MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 11

Henry is three times as old as Truman. Two years ago, Henry was five times as old as Truman. How old is Henry now?

0%
4
0%
8
0%
12
0%
16
0%
20

The solution of the system of equations

$\displaystyle \frac{2x+5y}{xy} = 6$ and

$\displaystyle \frac{4x-5y}{xy} + 3 = 0$ (where

$x \neq 0, y \neq 0$ ), respectively is ___________.

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

Explanation

Equation

$1 : \dfrac{2x+5y}{xy} = 6$

$\Rightarrow 2x+5y = 6xy$ ........

$(1)$

Equation

$2 : \dfrac{4x-5y}{xy} + 3 = 0$

$\Rightarrow 4x-5y = -3xy$ ........

$(2)$

Adding both the equations, we get

$6x = 3xy$

$\therefore y = 2$

Substituting value of

$y$ in

$(1)$ we get

$2x + 10 = 12x$

$\therefore x = 1$

$\left| x \right| +x+y=10$ and

$x+\left| y \right| -y=12$ then find the value of '

$x+y$ '.

0%
$\dfrac { 18 }{ 5 }$
0%
$\dfrac { 17 }{ 5 }$
0%
$\dfrac { 8 }{ 5 }$
0%
None of these

Explanation

Assuming

$y>0$ leads to

$x = 12$ from 2nd equation

as:

$x+y-y = 12$

$x= 12$

and then from 1st Equation

$|x|+x+y = 10$

$2x+y = 10$ or

$y = 14$

which is contradiction as

$y>0$

Assuming

$y \leq 0$ leads to two cases

(i) if

$x\leq 0$ then from 1st Equation

$y = 10$ and

that contradictory again

(2) if

$x>0$ then

$2x+y = 10...(1), x-2y = 12 ...(2)$

$(1)\times 2+(2)$

$4x+2y=20$

$x-2y = 12$

__________

$5x = 32$

$x = \dfrac{32}{5}$

So on subtituting the value of

$x$ in equ (2)

$\dfrac{32}{5}-2y = 12$

$2y = \dfrac{32}{5}-12$

$y = \dfrac{-14}{5}$

$\therefore x+y=\dfrac{32}{5}+\dfrac{-14}{5}$

$\therefore x+y=\dfrac{18}{5}$

Some students are divided into two groups A & B. If

$10$ students are sent from A to B, the number in each is the same. But if

$20$ students are sent from B to A, the number in A is double the number in B. Find the number of students in each group A & B.

0%
$100, 80$
0%
$80, 100$
0%
$110, 70$
0%
$70, 110$
0%
None of these

Explanation

Let the number of students in

$A$ and

$B$ be

$a \& b$ respectively.
As per the question:

$a-10 = b + 10$

$a- b = 20 ... (i)$
and

$a + 20 = 2 (b -20)$

$a - 2b = -60..... (ii)$
Since coefficient of

$a$ is same in both equations, we can subract

$(ii)$ from

$(i)$ .

$(i)-(ii) :$

$a- b -(a-2b) = 20-(-60)$

$\implies b =80$

$a =80+20 =100$

A sum of

$Rs. 6.25$ is made up of

$80$ coins which are either

$10$ paise or

$5$ paise coins. The number of

$10$ paise and

$5$ paise coins are _____ and ____ respectively.

0%
$45 , 35$
0%
$42 , 38$
0%
$39, 41$
0%
$37 , 43$

Solve the following pairs of equations.

$3x-y=0$ ;

$x-2=0$

0%
$x=2,y=6$
0%
$x=3,y=6$
0%
$x=4,y=6$
0%
None of these

Explanation

As we are give

$3x-y=0$ &

$x-2= 0$

$\therefore$

$x=2$ &, on substituting the value of

$x$ in

$3x-y=0$

$\therefore y=6$

The graph of the equations

$x + y = 5$ and

$x -y = 1$ will be.....

0%
Parallel lines
0%
Intersecting lines
0%
Lines coincide
0%
Concurrent lines

Explanation

Take few points on bothe lines and draw the graphs of both lines.

For

$x + y = 5$ :

x	0	3	5
y	5	2	0

For

$x - y = 1$ :

x	0	1	3
y	-1	0	2

From the graphs, we can say that the lines are intesecting at

$(3,2)$ .

Srikant bought two cows for Rs.

$30000$ . By selling one at a loss of

$15\%$ and the other at a gain of

$19\%$ , he found that selling price of both cows is the same. Find the cost price of each ( in Rs).

0%
$10000, 20000$
0%
$15000, 15000$
0%
$17500, 12500$
0%
$17000, 12500$

Explanation

Let

$x,y$ be the

$C.P$ of the

$2$ cows

$x+y=3000-1$ (given)

One is sold at

$15$ % loss and other at gain of

$19$ % S.P of these

$2$ are equal

$x-\cfrac { 15 }{ 100 } x=y+\cfrac { 19 }{ 100 } y$

$85x=119y\Rightarrow x=\cfrac { 119 }{ 85 } y$

Using

$1$ and above result

$\cfrac { 119 }{ 85 } y+y=30000$

$204y=85\times 30000$

$y=12,500$

$x=30000-12500=17500$

Cost price of each cow is

$17500,12500$

Solve the following pairs of equations.

$4x-y-5=0$ ;

$x+y-5=0$

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

Explanation

$x+y=5$ .......eq1;

$4x-y=5$ ......eq2

Add both the equations, we will get

$x=2$

Put

$x= 2$ in eq1, we get

$y=3$

Solve the following pairs of equations.

$2x-4=0$ ;

$4x+y+4=0$

0%
$\left( 2,-12 \right)$
0%
$\left( 1,-12 \right)$
0%
$\left( 2,12 \right)$
0%
None of these

Explanation

$2x-4= 0 \implies x=2$

and

$4x+y =-4$

$4(2)+y=-4$

$8+y =-4$

$y =-12$

$\Rightarrow x=2, y = -12$

Solve the following pairs of equations.

$x-y=0$ ;

$y+3=0$

0%
$\left( -3,-3 \right)$
0%
$\left( 3,-3 \right)$
0%
$\left( 2,-3 \right)$
0%
None of these

Explanation

Consider

$x-y=0$

$\Rightarrow x=y$

$y+3=0 \implies y =-3$

$\Rightarrow x= y =-3$

Hence solution for given equations is

$(-3,-3)$

Solve the following pairs of equations.

$2x=y+1$ ;

$x+2y-8=0$

0%
$x= 2,y=3$
0%
$x= 3,y=3$
0%
$x= 2,y=2$
0%
None of these

Explanation

$2x=y+1$ .....(ii) ;

and

$x+2y=8$

$x=8-2y$ .....(i)

Put (i) in (ii)

$16-4y=y+1$

$y=3$

$y=3$ &

$x=2$

Solve the following pairs of equations.

$x+y=5$ ;

$x-y=1$

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

Explanation

$x-y = 1$ ....eq1

$x+y=5$ .....eq2

Add & subtract both the equations, we will get

$2X=6$

$\implies x=3$ &

Put the value of x in equ 1

$y=2$

Solve the following pairs of equations.

$y=2x+1$ ;

$y+3x-6=0$

0%
$\left( 1,3 \right)$
0%
$\left( 2,3 \right)$
0%
$\left( 4,3 \right)$
0%
None of these

Explanation

Here given equations are

$y-2x= 1$ ;

$y+3x=6$

Subtract both the equations, we will get

$x=1$ .

On substituting the value of

$x$ in any given equations we get

$y=3$

Solve the following puzzles using the equations:

In a village , the number of women is 89 more than the number of men.The number of children is 400 more than the number of men.If the total population of the village isFind the number of men , women and children.

0%
Men-1500 , Women 1589 , Children 1900.
0%
Men-1500 , Women 1600 , Children 1900.
0%
Men-1500 , Women 1589 , Children 1200.
0%
Men-1400 , Women 1589 , Children 1900.

Solve the following equations by substitution method.

$x + 3y = 10; 2x + y = 5$

0%
$x = 1, y = 3$
0%
$x = 2, y = 3$
0%
$x = 1, y = 1$
0%
None of these

Explanation

$x+3y=10$ .....

$(1)$ ;

$2x+y=5$ ........

$(2)$

$y=5-2x$ ......from

$(2)$

putting in

$(1)$ , we get

$\Rightarrow x+3(5-2x)=10$

$\Rightarrow x+15-6x=10$

$\Rightarrow -5x=-5$

$\Rightarrow x=1$

$1+3y=10$

$y=3$

Solve the following equations by substitution method.

$2x + y = 1; 3x - 4y = 18$

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

Explanation

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

$3x-4y=18$ ........(2)

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

putting value of

$y$ in (1)

$\Rightarrow 3x-4(1-2x)=18$

$\Rightarrow 3x-4+8x=18$

$\Rightarrow 11x=22$

$\Rightarrow x=2$

Using value of

$x$

$\Rightarrow 4+y=1$

$\Rightarrow y=-3$

Solve the system of equations

$65x-33y=97,$ and

$33x-65y=1$ by substitution method.

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

Solve the following equations by substitution method.

$\dfrac{1}{x} + \dfrac{2}{y} = 9; \dfrac{2}{x} + \dfrac{1}{y} = 12 (x \neq 0, y \neq 0)$

0%
$x = \dfrac{1}{5} , y = \dfrac{1}{2}$
0%
$x = \dfrac{2}{5} , y = \dfrac{1}{2}$
0%
$x = \dfrac{1}{5} , y = \dfrac{-1}{2}$
0%
None of these

Explanation

Given equations,

$\dfrac{1}{x} + \dfrac{2}{y} = 9$ ..............(1)

$\dfrac{2}{x} + \dfrac{1}{y} = 12$ ..............(2)

$\dfrac{1}{x}=9-\dfrac{2}{y}$ .........from (1)

putting value of

$\dfrac {1}{x}$ in (2), we get

$2\left[9-\dfrac{2}{y}\right]+\dfrac{1}{y}=12$

$18-\dfrac{3}{y}=12$

$6=\dfrac{3}{y}$ ......

$\boxed{y=\dfrac{1}{2}}$

putting value of

$y$ in (1), we get

$\dfrac {1}{x}=9-4$

$\dfrac{1}{x}=5$ ......

$\boxed{x=\dfrac{1}{5}}$

Solve the following equations by substitution method.

$\dfrac{3}{x} + \dfrac{1}{y} = 7; \dfrac{5}{x} - \dfrac{4}{y} =(x \neq 0, y \neq 0)$

0%
$x = \dfrac{1}{2}, y = 1$
0%
$x = \dfrac{-1}{2}, y = 1$
0%
$x = \dfrac{1}{2}, y = -1$
0%
None of these

Explanation

Given equations are,

$\dfrac{3}{x}+\dfrac{1}{y}=7$

$\dfrac{5}{x}-\dfrac{4}{y}=6$

$\dfrac{1}{x}=\dfrac{1}{3}\left(7-\dfrac{1}{y}\right)$ .................from (1)

Putting value of

$\dfrac {1}{x}$ in (1), we get

$5\left[\dfrac{1}{3}(7-\dfrac{1}{y})\right]-\dfrac{4}{y}=6$

$\dfrac{35}{3}-\dfrac{5}{3y}-\dfrac{4}{y}=6$

$\dfrac{17}{3}=\dfrac{5+12}{3y}$

$\boxed {y=1}$

Putting value of

$y$ in (1)

$\dfrac{1}{x}=\dfrac{1}{3}\times 6\Rightarrow \dfrac {1}{x}=2=\boxed{x=\dfrac{1}{2}}$

Solve following system of equations by elimination method.

$x+\dfrac{y}{2}=4, \dfrac{x}{3}+2y=5$

0%
$(3, 2)$
0%
$(3, 1)$
0%
$(2, 2)$
0%
None of these

Explanation

Given equations are

$x+\dfrac{y}{2}=4$ .....(i)

$\dfrac{x}{3}+2y=5$ ......(ii)

Multiply equation (i) by 4 we get

$4x+2y=16$ ....(iii)

Subtract equation (ii) from (iii) we get

$\dfrac{11}{3}x=11$

$\therefore x=3$

Put value of

$x$ in equation (iii)

$(4\times3)+2y=16$

$\Rightarrow 2y=4$

$\Rightarrow y=2$

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

Solve the following system of equations by elimination method.

$\dfrac {2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}, \dfrac{3}{x}+\dfrac{2}{y}=0, x\neq 0, y\neq 0$

0%
$(6, -4)$
0%
$(6, 4)$
0%
$(-6, -4)$
0%
None of these

Explanation

Consider the given equations.

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6}$

$4x+12y=xy$

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

$\dfrac{3}{x} +\dfrac{2}{y} = 0$

$2x+3y=0$

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

From

$(1)$ and

$(2)$

Multiply by

$2$ in

$(2)$ and by

$1$ in

$( 1 )$ and subtract

$4x + 12y = xy$

$4x + 6y = 0$

—————————

$6y = xy$

$x = 6$

$[$ put in

$( 2 )\ ]$

$2\times 6+3y=0$

$12+3y=0$

$3y=-12$

$y=-4$

Hence, the value of

$x$ is

$6$ and the value of

$y$ is

$-4$ .

Solve following system of equations by elimination method.

$x +2y=7, x-2y=1$

0%
$\left ( 4, \dfrac{3}{2} \right )$
0%
$\left ( 4, \dfrac{2}{2} \right )$
0%
$\left ( 3, \dfrac{3}{2} \right )$
0%
None of these

Explanation

Given equations are

$x+2y=7$ .......(i)

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

Subtract equation (ii) from (i), we get

$2x=8$

$\Rightarrow x=4$

put value of

$x$ in equation (ii)

$4-1=2y$

$3=2y$

$\therefore y=\dfrac{3}{2}$

$\therefore x=4 , \ y=\dfrac{3}{2}$

Solve each of the following system of equations by substitution method.

$13x +11y=70, 11x+13y=74$

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

Explanation

Since,

$13x+11y=70$ ...eq1

$11x+13y=74$ ......eq2

$x=\dfrac{70-11y}{13}$ , put this value in eq2, we get

$11\left[\dfrac{70-11y}{13}\right]+13y=74$

$770-121y+169y=74\times 13$

$\Rightarrow y=4$

and

$x=\dfrac{70-11(4)}{13}$

$x=2$

A number is

$\dfrac{2}5{}$ times another number. If their sum is

$70$ , Find the numbers.

0%
$70$ and $50$
0%
$20$ and $30$
0%
$40$ and $50$
0%
$20$ and $50$

Explanation

Let the numbers be

$x$ and

$y$

According to the question

$x=\dfrac{2}{5}y............(i)$

$x+y=70...............(ii)$

$\dfrac{2}{5}y+y=70$

$\dfrac{2y+5y}{5}=70$

$7y=350$

$y=50$

$x+50=70$

$x=20$

So the numbers are

$20$ and

$50$

Solve the following system of equations by substitution method.

$\dfrac{15}{x}+\dfrac{2}{y}=17, \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}, x\neq 0, y\neq 0$

0%
$\left ( 5, \dfrac{1}{7} \right )$
0%
$\left ( 2, \dfrac{1}{7} \right )$
0%
$\left ( 5, \dfrac{2}{7} \right )$
0%
None of these

Explanation

Given equations are

$\dfrac{15}{x}+\dfrac{2}{y}=17,$

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}$

$Let \ \dfrac{1}{x}=a, \ \ \dfrac{1}{y}=b$

$\therefore \ 15a+2b=17$ ......(i)

and

$a+b=\dfrac{36}{5}$

$\Rightarrow a=\dfrac{36}{5}-b$ ....(ii)

Putting equation (ii) in (i) we get,

$15\left[\dfrac{36}{5}-b\right]+2b=17$

$\Rightarrow 108-13b=17$

$\Rightarrow b=7$

$\therefore \ y=\dfrac{1}{7}$

Putting value of

$b$ in equation (i), we get,

$15a+(2\times7)=17$

$\Rightarrow a=\dfrac{1}{5}$

$\therefore \ x=5$

Hence,

$\left(5,\dfrac17\right)$ is the solution of the given system of equations.

Solve following system of equations by elimination method.

$3x+y=8, 5x+y=10$

0%
$(1, 5)$
0%
$(2, 5)$
0%
$(1, 6)$
0%
None of these

Explanation

Given equations are

$3x+y=8$ .....(i)

$5x+y=10$ ......(ii)

Subtract equation (i) from (ii), we get

$5x+y-3x-y=10-8$

$\Rightarrow 2x=2$

$\therefore x=1$

Put value of

$x$ in equation (i)

$(3\times 1)+y=8$

$\Rightarrow y=5$

$\therefore x=1, \ y=5$

The sum of two numbers is

$25$ and their difference is

$7$ , then the numbers are.

0%
$20$ and $5$
0%
$18$ and $7$
0%
$15$ and $10$
0%
$9$ and $16$

Explanation

let the two numbers be

$A$ and

$B$

so it is given that

$A + B = 25$ ...(1)

$A - B = 7$ ...(2)

Adding the equations we get

$A + B + A - B = 25 +7$

$2A = 32$

$A = 16$

Using the above value of A in (1)

$16 + B = 25$

$=>B = 9$

Thus

$A =16$ and

$B =9$

For what value of

$'k'$ the system of equation

$kx + 2y = 5$ and

$3x + y = 1$ has no solution?

0%
$k = 3$
0%
$k = 6$
0%
$k \neq 6$
0%
$k = 4$

Explanation

$kx + 2y = 5 \dots (1)$

$3x + y = 1\ \dots (2)$

First multiply the second equation by

$2$

We get:

$6x + 2y = 2$

Now subtracting

$(2)$ from

$(1)$ , we get:

$kx + 2y - 6x - 2y = 5 -2$

$(k-6)x = 3$

$x =\dfrac{3}{k-6}$

So, here if we put any value of

$k$ , then we will get a corresponding value of

$x$ but if we put

$k= 6$ , then we will not get any value of

$x$

So, for

$k = 6$ this equation has no solution.

Solve the following systems of equations using elimination method.

$0.5x+0.8y=0.44, 0.8x+0.6y=0.5$

0%
$(0.4, 0.3)$
0%
$(0.2, 0.3)$
0%
$(0.4, 0.1)$
0%
None of these

Explanation

Consider the given equations.

$0.5x+0.8y=0.44$

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

$0.8x+0.6y=0.5$

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

From

$(1)$ and

$(2)$

On multiplying by

$0.8$ in equation

$(1)$ and multiplying by

$0.5$ in equation

$(2)$ , and subtracting we get

$0.4x + 0.64y = 0.352$

$0.4x +0.3y =0.25$

—————————

$0.34y = 0.102$

$y=0.3$

$[$ put in

$( 2 )\ ]$

$0.8x+0.6\times 0.3=0.5$

$0.8x+0.18=0.5$

$0.8x=0.32$

$x=0.4$

Hence, the value of

$x$ is

$0.4$ and the value of

$y$ is

$0.3$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 11 - 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 11 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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