Let the capital of one is x and that of another is y .
According to the question.
x + 100 = 2(y - 100) and
y + 10 = 6(x - 10)

Correct Option: A

Let the capital of one is x and that of another is y .

According to the question.
x + 100 = 2(y - 100)
x + 100 = 2y - 200
or x - 2y = -300 ...(i)

Again, according to the question
y + 10 = 6(x - 10)
⇒ y + 10 = 6x - 10
⇒ 6x - y = 70 ..(ii)

On multiplying Eq. (ii) by 2 and subtracting from Eq. (i) , we get
x - 2y = -300
12x - 2y = 140
----------------
-11x = -440
∴ x = 40

On putting the value of x in Eq. (i) , we get
40 - 2y = -300
⇒ 2y = 340
∴ y = 170
So, their capital are ₹ 40 and ₹ 170.

(p/x) + (q/y) = m ..(i)
(q/x) + (p/y) = n ...(ii)

On multiply Eq (i) by q and Eq. (ii) by p and subtracting, we get
(pq/x) + (q²)/y = mq
(pq/x) + (p²)/y = np
-----------------------------------------
(q²/y) - (p²/y) = mq - np
∴ (q² - p²) = y(mq - np)
∴ y = (q² - p²)/mp - np = (p² - q²)/np - mq

Again, on multiplying Eq. (i) by p and Eq. (ii) by q and subtracting, we get
(p²/x) + (pq/y) = mp
(q²/x) + (pq/y) = nq
-----------------------------------------
(p²/x) - (q²/x) = mp - nq

Correct Option: C

(p/x) + (q/y) = m ..(i)
(q/x) + (p/y) = n ...(ii)

On multiply Eq (i) by q and Eq. (ii) by p and subtracting, we get
(pq/x) + (q²)/y = mq
(pq/x) + (p²)/y = np
-----------------------------------------
(q²/y) - (p²/y) = mq - np
∴ (q² - p²) = y(mq - np)
∴ y = (q² - p²)/mp - np = (p² - q²)/np - mq

Again, on multiplying Eq. (i) by p and Eq. (ii) by q and subtracting, we get
(p²/x) + (pq/y) = mp
(q²/x) + (pq/y) = nq
-----------------------------------------
(p²/x) - (q²/x) = mp - nq
⇒ (p² - q²) = x (mp - nq)
⇒ x = (p² - q²)/(mp - nq)
∴ x = (p² - q²)/(mp - nq)
and y = (p² - q²) / (np - mq)

Given, 2^a + 3^b = 17
and 2^{a + 2} - 3^{b + 1} = 5
⇒ 2² x 2² - 3^b x 3¹ = 5
⇒ 4.2^a - 3.3^b = 5
Let 2^a = x amd 3^b = y
Then, x + y = 17 ...(i)
4x - 3y = 5 ...(ii)

Now find and x and y and finally a and b using x and y.

Correct Option: D

Given, 2^a + 3^b = 17
and 2^{a + 2} - 3^{b + 1} = 5
⇒ 2² x 2² - 3^b x 3¹ = 5
⇒ 4.2^a - 3.3^b = 5
Let 2^a = x amd 3^b = y
Then, x + y = 17 ...(i)
4x - 3y = 5 ...(ii)
On multiplying Eq. (i) by 3 and adding to Eq (ii), we get
3x + 3y = 51
4x - 3y = 5
------------------
7x = 56
⇒ x = 8

On putting the value of x in Eq. (i), we get
8 + y = 17
∴ y = 9

Now, 2^a = x
⇒ 2^a = 8 (2)³
∴ 3^b = y = 9
⇒ 3^b = 3²
∴ b = 2
Hence, a = 3 and b = 2

Let the cost of one chair be ₹ x
and cost of one table be ₹ y.
By given condition,
10x + 6y = 6200 ..(i)
and 3x + 2y = 1900
⇒ 9x + 6y = 5700 ...(ii)

Correct Option: A

Let the cost of one chair be ₹ x
and cost of one table be ₹ y.
By given condition,
10x + 6y = 6200 ..(i)
and 3x + 2y = 1900
⇒ 9x + 6y = 5700 ...(ii)

On subtracting Eq. (ii), we get
x = ₹ 500

From Eq (i),
5000 + 6y = 6200
⇒ 6y = 1200
∴ y = ₹ 200

The cost of 4 chair and 5 tables
= 4x + 5y
= 2000 + 1000
= ₹ 3000

ax + by = c and dx + ey = f
a₁/a₂ = a/d, b₁/b₂ = b/e, c₁/c₂ = c/f
∵ b₁/b₂ ≠ c₁/c₂
∴ b/e ≠ c/f
it represent a pair of parallel lines.
∵ a₁/a₂ ≠ b₁/b₂
∴ a/d ≠ b/e
Therefore, system has unique solutions and represents a pair of intersecting lines.

Correct Option: C

ax + by = c and dx + ey = f
a₁/a₂ = a/d, b₁/b₂ = b/e, c₁/c₂ = c/f
∵ b₁/b₂ ≠ c₁/c₂
∴ b/e ≠ c/f
it represent a pair of parallel lines.
∵ a₁/a₂ ≠ b₁/b₂
∴ a/d ≠ b/e
Therefore, system has unique solutions and represents a pair of intersecting lines.