Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables Important Questions and Answers.

## RBSE Class 10 Maths Chapter 3 Important Questions Pair of Linear Equations in Two Variables

Objective Type Questions—

Question 1.

If 2A + y = 6 then the pair satisfying it is—

(A) (1, 2)

(B) (2, 1)

(C) (2, 2)

(D) (1, 1)

Answer:

(C) (2, 2)

Question 2.

If \(\frac{4}{x}\) + 5y = 7 and x = \(\) then the value of y will be—

(A) \(\frac{37}{15}\)

(B) 2

(C) \(\frac{1}{2}\)

(D) \(\frac{1}{3}\)

Answer:

(B) 2

Question 3.

In the equation \(\frac{y-3}{7}\) – \(\frac{x}{2}\) = 1 if y = 10, then A is equal to—

(A) 0

(B) 1

(C) – 2

(D) 2

Answer:

(A) 0

Question 4.

The age of the father is three times the age of the son, if the age of father is A years, then age of the son after 5 years will be—

(A) 3A + 5

(B) A + 5

(C) \(\frac{x}{3}\) + 5

(D) \(\frac{x+5}{3}\)

Answer:

(C) \(\frac{x}{3}\) + 5

Question 5.

The point on A-axis is—

(A) (2, 3)

(B) (2, 0)

(C) (0, 2)

(D) (2, 2)

Answer:

(B) (2, 0)

Question 6.

The quadrant, in which the point P(3, – 4) lies, is—

(A) first

(B) second

(C) third

(D) fourth

Answer:

(D) fourth

Question 7.

If in any number the unit’s digit is a and the ten’s digit is b, then that number is—

(A) 10a + b

(B) a + 10b

(C) a + b

(D) ab

Answer:

(B) a + 10b

Very Short Answer Type Questions—

Question 1.

For what value of K there exists no solution of the pair of equation

2x + Ky = 1,

3x – 5y = 7

Solution:

For no solution

Hence, for K = \(\frac{-10}{3}\), the system will have no solution.

Question 2.

On comparing the ratios \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the pair of linear equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 are consistent, or inconsistent.

Solution:

Comparing the equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 with a_{1}x + b_{1}y + c_{1}x = 0 and a_{2}x + b_{2}y + c_{2} = 0

Hence the pair of equations is Inconsistent.

Question 3.

Show that the lines x – 4y + 5 = 0 and 3r – 12y +8 = 0 are parallel.

Solution:

The equations of the given lines are

x – 4y + 5 = 0

and 3x – 12y +8 = 0

Comparing the above pair of equations with general pair of equations

a_{1} = 1, b_{1} = – 4, c_{1} = 5

a_{2} = 3, b_{2} = – 12, c_{2} = 8

Hence the given pair of equations is inconsistent. Hence the given lines are parallel.

Question 4.

Twice a number x is 24 more than y. Write an equation representing this statement.

Solution:

2x – y = 24

Question 5.

The age of Ram is x years and the age of Shyam is y years. Five years ago the age of Ram was twice the age of Shyam. Write the equation representing this statement in the form of ax + by + c = 0.

Solution:

x – 2y + 5 = 0

Question 6.

Write the methods of representing and presenting the solution of pair of linear equations in two variables.

Solution:

- Graphical Method,
- Algebraic Method.

Question 7.

What do you understand by the inconsistent pair of linear equations?

Answer:

If both the lines are parallel, then there exists no solution of this pair of linear equations. In this case the pair of linear equations is called an inconsistent pair.

Question 8.

Find the nature of solution of the following system of equations—

2x + 4y = 7, 3x + 6y = 10

Solution:

2x + 4y – 7 = 0

3x + 6y – 10 = 0

Comparing with a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0.

Hence the pair of equations is inconsistent and the system has no solution.

Question 9.

If we add 1 in the numerator and the denominator each of a fraction, then the value of the fraction becomes \(\frac{1}{2}\). How will you write it in the form of an equation.

Solution:

Let the fraction be \(\frac{x}{y}\)

Then, \(\frac{x+1}{y+1}\) = \(\frac{1}{2}\)

Solve for x means find the value of x that would make the equation you see true.

Question 10.

Find the value of x in the following system of equations—

2x + 3y = 4

3x + 4y = 5

Solution:

From given equations

2x + 3y – 4 = 0 …. (1)

3x + 4y – 5 = 0 …. (2)

Multiplying equation (1) by 4 and equation (2) by 3

Question 11.

Solving the equations x + y = 2xy and \(\frac{1}{x}+\frac{2}{y}\) = 10, find the value of y only.

Solution:

x + y = 2xy

Dividing both sides by xy

\(\frac{x}{x y}+\frac{y}{x y}\) = \(\frac{2 x y}{x y}\)

\(\frac{1}{y}+\frac{1}{x}\) = 2 …(1)

From second equation

\(\frac{1}{x}+\frac{2}{y}\) = 10 …(2)

Subtracting equation (2) from equation (1)

\(-\frac{1}{y}\) = -8

∴ y = \(\frac{1}{8}\)

Question 12.

Write the names of the methods of solving pair of linear equations in algebraic form.

Solution:

- Substitution method
- Elimination method
- Cross-multiplication method.

Question 13.

In the 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 explain the meaning of this situation.

Solution:

If \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\), then the pair of linear equations is inconsistent.

In this case lines are parallel and the pair of equation has no solution.

Question 14.

In the equation 5y – 3x – 10 = 0, express y in terms of x. Find the point where the line represented by the equation 5y – 3x – 10 = 0 intersects the y- axis.

Solution:

The given equation is 5y – 3x – 10=0

or 5y = 3x + 10

∴ y = \(\frac{3 x+10}{5}\)

The line represented by the equation 5y – 3x – 10 = 0 will intersect y-axis when x = 0. So y = \(\frac{3 \times 0+10}{5}\) = 2

Hence that point will be (0, 2)

Question 15.

In the linear equations a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0. If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\), then explain the meaning of this situation.

Solution:

If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\), then the pair of linear equations is consistent.

Question 16.

Write the solution of the pair of linear equations

\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 and \(\sqrt{3}\)x – \(\sqrt{2}\)y = 0

Solution:

The given pair of linear equations is

\(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 ….(1)

\(\sqrt{3}\)x – \(\sqrt{2}\)y = 0 ….(2)

From equation (2)

\(\sqrt{3}\)x = \(\sqrt{2}\)y

x = \(\frac{\sqrt{2}}{\sqrt{3}}\)y

Putting this value of x in equation (1)

Hence, x = 0, y = 0

Question 17.

The total cost of 7 pencils and 5 pens is Rs. 29. Write it in algebraic form.

Solution:

Let the cost of 1 pencil = Rs. x

and the cost of 1 pen = Rs. y

Then according to the question 7x + 5y = 29

Question 18.

Write the solution of the pair of linear equations 3x + 4y = 0

and 2x – y = 0

Solution:

x = 0 and y = 0

Question 19.

Write the solution of the pair of linear equations 4x + 2y = 5 and x – 2y = 0

Solution:

Pair of linear equations

4x + 2y = 5 …. (1)

and x – 2y = 0 …. (2)

From equation (2) we get

x = 2y

putting this value x = 2y in equation (1) then we get

4(2y) + 2y = 5

⇒ 8y + 2y = 5 ∴10y = 5

or y = \(\frac{5}{10}\) = \(\frac{1}{2}\)

putting this value y =1/2 in equation (2) then we get

x – 2 × \(\frac{1}{2}\) = 0 ∴ x – 1 = 0

or x = 1

so the required solution of the pair of linear equations is

x = 1 and y = \(\frac{1}{2}\)

Short Answer Type Questions—

Question 1.

A two-digit number in such that the product of its digits is 12. When 36 is added to this number, then the digits of the number are interchanged. Find the number.

Solution:

Let the ten’s digit and the unit’s digit be x and y respectively. So the number = 10x + y

According to the question

x × y = 12 ….(1)

and 10x + y + 36 = 10y + x ….(2)

or 9x – 9y = – 36

⇒ x – y = – 4 …(3)

From equation (3)

(x – y)^{2} = x^{2} + y^{2} – 2xy

⇒ (-4)^{2} = x^{2} + y^{2} – 2xy

⇒ x^{2} + y^{2} = 16 + 24 = 40

Again (x + y)^{2} = x^{2} + y^{2} + 2xy

= 40 + 2 × 12

⇒ (x + y)^{2} – 64

⇒ x + y = \(\sqrt{64}\) = ± 8 ….(4)

From equation (3) and (4)

x = 2 and y = 6

Hence the required number will be 26.

Question 2.

The sum of the digits of a two-digit number is 9. Four times this number is seven times the number formed by reversing the digits of the number. Find that number.

Solution:

Let the ten’s digit be x and the unit’s digit be y

Then required number = 10x + y

According to the question

x + y = 9 ….(1)

Again according to the question

4(10x + y) = 7(10y + x)

⇒ 40x + 4y = 70y + 7x

⇒ 33x – 66y = 0

⇒ x – 2y = 0 ….(2)

Subtracting equation (2) from equation (1)

Putting the value of y in equation (2)

x – 2 × 3 = o

⇒ x = 6

Hence required number = 10x + y

= 10 × 6 + 3 = 60 + 3

= 63

Question 3.

The number of teachers in a government college and a private college in a town is 210. When 10 teachers from private college resign and join government college then the number of teachers in both the colleges becomes the same. Find the number of teachers in each college.

Solution:

Let the number of teachers in government college = x

and the number of teachers in private college = y

Then according to the question

x + y = 210 …(1)

and (x + 10) = (y – 10)

⇒ x – y = – 20 ….(2)

Adding equations (1) and (2)

2x = 190

⇒ x= \(\frac{190}{2}\) = 95

Putting the value of x in equation (1)

93 + y = 210

⇒ y = 210 – 95 = 115

Hence the number of teachers in government college is 95 and in private college is 115.

Question 4.

The sum of the digits of a two-digit number is 7. On reversing the order of the digits the number obtained is a more than the original number. Find that number.

Solution:

Let in the number the ten’s digit = x

and the unit’s digit = y

Then the given number = (10x + y)

The number on reversing the order of the digits = (10y + x)

According to the question

x + y = 7

or x + y – 7 = 0

and (10x + y) + 9 = 10y

or 10x + y + 9 = 10y

or 9x + 9 = 9y

or x – y + 1 = 0

Solving equations (1) and (2)

Putting the value y = 4 in equation (1)

x + 4 – 7 = 0

⇒ x – 3 = 0

⇒ x = 3

∴ Required number = 10x + y

= 10 × 3 + 4

= 30 + 4 = 34

Question, 5.

6 years hence the age of a man will be 3 times the age of his son and 3 years ago he was 9 times the age of his son. Find their present ages.

Solution:

Let the present ages of the man and his son be reprectively x years and y years.

Age of father 6 years hence = (x + 6) years

and age of son 6 years hence = (y + 6) years

Hence according to the question

x + 6 = 3(y + 6)

⇒ x + 6 = 3y + 18

or x – 3y = 12 ….(1)

Again 3 years ago the ages of the father and the son will be respectively (x – 3) years and (y – 3) years.

Hence according to the question x – 3 = 9(y – 3)

⇒ x – 3 = 9y – 27

⇒ x – 9y = – 24 ….(2)

Now writing both the equations

Putting the value of y in equations (1)

x – 3 × 6 = 12

⇒ x – 18 = 12

⇒ x = 12 + 18 = 30

Hence age of father = 30 years

and age of son = 6 years

Question 6.

Aftab says to his daughter, ‘seven years ago I was of an age seven times of you. 3 years hence from now I shall be of an age only three times of you.’ (Is it interesting?) Express this situation in algebraic and graphical form.

Solution:

Let the age (in years) of Aftab and his daughter be respectively s and t. Then the pair of linear equations are

s – 7 = 7 (t – 7)

i.e., s – 7t + 42 = 0 ….(1)

and s + 3 = 3 (t + 3)

i.e., s – 3t = 6 …(2)

Using equation (2)

s = 3t + 6

Putting the value of s in equation (1)

(31 + 6) – 7t + 42 = 0

i.e., 4t = 48, so that t = 12

Putting this value of t in equation (2)

s – 3 × 12 = 6

⇒ s – 36 = 6

∴ s = 6 + 36 = 42

Hence, Aftab and his daughter are of age 42 years and 12 years respectively.

Question 7.

Two rails are represented by the equation x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails intersect each other?

Solution:

According to the question the linear equations are

x + 2y – 4 = 0 ….(1)

2x + 4y – 1 2 = 0 …(2)

From equation (1) expressing x in terms of y

x = 4 – 2y

Now, substituting this value of x in equation (2)

2 (4 – 2y) + 4y – 12 = 0

or 8 – 12 = 0

or -4 = 0

Which is a false statement.

Hence, there is no common solution of the given equations. Therefore, both the rails will not intersect each other.

Question 8.

For what values of p, the following pair of equations has a unique solution

4x + py + 8 = 0

2x + 2y + 2 = 0

Solution:

Here a_{1} = 4. a_{2} = 2, b_{1} = p, b_{2} = 2

Now for a unique solution of the given pair

Hence, for each value of p except 4, the given pair of equations will have a unique solution.

Question 9.

Show graphically that the following pair of equations 2x + 4y = 10; 3x + 6y = 12 has no solution.

Solution:

From equation

2x + 4y= 10

4y= 10 – 2x

or y = \(\frac{10-2 x}{4}\)

x |
1 | – 3 | 3 |

y |
2 | 4 | 1 |

From equation

3x + 6y = 12

y = \(\frac{12-3 x}{6}\)

x |
0 | 2 | – 2 |

y |
2 | 1 | 3 |

Plotting the points (1, 2), (- 3, 4) and (3, 1) and joining we get the graph AB of the equation 2x + 4y =10 and plotting the points (0, 2), (2, 1) and (-2, 3) and joining we get the graph CD of the equation 3x + 6y= 12. These two lines are mutually parallel. Hence the given pair of equations is inconsistent and it has no solution.

Question 10.

For what value of k will the following pair of linear equations have infinitely many solutions?

kx + 3y = k – 3

12x + ky = k

Solution:

For infinitely many solutions

⇒ 3 = k – 3

⇒ k = 6

Hence k = 6

Long Answer Type Questions—

Question 1.

Check graphically whether the pair of equations

x + 3y = 6

and 2x – 3y = 12

are consistent. It in then, solve graphically.

OR

Solve the following linear equations graphically

x + 3y = 6

2x – 3y = 12

Solution:

Let us form tables from the first equation

x + 3y = 6

or 3y = 6 – x

∴ y = \(\frac{6-x}{3}\)

x |
0 | 6 |

y |
2 | 0 |

From second equation

2x – 3y = 12

or 2x – 12 = 3y

or y = \(\frac{2x-12}{3}\)

x |
0 | 3 |

y |
– 4 | _ 2 |

Let us plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on a graph paper and joining the points lines AB and PQ are drawn as shown in the figure.

We see that the lines AB and PQ have a common point B(6, 0). Hence a solution of the pair of linear equations is x = 6,

y = 0, i.e., the pair of equations is consistent.

Question 2.

Find the solution of the pair of equations 2x – y = 1; x + 2y = 8 by graphical method and find the coordinates of these points where their corresponding lines meet y-axis.

Solution:

Equation 2x – y = 1

So y = 2x – 1

x |
1 | 2 | 3 |

y |
1 | 3 | 5 |

Equation x + 2y = 8

So y = \(\frac{8-x}{2}\)

x |
2 | 4 | 6 |

y |
3 | 2 | 1 |

Drawing the graph with the points (1, 1) (2, 3) (3, 5) and the points (2, 3) (4, 2) (6, 1) we find that the point (2, 3) is the point where both the graphs intersect. This is the required solution of the equations. Hence x = 2,y = 3.

The first line intersects the y-axis at point (0, – 1) and the second line intersects the y-axis at the point (0, 4).

Question 3.

The sum of a two digit number and the number formed by reversing the digits is 66. If the difference of the digits of the number is 2, then find the number. How many such numbers are there?

Solution:

Let the ten’s digit and the unit’s digit of the first number be A and y respectively. Therefore, the expanded form of the first number is 10x + y. When the digits are reversed, then x becomes the unit’s digit and y the ten’s digit. This number in the expanded form is 10y + x.

According to the given conditions

(10x + y) + (10y + x) = 66

or 11 (x + y) = 66

or x + y = 6 ….(1)

We are also given that the difference of the digits is 2.

Therefore,

either x – y = 2 …(2)

or y – x = 2 …(3)

If x – y = 2, then solving (1) and (2) by elimination method, we get A = 4 and y = 2. In this case, we get the number 42.

If y – x = 2, then solving (1) and (3) by elimination method, we get A = 2 and y = 4. In this case, we get the number 24.

Thus there are two such numbers 42 and 24.

Verification :

Here 42 + 24 = 66 and 4 – 2 – 2

and 24 + 42 = 66 and 4 – 2 = 2

Question 4.

For what value of k, will there be infinitely many solutions of the following pair of linear equations?

kx + 3y – (k – 3) = 0

12A + ky – k = 0

Solution:

For infinitely many solutions of the pair of linear equations, we should have

Which gives

3k = k^{2} – 3k

i.e., 6k = k^{2}

which means k = 0 or k = 6

Therefore the value of k which satisfies both the conditions is k = 6. For this value there are infinitely many solutions of the pair of equations.

Question 5.

From a bus stand of Bangalore if we purchase two tickets of Malleshwaram and 3 tickets of Yashwantpur, then the total cost is Rs. 46. But if we purchase 3 tickets of Malleshwaram and 5 ticket of Yaswantpur, then the total cost is Rs. 74. Find the fare from Bus stand to Malleshwaram and from Bus stand to Yashwantpur.

Solution:

Let the fare from bus stand of Bangalore to Malleshwaram be Rs. x and to Yashwantpur be Rs. y.

From the given informations

2x + 3y = 46

i.e., 2x + 3y – 46 = 0 ….(1)

3x + 5y = 74

i.e., 3x + 5y – 74 = 0 ….(2)

Solving these equations by cross-multiplication method

i.e., x = 8 and y = 10

Hence, from bus stand of Bangalore, the fare of Malleshwaram is Rs. 8 and the fare of Yashwantpur is Rs. 10.

Question 6.

Solve the following pair of equations by converting into pair of linear equations

Solution:

Putting \(\frac{1}{x-1}\) = p and \(\frac{1}{y-1}\) = q the given equations become

5p + q = 2 …(3)

6p – 3q = 1 …(4)

Equations (3) and (4) in general form a pair of linear equations. Now to solve these equations, we can use any method. We find,

p = \(\frac{1}{3}\) and q = \(\frac{1}{3}\)

Now substituting \(\frac{1}{x-1}\) for p

⇒ y – 2 = 3

⇒ y = 5

Hence, the required solution of the given pair of equations is x = 4, y = 5.

Question 7.

The ratio of the incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each person saves ₹ 2000 each month, then find their monthly incomes.

Solution:

Let the monthly incomes of both the persons be ₹ 9x and ₹ 7x respectively and their monthly expenditures be ₹ 4y and ₹ 3y respectively. Then, in that

situtation the equations formed are

9x – 4y = 2000 ….(1)

and 7x – 3y = 2000 …(2)

Step 1 : To make the coefficients of y equal multiplying equation (1) by 3 and equation (2) by 4

27x – 12y = 6000 ….(3)

28x – 12y = 8000 …(4)

Step 2 : To eliminate y subtracting equation (3) from equation (4) since the coefficients of y are equal, we get

(28x – 27x) – (12y- 12y) = 8000 – 6000

⇒ x = 2000

Step 3 : Substituting the value of x in (1)

9 (2000) – 4y= 2000

⇒ y = 4000

Hence the solution of the pair of equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are respectively ₹ 18,000 and ₹ 14,000.

Question 8.

A boat gives 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Solution:

Let the speed of the boat in still water be x km/h and speed of the stream be y km/h. Tfyen the speed of the boat downstream = (x + v) km/h and the speed of the boat upstream = (x – y) km/h

Also, time = \(\frac{\text { distance }}{\text { speed }}\)

In the first case, when the boat goes 30 km upstream. Let the time taken in hours be t_{1}. Then,

t_{1} = \(\frac{30}{(x-y)}\)

Let t_{2} be the time in hours taken by the boat to go 44 km downstream. Then,

t_{2} = \(\frac{44}{(x+y)}\)

The total time taken t_{1} + t_{2} = 10 hours

Therefore, \(\frac{30}{(x-y)}\) + \(\frac{44}{(x+y)}\) = 10 ….(1)

In the second case, in 13 hours it can go 40 km upstream and 55 km downstream. Then

\(\frac{40}{(x-y)}\) + \(\frac{55}{(x+y)}\) = 13 ….(2)

Putting \(\frac{1}{(x-y)}\) = u and \(\frac{1}{(x+y)}\) = v …(3)

Substituting these values in equations (1) and (2)

30u + 44v = 10

or 30u + 44v – 10 = 0 ….(4)

40u + 55v = 13

or 40u + 55v – 13 = 0 ….(5)

Using cross-multiplication method,

or x – y = 5 and x + y = 11 …(6)

Adding these equations

2x = 16

or x = 8

Subtracting equation given in (6)

2y = 6

or y = 3

Hence, the speed of the boat in still water in 8 km/h and the speed of the stream water is 3 km/h.

Question 9.

The coach of a cricket team buys one bat and 3 balls for ₹ 300. Later, she buys another 2 bats of the samekind and 3 balls for ₹ 525. Represent this situation algebraically and solve it by graphical method. Also find for how many rupees can be coach buy one bat and one ball.

Solution:

Let the cost of one bat = ₹ x

and the cost of one ball = ₹ y

Algebraic Form—

According to the first condition of the « problem

x + 2y = 300

According to the second condition of the problem

2x + 3y = 525

∴ Pair of linear equations in two variables

x + 2y = 300

and 2x + 3 y = 525

Graphical Form—

x + 2y = 300

⇒ x = 300 – 2y ….(1)

Substituting y = 0 in (1)

x = 300 – 2 × 0

⇒ x – 300

Substituting y = 75 in (1)

x = 300 – 2 × 75

= 300 – 150

⇒ x = 150

Substituting y = 100 in (2)

x = 300 – 2 × 100

= 300 – 200

⇒ x = 100

x |
300 | 150 | 100 |

y |
0 | 75 | 100 |

2x + 3y = 525 …(2)

2x = 525 – 3y

x = \(\frac{525-3 y}{2}\)

Substituting y = 25 in (2)

x = \(\frac{525-3 \times 25}{2}\)

x |
225 | 150 | 75 |

y |
25 | 75 | 125 |

On solving graphically x = 150 and y = 75 and the coach will be able to buy a bat and a ball for

= 150 + 75

= ₹ 225

Question 10.

Ashok scored 65 marks in a test, getting 5 marks for each right answer and losing 2 marks for each wrong answer. That 3 marks been awarded for each correct answer and 1 mark been deducted for each incorrect answer, then Ashok would have scored 40 marks. Represent this situation algebraically and solve it graphically. How many questions were there in the test?

Solution:

Let the right answer question be x and the wrong answer questions be y.

Then, according to the first condition of the problem

5x – 2y = 65 ….(1)

According to the second condition of the problem

3x – y = 40 ….(2)

From equation (1) 5x = 65 + 2y

∴ x = \(\frac{65+2 y}{5}\)

x |
13 | 15 | 11 |

y |
0 | 5 | -5 |

Putting y = 0 x = \(\frac{65+0}{5}\) = 13

Putting y = – 5 x = \(\frac{65+10}{5}\) = 15

Putting y = -5 x = \(\frac{65-10}{5}\) = 11

From equation (2)

3x – y = 40

⇒ -y = 40 – 3x

⇒ y = 3x – 40

x |
13 | 15 | 20 |

y |
-10 | 5 | 20 |

Putting x = 10 y – 30 – 40 = – 10

Putting x = 15 y = 45 – 40 = 5

Putting x = 20 y = 60 – 40 = 20

∵ The point of intersection of both the graphs is (15, 5)

So total number of questions = 15 + 5 = 20

Question 11.

The cost of 5 apples and 3 oranges is Rs. 35 and the cost of 2 apples and 4 oranges is Rs. 28. For mulati the problem algebraically and solve it graphically.

Solution:

Let the number of apples denote by x and the number of oranges denote by y according to 1st condition of question.

5x + 3y = 35

3y = 35 – 3x

∴ y = \(\frac{35-5 x}{3}\) …(1)

Apples (x) |
1 | 4 | – 2 | 7 |

Oranges (y) |
10 | 5 | 15 | 0 |

According to Ilnd condition of question

2x + 4y – 28

4y = 28 – 2x

∴ y = \(\frac{28-2 x}{4}\) …(2)

Apples (x) |
0 | 14 | 4 | -2 |

Oranges (y) |
7 | 0 | 5 | 8 |

Drawing the graph with the points (1, 10), (4, 5), (-2, 15), (7, 0) and the points (0, 7), (14, 0) (4, 5) and (-2, 8). We find that the point (4, 5) in the point where both the graphs intersect. This is the required solution of the equations.

Hence x = 4 Apples and y = 5 Oranges.

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