Graphs of Linear Equations - 2 - Free MCQ Practice Test with solutions,

the given equations becomes: 

or 7u + 24v = 7 and 7u + 4v = 2.
Solving these, we get: 

∴ 2x + 3y = 7 and 3x - 2y = 4.
Now solving these equations again, we get: x = 2 and y = 1.
∴ Solution set = {(2,1)}.
Hence, option D is correct.

By cross multiplication, 2nd equation becomes 2x + 2y = 11.
Now multiplying 1st equation by 2 and adding to this equation, we get 4x = 12.8 or x = 3.2.
Putting x = 3.2 in 1st equation, we get y = 3.2 - 0.9 = 2.3.
Hence, option A is correct.

2^{x + y} = 2^{x – y}
2^{x + y} = 2^{x - y} = 8 = 2^3/2

Solving these equation we get: 

Hence, option A is correct.

Putting 1/x = u and 1/y = v, The given equation become: 
6u + 3v = 7 and 3u + 9y = 11.
On solving these equations, we get: u = 2/3 and v = 1.

Hence, option A is correct.

To eliminate x, we must make the coefficients of x equal in both of the given equation.
So, multiplying the first equation by 8, second equation by 3 and subtracting, we get: y = 5.
Putting y = 5 in first equations, we get 3x + 10 = 4 or 3x = – 6 or x = – 2.
∴ Solution set is {( – 2, 5)}.
Hence, option B is correct.

Putting y = 0 in 4x + 3y = 12
we get x = 3
Putting x = 0 in 4x + 3y = 12, we get y = 4

Hence, option B is correct.

x-axis ⇒ y = 0, putting in equation 3x + 4y = 12
3x = 12 ⇒ x = 4
⇒ Co-ordinates of point of intersection on x-axis = (4, 0)
Putting on y-axis = (0, 3)
∴ (0, 3)
OA = 4
OB = 3

Hence, option C is correct.

x = 0 is the equation of y-axis.
y = 0 is the equation of x-axis.
Putting x = 0 in x + y = 1, y = 1
Putting y = 0 in x + y = 1, x = 1
Putting x = 0 in 2x + 3y = 6
3y = 6 ⇒ y = 2
Putting y = 0 in 2x + 3y = 6 ⇒ 2x = 6 ⇒ x = 3
∴ OB = 1; OA = 1 OD = 3; OC = 2 ∴ Required area = ΔOCD – ΔOAB


Hence, option C correct.

Given,
5x + 9y = 5 .....(i)
125x³ + 729y³ = 120 ....(ii)
Make whole cube of 1st equation, we get
(5x + 9y)³ = (5)³
⇒ 125x³ + 729x³ + 3 × 5x × 9y (5x + 9y) = 125
Put the value of given terms in equation
⇒ 125x³ + 729x³ + 3 × 5x × 9y (5x + 9y) = 125
⇒ 120 + 135xy × 5 = 125
⇒ 135xy × 5 = 5
⇒ xy = 1/135.
Hence, option D is correct.

x + y + z = 13
x² + y² + z² = 69
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
⇒ (13)² = 69 + 2[xy + z(x + y)]
⇒ 2[xy + z(x + y)] = 169 – 69 = 100
⇒ [xy + z(x + y)] = 100/2 = 50.
Hence, option C is correct.

6x – 5y = 13 ...... (i)
7x + 2y = 23 ...... (ii)
By equation (i) × 2 & (ii) × 5,
12x – 10y = 26 ..... (iii)
35x + 10y = 115 ...... (iv)
by solving these equation we get
x = 3, y = 1
∴ 11x + 18y = 11 × 3 + 18 × 1 = 33 + 18 = 51.
Hence, option B is correct.

Let the fare from station P to station Q is Rs. x and the fare from
station P to station R is Rs. y.
By given condition, x + y = 42 .....(i)
and 5x + 10y = 350 .... (ii)
On solving Eqs. (i) and (ii), we get
x = 14 and y = 28
Hence, fare from station P to station Q is Rs. 14.
Hence, option B is correct.

Let the number be y,

⇒ y = 12 × 8 = 96
∴ Sum of digits = 9 + 6 = 15.
Hence, option C is correct.

Let the greater number be x and smaller number be y
∴ x – y = 45 ..... (i)
and x = 4y ..... (ii)
From eqs. (i) and (ii)
4y – y = 45
⇒ y = 45/3 = 15
On putting the value of y in Eq. (ii), we get
x = 4 × 15 = 60
Hence, required sum = x + y = 60 + 15 = 75.
Hence, option D is correct.

Let a two-digit number be (10x + y) and reversing number be (10y + x)
∴ Required sum = 10x + y + 10y + x = 11x + 11y = 11(x + y)
Hence, it's divisible by 11.
Hence, option D is correct.