitution to solve the equations

$x+2y=-4$ and

$4x+5y=2$

0%
$\dfrac{-16}{3}$ and $\dfrac{-2}{3}$
0%
$\dfrac{-16}{3}$ and $\dfrac{2}{3}$
0%
$8$ and $-6$
0%
$\dfrac{16}{3}$ and $\dfrac{2}{3}$

Explanation

$x+2y=-4$

$\implies x=-4-2y$

Substituting

$x=-4-2y$ in

$4x+5y=2$

$\implies 4(-4-2y)+5y=2$

$\implies -16-8y+5y=2$

$\implies -16-3y=2$

$\implies -16-2=3y$

$\implies y=\dfrac{-18}{3}=-6$

Substituting

$y=-6$ in

$x=-4-2y$ , we get,

$x=-4-2(-6)$

$=-4+12=8$

$\therefore x=8, y=-6$

The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages, in years, of the son and the father respectively are

0%
$4$ and $24$
0%
$5$ and $30$
0%
$6$ and $36$
0%
$3$ and $24$

Explanation

$\textbf{Step1: Convert given data into equations .}$

$\text{Let, the present age of father = x years}$

$\text{Let, the present age of son = y years}$

$\text{Given that fathers Age is $$6$$ time of sons age}$

$\Rightarrow x=6y \ldots (1)$

$\text{After four years,}$

$\text{Age of the father will be four times of his son's age}$

$\Rightarrow x+4=4 (y+4)$

$\Rightarrow x+4=4y+16$

$\Rightarrow x=4y+16-4$

$\Rightarrow x=4y+12 \ldots(2)$

$\textbf{Step2: Solving the equations.}$

$\text{Substitute equation $$(1)$$ in equation $$(2)$$}$

$\Rightarrow 6y= 4y+12$

$\Rightarrow 2y=12$

$\Rightarrow y=6$

$\text{Substitute $$y=6$$ in equation (1)}$

$\Rightarrow x=36$

$\textbf{Hence, the solution of the father and his son ages are $$x=36, y=6$$.}$

$x=a, y=b$ is the solution of the equations

$x-y=2$ and

$x+y=4$ , then the values of

$a$ and

$b$ are, respectively:

0%
$3$ and $5$
0%
$5$ and $3$
0%
$3$ and $1$
0%
$-1$ and $-3$

Explanation

Given,

$x-y=2$ and

$x+y=4$ .

Adding both the equations, we get

$2x=6$

$\Rightarrow x=3$

Putting the value of

$x$ in the first equation, we get

$3-y=2$

$\Rightarrow y=1$

As stated in the question,

$x=a$ and

$y=b$ .

Thus,

$a=3$ and

$b=1$

The values of x and y satisfying the two equations

$32x + 33y = 31$ ,

$33x + 32y = 34$ respectively will be

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

Explanation

Multiplying

$32x+33y=31$ with

$33$ , we get

$1056x + 1089y = 1023$ ---- (1)

Multiplying

$33x+32y=34$ with

$32$ , we get

$1056x + 1024y = 1088$ ---- (2)

Subtracting (2) from (1),

$1056x + 1089y - 1056x - 1024y = 1023 - 1088$

$65y = - 65$

$y = -1$

Substituting

$y = - 1$ in

$32x+33y=31$ , we get $$ 32x - 33 = 31

$32x = 64$

$x = 2$

$p+q=k,\ p-q=n$ and

$k > n$ , then

$q$ is ________ .

0%
positive
0%
negative
0%
$0$
0%
none of the above

Explanation

$p+q=k$ ------- (1)

$p-q=n$ ------- (2)
Subtracting (2) from (1),

$2q=(k-n)$
Since

$k>n$ ,

$(k-n)$ is positive, therefore,

$q=\dfrac{k-n}{2}$ , which is positive

Solution of the equations

$\cfrac{x + 3}{4} + \cfrac{2y + 9}{3} = 3$ and

$\cfrac{2x - 1}{2} - \cfrac{y + 3}{4} = 4 \cfrac{1}{2}$ is

0%
$x = - 5, y = - 3$
0%
$x = - 5, y = 3$
0%
$x = 5, y = 3$
0%
$x = 5, y = -3$

Explanation

$\dfrac{x+3}{4}+\dfrac{2y+9}{3}=3$

Multiplying both sides by

$12$ we get,

$\Rightarrow$

$3x+9+8y+36=36$

$\Rightarrow$

$3x+8y=-9$ ----- ( 1 )

$\dfrac{2x-1}{2}-\dfrac{y+3}{4}=4\dfrac{1}{2}$

$\dfrac{2x-1}{2}-\dfrac{y+3}{4}=\dfrac{9}{2}$

Multiplying both sides by

$4$ , we get

$\Rightarrow$

$4x-2-y-3=18$

$\Rightarrow$

$4x-y=23$ ---- ( 2 )

Multiplying equation ( 1 ) by

$4$ ,

$12x+32y=-36$ ----- ( 3 )

Multiplying equation ( 2 ) by

$3$ ,

$12x-3y=69$ ---- ( 4 )

Subtracting equation ( 4 ) from ( 3 ) we get,

$\Rightarrow$

$35y=-105$

$\therefore$

$y=-3$

Substituting

$y=-3$ in equation ( 1 )

$3x+8(-3)=-9$

$3x-24=-9$

$3x=15$

$\therefore$

$x=5$

$\therefore$

$x=5$ and

$y=-3$

The solution of

$37x+41y=70$

$41x+37y=86$ is

0%
$x=3, y=1$
0%
$x=3, y=-1$
0%
$x=-3, y=1$
0%
$x=1, y=3$

Explanation

The given equations are

$37x+41y=70$ and

$41x+37y=86$

According to the given passage

$a=37, b=41, c=70$ and

$d=86$

$\therefore x=\dfrac { 1 }{ 2 } \left( \dfrac { c+d }{ a+b } +\dfrac { c-d }{ a-b } \right) =\dfrac { 1 }{ 2 } \left( \dfrac { 70+86 }{ 37+41 } +\dfrac { 70-86 }{ 37-41 } \right)$

$=\dfrac { 1 }{ 2 } \left( \dfrac { 156 }{ 78 } +\dfrac { 16 }{ 4 } \right) =\dfrac { 1 }{ 2 } \left( 2+4 \right) =3$

and

$y=\dfrac { 1 }{ 2 } \left( \dfrac { c+d }{ a+b } -\dfrac { c-d }{ a-b } \right) =\dfrac { 1 }{ 2 } \left( \dfrac { 70+86 }{ 37+41 } -\dfrac { 70-86 }{ 37-41 } \right)$

$=\dfrac { 1 }{ 2 } \left( \dfrac { 156 }{ 78 } -\dfrac { 16 }{ 4 } \right) =\dfrac { 1 }{ 2 } \left( 2-4 \right) =-1$

$\therefore \left( x,y \right) =\left( 3,-1 \right)$

The solution of

$217x+131y=913$

$131x+217y=827$ is

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

Explanation

Let

$217x+131y=913$ .....(1)

and

$131x+217y=827$ ......(2)

Here, according to the given passage

$a=217, b=131, c=913$ and

$d=827$

$\therefore 2x= \left[ \dfrac { c+d }{ a+b } +\dfrac { c-d }{ a-b } \right]$

$=\dfrac { 1740 }{ 348 } +\dfrac { 86 }{ 86 } =\dfrac { 2088 }{ 348 }$

$\therefore x=\dfrac { 2088 }{ 348\times 2 } =3$

and

$2y= \left[ \dfrac { c+d }{ a+b } -\dfrac { c-d }{ a-b } \right]$

$=\dfrac { 1740 }{ 348 } -1=\dfrac { 1392 }{ 348 }$

$\therefore y=\dfrac { 1392 }{ 348\times 2 } =2$

$\therefore x=3, y=2$

The solution of

$x+2y = \displaystyle \frac{3}{2}$

$2x+y=\displaystyle \frac{3}{2}$ is

0%
$x=3, y=1$
0%
$x = \displaystyle \frac{1}{2}, y=\frac{1}{2}$
0%
$x=\displaystyle \frac{1}{2}, y=0$
0%
$x=0, y=\frac{1}{2}$

Explanation

$The\quad given\quad equations\quad are\quad \\ x+2y=\frac { 3 }{ 2 } \quad and\quad 2x+y=\frac { 3 }{ 2 } .\\ According\quad to\quad the\quad given\quad passage\\ a=1,\quad b=2,\quad c=\frac { 3 }{ 2 } \quad and\quad d=\frac { 3 }{ 2 } .\\ \therefore \quad x=\frac { 1 }{ 2 } \left( \frac { c+d }{ a+b } +\frac { c-d }{ a-b } \right) =\frac { 1 }{ 2 } \left( \frac { \frac { 3 }{ 2 } +\frac { 3 }{ 2 } }{ 1+2 } +\frac { \frac { 3 }{ 2 } -\frac { 3 }{ 2 } }{ 1-2 } \right) \\ =\frac { 1 }{ 2 } \left( \frac { 3 }{ 3 } +0 \right) =\frac { 1 }{ 2 } \quad and\quad \\ y=\frac { 1 }{ 2 } \left( \frac { c+d }{ a+b } -\frac { c-d }{ a-b } \right) =\frac { 1 }{ 2 } \left( \frac { \frac { 3 }{ 2 } +\frac { 3 }{ 2 } }{ 1+2 } -\frac { \frac { 3 }{ 2 } -\frac { 3 }{ 2 } }{ 1-2 } \right) \\ =\frac { 1 }{ 2 } \left( \frac { 3 }{ 3 } -0 \right) =\frac { 1 }{ 2 } .\\ \therefore \quad \left( x,y \right) =\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right) .\\ Ans-\quad Option\quad B.\\$

$2^{2x - y} = 32$ and

$2^{x + y} = 16$ then

$x^{2} + y^{2}$ is equal to

0%
$9$
0%
$10$
0%
$11$
0%
$13$

Explanation

$Given\quad { 2 }^{ 2x-y }=32\\ \Rightarrow { 2 }^{ 2x-y }={ 2 }^{ 5 }\\ \Rightarrow 2x-y=5----(1)\\ Again\quad { 2 }^{ x+y }=16\\ \Rightarrow { 2 }^{ x+y }={ 2 }^{ 4 }\\ \Rightarrow x+y=4----(2)\\ Adding\quad (1)\quad and\quad (2)\quad we\quad get\\ 3x=9\\ \Rightarrow x=3\\ Substituting\quad x=3\quad in\quad (2)\quad we\quad get\\ y=1.\\ \therefore \quad { x }^{ 2 }+{ y }^{ 2 }={ 3 }^{ 2 }+{ 1 }^{ 2 }=10\quad (Ans)$

Suresh is half his father's Age. After

$20$ years, his father's age will be one and a half times the Suresh's age. What is his father's age now?

0%
$40$
0%
$20$
0%
$26$
0%
$30$

Explanation

Let Suresh's present age be

$x$
Let Father's present age be

$y$
Therefore applying condition no.

$1$

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

$y-2x=0$ ...............................................................(i)

Also applying condition no. 2

$\dfrac{3(x+20)}{2}=y+20$

$2y-3x=20$ ......................................................(ii)
Multiplying equation (i) by

$2$

$2y-4x=0$ ......................................................(iii)

Subtracting equation (iii) from equation (ii), we get

$(2y-3x-2y+4x)=20$
Therefore

$x=20$ years.
Substituting in equation (i), we get

$y=40$ years
Hence, the age of the father is

$40$ years.

$3^{x - y} = 27$ and

$3^{x + y} = 243$ , then

$x$ is equal to

0%
$0$
0%
$4$
0%
$2$
0%
$6$

Explanation

Firstly make the bases same on both the sides of equation. Apply the concept if

${ a }^{ m }={ a }^{ n }\quad then\quad m=n$ .

${ 3 }^{ x-y }=27\Rightarrow { 3 }^{ x-y }={ 3 }^{ 3 }\Rightarrow x-y=3---(1)\\ again\quad { 3 }^{ x+y }=243\Rightarrow { 3 }^{ x+y }={ 3 }^{ 5 }\Rightarrow x+y=5---(2)\\ Adding\quad (1)\quad and\quad (2)\quad we\quad get\\ 2x=8\\ or\quad x=4\quad \quad (Ans)$

$6$ kg of sugar and

$5$ kg of tea together cost Rs.

$209$ and

$4$ kg of sugar and

$3$ kg of tea together cost Rs.

$131$ , then the cost of

$1$ kg sugar and

$1$ kg tea are respectively

0%
Rs. $11$ and Rs. $25$
0%
Rs. $12$ and Rs. $20$
0%
Rs. $14$ and Rs. $20$
0%
Rs. $14$ and Rs. $25$

Explanation

Let the price of sugar be

$x$ and tea be

$y$ .
Then,

$6x + 5y = 209....(1)$

$4x + 3y = 131....(2)$

Multiply equation

$(1)$ by

$4$ and equation

$(2)$ by

$6$ and then subtract:

$(6x + 5y = 209 )4$

$(4x + 3y = 131)6$

$.............................................$

$24x + 20y - 24x - 18y = 209\times4 - 131\times 6$

$.............................................$

$2y = 836-786$

$2y = 50$

$y = 25$
Therefore,

$x = 14$
Price of sugar is

$Rs.14$ and price of tea is

$Rs.25$ .

Solve the systems of equations:

$\displaystyle \frac{x-2}{4} + \frac{y+1}{3} = 2 ; \ \frac{x+1}{7}+ \frac{y-3}{2} = \frac{1}{2}$

0%
$(6, 2)$
0%
$(2, 2)$
0%
$(2, 3)$
0%
$(3, 4)$

Explanation

$\displaystyle \frac{x-2}{4} + \frac{y+1}{3} = 2$ ......... (1)

$\displaystyle \frac{x+1}{7} + \frac{y-3}{2} = \frac{1}{2}$ ..... (2)

First, we need to simplify the two equations. Let's multiply both sides of equation (1) by 12 and simplify.

$\displaystyle 12 \left ( \frac{x-2}{4} + \frac{y+1}{3} \right ) = 12 (2)$

$3(x-2) + 4(y+1) = 24$

$3x - 6 + 4y + 4 =24$

$3x + 4y - 2 =24$

$3x + 4y = 26$

Let's multiply both sides of eq. (2) by 14

$\displaystyle 14 \left ( \frac{x+1}{7} + \frac{y-3}{2} \right ) = 14 \left ( \frac{1}{2} \right )$

$2(x+1)+7(y-3)=7$

$2x + 2 + 7y - 21 = 7$

$2x + 7y - 19 = 7$

$2x+7y = 26$

Now we have the following system to solve

$3x+4y = 26$ .............. (3)

$2x+7y = 26$ ............... (4)

Because changing the form of their eq (3) or eq. (4) in preparation for the substitution method would produce a fractional form, let's use the elimination-by-addition method. We can start by multiplying equation (4) by -3

$3x+4y=26$ ............. (5)

$-6x - 21y =-78$ .......... (6)

No, we can replace eq. (6) with an eq. we form by multiplying eq. (5) by 2 then adding that result to eq. (6)

$3x+4y=26$ ............ (7)

$-13y = -26$ ............ (8)

From eq. (8) we can find the value of y.

$-13 y = -26 \Rightarrow y=2$

Now we can substitute 2 for y in eq. (7)

$3x+4y = 26$

$3x + 4(2) = 26$

$3x= 18 \Rightarrow x=6$

The solution set is

$(6, 2).$

Choose the correct matching(s) for solving questions of the system of linear equation in two variables

	Methods	Uses/Disadvantages
$(a)$	Graphical	$(i)$ Use: When the coefficients of the variables and the solutions are integers. Disadvantage: If the solutions are not integers, they are hard to plot and read on the graph.
$(b)$	Substitution	$(ii)$ Use: When variables with coefficients that are the same or additive inverse of each other ( for example, $2x$ and $-2x$ ) are present. Disadvantage: If fractions are involved, you may have much computation.
$(c)$	Elimination	$(iii)$ Use: When one of the variables is isolated (alone) on one side of the equation Disadvantage: You may have lots of computations involving signed numbers.

0%
$(a) \rightarrow(i)$
0%
$(b) \rightarrow (iii)$
0%
$(c) \rightarrow (ii)$
0%
$(b) \rightarrow (i)$

Explanation

The correct uses and disadvantages of various methods are:

$Graphical:$

$Use -$ When the coefficient of the variables and the solutions are integers.

$Disadvantage -$ If the solutions are not integers, they are hard to read on the graph.

$Substitution:$

$Use -$ When one of the variables is isolated (alone) on one side of the equation

$Disadvantage -$ If fractions are involved, you may have much computation

$Elimination:$

$Use -$ When fractions, decimals or variables with coefficients that are the same or negative inverse of each other

$(eg: 2x\ and\ -2x)$ are present.

$Disadvantage -$ You may have lots of computations involving signed numbers.

Solve the following simultaneous equations.

$2x + y = 5, \, \, 3x-y=5$

0%
$x=2,$ $y=1$
0%
$x=-2,$ $y=1$
0%
$x=2,$ $y=-1$
0%
$x=-2,$ $y=-1$

Explanation

Let

$2x + y = 5$ --- (1)
and

$3x -y = 5$ --- (2)

Adding equations

$(1)$ , and

$(2)$ , we get

$5x = 10$

$\Rightarrow x = 2$

Now, putting

$x = 2$ in equation

$(1)$ we get,

$4 + y = 5$

$\Rightarrow y = 1$

Solve the following simultaneous equations by substitution method.

$2x +3y= -4; \, \, x-5y =11$

0%
$x=-1, \, y=2$
0%
$x=1, \, y=-2$
0%
$x=-1, \, y=-2$
0%
$x=1, \, y=2$

Explanation

$2x +3y= -4$ ..........(1)

$x-5y =11$

$\therefore x=11+5y$ .............(2)
Substituting eq. (2) in eq. (1), we get,

$2(11+5y)+3y = -4$

$22+10y+3y=-4$

$22+4 = -13y$

$\therefore y = -\dfrac {26}{13}$

$y=-2$
Substituting

$y=-2$ in (1) we get,

$x=1$

$\left (x+y,1 \right )$

$=$

$\left (3,y-x \right )$ , then

$x$

$=$ ,

$y$

$=$

0%
2, 1
0%
-1, -2
0%
1, 2
0%
2, -1

Explanation

We have,

${(x+y,1)}={(3,y-x)}$

$\implies x+y=3$ and

$y-x=1$

$\Rightarrow x+y=3$ and

$-x+y=1$

Adding both equations, we get

$2y=4$

$\Rightarrow y=2$

Put the value of

$y$ in

$x+y=3$ , we get

$x+2=3$

$\Rightarrow x=1$

So,

$\text{C}$ is the correct option.

Solve :

$3x-2y=1$

$2x+y=3$

0%
${1,1}$
0%
${3,3}$
0%
${4,4}$
0%
None of thse

Explanation

Given:

$3x-2y=1$

$2x+y=3$

$y=3-2x$

$3x-2(3-2x)=1$

$3x-6+4x=1$

$7x=7$

$x=1$

$3(1)-2y=1$

$3-2y=1$

$2y=2$

$y=1$

Sum of two numbers isIf the larger number is divided by the smaller, the quotient is 7 and the remainder isFind the numbers.

0%
$75, 14$
0%
$90, 14$
0%
$75, 18$
0%
$85, 12$

Explanation

$\textbf{Step-1: Assume the numbers to be equal to some variables& then apply all information's given.}$

$\text{Let the two numbers be}$

$x$

$\text{and}$

$y,$

$\text{where}$

$x$

$\text{is the larger number}$

$\text{Given, Sum of two numbers is 97}$

$x + y = 97$

$\text{...(1)}$

$\text{As per the given condition,}$

$\text{If the larger number is divided by the smaller, the quotient is 7 and the remainder is 1}$

$x = 7y +1$

$\text{...(2)}$

$\textbf{Step-2: Solve above equations.}$

$\text{Substitute eq(2) in eq(1) we get,}$

$7y+1 + y = 97\\ 8y = 96\\ y = 12\\ \therefore x = 7(12) +1 = 85$

$\textbf{Hence, option - D is the answer.}$

A man starts his job with a certain monthly salary and a fixed increment every year. If his salary will be Rs.

$11000$ after

$2$ years and Rs.

$14000$ after

$4$ years of his service. What is his starting salary and what is the annual increment?

0%
Starting salary is Rs. $7500$ and increment is Rs. $2500$
0%
Starting salary is Rs. $5000$ and increment is Rs. $1800$
0%
Starting salary is Rs. $8000$ and increment is Rs. $1500$
0%
Starting salary is Rs. $7000$ and increment is Rs. $1300$

Explanation

Let starting salary

$=\ x$

Increment every year

$=\ y$

$x\ + 2y\ =\ 11,000\quad$ ....... eq(1)

$x\ +\ 4y\ =\ 14,000$ ......eq(2)

Substract eq(1) from eq(2)

$2y\ =\ 3,000\\ y\ =\ 1500$

Substitute

$y =\ 1500$ in eq(1), we get

$x\ =\ 11,000\ -\ 3,000$

$x\ =\ 8,000$

starting salary

$\ =\ 8,000$

Increment

$\ = \ 1500$

Solve the following simultaneous equations:

$4m+3n=18;\, \, 3m-2n=5$

0%
$m = 3,\, n= 2$
0%
$m = -3,\, n= -2$
0%
$m = 3,\, n= -2$
0%
$m = -3,\, n= 2$

Explanation

Let

$4m + 3n = 18$ --- (1)
and

$3m -2n = 5$ --- (2)
Multiplying the first equation by

$2$ , we get

$8m + 6n = 36$ -- (3)
Multiplying the second equation by

$3$ , we get

$9m - 6n = 15$ -- (4)

Adding equations 3, and 4 , we get

$17m = 51$

$=> m = 3$

Now, putting

$m = 3$ in equation 1 we get,

$12 + 3n = 18$

$=> 3n = 6$

$=> n = 2$

Find the values of the

$x+y$ and

$x-y$ from the examples given below without solving for x and y

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

0%
$-2$ , $6$
0%
$2$ , $6$
0%
$6$ , $2$
0%
$6$ , $-2$

Explanation

$\text{The given equations are}$

$5x-3y=14$ ...(i)

$3x-5y=2$ ...(ii)

On adding (i) and (ii), we get

$8x-8y=16\Rightarrow x-y=2$

On subtracting (ii) from (i), we get

$2x+2y=12\Rightarrow x+y=6$

$\therefore x+y=6 \space and \space x-y=2$

Find the values of the

$x+y$ and

$x-y$ from the examples given below without solving for x and y

$5x+7y=17; \, \, 7x+5y=19$

0%
$-3$ , $-1$
0%
$3$ , $1$
0%
$1$ , $3$
0%
$-1$ , $3$

Explanation

$\text{The given equations are}$

$5x+7y=17$ ...(i)

$7x+5y=19$ ...(ii)

$\text{On adding (i) and (ii), we get}$

$12x+12y=36\Rightarrow x+y=3$

$\text{On subtracting (i) from (ii), we get}$

$2x-2y=2\Rightarrow x-y=1$

$\therefore x+y=3 \space and \space x-y=1$

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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