# RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4

RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4.

 Board RBSE Class Class 9 Subject Maths Chapter Chapter 4 Chapter Name Linear Equations in Two Variables Exercise Ex 4.4 Number of Questions Solved 11 Category RBSE Solutions

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4

Question 1.
In a two digit number, the unit digit (RBSESolutions.com) is three times of tens digit. If 10 is added to two times the number, the digits are reversed. Find the number.
Solution.
Let a two digit number be 10x + y, where x is tens place digit and y is unit place digit.
According to question
y = 3x …(i)
and 2(10x + y) + 10 = 10y + x
⇒ 20x + 2y + 10 = 10y + x
⇒ 19x – 8y = -10 …(ii)
Putting y = 3x in (ii), we get
19x – 8 × 3x = – 10
⇒ 19x – 24x = – 10
⇒ -5x = -10
⇒ x = 2
y = 3 × 2 = 6
Hence, the required number is
10x + y = 10 × 2 + 6 = 26. Question 2.
The perimeter of a rectangle is 56 cm. The ratio (RBSESolutions.com) of their length and breadth is 4:3. Find the length and breadth of the rectangle.
Solution.
Let length of the rectangle be l and breadth be b
∴Perimeter = 2(l + b)
⇒ 56 = 2(l + b)
⇒ l + b = 28 …(i)
and l : b = 4 : 3
⇒ l = 4x, b = 3x …(ii)
Substituting l = 4x and b = 3x in (i), we
get 4x + 3x = 28
⇒ 7x = 28 ⇒ x = 4
∴ l = 4 x 4 = 16 cm
and b = 3 x 4 = 12 cm
Hence, length of the rectangle = 16 cm and breadth = 12 cm.

Question 3.
Two numbers are in the ratio of 3 : 4. If 5 is subtracted (RBSESolutions.com) from each number then their ratio become 5 : 7. Find the numbers.
Solution.
Let one number be x and another number be y,
According to question x : y = 3 : 4
x = 3k, y = 4k …(i)
Again $$\frac { x-5 }{ y-5 }$$ = $$\frac { 5 }{ 7 }$$
⇒ 7x – 35 = 5y – 25
(by cross multiplication)
⇒ 7x – 5y – 10 ….(ii)
Substituting the value of x and y in (RBSESolutions.com) equation (ii), we get
7 × 3k – 5 × 4k = 10
⇒ 21k – 20k = 10
⇒ k = 10
Hence, x = 3k = 3 × 10 = 30
and y = 4k = 4 × 10 = 40.

Question 4.
The present age of a man is (RBSESolutions.com) five years more than six years more than six times the age of his son. Seven years later, the man’s age will be three years more than thrice the age of the sun, then determine their present age.
Solution.
Let present age of the man be x years and his son’s age be y years.
Then according to problem x = 6y + 5
⇒ x – 6y = 5 …(i)
After 7 years
their ages will be (x + 7) years and (y + 7) years
∴ x + 7 = 3(y + 1) + 3
⇒ x + 7 = 3y + 21 + 3
or x – 3y = 17 …(ii)
Subtracting (i) from (ii), we get
3y = 12 ⇒ y = 4
Substituting y = 4 in (i), we get
x – 6 × 4 = 5
⇒ x = 5 + 24 = 29 years
Hence, father’s present age = 29 years and his son’s age = 4 years.

Question 5.
Ram said to Shyarn, if you will give ₹ 100 to me them it (RBSESolutions.com) will be doubles of yours then Shyam said to Ram if you will give me ₹ 10 then my amount will become 6 times. Find what amount they both have?
Solution.
Let Ram is having ₹ x and Shyam be ₹ y
x + 100 = 2(y – 100)
⇒ x – 2y = – 300 …(i)
Now y + 10 = 6(x – 10)
y – 6x = – 70 …(ii)
Multiplying equation (ii) by 2 and then adding in (i), we get
11x = – 440
⇒ x = 40
Putting x = 40 in (i), we get
40 – 2y = – 300
⇒ 340 = 2y ⇒ y = 170
Hence, Ram is having ₹ 40 and Shyam is having ₹ 170. Question 6.
The cost of 4 chairs and 3 tables is ₹ 2,100 and cost of 5 chairs and 2 tables are ₹ 1,750. Find the (RBSESolutions.com) cost of 1 chair and 1 table.
Solution.
Let cost of a chair be ₹ x and cost of a table be ₹ y
4x + 3y = 2100 …(i)
and 5x + 2 y = 1750 …(ii)
Multiplying equation (i) by 2 and equation
(ii) by 3 and then subtracting, we get
– 7x = – 1050
⇒ x = 150
Putting the (RBSESolutions.com) value of x in (i), we get
4 x 150 + 3y = 2100
⇒ 600 + 3y = 2100
⇒ 3y = 1500
⇒ y = 500
Hence, cost of chair is ₹ 150 and cost of table is ₹ 500.

Question 7.
Two numbers are such that if we divide the 3 times the greater number by the smaller number we obtained the quotient is 4 and remainder 3 and if 7 times the smaller number is (RBSESolutions.com) divided by bigger number then quotient and remainder are 5 and 1 respectively. Find the numbers.
Solution.
Let x and y are two numbers such that x > y
3x = 4 x y + 3
⇒ 3x – 4y = 3 …(i)
7y = 5x + 1
⇒ 7y – 5x = 1 …(ii)
Multiplying (i) by 5 and (ii) by 3 and then adding, we get
y = 18
Putting the (RBSESolutions.com) value of y in (i), we get
3x – 4 x 18 = 3
⇒ 3x – 72 = 3
⇒ 3x = 75
⇒ x = 25
Hence, bigger number is 25 and smaller number is 18.

Question 8.
A two digit number is 4 times its sum of digits and 2 times the (RBSESolutions.com) product of its digits. Find the number.
Solution.
Let a two digit number be 10x + y, where y is unit place digit and x is tens place digit
10x + y = 4(x + y)
⇒ 10x + y = 4x + 4y
⇒ 6x – 3y = 0
⇒ 2x = y …(i)
and 10x + y = 2xy …(ii)
Substituting y = 2x in (ii), we get
10x + 2x = 2x x 2x
⇒ 4x² – 12x = 0
⇒ 4x(x – 3) = 0
⇒ x = 0, 3
where x = 0, y = 0
when x = 3, y = 2 x 3 = 6
∴ Required number is
10x + y = 10 x 3 + 6 = 36

Question 9.
If 1 is added to both numerator and denominator of a fraction, the (RBSESolutions.com) fraction becomes $$\frac { 4 }{ 5 }$$. When 5 is subtracted both from numerator and denominator then it become $$\frac { 1 }{ 2 }$$. Find the fraction.
Solution.
Let fraction be $$\frac { x }{ y }$$
According to question, we have Hence, required fraction is $$\frac { 7 }{ 9 }$$

Question 10.
5 years ago, Geeta’s age was 3 times of Kamla’s age. 10 years hence, Geeta’s age will be 2 times Kamala’s age. Find (RBSESolutions.com) their present age.
Solution.
Let Geeta’s present age be x years and Kamla’s be y years
5 years ago their ages were x – 5, y – 5 respectively.
According to question, we have
x – 5 = 3(y – 5)
⇒ x – 5 = 3y – 15
x – 3y = – 10 ..(i)
10 years hence (RBSESolutions.com) their ages will be
x + 10, y + 10
x + 10 = 2 (y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 10 …(ii)
Subtracting (i) from (ii), we get
y = 20
Putting y = 20 in (ii), we get
x – 2 x 20 = 10 ⇒ x = 50
Hence, Geeta’s present age is 50 years and Kamla’s present age is 20 years. Question 11.
A man travels 370 km partly by train and partly by car. He takes 4 hours if he (RBSESolutions.com) travels 250 km by train and the remaining by car. If he travels 130 km by train and the remaining by car, he takes 18 minutes longer. Find the speed of the train and the car separately.
Solution.
Let the speed of tram be x km/h and the speed of car be y km/h
Total distance travelled by man = 370 km
If he travels 250 km by train, then distance travelled by car (370 – 250) = 120 km
Time taken to travel 250 km by train
$$=\frac { 250 }{ x }$$
and the time taken to travel 120 km by car   Hence, speed (RBSESolutions.com) of train is 100 km/h and speed of car is 80 km/h. We hope the given RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4, drop a comment below and we will get back to you at the earliest.