MSBSHSE Solutions For SSC (Class 10) Maths Part 1 Chapter 1 - Linear Equations in Two Variables

MSBSHSE Solutions For SSC (Class 10) Maths Part 1 Chapter 1 Linear Equations in Two Variables are provided here for students to practise and prepare for their exam. BYJU’S brings you Maharashtra Board Solutions for Class 10, designed by our subject experts to facilitate smooth and precise understanding of concepts. These solutions of MSBSHSE for Class 10 (SSC) have detailed step-by-step explanations of problems given in the Maharashtra Board Textbooks for SSC Part 1. The Maharashtra State Board Solutions for Chapter 1 can be downloaded in the form of a PDF and students can use it as a reference tool to quickly review all the topics. This chapter mainly focuses on the concept of linear equations and methods to solve them. The chapter deals with the definition, meaning, and graphical methods related to linear equations. The Cramer’s Rule as well as application of simultaneous equations is also discussed in this chapter.

Download the PDF of Maharashtra Solutions For SSC Maths Part 1 Chapter 1 Linear Equations in Two Variables

 

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Access answers to Maths MSBSHSE Solutions For SSC Part 1 Chapter 1 – Linear Equations in Two Variables

Practice set 1.1 Page no: 4

1. Complete the following activity to solve the simultaneous equations.

5x + 3y = 9 ……. (i)
2x + 3y = 12 ……… (ii)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 1

Solution:

Given

5x + 3y = 9 ……. (i)

2x + 3y = 12 ……… (ii)

Subtracting equation (ii) from (i), we get,

(5x + 3y) – (2x + 3y) = 9 – 125x – 2x + 3y – 3y

= -33x = -3x = -1

Putting the value of x in equation (i),

5(-1) + 3y = 9-5 + 3y = 93y = 14y = 14/3

Let’s add equations (I) and (II).

Hence, x = -1 and y = 14/3 is the solution of the equation.

2. Solve the following simultaneous equation.

(1) 3a + 5b = 26; a + 5b = 22

Solution:

3a + 5b = 26 …… (i)

a + 5b = 22 ……… (ii)

Now by changing the sign of equation (ii) we get

– a – 5b = – 22

Subtracting equation (ii) from (i) we get

2a = 4

a = 4/2

a = 2

Substituting a = 2 in equation (ii) we get

2 + 5b = 22

5b = 22 – 2

5b = 20

b = 20/5

b = 4

∴ solution is (a, b) = (2, 4)

(2) x + 7y = 10; 3x – 2y = 7

Solution:

Given

x + 7y = 10 ……. (i)

x – 2y = 7 ……… (ii)

Multiply equation (i) by 2 and equation (ii) by 7

2x + 14 y = 20

21x – 14 y = 49

Which implies

23 x = 69

x = 69/23

x = 3

Substituting x = 3 in equation (i)

3 + 7y = 10

7y = 10 – 3

7y = 7

y = 7/7

y = 1

∴ Solution is (x, y) = (3, 1)

(3)
2x – 3y = 9; 2x + y = 13

Solution:

Given

2x – 3y = 9 …… (i)

2x + y = 13 …… (ii)

To subtract equation (ii) from (i)

Change the sign of equation (ii)

2x – 3y = 9

-2x – y = – 13

Which implies

– 4y = – 4

y = 4/4

y = 1

Substituting y = 1 in equation (ii)

2x + 1 = 13

2x = 13 – 12x = 12x = 6

∴ solution is (x, y) = (1, 6)

(4) 5m – 3n = 19; m – 6n = –7

Solution:

Given

5m – 3n = 19 …… (i)

m – 6n = –7 …. (ii)

Multiply equation (ii) by 5

5m – 30n = -35 …. (iii)

Equating (i) and (iii), change the sign of equation (iii)

5m – 3n = 19

– 5m + 30 n = 35

Adding both we get

27n = 54

n = 54/27

⇒ n = 2

Substituting n = 2 in equation (i)

⇒ 5m – 3(2) = 19

⇒ 5m – 6 = 19

⇒ 5m = 25

⇒ m = 5

∴ Solution is (m, n) = (5, 2)

(5) 5x + 2y = –3; x + 5y = 4

Solution:

5x + 2y = – 3 …. (i)

x + 5y = 4 …… (ii)

Multiply equation (i) by 5 and equation (ii) by 2

25 x + 10 y = -15 …. (iii)

2x + 10 y = 8 …. (iv)

Change sign of equation (iv)

25 x + 10 y = -15

– 2x – 10 y = – 8

23 x = -23

x = -1

Substituting x = –1 in equation (ii)

– 1 + 5y = 4

5y = 4 + 1

5y = 5

Y = 1

∴ solution is (x, y) = (–1, 1)


Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 2

(6)

Solution:


Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 3

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 4

(7) 99x + 101 y = 499; 101x + 99y = 501

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 5

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 6

(8) 49x – 57y = 172; 57x – 49y = 252

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 7

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 8

Practice set 1.2 Page no: 8

1. Complete the following table to draw graph of the equations

(I) x + y = 3

(II) x – y = 4

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 9

Solution:

(I) Given

x + y = 3 …. (i)

(i) Put value x=3 in equation (i)

We get, y = 3 – 3

⇒ y = 0

ii. Put value y = 5 in equation (i)

We get, x = 3 – 5

⇒ x = -2

iii. Put value y = 3 in equation (i)

We get, x = 3 – 3

⇒ x = 0

Now the table becomes,

x 3 -2 0
y 0 5 3
(x, y) (3, 0) (-2, 5) (0, 3)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 10

(2) Given

x – y = 4 ……. (ii)

i. Put value y = 0 in equation (ii)

we get, x = 4 – 0

⇒ x = 4

ii. Put value x = –1 in equation (ii)

we get, – y = 5

iii. Put value y = –4 in equation (ii)

we get, x = 4 + 4

⇒ x = 8

x 4 -1 0
y 0 -5 -4
(x, y) (4, 0) (-1, -5) (0, -4)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 11

2. Solve the following simultaneous equations graphically.

(1) x + y = 6; x – y = 4

Solution:

Given x + y = 6 …. (i)

x 0 6 5
y 6 0 1
(x, y) (0, 6) (6, 0) (5, 1)

x – y = 4 …… (ii)

x 0 2 5
y -4 -2 1
(x, y) (0, -4) (2, -2) (5, 1)

Calculating intersecting point

x + y = 6

x – y = 4

2x = 10

x = 10/2

x = 5

Putting x= 5 in equation (i)

5 + y = 6

y = 6 – 5

y = 1

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 12

(2) x + y = 5; x – y = 3

Solution:

x + y = 5 …. (i)

x 0 2 4
y 5 3 1
(x, y) (0, 5) (2, 3) (4, 1)

x – y = 3 …… (ii)

x 0 2 4
y -3 -1 1
(x, y) (0, -3) (2, -1) (4, 1)

Calculating intersecting point

x + y = 5

x – y = 3

which implies

2x = 8

x = 8/2

x = 4

Putting x= 4 in equation (i)

4 + y = 5

y = 5 – 4

y = 1

Intersection Point (4,1)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 13

(3) x + y = 0; 2x – y = 9

Solution:

x + y = 0 …. (i)

x 1 3 5
y -1 -3 -5
(x, y) (1, -1) (3, -3) (5, -5)

2x – y = 9 …. (ii)

x 2 3 4
y -5 -3 -1
(x, y) (2, -5) (3, -3) (4, -1)

Calculating intersecting point

x + y = 0

2x – y = 9

3x = 9

x = 9/3

x = 3

Putting x= 3 in equation (i)

3 + y = 0

y = 0 – 3

y = -3

Intersection point (3, –3)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 14

(4) 3x – y = 2; 2x – y = 3

Solution:

3x – y = 2 …… (i)

x 0 1 -1
Y -2 1 -5
(x, y) (0, -2) (1, 1) (-1, -5)

2x – y = 3 …… (ii)

x 3 2 -1
y 3 1 -5
(x, y) (3, 3) (2, 1) (-1, -5)

Calculating intersecting point

3x – y = 2

– 2x + y = -3

x = -1

Putting x= –1 in equation (i)

3× –1 – y = 2

– 3 – y = 2

– y = 2 + 3

y = -5

Intersection point (–1, –5)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 15

(5) 3x – 4y = –7; 5x – 2y = 0

Solution:

3x – 4y = 7 …. (i)

When x = 0, 4y = 7, y = 7/4

When y = 0, 3x = -7, x = -7/3

5x – 2y = 0 …… (ii)

When x = 0, y = 0

When x = 1, y = 5/2

Plotting both the graphs we get,

Calculating intersecting point

3x – 4y = -7

5x – 2y = 0

x = -1

Putting x= –1 in equation (i)

3 × –1 – y = 2

– 3 – y = 2

– y = 2 + 3

y = – 5

Intersection point (–1, –5)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 16

 

practice set 1.3 Page no: 16

1. Fill in the blanks with correct number

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Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 18

2. Find the values of following determinants.

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Solution:

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Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 21

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 22

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 23

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 24

3. Solve the following simultaneous equations using Cramer’s rule.

(1) 3x – 4y = 10; 4x + 3y = 5

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 25

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 26

(2) 4x + 3y – 4 = 0; 6x = 8 – 5y

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 27

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 28

(3) x + 2y = –1; 2x – 3y = 12

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 29

(4) 6x – 4y = –12; 8x – 3y = –2

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 30

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 31

(5) 4m + 6n = 54; 3m + 2n = 28

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 32

(6) 2x + 3y = 2; x – y/2 = ½

 

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 33

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 34

Practice set 1.4 Page no: 19

1. Solve the following simultaneous equations.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 35

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 36

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Solution:

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Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 42

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Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 47

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Practice set 1.5 Page no: 26

1. Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 51

2. Complete the following.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 52

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 53

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 54

3. The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 55

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 56

4. The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 57

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 58

5. Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighs 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 59

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 60

6. Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 61

Practice set 1.6 Page no: 27

1. Choose correct alternative for each of the following question
To draw graph of 4x+5y=19, Find y when x = 1.
A. 4
B. 3
C. 2
D. –3

Solution:

B. 3

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 62

2. For simultaneous equations in variables x and y, DX = 49, DY = –63, D = 7, then what is x?
A. 7
B. –7
C. 1/7
D. -1/7

Solution:

A. 7

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 63

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 64

Solution:

D. 1

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 65

4. To solve x + y = 3; 3x–2y – 4 = 0 by determinant method find D.
A. 5
B. 1
C. –5
D. –1

Solution:

C. -5

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 66

5. ax + by = c and mx + ny = d and an ≠ bm then these simultaneous equations have –
A. Only one common solution.
B. No solution.
C. Infinite number of solutions.
D. Only two solutions.

Solution:

A. Only one common solution.

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 67

2. Complete the following table to draw the graph of 2x – 6y = 3

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 68

Solution:

Put x = –5, then

2 × -5 – 6y = 3

⇒ 3 + 10 = -6y

y = -13/6

Put y = 0, then

2x – 0 = 3

x = 3/2

x -5 3/2
y -13/6 0
(x, y) (-5, -13/6) (3/2, 0)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 69

Where A = (-5, -13/6) and B = (3/2,0)

3. Solve the following simultaneous equation graphically.
(1) 2x + 3y = 12; x – y = 1

Solution:

Given

2x + 3y = 12

Substitute the values for x as shown in the table

x 0 6 3
y 4 0 2

Consider x – y = 1

x 1 0 2
y 0 -1 1

Now plot the graph as shown below:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 70

(2) x – 3y = 1; 3x – 2y + 4 = 0

Solution:

Given

x – 3y = 1

Substitute the values for x as shown in the table

x -2 4 1
y -1 1 0

Also given

3x – 2y + 4 = 0

x 0 -2 -4
y 2 -1 -1

Now plot the graph as shown below:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 71

(3) 5x – 6y + 30 = 0; 5x + 4y – 20 = 0

Solution:

Given

5x – 6y + 30 = 0

Substitute the values for x as shown in the table

x 0 -6 6
y 5 0 10

Also given

5x + 4y – 20 = 0

x 0 4 8
y 5 0 -5

Now plot the graph as shown below:

 

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 72

(4) 3x – y – 2 = 0; 2x + y = 8

Solution:

For equation 1, let’s find the points for graph

3x – y – 2 = 0

At x = 0

3(0) – y – 2 = 0

⇒ y = -2

At x = 13

(1) – y – 2 = 0

⇒ y = 1At x = 2

3(2) – y – 2 = 0

⇒ 6 – y – 2 = 0⇒ y =4

Hence, points for graph are (0, -1) (1, 1) and (2, 4)

For equation 22x + y = 8

At x = 0y = 8

at x = 12

(1) + y = 8

⇒ y = 6

at x =42(4) + y = 8

⇒ y = 0

Hence, points for graph are (0, 8) (1, 6) and (4, 0)

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 73

From graph, we observe both lines intersect at (2, 4)

Hence, x = 2 y = 4 is the solution of given pair

(5) 3x + y = 10; x – y = 2

Solution:

Given 3x + y = 10

x 1 2 3
y 7 4 1

Also, we have

x – y = 2

x 0 2 3
y -2 0 1

Solving Both equations

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 74

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 75

4. Find the values of each of the following determinants.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 76

Solution:

Given

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 77

= (4 × 7) – (3 × 2) = 28 – 6 = 22

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 78

Solution:

Given

D =
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Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 80

Solution:

Given

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 81

5. Solve the following equations by Cramer’s method.

(1) 6x – 3y = –10; 3x + 5y – 8 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 2

(2) 4m – 2n = –4; 4m + 3n = 16

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 83

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 84

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 85

∴ (x, y) = (1/2, -1/2)

(4) 7x + 3y = 15; 12y – 5x = 39

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 86

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 87

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 88

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6. Solve the following simultaneous equations.

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

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7. Solve the following word problems.

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Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 108

2. Kantabai bought Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 109 kg tea and 5 kg sugar from a shop. She paid Rs 50 as return fare for rickshaw. Total expense was Rs 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid Rs 880 for that. Find the rate of sugar and tea per kg.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 110

3. To find number of notes that Anushka had, complete the following activity

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 111

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 112

4. Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 113

5. In a factory the ratio of salary of skilled and unskilled workers is 5: 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 114

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 115

6. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter1 - Image 116

A linear equation in two variables is an equation that can be expressed in the form ax + by + c = 0, such that, b and c are real numbers and also, a, b not equal to zero. Learn more about the concepts covered in the chapter Linear Equations in Two Variables at BYJU’S.

 

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