Linear Equations Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

x+ y ≥ 5

let first draw graph of x + y = 5 ... (i)

put y = 0 in (i)

x = 5

put x = 0 in (i)

y = 5

Checking for (0, 0)

x + y ≥ 5

0 ≥ 5, which is false

Hence, origin does not lie in x + y ≥ 5

Now,

x – y ≤ 3

let first drawn graph of x – y = 3 ...(ii)

put x = 0 in (ii)

y = –3

put y = 0 in (ii)

x = 3'

Checking for (0, 0)

x – y ≤ 3

0 ≤ 3, which is true

Hence, origin lies in x – y ≤ 3

So, shaded region will be the solution of these linear inequalities lies in the first and second quadrant.

Answer: (a)

The equations kx – y = 2 and 6x – 2y = 3 have a unique solution. Then,

∴ $k/6 ≠ 1/2$ ⇒ k ≠ 3

Answer: (c)

Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts.

Let the number of spades be x. Then, number of clubs = (7 – x).

Number of diamonds = 2 x number of spades = 2x;

Number of hearts = 2 x number of diamonds = 4x.

Total number of cards = x + 2x + 4x + 7 – x – 6x + 7.

Therefore 6x + 7 = 13⇔6x = 6⇔x – 1.

Hence, number of clubs = (7 – x) = 6.

Answer: (c)

Given that x + $1/x$ = 2 ... (i)

Squaring both sides, we get

$(x + 1/x)^2$ = 4

⇒ $x^2 + 1/{x^2}$ + 2 = 4

⇒ $x^2 + 1/{x^2}$ = 2 ... (ii)

Now, $(x - 1/x)^2 = (x^2 + 1/{x^2})$ – 2

= 2 – 2 = 0 [from equation (ii)]

∴ x – $1/x$ = 0

Answer: (b)

Given,

${5x - 7y + 10}/1 = {3x + 2y + 1}/8 = {11x + 4y - 10}/9$

= ${(5x + 10 - 7y) + (3x + 2y + 1) - (11x + 4y - 10)}/{(1 + 8) - 9}$

= ${- 3x - 9y + 21}/0$

⇒x + 3y = 7 ... (i)

On taking first two terms,

8(5x – 7y + 10) = 3x + 2y + 1

37x – 58y + 79 = 0 ... (ii)

From equation (i), on putting the value of x in equation (ii), we get

37(7 – 3y) – 58y + 79 = 0

⇒259 – 111y – 58y + 79 = 0

⇒169y = 338

⇒y = 2

From equation (i),

x = 7 – 3(2) = 1

∴ x + y = 1 + 2 = 3