# Free Pair of Linear Equations in Two Variables 02 Practice Test - 10th Grade

### Question 1

Solve the following pair of equations:

2x+y=7

3x+2y=12

Choose the correct answer from the given options.

(2,3)

(3,2)

(1,0)

(-3,2)

#### SOLUTION

Solution :A

We have,

2x+y=7 ...(1)

3x+2y=12...(2)Multiply equation (1) by 2, we get:

2(2x+y)=2(7)

⇒4x+2y=14...(3)Subtracting (2) from (3) we get ,

x=2

Substituting the value of x in (1) we get,

2(2)+y=7⟹y=3Thus, the solution for the given pair of linear equations is (2,3).

### Question 2

The time taken to travel 30 km upstream and 44 km downstream is 14 hours. If the distance covered in upstream is doubled and distance covered in downstream is increased by 11 km then the total time taken is 11 hours more than earlier. Find the speed of the stream.

4 km/hr

7 km/hr

3 km/hr

6 km/hr

#### SOLUTION

Solution :A

Let's assume that the speed of the boat in still water is x km/hr and speed of the stream is y km/hr.

So, the speed of the boat in upstream will be (x-y) km/hr.

Similarly, the speed of the boat downstream will be (x+y) km/hr.

We know time=(distancespeed).

Using the above formula we can form the equations in two variables.

Taking the first case,

30x - y+44x + y=14.

Taking the second case,

60x - y+55x + y=25.

Now, we have the equations in two variables but the equations are not linear.

So, we will assume 1x - y=u and 1x + y=v.

So on substituting u and v in the above two equations, we get

30u+44v=14 ...(1)

60u+55v=25 ...(2)

We can solve the above two equations using the elimination method.

60u+88v=28 ...(3)

(by multiplying equation (1) by 2)

On subtracting equation (2) from (3), we get v=111

On substituting v in equation (2) we get u=13

Now as we have assumed

1x - y=u and 1x + y= v

On substituting the values of u and v,

we get a pair of linear equations in x and y

x - y=3...(4)

x + y=11...(5)

On adding (5) from (4), we have

2x=14

x=7

On subsituting the value of x in x−y=3, we get y = 4.

So, the speed of the boat in still water is 7 km/hr and the speed of the stream is 4 km/hr.

### Question 3

For what value of *k*, the pair of linear equations 3x+ky=9 and 6x+4y=18 has infinitely many solutions?

-5

6

1

2

#### SOLUTION

Solution :D

Given equations gives infinitely many solutions if,

a1a2=b1b2=c1c2

The given linear equations are:

3x+ky=9;6x+4y=18.

⇒a1=3,b1=k,c1=−9 and a2=6,b2=4,c2=−18

⇒36=k4=−9−18

⇒12=k4

⇒k=2

### Question 4

Find the solution of the given system of equations.

x+y−82=x+2y−148=3x+y−1211

(1, -1)

(2, 6)

(2, 2)

(0, 1)

#### SOLUTION

Solution :B

Take first two components,

x+y−82=x+2y−148

⇒8(x+y−8)=2(x+2y−14)

⇒8x+8y−64=2x+4y−28

⇒6x+4y−36=0

⇒3x+2y−18=0.....(i)Take last two components,

x+2y−148=3x+y−1211

⇒11(x+2y−14)=8(3x+y−12)

⇒11x+22y−154=24x+8y−96

⇒−13x+14y−58=0.....(ii)On multiplying equation (i) by 7, we get

⇒21x+14y−126=0...(iii),

On subtracting equation (ii) from (iii), we get

⇒34x=68

⇒x=2

On substituting value of x=2 in equation (i), we get

3×2+2y−18=0

⇒y=6

∴ The solution is (2, 6).

### Question 5

Given graph represents _______________ pair of linear equation having _____________solution(s).

Consistent, unique

Inconsistent ,zero

Dependent, infinite

None of these

#### SOLUTION

Solution :B

In the given graph, lines of the equations 2x + 4y - 12 = 0 and x + 2y - 4 = 0 are parallel to each other which gives no solution; hence it is an inconsistent pair of linear equation.

### Question 6

True

False

#### SOLUTION

Solution :A

Let, the cost of a bike be x.

The cost of scooter be y.

According to question x=2y

Therefore linear equation will be:

⇒x−2y=0

### Question 7

If (1,3k) lies on kx+4y=26, the value of k is

#### SOLUTION

Solution :If the point (1,3k) lies on the given equation, then it should satisfy the equation.

Substituting the values of x and y in the given equation:

⇒k(1)+4(3k)=26

⇒13k=26

⇒k=2

Thus, the value of k is 2.

### Question 8

The pairs of linear equations which have the unique solution x=2,y=–3 are,

x+y= –1;2x+3y= –5

2x+5y= –11;4x+10y= –22

2x–y=1;3x+2y=0

x–4y–14=0;5x–y–13=0

#### SOLUTION

Solution :A and D

Option A:

x+y= –1−−−−−(i)

2x+3y= –5−−−−(ii)

(i)×2→ 2x+2y=−2−−−−(iii)

Solving by elimination((iii)−(ii)) we get,

x=2 and y=−3.

Option D:

x−4y=14−−−−(i)5x−y=13−−−−−(ii)

(ii)×4→20x−4y=52−−−−(iii)Solving by elimination method ((iii)−(i)) we get,

x=2 and y=−3

### Question 9

A man starts his job with a certain salary and earns a fixed increment every year. If his salary was ₹ 15000 after 4 years of service and ₹ 18000 after 10 years of service, then find his starting salary and annual increment respectively.

₹ 11000, ₹ 700

₹ 13000, ₹ 500

₹ 13000, ₹ 700

₹ 11000, ₹ 500

#### SOLUTION

Solution :B

Let starting salary be ₹ x

and annual increment be ₹ y

According to the first condition:

x + 4y = 15000 ......(i)

According to the second condition:

x + 10y = 18000 ......(ii)

Subtracting (i) from (ii), we get

6y = 3000

y = 500

Substituiting y = 500 in equation (i), we get

x + (4×500) = 15000

x = 15000 - 2000

x = 13000

Hence starting salary is ₹ 13000 and annual increment is ₹ 500.

### Question 10

If the numerator of a fraction is increased by 2 and its denominator is decreased by 1, it becomes 23. If the numerator is increased by 1 and the denominator is increased by 2, it becomes 13. Find the fraction.

x=2 and y=7

x=−2 and y=−7

x=3 and y=7

x=2 and y=6

#### SOLUTION

Solution :A

Let the fraction be xy

∴ x+2y−1=23 and x+1y+2=13

⇒ 3x−2y=−8....(i)

and 3x−y=−1......(ii)

⇒3x−2y=−8

−(3x−y=−1)

⇒3x−2y=−8

−3x+y=1

______________________

−y=−7

______________________

⇒y=7

Substituting y = 7 in equation (ii), we get

3x−(14)=−8

3x=−8+14=6

∴x=2⇒x=2 and y=7