Pair of Linear Equations in Two Variables

20 Questions | 90 Marks

A shopkeeper sells two types of notebooks: a plain one and a ruled one. He notices that he sold 15 more ruled notebooks than plain notebooks in a week. The total number of notebooks sold was 85.(a) Formulate a pair of linear equations to represent this situation.(b) Find the number of plain notebooks and ruled notebooks sold. 3marks

The sum of a two-digit number and the number formed by reversing its digits is 110. If the difference between the digits is 2, find the number. 5marks

Assertion: The pair of linear equations

2 x + 3 y = 5

and

4 x + 6 y = 10

has infinitely many solutions.Reason: For a pair of linear equations

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

, if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

, then the pair has infinitely many solutions.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true, but reason is false.(d) Assertion is false, but reason is true. 1mark

A farmer has a certain number of cows and chickens. He counts a total of 70 heads and 190 legs.(a) Set up a system of linear equations to find the number of cows and chickens.(b) Determine the number of cows and chickens the farmer has. 3marks

Consider the following scenario: A boat moves downstream in a river. The speed of the boat in still water is

x

km/hr and the speed of the stream is

y

km/hr. The boat covers a distance of 30 km downstream in 3 hours. The same boat takes 5 hours to cover the same distance upstream.(a) Formulate algebraic equations representing the given information.(b) Calculate the speed of the boat in still water and the speed of the stream. 5marks

Two angles are complementary. The larger angle is

10^{\circ}

more than the smaller angle. Find the measure of each angle. 3marks

Assertion: The pair of linear equations

x - 2 y = 0

and

3 x + 4 y - 20 = 0

has a unique solution.Reason: If

\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}

, the pair of linear equations has a unique solution.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true, but reason is false.(d) Assertion is false, but reason is true. 1mark

A test consists of multiple-choice questions and long-answer questions. Each multiple-choice question carries 4 marks and each long-answer question carries 6 marks. A student attempted all questions and scored 80 marks. If the number of multiple-choice questions attempted was 10 more than the number of long-answer questions, find the number of each type of question attempted by the student. 5marks

In a classroom, there are some benches and desks. If 3 students sit on each bench, 2 students are left standing. If 4 students sit on each bench, 2 benches are empty. Find the total number of students and benches. 5marks

The graph of the linear equation

2 x - y = 1

and

x + y = 5

intersect at a point.(a) Find the coordinates of the intersection point.(b) What does this point represent in terms of the solution of the system of these equations? 3marks

A fraction becomes

\frac{1}{3}

when 1 is subtracted from its numerator and it becomes

\frac{1}{2}

when 8 is added to its denominator. Find the fraction. 3marks

Assertion: The lines represented by the equations

x + 2 y = 3

and

2 x + 4 y = 12

are parallel.Reason: For a pair of linear equations

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

, if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

, the lines are parallel.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true, but reason is false.(d) Assertion is false, but reason is true. 1mark

A car rental company charges a fixed rental fee plus a charge per kilometer driven. For a rental of 200 km, the total charge was 4000. For a rental of 350 km, the total charge was 5500.(a) Set up a system of linear equations to represent this situation.(b) Find the fixed rental fee and the charge per kilometer. 3marks

Consider two points A(1, 2) and B(3, 4). These points can be considered as the intersection of two lines.(a) Find a linear equation passing through A.(b) Find a linear equation passing through B.(c) Find a pair of linear equations such that their intersection point is (1, 2). 3marks

Mr. Sharma is choosing between two electricity plans. Plan A charges a fixed monthly fee of 300 plus 2 per unit of electricity consumed. Plan B charges a fixed monthly fee of 500 plus 1 per unit of electricity consumed.(a) Write down the linear equations representing the cost for each plan.(b) For what number of units consumed will the cost under both plans be the same? 5marks

The sum of two numbers is 150. If the difference between the numbers is 20, find the numbers. 3marks

Assertion: The pair of linear equations

x + y = 7

and

2 x + 2 y = 14

has infinitely many solutions.Reason: If

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

, the pair of linear equations represents coincident lines.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true, but reason is false.(d) Assertion is false, but reason is true. 1mark

A library has a fixed charge for the first three days and an additional charge for each day thereafter. Ravi paid 35 for a book kept for 6 days, while Sunita paid 30 for a book kept for 5 days. Find the fixed charge and the charge for each extra day. 5marks

The bottom half of a rectangular field is a square. The perimeter of the entire field is 160 meters. Find the dimensions of the field. 3marks

A bus travels a distance of 100 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the journey. Find the original speed of the bus. 5marks

Answers

(a) Let

p

be the number of plain notebooks and

r

be the number of ruled notebooks.Equation 1:

r = p + 15

Equation 2:

p + r = 85

(b) Substitute (1) into (2):

p + (p + 15) = 85 \Rightarrow 2 p + 15 = 85 \Rightarrow 2 p = 70 \Rightarrow p = 35

.Then,

r = 35 + 15 = 50

.So, 35 plain notebooks and 50 ruled notebooks were sold.

Let the two-digit number be

10 x + y

. The number formed by reversing the digits is

10 y + x

.Sum of the number and its reverse:

(10 x + y) + (10 y + x) = 110 \Rightarrow 11 x + 11 y = 110 \Rightarrow x + y = 10

. (Equation 1)Difference between the digits is 2:

x - y = 2

(Equation 2) or

y - x = 2

(Equation 3).Case 1: Using Equation 1 and 2.Add (1) and (2):

2 x = 12 \Rightarrow x = 6

.Substitute

x = 6

in (1):

6 + y = 10 \Rightarrow y = 4

.The number is

10 (6) + 4 = 64

.Case 2: Using Equation 1 and 3.Add (1) and (3):

2 y = 12 \Rightarrow y = 6

.Substitute

y = 6

in (1):

x + 6 = 10 \Rightarrow x = 4

.The number is

10 (4) + 6 = 46

.The numbers can be 64 or 46.

(a) Assertion: The pair of linear equations

2 x + 3 y = 5

and

4 x + 6 y = 10

has infinitely many solutions.For the given equations:

a_{1} = 2, b_{1} = 3, c_{1} = - 5

and

a_{2} = 4, b_{2} = 6, c_{2} = - 10

\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}

\frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}

\frac{c_{1}}{c_{2}} = \frac{- 5}{- 10} = \frac{1}{2}

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ , the assertion is true.
Reason: For a pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ , if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ , then the pair has infinitely many solutions. This reason correctly explains the condition for infinitely many solutions.
Therefore, the correct option is (a).

(a) Let

c

be the number of cows and

h

be the number of chickens.Each animal has one head:

c + h = 70

(Equation 1)Cows have 4 legs, chickens have 2 legs:

4 c + 2 h = 190

(Equation 2)(b) From Equation 1,

h = 70 - c

. Substitute this into Equation 2:

4 c + 2 (70 - c) = 190

4 c + 140 - 2 c = 190

2 c = 190 - 140

2 c = 50

$c = 25$ .
Substitute $c = 25$ into Equation 1: $25 + h = 70 \Rightarrow h = 45$ .
The farmer has 25 cows and 45 chickens.

(a) Let

x

be the speed of the boat in still water and

y

be the speed of the stream.Speed downstream =

(x + y)

km/hr.Speed upstream =

(x - y)

km/hr.Distance = Speed

\times

TimeDownstream:

30 = (x + y) \times 3 \Rightarrow x + y = 10

(Equation 1)Upstream:

30 = (x - y) \times 5 \Rightarrow x - y = 6

(Equation 2)(b) Add Equation 1 and Equation 2:

(x + y) + (x - y) = 10 + 6

2 x = 16

$x = 8$ km/hr.
Substitute $x = 8$ into Equation 1:

8 + y = 10

$y = 2$ km/hr.
The speed of the boat in still water is 8 km/hr and the speed of the stream is 2 km/hr.

Let the smaller angle be

x

degrees and the larger angle be

y

degrees.Since the angles are complementary, their sum is

90^{\circ}

x + y = 90

(Equation 1)The larger angle is

10^{\circ}

more than the smaller angle:

y = x + 10

(Equation 2)Substitute Equation 2 into Equation 1:

x + (x + 10) = 90

2 x + 10 = 90

2 x = 80

$x = 40^{\circ}$ .
Substitute $x = 40$ into Equation 2:

y = 40 + 10

$y = 50^{\circ}$ .
The angles are $40^{\circ}$ and $50^{\circ}$ .

Assertion: The pair of linear equations

x - 2 y = 0

and

3 x + 4 y - 20 = 0

has a unique solution.For the given equations:

a_{1} = 1, b_{1} = - 2, c_{1} = 0

and

a_{2} = 3, b_{2} = 4, c_{2} = - 20

\frac{a_{1}}{a_{2}} = \frac{1}{3}

\frac{b_{1}}{b_{2}} = \frac{- 2}{4} = \frac{- 1}{2}

Since $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , the assertion is true.
Reason: If $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , the pair of linear equations has a unique solution. This reason correctly explains the condition for a unique solution.
Therefore, the correct option is (a).

Let

m

be the number of multiple-choice questions and

l

be the number of long-answer questions.Marks scored:

4 m + 6 l = 80

(Equation 1)Number of questions:

m = l + 10

(Equation 2)Substitute Equation 2 into Equation 1:

4 (l + 10) + 6 l = 80

4 l + 40 + 6 l = 80

10 l = 40

$l = 4$ .
Substitute $l = 4$ into Equation 2:

m = 4 + 10

$m = 14$ .
The student attempted 14 multiple-choice questions and 4 long-answer questions.

Let

s

be the number of students and

b

be the number of benches.If 3 students sit on each bench, 2 are left standing:

s = 3 b + 2

(Equation 1)If 4 students sit on each bench, 2 benches are empty, meaning

4

students occupied

(b - 2)

benches:

s = 4 (b - 2)

(Equation 2)Equate Equation 1 and Equation 2:

3 b + 2 = 4 (b - 2)

3 b + 2 = 4 b - 8

$b = 10$ .
Substitute $b = 10$ into Equation 1:

s = 3 (10) + 2

s = 30 + 2

$s = 32$ .
There are 32 students and 10 benches.

(a) The intersection point is the solution to the system of equations:Equation 1:

2 x - y = 1

Equation 2:

x + y = 5

Add Equation 1 and Equation 2:

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

3 x = 6

$x = 2$ .
Substitute $x = 2$ into Equation 2:

2 + y = 5

$y = 3$ .
The coordinates of the intersection point are (2, 3).
(b) This point (2, 3) represents the unique solution to the system of linear equations, meaning it is the only pair of values for $x$ and $y$ that satisfies both equations simultaneously.

Let the fraction be

\frac{x}{y}

.If 1 is subtracted from the numerator, it becomes

\frac{1}{3}

\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3 (x - 1) = y \Rightarrow 3 x - 3 = y

(Equation 1)If 8 is added to the denominator, it becomes

\frac{1}{2}

\frac{x}{y + 8} = \frac{1}{2} \Rightarrow 2 x = y + 8 \Rightarrow 2 x - 8 = y

(Equation 2)Equate Equation 1 and Equation 2:

3 x - 3 = 2 x - 8

3 x - 2 x = - 8 + 3

$x = - 5$ .
Substitute $x = - 5$ into Equation 1:

y = 3 (- 5) - 3

y = - 15 - 3

$y = - 18$ .
The fraction is $\frac{- 5}{- 18} = \frac{5}{18}$ .

Assertion: The lines represented by the equations

x + 2 y = 3

and

2 x + 4 y = 12

are parallel.For the given equations:

a_{1} = 1, b_{1} = 2, c_{1} = - 3

and

a_{2} = 2, b_{2} = 4, c_{2} = - 12

\frac{a_{1}}{a_{2}} = \frac{1}{2}

\frac{b_{1}}{b_{2}} = \frac{2}{4} = \frac{1}{2}

\frac{c_{1}}{c_{2}} = \frac{- 3}{- 12} = \frac{1}{4}

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ , the lines are parallel. The assertion is true.
Reason: For a pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ , if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ , the lines are parallel. This reason correctly explains the condition for parallel lines.
Therefore, the correct option is (a).

(a) Let

f

be the fixed rental fee and

k

be the charge per kilometer.For 200 km:

f + 200 k = 4000

(Equation 1)For 350 km:

f + 350 k = 5500

(Equation 2)(b) Subtract Equation 1 from Equation 2:

(f + 350 k) - (f + 200 k) = 5500 - 4000

150 k = 1500

$k = 10$ .
Substitute $k = 10$ into Equation 1:

f + 200 (10) = 4000

f + 2000 = 4000

$f = 2000$ .
The fixed rental fee is 2000 and the charge per kilometer is 10.

(a) A linear equation passing through A(1, 2). Many possibilities exist. One example is

x + y = 3

.(b) A linear equation passing through B(3, 4). Many possibilities exist. One example is

x + y = 7

.(c) To find a pair of linear equations whose intersection point is (1, 2), we need two distinct linear equations that are satisfied by

x = 1

and

y = 2

.Equation 1:

x + y = 1 + 2 = 3

.Equation 2:

x - y = 1 - 2 = - 1

.The pair of linear equations could be

x + y = 3

and

x - y = - 1

(a) Let

C_{A}

be the cost for Plan A and

C_{B}

be the cost for Plan B. Let

u

be the number of units consumed.Plan A:

C_{A} = 300 + 2 u

Plan B:

C_{B} = 500 + 1 u

(b) For the cost to be the same,

C_{A} = C_{B}

300 + 2 u = 500 + u

2 u - u = 500 - 300

$u = 200$ .
The cost under both plans will be the same for 200 units of electricity consumed.

Let the two numbers be

x

and

y

.Sum:

x + y = 150

(Equation 1)Difference:

x - y = 20

(Equation 2)Add Equation 1 and Equation 2:

2 x = 170

$x = 85$ .
Substitute $x = 85$ into Equation 1:

85 + y = 150

y = 150 - 85

$y = 65$ .
The numbers are 85 and 65.

Assertion: The pair of linear equations

x + y = 7

and

2 x + 2 y = 14

has infinitely many solutions.For the given equations:

a_{1} = 1, b_{1} = 1, c_{1} = - 7

and

a_{2} = 2, b_{2} = 2, c_{2} = - 14

\frac{a_{1}}{a_{2}} = \frac{1}{2}

\frac{b_{1}}{b_{2}} = \frac{1}{2}

\frac{c_{1}}{c_{2}} = \frac{- 7}{- 14} = \frac{1}{2}

Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ , the assertion is true.
Reason: If $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ , the pair of linear equations represents coincident lines, which implies infinitely many solutions. The reason correctly explains the condition.
Therefore, the correct option is (a).

Let the fixed charge for the first three days be

f

and the charge for each extra day be

e

.Ravi paid 35 for 6 days (3 fixed + 3 extra): $f + 3e = 35$ (Equation 1) Sunita paid 30 for 5 days (3 fixed + 2 extra):

f + 2 e = 30

(Equation 2)Subtract Equation 2 from Equation 1:

(f + 3 e) - (f + 2 e) = 35 - 30

$e = 5$ .
Substitute $e = 5$ into Equation 2:

f + 2 (5) = 30

f + 10 = 30

$f = 20$ .
The fixed charge is 20 and the charge for each extra day is 5.

Let the dimensions of the rectangular field be length

l

and width

w

.The bottom half is a square, which means

w = l / 2

.The perimeter of the entire field is

2 (l + w) = 160

.Substitute

w = l / 2

into the perimeter equation:

2 (l + \frac{l}{2}) = 160

2 (\frac{3 l}{2}) = 160

3 l = 160

$l = \frac{160}{3}$ meters.
Now find the width:
$w = \frac{l}{2} = \frac{160 / 3}{2} = \frac{160}{6} = \frac{80}{3}$ meters.
The dimensions of the field are length $\frac{160}{3}$ m and width $\frac{80}{3}$ m.

Let the original speed of the bus be

s

km/hr. The distance is 100 km.Time taken at original speed =

\frac{100}{s}

hours.If the speed had been 5 km/hr more, the speed would be

(s + 5)

km/hr.Time taken at increased speed =

\frac{100}{s + 5}

hours.The new time is 1 hour less than the original time:

\frac{100}{s + 5} = \frac{100}{s} - 1

.Multiply by

s (s + 5)

to clear denominators:

100 s = 100 (s + 5) - s (s + 5)

100 s = 100 s + 500 - s^{2} - 5 s

0 = 500 - s^{2} - 5 s

$s^{2} + 5 s - 500 = 0$ .
Using the quadratic formula $s = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

s = \frac{- 5 \pm \sqrt{5^{2} - 4 (1) (- 500)}}{2 (1)}

s = \frac{- 5 \pm \sqrt{25 + 2000}}{2}

s = \frac{- 5 \pm \sqrt{2025}}{2}

$s = \frac{- 5 \pm 45}{2}$ .
Since speed must be positive, we take the positive root:
$s = \frac{- 5 + 45}{2} = \frac{40}{2} = 20$ .
The original speed of the bus was 20 km/hr.

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Pair of Linear Equations in Two Variables — Competency Based Questions

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