Problem Set 1 Linear Equations In Two Variables Class 10th Mathematics Part 1 MHB Solution

Problem Set 1
  1. Choose correct alternative for each of the following question To draw graph of…
  2. For simultaneous equations in variables x and y, DX = 49, DY = -63, D = 7, then what…
  3. Find the value of | cc 5&3 -7&-4 | Choose correct alternative for each of the…
  4. To solve x + y = 3; 3x-2y - 4 = 0 by determinant method find D. Choose correct…
  5. ax + by = c and mx + ny = d and an ≠ bm then these simultaneous equations have -…
  6. Complete the following table to draw the graph of 2x - 6y = 3
  7. 2x + 3y = 12; x - y = 1 Solve the following simultaneous equation graphically.…
  8. x - 3y = 1; 3x - 2y + 4 = 0 Solve the following simultaneous equation graphically.…
  9. 5x - 6y + 30 = 0; 5x + 4y - 20 = 0 Solve the following simultaneous equation…
  10. 3x - y - 2 = 0; 2x + y = 8 Solve the following simultaneous equation graphically.…
  11. 3x + y = 10; x - y = 2 Solve the following simultaneous equation graphically.…
  12. Find the values of each of the following determinants. (1) | ll 4&3 2&7 | (2) | cc 5&-2…
  13. 6x - 3y = -10; 3x + 5y - 8 = 0 Solve the following equations by Cramer’s method.…
  14. 4m - 2n = -4; 4m + 3n = 16 Solve the following equations by Cramer’s method.…
  15. 3x-2y = 5/2 1/3 x+3y = - 4/3 Solve the following equations by Cramer’s method.…
  16. 7x + 3y = 15; 12y - 5x = 39 Solve the following equations by Cramer’s method.…
  17. x+y-8/2 = x+2y-14/3 = 3x-y/4 Solve the following equations by Cramer’s method.…
  18. 2/x + 2/3y = 1/6 3/x + 2/y = 0 Solve the following simultaneous equations.…
  19. 7/2x+1 + 13/y+2 = 27 13/2x+1 + 7/y+2 = 33 Solve the following simultaneous equations.…
  20. 148/x + 231/y = 527/xy 231/x + 148/y = 610/xy Solve the following simultaneous…
  21. 7x-2y/xy = 5 8x+7y/xy = 15 Solve the following simultaneous equations.…
  22. 1/2 (3x+4y) + 1/5 (2x-3y) = 1/4 5/(3x+4y) - 2/(2x-3y) = - 3/2 Solve the following…
  23. Solve the following word problems. A two digit number and the number with digits…
  24. Kantabai bought 1 1/2 kg tea and 5 kg sugar from a shop. She paid Rs 50 as return fare…
  25. To find number of notes that Anushka had, complete the following activity…
  26. Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4…
  27. In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total…
  28. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A…

Problem Set 1
Question 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


Answer:

Put x= 1 in Eq. 



⇒ 


⇒ 


⇒ 


Hence, option B is correct.


Question 2.

Choose correct alternative for each of the following question

For simultaneous equations in variables x and y, DX = 49, DY = –63, D = 7, then what is x?
A. 7

B. –7

C. 

D. 


Answer:


Hence option A is correct.


Question 3.

Choose correct alternative for each of the following question

Find the value of 
A. –1

B. –41

C. 41

D. 1


Answer:

 = 


Hence, option D is correct.


Question 4.

Choose correct alternative for each of the following question

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

B. 1

C. –5

D. –1


Answer:



 = 


Hence, Option C is correct.


Question 5.

Choose correct alternative for each of the following question

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.


Answer:

Given: ax + by = c and mx + ny = d
Then,

Now, we know that when the ratio of coefficients is not equal.
Equations will have unique solution.

Hence, A is the correct answer.


Question 6.

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



Answer:

Put x= –5, then  ⇒ 


Put y = 0, then 





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

Question 7.

Solve the following simultaneous equation graphically.

2x + 3y = 12; x – y = 1


Answer:







Question 8.

Solve the following simultaneous equation graphically.

x – 3y = 1; 3x – 2y + 4 = 0


Answer:







Question 9.

Solve the following simultaneous equation graphically.

5x – 6y + 30 = 0; 5x + 4y – 20 = 0


Answer:





Question 10.

Solve the following simultaneous equation graphically.

3x – y – 2 = 0; 2x + y = 8


Answer:

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 = 1
3(1) - y - 2 = 0
⇒ y = 1

At 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 2
2x+ y = 8

at x = 0
y =8

at x = 1
2(1) + y = 8
⇒ y = 6

at x =4
2(4) + y = 8
⇒ y = 0

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


From graph, we observe both lines intersect at (2, 4)
hence, x =2 y = 4 is the solution of given pair

Question 11.

Solve the following simultaneous equation graphically.

3x + y = 10; x – y = 2


Answer:


3x+y=10


x-y=2


Solving Both equations



Question 12.

Find the values of each of the following determinants.
(1) 

(2) 

(3) 


Answer:

(1)  = 


(2) D = 


(3) 



Question 13.

Solve the following equations by Cramer’s method.

6x – 3y = –10; 3x + 5y – 8 = 0


Answer:







∴ (x , y) = 



Question 14.

Solve the following equations by Cramer’s method.

4m – 2n = –4; 4m + 3n = 16


Answer:





∴ (x , y) = 



Question 15.

Solve the following equations by Cramer’s method.



Answer:






∴ (x , y) = (1/2, -1/2)


Question 16.

Solve the following equations by Cramer’s method.

7x + 3y = 15; 12y – 5x = 39


Answer:





∴ (x , y) = 



Question 17.

Solve the following equations by Cramer’s method.



Answer:

Let,



⇒ 3x + 3y – 24 = 2x + 4y – 28


⇒ x – y = –4 …(1)


Also,


Let 


⇒ 4x + 8y – 56 = 9x – 3y


⇒ 5x –11y = – 56 …(2)


Hence the two equations are:


x – y = –4 …(1)


5x – 11y = – 56 …(2)


Now,



⇒ D = (–11 -(- 5)) = -6


Also, 


Dx = 44 - 56 = -12


And,



⇒ Dy = –56 + 20 = – 36


Now, 


And, 


Hence, (2, 6) is the solution


Question 18.

Solve the following simultaneous equations.



Answer:

Let 



Multiply Eq. II by 2



Subtract Eq.III from Eq. I






Substitute m=1/6 in Eq. I







∴ 


∴ 


Hence, 



Question 19.

Solve the following simultaneous equations.



Answer:

Let 



Adding Eq. I and II



Subtract Eq. I and II



Equating Eq. III and IV






Substituting n=1 in Eq. III





∴ 


⇒ 


∴ 


Hence, 



Question 20.

Solve the following simultaneous equations.



Answer:



Adding Eq. I and II



 …(III)


Subtracting Eq. I and II



 …(IV)


Equating I and II







Substituting x=1 in Eq. I





Hence, 



Question 21.

Solve the following simultaneous equations.



Answer:



Let




Multiply Eq. 1 by 7 and Eq.II by 2






Substituting value in Eq.VI







∴ 


∴ 


Hence, 



Question 22.

Solve the following simultaneous equations.



Answer:

Let 

…(I)


… (II)


Equating Eq. I and II







Substituting  in Eq. I








∴ 


∴ 


Multiply Eq. III by 3 and Eq. IV by 4 and Equate







Substituting x=2 in Eq. V








Hence, 



Question 23.

Solve the following word problems.

A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.

Let the digit in unit’s place is x

and that in the ten’s place is y



The number obtained by interchanging the digits is 

According to first condition two digit number + the number obtained by interchanging the digits = 143



From the second condition,

digit in unit’s place = digit in the ten’s place + 3



Adding equations (I) and (II)



Putting this value of x in equation (I)

x + y = 13



The original number is 10



Answer:

Let the digit in unit’s place is x

and that in the ten’s place is y


∴ the number = 10y+ x


The number obtained by interchanging the digits is 


According to first condition two digit number + the number obtained by interchanging the digits = 143


∴ 


∴ 


∴ 


From the second condition,


digit in unit’s place = digit in the ten’s place + 3


∴ 


∴  ….. (II)


Adding equations (I) and (II)




Putting this value of x in equation (I)


x + y = 13



∴ 


The original number is 10


⇒ 50+8


⇒ 58



Question 24.

Kantabai bought  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.


Answer:

Let x be the cost of tea and y be the cost of sugar

As she paid ₹50 as return fare



∴ 


According to second situation,



Multiplying Eq. I by 2 and Eq. II by 3


…..(III)



Subtracting Eq. III from IV





Substituting y=40 in Eq. I





3x=900




Tea; ₹ 300 per kg.


Sugar ; ₹ 40 per kg.



Question 25.

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



Answer:

According to 1st situation ,



According to 2nd situation,



Adding I and II,



 …III


Subtracting I from II



 ….IV


Equating Eq. III with Eq. IV






Substituting x=10 in Eq. III




₹100 notes = 10


₹50 notes = 20



Question 26.

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.


Answer:

Let Manish’s present age be x

Let Savita’s present age be y


According to 1st situation,



According to second situation,






Subtracting Eq. II from I







Substitute y=8 in eq. I





Manisha's age 23 years


Savita's age 8 years.



Question 27.

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.


Answer:

Ratio of skilled and unskilled worker’s salary = 5:3

Let it be 5x and 3x


Total of one day’s salary = ₹720


So, 





Skilled worker's wages =  = ₹450.


unskilled worker's wages  = ₹270



Question 28.

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.


Answer:

Let the speed of Joseph = x km/h

Let the speed of Hamid be = y km/h


When approaching each other, combined speed = 


Time taken to meet = 


∴ 


When moving away from each other, combined speed = 


Time taken for Hamid to catch up = 


∴ 


Equating I and II,






Substituting x=50in eq. I





Hamid's speed 50 km/hr.


Joseph's speed 40 km/hr.


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