Linear Equations Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Given that,

2x – 3y = 5 ... (i)

and –4x + 6y = 11 ... (ii)

Also, ${a_1}/{a_2} = 2/{-4} = {-1}/2$

${b_1}/{b_2} = {-3}/6 = {-1}/2$

⇒${a_1}/{a_2} ={b_1}/{b_2} = {-1}/2 ≠ {c_1}/{c_2}$

So, both equations are in parallel. So, it is not solved graphically.

Hence, A and R are individually true and R is correct explanation of A.

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: (b)

Let the two-digit number be 10y + x.

According to question,

x + y = 10 ... (i)

and 10y + x – 18 = 10x + y

⇒9x – 9y = –18

⇒x – y = –2 ... (ii)

On solving equations (i) and (ii), we get

x = 4 and y = 6

∴ Required product = xy = 4 × 6 = 24.

Answer: (a)

Since a, b are negative numbers.

a < 0 and b < 0

C is a positive real number

⇒ c > 0

(1) a – b < a – c

⇒ – b < – c

⇒ b > c

It is not true as b < c

(2) if a < b ⇒ $a/c < b/c$ this is true.

(3) $1/b < 1/c$

Since c > b ⇒ $1/c < 1/b$ this is not true.

∴ option (a) is correct.

Answer: (a)

Let the positive number be x.

According to the question,

x + 10 = ${200}/x$

⇒ $x^2$ + 10x = 200

⇒ $x^2$ + 10x – 200 = 0

⇒ (x – 10) (x + 20) = 0

∴ x = 10, –20

But x ≠ –20, since x is a positive number

The required number is 10.