Class 9 Maths Ganita Manjari Chapter 13 Exercise 13.5 Solutions Two Variables, One Line

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise Set 13.5

(i) The sum of two integers is +5 and their difference is −21.

Let the two integers be x and y.

x + y = 5
x − y = −21

Adding the equations:

2x = −16
x = −8

Substituting in x + y = 5:

−8 + y = 5
y = 13

Therefore, the two integers are −8 and 13.

(ii) The difference between two numbers is 26 and one number is three times the other.

Let the larger number be x and the smaller number be y.

x − y = 26
x = 3y

Substituting x = 3y:

3y − y = 26
2y = 26
y = 13

Therefore,

x = 3 × 13 = 39

Hence, the numbers are 39 and 13.

(iii) The coach buys 7 bats and 6 balls for ₹8880 and 3 bats and 5 balls for ₹4000.

Let the cost of one bat be ₹x and one ball be ₹y.

7x + 6y = 8880
3x + 5y = 4000

Multiplying the first equation by 5:

35x + 30y = 44400

Multiplying the second equation by 6:

18x + 30y = 24000

Subtracting:

17x = 20400

x = 1200

Substituting in 3x + 5y = 4000:

3600 + 5y = 4000

5y = 400

y = 80

Therefore, the cost of each bat is ₹1200 and each ball is ₹80.

(iv) Taxi charges

Let the fixed charge be ₹x and the charge per kilometre be ₹y.

For 10 km:

x + 10y = 155

For 15 km:

x + 15y = 220

Subtracting the first equation from the second:

5y = 65

y = 13

Substituting in x + 10y = 155:

x + 130 = 155

x = 25

Therefore:

Fixed charge = ₹25
Charge per km = ₹13

For 25 km:

Fare = 25 + 25(13)
= 25 + 325 = ₹350

Hence, the fare for travelling 25 km is ₹350.

(v) Find the fraction.

Let the numerator be x and the denominator be y.

According to the first condition:

(x + 2)/(y + 2) = 9/11

Cross-multiplying:

11x + 22 = 9y + 18

11x − 9y = −4 ...(1)

According to the second condition:

(x + 3)/(y + 3) = 5/6

Cross-multiplying:

6x + 18 = 5y + 15

6x − 5y = −3 ...(2)

Solving (1) and (2), we get:

x = 7, y = 9

Therefore, the fraction is 7/9.

(vi) Find the fraction.

Let the numerator be x and the denominator be y.

From the first condition:

(x + 1)/(y − 1) = 1

Therefore,

x + 1 = y − 1

x − y = −2 ...(1)

From the second condition:

x/(y + 1) = 1/2

Therefore,

2x = y + 1

2x − y = 1 ...(2)

Subtracting (1) from (2):

x = 3

Substituting in (1):

3 − y = −2

y = 5

Therefore, the fraction is 3/5.

(vii) Ages of Nuri and Sonu

Let Nuri's present age be x years and Sonu's present age be y years.

Five years ago:

x − 5 = 3(y − 5)

x − 3y = −10 ...(1)

Ten years later:

x + 10 = 2(y + 10)

x − 2y = 10 ...(2)

Subtracting (1) from (2):

y = 20

Substituting in (2):

x − 40 = 10

x = 50

Therefore, Nuri is 50 years old and Sonu is 20 years old.

(viii) Find the two-digit number.

Let the tens digit be x and the units digit be y.

The sum of the digits is 9:

x + y = 9 ...(1)

The number is 10x + y, and the reversed number is 10y + x.

According to the question:

9(10x + y) = 2(10y + x)

90x + 9y = 20y + 2x

88x − 11y = 0

8x − y = 0 ...(2)

From (2):

y = 8x

Substituting in (1):

x + 8x = 9

9x = 9

x = 1, y = 8

Therefore, the number is 18.

(ix) ₹50 and ₹100 notes

Let the number of ₹50 notes be x and the number of ₹100 notes be y.

Total number of notes:

x + y = 25 ...(1)

Total amount:

50x + 100y = 2000

Dividing by 50:

x + 2y = 40 ...(2)

Subtracting (1) from (2):

y = 15

Therefore,

x = 25 − 15 = 10

Hence, Meena received 10 notes of ₹50 and 15 notes of ₹100.

(x) Lending library charges

Let the fixed charge for the first three days be ₹x and the additional charge per extra day be ₹y.

Saritha kept the book for 7 days, so she paid for 4 extra days:

x + 4y = 27 ...(1)

Susy kept the book for 5 days, so she paid for 2 extra days:

x + 2y = 21 ...(2)

Subtracting (2) from (1):

2y = 6

y = 3

Substituting in (2):

x + 6 = 21

x = 15

Therefore, the fixed charge is ₹15 and the charge for each extra day is ₹3.

Let the number of boys be x and the number of girls be y.

Since there are 10 students in total:

x + y = 10 ...(1)

Since the number of girls is 4 more than the number of boys:

y = x + 4

or,

y − x = 4 ...(2)

To represent the equations graphically:

• Equation (1): x + y = 10
• Equation (2): y − x = 4

The two lines intersect at the point (3, 7).

For a pair of linear equations

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

we use:

• If a₁/a₂ ≠ b₁/b₂, the lines intersect at one point.
• If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
• If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident.

(i) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0

Here,

a₁/a₂ = 5/7

b₁/b₂ = −4/6 = −2/3

Since,

5/7 ≠ −2/3

the lines intersect at one point.

(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0

Here,

a₁/a₂ = 9/18 = 1/2

b₁/b₂ = 3/6 = 1/2

c₁/c₂ = 12/24 = 1/2

Since,

a₁/a₂ = b₁/b₂ = c₁/c₂

the lines are coincident.

(iii) 6x − 3y + 10 = 0 and 2x − y + 9 = 0

Here,

a₁/a₂ = 6/2 = 3

b₁/b₂ = −3/−1 = 3

c₁/c₂ = 10/9

Since,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

the lines are parallel.

Therefore:

(i) Intersecting lines
(ii) Coincident lines
(iii) Parallel lines

(i) x + y = 5 and 2x + 2y = 10

Dividing the second equation by 2:

x + y = 5

Thus, both equations represent the same line. Therefore, the lines coincide and have infinitely many common points.

Hence, the pair has infinitely many solutions.

(ii) x − y = 8 and 3x − 3y = 16

The equations can be written as:

y = x − 8

and

y = x − 16/3

Both lines have the same slope but different intercepts. Therefore, they are parallel and never intersect.

Hence, the pair has no solution.

(iii) 2x + y − 6 = 0 and 4x − 2y − 4 = 0

The equations are:

2x + y = 6

4x − 2y = 4

From the first equation:

y = 6 − 2x

From the second:

y = 2x − 2

The two lines intersect at (2, 2).

Therefore, the pair has a unique solution:

x = 2, y = 2

(iv) 2x − 2y − 2 = 0 and 4x − 4y − 5 = 0

The first equation becomes:

x − y = 1

The second equation becomes:

x − y = 5/4

The two lines have the same slope but different intercepts. Therefore, they are parallel and do not intersect.

Hence, the pair has no solution.

Pair	Graphical relationship	Solution
(i)	Coincident lines	Infinitely many solutions
(ii)	Parallel lines	No solution
(iii)	Intersecting lines	(2, 2)
(iv)	Parallel lines	No solution

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer:

Let the width of the garden be x m.

Then the length is:

x + 4 m

Half the perimeter of a rectangle is:

Length + Width

So,

x + (x + 4) = 36

2x + 4 = 36

2x = 32

x = 16

Therefore,

Width = 16 m

Length = 16 + 4 = 20 m

Hence, the dimensions of the garden are 20 m × 16 m.

Given the equation:

2x + 3y − 8 = 0

Here, a₁ = 2, b₁ = 3, c₁ = −8.

(i) Intersecting lines

For intersecting lines:

a₁/a₂ ≠ b₁/b₂

One suitable equation is:

x + y − 4 = 0

Here,

2/1 ≠ 3/1

Therefore, the lines intersect at one point.

(ii) Parallel lines

For parallel lines:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

One suitable equation is:

4x + 6y − 10 = 0

Here,

2/4 = 3/6 = 1/2

but

−8/−10 = 4/5

Therefore, the lines are parallel.

(iii) Coincident lines

For coincident lines:

a₁/a₂ = b₁/b₂ = c₁/c₂

Multiplying the given equation by 2 gives:

4x + 6y − 16 = 0

Here,

2/4 = 3/6 = −8/−16 = 1/2

Therefore, the lines are coincident.

Hence, one possible set of answers is:

(i) x + y − 4 = 0
(ii) 4x + 6y − 10 = 0
(iii) 4x + 6y − 16 = 0

Let the price of one citron be x and the price of one fragrant wood-apple be y.

The equations are:

9x + 7y = 107 ...(1)
7x + 9y = 101 ...(2)

Since the coefficients of x and y are interchanged, add and subtract the equations.

Adding (1) and (2):

16x + 16y = 208

Dividing by 16:

x + y = 13 ...(3)

Subtracting (2) from (1):

2x − 2y = 6

Dividing by 2:

x − y = 3 ...(4)

Now, adding (3) and (4):

2x = 16

x = 8

Substituting in x + y = 13:

8 + y = 13

y = 5

Therefore, the price of each citron is 8 and the price of each fragrant wood-apple is 5.

8. 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of each pencil and pen.

Answer: