Q. The cost of 2 kg apples and 1 kg grapes was โ‚น 160. After a month, the cost of 4 kg apples and 2 kg grapes was โ‚น 300. Represent the situation algebraically.

  • Sol.Let the cost of 1 kg of apples be โ‚น a
  • and the cost of 1 kg of grapes be โ‚น g.
  • Thus, 2a + g = 160 โ€ฆ(i)
  • and 4a + 2g = 300
  • or 2a + g = 150 โ€ฆ(ii)
  • Thus, equations are 2a + g = 160
  • and 2a + g = 150 Ans.

Q. Solve for x and y : 152x โ€“ 378y = โ€“ 74 and โ€“ 378x + 152y = โ€“ 604.

Sol. Given, 152x โ€“ 378y = โ€“ 74 โ€ฆ(i)

and โ€“ 378x + 152y = โ€“ 604 โ€ฆ(ii)

Adding (i) and (ii), we get

โ€“ 226x โ€“ 226y = โ€“ 678

โ‡’ x + y = 3 โ€ฆ(iii)

Subtracting (ii) from (i), we get

530x โ€“ 530y = 530

โ‡’ x โ€“ y = 1 โ€ฆ(iv)

Adding (iii) and (iv), we get

2x = 4

Substituting x = 2 in (iii), we get

2 + y = 3

โ‡’ y = 1

Thus, x = 2 and y = 1. Ans.

Q. Determine the value of k, for which the given system of equations has infinitely many solutions:

kx + 3y = k โ€“ 3 and 12x + ky = k.

Sol. Given,

kx + 3y โ€“ (k โ€“ 3) = 0 โ€ฆ(i)

and 12x + ky โ€“ k = 0 โ€ฆ(ii)

For this pair of equations to have infinite number of solutions, it must satisfy the condition of

k/12=3/k=(k-3)/k

โ‡’ k2 = 12(k โ€“ 3)
โ‡’ k2 โ€“ 12k + 36 = 0
โ‡’ k2 โ€“ 6k โ€“ 6k + 36 = 0

โ‡’ k(k โ€“ 6) โ€“ 6(k โ€“ 6) = 0

โ‡’ (k โ€“ 6)2 = 0

โ‡’ k = 6

Also k2 = 36

k = ยฑ 6

Common value k = 6

Thus, the pair of equations will have infinite number of solutions for k = 6. Ans.

Q. Determine the value of k, for which the given system of equations has no solution: 3x + y = 1 and (2k โ€“ 1)x + (k โ€“ 1)y = 2k + 1.

  • Sol. Given, 3x + y = 1 โ€ฆ(i)
  • and (2k โ€“ 1)x + (k โ€“ 1)y = 2k + 1 โ€ฆ(ii)
  • For this pair of equations not to hav e any solution, it must satisfy the condition of
  • 3/(21-k)=1/(k-1)โ‰ 1/(2k=1)
  • โ‡’ 3(k โ€“ 1) = 2k โ€“ 1
  • โ‡’ 3k โ€“ 3 โ€“ 2k + 1 = 0
  • โ‡’ k โ€“ 2 = 0
  • โ‡’ k = 2
  • Thus, the pair of equations will hav e no solution for k = 2. Ans.

Q. For what value of k will the system of equations 4x + ky + 8 = 0 and 2x + 2y + 2 = 0 has a unique solution?

  • Sol. The system of equations will have a unique solution, if
  • 4/2โ‰ k/2
  • or kโ‰ 4

Q. Determine the values of a and b for which the given system of equations has infinitely many solutions:ย 

2x + 3y = 7 and (a โ€“ b)x + (a + b)y = 3a + b โ€“ 1.

  • Sol. Given, 2x + 3y = 7 โ€ฆ(i)
  • and (a โ€“ b)x + (a + b)y = 3a + b โ€“ 1 โ€ฆ(ii)
  • For this pair of equations to have infinite number of solutions, it must satisfy the condition of
  • 2/(a-b)=3/(a+b)=7/(3a+b-1)
  • โ‡’ 2(a + b) = 3(a โ€“ b)
  • โ‡’ 2a + 2b = 3a โ€“ 3b
  • โ‡’ a = 5b โ€ฆ(iii)
  • and 2(3a + b โ€“ 1) = 7(a โ€“ b)
  • โ‡’ 6a + 2b โ€“ 2 = 7a โ€“ 7b
  • โ‡’ a โ€“ 9b = โ€“ 2 โ€ฆ(iv)
  • Substituting a = 5b in equation (iv), we get
  • 5b โ€“ 9b = โ€“ 2
  • โ‡’ โ€“ 4b = โ€“ 2
  • โ‡’ 2b = 1
  • โ‡’ b=1/2
  • Thus, a = 5(1/2)=5/2
  • Thus, the pair of equations will have infinite
  • number of solutions for a =5/2 and b = 1/2. Ans.

Q. The path of a train A is given by the equation x + 2y โ€“ 4 = 0 and of B by the equation 2x + 4y โ€“ 12 = 0. Represent this situation graphically.

  • Sol. Let XOXโ€ฒ and YOYโ€ฒ be the X-axis and Y-axis respectively.
  • Now, x + 2y โ€“ 4 = 0
  • โ‡’ x = 4 โ€“ 2y
  • If y = 2, x = 0
  • If y = 1, x = 2
  • If y = 0, x = 4
  • Therefore,
x 0 2 4
y 2 1 0
  • Now, 2x + 4y โ€“ 12 = 0
  • โ‡’ 4y = 12 โ€“ 2x
  • โ‡’ 2y = 6 โ€“ x
  • โ‡’ y = (6โˆ’x) / 2
  • If x = 6, y = 0
  • If x = 4, y = 1
  • If x = 2, y = 2
  • Therefore,
x 6 4 2
y 0 1 2

Thus, plotting the points S(6, 0), T(4, 1) and U(2, 2) on the graph paper, we get the graph of 2x + 4y = 12, which is represented by SU.

Q. Given the linear equation 2x + 3y โ€“ 8 = 0, write another equation in two variables such that the graphical representation of the pair so formed is

(i) intersecting,ย 

(ii) parallel andย 

(iii) coincident.

  • Ans. (i) 2x + 4y = 6
  • (ii) 4x + 6y = 8
  • (iii) 6x + 9y = 24

So, the system is consistent and thus the reason is also correct.