[√3 + x + √3 - x] / [√3 + x - √3 - x] = 2

Let √3 + x = a and √3 - x = b
Then, (a + b) / (a - b) = 2/1
∴ a + b = 2a - 2b
⇒ a = 3b

∴ √3 + x = 3√3 - x

Correct Option: B

[√3 + x + √3 - x] / [√3 + x - √3 - x] = 2

Let √3 + x = a and √3 - x = b
Then, (a + b) / (a - b) = 2/1
∴ a + b = 2a - 2b
⇒ a = 3b

∴ √3 + x = 3√3 - x

On squaring bothv sides, we get
(√3 + x)² = (3√3 - x)²
⇒ 3 + x = 9(3 - x)
⇒ 3 + x = 27 - 9x
⇒ 10x = 24
∴ x = 12/5

Given,
(x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11
⇒ (x + y - 8)/2 = (x + 2y - 14)/3
⇒ 3x + 3y - 24 = 2x + 4y - 28
⇒ 3x + 3y - 2x - 4y = -28 + 24
x - y = -4 ...(i)

Again, (x + 2y - 14)/3 = (3x + y - 12)/11
⇒ 11x + 22y - 154 = 9x + 3y - 36
⇒ 2x + 19y = 118 ..(ii)

solve these 2 equations and find the value of x and y.

Correct Option: A

Given,
(x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11
⇒ (x + y - 8)/2 = (x + 2y - 14)/3
⇒ 3x + 3y - 24 = 2x + 4y - 28
⇒ 3x + 3y - 2x - 4y = -28 + 24
x - y = -4 ...(i)

Again, (x + 2y - 14)/3 = (3x + y - 12)/11
⇒ 11x + 22y - 154 = 9x + 3y - 36
⇒ 2x + 19y = 118 ..(ii)


On multiplying Eq. (i) by 2 and subtracting from Eq., we get
2x - 2y = -8
2x + 19y = 118
-------------------
-21y = -126
∴ y = 6

On putting the value of y in Eq. (i), we get
x - 6 = -4
∴ x = 2
∴ x = 2 and y = 6

Let the number of gold coins initially be x and the number of non-gold coins be y.
According to the question,
3x = y

When 10 more gold coins, total number of gold coins becomes x + 10 and the number of non-gold coins remain the same at y.
Now, we have 2(x + 10) = y

Solving these two equations, we get
x = 20 and y = 60.

Correct Option: A

Let the number of gold coins initially be x and the number of non-gold coins be y.
According to the question,
3x = y

When 10 more gold coins, total number of gold coins becomes x + 10 and the number of non-gold coins remain the same at y.
Now, we have 2(x + 10) = y

Solving these two equations, we get
x = 20 and y = 60.
Total number of coins in the collection at the end is equal to
x + 10 + y = 20 + 10 + 60 = 90.

Given equations 2x + 4y = 6 and 4x + 8y = 6

then,
a₁/a₂ = 2/4 = 1/2;
b₁/b₂ = 4/8 = 1/2;
c₁/c₂ = 6/6 = 1
∴ a₁/b₂ = b₁/b₂ ≠ c₁/c₂

Correct Option: B

Given equations 2x + 4y = 6 and 4x + 8y = 6

then,
a₁/a₂ = 2/4 = 1/2;
b₁/b₂ = 4/8 = 1/2;
c₁/c₂ = 6/6 = 1
∴ a₁/b₂ = b₁/b₂ ≠ c₁/c₂

So there is no solution for these equations.

Given, 6x - 10y = 10 ..........(i)
and x/(x + y) = 5/7
⇒ 7x = 5x + 5y
⇒ 2x - 5y = 0 ...(ii)

Multiplying Eq. (ii) by 2 and subtracting from Ed.(i),

Correct Option: D

Given, 6x - 10y = 10 ..........(i)
and x/(x + y) = 5/7
⇒ 7x = 5x + 5y
⇒ 2x - 5y = 0 ...(ii)

On multiplying Eq. (ii) by 2 and subtracting from Ed.(i), we get
6x - 10y = 10
4x - 10y = 0
---------------------
2x = 10
∴ x = 5
Putting the value of x in Eq. (i), we get
30 - 10y = 10
⇒ 10y = 20
⇒ y = 2
∴ (x - y) = 5 - 2 = 3