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$$.$$

$$A=1,275\:Rs.$$ and $$B=1,500\:Rs.$$

$$A=1,873\:Rs.$$ and $$B=1,600\:Rs.$$

$$A=1,647\:Rs.$$ and $$B=1,900\:Rs.$$

$$A=1,450\:Rs.$$ and $$B=1,300\:Rs.$$

MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 16

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

Solve :

3 x - 2 y = 1

$3x-2y=1$

2 x + y = 3

$2x+y=3$

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

Explanation

Given:

3 x - 2 y = 1

$3x-2y=1$

2 x + y = 3

$2x+y=3$

y = 3 - 2 x

$y=3-2x$

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

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

3 x - 6 + 4 x = 1

$3x-6+4x=1$

7 x = 7

$7x=7$

x = 1

$x=1$

3 (1) - 2 y = 1

$3(1)-2y=1$

3 - 2 y = 1

$3-2y=1$

2 y = 2

$2y=2$

y = 1

$y=1$

Choose the correct alternative answers for the following questions.

3 x + 5 y = 9

$3x + 5y = 9$ and

5 x + 3 y = 7

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

x + y ?

$x + y~ ?$

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

The solutions of the pair of equations

x - y = 0, x + y = 0

$x- y = 0 , x + y = 0$

0%
$x = 3 , y = 5$
0%
$x = 3 , y = 2$
0%
$x = 0 , y = 0$
0%
$x = 5 , y = 5$

Explanation

Clearly from the graph , the intersection point of

x - y = 0

$x-y=0$ (blue line) and

x + y = 0

$x+y=0$ (brown line) is

(0, 0)

$(0,0)$ . Hence,

x = 0, y = 0

$x =0 , y=0$ is the solution of given linear equations.
1944681_1867743_ans_e5fbe6382643421e968eee34682f6079.jpg

Based on equations reducible to linear equations
Solve for x and y:

\frac{16}{x + 3} + \frac{3}{y - 2} = 5; \frac{8}{x + 3} - \frac{1}{y - 2} = 0

$\dfrac {16}{x+3}+\dfrac {3}{y-2}=5; \dfrac {8}{x+3}-\dfrac {1}{y-2}=0$

0%
$x = 5, y = 3$
0%
$x = 7, y = 2$
0%
$x = 1, y = 5$
0%
None of these

Explanation

The given equations are
$\Rightarrow \dfrac{16}{x+3}+\dfrac{3}{y-2}=5....eq\left ( 1 \right )$
$\Rightarrow \dfrac{8}{x+3}-\dfrac{1}{y-2}=0....eq\left ( 2\right )$
$Put \dfrac{1}{x+3}=u \ and \ \ \dfrac{1}{y-2}=v \ in \ eq\left (1 \right ) and \left ( 2\right )$
$\Rightarrow 16u+3v=5....eq\left ( 3 \right )$
$\Rightarrow 8u-v=0 ....eq\left ( 4\right )$
$multiply \ eq\left (4 \right ) \ by \ 3 \ and \ subtract\ eq \left (3 \right ) and \ eq\left ( 4 \right )$
$\Rightarrow \left (16u+3v=5 \right )-\left ( 24u-v=0 \right )$
$\Rightarrow 40u=5\Rightarrow u=\dfrac{1}{8}$
$Put u=\dfrac{1}{8} \ in \ eq\left ( 3\right )$
$\Rightarrow 16\left (\dfrac{1}{8} \right )+3v=5\Rightarrow 3v=3\Rightarrow v=1$
$Hence ,\dfrac{1}{x+3}= u=\dfrac{1}{8}\Rightarrow x+3=8\Rightarrow x=5$
$\dfrac{1}{y-2}= v=1\Rightarrow y-2=1\Rightarrow y=3$

Based on equations reducible to linear equations
Solve for x and y:

\frac{24}{2 x + y} - \frac{13}{3 x + 2 y} = 2; \frac{26}{3 x + 2 y} + \frac{8}{2 x + y} = 3

$\dfrac {24}{2x+y}-\dfrac {13}{3x+2y}=2; \dfrac {26}{3x+2y}+\dfrac {8}{2x+y}=3$

0%
$x = -2, y = 6$
0%
$x = -5, y = 3$
0%
$x = 3, y = 2$
0%
$x = 1, y = 0$

Explanation

The given equations are
$\Rightarrow \dfrac{24}{2x+y}-\dfrac{13}{3x+2y}=2...eq\left ( 1 \right )$
$\Rightarrow \dfrac{26}{3x+2y}+\dfrac{8}{2x+y}=3...eq\left ( 1 \right )$
$Put \dfrac{1}{2x+y}=u \ and \ \dfrac{1}{3x+y}=v \ in \ eq\left (1 \right ) and \left ( 2 \right )$
$\Rightarrow 24u-13v=2.....\left ( 3 \right )$
$\Rightarrow 8u+26v=3......\left ( 4 \right )$
$Multiply \ eq\left ( 3\right ) \ by\ 26 \ and \ eq\left (4 \right ) \ by\ 13 \ and\ subtract \ both$
$\Rightarrow \left (624u-338v=52 \right )-\left ( 104u+338v=39 \right )$
$\Rightarrow 728u=91\Rightarrow u=\dfrac{1}{8}$
$Put u=\dfrac{1}{8} \ in \ eq\left ( 4 \right )$
$\Rightarrow 8\times\dfrac{1}{8}+26v=3\Rightarrow 26v=2\Rightarrow v=\dfrac{1}{13}$
$Hence, u=\dfrac{1}{2x+y}=\dfrac{1}{8} \Rightarrow 2x+y=8.....eq\left ( 5 \right )$
$v=\dfrac{1}{3x+2y}=\dfrac{1}{13}\Rightarrow 3x+2y=13...eq\left ( 6 \right )$
$Multiply \ eq\left (5 \right ) \ by \ 2 \ and \ subtract \ from \ eq\left ( 6 \right )$
$\Rightarrow \left ( 4x+2y=16 \right )-\left ( 3x+3y=13 \right )$
$\Rightarrow x=3$
$put \ x=3 \ in \ eq\left (5 \right )$
$\Rightarrow 2\times 3+y=8 \Rightarrow y=2$

The cost of an article

A

$A$ is

15

$15$ % less than that of article

B .

$B.$ If their total cost is

2, 775 R s;

$2,775\:Rs\:;$ 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

$x$ and of B be Rs

y

$y$ .

Given, Total cost = Rs

2775

$2775$

=> x + y = 2775

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

Also,cost of an article A is

15

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

x = y - 0.15 y => x = 0.85 y

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

From

(1)

$(1)$ and

(2)

$(2)$ , we get

0.85 y + y = 2775

$0.85y + y = 2775$

1.85 y = 2775

$1.85y = 2775$

=> y = 1500

$=> y = 1500$

And,

x = 0.85 y = 1275

$x = 0.85y = 1275$

Hence, cost of A is Rs

1275

$1275$ and of B is Rs

1500

$1500$

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

\frac{1}{2}

$\dfrac {1}{2}$ 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

Step 1: Form equation using the details.

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

Let numerator be x

$\text{Let numerator be }x$

And denominator be y

$\text{And denominator be }y$

So ratio is \frac{x}{y}

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

Now add 1 to numerator and subtract 1 from denominator as per question

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

So, numerator = x + 1

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

And, denominator = y - 1

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

So ratio is \frac{x + 1}{y - 1}

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

So, as per question

$\text{So, as per question }$

\Rightarrow \frac{x + 1}{y - 1} = 1

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

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

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

\Rightarrow x + 1 = y - 1

$\Rightarrow x+1= y-1$

\Rightarrow x - y = - 2 \dots (i)

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

Now add 1 to denominator as per question

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

So, numerator = x

$\text{So, numerator }=x$

And, denominator = y + 1

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

So ratio is \frac{x}{y + 1}

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

So, as per question

$\text{So, as per question }$

\Rightarrow \frac{x}{y + 1} = \frac{1}{2}

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

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

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

\Rightarrow 2 x = y + 1

$\Rightarrow 2x= y+1$

\Rightarrow 2 x - y = 1 \dots (i i)

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

Step 2: Solve the equation by elimination method

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

Subtracting equation (i i) from equation (i)

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

x - y = - 2

$\; x-y= -2$

2 x - y = 1

$2x-y= 1$

- + -

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

______________

- x + 0 = - 3

$-x+0 =-3$

________________

\Rightarrow x = 3

$\Rightarrow x= 3$

Substituting value of x in equation (i)

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

\Rightarrow x - y = - 2

$\Rightarrow x-y= -2$

\Rightarrow 3 - y = - 2

$\Rightarrow 3-y= -2$

\Rightarrow - y = - 2 - 3

$\Rightarrow -y= -2 -3$

\Rightarrow y = 5

$\Rightarrow y= 5$

So ratio is \frac{3}{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 10 Maths Pair Of Linear Equations In Two Variables Quiz 16 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 16 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

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