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

$8x-9y=20$ and

$7x-10y=9$ , then what is

$2x-y$ equal to?

0%
$10$
0%
$11$
0%
$12$
0%
$13$

Explanation

Given:

$8x-9y=20$ ...........

$(1)$

$7x-10y=9$ .........

$(2)$

Multiplying

$(1)$ by

$7$ and

$(2)$ by

$8$ and then subtract

$(2)$ from

$(1)$ , we get

$(56x-63y)-(56x-80y)=140-72$

$\Rightarrow 17y=68$

$\Rightarrow y=4$

Now, put

$y=4$ in (1)

$8x-9\times 4=20$

$\implies 8x=20+36=56$

$\implies x=7$

Now

$2x-y=2\times 7-4=10$

Sahil has Rs.12 less than three times what Sohan has. If both have a total of Rs.88, how many rupees each must have got?

0%
Sohan Rs.25 , Sahil-Rs.63
0%
Sohan Rs.25 , Sahil-Rs.33
0%
Sohan Rs.55 , Sahil-Rs.63
0%
Sohan Rs.25 , Sahil-Rs.55

Explanation

Let Sahil have rupees

$x$ and Sohan has rupees

$y$ .

Sahil has Rs.12 less than 3 times Sohan

$\implies x=3y-12$

$\implies x-3y=-12 ---eqn(1)$

Total both have Rs.88

$x+y=88 --- eqn(2)$

Substracting equation 1 from 2 we get

$y=25$

Then substituting in any one of the equation we get

$x=63$

Solve the following puzzles using the equations:

The present age of Beena is 2 years more than Reeta.The present age of Teena is 3 years more than Beena.If the sum of the ages of Reeta , Beena , and Teena is 79 years, find the present age of all the three of them.

0%
Reeta 24 years , Beena 26 years , Teena 29 years.
0%
Reeta 24 years , Beena 21 years , Teena 29 years.
0%
Reeta 24 years , Beena 29 years , Teena 29 years.
0%
Reeta 24 years , Beena 26 years , Teena 30 years.

Explanation

Let Beena's present age be

$b$ , Reeta's present age be

$r$ and Teena's present age be

$t$

Present age of Beena is 2 years more than of Reeta

$\implies b=r+2$

$\implies b-r=2 --eqn(1)$

Present age of Teena is 3years more than Beena

$\implies t=b+3$

$\implies t-b=3 --eqn(2)$

Sum of all the present ages is 79

$\implies b+t+r=79--eqn(3)$

Adding Equation 1 and 2 we get

$t-r=5 --eqn(4)$

Adding equation 3 and 4 we get

$b+2t=84--eqn(5)$

Adding equation 2 and 5 we get

$3t=87$

$t=29$

Thus substituting

$t$ in equation 2 and equation 4 we will get

$b=26$

and

$r=24$

When a bucket is half full, the weight of the bucket and the water is

$10\text{ kg}$ . When the bucket is two-thirds full, the total weight is

$11\text{ kg}$ . What is the total weight, in kg, when the bucket is completely full?

0%
$12 \text{ kg}$
0%
$12\displaystyle\frac{1}{2} \text{ kg}$
0%
$12\displaystyle\frac{2}{3}\text{ kg}$
0%
$13\text{ kg}$

Explanation

Let the weight of the empty bucket be

$B$ kg and total weight of water it can accommodate be

$W$ kg.

$\therefore B + \cfrac{W}{2} = 10 \ \ ...(1)$

Also,

$B + \cfrac{2W}{3} = 11 \ \ ...(2)$

Subtracting (1) from (2), we have

$\cfrac{2W}{3} - \cfrac{W}{2} = 1$

$\therefore \cfrac{4W - 3W}{6} = 1$ or

$W = 6$

Substituting

$W=6$ in (1)

$\implies B+\dfrac{6}{2}=10$

$\therefore B = 7$

Total weight when completely filled

$= B + W = 13$ kg

Let

$f(x)$ be a quadratic polynomial with

$f(2)=10$ and

$f(-2)=-2$ . Then the coefficient of x in

$f(x)$ is

$?$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Let the quadratic equation be

$f(x)=x^{2}+ax+b$

Given

$f(2)=10$ and

$f(-2)=-2$

$\Rightarrow 4+2a+b=10$ and

$4-2a+b=-2$

$\Rightarrow 2a+b=6$ and

$2a-b=6$

on solving both equations , we get

$a=3$ and

$b=0$

Therefore, the coefficient of

$x$ in

$f(x)$ is

$3$ .

$2x + 3y = 24$ and

$2x - 3y = 12$ , then the value of

$xy$ is ___________.

0%
$10$
0%
$12$
0%
$18$
0%
$14$

Explanation

The given equations are,

$2x + 3y = 24$ ....(1)

$2x - 3y = 12$ ....(2)

Adding Equations (1) and (2), we get

$2x + 3y+2x-3y = 24+12$

$\Rightarrow 4x = 36$

$\Rightarrow x = 9$

Putting

$x = 9$ in Equation (1), we get

$2\times9 + 3y = 24$

$\Rightarrow 18 + 3y = 24$

$\Rightarrow 3y = 6$

$\Rightarrow y =2$

Thus

$xy = 9\times2 = 18$

Solve the following equations:

$\dfrac {x}{2} + \dfrac {y}{5} = 5$ .

$\dfrac {2}{x} + \dfrac {5}{y} = \dfrac {5}{6}$ .

0%
$x=6; y=10$
0%
$x=5; y=6$
0%
$x=4; y=15$
0%
$x=3; y=9$

Explanation

Given equations are $\dfrac { x }{ 2 } +\dfrac { y }{ 5 } =5$ and $\dfrac { 2 }{ x } +\dfrac { 5 }{ y } =\dfrac { 5 }{ 6 }$

Let $\dfrac { x }{ 2 } =u$ and $\dfrac { y }{ 5 } =v$

$u+v=5$ .....(i)

$\dfrac { 1 }{ u } +\dfrac { 1 }{ v } =\dfrac { 5 }{ 6 }$ .....(ii)

$\Rightarrow \dfrac { u+v }{ uv } =\dfrac { 5 }{ 6 } \\ \Rightarrow \dfrac { 5 }{ uv } =\dfrac { 5 }{ 6 } \\ \Rightarrow uv=6$ .........(iii)

From (i) $v=5-u$

$(5-u)u=6\\ \Rightarrow 5u-{ u }^{ 2 }=6\\ \Rightarrow { u }^{ 2 }-5u+6=0\\ \Rightarrow { u }^{ 2 }-2u-3u+6=0\\ \Rightarrow u(u-2)-3(u-2)=0\\ \Rightarrow (u-3)(u-2)=0\\ \Rightarrow u=2,3$

But $\dfrac { x }{ 2 } =u$

$\Rightarrow x=2u\\ \Rightarrow x=4,6$

Also, $\dfrac { y }{ 5 } =v$

$\Rightarrow y=5v\\ \Rightarrow y=15,10$

The sum of a two digit number and the number obtained by reversing the order of its digits is

$121$ , and the two digits differ by

$3$ . Find the number.

0%
$47$
0%
$74$
0%
$44$
0%
$34$

Explanation

Let the digit in the unit's place be

$x$ and the digit at the ten's place be

$y$ . Then,

Number

$=10y+x$

The number obtained by reversing the order of the digits is

$10x+y$ .

According to the given conditions, we have

$(10y+x)+(10x+y)=121$

$\Rightarrow 11(x+y)=121$

$\Rightarrow x+y=11$

and,

$x-y=\pm 3$ [

$\because$ Difference of digits is

$3$ ]

Thus, we have the following sets of simultaneous equations and,

$\left.\begin{matrix} x+y=11 & (i)\\ x-y=3 & (ii)\end{matrix}\right.$ or

$\left\{\begin{matrix} x+y=11 & .(iii) \\ x-y=-3 & (iv) \end{matrix}\right.$

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

$2x=14,x=7$ , Putting this in (ii)

$y=7-3=4$

$x=7, y=4$

On adding equations (iii) and (iv), we get

$2x=8,x=4$ ,

Putting this in (iv)

$y=4+3=7$

$x=4, y=7$

When

$x=7, y=4$ , we have

Number

$=10y+x=10\times 4+7=47$

When

$x=4, y=7$ , we have

Number

$=10y+x=10\times 7+4=74$

Hence, the required number is either

$47$ or

$74$ .

Choose the correct answer from the alternatives given.
If

$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{x \, + \, 2y \, - \, 14}{3} \, = \, \frac{3x \, + \, y \, - \, 12}{11}$ then find the values of x and y, respectively.

0%
$2,6$
0%
$4,8$
0%
$3,5$
0%
$4,5$

Explanation

Given that

$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{x \, + \, 2y \, - \, 14}{3}$

$3x + 3y - 24 = 2x + 4y - 28$

$x - y = -4$ ....... (1)
and

$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{3x \, + \, y \, - \, 12}{11}$

$11x + 11y - 88 = 6x + 2y - 24$

$5x + 9y = 64$ ....... (2)

Multiply (1)by 9

$9x-9y=-36$ ..................(3)

adding (2) and (3)

$14x=28$

$x=\dfrac{28}{14}=2$

put value of x in (1)

$2-y=-4$

$-y=-4-2$

$-y=-6$

$y=6$

we get

$x = 2, y = 6$

Ravish tells his daughter Aarushi, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically and find their present ages.

0%
$x=12,y=49$
0%
$x=12,y=42$
0%
$x=10,y=49$
0%
$x=10,y=42$

Explanation

Present Age of Ravish

$= y$ years

Present Age of Aarushi

$= x$ years

According to Question

$7$ Years ago,

$y-7=7(x-7)\Rightarrow 7x-y-42=0.............(1)$

and

$3$ Years from now

$y+3=3(x+3)\Rightarrow 3x-y+6=0............(2)$

From (1) and (2)

$(1)-(2)\Rightarrow 7x-3x-y+y-42-6=0\Rightarrow 4x=48\Rightarrow x=12$

putting

$x=12$ in Equation (2)

we get,

$3 \times 12-y+6=0\Rightarrow y=42$

The liquids

$X$ and

$Y$ are mixed in the ratio of

$3:2$ and the mixture is sold at

$Rs\;11$ per liter at a profit of

$10\%$ . If the liquid

$X$ costs

$Rs\;2$ more per liter than

$Y$ , the cost of

$X$ per liter is (in Rs.):

0%
$9.50$
0%
$10.80$
0%
$11.75$
0%
$11$

Explanation

Let Cost of

$X$ be

$x$

Cost of

$Y$ be

$y$

and their volumes be

$3K$ and

$2K$

Now,

$x=y+2$

Let original price be

$P$

So,

$P+10$ % of

$P=11$

$P+0.1P=11$

$1.1P=11$

$P=10$

So,

$\dfrac { 3Kx+2Ky }{ 3K+2K } =\dfrac { 3x+2y }{ 5 } =10$

$3x+2y=50$ ....(1)

and

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

Using (2) in (1), we get

$\Rightarrow 3y+6+2y=50\Rightarrow 5y=44\Rightarrow y=\dfrac { 44 }{ 5 } =8.8$ Rs.

$x=10.8$ Rs.

Find the solution of pair of equations

$\dfrac{x}{10}+\dfrac{y}{5}-1=0$ and

$\dfrac{x}{8}+\dfrac{y}{6}=15.$ Hence, find

$\lambda$ , if

$y=\lambda x+5$ .

0%
$\dfrac{-165}{17}$
0%
$\dfrac{-1}{2}$
0%
$\dfrac{361}{696}$
0%
$\dfrac{340}{17}$

Explanation

$\dfrac{x}{10}+\dfrac{y}{5}-1=0$

$\dfrac{x}{10}+\dfrac{y}{5}=1$

$\dfrac{x+2y}{10}=1$

$x+2y=10$ .....................(1)

now consider, $\dfrac{x}{8}+\dfrac{y}{6}=15$

$\dfrac{6x+8y}{48}=15$

$3x+4y=24\times{15}=360$ .................................(2)

now multiply (1)by 3

$3x+6y=30$ ........................(3)

subtracting (2) and (3)

$2y=-330$

$y=-165$

put in(1)

$x+2(-165)=10$

$x-330=10$

$x=340$

now, $y=\lambda{x}+5$

$\implies{-165=\lambda{(340)}+5}$

$-170=\lambda{(340)}$

$\lambda{=\dfrac{-170}{340}=\dfrac{-1}{2}}$

Solve by using substitution method.

$3x +4y =10$ &

$2x - 2y =2$

0%
$x=2, y=1$
0%
$x=-1, y=-2$
0%
$x=1, y=0$
0%
$x=-2, y=3$

Explanation

$3x+4y= 10 \ .....(1)$

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

equation $(2)$ can be written as

$\Rightarrow x-y=1$

$\Rightarrow x=y+1\ .....(3)$

substitute $x$ value in equation $(2)$

$\Rightarrow 3x+4y= 10$

$\Rightarrow 3(y+1)+4y=10$

$\Rightarrow 3y+3+4y=10$

$\Rightarrow 7y=7$

$\Rightarrow y=1$

then, substitute $y$ value in equation $(3)$

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

$\therefore x=2, y=1$

Solve each pair of equation by using the substitution method.

$x+\dfrac{6}{y}=6$

$3x-\dfrac{8}{y}=5$

0%
$x=3$ and $y=-2$
0%
$x=30$ and $y=2$
0%
$x=3$ and $y=2$
0%
None of these

Explanation

$x+\dfrac {6}{y}=6......(i)$

$3x-\dfrac {8}{y}=5.....(ii)$

From equation

$(i)$ , we get

$\Rightarrow \ x=6-\dfrac {6}{y}...(iii)$

Substituting value of

$x$ in equation

$(ii)$ , we get

$3\left (6-\dfrac {6}{y}\right)-\dfrac {8}{y}=5$

$\Rightarrow \ 18-\dfrac {18}{y}-\dfrac {8}{y}=5$

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

$\Rightarrow \ y=2$

Substituting the value of

$y$ in (iii)

$\ x=6-\dfrac {6}{2}=3$

$\therefore x = 3, y = 2$

Classes A and B have

$35$ students each. If seven girls shift from class

$A$ to class

$B$ , then the number of girls in the classes would interchange. If four girls shift from class

$B$ to class

$A$ , then the number of girls in class

$A$ would be twice the original number of girls in class

$B$ . What is the number of boys in Class

$A$ and in Class

$B$ ?

0%
$18$ and $11$
0%
$24$ and $19$
0%
$18$ and $27$
0%
$17$ and $24$

Explanation

$\textbf{Step -1: Form the required equations.}$

$\text{Let the number of girls in class }A\text{ be }x.$

$\text{Let the number of girls in class }B\text{ be }y.$

$\text{Let the number of boys in class }A\text{ be }35-x.$

$\text{Let the number of boys in class }B\text{ be }35-y.$

$\text{If seven girls are shifted from class }A\text{ to }B,$

$\text{Number of girls in class }A\text{ would become }x-7.$

$\text{Number of girls in class }B\text{ would become }y+7.$

$\text{But according to the question,}$

$y=x-7$

$\Rightarrow x-y=7\ldots(i)$

$\text{If four girls are shifted from class }B\text{ to }A,$

$\text{Number of girls in class }A\text{ would become }x+4.$

$\text{Number of girls in class }B\text{ would become }y-4.$

$\text{But according to the question,}$

$x+4=2y$

$\Rightarrow x-2y=-4\ldots(ii)$

$\textbf{Step -2: Solve the above formed two equations simultaneously.}$

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

$-y+2y=7+4$

$\Rightarrow y=11$

$\text{On putting the value of }y\text{ in equation }(i),\text{ we get}$

$x-11=7$

$\Rightarrow x=18$

$\textbf{Step -3: Find the number of boys in class A and B.}$

$\text{Number of boys in class }A=35-18$

$=17$

$\text{Number of boys in class }B=35-11$

$=24$

$\textbf{Therefore, option D is correct.}$

Solve each pair of equation by using the substitution method.

$0.2x+0.3{y}=1.3$

$0.4x+0.5{y}=2.3$

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

Explanation

$0.2x+0.3y=1.3.......(i)$

$0.4x+0.5y=2.3......(ii)$

From equation

$(i)$ , we get

$0.2x=1.3-0.3y$

$\Rightarrow \ x=\dfrac {13-3y}{2}.....(iii)$

From equation

$(iii)$ and equation

$(ii)$

$\therefore \ 0.4\left(\dfrac {13-3y}{2}\right)+0.5y=2.3$

$\Rightarrow \ 2.6-0.6y+0.5y=2.3$

$\Rightarrow \ -0.1y=-0.3\\\ \Rightarrow y=3$

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

Solve each pair of equation by using the substitution method.

$2x+3y=9$

$3x+4y=5$

0%
$x=-21$ and $y=7$
0%
$x=1$ and $y=17$
0%
$x=-21$ and $y=17$
0%
None of these

Explanation

$2x+3y=9---(i)$

$3x+4y=5--(ii)$

From equation

$(i)$ , we get

$\Rightarrow \ 2x=9-3y\ \Rightarrow x=\dfrac {9-3y}{2}---(iii)$

Substituting equation

$(iii)$ in equartion

$(ii)$ , we get

$\Rightarrow \ 3\left(\dfrac {9-3y}{2}\right)+4y=5$

$\Rightarrow \ 27-9y+8y=10$

$\Rightarrow \ y=27-10=17$

$\therefore \ x=\dfrac {9-3\times 17}{2}=\dfrac {-42}{2}=-21$

Solve the following pairs of linear equations by elimination method

$3x+4y=25$

$5x-6y=-9$

0%
$x=3$ and $y=3$
0%
$x=4$ and $y=4$
0%
$x=3$ and $y=4$
0%
None of these

Explanation

$3x+4y=25\_\_\_\_\_\_(i)\times 3$

$5x-6y=-9\_\_\_\_\_\_(ii)\times 2$

$9x+12y=75$

$10x-12y=-18$

(+)

$\quad$ (+)

$\quad\quad\quad$ (+)

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

$19x\quad=57$

$\Rightarrow x=3$

$\Rightarrow 3 \times 3+4y=25$

$\Rightarrow 4y=16\Rightarrow y=4$

Sum of two numbers is

$407$ . The sum and difference of their LCM and HCF are

$925$ and

$851$ respectively. The difference of two numbers is

0%
$518$
0%
$185$
0%
$158$
0%
$175$

Explanation

Let the two numbers: be

$x$ &

$y$

$\therefore x+y=407$ .........

$(1)$

Acc to the question,

$LCM+HCF=925$ ..........

$(2)$

$LCM-HCF=851$ .........

$(3)$

On subtracting

$(2)$ &

$(3)$

$LCM+HCF=925\\ -LCM+HCF=-851\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \ \ \ \ \ 2HCF=74\\ \quad \quad \quad \ \ \ \ \ \ \ HCF=37$

$\therefore LCM -(37)=857$

$LCM=857+37$

$=888$

As we know,

$LCM\times HCF=$ product of two numbers

$\therefore 888\times 37=x xy$ .............

$(4)$

From

$(1)$ ,

$x=(407-y)$

$\therefore 888\times 37 = (407-y)y$

$\Rightarrow y^{2}-407y+832856$

$(y-111)(y-296)$

$\therefore x=296$ and

$y=111$

$\therefore$ Difference of two numbers

$=296-111$

$=185$

$\therefore$ Option

$B$ is correct.

At a zoo, there were parrots and rabbits in the same enclouser. Sravanthi counted 30 heads and 100 legs altogether. how many animals of each type were in the enclouser

0%
$9$ parrots $21$ rabbits
0%
$10$ parrots and $20$ rabbits
0%
$20$ parrots and $19$ rabbits
0%
$11$ rabbits and $19$ parrots

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

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

Explanation

Let the present age of father be

$x$ years
and the present age of son be

$y = years$

$\therefore$ According to the question ,

$x = 6y ....(i)$
Age of the father after four years =

$(x +4) years$
And the age of son after four years

$= (y +4) years$
Now according to the question,

$x +4 = 4(y +4) ...(ii)$

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

$\Rightarrow 6y-4y =16 -4 [ Rs x = 6y]$

$\Rightarrow 2y =12$

$\Rightarrow y = 6$

$\therefore x =6 \times 6 = 36years$ [From (i) ]
and

$y = 6$ years
So the present ages of the son and the father are

$6$ years and

$36$ years respectively.

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

$x- y=2$ and

$x+y=4$ then the value of

$a$ and

$b$ are respectively

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

Explanation

$(a.b)$ is the solution of the given equations, then it mus satisfy the given equations so,

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

$a+b = 4 .....(ii)$

$\Rightarrow 2a = 6$ [ adding (i) and (ii) ]

$\Rightarrow a= 3$

Now,

$3 +b = 4 [ from (ii) ]$

$\Rightarrow b = 1$
So,

$(a,b) = (3,1)$

Aruna has only

$Rs 1$ and

$Rs 2$ coins with her. if the total number of coins that she has is

$50$ and the amount of money with her is

$Rs 75$ then the number of

$Rs 1$ and

$Rs 2$ coins are respectively

0%
$35 \quad and \quad 15$
0%
$35 \quad and \quad 20$
0%
$15 \quad and \quad35$
0%
$25 \quad and \quad 25$

Explanation

Let the number of

$Rs 1$ coins = x
And the number of

$Rs 2$ coin = y
So, according to the question

$x+y= 50...(i)$

$1x +2y= 75....(ii)$
Subtracting equations

$(i)$ From

$(ii)$

$y =25$
Substituting value of

$y$ in (i)

$x = 25$
So,

$y = 25$ and

$x = 25$

Choose the correct alternative answers for the following questions.

$3x + 5y = 9$ and

$5x + 3y= 7$ then What is the value of

$x + y~ ?$

0%
$2$
0%
$16$
0%
$9$
0%
$7$

The cost of an article

$A$ is

$15$ % less than that of article

$B.$ If their total cost is

$2,775\:Rs\:;$ find the cost of each article

$.$

0%
$A=1,873\:Rs.$ and $B=1,600\:Rs.$
0%
$A=1,275\:Rs.$ and $B=1,500\:Rs.$
0%
$A=1,647\:Rs.$ and $B=1,900\:Rs.$
0%
$A=1,450\:Rs.$ and $B=1,300\:Rs.$

Explanation

Let cost of A be Rs

$x$ and of B be Rs

$y$ .

Given, Total cost = Rs

$2775$

$=> x + y = 2775$ --- (1)

Also,cost of an article A is

$15$ % less than that of article B
=>

$x = y - 0.15y => x = 0.85y$ --- (2)

From

$(1)$ and

$(2)$ , we get

$0.85y + y = 2775$

$1.85y = 2775$

$=> y = 1500$

And,

$x = 0.85y = 1275$

Hence, cost of A is Rs

$1275$ and of B is Rs

$1500$

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces toIt becomes

$\dfrac {1}{2}$ if we only add 1 to the denominator. What is the fraction?

0%
$\dfrac{2}{7}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{6}{7}$
0%
$\dfrac{1}{9}$

Explanation

$\textbf{Step 1: Form equation using the details.}$

$\text{Let numerator be }x$

$\text{And denominator be }y$

$\text{So ratio is }\dfrac{x}{y}$

$\text{Now add 1 to numerator and subtract 1 from denominator as per question}$

$\text{So, numerator }=x+1$

$\text{And, denominator }=y-1$

$\text{So ratio is }\dfrac{x+1}{y-1}$

$\text{So, as per question }$

$\Rightarrow\dfrac{x+1}{y-1}=1$

$\Rightarrow (x+1)= (y-1)$

$\Rightarrow x+1= y-1$

$\Rightarrow x-y= -2 \ldots(i)$

$\text{Now add 1 to denominator as per question}$

$\text{So, numerator }=x$

$\text{And, denominator }=y+1$

$\text{So ratio is }\dfrac{x}{y+1}$

$\text{So, as per question }$

$\Rightarrow\dfrac{x}{y+1}=\dfrac{1}{2}$

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

$\Rightarrow 2x= y+1$

$\Rightarrow 2x-y= 1 \ldots(ii)$

$\textbf{Step 2: Solve the equation by elimination method}$

$\text{Subtracting equation }(ii) \text{ from equation } (i)$

$\; x-y= -2$

$2x-y= 1$

$-\; \;\; + \; \; \;-$

______________

$-x+0 =-3$

________________

$\Rightarrow x= 3$

$\text{Substituting value of }x \text{ in equation } (i)$

$\Rightarrow x-y= -2$

$\Rightarrow 3-y= -2$

$\Rightarrow -y= -2 -3$

$\Rightarrow y= 5$

$\text{So ratio is }\dfrac{3}{5}$

$\textbf{Hence the correct option is B }$

The pair of linear equation px +2y=5 and 3x+y=1 has unique solution of

0%
$p\neq 6$
0%
p=6
0%
p=5
0%
$p\neq 5$

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

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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